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gannon
2006-Jul-16, 04:52 PM
hi, the one message keeps on saying for me to post so I will.

Say you where on a ship accelerating at 1g. Would it always fell like 1g the closer to the speed of light you got?
Or is it that you can not accelerate at 1g at those types of speeds? And if that is the case what would happen to a planet at those speeds (gravity wise)?

ToSeek
2006-Jul-16, 06:11 PM
It would always feel like 1g. Without looking out the windows (at something going a different speed), there is no way of telling that you're approaching the speed of light.

astromark
2006-Jul-16, 07:39 PM
If we were to be accelerating at 1 g. We would only be aware of any sudden changes. As it is we have no sensation of movement., and we know we are orbiting our sun., and that it is orbiting the galactic center., and it is . . .you get the idea?

gannon
2006-Jul-16, 08:07 PM
I was under the impression as you got closer to the speed of light it was harder too accelerate was that wrong?

Dragon Star
2006-Jul-16, 08:15 PM
I was under the impression as you got closer to the speed of light it was harder too accelerate was that wrong?

Right, you can feasibly reach 99.9%, But true 100% is extremely improbable.

ToSeek
2006-Jul-16, 08:20 PM
I was under the impression as you got closer to the speed of light it was harder too accelerate was that wrong?

That's correct, but only from an external perspective. If you're onboard the ship, you don't experience that. You'll still experience the acceleration, but the effect will be more to reduce the apparent distance to your destination (due to time and length dilation effects) than to change your speed as measured against an external object that's not being accelerated.

Ken G
2006-Jul-16, 10:22 PM
Yes that's it exactly. As you approach c, it is like you are an ant crawling on a rubber ruler at close to the maximum speed an ant can crawl. If you accelerate, what "really" happens is the rubber ruler shrinks, so you reach the end faster even though you can only crawl along the ruler at the speed of an ant. If at any point you stop accelerating at g, the ruler stops shrinking, and you continue at a speed that is only a teensy bit closer to the maximum value of c.

Now you might ask, what about the people on the planets you zing past, surely they don't think the "ruler" is shrinking, so what do they see? They don't think you are accelerating at anything close to g, they think your clock is running incredibly slowly so it is actually taking you much longer to accelerate than you think (so they think your acceleration is much smaller than g, you are grossly overestimating it from your own perspective. The gall of some people...). So that's why they would say it is very hard for you to maintain g as your acceleration as you get closer to c, even though it does not seem that hard to you at all. (Welcome to the forum gannon.)

publius
2006-Jul-17, 05:00 AM
What's interesting is what we mean by "constant acceleration". What we really mean is the accelerating observer feels a constant force, F, and via the Equivalence Principle can assume he is a stationary in a gravitational field with acceleration g = F/m, where m is his mass.

From the point of view of an inertial observer watching the show, his accleration becomes smaller and smaller as he aproaches c.

However, suppose the accelerating observer drops a ball in his frame and measures its acceleration. In his (non-inertial) frame, he will measure that "free fall" acceleration to always be g (well almost, as we were discussing before, there will be a gradient, but for small g it will be small and can be ignored over short distances in the accelerating observer's frame).

The inertial observer will not see anything like that. He will see the dropped ball stop accelerating at some speed close to c, and see the accelerating observer slowly moving away as his speed only slightly increases over a long time in the inertial frame.

And another interesting thing is the lack of "reciprocity" between the accelerating and inertial frame. The inertial observer can always see the accelerating observer inching closer and closer to c but never getting there, and his clock getting ever slower and his ruler getting ever shorter.

But there is an even horizon from the POV of the accelerating observer, a region past which he can't see the inertial observer. From the inertial observer's frame, this is a line past which light will never reach the accelerating observer if he continues to accelerate. He will never reach c, but because he speed continues to increase, ever so slowly, light past that line will take an infinite time to reach him. It will get asympotically close, but never quite make it.

From the non-inertial frame's POV, a free falling object's speed would be greater than c past that horizon, and so it is "censored" from him. :) But the inertial observer can always see the accelerating observer, and that shows how "reciprocity" does not hold in non-inertial frames (or gravitational fields).
-Richard

Ken G
2006-Jul-17, 08:55 AM
That's also related to the weird lack of reciprocity when someone falls into a black hole. A distant observer sees them frozen in time, and never actually reaching the event horizon, while the falling observer crosses the event horizon in short order and loses contact with the distant observer (who has grown old and died anyway). But note there is still a kind of reciprocity there, in that both observers agree that the distant observer passed away before the falling observer reached the event horizon. Put differently, there are some things that both observers must agree on, and one of them is that if the Earth sends out light pulses toward the falling observer, everyone must agree that every pulse emitted by Earth, no matter how long they are continuously emitted, will reach the falling observer before they cross the event horizon, while only a finite number of pulses emitted by the falling observer will ever reach Earth. So I think the form of reciprocity that survives GR is not that all the results are symmetric, it's that there still has to be an agreed-upon reality for all, and each can use GR to infer what the other will perceive from that reality. That's the important sense to which relativity does not eliminate, but rather preserves, the concept of objective reality, despite all the subjectiveness of the various reference frames.

gannon
2006-Jul-17, 10:15 AM
If the ship is near the speed of light and you started accelerating the opposite way would the "rubber ruler" stretch out again? If so are there any conditions that it would seem that the destination would be retreating even though you would still be approaching it (not counting the case where you really are going away from it)?

Ken G
2006-Jul-17, 10:28 AM
Yes, that's an excellent thought and you are absolutely right. If you are going close to c, and you slow down, the destination will seem farther away even though you have always been approaching it. Relativistic motion is very subtle this way!

jseefcoot
2006-Jul-17, 08:16 PM
That's why I hate this relativity stuff. Makes my brain hurt.

trinitree88
2006-Jul-17, 10:40 PM
It would always feel like 1g. Without looking out the windows (at something going a different speed), there is no way of telling that you're approaching the speed of light.

ToSeek. Actually, with all due respect,I disagree. The current work at the Minos experiment is to determine the cross-sections for fairly energetic neutrinos, and very energetic neutrinos. Trickier work in the future should yield the poorly known low-energy bandpass.
Any acceleration in what is presumably..(at this point)...a relatively isotropic neutrino sea will change that isotropy to an anisotropic flow. The differences in cross-section for the blue-shifted,fore, and red-shifted aft fluxes should be detectable...(if not now, soon). There is as yet no way to conduct a ship acceleration experiment without invoking the consequences of neutrino sea isotropy. If your ship were large enough to accommodate a Minos setup...you'd "see" the acceleration in the scattering data, without looking out the windows. Pete.

Ken G
2006-Jul-17, 10:59 PM
Actually, not looking out the windows is just like not conducting a Minos experiment, so ToSeek's analogy does hold in the extended form. Note he did not say that there were no windows, he said you were simply not looking out them. I realize that neutrinos are harder to block than light, so it is much harder to experimentally rule out the possibility that they may play an important role in relativity. But there is also no hard evidence that they do play a role in relativity, so in the best tradition of Occam's Razor there is as yet no need to invoke such a role. Future experiments might of course require it, so it is interesting to be reminded of this, but it doesn't really contradict ToSeek's statement, or the way the principle of equivalence is currently applied.

Grey
2006-Jul-17, 11:12 PM
That's also related to the weird lack of reciprocity when someone falls into a black hole. A distant observer sees them frozen in time, and never actually reaching the event horizon, while the falling observer crosses the event horizon in short order and loses contact with the distant observer (who has grown old and died anyway). But note there is still a kind of reciprocity there, in that both observers agree that the distant observer passed away before the falling observer reached the event horizon.Not quite. The observer falling in will do so in a very short period of time, and will also see only a short period of time elapse outside. So he won't agree that the distant observer grows old before the falling observer reaches the event horizon, though he will agree that the distant observer will grow old and die before he sees the falling observer reach the event horizon.

Ken G
2006-Jul-18, 02:00 AM
I think I've gotten this wrong before, as we've discussed on another thread. It must come down to the question, how blueshifted would light from the Earth appear to be for the falling observer at the instant he crosses the event horizon? That will tell us how "compactified" Earth history is for that observer at that moment, and the answer is not going to be infinite, is it. So it's not an infinite amount of time going by on the Earth for the falling observer. However, as you say, the light pulses sent out by the falling observer will be infinitely redshifted before reaching Earth, so in that sense all of Earth's future history will be compressed into an instant of time for the falling observer. Another example of a failed reciprocity, it would seem. But once he corrected for light travel effects, what would the falling observer use as the most natural way to conceptualize the passage of time on Earth as he is passing the event horizon?

publius
2006-Jul-18, 02:40 AM
I think I've gotten this wrong before, as we've discussed on another thread. It must come down to the question, how blueshifted would light from the Earth appear to be for the falling observer at the instant he crosses the event horizon?

Ken,

This gets very tricky. Why do you think the light would be blueshifted. Because you're thinking about how it works for *stationary observers* in the gravitational field (or coaccelerating observers such that the distance between them in their own frame remains fixed). :)

To modify Obi-Wan's "Use the Force, Luke!", I say "Use the Equivalence Principle, Ken".

The gravitational field of a black hole is different from the Equ. P's psuedo-field for the accelerating observer, but they have enough in common to get a handle on what's going on. The free-falling black hole observer has a date with a singularity, while the acceleration field free-faller has nothing to worry about and sees a normal universe as always.

Remember the free-falling observer in the black-hole field is equivalent to the inertial observer watching the accelerating observer. The inertial observer sees the accelerating observer's light *red shifted* greatly as he approches c, and his clock slowing down. While I don't know how the SR effects are wrapped up in GR, in this case it looks the SR time dilation due to high relative velocity overwhelms the clock speed up to being at a higher gravitational potential.

An observer near the event horizon would only see the high blue shift if were stationary, or co-accelerating by the EP.

So it seems (and I say seems because the black hole field is different) that a free-falling observer would see distant light red-shifted as he crossed the event horizon.

-Richard

publius
2006-Jul-18, 03:11 AM
Thinking about this some more, I should clarify that what I said above applies only to a free-fall into a black hole with no angular momentum. That is a straight down free fall, to compare to an inertial observer watching another accelerate away.

If the free-faller is in an orbital path around the black hole, things will be very different, I'm certain. From the frame of an orbit deep in a black hole gravity well, things would look different. He would see far away objects spinning around at high speed, maybe close to c -- in fact, recalling our discussion about coordinate speeds and local vs global coordinates, he might see objects moving much faster than local c, but of course he sees the coordinate speed of light as much greater high up the well so he sees nothing moving faster than coordinate speed of light whereever he looks.



Sometimes objects would be receding away at high speed, then turn quickly and head toward him at very high speed. But light itself will be following a very curved path, and even almost orbit the black hole itself, and I'm sure that would change what the orbiting observer actually sees as well.

The accelerating observer's event horizon is a massive global thing, an infinite plane (well, it might actually be a curved surface) blocking off all of space-time at some distance behind him. For the black hole field, that event horizon is a small thing. A free faller into that, at least some distance away before the event horizon looms large in his field of view would see objects on the other side accelerating toward him.

If you gave it some thought, you might be able to come up with some similiar set of accelerating observers with different initial velocities so our inertial observer might for a time see something close via the EP to what the black hole free-faller would see for a time. And then, you could use SR alone to at least explain how that inertial observer would see all the different clock rates and red vs blue shifts and all that.

ETA: And above I was falling in the same trap as I was warning you about, thinking about things from a stationary perpsective. The light curves and even orbits from the stationary perspective, and I have no idea if a pure free-faller, even an orbiting one, would have to conclude light was curving. If he doesn't notice tidal effects, then he's inertial for his own frame, and could see everything else as doing some funky accelerating.


-Richard

Squashed
2006-Jul-18, 01:47 PM
Thinking about this some more, I should clarify that what I said above applies only to a free-fall into a black hole with no angular momentum.

This is an interesting topic, to me, as a few of you can attest.

Gravity redshifts light if the light is moving out of the gravitational field and gravity blueshifts light if the light is moving into the gravitational field.

So if one were freefalling, as publius describes above, then by the time the light reaches the freefaller it has undergone the same acceleration via gravity as the freefaller and so there would be no blueshifting due to gravity.

On the other hand the velocity of the freefaller is increasing and it is the velocity which causes the light from afar to be redshifted.

If you were orbiting the blackhole then the light would be blueshifted because the orbiter is essentially stationary with respect to the gravitational field and the orbiter has zero radial velocity to redshift the incoming light.

As far as time dilation for blackholes, if it is true according to Einstein, then I still maintain that time stops at the event horizon and the freefaller never falls in. As for the freefaller the person will look at their clock and forever wait for the next second to tick but will never realize that time has stopped for him. If the freefaller could see back into the universe the person would see time flying infinitely fast compared to his/her clock.

Ken G
2006-Jul-18, 02:38 PM
So if one were freefalling, as publius describes above, then by the time the light reaches the freefaller it has undergone the same acceleration via gravity as the freefaller and so there would be no blueshifting due to gravity.
I was tempted to reach this conclusion as well, after publius pointed out my mistake in not including the motion of the falling observer. One thing that's sure is that it is very hard to get things right in GR without actually doing the calculation! So the questions remain:
1) At the event horizon, for a stationary observer held up by some force, what is the blueshift of light from Earth? (That's the question that I meant to ask, not involving the motion of the falling observer, just the comparison of the flow of time at that point.)
2) For the falling observer (whose motion is as though he/she started with zero speed at infinity-- note that publius' redshift might be due to propelled motion by the observer's ship as it speeds toward the black hole in an effort to speed up the trip) is the light redshifted or do the effects compensate and give no redshift? (Which is also a measure of the flow of time for the falling observer compared to back on Earth.)

As for an orbiting observer or a spinning black hole, let's worry about these other issues first!



As for the freefaller the person will look at their clock and forever wait for the next second to tick but will never realize that time has stopped for him.
This statement treats the time of the distant observer as if it were somehow absolute. In fact the time of the freefaller is equally valid, and is more appropriate to use when considering the freefaller. Hence the freefaller does not "wait" a long time and not notice his waiting-- he just perceives time proceeding normally and crosses the event horizon in short order.

If the freefaller could see back into the universe the person would see time flying infinitely fast compared to his/her clock.
This is not consistent with your claim that light from the Earth would not be redshifted. Imagine a pulse of light is emitted by the Earth every second. If that light arrives one pulse per second to the freefaller, and if he "sees" time go by infinitely fast on Earth, then how could all of Earth history occur between two pulses of light that were only emitted a second apart from Earth? The redshift is the rate of flow of time on Earth that the freefaller sees. But even that isn't complete, because we never base our idea of how time is proceeding based on what we see, we always know we have to correct for a varying light travel time. So if the freefaller sees a pulse a second, and the Earth is emitting a pulse a second, note that the freefaller will also conclude that the distance to Earth is increasing as he/she approaches the event horizon (since the event horizon is farther than points leading to it). So an increasing distance and an observed pulse a second means that the freefaller will conclude that time was indeed passing faster on Earth and the pulses were emitted more rapidly than 1 per the freefaller's concept of a second, i.e., it must have been redshifted. So ultimately publius may be correct that a redshift is inferred (although not actually seen, since it would seem to be correct that the light "falls" in the same way as a freefaller with no excess speed). But again, I would defer all this reasoning to the result of an actual calculation. Grey?

Squashed
2006-Jul-18, 03:36 PM
2) For the falling observer is the light redshifted or do the effects compensate and give no redshift?

I would think that, at the event horizon, the blueshifting due to gravity would be equivalent to the redshifting due to velocity and so the effect would be null.

If the freefaller started from infinity then the freefaller's velocity would equal the speed of light at the event horizon.

Light that started from infinity would be blueshifted by gravity to a zero wavelength but the velocity of the freefaller would redshift the light to infinite wavelength - do these two mathematically cancel to equal the original wavelength that the light was created at?

Ken G
2006-Jul-18, 08:48 PM
That is indeed the key question, we need a GR calculation to be sure.

ToSeek
2006-Jul-18, 10:15 PM
ToSeek. Actually, with all due respect,I disagree. The current work at the Minos experiment is to determine the cross-sections for fairly energetic neutrinos, and very energetic neutrinos. Trickier work in the future should yield the poorly known low-energy bandpass.
Any acceleration in what is presumably..(at this point)...a relatively isotropic neutrino sea will change that isotropy to an anisotropic flow. The differences in cross-section for the blue-shifted,fore, and red-shifted aft fluxes should be detectable...(if not now, soon). There is as yet no way to conduct a ship acceleration experiment without invoking the consequences of neutrino sea isotropy. If your ship were large enough to accommodate a Minos setup...you'd "see" the acceleration in the scattering data, without looking out the windows. Pete.

Sounds as if you could be helpful in answering this question. (http://www.bautforum.com/showthread.php?t=44463)

RussT
2006-Jul-18, 11:32 PM
Neutrinos get created through nuclear fusion..


Yes, I would definitely like to know the answer to this one.

grav
2006-Jul-19, 12:29 AM
Not quite. The observer falling in will do so in a very short period of time, and will also see only a short period of time elapse outside. So he won't agree that the distant observer grows old before the falling observer reaches the event horizon, though he will agree that the distant observer will grow old and die before he sees the falling observer reach the event horizon.
Shouldn't the time that elapses for the distant observer and the one falling in be the same in both of their respective cases? That is, if in one second of the faller's time, he sees the other age by, say, one minute, then in one minute of the distant observer's time, he should see the faller age by only one second. In other words, the faller will seem to be moving very slowly (although falling rapidly) to the distant observer, and the distant observer will be moving around (and aging) very rapidly to the faller's point of view. Also, the distant observer wouldn't grow old and die before the falling one reached the horizon. Only that amount of time that has actually elapsed for the distant observer as he watches the faller cross the event horizon will actually be observed by the faller. Since the faller is travelling close to the speed of light to the distant observer, this time is small. The falling observer will only see those events which actually take place during this same time, but to him, they will appear to occur much more rapidly. If he falls in within one minute of the distant observer's time, then only one minute of aging will be seen by the falling one before he crosses, but it will seem to occur within only a second to the faller.

Furthermore, doesn't all of what we are talking about go against the time dilation concept, at least as far as the pure velocity part of it is concerned? When accelarating in or out of a gravitational field, which according to the equivalence principle would be the same as "normal" acceleration, we say one grows old while the other stays young. But for uniform velocities, we say that both observers see the other as moving more slowly. But shouldn't the same thing apply to acceleration? Each would see the other as accelerating away, and the time elapsed should seem smaller to both observers. Imagine, for instance, two observers moving away from each other at equal accelerations. How would each observe the other? After all, acceleration is just as relative as velocity. If it weren't, we couldn't add and subtract forces or use vectors for them the same as we do for distance and velocity.

Since I know someone will say it depends on which one is actually doing the accelerating, imagine then, both observers falling into black holes of equally mass from equal distances. How would they then view each other?

grav
2006-Jul-19, 01:37 AM
1) At the event horizon, for a stationary observer held up by some force, what is the blueshift of light from Earth? (That's the question that I meant to ask, not involving the motion of the falling observer, just the comparison of the flow of time at that point.)
My take on this is rather simplistic. The theories of relativity depend on relative values, and since there is no motion (the observer is stationary), it should not matter that they are standing on the edge of a black hole. Any mass is the same as far as the gravity is concerned. So this is all we need consider. The light is redshifted from Earth's gravitational well according to its mass and radius and then blueshifted by the black holes gravity according to its mass and the radius of the event horizon.I'm sure, however, that some of you would think that a lot more is going on than this, but I don't think so.


2) For the falling observer (whose motion is as though he/she started with zero speed at infinity-- note that publius' redshift might be due to propelled motion by the observer's ship as it speeds toward the black hole in an effort to speed up the trip) is the light redshifted or do the effects compensate and give no redshift? (Which is also a measure of the flow of time for the falling observer compared to back on Earth.)
Whether we are considering a relativistic gravity well or not, the results should be the same for both. But instead of light speeding up, it simply shifts frequency. It gains the same amount of energy coming from an infinite distance as one would gain from falling from infinity, so within the same frame of reference, one should see no redshift or blueshift.

grav
2006-Jul-19, 02:13 AM
If the freefaller started from infinity then the freefaller's velocity would equal the speed of light at the event horizon.

I have wondered about this. Do we define the event horizon by where the escape velocity is the speed of light or that point where light travelling perpendicularly to it (across its tangent) will not escape, but orbit it instead. The first definition seems more likely. But if that is the case, then objects could still orbit at the event horizon at much less than the speed of light. The escape velocity, where one is travelling straight out from some initial distance, is typically 21/2 times greater than the orbital velocity at that same distance.

All of this brings up another interesting question about the bending of light in a gravitational field. If the path of light runs tangent to a massive body, it will bend at twice the predicted Newtonian value. But isn't this simply because the speed of light does not change, only its direction? In other words, for an object travelling toward a massive body, at an initially large distance from the body, even though it eventually passes tangent to it, it is travelling almost directly toward the body. It will then gain velocity due to gravity. By the time it passes tangent to it, it will have accelerated greatly to a larger velocity than it originally had. During the small time that it actually travels tangent to the body, it will change its direction slightly toward the body. It will then decelerate after passing it as it travels almost directly away from the body, but with a changed direction.

The thing is here that light will not accelerate as it approaches the body nor decelerate after, since its velocity remains constant. It will only blueshift and then redshift. But while it is travelling at the tangent, it will be pulled toward the body and it changes direction. So light is not effected by gravity except perpendicularly. It only blueshifts or redshifts when travelling toward or away. This means that the light, since it did not accelerate upon approach, will take longer than it otherwise should to cross the tangent (the Newtonian version would have it accelerate upon approach and cross the tangent more quickly). Since it takes longer to cross the tangent, which is where the actual "bending" would take place, gravity will have more time to effect it, and it will bend much further than the Newtonian value.

publius
2006-Jul-19, 04:11 AM
If the freefaller started from infinity then the freefaller's velocity would equal the speed of light at the event horizon.


Squashed,

All this gets very tricky. Relative to what will the free-faller's velocity be equal to c? In SR, we knows "things are relative" plus simultaneity plays games as well. Well, in GR and non-inertial frames, these tricks get even trickier.

We all hear this business about free fall velocity reaching light speed at the event horizon (and that holds true for the acceleration field event horizon as well as the black hole), or that escape velocity is equal to c. In fact, the GR expression for the Schwarzchild radius is just the Netwonian expression for a classical escape velocity of c.

But again, I ask, relative to whom -- what frame -- will that velocity be c? You might say, to an observer a long distance away watching the free fall, as that would be the classical picture. But not so. That distant frame sees the free-faller *stop dead*, with his clock stopping completely at the event horizon. In his global coordinates, the free-faller *SLOWS DOWN* to a crawl and his image is redshifted to DC, nothing!

That blows your mind, but at the same time, you hear the "speed reaches c", yet appears to stop. What is meant by the speed approaching c is relative to a *local stationary* frame right at the event horizon. (But to remain stationary at the event horizon, you would have to exert infinite force, or expend infinite energy or something like that -- there are no stationary frames inside the horizon -- you can't think of an observer being stationary there). That frame will see the free-faller reaching lightspeed. In GR, you work out the "metric" in terms of some global coordinates, and this gives a stationary frame picture. You then calculate local frames at some stationary point.

And remember free-fallers, and that includes orbiters as well, are not stationary. Things are going to look very different to them. In fact, I think a free-faller, so long as he was small enough not to notice tidal forces, could use SR entirely to calculate clock rates and doppler shifts of everyone else, intergrating any acclerations he sees. And this because, via the Equivalence Principle and the very basic tenets of GR, that free-faller is inertial, follwing geodesics.

Now, an observer who is feeling a force, stationary, or otherwise resisting gravity by not following geodesics(and that would tidal free-fallers, resisting the tidal forces stretching them) could not use SR but would have to invoke the full GR to calculate what other observers see.

You can see how things get very weird quickly with "strong gravitational fields" (and that means very high differences in potential from the starting point, not so much the value of g at any point as illustrated by the supermassive black holes).

-Richard

publius
2006-Jul-19, 04:54 AM
Grav,

Acceleration is not relative like velocity is in SR. It is far from it. Sometimes the Equivalence Principle is misunderstood to mean this. An accelerating observer "feels a force". In the language of GR, that means one is deviating from geodesics.

Sitting at your chair reading this, in GR you are accelerating. The earth below is pushing up against you, preventing you following the geodesics. I flat space-time, geodesics are straight lines. If we aren't feeling a force, and observer some other object to be following a curved path (v(t) is not constant), then that object is accelerating.

However, gravitational fields curve geodesics from straight lines. Inertial objects *appear* to accelerate, just like an acclerating observer sees inertial objects appear to accelerate in his coordinates. And since those geodesics are curved, a actual straight line path, the 3D part of 4D space-time is actually an accelerated path.

That is what is being equated with the Equivalence Principle. The free faller is equivalent to an inertial observer, and someone stationary in a gravitational field, resisting gravity is equated to the accelerating observer actually feeling a force.

Following geodesics = inertial, feeling no force. Deviating from geodesics = accelerated, and feeling a force.

As we were discussing, gravitational fields and accelerated frames of reference "break" the reciprocity and symmetries you see in SR. If one observer see another observer moving at a certain speed, and a clock on that ticking at a certain rate, that does not mean that other observer has to see the "reciprocal" picture in the SR sense.

And that is just what you're trying to invoke in your reasoning, making one observer see the "inverse/reciprocal" or whatever the proper term would be for the other. This does not hold in accelerated frames.

-Richard

grav
2006-Jul-19, 05:26 AM
Publius,

I have a question for you. A free-faller in an accelerating field of a black hole would see a distant observer aging more rapidly, correct? Now what if we suddenly removed the gravity before he reached the event horizon (the black hole disappeared)? I know this could never really happen, but hypothetically. The free-faller would no longer accelerate, nor would he decelerate. He would just continue along at his present velocity, close to the speed of light. No force has been acted upon him, just shut off. Yet, whereas before he viewed the distant observer as aging very rapidly, now he views him as moving very slowly, the same as the distant observer views the free-faller. So my question is, how do we jump from one frame to the other when the same velocity is present with or without the acceleration (before or after the moment of shut-off)? After all, all acceleration really is is just an increase in velocity, so SR should hold for any point in the instantaneous velocity at any particular moment. In other words, acceleration is not some mystical entity that effects bodies in some strange way. And it is not the same as gravity when one is stationary on a surface (because the surface also pushes back, and no real acceleration takes place since the forces cancel). But pure acceleration is really just a pure velocity that is constantly changing. So we should need only to consider this velocity at some particular point to relate the relativistic effects. (Okay, go ahead. Let me have it. :whistle: )

[EDIT-Okay. Before you answer, I think I figured part of this out. After the gravity is shut off, SR is all that remains. But while gravity is still in effect, the light recieved by the distant observer is redshifted due to gravity, so that the pulses are more spread out. That would mean that they are received over a longer interval of time, so the falling observer appears to move even slower than SR only would imply (which would also make it a mere illusion, not just time dilation). But then, going by this same line of reasoning, the falling observer would have to see blueshifted light in order for the distant observer to appear to be moving (aging) more quickly. If the falling observer were supported near the event horizon, this might be the case. But since he is falling, this should also cancel out the blueshift, which brings us back to the original question. Sorry, I guess the question still stands. (In case you can't tell, I'm thinking about this as I write. :D)]

publius
2006-Jul-19, 05:50 AM
Grav,

I'm going to defer your question about the free-faller in the black hole to someone who actually can do the GR calculation. However, a free faller in some accelerated psuedo-gravitational field (who is just an inertial observer) would not see the accelerated observer's clock speed up and age rapidly. He would see the accelerated observer receding away closer and closer to c and his clock moving very slowly. However, a coaccelerating observer "deep down" in that acceleration field would see the clock above him ticking very fast. But a black hole field is different, but for a supermassive black hole, g is small and gradients in g small near the event horizon, and that field is very near the accelerated psuedo field. So I think there, it would be the same.

Now, to the other points. What force is being cancelled standing on the surface of the earth? The only "force" I can think of that would cancel the real force of the earth pushing up is this "mg" thing. But I don't feel that at all. Looks exactly like the fictitious forces I need to invoke to calculate things in a Newtonian non-inertial frame.

That force I feel sitting here pushing my hind end is unbalanced. I am accelerating. In fact, without looking out the window, that's exactly what I would feel if I were in a rocket ship thrusting at constant thrust so that I felt 1g of force. I cannot tell the difference. That's what the Equivalence Principle is all about. If I drop a ball, I'll see it free fall at 1g worth of acceleration. All objects will fall at the same rate. I cannot tell the difference, and have no clue if I'm accelerating in 3-space or sitting stationary in a gravitational field.

In GR, the force of the earth on our hind ends is indeed unbalanced. It is keeping us from following our geodesics, pushing us in a non-inertial orbit around the center of the earth. And just like the centripetal force (the real force) that keeps an object moving in a circle, that real, unbalanced force is doing no work, always perpendicular to the velocity (and this is 4D, 4-vector velocities and directions).

-Richard

Grey
2006-Jul-19, 01:19 PM
Wow, I have catching up to do!


I was tempted to reach this conclusion as well, after publius pointed out my mistake in not including the motion of the falling observer. One thing that's sure is that it is very hard to get things right in GR without actually doing the calculation!Absolutely! However, using the equivalence principle as a guide, we can often make good judgments, as publius has suggested. All of my comments here assume that general relativity is the correct way to describe what happens near a black hole, or any other strong source of gravity.


1) At the event horizon, for a stationary observer held up by some force, what is the blueshift of light from Earth? (That's the question that I meant to ask, not involving the motion of the falling observer, just the comparison of the flow of time at that point.)For a stationary observer, the blueshift can be arbitrarily large depending on how close the observer is to the event horizon, and such an observer will see. Yes, that means that, by approaching close to a black hole, staying there for a while, and then leaving (expending truly immense amounts of fuel in the process), you can effectively move forward in time.


2) For the falling observer (whose motion is as though he/she started with zero speed at infinity-- note that publius' redshift might be due to propelled motion by the observer's ship as it speeds toward the black hole in an effort to speed up the trip) is the light redshifted or do the effects compensate and give no redshift? (Which is also a measure of the flow of time for the falling observer compared to back on Earth.)I've seen the calculation, but I can't find it now. However, publius pointed the way. An observer freefalling into a black hole can be considered (via the equivalence principle) to be an observer in an inertial frame. The light is coming from a source moving away, and so is redshifted.


This statement treats the time of the distant observer as if it were somehow absolute. In fact the time of the freefaller is equally valid, and is more appropriate to use when considering the freefaller. Hence the freefaller does not "wait" a long time and not notice his waiting-- he just perceives time proceeding normally and crosses the event horizon in short order.Correct.


As far as time dilation for blackholes, if it is true according to Einstein, then I still maintain that time stops at the event horizon and the freefaller never falls in. As for the freefaller the person will look at their clock and forever wait for the next second to tick but will never realize that time has stopped for him. If the freefaller could see back into the universe the person would see time flying infinitely fast compared to his/her clock.As Ken G points out, this is absolutely not the case in general relativity. An observer freely falling into a black hole will measure a (short!) proper time. Moreover, such an observer can see out into the outside universe, and will only see a short time elapse for people distant from the black hole.


Shouldn't the time that elapses for the distant observer and the one falling in be the same in both of their respective cases? That is, if in one second of the faller's time, he sees the other age by, say, one minute, then in one minute of the distant observer's time, he should see the faller age by only one second. In other words, the faller will seem to be moving very slowly (although falling rapidly) to the distant observer, and the distant observer will be moving around (and aging) very rapidly to the faller's point of view.Nope. That's not what general relativity says. It may seem natural, but it's just not what the calculations show. And in any case, it should be no surprise that two observers do not always see the inverse of each other. Even in special relativity, for two observers moving past each other with some relative velocity, one doesn't see the other's clock slowed while the second see's the first's clock moving faster. Both see the other's clock slowed. It's true that, if you have two observers that somehow reunite, all observers will agree on what both clocks read at the point that they do so, but they may disagree on just how those clocks got to read those values. And for two observers who do not reunite (such as when one falls into a black hole!), they can never directly compare their clocks again, so there's no requirement that the clocks have any specific relation to each other. Remember, for observers that are moving differently, they will have different but equally valid views about which distant events are considered simultaneous with "now".


Since I know someone will say it depends on which one is actually doing the accelerating, imagine then, both observers falling into black holes of equally mass from equal distances. How would they then view each other?In this case, they should both see each other redshifted into oblivion.


Whether we are considering a relativistic gravity well or not, the results should be the same for both. But instead of light speeding up, it simply shifts frequency. It gains the same amount of energy coming from an infinite distance as one would gain from falling from infinity, so within the same frame of reference, one should see no redshift or blueshift.Well, no actually. If general relativity gave all the same results as Newtonian gravity, we wouldn't need to bother with it. It's precisely in cases where gravity is particularly strong that the results from general relativity can differ significantly from the Newtonian equivalent.


I have wondered about this. Do we define the event horizon by where the escape velocity is the speed of light or that point where light travelling perpendicularly to it (across its tangent) will not escape, but orbit it instead. The first definition seems more likely. But if that is the case, then objects could still orbit at the event horizon at much less than the speed of light. The escape velocity, where one is travelling straight out from some initial distance, is typically 21/2 times greater than the orbital velocity at that same distance.This is one of those places where gneral relativity and Newtonian mechanics differ. There is a location (called the photon sphere) where the orbital velocity is exactly the speed of light. However, this is actually outside the event horizon. Below the photon sphere, there are no stable orbits. That is, it's theoretically possible to cross the photon sphere, travel closer to the event horizon, and still make it back out(by expending massive amounts of fuel), but there are no free orbits smaller than this.


A free-faller in an accelerating field of a black hole would see a distant observer aging more rapidly, correct?This is not correct. An observer freely falling into a black hole will see the outside universe redshifted and time moving more slowly. It's only an observer accelerating in order to remain stationary near a black hole that will see a distant observer aging more rapidly.

[edited to correct attribution]

Ken G
2006-Jul-19, 02:25 PM
OK, that straightens things up, thanks Grey and publius. I think publius' initial point has been firmly established-- there are very few vestiges of "reciprocity" remaining in extremely general-relativistic situations like falling into a black hole!

Squashed
2006-Jul-19, 03:20 PM
As Ken G points out, this is absolutely not the case in general relativity. An observer freely falling into a black hole will measure a (short!) proper time. Moreover, such an observer can see out into the outside universe, and will only see a short time elapse for people distant from the black hole.

Ok, the freefaller into a blackhole is essentially inertial and so only the velocity of the freefaller would affect time, according to Special Relativity, which means that the time dilation is reciprocal: the distant observer's clock has stopped in the freefaller's perspective whereas the freefaller's clock has stopped in the distant observers perspective.

But according to the time dilation formula for special relativity the only inputs are "proper time" and "velocity" and so the actual "seeing" of the participants is not of clocks but of velocities. This means that each participant sees the other participant as the moving subject and according to the formula their relativitstic time is slow, or stopped, but in reality time moves the same for both and so the freefaller plummets to their fate while the distant observer watches helplessly.

General relativity only affects time dilation if the clock/participant "feels" the force of gravity.

Grey
2006-Jul-19, 03:49 PM
This means that each participant sees the other participant as the moving subject and according to the formula their relativitstic time is slow, or stopped, but in reality time moves the same for both and so the freefaller plummets to their fate while the distant observer watches helplessly.What do you mean by the bolded statement? Remember, in relativity, there is no absolute reality. Each observer will always consider his or her own clocks and rulers to be behaving normally, and can use the mathematics of relativity to determine what lengths or times some other observer might measure. Either view is considered just as valid. It's not "each observer sees the other's time slowed, so they're really moving at the same rate". It's that different observers see things differently, and that both view are valid. So, from the falling observer's point of view, the fall takes a short period of time, and only a small amount of time passes in the outside universe. From the distant observer's point of view, the falling observer takes an infinite amount of time to reach the event horizon. There's no sense in which one of these views is correct and the other is wrong.


General relativity only affects time dilation if the clock/participant "feels" the force of gravity.No, in a situation where spacetime is curved that strongly, both observers will need to use general relativity to accurately calculate what the other will see. Remember that general relativity only says that a freely falling observer can be considered to be inertial, within a reference frame that covers a small region that can be considered locally flat. (How small of a reference frame depends on how accurate you want your calculations to be). To determine what you'd see in regions beyond such a local inertial frame, you'd have to essentially integrate over a series of such inertial frames, each covering a small region. In a region where the curvature of space is not negligible, there are no globally valid special relativistic reference frames.

Squashed
2006-Jul-19, 05:31 PM
... in reality time moves the same for both ...

What do you mean by the bolded statement?

CALC-A:
tB = TA*sqrt(1-vB2/c2) <--Relativistic time calculation for "B" from "A" perspective

CALC-B:
tA = TB*sqrt(1-vA2/c2) <--Relativistic time calculation for "A" from "B" perspective

The velocities viewed from either perspective are equal and so: vA=vB and therefore I can drop the subscripts from the velocity variable: "v".

From CALC-A, solving for TA:

TA=tB/sqrt(1-v2/c2)

The above is the ratio of time as it ellapses on the "B" clock with respect to the "A" clock. Since the "B" clock is the proper time for the "B" perspective then

TB=tB/sqrt(1-v2/c2)

substituting this calculated proper time for the "B" perspective into CALC-B for the relativistic time for "A" yields:

tA = tB/sqrt(1-v2/c2)*sqrt(1-v2/c2)

cancelling out the sqrt(1-v2/c2) gives:

tA = tB

Therefore, relative velocity should not affect the rate of flow of time in either perspective and; thusly, both clocks run at the same rate.

grav
2006-Jul-19, 10:21 PM
Grey,

This thread has confused me immensely. I seem to be getting conflicting answers. Can you help me out with this? Please allow me to start over and see what we can determine. I will try the scenario according to what I think would be observed and see what you think of it.

Now, let's start by saying that the black hole has a solid surface to stand on near the event horizon. Since the observer at the event horizon is stationary and so is the distant observer, SR doesn't apply. But GR says that light received by the distant observer would be redshifted by the black hole's gravity. This means that the apparent time between pulses will be much greater, so that the person standing near the black hole will appear to move very slowly. But the light received by the observer at the black hole will be blueshifted because of the enormous gravity of the black hole as the light enters it, whereas it gains energy. So for that observer, the distant one will appear to move around (or age) much more quickly. Now, the redshift that the distant observer will see should be precisely equal to the amount of blueshift from the other direction since the same amount of energy will be lost or gained by light depending on which direction it is travelling relative to the black hole. Whether or not this means that the times observed by each will be reciprocals or not, I cannot say for sure, but they are certainly opposite (one faster, one slower). This much of it is obviously just an illusion, however, afforded by the gain or loss of energy by light as it travels in or out of the gravity well. It is not a real time dilation.

Now let's say that the observer at the black hole is freefalling into it. Now the light that is received by the freefaller is again redshifted according to SR (the relative velocities between the two observers) and blueshifted by the black hole in equal amounts. This is because the distant observer is stationary to the black hole (at least in this example). Therefore, the instantaneous velocity at any point in freefall will create a redshift relative to the distant observer according to this velocity that will equal the amount that the light should have gained equally in the gravity well, but is blueshifted instead. These effects will cancel out, so no resulting redshift or blueshift is observed. If the falling observer had some initial velocity toward the black hole (and/or the distant observer away), however, this would add to the relative velocity between the two observers, and the light for him would be redshifted, and vice versa (blueshifted) if he was initially moving away from the black hole (and/or the distant observer toward). In the case of the stationary distant observer (when the other is freefalling), he will see the light redshifted due to GR (the gravity well) and redshifted some more due to SR (their relative velocities away from each other), so from the stationary distant observer's point of view, the light received from the freefaller will always be redshifted.

grav
2006-Jul-20, 12:16 AM
Publius,

Thank you for your response. I am still trying to determine some of what you are saying, however. Your knowledge on such things may be a little above my comprehension. I don't believe relativity to be, however. But you're right, I do tend to try to place it within a three dimensional geometry. This is partly because this is the only way I can truly imagine it, but also because I believe this is where it truly belongs. But until I can show otherwise, I will try to keep an open mind on such things. Could you also look at the preceding post and see if you can spot any specific errors on my part?

publius
2006-Jul-20, 05:41 AM
Grav,

There is nothing here above your comprehension, you just need to, well, expand your horizons a bit. I've read some of your posts in the ATM section. I get the sense of someone trying to reinvent a lot of wheels and going off down wrong directions in the process. Get some textbooks and some math books and study how the wheels of modern physics were invented and discovered.

Forget about relativity and all the weirdness there and think Newtonian for a bit. Imagine a simple linear accelerating reference frame, a simple coordinate system whose origin is accelerating. What is Newton's second law of motion in that frame?

You will find you have to introduce a mass dependent force, proportional to the acceleration of the frame to keep Netwon's F = ma. This force is not real, just an "artificat" of the accelerating frame.

Mass dependent force.......mass dependent force. Does that ring a bell? All objects will accelerate at the same rate unless acted on by another force (a real force).....all objects will accelerate at the same rate. Does that ring that bell a little louder?

That is where the Equivalence Principle comes from. The fact we feel no force in free fall (nor in any orbital path, neglecting tidal forces) strongly suggests that gravity is not a force at all, just something like the psuedoforces that have to be introduced in accelerating frames of reference.

Einstein is supposed to have quipped that he realized this when he saw someone fall off a tall building, get up and dust himself off and declare he felt no force of gravity.

The apple hit Newton on the head, and years later Einstein realized that apple felt no force (until it hit Newton on the head). Those were the two giant leaps of gravitational theory. There may be other leaps. I hope there is, but that next leap will certainly have to include something that looks a lot like GR, just as GR reduces to Newton for weak fields.

Me, I hope for a coupling between gravity and EM or some other way that I can make an "artifical" gravitational field and accelerate myself at some ridiculous rate without feeling a force. Some recent experiments show certain superconductors are making a larger gravitomagnetic field than expected from pure GR. That I think is the first serious evidence of something that will go beyond GR

-Richard

grav
2006-Jul-20, 12:06 PM
Publius,

Reinventing the wheel is exactly what I'm attempting to do. I believe physics has taken a wrong turn somewhere (although I'm sure many will disagree) and I'm trying to see if it can be put back on track. But I have to do this within my own limited abilities. I am only one man, and I am bound to go down many of the wrong roads myself. This makes my learning process more of a process of elimination. I consider all of the possibilities (which also allows me to keep an open mind about things) and then scratch off the ones that result in paradoxes and impossibilities. Relativity appears to carry many of these. But I have also learned that there are many ways to look at things. So I am trying to see if some of them can be resolved. My only brawback is my own impatience. I cannot stand to be "stuck" for too long. That is why I am sticking this one out, in order to bring it to its full conclusion. Once I can see precisely where I have erred, I can then once again move forward, either accepting or altering my concept of the theories of relativity. I don't suppose I should reject it altogether.

I not sure why you keep insisting on using the equivalence principle. I thought that was what I was doing. I do have issues with it, however, as you know. I can see perfectly well how it is applied to the effects of gravity. There appear to be two ways of thinking about gravity, one is as the cause (the warping of space-time) and the other the effect (acceleration). As far as the effect goes, gravity is acceleration, but that to me is only common sense since gravity creates acceleration. I realized the similarity as a teenager, before I'd even read about relativity. One of my first concepts for the cause of gravity was an accelerating expanding surface for the Earth. We would not notice the expansion because all objects within the Earth's "space-time" would also be expanding at the same rate. But we could see this expansion in other massive bodies with different expansion rates. I suppose this could be considered my first official delve into relativity, and unknowingly using the equivalence principle.

Let's try something. If gravity is acceleration, cause as well as effect, then I suppose this would be where the warping of space-time comes in, right? In other words, acceleration is a change is space and time (distance and velocity with time), so gravity causes this by warping (changing) space-time itself, is that correct? But if this is the case, then a rocket propulsion engine, which also creates acceleration, must also warp space-time. The equivalence principle holds for it as well. After all, this is what we are really comparing gravity to in our thought experiment, aren't we? This is what is causing our uniformly accelerating platform to accelerate in the first place. So by saying the effects of gravity are the same as the acceleration in both cases, then the causes must both produce those same results precisely for this to occur, which is the warping of space-time. This would mean that we can not tell the difference between standing on Earth in one g or accelerating in a ship at one g. If I were in such a box, where I could not tell the difference between gravity and normal acceleration when I cannot see out of it, it's also true that I would't know if I were standing in my own bedroom or the Oval office. Many such conclusions can be made with the conditions that our eyes are closed, but many differences exist as well, we just can't directly see them this way, but that is only common sense. So my question here is this. How does the ship warp space-time to produce acceleration in the same way as a massive body such as the Earth? Why isn't the acceleration felt outside the ship as it is with gravity? How does the inverse square law apply to rocket propulsion? A wall that exhibits 1 g of gravity will allow us to stand sideways upon it because the force is distributed throughout our body while a platform pushing us into the wall at 1 g will crush us in a second because the force is only distributed over the surface of our body. Where is the equivalence principle in that? It certainly wouldn't "feel" the same. Much more painful, I would imagine. I know I'm going a little overboard with this, but you can see my point. My real question is, if the ship doesn't require warping space-time to produce acceleration, then why does gravity necessarily have to?

Just so you know, I also understand how freefall would feel as if no forces were acting. This is the kind of relativistic effect I am currently looking into. In my opinion, this is where the real relativity lies. But a force must still be present, otherwise a falling object would not continue to accelerate (according to the gravitational body's frame of reference). The concept of the relativity of it also applies to centrifugal forces. With it, I have determined that a perpendicular velocity acts to diminish the centripetal force (gravity). In the case of an orbit, the centrifugal force has completely balanced gravity and the orbitting body is in equilibrium. But this does not mean that gravity is not a real force. Just the opposite, in fact. It is the centrifugal force that is not real, but only the difference between the full force of gravity at that point (for a stationary object) and its diminished value caused by the relativity of velocities perpendicular to the lines of force. Since you mentioned it, I also believe magnetism to be caused by similar means when applied to an electric field and I will explore that possibility as well (with much help from you, of course).

Grey
2006-Jul-20, 02:20 PM
From CALC-A, solving for TA:

TA=tB/sqrt(1-v2/c2)

The above is the ratio of time as it ellapses on the "B" clock with respect to the "A" clock. Since the "B" clock is the proper time for the "B" perspective then

TB=tB/sqrt(1-v2/c2):eh: What justification do you have for replacing the time as seen by A (in A's reference frame) with the time as seen by B (in B's reference frame)?

Squashed
2006-Jul-20, 02:27 PM
...because the force is distributed throughout our body while a platform pushing us ... because the force is only distributed over the surface of our body...

I have noted the same differences between gravity and acceleration in this post (http://www.bautforum.com/showpost.php?p=773345&postcount=11) and also others in this post (http://www.bautforum.com/showpost.php?p=773180&postcount=6).

Grey
2006-Jul-20, 03:00 PM
Grey,

This thread has confused me immensely. I seem to be getting conflicting answers. Can you help me out with this? Please allow me to start over and see what we can determine. I will try the scenario according to what I think would be observed and see what you think of it.General relativity can be pretty complex, and black holes are specifically places where it gives some pretty unexpected results. I'm not certain what all of these results would be either, but I'll do my best. Hopefully, for any of the questions that we really need to calculate, we can find a paper or text where someone else who has done them, and we can just follow along. That's both to avoid a lot of work, and because if it's been published it's less likely that they've made a mistake than if I worked it out. :)


Now, let's start by saying that the black hole has a solid surface to stand on near the event horizon. Since the observer at the event horizon is stationary and so is the distant observer, SR doesn't apply. But GR says that light received by the distant observer would be redshifted by the black hole's gravity. This means that the apparent time between pulses will be much greater, so that the person standing near the black hole will appear to move very slowly. But the light received by the observer at the black hole will be blueshifted because of the enormous gravity of the black hole as the light enters it, whereas it gains energy. So for that observer, the distant one will appear to move around (or age) much more quickly.We're good so far. This is definitely the case.


Now, the redshift that the distant observer will see should be precisely equal to the amount of blueshift from the other direction since the same amount of energy will be lost or gained by light depending on which direction it is travelling relative to the black hole.I'm not actually certain about this one. I'll see if I can find out. It seems sensible enough, but near a black hole, spacetime cannot even be considered static, let alone flat. So there may be differences between what happens if a photon is leaving (moving against the "flow" of spacetime), or moving toward the black hole (travelling "with the flow").


This much of it is obviously just an illusion, however, afforded by the gain or loss of energy by light as it travels in or out of the gravity well. It is not a real time dilation.Saying it's not real is a bit of a mistake. If you travel close to a black hole, stay there a while, and then come back to meet up with someone that stayed far away (all the while keeping your speed low enough that special relativity won't have signficant effects) you really will have aged less. You can use that technique to effectively travel as far into the future as you'd like. It's not just an illusion.


Now let's say that the observer at the black hole is freefalling into it. Now the light that is received by the freefaller is again redshifted according to SR (the relative velocities between the two observers) and blueshifted by the black hole in equal amounts. This is because the distant observer is stationary to the black hole (at least in this example).If you want to know what someone sees, you really need to work from their reference frame, not someone else's. Here, you're trying to figure out what happens from a reference frame stationary with respect to the distant observer, and then work out what someone accelerating will see. So without working out the math in detail, you can't be sure that these two effects exactly cancel. When I saw the math worked out (years ago, I'm afraid), the result was a net redshift, and, as publius pointed out, when you realize that such an observer can consider themselves to be inertial with the distant source accelerating away, it's possible to see why that redshift would result.


Therefore, the instantaneous velocity at any point in freefall will create a redshift relative to the distant observer according to this velocity that will equal the amount that the light should have gained equally in the gravity well, but is blueshifted instead. These effects will cancel out, so no resulting redshift or blueshift is observed.Not according to general relativity, actually.


In the case of the stationary distant observer (when the other is freefalling), he will see the light redshifted due to GR (the gravity well) and redshifted some more due to SR (their relative velocities away from each other), so from the stationary distant observer's point of view, the light received from the freefaller will always be redshifted.This is correct.

Grey
2006-Jul-20, 03:23 PM
I have noted the same differences between gravity and acceleration in this post (http://www.bautforum.com/showpost.php?p=773345&postcount=11) and also others in this post (http://www.bautforum.com/showpost.php?p=773180&postcount=6).And publius explained your misunderstanding about that in the post (http://www.bautforum.com/showthread.php?p=773367#post773367) immediately afterward. :)

Squashed
2006-Jul-20, 04:03 PM
:eh: What justification do you have for replacing the time as seen by A (in A's reference frame) with the time as seen by B (in B's reference frame)?

I think I may have messed up in that calculation because it is not real clear how I got to where I did and so I will try to explain it in another way:

CALC-A determines the amount of proper time that the "B" clock registers with respect to the "A" clock which is a ratio of "B" proper time to "A" proper time. This means that tB=TB.

If I put the tB calculation which is TA*sqrt(1-v2/c2) into CALC-B then I get nonsense because the tA does not equal the original TA.

What I am trying to get at is that I can calculate the amount of proper time that ellapses for "B" from the "A" perspective but when I input that calculated proper time into the "B" perspective equation to backcheck the proper time that "A" should be experiencing, the value is less than the value I started with.

Putting numbers in will be more explanatory, I hope:

If the relative velocity between "A" and "B" is 50% the speed of light or v = 0.5c then for one second of proper time in the "A" perspective then the "B" perspective will experience:

tB = TA*sqrt(1-v2/c2)

for:

v = 0.5c
TA = 1.0 second

then

tB = 1.0 second*sqrt(1-(0.5c)2/c2)
tB = 1.0 second*sqrt(1-0.25*c2/c2)
tB = 1.0 second*sqrt(1-0.25) <-- the c2 cancels out of both the numerator and denominator
tB = 1.0 second*sqrt(0.75)
tB = 1.0 second*0.866025403784439
tB = 0.866025403784439 second

From the "B" perspective 0.866025403784439 second passes for every 1.0 second in the "A" perspective but using the same formula from the "B" perspective yields:

tA = TB*sqrt(1-v2/c2)

for:

v = 0.5c
TB = 0.866025403784439 second

then

tA = 0.866025403784439 second*sqrt(1-(0.5c)2/c2)
tA = 0.866025403784439 second*sqrt(1-0.25*c2/c2)
tA = 0.866025403784439 second*sqrt(1-0.25) <-- the c2 cancels out of both the numerator and denominator
tA = 0.866025403784439 second*sqrt(0.75)
tA = 0.866025403784439 second*0.866025403784439
tA = 0.75 second

Since tA = 0.75 second does not equal TA = 1.0 second ... then I conclude that the formula is in error.

To correct the error requires that the formula be written as:

t = T*((1-v2/c2)0.5)(v/|v|)

where |v| = the absolute value of v

Which requires the velocity vector direction to be included into the calculation.

Requiring the direction to be included means that relative velocity alone is insufficient for determining relativistic time because one perspective must consider itself to be the base away from which all velocities are positive and towards which all velocities are negative.

In other words: In the current special relativity formula there is no provision to calculate from a slower time rate perspective to a faster time rate perspective.

The only way to make the current special relativity formula work is by taking a "god's eye view" of the proper time and so for both CALC-A and CALC-B the proper time entered equals 1.0 second which ends up that:

tA = 0.866025403784439 second
tB = 0.866025403784439 second

which means that tA = tB and therefore there is no velocity-induced time dilation between perspective "A" and perspective "B".

Grey
2006-Jul-20, 04:22 PM
If I put the tB calculation which is TA*sqrt(1-v2/c2) into CALC-B then I get nonsense because the tA does not equal the original TA.

What I am trying to get at is that I can calculate the amount of proper time that ellapses for "B" from the "A" perspective but when I input that calculated proper time into the "B" perspective equation to backcheck the proper time that "A" should be experiencing, the value is less than the value I started with.And that's expected, because special relativity does not predict that the two will be reciprocal. What you're expecting would only be the case if, when A sees B's clock slowing down, B saw A's clock speeding up, and that's nto what relativity predicts. In particular, the problem you're going to run into is that your CALC-A is only valid in A's reference frame, while CALC-B is only valid from B's reference frame. If you try to solve for some of these variables in one reference frame, and then put them into equations that are valid in another reference frame, you're guaranteed to get nonsense, because under special relativity, A and B measure time and space differently.


The only way to make the current special relativity formula work is by taking a "god's eye view" of the proper time...No, the current formulae in special relativity work just fine, as long as you don't try to apply them in reference frames where they are not valid. It is also possible to pick a single inertial reference frame, as you're doing here, and decide that it is the absolute reference frame, and measure all distances, velocities, and times relative to that frame, and everything will still work out. However, any inertial reference frame will suffice for this purpose; i.e., there is no measurable way to distinguish between inertial reference frames, or to give us a reason to select any particular one over any other. Moreover, if we do pick a specific reference frame as the "one true frame", then to get the real values for distances and times, we can't just use rulers and clocks. We'd have to add in a correction factor based on our "absolute" velocity.

Einstein instead choose to say that distances and times are defined as what is measured by locally stationary rulers and clocks, and that since we cannot find a way to choose a specific reference frame, that we should consider all inertial frames to be equally valid. Doing so results in some strange results, but it matches the experimental results very well.

Squashed
2006-Jul-20, 07:18 PM
No, the current formulae in special relativity work just fine, as long as you don't try to apply them in reference frames where they are not valid...

If the rate of flow of time is dilated then even though the relative velocity should be the same for both participants wouldn't the velocity actually be different because velocity is a time dependent observation?

In my example, perspective "A" measures "B" moving at 0.5c but since "B" is in a slower time then wouldn't "B" measure the "A" velocity as 0.5c/0.866025403784439?

So if I input this faster velocity into the "B" perspective formula maybe it will work out that tA = TA?????????????

publius
2006-Jul-20, 08:33 PM
Grav,


But a force must still be present, otherwise a falling object would not continue to accelerate (according to the gravitational body's frame of reference).

This is the heart of what I'm trying to get you to appreciate and the heat of the Equivalence Principle.

In the GR "geometric view" of gravity, a free falling body is not accelerating. It is completely inertial and feeling no force. Only bodies that are not in free fall must feel a force.

In flat space-time inertial paths are straight lines. v(t) = constant, and r(t) (position)is some straight line. Now subject a body to a real force. v(t) is not constant and r(t) is a curve rather than straight line.

However, the mass of a gravitating body "curves space-time" and makes those straight line r(t) become curves, some of them the orbital ellipses (depending on the angular momentum, or the component of velocity normal to the direction of the gravitating body). In 3-D space, it looks like a curve. But that curve is just an inertial path in a gravitational field. And conversely, in a gravitational field, if you see a body following a straight line path in 3D space, it is feeling a force because it is deviating from geodesics.

A free falling (and that includes orbiting) body is not accelerating, it is just following those curved geodesics. That is the overriding theme of General Relativity. All the complexity is just the geometry of how those geodesics curve and all that mess of particular coordinate systems used to describe it.

Sitting on the earth (and becase of its rotation, we are really in a non inertial orbit about the axis of rotation, requiring slightly force to push us up that would be required if the earth were not rotating) we are being accelerated in the view of General Relativity. We are being pushed off geodesics.

Is the geometric view of gravity absolutely proven? No, but it sure is a rather sastifying explanation for why we "feel no force of gravity". And the predictions of GR have stood up to every experiment and measurement so far conducted.

Gravity Probe B's data and results will be very interesting to analyze, looking for the frame dragging (B_g) and geodetic effects (ie because of space curvature, a gyro's spin axis should change slightly as completes orbit after orbit around the earth).

-Richard

grav
2006-Jul-20, 10:59 PM
Grey,

Thank you very much for that part-by-part analysis. It was just what I was hoping for. It may have cleared something up for me also that I believe Publius was trying to get across.



Quote:
Originally Posted by grav
This much of it is obviously just an illusion, however, afforded by the gain or loss of energy by light as it travels in or out of the gravity well. It is not a real time dilation.

Saying it's not real is a bit of a mistake. If you travel close to a black hole, stay there a while, and then come back to meet up with someone that stayed far away (all the while keeping your speed low enough that special relativity won't have signficant effects) you really will have aged less. You can use that technique to effectively travel as far into the future as you'd like. It's not just an illusion.


Actually, in this respect (the way light shifts), it is just an illusion. But I think what Publius meant and what you are saying is that at the vicinity of a black hole, one must also think about how real time dilation (and length contraction or lengthening) caused by gravity effects the observer near the black hole, right? This might also mean that although the redshift seen by a freefalling observer is precisely matched by the blueshift of the gravitational field and so is cancelled out, one might indeed always see a redshift due to the real time dilation from being in a gravitational field in the first place. But in this case, wouldn't the distant observer age less according to GR, since the observer near the black hole is in a gravitational field? In other words, doesn't the twin paradox say that the one furthest away from the gravitational field ages less?

[EDIT-Well, it appears the twin paradox for GR is somewhat difficult to find. But I just read a link that says that in SR, although two observers moving away from each other will see a lesser time for the other, the one that turns around will have experienced less aging than the other when they meet up again. Since it requires acceleration to turn around, I suppose that according to the equivalence principle, that would be the same as being in a momentary gravitational field. So the one in the gravitational field ages less.]

grav
2006-Jul-20, 11:37 PM
Squashed,

Thank you for the links to those posts. They were very enlightening. I'm glad you see also that there are very specific differences between the causes of gravity and normal acceleration, even if their results are the same. We cannot just blindly consider them both to be equal just because their effects are felt the same. Actually, acceleration does not really even have an effect, but is the measurement of it. Acceleration is purely geometrical, a measurement of an effect. Gravity, on the other hand, although resulting in such a geometrical solution, cannot itself also be caused by this same geometry. It's a "Which came first, the chicken or the egg?" kind of thing.

grav
2006-Jul-21, 12:40 AM
In the GR "geometric view" of gravity, a free falling body is not accelerating. It is completely inertial and feeling no force. Only bodies that are not in free fall must feel a force.

This is a very interesting way of looking at things. It is true, a body in freefall will feel no force. It would be as if they were "floating" in free space. I guess this is what the equivalence principle is saying as well. So in free space, there is no gravity, and so no real force. I not sure how to think about this yet. I would think that a force is still acting in order for an object to accelerate, but is counterbalanced somehow. The counterbalance, however, would only apply to the actual acceleration, not to the "feel" of it. This might be similar to the centrifugal force, which is also not real, but only a lessening of the effect of gravity. Gravity itself, however, I still believe to be very real.


However, the mass of a gravitating body "curves space-time" and makes those straight line r(t) become curves, some of them the orbital ellipses (depending on the angular momentum, or the component of velocity normal to the direction of the gravitating body). In 3-D space, it looks like a curve. But that curve is just an inertial path in a gravitational field. And conversely, in a gravitational field, if you see a body following a straight line path in 3D space, it is feeling a force because it is deviating from geodesics.
Well, the lines of force of gravity are still straight and act in all directions from the center of mass, spreading out with the square of the distance. It is only the objects that are effected by gravity that move in curved paths according to its force and their velocities. An object starting with no initial velocity will move in a straight line toward the mass. So once again, it is the effect that is a curved path, not the cause. Also, this might explain the apparent length contraction (or lengthening) in a gravitational field. As light moves toward a massive body, all of the rays will converge since light is also effected by gravity. That is to say, light that would normally spread out as it approached the Earth will instead be pulled together slightly as it nears the Earth, due to its gravitational field, sort of a gravitational lensing. This will make objects appear further away than they really are. A sort of GR lengthening of distance, if you will. But this, however, would also be just an illusion.


A wall that exhibits 1 g of gravity will allow us to stand sideways upon it because the force is distributed throughout our body while a platform pushing us into the wall at 1 g will crush us in a second because the force is only distributed over the surface of our body. Where is the equivalence principle in that? It certainly wouldn't "feel" the same. Much more painful, I would imagine.
Well, looks like you answered this one quite well in your post to Squashed that Squashed linked to. The platform (or ground) acts like a real force that acts against gravity, and is transmitted through our feet atom by atom throughout our bodies. So the ground stopping us from following the lines of gravity would be exactly the same as the platform stopping us. They are both caused by the same "real" force. That is the equivalence principle. Well, my question was still fun to think about, though, wasn't it?

Seriously, though, I see a serious flaw in this reasoning. The real "stopping" force you are referring to is the electromagnetic force within atoms. The electric force, at least, also follows similar rules as gravity, only much, much stronger. If you really believe that gravity and EM can eventually be related in the same way, how can you refer to EM as a real force and gravity as not? The "real" EM force, as you call it, causes a stress in our feet that works it way through our bodies as atoms push against atoms. This is the stress that we feel. It is the same for a platform accelerating us or the ground stopping us, which are themselves two very different things. So gravity is not the same as uniformly accelerated motion, but a "platform accelerating" is the same as the "ground stopping" us because of the stress we feel on our bodies which is the same in both cases.

This, by the way, answers your confusion about gravity being a real force. Gravity acts on all of the atoms in our bodies equally. So a person in freefall will not feel a stress because the atoms in their body are not pushing against each other as they would if they were standing on the ground. That is why an accelerating platform feels the same as gravity on the ground, and freefall feels the same as "floating" in free space. In the first set of cases, we have a stress that is felt through our bodies, and in the second set there is none.

publius
2006-Jul-21, 01:23 AM
Grav,

I should have clarified "curved paths", and thought about it, but didn't want to type more. :)

Take the straight line free fall. What is r(t), the position of the body vs time. Well, in the simple case of a constant g field, as we can assume for short heights above the surface of the earth, r(t) = r0 - 1/2gt^2, where r0 is the initial height the object is dropped from. (If you want a little more mathematical exercise, try solving the actual equation from straight down free fall: d^2r/dt^2 = -k/r^2)

That is a parabola, a *curve*. :) r(t) is a curve, not a straight line. For orbits, where we go into 3D space, the paths become curves in space as well as space vs time. Drop a ball with some initial sideways velocity, and the trajectory is a parabola (ignoring air resistance) as anyone who watches a stream of water out of a garden hose can see.

So the straight line free-fall, while a straight line in 3-space, is a curve when plotted against time. And that is, according to GR, is how the gravitational field of mass "warps" the straight line paths of inertial motion.

And yes, the "geometric interpretation" is not the only explanation for feeling no force of gravity. A force acting on all bits of mass at the same time, rather than having to be transmitted "atom through atom" would not cause any internal stresses. That is certainly true.

However, the other point in favor of the geometric view is the mass dependence of the force, how come all objects accelerate at the same rate. That is, how is the gravitional mass, the "charge" (EM analogy) of the gravitational field exactly proportional to the inertial mass?

In the geometric view of GR, this is not a mystery at all, it's just like the mass dependent "force" of accelerated reference frames. Every object follows those same curved geodesics, regardless of inertial mass.

And in the proper GR view, when an object stops, hits the ground, it is being accelerated then. It is like the accelerating platform hitting an inertial object moving at constant velocity. That is object is suddenly and rapidly accelerated.

Again, in the GR picture, those who are stationary in a gravitational field are the ones who are being accelerated, not those in free fall.

-Richard

gannon
2006-Jul-21, 02:03 AM
hmm this leads me to a thought
Lets say you have a ball in your ship and drop it in a hole in the ship while the ship is accelerating (lets ignore air). You could say it stopped feeling that acceleration and would continue going at its velocity.

So if you could drop the ball on earth and removed the gravity it would still go at the same velocity it was when gravity was removed. (even if it was zero)

Very interesting how that works. Please tell me if I am misunderstanding something.

grav
2006-Jul-21, 02:18 AM
So the straight line free-fall, while a straight line in 3-space, is a curve when plotted against time. And that is, according to GR, is how the gravitational field of mass "warps" the straight line paths of inertial motion.
Okay. That makes more sense. But then, the only thing that would not be a curve on such a spacetime graph would indeed be uniform motion, where v=d/t, so d and t are always in proportion. And again, even though the inertial motion is altered (curved), this does not mean that the lines of force of gravity are. The resultant curved motion is the effect. Gravity is the cause. We could also apply this to a rocket that is constantly accelerating at an angle so that its motion is curved. In this case, an acceleration is also felt which also curves the inertial motion, but without the curvature of spacetime. If gravity is acceleration, then why isn't the curved motion of the rocket also created by the warping of spacetime, or is it? But then, if it was, this should also be felt immediately outside of the ship as well.


However, the other point in favor of the geometric view is the mass dependence of the force, how come all objects accelerate at the same rate. That is, how is the gravitional mass, the "charge" (EM analogy) of the gravitational field exactly proportional to the inertial mass?
This is because of the same reason as before. Gravity effects all atoms in a body separately. So it would not matter if they were all falling together as a solid mass or separately as individual parts. I can show easily how all of this is done. But instead of writing it all down here, why don't you have a look at my website which explains it all in detail? The link for it is at the bottom of this post. Please let me know what you think. :)

grav
2006-Jul-21, 02:37 AM
gannon,

That sounds about right. Unless you are saying that they are equivalent. The ball would be accelerating with the ship while you were holding it, but would continue at a constant velocity at the moment you let go. The ship, however, would continue to accelerate past the ball as it went through the hole. With gravity, the velocity depends on when gravity is shut off. The interesting thing here is that if it were shut off at the moment the ball is released (to match the case of the accelerating ship), it would have no velocity in respect to you or the Earth and would not fall at all. The reason these two scenarios would not be equal, then, is because the relative distance between the ball and the ship still increases as if the ball is actually falling, even though it is not, but it is because the ship's acceleration has not been shut off.

publius
2006-Jul-21, 02:39 AM
Grav,

The lines of force of gravity (the 'g' field lines) can indeed be curved around, just like E and B fields lines can be. The differential form, it the contribution of each bit of mass which is then integrated is inverse square, just like Coulomb for the E field and just like Biot-Savart for the B-field. It is only when that distribution of mass is spherically symmetric does the total field become inverse square. That is close enough for planets and stars.

However, look at the total g-field of say, the Earth-Moon system, looking at the total field of both bodies together. This makes the "restricted
3-body", where you consider the motion of a small test mass in the field of two larger masses. The test mass's contribution to the total field is small enough to be ignored.

You'll see those g field lines do indeed curve around a bit.

And remember, gravity has a magnetic-like components as well, the so-called "frame dragging", or gravitomagnetic effect: moving mass, mass currents exert velocity dependent "forces" on each other just like charge currents do. But this is very small -- you can "break out" something like a mu, a gravitomagnetic permeability of space and see that is very small. Big 'G' plays the role of epsilon (actually it is the gravitational 1/4pi*epsilon), and there is corresponding "mu" for gravity as well. But in GR, this is usually wrapped up in a factor of 'c', because it is taken that gravity must propagate at 'c', just like EM -- you could write Maxwell entirely in terms of a Coulomb force constant plus 'c' rather than mu and epsilon.

This is actually done in so-called Guassian units, which is actually the unit system the EM highbrows prefer. Jackson switched to SI units in the latest edition of his "Classical Electrodynamics", the "bible" of Classical EM theory, but the switch wasn't complete, and he still left Guassian units in the Relativity part. The highbrows didn't like that at all. But me, I prefer SI and mus and epsilons.

Anyway, GR is formulated like that, with G and c. Well, enough of that unit tangent. Anyway, in GR's geometric view, gravitomagnetism is another geometric effect. A moving mass "drags" the geodesics along with it as well as statically curving them, and that produces something likes just a velocity dependent, sideways, magnetic-like force.

And Einstein did indeed try to make EM be geometric just like gravity. He was working on unifying the two in a common geometric framework, but wasn't successful. The geometry he was trying to work that in was much more complicated, with more dimensions and ways for that space-time to curve and even twist.

However, there are conspiracy theories that he was successful, but the US government covered it up to keep secret all the amazing technology that his Unified Field Theory would allow. :)

-Richard

grav
2006-Jul-21, 03:10 AM
The lines of force of gravity (the 'g' field lines) can indeed be curved around, just like E and B fields lines can be. The differential form, it the contribution of each bit of mass which is then integrated is inverse square, just like Coulomb for the E field and just like Biot-Savart for the B-field. It is only when that distribution of mass is spherically symmetric does the total field become inverse square. That is close enough for planets and stars.
Well, putting it that way, I agree with you one hundred percent. Gravity depends on the gradient of the body. For an evenly distributed spherical mass, it acts the same as a point mass at its center with the lines of force pointing straight out. But then, this is why most bodies are spherical. It is where the force (and pressure) become evenly distributed in all directions. If it were not, the resulting force from any side would eventually make it that way.

In any case, if we took two massive bodies and placed them at some distance from each other, the resulting lines of force would indeed be curved. But this is only because of how the original straight line vectors cross at every point between them, varying from one or the other with the square of the distance. Two unequal masses will create a difference in potential between them along this path. Maybe this would be similar to the workings of EM (any ideas?). Except that a plane exists between the masses (curved) where the forces from either mass cancels out. Actually, not a plane, only a single point. The plane would exist where no force is felt toward or away from either mass (because they are felt equally in each direction), but an object would still accelerate perpendicularly toward the equilibrium point. Does this sound similar to EM in any way to you? Can you think of any ways this might be applied?

gannon
2006-Jul-21, 04:22 AM
grav how does that make a difference?
The ball doesn't care what its velocity is it just wants to keep going at that same velocity.
If you let the ball go for one sec then turned off the gravity or acceleration of the ship. (assuming that they are both 1g) The ball would be going at the same velocity in both cases. So if there was a hole through the earth the ball would fall through the earth the same way it fell through the hole in the ship. With the ball going at a velocity away from both.

grav
2006-Jul-21, 04:34 AM
grav how does that make a difference?
The ball doesn't care what its velocity is it just wants to keep going at that same velocity.
If you let the ball go for one sec then turned off the gravity or acceleration of the ship. (assuming that they are both 1g) The ball would be going at the same velocity in both cases. So if there was a hole through the earth the ball would fall through the earth the same way it fell through the hole in the ship. With the ball going at a velocity away from both.
The ball will only fall through the Earth when some initial velocity has already been achieved. That is, if the ball has been allowed to fall some distance before gravity is switched off. At this point, it will have gained some velocity relative to the Earth, which we can consider stationary. This relative velocity will continue this way. But with the accelerating ship, their relative velocities will continue to increase, unless (as you implied in your last post) the ship's acceleration is also turned off. In that case, both scenarios would be exactly the same.

publius
2006-Jul-21, 05:16 AM
Grav,

You asked if ("regular") acceleration, say the rocketship, warps space-time just like gravity, and this question is tied strongly with the Equivalence Principle.

Well, the answer is yes or no depending on how you look at things. The reference frame of an accelerating observer is very much "warped" just like a gravitational field. But to an inertial observer watching it, or flying past the accelerating observer, space-time is flat, and the "warping" is only from the POV of the accelerating observer, and is entirely an artifact of his acceleration.

There is a sort of "relativity of curvature" in GR. But the accelerating observer can say, no, space-time is really curved here, and I'm just firing my rocket to keep from free-falling, and the reason you so-called inertial guys think space-time is flat is because you're all free-falling.

So, it is possible to declare a global, massive (nearly) uniform g-field exists and everyone is actually free-falling in it (since you feel no force no matter what the value of g is, heck it could be anything). One observer can declare that global field is 1g, and accelerate accordingly. Another observer can say, no, it's actually 10g, and accelerate 10 times faster.

There is no way to tell the difference, and so that difference doesn't matter. What can be detected is local variations in curvature, local variations in g, like an inverse square field.

Two observers feeling no force would notice they are actually accelerating relative to each other. They could radio back and forth and learn neither was feeling a force. They would have to conclude they were in a gravitational field.

However, they could only determine there were gradients in that field, not it's absolute value. This is a twist above the arbitrariness of a potential -- the field comes from differences in potential, and the absolute value doesn't matter -- here the field itself can take an arbitrary constant. Why can't you do that with an electric field? :) Think about adding an arbitrary, constant field vector to a system of charges, both positive *and* negative, plus a neutral body. Now what if there were only one sign of charge present. Would it work then? Think about inertial mass. Now consider if the ratio of charge to inertial mass was constant. You see, the equivalence of inertial mass to gravitional mass (the "charge" of the gravitational field), is what allows us to add this arbitrary constant to the field.

Consider the earth-moon field. You can add any (nearly) uniform g-field to that, and have the whole shebang free falling at any rate you desire, and do it to the whole solar system, or the galaxy or even the entire universe. You could only plot the relative field.

Indeed one inertial observer could choose that value of global g so it was exactly 0 at his location, declare his initial velocity was 0 as well, and declare he wasn't really moving at all, but it was the other inertial-feeling observer who was at some non-zero g. And that observer could choose a different global 'g' and do the same thing.

Gravity and EM have a lot in common, both those identical inverse-square Coulomb expressions, but when you get into the details, they are very different, and that's why Einstein couldn't make EM geometric nor unify the two together.

-Richard

grav
2006-Jul-21, 12:52 PM
Sorry, Publius. I just can't see where you are coming from with this. I waited till morning to see if it read any differently to me, bu I've gone over it a couple of times now and I just don't get it. :eh: There is equivalence for standing stationary in gravity to an accelerating ship and freefalling to "floating" in free space, but only from the effects and only in the way it feels. The differences in each set are that one is really accelerating and the other is not (relative to the fixed stars or a body in uniform motion). So we cannot compare freefalling to a relativity of accelerations, which is what it sounds like you are trying to do. There is such a thing as relative accelerations, otherwise we could not produce vectors for forces, or add and subtract them to gain the resultant (relative) force (or acceleration). But a definite force is still necessary to produce such accelerations, so we cannot assume any arbitrary values simply because we do not feel the effects in freefall. I think you might be taking the equivalence principle a little too literally. I'll read your post again later and see if it reads any differently to me. I would like to work on your ideas of how E, M, and G might be comparable or different. I definitely believe that a relationship can be found through relativity. But it appears the way you view things is much different than mine. Let's see if we can find some common ground. ;) :D

<<<<<>>>>>

Moderators, could you please move or copy part of this to the ATM section by any chance? It is getting quite interesting to me and I think we might actually get somewhere with it, but we might be moving off topic at this point. I think I made some good points about the equivalence principle and I would like to discuss my and publius' ideas further, but it may get quite involved and I would also like to hear the opinions of others that may not read this in the questions and answers section.

Grey
2006-Jul-21, 01:31 PM
Actually, in this respect (the way light shifts), it is just an illusion. But I think what Publius meant and what you are saying is that at the vicinity of a black hole, one must also think about how real time dilation (and length contraction or lengthening) caused by gravity effects the observer near the black hole, right?Nope, the observed redshift of light can be thought of either as a loss of energy for photons escaping or as an overall change in the rate of time (i.e., the fundamental processes that produce the radiation are slowed down by an equal amount). They are the same thing. And, since those time effects have actually been measured, and affect clocks as well as redshifts, I don't think it makes sense to call them an illusion. It's not something that just happens to the light but doesn't affect anything else.


In other words, doesn't the twin paradox say that the one furthest away from the gravitational field ages less?
...
Since it requires acceleration to turn around, I suppose that according to the equivalence principle, that would be the same as being in a momentary gravitational field. So the one in the gravitational field ages less.You've already figured this out. Being closer to a gravitational field slows time down. If both observers are remaining stationary relative to each other, both will agree that the one at a lower gravitational potential will see shorter time period elapse. Here (http://mentock.home.mindspring.com/twin2.htm)'s a document that talks about the classic special relativistic twin paradox from the point of view of general relativity.

Squashed
2006-Jul-21, 03:26 PM
Nope, the observed redshift of light can be thought of ... as an overall change in the rate of time (i.e., the fundamental processes that produce the radiation are slowed down by an equal amount). They are the same thing. And, since those time effects have actually been measured, and affect clocks as well as redshifts, I don't think it makes sense to call them an illusion. It's not something that just happens to the light but doesn't affect anything else...

Wow, its amazing that you brought that up because I was just thinking, yesterday, about the relativistic effects of time dilation on light waves. I was gonna post it but did not get it written in the way that, I thought, would clearly convey what I was intending - so I did not post - but I will think about it and try to post my thoughts eventually.

Ken G
2006-Jul-21, 03:47 PM
And I would add that I think it may be even easier to think about the twin pradox using pure special relativity. General relativity seems a lot more difficult, so I'm not sure why people often employ the equivalence principle to turn a special relativity problem in to general relativity. The trick with pure special relativity is to recognize that there are two processes going on in the twin paradox, one is time dilation, which is symmetric between the twins and so by itself would create a paradox. The second process is clock synchronicity issues, and that resolves the paradox. Two clocks that are next to each other are easy to synchronize at some point, but when the clocks are separated synchronicity becomes more an issue of convention. Einstein's convention involves light beams to synchronize displaced clocks (it's the convention associated with time dilation), but when one twin changes direction, this convention immediately induces a sudden shift in the synchronized result (what breaks the symmetry is that the twin who feels the acceleration is the one who must resynchronize his sense of "now" with all distant objects. If he doesn't feel any acceleration, then it is true gravity and we have a true general relativity problem that gives the same answer). According tot his convention, the twin that changes direction perceives his other twin's clock suddenly change from being way behind his clock to being way ahead. But the interesting part is, this change in synchronization does not involve anyone moving the hands of the clock separately from the time it is measuring, it happens to time itself. So it happens to the age of the other twin, and to that chain of experiences that twin calls his life. If it seems strange that the way time flows for the other twin should be simply a function of the convention of synchronization, remember that time is just a coordinate for keeping track of the order of things, so when considered by someone who is not actually experiencing it, it is no different from a timeline in a history book. (Edited to add the parenthetical remarks.)

publius
2006-Jul-21, 05:25 PM
Ken,

I'll agree it's easier to *solve* the twin paradox using SR (and this must be done from the inertial frame of the twin who does not accelerate -- if we wanted to calculate the view of the accelerating twin from his own frame, we'd have to use GR) and see the twin who accelerates is the one who will be younger.

But, by invoking the psuedo gravitational field of acceleration, it allows you to see how that accelerating twin will see the other's clock speeding up greatly for a short burst during the acceleration. Thinking in terms of gravity sort of makes it easier for me at least.

And, as I'm trying to get Grav to apppreciate, the twin who accelerates can indeed declare there is a massive uniform gravitational field which they've been free-falling in the whole time. :) He just fires his rocket to resist that free-fall for a while, allowing the other twin to behave just like a ball thrown up. It slows, stops, then falls back. When there is sufficent relative closing velocity, he turns off his rocket and they both free-fall again.

Grav is protesting that we can add such an arbitrary field, and I'm gonna try to convince him. :)

-Richard

Squashed
2006-Jul-21, 05:52 PM
publius,

I could be wrong but I think it was you who stated that you wanted to be able to accelerate with a high g-force without feeling the force so I'm thinking the only way to do so is to duplicate the gravitational mechanism of acceleration (changing velocity) because it accelerates people without their sensing it.

That would be the ideal method because the strength of materials would not be a limiting factor in the ship design (except if the tidal forces are too large).

Your contention of "standing on the earth is when the real acceleration takes place" had an interesting counterpoint in Grav's reply that there is no change in velocity with respect to distant stars. Your analogy makes it seem to "fall in line" with the equivalence principle but Grav's point makes me wonder.

Grey
2006-Jul-21, 06:18 PM
Your contention of "standing on the earth is when the real acceleration takes place" had an interesting counterpoint in Grav's reply that there is no change in velocity with respect to distant stars. Your analogy makes it seem to "fall in line" with the equivalence principle but Grav's point makes me wonder.The equivalence principle specifically ignores the "fixed stars". After all, those are just other objects to which you can have relative motion, and if you take relativity seriously, the frame in which they are as close to motionless as you can get (it will never be zero for all of them at once, of course, since they're not really fixed and are all moving differently) is no better than any other reference frame. From the point of view of relativity and the equivalence principle, if there is no measurable difference, then there is no difference.


There is equivalence for standing stationary in gravity to an accelerating ship and freefalling to "floating" in free space, but only from the effects and only in the way it feels. The differences in each set are that one is really accelerating and the other is not (relative to the fixed stars or a body in uniform motion).But since all the locally measurable effects are the same, there is no local experiment that you could do to tell the difference. Sure, you can look at the stars, but as I pointed out above, a distant star is just another object. Why should you choose that as your preferred frame? As far as choosing a body that's in uniform motion, the only reliable way to tell if something is inertial is to look for the absence of inertial forces, since again, any other definition where we look at the path and try to see if it's straight is imposing an arbitrary coordinate system.

Now, it's true that general relativity is a pretty radical departure from earlier ideas, and it's pretty natural to want to say that something is "really" accelerating or not. The impressive thing is that, if you instead take the viewpoint Einstein adopted, that there is no local difference and that whether something is "really" accelerating is a matter of the coordinate system you choose, you get all the right answers when you do the experiments. That doesn't mean that a geometric interpretation of gravity is the only possibility you can use, but it certainly works well if you do.

publius
2006-Jul-21, 11:06 PM
Grav and Squashed,


You guys are sticking to a 3D definition of acceleration. That is defining r(t), the position of a particle or body vs time as accelerated if it's not a straight line (as measured from an inertial frame, of course).

In GR, acceleration is *deviation from geodesics*, and a gravitational field *curves* those straight line geodesics, which makes an inertial path be a curve relative to a stationary coordinate system. It takes a force, proportional to the inertial mass to make a body deviate from geodesic, curved or straight. And this inertial mass is responsible for feeling the force of deviation from geodesics.

So sitting on a non-rotating planet, your position vs time will be straight line, and you're declaring that is therefore unaccelerated (using the fixed distant star backdrop as your stationary reference frame, or any other such frame). But the geodesics are curved, and that straight line path is therefore deviated from them, and that straight line path is accelerated.

And one certainly feels a force on that planet, confirming one is accelerated. :) That is the view of GR. Inertial paths, geodesics, are curved by gravitational fields, making them appear accelerated from a Newtonian perspective. Acceleration, which requires a force proportional to the inertial mass, is a deviation from those geodesics, no matter if they are curved or straight.

And, in curved space-time, that makes a straight-line path actually be accelerated ones, as the inertial paths are curved. What defines acceleration is deviation from geodesics, and the force required to cause that. And in curved space-time, sitting still on the surface of a planet is indeed an accelerated path.

-Richard

grav
2006-Jul-21, 11:23 PM
Originally Posted by grav
Actually, in this respect (the way light shifts), it is just an illusion. But I think what Publius meant and what you are saying is that at the vicinity of a black hole, one must also think about how real time dilation (and length contraction or lengthening) caused by gravity effects the observer near the black hole, right?

Nope, the observed redshift of light can be thought of either as a loss of energy for photons escaping or as an overall change in the rate of time (i.e., the fundamental processes that produce the radiation are slowed down by an equal amount). They are the same thing. And, since those time effects have actually been measured, and affect clocks as well as redshifts, I don't think it makes sense to call them an illusion. It's not something that just happens to the light but doesn't affect anything else.
But then, in what way is the redshift of a photon connected to the time dilation of the observer? The photons are only a source of measurement, and so can cause an optical illusion that plays tricks on our eyes and sensors, but that would not be a real effect since the photon is not physically effecting the observer otherwise. The only way I can see that the gravity well would have a direct effect on the observer is if he were attempting to escape the gravity well himself. This would cause a loss of energy on the observer's part. But even this would only be inertial energy.

For instance, let's say the observer was trying to escape the surface of the sun. He would begin with some initial energy away from it. But the gravity well diminishes this energy the further he attempts to travel. The formula for this is Eo-Ef=GMSMobs(1/R-1/S), where R is the initial distance from the center of the body (the radius) and S is the final distance achieved. If S is infinity, the formula is simply GMSMobs/R. Once again, this is only inertial energy that is lost, so he would lose this energy in the form of his velocity. So if the initial velocity is not great enough to overcome this, he will fall back in.

Now let's look at light. Light also will lose energy as it climbs out of the gravity well, and it will do so in the same way that our observer did, with Eo-Ef=GMSMphoton/R. But light doesn't have mass, you say? The energy of a photon is E=hf. So since hf=E=mc2, the mass of a photon can be "represented" by M=hf/c2. Substituting this into the equation, we get a loss of energy of GMShf/Rc2. Now, the measure of the redshift is the ratio of the loss of energy to the initial energy, which is E=hf. So the total redshift should be [GMShf/Rc2]/[hf]=GMS/Rc2. Plugging in the values for the mass and radius of the sun (Ms=1.9891*1030 kg and R=6.96*108 m), we get the resulting redshift for the sun of 2.12*10-6. Look familiar, anyone? It is the gravitational redshift of our sun.

This was done with purely Newtonian concepts. But relativity also plays a small part. The energy lost by an object effects its inertial velocity, so it stays closer to the body longer as it loses energy. Light, however, will not lose velocity (its speed is maintained), so will stay close to the body for a lesser amount of time and less redshift will be observed than expected. This is the real time dilation, although I have not performed the calculations for it (and so cannot be absolutely sure if it would match that of relativity). The difference will only be appreciable, however, with very large masses, such as black holes.

grav
2006-Jul-21, 11:56 PM
Quote:
Originally Posted by Squashed
Your contention of "standing on the earth is when the real acceleration takes place" had an interesting counterpoint in Grav's reply that there is no change in velocity with respect to distant stars. Your analogy makes it seem to "fall in line" with the equivalence principle but Grav's point makes me wonder.

The equivalence principle specifically ignores the "fixed stars". After all, those are just other objects to which you can have relative motion, and if you take relativity seriously, the frame in which they are as close to motionless as you can get (it will never be zero for all of them at once, of course, since they're not really fixed and are all moving differently) is no better than any other reference frame. From the point of view of relativity and the equivalence principle, if there is no measurable difference, then there is no difference.

Okay. You got me there. The entire system could actually be moving in any direction at any speed in relation to the stars and it wouldn't make any difference. What I should have said (and really meant) was that the acceleration is relative to the body doing the accelerating, not the fixed stars. Real acceleration shows a change in velocity over time. This is the mathematical definition of it. It can also be a change in position per time over time. A stationary body at rest on the Earth, therefore, is not really accelerating in respect to it. This is only true with freefall. Expressing gravity as 1 g is really only a convenient way to think about what would occur in freefall. Otherwise, only a force is felt. But this force creates acceleration, you say? Only in freefall (or with some unequal resistance applied, whereby it becomes the resulting difference between the forces). Otherwise it is counterbalanced by the force of the ground pushing back. In this respect, the measurable difference is tremendous. An object freefalling toward the Earth would be immensely different than floating in free space. It might "feel" the same with our eyes closed, but I prefer to do physics with my eyes wide open.

Ken G
2006-Jul-21, 11:58 PM
But, by invoking the psuedo gravitational field of acceleration, it allows you to see how that accelerating twin will see the other's clock speeding up greatly for a short burst during the acceleration. Thinking in terms of gravity sort of makes it easier for me at least.That's certainly true if you are used to general relativity. If you aren't, it brings in a bunch of new stuff. But you are right, it's best to look at it both ways, and learn something from each.


When there is sufficent relative closing velocity, he turns off his rocket and they both free-fall again. Cute, now we have three ways to think about it! Interesting, ain't relativity grand.

grav
2006-Jul-22, 12:31 AM
Grav,

You asked if ("regular") acceleration, say the rocketship, warps space-time just like gravity, and this question is tied strongly with the Equivalence Principle.

Well, the answer is yes or no depending on how you look at things. The reference frame of an accelerating observer is very much "warped" just like a gravitational field. But to an inertial observer watching it, or flying past the accelerating observer, space-time is flat, and the "warping" is only from the POV of the accelerating observer, and is entirely an artifact of his acceleration.

There is a sort of "relativity of curvature" in GR. But the accelerating observer can say, no, space-time is really curved here, and I'm just firing my rocket to keep from free-falling, and the reason you so-called inertial guys think space-time is flat is because you're all free-falling.
Okay. This much I agree with (I think). There is a relativity for acceleration just as there is for distance and velocity. For acceleration, we can think of it as the same for velocity (SR), but changing with the relative difference of the instantaneous velocities. So we would not know whether we were accelerating or everything else was accelerating past us.


So, it is possible to declare a global, massive (nearly) uniform g-field exists and everyone is actually free-falling in it (since you feel no force no matter what the value of g is, heck it could be anything). One observer can declare that global field is 1g, and accelerate accordingly. Another observer can say, no, it's actually 10g, and accelerate 10 times faster.

There is no way to tell the difference, and so that difference doesn't matter. What can be detected is local variations in curvature, local variations in g, like an inverse square field.

Two observers feeling no force would notice they are actually accelerating relative to each other. They could radio back and forth and learn neither was feeling a force. They would have to conclude they were in a gravitational field.
Here's where you lost me. I thought that maybe you were just getting really tired or something. I know you know better than this. While it is true that a relativity of acceleration exists, they cannot just declare themselves to be at any arbitrary acceleration simply because they are in freefall. They would all fall according to the laws of gravity, which includes their initial positions (the distance from a gravitating mass) as well as their initial velocities and directions and the value of the gravitating mass. To say one is at 1g and another at 10 g's could easily be verified with the difference of 9 g's between them. Besides, the acceleration also changes with respect to the position in the gravitational field as well. All that the relativity of accelerations really says here is that we could say one is accelerating at 9 g's as compared to the other. Even if we couldn't see the gravitating mass, we could still determine a formula for gravity from the value GM and the direction of the resulting gravitational field.

grav
2006-Jul-22, 01:12 AM
Grav and Squashed,


You guys are sticking to a 3D definition of acceleration. That is defining r(t), the position of a particle or body vs time as accelerated if it's not a straight line (as measured from an inertial frame, of course).

In GR, acceleration is *deviation from geodesics*, and a gravitational field *curves* those straight line geodesics, which makes an inertial path be a curve relative to a stationary coordinate system. It takes a force, proportional to the inertial mass to make a body deviate from geodesic, curved or straight. And this inertial mass is responsible for feeling the force of deviation from geodesics.

So sitting on a non-rotating planet, your position vs time will be straight line, and you're declaring that is therefore unaccelerated (using the fixed distant star backdrop as your stationary reference frame, or any other such frame). But the geodesics are curved, and that straight line path is therefore deviated from them, and that straight line path is accelerated.

And one certainly feels a force on that planet, confirming one is accelerated. :) That is the view of GR. Inertial paths, geodesics, are curved by gravitational fields, making them appear accelerated from a Newtonian perspective. Acceleration, which requires a force proportional to the inertial mass, is a deviation from those geodesics, no matter if they are curved or straight.

And, in curved space-time, that makes a straight-line path actually be accelerated ones, as the inertial paths are curved. What defines acceleration is deviation from geodesics, and the force required to cause that. And in curved space-time, sitting still on the surface of a planet is indeed an accelerated path.

-Richard
Well, I know perfectly well what you are saying here. I have also thought about it all before, a long time ago. You are saying that in a gravitational field, no force is "felt" if one is freefalling or in orbit. Let's say we are in orbit. That way we don't have the option of falling to our doom in our thought experiment. Okay, so we are in orbit (I'm starting to feel like it, anyway :D ) and from our frame of reference, we feel no force. It is as if no force is acting. We are simply following the geosedic curves provides by the gravitational mass. And this means that the mass is curving spacetime, right? The orbit will only change if acted on by another force, or in this case, a force to begin with. I suppose, also, that you could define what you feel when you are standing on the surface of the Earth as acceleration, just as gravity can be defined that way, but in my opinion, that cannot really be the case. It all depends on how you think about it, and how you precisely define it.

But a ship can be constantly moving in a curve and the same things can be said, except one. This one makes all the difference in the equivalence principle. It is that the curved motion of the ship will experience a centrifugal force, whereas curved motion due to gravity does not. But this is because the centrifugal force of orbit perfectly cancels the gravitational force. If it did not, we would still feel its force, only weakened. In the case of the ship, we feel the force pushing out. In orbit, it is balanced. And we could also say that this is the natural geosedic curve for the ship, and that it will not change unless acted upon by another force.

publius
2006-Jul-22, 01:13 AM
Grav,

What is this "GM" business? See, I knew your problem is you're a square, an inverse square. :dance: Consider an infinite sheet of mass, of some fixed density. What kind of (Newtonian) g-field does that produce? You'll find it produces a uniform gravitational field, one with constant magnitude and direction through all space on either side of the sheet. Far from inverse square.

Seriously, I may not be doing a good job of explaining what I'm trying to get across.

Now, imagine your inverse square gravitating bodies in that flat-sheet field. At any point in space, we add the constant, g_sheet to the g's of the other masses.

The look at the equation of motion. Consider the center of mass of the gravitating inverse square bodies. That center of mass will accelerate toward the sheet at a rate of g_sheet. In a reference frame centered on the center of mass, the relative motion looks exactly like it would if there was not g_sheet at all. The whole shebang is free-falling at g_sheet toward the sheet, which could be infinitely far away (the field is uniform and constant through all space, out to infinity).

If the sheet is infinitely far away, there is no way to detect its presence. g_sheet could be anything. One observer could say g_sheet points this way at 10g. Another could say, no it points thataway at 1g. And a third could say, not it points in yet a third direction at 100g.

No matter what g_sheet is, there *relative acceleration*, due to your inverse square bodies orbiting around will be the same.

Consider the earth-moon system, and forget about the rest just to keep in simple. We plot the actual g field map in the space around the two, and have different observers at different initial points in that field. They free fall, and they notice relative acceleration between them. They radio back and forth and confirm that none are feeling a force.

Now, what I was saying is that each one can invoke a g_sheet so that it exactly cancels out the g due to the earth and moon at his location. He can then declare he is not moving, there is no g at his location, and it's the other observers who are actually moving and free-fall accelerating. We can call this "coordinate acceleration" = curved r(t) path relative to a local coordinate system centered on each observer.

Each observer can do the same, invoking a different uniform g_sheet so that the total g at his location is 0. All observed motion will be the same. Each observer can say the earth and moon are dancing around him and he's not even moving, much less accelerating.

Remember, no matter what g_sheet is, the motion of the system about its center of mass will be the same, even though from some observer stationary to the infinite sheet will see that whole shebang freefalling away. The coordinate accelerations viewed by each observer will the same no matter whose g_sheet they use.

Now, my other point was the comparison with EM. I cannot invoke an arbitrary E_sheet (or B_sheet) because that would indeed alter the motion relative to the center of mass. And the reason is there is both positive and negative charge, and that inertial mass and electric charge are not tied together (and radiation is something much more complex that would destroy E_sheet invariance even the charge to mass ratio were constant!).

-Richard

grav
2006-Jul-22, 01:23 AM
publius,

I could be wrong but I think it was you who stated that you wanted to be able to accelerate with a high g-force without feeling the force so I'm thinking the only way to do so is to duplicate the gravitational mechanism of acceleration (changing velocity) because it accelerates people without their sensing it.

And there you have it. If we could accelerate people through every atom in their bodies, we would feel no force. Well, actually it would be minimized. It is because what we are really "feeling" is pressure (or stress), not force at all. It all depends on the surface area which the force is distributed across. It would not feel the same as an accelerating platform because the force would be distributed throughout their bodies instead of a surface pressure acting directly against their feet. It would not feel the same as standing on the ground on Earth either for the same reason.

This link (http://en.wikipedia.org/wiki/Stress_(physics)) appears to have every definition of stress available.

grav
2006-Jul-22, 02:15 AM
publius,

Okay. You got me with your last post. But it seems you may be trying to force a round peg into a square hole :p . I get what you mean this time for the most part. The g-sheet thing is ingenious. It would indeed produce a constant acceleration. You would still be able to see it at an infinite distance, though, if had infinite dimensions (and radiated visible light, due to the luminosity law, similar to Olber's paradox). If not, you wouldn't still have a constant acceleration because it could not really be infinitely large. But even if you could see it, you could still not tell an acceleration toward it if it is infinitely large and its surface is very smooth, until you were right up on it and could see some sort of defect to use as a point of reference, but that is beside the point. If such a thing did exist, you would not know you were accelerating toward it because everything else would be accelerating at the same rate also. But then, we are right back where we started. Acceleration is relative, just like velocity. All we can really know is the difference in accelerations. We still could not arbitrarily define an acceleration because we have nothing to compare it to. Compared to everything else, it remains zero. If others were to say they are accelerating at 1g or 10 gs or 100 gs, then they should have a reference point by which to compare. If they are comparing these arbitrary relative accelerations to each other, then there had better be 9 gs between the first and the second, and 90 gs between the second and the third. This is all relative and depends on the frame of reference. If each are considering themselves to be that frame of reference, then their own acceleration should be considered zero, and all others measured in comparison. This only pertains to the first half of your post. It was difficult to read further with this point in mind, because I could not then comprehend what you were trying to say. I will read it again.

grav
2006-Jul-22, 03:26 AM
Let's try something. The equivalence principle says that one cannot tell the difference between freefall and floating in free space. This is because with all other senses obscured, they would "feel" the same, and therefore are. I think we can all agree that a person floating in free space (away from all influences of mass) will feel no forces acting upon him. In other words, the force is exactly zero. So, according to the equivalence principle, this must also be true for freefall. It is my contention that it is not, that a force will still be present and felt.

Whether we were standing on the ground on Earth or a platform accelerating at 1g, the forces applied will feel the same because we are feeling them applied through the same surface, our feet. This makes the counteracting force (the ground or platform) a surface pressure. Let's say our feet are about .1 m2. The force will be felt over this area, creating a stress, or pressure, which we feel. Now let's say our back and the back of our legs and so forth are .6 m2. So if we lie down, we reduce the pressure sixfold, and we can feel the difference (neglecting air pressure). This is what we mean by "take a load off".

Gravity, on the other hand, acts on every particle in our bodies. The force is then distributed over the sum of the areas of all of the particles put together. Now, the radius of an atom is about 5*10-11 m. The area of the the nucleus is about 1/10000 of this. So we can estimate the radius of a single baryon, which makes up most of the mass of a body, at about 5*10-13 m. The mass of a baryon is about 1.673*10-27 kg. So a 75 kg person contains about Nbar=M[sub]person/Mbar=4.483*1028 baryons. The cross-sectional area of a baryon would be about Abar=pi rbar2=7.854*10-25 m2. So the total internal area that the force of gravity is distributed across is NbarAbar=35,210 m2. This is much greater than that which would be applied across our feet. The total pressure felt while standing up on a platform or on the ground would be 35,210 m2/.1 m2=352,100 times greater than what it would feel like in freefall. And although lying on our back may relieve some of this pressure, it would still also be 58,683 times greater. The pressure felt through our feet on the Earth would be about 7530 N/m2. With freefall, it would be only be about .02 N/m2, as applied to each baryon separately. So although the pressure (stress) felt in freefall may be neglible, it would not be zero as it is when floating in free space.

publius
2006-Jul-22, 05:11 AM
Grav,

Think about the "pressure" thing some more -- if gravity is reaching out and through your body and grabbing every atom, every bit of mass equally, then I don't see how any pressure will be felt. You will feel tidal forces, however -- in the view of GR, it is impossible for a body to remain rigid with all parts following geodesics in high g gradients. To remain rigid, internal forces must keep parts off of geodesics, and you certainly feel that stress. And indeed, that stress is a very real force.

Back to this relativity of free-fall. Coordinate acceleration (as I've taken to calling it) is indeed relative. However, one can tell if one is deviating from geodesics, following non-inertial paths, because one feels the force.

My point about the g-sheet (arbitrary additional uniform g-field added in) was to show there is no way to tell absolute curvature. We can see relative curvature (tidal forces) like in an inverse square field. But we can still add some arbitrary free-fall acceleration to the global picture and not tell the difference.

But we can tell is we're deviating from geodesics, just not how much. For example, suppose we're floating in free space and fire our rocket so we feel exactly 1g of force (say we've got a very accurate accelerometer mounted on the dash). Well, in the view of GR, we are really accelerating at
1g. We are deviating from our geodesic at 1g's worth of acceleration.

Now, we could be in perfectly flat space-time, 0g gravity, and so our coordinate acceleration (relative to some fixed frame we won't worry about defining. :) ) is actually 1g. However, we could be in g_sheet field of 1 g pointing exactly opposite our rocket thrust vector. Our coordinate acceleration would then be 0g. That g_sheet field could be 1g in the same direction as g_sheet, and we're accelerating faster than free fall, and our coordinate acceleration would then be 2g.

Or g_sheet could be 100g or anything else in any direction, and our coordinate acceleration could be actually anything itself. But our deviation from the geodesics, our coordinate acceleration *less the ambient g field* will always remain 1g along our rocket's thrust vector.

No matter what our coordinate acceleration is, our "real acceleration" is just what we feel, and what our accelerometer measures. An accelerometer actually measures how many g's we feel, but it doesn't register free-fall acceleration. In fact, an accelerometer sitting on the surface of the earth will claim we are accelerating at 1g. Believe the accelerometer. It is telling you the truth according to GR.

A better term for "real acceleration" might be "inertial acceleration".

-Richard

grav
2006-Jul-22, 01:26 PM
Grav,

Think about the "pressure" thing some more -- if gravity is reaching out and through your body and grabbing every atom, every bit of mass equally, then I don't see how any pressure will be felt. You will feel tidal forces, however -- in the view of GR, it is impossible for a body to remain rigid with all parts following geodesics in high g gradients. To remain rigid, internal forces must keep parts off of geodesics, and you certainly feel that stress. And indeed, that stress is a very real force.

I think you might be right about that. The stress felt in our bodies on the ground or on a platform is still caused by the forces of gravity directed downward and the ground (or platform) pushing up. But in freefall, the pressure is unbalanced (one directional only). The atoms on the top of an object will observe the same pressure as those on the bottom, so while the ones on top are being pushed toward the lower ones, the lower ones are also moving and accelerating with them. It causes the object to aquire a real acceleration toward a gravitational body. But although this force and pressure are both very real, since there is no counteracting pressure (upward), no actual stress is actually felt within our bodies.

It appears we'll probably keep going in "circles" on this one. I will always maintain that the cause and force of gravity are real and separate from the effects, that one isn't equivalent to the other. Only in the effects between the stress felt in an inertial system and a stationary one where some counterforce is or is not being applied can the similarities be felt, and otherwise is not so except with our eyes closed.

So let's concentrate on what similarities do exist. We can both agree that the effects of gravity will create a curved path between two massive bodies. I still maintain that this is caused by the resultant straight line vectors through a changing gravitational field (changing with the square of the distance from the center of each body), but I have come to realize through this forum that there are indeed many ways things can be perceived.

At first I thought the lines of force for the Earth-moon system would follow a path similar to that of a magnetic field, with the Earth and moon as the poles. But this cannot be correct because both poles flow inward and are broken up at some point between the bodies where equilibrium is maintained. So it seems to me now that the lines would follow a path along the equilibrium plane (which is curved) toward the line between the centers of the bodies, and then quickly jut one way other the other toward each of the bodies on either side of the plane as it approaches this center line. This looks more like the field lines that arise for two electrically charged particles that are identical in charge. Of course, two gravitating bodies are identical, they both attract, but identically charged particles would push each other away. Not so with the Earth-moon system. But the resulting field lines look exactly the same as far as I can tell. Let's see if we can find the similarities and differences there.

I believe, as I think you also do, that some similarities can be found between gravity and E fields. I believe only the differences lead to relativity for bodies in orbit with gravity, where the centrifugal force is really just a weakened gravitational field, and to a similar but very different result being created for the E field, where the same relativistic effects lead to the creation of a B field. What are your thoughts on all of this?

publius
2006-Jul-22, 05:22 PM
Grav,

I've said this before in some of our EM discussions, and I'll say it again now. Get some textbooks on this and study. For example, a good undergraduate text on Classical Mechanics will go into great detail on Newtonian gravity, showing you various methods and insights into calculating orbits, and even will have plots of the 2-body field like the earth-moon system. With an EM text nearby you can see the similarities.

The infinite sheet I brought up is a standard example of how to produce a uniform field, whether gravitational, electric, or magnetic (current sheet), for example.

All these various wheels have been invented, and it will help you understand (and appreciate) all this much better.

A good mechanics text will also go into great detail on rotating coordinate systems and other accelerating coordinate systems. The centrifugal and coriolis forces are just "fictitious", artifacts of an accelerating, non-inertial frame. They are exactly the same as the apparent "free fall" acceleration viewed by a linear accelerating frame.

A "free-falling" observer in a rotating frame (which I call a Coriolis frame) does not feel the centrifugal or coriolis forces at all. Indeed, from an inertial frame watching, that free-faller is just moving in a straight line.

The force you feel spinning around in a circle is entirely the real force, a centripetal force that is actually accelerating you in that circle. But in this frame, you invoked the centrifugal force as cancelling out the real force so the object remains stationary in your rotating frame. That's all that's going on.

Going to GR and the Equivalence Principle, that tells us there is always some gravitational field that can mimick any accelerating frame. The rotating observer can invoke this and declare he is the one who is "stationary", not the free-falling, inertial observers.

And the source mass distribution to make such a Coriolis gravitational field is a massive rotating cylinder. This is taking our mass sheet a long distance away and folding it into a big cylinder and then rotating it fast its axis. It produces a centrifugal g field that looks like a centrifugal force, and the mass current produces a massive gravitomagnetic field that acts like the coriolis force. And one doesn't feel these either.

Geocentrists, who for religious reasons declare the earth has to be stationary and non-rotating and the center of the universe, will invoke such a GR situation (at least the ones who try to be scientifically plausable about it, others just say everything we know is wrong), to explain the Focault pendulum and other effects.

-Richard

grav
2006-Jul-23, 12:04 AM
Grav,

I've said this before in some of our EM discussions, and I'll say it again now. Get some textbooks on this and study. For example, a good undergraduate text on Classical Mechanics will go into great detail on Newtonian gravity, showing you various methods and insights into calculating orbits, and even will have plots of the 2-body field like the earth-moon system. With an EM text nearby you can see the similarities.
That is what I am doing. I imagined the Earth-moon system as stationary (we are moving with it as a frame of reference) with the line between their centers horizontal to our piont of view and the moon on the left. I imagined the distance between them to be constant, which is about the case with the Earth-moon system. I tried to think of the simplest situation where the inertial motion is not a consideration. That would be when we plot points randomly around the two masses and consider the velocity at those points to be zero (in the same frame of reference as the masses). We then find the vector of the forces acting at those points and move to a point that is slightly in that direction. We then consider the same thing until we have the lines of force (or acceleration). The resulting field would look exactly like that for repulsive charges (identically charged particles). If we were to place a positive charge between two negative charges, it would behave the same way. So everything is the same except that the two initial bodies will attract for gravity and repel for electric charges. I want to know why.


The infinite sheet I brought up is a standard example of how to produce a uniform field, whether gravitational, electric, or magnetic (current sheet), for example.
I know about Guassian surfaces, but I just thought your application of it with the g-sheet was a good way to demonstrate your point. However, wouldn't a magnetic sheet act entirely differently? First of all, there would have to be two of them, one for the north pole and one for the south, on opposite sides of the universe, I suppose. But even then, the direction of the charges between the sheets would accelerate perpendicularly to them, causing the the entire universe (at least the part between the sheets) to rotate. But I have worked it out and found that the rotation will be the same for all charged particles between them. That is, they will all rotate with the same period, so a relativity of velocities still exists. And since F=Bqv=mv2/r, where B and q and v/r are all constant, then m is proportional to v, and since v is proportional to r, then particles with larger m/q ratios will rotate at a smaller radius. I'm not sure how this would effect the relativity of accelerations.

[EDIT-I just realized that a constant magnetic field would not rotate the entire universe as a whole, but only cause particle to circle individually to each other, so no relativity of velocities or accelerations could be present in this case, unless you know of some other way it could all be set up.]


All these various wheels have been invented, and it will help you understand (and appreciate) all this much better.
But those are the simple relationships. I am attempting to find out how it all relates on a grander scale. That means I cannot accept anything at face value. I must break down everything already known to its finest degree and build them back up on my own to fully understand its potential and then see how it can be applied to other phenomena. If this means reinventing every wheel known to man, then so be it. I will spend the rest of my life doing so, even if I die without producing any new results. I made this decision long ago. But with one major difference. I am doing it in 3D. By the way, I have been reading every link and every book I can get my hands on that relates to any of this. This forum has also been extremely helpful in letting me know what I should be looking for.


A good mechanics text will also go into great detail on rotating coordinate systems and other accelerating coordinate systems. The centrifugal and coriolis forces are just "fictitious", artifacts of an accelerating, non-inertial frame. They are exactly the same as the apparent "free fall" acceleration viewed by a linear accelerating frame.

A "free-falling" observer in a rotating frame (which I call a Coriolis frame) does not feel the centrifugal or coriolis forces at all. Indeed, from an inertial frame watching, that free-faller is just moving in a straight line.
Well of course these forces don't effect a freefaller. As far as the Coriolis frame goes, that is the point. The Earth is turning under a "falling" body, but the body do not care which way the Earth turns. It will only fall to its center point, regardless of the Earth's spin, while the Earth turns underneath it. With centrifugal force, if a body in freefall has no inertial velocity perpendicular to the lines of force, then it will not be effected by this either.


The force you feel spinning around in a circle is entirely the real force, a centripetal force that is actually accelerating you in that circle. But in this frame, you invoked the centrifugal force as cancelling out the real force so the object remains stationary in your rotating frame. That's all that's going on.
If I am reading this correctly, then that would mean gravity is the real force. A centrifugal force cannot exist without a centripetal one, but the centripetal force can exist without the centrifugal. In other words, some real force must always be present that keeps a body that is spinning in a circle from flying off to begin with.


Going to GR and the Equivalence Principle, that tells us there is always some gravitational field that can mimick any accelerating frame. The rotating observer can invoke this and declare he is the one who is "stationary", not the free-falling, inertial observers.
That is true to a point, since acceleration is relative. We can see some similarities because gravity produces acceleration, but we can also (when our eyes are open) see some major differences, such as the actual inertial acceleration, which is exactly opposite for each set in the equivalence principle, rendering them unequivocal.


And the source mass distribution to make such a Coriolis gravitational field is a massive rotating cylinder. This is taking our mass sheet a long distance away and folding it into a big cylinder and then rotating it fast its axis. It produces a centrifugal g field that looks like a centrifugal force, and the mass current produces a massive gravitomagnetic field that acts like the coriolis force. And one doesn't feel these either.
A centrifugal force would not be produced. Bodies would not know the difference whether the cylinder is rotating or not. They are only pulled toward that general direction. A difference would only be observed for bodies that are turning with the cylinder (they are resting on its surface) as is the case with the spin of the Earth.

[EDIT-Actually, if the sheet is made cylindrical, no acceleration or force will be felt in any direction. Its effects will be cancelled out at all points in all directions so its net force is zero. It would be the same as that for a "hollow Earth", where all points from the center to the edge of the cavity experience zero gravitational pull.]


Geocentrists, who for religious reasons declare the earth has to be stationary and non-rotating and the center of the universe, will invoke such a GR situation (at least the ones who try to be scientifically plausable about it, others just say everything we know is wrong), to explain the Focault pendulum and other effects. I'm not sure how geocentrists could use GR to prove anything as far as this goes, since they are claiming an absolute frame of reference, which is contrary to the very foundations of relativity. I suppose it would act as an "anti-GR", where they are trying to show that our observations of orbits are really its effects. The effects of the Foucault pendulum would seem to me to be the same as the coriolis effect (although the coriolis effect would be more along the lines of a larger relative motion toward the equator, which causes systems to rotate faster on that edge. But then, this is only because air is not perfectly fluid in the case of hurricanes and such), where the Earth is turning underneath it, but it does not care about this, since it only gravitates toward the center of the Earth. It is not otherwise physically attached, except by the string that holds it up. Since the string is pliable, it will twist under the effect. This might also relate to Mach's principle in a negative way. Even though the pendulum does not care about the spin of the Earth per say, so Mach might say it is actually related to "the fixed stars", this is only because it does not move directly with the spin but with the center of gravity. It still travels with the same inertial velocity as the Earth, however, and so still only exists within the Earth's frame of reference, not the fixed stars.

publius
2006-Jul-23, 01:01 AM
Grav,

Think about this more carefully. A cylinder is very different symmetry from a spherical shell (of uniform density). An infinitely long cylinder will indeed produce a centrifugal g field about its axis. That is, g(r) = k*r, increasing radially, just like the centrifugal force. You will also find a ring of mass produces the same centrifugal field in the plane containing it. Above or below the field is different. It's very easy to imagine such a ring has spherical symmetry, but it doesn't, and if you look at the integrations to get the field in both cases, you'll see what's happening -- with the spherical shell you are integrating a surface area density, which is proportional to r^2. But with a ring, you are intergrating a circular line density, which is propotional to r.

Now, I mentioned gravitomagnetism twice now, and was wondering if you'd catch that. Apparently you're not aware of it. Moving mass produces a magnetic-like effect that causes forces on other moving mass just like charge currents do. This is sometimes called "frame dragging" in GR, but I prefer gravitomagnetism.

Now we can write a Lorentz-like expression for the total gravitational acceleration felt by a mass:

a = g + v x B_g, where 'g' is the familiar (gravitoelectric) field, and B_g is the gravitomagnetic field. Actually, sometimes a GR field is written. IOW, gravity has velocity dependent components just like EM fields do.

a = g + v x B_g + SM_stuff. SM = Spatial metric, and wraps up all the weird stuff due to curvature and clock differences and all that. But for most familiar systems (like the solar system), B_g is very small, and SM_stuff outweighs it.

Now, if that large cylinder is very dense, and rotating as fast as we can (near light speed, say), we can make a very large B_g field that will look just like the Coriolis force.

And that will make a Foucault Pendulum precess just like it would on a rotating body. This g/B_g field produces forces just like a rotating reference frame. The Equivalance Principle means what it says, and you can indeed construct a gravitational field that will mimick any accelerating reference frame you can come up with.

Now, for the "Geocentric" case, of the Earth centered right on the axis of that cylinder, remember the earth's gravity adds to it. Here a geostationary orbit would indeed be stationary. The centrifugal gravity of the cylinder would exactly cancel the earth's g, and if the satellite has no velocity, it just sits still. :)


-Richard

publius
2006-Jul-23, 01:24 AM
Grav,

GR gravity has a lot more to it that simple differential Netwonian inverse square components. A lot more. It's like Coulomb without Ampere, but GR gravity goes beyond Maxwell-like, although you can "linearize" the field equations to make something that looks exactly like Maxwell. Only problem is that drops out the "SM_stuff" components, which generally are much stronger than the B_g components, and so the so-called gravitoelectromagnetic (Maxwell_Einstein) is pretty useless other than a fascinating mathematical exercise, and allow you to see how gravity can act like EM in many ways.

Now, EM can be expressed in terms of two vector fields, E and B. In the more elegant 4-vector formulation, you can put both into a single tensor field, but no matter, there are just 6 components, which is equivalent to two vectors.

GR gravity, however, has more than 6 components (8 or 10, I forget), and cannot be fully expressed as two vectors. The gravitational field is a full tensor field.

Have you ever seen the main governing equation for GR gravity? It's called Einstein's Field Equation (EFE). In tensor notation it's a rather simple looking equation, but that elegant tensor notation wraps up a lot of complexity. It's actually 8 or 10 sets of coupled partial differential equations. That is why Einstein's hair looked like it did.

In Netwonian gravity, the source term is just mass. However, the source term for GR gravity, the thing on the right hand side of the EFE, is something called the Stress-energy tensor. Gravity depends on mass (although written in terms of the energy density via E=mc^2), but depends on the momentum (ie mass currents) plus the "transports" of both of them. That is, it depends on the flow of energy (or mass current), and the flow of momentum too.

The stress part of the Stress-energy tensor is just the full tensor notion of "pressure" (in the general case, "pressure" is not a simple scalar, but a 3x3 tensor). So it sometimes said that gravity depends on the "pressure" as well as the mass and mass currents. Indeed, in the simple case where the pressure tensor reduces to a scalar pressure, you can write something simple about the effect of the pressure on the gravitational field.


-Richard

grav
2006-Jul-23, 01:45 AM
Grav,

Think about this more carefully. A cylinder is very different symmetry from a spherical shell (of uniform density). An infinitely long cylinder will indeed produce a centrifugal g field about its axis. That is, g(r) = k*r, increasing radially, just like the centrifugal force. You will also find a ring of mass produces the same centrifugal field in the plane containing it. Above or below the field is different. It's very easy to imagine such a ring has spherical symmetry, but it doesn't, and if you look at the integrations to get the field in both cases, you'll see what's happening -- with the spherical shell you are integrating a surface area density, which is proportional to r^2. But with a ring, you are intergrating a circular line density, which is propotional to r.
Think about it this way. For a body that is in such a cylinder (directed vertically), it will feel the same forces directed up as down. As far as its sideways acceleration, the same constant acceleration that is felt in one direction is also felt in the opposite direction, so no force is felt this way either. Its net acceleration is 0g.


Now, I mentioned gravitomagnetism twice now, and was wondering if you'd catch that. Apparently you're not aware of it. Moving mass produces a magnetic-like effect that causes forces on other moving mass just like charge currents do. This is sometimes called "frame dragging" in GR, but I prefer gravitomagnetism.

Now we can write a Lorentz-like expression for the total gravitational acceleration felt by a mass:

a = g + v x B_g, where 'g' is the familiar (gravitoelectric) field, and B_g is the gravitomagnetic field. Actually, sometimes a GR field is written. IOW, gravity has velocity dependent components just like EM fields do.

a = g + v x B_g + SM_stuff. SM = Spatial metric, and wraps up all the weird stuff due to curvature and clock differences and all that. But for most familiar systems (like the solar system), B_g is very small, and SM_stuff outweighs it.
This all sounds very interesting. I will have to look into it. Do you know of any links about it?

Now, if that large cylinder is very dense, and rotating as fast as we can (near light speed, say), we can make a very large B_g field that will look just like the Coriolis force.

And that will make a Foucault Pendulum precess just like it would on a rotating body. This g/B_g field produces forces just like a rotating reference frame. The Equivalance Principle means what it says, and you can indeed construct a gravitational field that will mimick any accelerating reference frame you can come up with.

Now, for the "Geocentric" case, of the Earth centered right on the axis of that cylinder, remember the earth's gravity adds to it. Here a geostationary orbit would indeed be stationary. The centrifugal gravity of the cylinder would exactly cancel the earth's g, and if the satellite has no velocity, it just sits still. :)
Again, even if the acceleration was not zero, the Coriolis effect would only be observed by bodies that are at rest on the surface of the g-sheet. That is, they would have to be physically attached in some way, like a string as with the Foucault pendulum, otherwise bodies do not care if the sheet is turning. You should know this because you stated it earlier when you said that a body in freefall does not feel the Coriolis effect. That is indeed true. So why should they feel it in this case? The same thing goes for centrifugal force. The only bodies that would feel this force are those that lie on the surface of the cylinder and are therefore turning with it within its own frame of reference.

publius
2006-Jul-23, 01:59 AM
Grav,

No, inside the cylinder there will a centrifugal g field. Do the integration yourself. Start with the simple ring of mass, and integrate the potential, then take the gradient -- it will be simpler than doing the field integration directly. Trust me, I know what I'm talking about here. A cylinder will produce an axial, centrifugal g field. 0 at the center, and increasing linearly with radius.

Now, the B_g field. When we set that cylinder to rotating, we get a B_g field. The gravitational acceleration is the Lorentz-like

a_total = g + v x B_g, where v is the velocity of the test mass, and g is some k*r, centrifugal like.

This is exactly the same as the Coriolis psuedo-forces:

a = -2w x v + w^2 r

This also demonstrates why gravitomagnetism is called "frame dragging". Moving mass "drags" geodesics along with it, so to speak. So the massive rotating cylinder "drags" space-time along with it, and one can say that space-time itself is "rotating" with the cylinder.

That is, inertial frames are "dragged" into a rotating thing, and resisting that inertial path, acts just like being in a rotating coordinate system.

This is just a bit more complex use of the Equivalence Principle. An accelerating frame is equivalent to a stationary frame in a gravitational field. The rotating cylinder gives us the gravitational field that mimicks a rotating frame.

Grav, trust me, the cylinder makes a field like this. It's in many textbooks on GR and more advanced gravity. It's not just me. :)

-Richard

publius
2006-Jul-23, 02:09 AM
Grav,

Bodies are not experiencing a Coriolis force from the point of view of a stationary frame about the rotating cylinder. They are experiencing a gravitomagnetic accelerationg, v x B_g, which looks like a Coriolis force.

You're not as familiar with B_g and velocity dependent forces as you are with regular g fields, but this is just the Equivalence Principle at work in a more complex, general situation, finding a gravitational field that acts like an accelerating frame of reference.

You can do this with any dizzingly complex accelerated motion you desire, coming up with some dizzingly complex gravitational field. Trust old Richard here, he knows what he's talking about.

ETA: and another important point about this: In a frame rotating with the cylinder, there are no forces whatsoever. That frame is completely inertial. There is no centrifugal g field, no gravitomagnetic field (actually the two cancel each other out so to speak). That's what the rotating cylinder does -- It makes space-time and intertial frames in it simple rotate about the central axis. An object at rest with respect to the axis in that rotating inertial frame is in an orbit from the POV of the stationary frame, where the v x B_g just cancels the g field. (which acts like a regular rotating frame -- this gets confusing, I know), but it is sitting perfectly still in the inertial frame.

An object moving in straight line in the inertial frame appears to be spiralling outward from the stationary frame, the centrifugal g field being a bit stronger than v x B_g. In the inertial frame, it just goes in a straight line until it eventually hits the walls of the cylinder.

-Richard

grav
2006-Jul-23, 02:35 AM
publius,

I'm sorry, but I did that calculation a while back and even similated it with a computer program to be sure (I forget what I was working on at the time. Something to do with electric current, I believe). In the program, I found the gravitational attraction from some arbitrary point to every point on the ring by working my way around it little by little. The net force is zero. I also could not determine why this should be, but then it hit me (like you probably want to at this point :o :p ;) ). If you run a line perpendicularly to the center of the ring to a point within it (through the point), you will see that there is more of the ring on one side than the other. The point is further from the side with more mass and is inversely proportional to the distance. But the side with less mass is closer. No matter where you place the point within the ring, the force cancels out. In other words, the inside of the ring where the force is inversely proportional to the distance is no different than a sphere where it is inversely proportional to the square of the distance. The net force in both cases is zero.

Now rotating frame drag is another story (although it would still require a net force). But I do not agree with it either. In order for such a frame to be "dragged", there must be some difference in the gravitating field that a body can feel and be dragged with. For a rotating body, the lines of force eminate from the center of the body, and no other determination can otherwise be made. The only way another body can know any different is if some difference is applied across its gradient, like a bulge on one side, or some other "defect". Now, this would make it a velocity frame drag, where the defect is moving across its frame of reference. This would make a difference. I know Einstein predicts that the spacetime field might twist around a rotating body. But this is also based on Mach's principle and the concept behind the Foucault pendulum. If this were the case, then that would mean that spacetime is also effected by gravity. That makes it a real substance that exists within space and time, not spacetime itself. That sounds more like an ether to me.

publius
2006-Jul-23, 02:41 AM
Grav,

No, I don't want to hit you, though I am :wall: a little bit. You made an error in your computer program somewhere.

With all my rambling, everyone else may be asleep by now, but I'll ask the others to back me up if you still don't believe me about the ring and the cylinder.


-Richard

publius
2006-Jul-23, 02:51 AM
Now rotating frame drag is another story (although it would still require a net force). But I do not agree with it either. In order for such a frame to be "dragged", there must be some difference in the gravitating field that a body can feel and be dragged with. For a rotating body, the lines of force eminate from the center of the body, and no other determination can otherwise be made. The only way another body can know any different is if some difference is applied across its gradient, like a bulge on one side, or some other "defect". Now, this would make it a velocity frame drag, where the defect is moving across its frame of reference. This would make a difference. I know Einstein predicts that the spacetime field might twist around a rotating body. But this is also based on Mach's principle and the concept behind the Foucault pendulum. If this were the case, then that would mean that spacetime is also effected by gravity. That makes it a real substance that exists within space and time, not spacetime itself. That sounds more like an ether to me.

Grav,

You see, you're thinking entirely in terms of Netwonian gravity. Just like thinking Coulomb without Ampere. Moving mass, mass currents produce the gravitomagnetic field just like electric currents make a magnetic field, and this produces a velocity dependent acceleration on other masses. Although with GR, it is much more complicated, but the basics are the same.

Newtonian gravity has nothing like that in it, and there's no way you can get the effect by using it. But it (and more) are there in GR gravity.

Gravity Probe B, a marvel of engineering, is up there in orbit testing the "frame drag" plus some other non-Newtonian aspects of GR gravity. The rotating earth produces a (very small) gravitomagnetic field, and it's measuring the effect. We await the results.

Tell you what, I'll bet you a dollar Einstein's GR comes through with flying colors, and observes the gravitomagnetic field of the rotating earth. It's also looks for another effect, the so-called "geodetic effect", which is one of those SM_stuff effects I was talking about.

-Richard

grav
2006-Jul-23, 02:56 AM
ETA: and another important point about this: In a frame rotating with the cylinder, there are no forces whatsoever. That frame is completely inertial. There is no centrifugal g field, no gravitomagnetic field (actually the two cancel each other out so to speak). That's what the rotating cylinder does -- It makes space-time and intertial frames in it simple rotate about the central axis. An object at rest with respect to the axis in that rotating inertial frame is in an orbit from the POV of the stationary frame, where the v x B_g just cancels the g field. (which acts like a regular rotating frame -- this gets confusing, I know), but it is sitting perfectly still in the inertial frame.
This seems like kind of a mixed up paragraph. The frame of the cylinder may be inertial, but this is what would cause the centrifugal acceleration (only on the surface) to begin with (a=v2/r). I am still trying to determine how you get centrifugal acceleration inside the cylinder, even if a net acceleration did exist. Is it frame drag? Also, if there was a constant acceleration toward it in this case, that would also add to the centrifugal acceleration (on the surface), since it would be directed outward, the same as on the ground or platform with the equivalence principle. But since there is no net acceleration, we need not consider this. If the cylinder made spacetime rotate as well, then this would be similar to a small rotating cylinder on Earth where air might rotate some with it (or if we filled it with water). This would make spacetime a real, touchable substance, like an ether. I do believe that such a thing exists, so I will look into that further as well. So instead of a rotating cylinder, can we discuss how this ether drag (oops, I mean frame drag :D ) might effect a spherical rotating body or one in motion?

publius
2006-Jul-23, 04:46 AM
Grav and all,

I should make something clear about the "frame dragging" of the cylinder, where I said inertial frames rotate with the cylinder. They do not necessarily rotate at the same speed as the cylinder. So the angular speed that "space-time is rotating" is not equal to the cylinder's rotational speed. For a "small cylinder", even rotating at very high speed, that effective 'w' will be very small.

The field depends on the density and speed of the cylinder, and you play with those two variables to make it look like some desired "w" of rotation. That is, B_g is equal to 2w, and g is equal to some w^2r. To get significant B_g, IIRC you've got to get some serious density (ie neutron star density or similiar) and some high speed. This is related to the difference between current and the velocity of charge carriers. You can get a given current by having few charge carriers moving fast, or lots of charge carriers moving slowly. The mass current goes as the density times the tip velocity.

But the point is, such a field can exist, and the rotating observer can claim he is actually stationary in such a field.

-Richard

grav
2006-Jul-23, 05:38 AM
Yes. I figured that if you were referring to rotational frame drag, the effects would be much, much smaller than the actual rotation of an object. Apparently this is what you are referring to. I am wondering, if such a drag were to be imposed on spacetime (or ether) near a massive and rapidly rotating body such as a pulsar, what kind of relativistic effects might we observe? Would this be where the idea of gravitational waves come from? Might it be responsible for the decay of orbit in binary systems? You put B_g in terms of inverse time. Would this be the frequency of rotation as felt by spacetime? In other words, should I consider it to be "points" in spacetime that are turning with the rotating object (but not at the same speed)? Or even if these points weren't turning very far, could they still trace out a curved path in spacetime?

publius
2006-Jul-23, 06:35 AM
Grav,

Neutron stars and rotating black holes should frame-drag like gangbusters, and the exotic superconducting states of neutron star innards, may even make invoked the enhanced gravitomagnetic moment of certain superconductors, and there is all sorts of speculation about that.

Google on "Neutron star frame drag" and you'll find a bunch of stuff. Consider a spinning magnet in an external field. The torque will make the axis of spin precess. Frame-dragging/B_g will have a similiar effect on spinning system, and they're looking for evidence of the precession of the accretion disks around neutron stars and black holes.

Gravitational radiation is, well, a big subject all of itself. It's getting late and I won't even try to start with that. And I'll wait in hopes that a real GR expert will come in tackle it, because all I can do is sort of speak in general terms about it.

And remember, as I mentioned before, EM can be expressed as two vector fields, 6 components, E and B. Gravity is a full tensor field, and has more components than that, and is more than a g-field and B_g-field. Gravitational radiation can involve all the additional components, so you can't see it as simply a g and B_g oscillating. There's more there than that.

ETA: Stumbling around with Google, I found a fascinating discussion about the possible effects of greatly enhanced neutron star B-g due to the T&M effect ( see http://www.bautforum.com/showthread.php?t=39692 ).

Anyway (and I assume these guys knew what they were talking about -- they knew GR), the were speculating that such a greatly enhanced B_g could be a source of gamma ray bursts! It had something to do some kind of "snapping" of B_g during some superconducting phase transition that might occur inside a neutron star of a binary pulsar system, which apparently would cause some sort of big regular magnetic field snap that would make a shower of gamma rays. The snapping of B_g in a binary system would "slap" the other neutron star somehow, and that would make one heck of a magnetic "slap" in the whole system.

-Richard

-Richard

grav
2006-Jul-23, 06:17 PM
Thank you for the link, publius. I think I am finally beginning to understand how it is all connected. Looks like I've got a lot of reading and catching up to do.

This might be a little premature, considering I have just barely began to look into these things, but I have been thinking about it, and have come up with a very simple model for binary systems. It's so simple, in fact, that it makes me think I just haven't put enough thought into it yet, or else someone else would have come up with it.

My thinking is (and this is only speculative at the moment) that as a star collapses, and its spin greatly increases, this would cause a large increase in electromagnetic effects. As far as I know, most large bodies have an electric and magnetic field around them. A fast rotation would act as though a large current were running through them and its magnetic field would also increase. So two such bodies that are close to each other will interact this way as well. The relativistic effects between their fields would add to the total force including the gravitation between them, so that their orbits will decay. But as there orbits decay, they will spin even faster because of the synodic rotation, where the period of revolution adds to their rotation relative to the barycenter. The effect, then, would not be caused by gravity or their magnetic fields alone, which I guess is where the term gravitomagnetism comes from. It is kind of a relatistic effect related to both of them.

In any case, this much aside, I also began to wonder how a neutron star could exist in the first place. I think of them as being composed almost entirely of neutrons. Is this correct? But neutrons decay rapidly on their own. So then I thought, what could keep them from decaying? I figured it must be the tremendous pressure that acts within the star and forces them together, so neutrons would only decay in free space. But then what about the ones on the suface? For them I figured the tremendous gravity that would act on the surface would create a similar stress as the ones within the star.

So what would be the only thing that could relieve this pressure enough for the star to explode? Centrifugal force. As the star spins faster and faster when the orbit of a binary system moves closer and closer, the gravity lessens due to centrifugal force. When the star spins fast enough, the pressure on the neutrons are relieved enough that they will in fact decay. Once this happens, it will create a chain reaction that is like an atomic blast, with gamma rays and neutrinos and beta decay (electrons) going in every which direction simultaneously. All that is left behind are protons and maybe a few surviving neutrons that might still be joined that will become the nuclei of elements. It would probably leave these behind in random clumps, both stable and unstable, which would eventually decay into mostly stable atoms, both heavy and light, and would probably be the process that leads to life as we know it.

As I said, this is all speculative, so don't take me too literally. But if this were the case, let's find out how fast the neutron star would have to spin to cause this reaction. As far as I can tell, most are estimated to be about 1.4 MS with a radius of about 20 km. This would put a force on a neutron at the surface of F=GMstarMneutron/R2. The energy required to escape the surface would be F*R. This would be the energy keeping the neutron from decaying. Let's say that the energy of decay is tied up in the gamma ray released. And let's say that the frequency of the gamma ray is the same as that for electron-positron annihilation. Now, I know the energy is also carried by the beta decay and neutrinos and even the recoil of the neutron, and the frequency for the gamma radiation would probably be completely different that that for annihilation, but I don't know what these would be exactly , so I will guesstimate with what I've got. Maybe someone can help me out with this.

Okay. In this example, the energy of decay would be released when the gravity minus the centrifugal force is equal to this amount. This would be E=[GMstarMneutron/R2-Mneutronv2/R]*R=hf, so GMstar/R-v2=hf/Mneutron. v is equal to 2piR/T, where T is the rotation period. T is what we want to find, so the formula becomes T=2piR/[GMstar/R-hf/Mneutron]1/2. Using the values given earlier, we come up with a minimum possible rotation period of 1.307*10-3 sec.

As it turns out, the decay energy (in this example) is only about one two hundredth of the gravitational value. If it were larger, the minimum rotation time would only increase. Even if the neutrons did not decay, they would still be thrown off the surface if it spun faster than 767 times per sec, or a period of 1.304*10-3 sec. This is not much different than what we found with the neutron decay, but I believe neutron stars can spin much faster than this. Is that correct? So this would be testimony to the incredible electromagnetic fields that would occur. Even though the neutron is neutral, they must still be effected in some way to stay in place, perhaps due to the relativistic effects involved.

This is my first take on this, so please don't judge it too critically, but let me know what your thoughts are.

publius
2006-Jul-24, 02:32 AM
Grav,

I don't know nuttin 'bout no neutron stars, other than they are very dense, at nucleon density, and have huge gravitational and magnetic fields very near them. The innards are not thought to pure neutrons, and there is all sorts of work being done on just what exotic phases of very dense matter exist there.

Now, if a neutron star has a magnetic field, there's got to be a current (and a big current) in there right? What carries that current is thought to be a superconducting fluid of some exotic nucleon states someway somehow.

Before I forget. A "current sheet" will indeed make a uniform B field. However that field will be parallel to the sheet and at a right angle to the current flow. The current sheet consists of a constant surface current density in an infinite plane. You don't have to worry about the field lines making closed loops because the source current is infinite in extent.

And magnetic fields do not necessarily have to be "dipolar" like that of a current loop or solenoid. Consider the field of an infinite wire. The B field lines are circular loops around the wire, and the symmetry is such that a clear "north" and "south" aren't that evident (depends on which way you're oriented), although when you bend a long wire around in a closed loop, which it will have to be eventually, you will get a N and S looking dipolar field forming.

-Richard

publius
2006-Jul-24, 02:57 AM
Grav,

Yes, the dimensions of B_g, the gravitomagnetic field are indeed frequency. By analogy with EM, B_g is a flux density, something per area, so the dimensions of gravitomagnetic flux are area per time. :) And that means the units of gravitomagnetic "monopoles" would be area per time as well (I'll leave it "as an exercise" to find the units of the gravitomagnetic H field, and you'll find that, taking g as the analog E, that g x H_g is power density, just like Poynting. And you can verify that mass "charge" times monopole "charge" is indeed action.)

Now, moving mass in a uniform B_g field would precess, in exact cyclotron analog to moving electric charge in a B field. And the angular speed is just equal to B_g. Well, I caution the gravitomagnetic Lorentz force expression is probably just an approximation since there is more to the full possible GR gravitational field than two E and B-like vectors can express. So in whatever limits it applies:

F = m v x B_g, or a = v x B_g, with mass cancelling as always with gravity.

That is just the centripetal acceleration, w^2r, and letting v = wr, and noting the acceleration vector is always perpendicular, we have

w^2 r = wr * B_g -> w = B_g [note the corresponding cyclotron expression has a q/m in front of B, this drops here because, gravitational "charge" is equal to inertial mass, which is the whole point of this long winded discussion!]

In other words, the angular speed of any mass, at any velocity (non-relativistic I'm sure, where the other terms would become strong as v/c became significant) is just equal to B_g. Note this is different from the Coriolis metric where we have a factor of 2, and that's the difference with the centrifugal 'g' in there as well -- you can work play around with it to see how the '2' plays out.

So what a gravitomagnetic field does is sort of curve geodesics into circles in a velocity dependent way.

-Richard

Squashed
2006-Jul-24, 04:25 PM
You guys are sticking to a 3D definition of acceleration. That is defining r(t), the position of a particle or body vs time as accelerated if it's not a straight line (as measured from an inertial frame, of course).

After thinking about this a while and re-reading your reply (http://www.bautforum.com/showpost.php?p=788810&postcount=44) that Grey linked me back to then I have to ask: What is acceleration?

Acceleration (http://en.wikipedia.org/wiki/Accelerate) is defined as the rate of change of velocity.

Feeling a force is an indication of a possible acceleration.

Gravity produces acceleration without the sensation of a force - always.

The force that we feel in our feet is the earth resisting the force of gravity, there is no acceleration occurring because the two forces are equally offset.

A rocket will not accelerate as long as it is held in place by the tower - even though a huge force, by the engines, is being applied. This is true because the tower exerts an equal but opposite force against the rocket.

In such a "tethered rocket" scenario there are two points where force is "felt" which are: 1.) at the engines and 2.) at the tether points.

On earth we only feel one of the opposing forces: the earth.

publius
2006-Jul-24, 04:39 PM
Squashed,

How about coordinate acceleration is defined as the rate of change of velocity relative to some coordinate system of some observer, using that observer's yardstick and that observer's clock.

How about "real acceleration" or inertial acceleration is deviation off geodesics, defined by the local geometry of spacetime, which or may or may not produce coordinate acceleration as seen by some observer's yardstick and clock.

Now, what force do the astronauts sitting atop that tethered rocket feel? 1g. An accelerometer registers 1g. Break the tehter and then they feel a lot more than 1g.

The tethered situation can be duplicated in free space with two rockets of equal force opposing each other. Same thing. The supports between the have to support large compressive force (or tension depending how the supports work). But these forces are truly balanced. You, sitting inside the frame feel nothing. Or rather than exactly balanced, 1 rocket could be thrusting with 10x force and the other 9x, so the net force was 1x which produced 1g of acceleration.

And those two opposing rockets could be freefalling at any rate and you'd never know the difference.

-Richard

publius
2006-Jul-24, 05:01 PM
And another thing to think of. We are sitting here feeling 1g. But what is that 1g of inertial acceleration we're feeling. It comes from resisting the gravity of the earth.

But the earth is free-falling around the sun (and the moon as well). Some of our intrepid astronomers and celestial mechanics might be able to tell us what our free fall acceleration actually is as measured from the barycenter of the solar system.

And then, the whole shebang is free falling around the center of the Galaxy, probably at a pretty good clip. I wonder what our free fall rate is relative to the center of mass of the Milky Way is? It might be rather impressive, but I'm not sure.

And then, the whole Milky Way is free falling itself toward the center of some huge "local lump" of mass.

So our actual geodesic sitting here is a rather complicated thing. We are only deviating from the component due to the earth's gravity, but freefalling with all the rest.

-Richard

Grey
2006-Jul-24, 06:21 PM
After thinking about this a while and re-reading your reply (http://www.bautforum.com/showpost.php?p=788810&postcount=44) that Grey linked me back to then I have to ask: What is acceleration?

Acceleration (http://en.wikipedia.org/wiki/Accelerate) is defined as the rate of change of velocity.

Feeling a force is an indication of a possible acceleration.

Gravity produces acceleration without the sensation of a force - always.These are Newtonian ideas. There's nothing wrong with that per se, but expecting Newtonian mechanics to remain unchanged under general relativity is kind of missing the point. General relativity says that gravity is not a force, and does not really produce any acceleration. Rather, it changes the shape of space such that an unaccelerated object will have a different path than it would have had space not been curved. Under general relativity, and object which experiences inertial forces can be said to be accelerating, and if it's not experiencing inertial forces, it's not accelerating. Any other definition you try to give is going to have to refer to some arbitrary coordinate system, and the point of general relativ ity is that it is arbitrary. That is, you could pick some other coordinate system, and it would work just as well.

It's a very different viewpoint than Newton's, and it completely changes how we look at gravity. Maybe it will turn out to be wrong. But at the moment, it's survived every test that's been thrown at it.

Squashed
2006-Jul-24, 07:37 PM
In a centrifuge or artificial gravity setup the force is a fictitious force which is derived from the inertia of whatever is experiencing the machine's effects.

An object travels in a straight line when released in the artificial gravity machine and it is actually the floor that accelerates to intercept the velocity vector of the object.

So a fictitious force, it would seem, is a force that is not felt, an inertial force.

Gravity seems like it would fall into that classification: a fictitious force; because it is never felt. Gravity is just the result of inertia sending us along a "force-less" vector, except the earth is impeding our travel.

It does seem that fictitious forces (http://en.wikipedia.org/wiki/Fictitious_force) always are paired with a real force which together result in the "apparent" force.

So the question becomes what is the actual force that causes the fictitious force to become apparent?

publius
2006-Jul-24, 08:20 PM
Squashed,

This can get very confusing. *Very* confusing. Inertial forces are real forces. They come from the inertia of a body resisting gedoesics. In a very rough, non-rigorous picture, they are the F required (or due to) the inertial mass, m, in F = ma. But we must use GR's definintion of 'a' here, a deviation from a geodesic. And indeed, we can *locally* measure the inertial forces we feel, and come up with a local definition of our 'a', the deviation from GR's geodesics.

And I stress this 'a' has nothing whatsover to do with the coordinate acceleration some observer might measure. That depends on the geometry of space-time, and that observer's motion through that space-time.

Fictitious forces are entirely the result, "artifacts" if you will of some particular coordinate system. In a Newtonian sense, one can state that the fictitious forces come about it in non-inertial, or accelerating (really accelerating) coordinate systems. We can still stick to trying to say the same thing applies in GR, but it gets tricky when there are gradients in the gravitational field -- ie, two free-falling observers, feeling no inertial forces, will measure coordinate accelerations between themselves. In a unifrom gravitational field, they would not, and this leads to the observation that differences in gravitational fields, or gradients in space-time curvature are observable, but the absolute field, or absolute curvature is not.

So we can (try) to say that fictitious come about from (GR defined) non-inertial frames as well as gradients in gravitational fields between inertial frames. But actually trying to do so becomes rather complicated on a global scale.

Back to Newtonian defined fictitious forces. There is no need for a real force. We can define an accelerating coordinate system, a rotating or translating-accelerating one or any combination of the two without having any real body moving with it.

Indeed, take a bunch of billiard balls floating in free-space, completely inertial. Now, define some rotating coordinate system in the middle of it. Those billard balls appear to be accelerating according to the centrifugal and coriolis forces in that rotating frame, going around in circles and/or spiralling.

There is no real force acting at all on anything, and all that accelerated motion is just an artificat of the non-inertial coordinate system we have chosen.

Now, if we put something real with inertia to rotating, then we're going to need real forces to make that sucker accelerate. In that frame, we see real forces balancing out the fictitious forces. And this is where your idea that fictitious forces must be paired with real forces come from. You're imagining a real body moving with your accelerated reference frame.

And a real force acting does not have to have anything to do with the fictitious forces of our chosen reference frame. Suppose we put a rocket on one of our billiard balls and have it accelerate off in a straight line.

Now, look at that from the rotating frame (which has nothing real rotating in it, just a mathematical device of a rotating coordinate system). In that frame, the real force will add to the fictitious forces we see, and that will change the motion into something even more spirally and crazy.

And one can forget about the fictitious forces being forces at all, just accelerations that have to be added to F= ma, ie

a = F_real/m + artifacts_of_accelerating_frame

If we want 'a' to be 0 in our frame, then those artifacts must be cancelled exactly by real forces. And, you'll note that artifacts_of_accelerating_frame look exactly like the expressions for (mass independent) gravitational accelerations, which is what Einstein noticed and began to say, "Hmmmmm..................". The result was GR.


-Richard

publius
2006-Jul-24, 08:42 PM
Squashed,

And another thing to be clear on. In the centrifuge, that force you'd feel slamming you against the wall is very real, indeed that is the real force making you go around in a tight circle.

That is, the wall is pushing you around in a circle, and your inertia is resisting. But from a coordinate system we set up rotating with the centrifugre, you have no coordinate acceleration. You are at rest in that frame. The real force of the wall pushing you flat is balanced by the fictitious centrifugal force.

Now, let the wall open a trap door and let you fly off (and imagine this is now in free space). From the POV of that rotating coordinate system, you are now accelerating under the influence of the centifrugal and coriolis forces, as the balancing real force is gone.

But you now feel nothing.................................you are in free fall of the fictitious forces of the accelerating coordinate system.

-Richard

grav
2006-Jul-25, 01:09 AM
Can we call non-moving acceleration something different than inertial acceleration, please? It is very confusing. When I think of inertia, I think of motion. I looked it up just to be sure. It can either be something that remains at rest or continues in a fixed direction. This definition does indeed match your definition, which makes me wonder if I've been thinking about it incorrectly for all this time, but seems to imply a relativity for velocities in its overall definition, not acceleration. Coordinate acceleration is okay for the moving kind. But perhaps stationary acceleration or potential acceleration would be better for the other. Maybe we should consider them the same as we do for energy and force, potential and kinetic.

grav
2006-Jul-25, 01:27 AM
Squashed,

And another thing to be clear on. In the centrifuge, that force you'd feel slamming you against the wall is very real, indeed that is the real force making you go around in a tight circle.

That is, the wall is pushing you around in a circle, and your inertia is resisting. But from a coordinate system we set up rotating with the centrifugre, you have no coordinate acceleration. You are at rest in that frame. The real force of the wall pushing you flat is balanced by the fictitious centrifugal force.

Now, let the wall open a trap door and let you fly off (and imagine this is now in free space). From the POV of that rotating coordinate system, you are now accelerating under the influence of the centifrugal and coriolis forces, as the balancing real force is gone.

But you now feel nothing.................................you are in free fall of the fictitious forces of the accelerating coordinate system.

-Richard
Are you doing this on purpose, or did I miss something somewhere? It sounds like you are refering to a spinning wall, where the centrifugal force would create acceleration (but is really only couterbalanced by the structure that holds it to the axis of rotation). If a trap door were to open, no acceleration would be applied at all any more. A person will just continue on at their previous angular velocity.

grav
2006-Jul-25, 01:44 AM
publius,

By the way, I see what you mean about a constant B field. It could be create by a bunch of wires that are lined up parallel to each other. The field lines would run perpendicularly to the sheet and to the current. The current would have to be closed, though, and for an infinite sheet, it is difficult to imagine where the wired would loop around. Also, the magnetic field lines would have to loop around the side "edges". But it's good enough if we consider it to really be just extremely large as compared to the body it is effecting. My question is, though, wouldn't such a field still cause bodies to move in circles, so that this acceleration is not relative to the motion of each other? Also, would a magnetic force which acts perpendicularly like this still produce the same acceleration at any distance from the sheet? I was also wondering where the poles would lie in this scenario. I am having trouble pin-pointing them.

[EDIT-Looks like you already considered some of this in the same post I read it in, #95. I must not have realized what you were saying exactly until I thought about it all myself.]

publius
2006-Jul-25, 01:57 AM
Grav,

:lol: Rotating frames do indeed cause a lot of confusion, especially about centrifugal vs. centripetal forces.

Squashed mentione a centrifuge, and I'm thinking about a big one, say a big wheel with walls on the sides against which an occupant would be, well, "Squashed".

The centrifugal (and coriolis) acceleations are just artificats of a frame rotating with that centrifuge. The only real force is that supplied to keep all the mass accelerating in a circle. That's the force you feel. It is unbalanced, because indeed, it is causing acceleration. A centrifugal force does not exist save in a rotating frame of reference. *Frame of reference*, a mathematical construct. Nothing has to be really accelerating, or rotating with such a frame. To get something with mass really rotating, you have to supply a real force, a centripetal one. That's the only real force at work, when you get something really rotating. The confusion comes because people don't generally separate the frame of reference from the actual object doing the rotating.

Now, *in a frame rotating with the centrifuge*, a non-inertial frame, the occupant isn't accelerating (or even moving). The "coordinate acceleration" is 0. But an accelerometer, as always, would tell us the real force and acceleration.

Now, open the trap door as I put it, or otherwise let the occupant go. Yes, in an inertial frame, he moves in a straight line. No real force is acting. But now, what of the rotating frame where he was originally stationary? It's still there, rotating along with the centrifuge.

That frame sees all sorts of coordinate acceleration. It sees the occupant accelerating in a spiral path, being acted on by the centrifugal and coriolis forces. The occupant is now in "free fall" relative to the fictitious (coordinate) forces of that non-inertial frame. But he's feeling no force, as his accelerometer would indicate.

Am I a stickler about "centrifugal force in not real" as some people are. No, because rotating frames of reference are very useful. I am being quite a stickler for purposes of this thread because I want to make clear what "is really going on" with non-inertial frames.

Now, you protest the guy who flies off the centrifuge is just going in a straight line. Here you are insisting on an inertial frame of reference, and only thinking non-inertially when the guy in the centrifuge is actually rotating, and actually accelerating.

But when we switch to gravity, you insist on a non-inertial frame of reference, one stationary with respect to the source of the field. :) That's your Newtonian (even "common sense") biases at work. I'm trying to remove those Newtonian blinders and let you see the GR view of acceleration and geodesics.

-Richard

grav
2006-Jul-25, 02:16 AM
Actually, I am considering myself as a distant observer to the whole situation, with the same frame of reference as the axis of rotation. It seems you are seeing yourself from the same original frame of reference as the person on the wall (maybe his buddy?). But even so, the spiral you would observe for your friend has nothing to do with acceleration. He would be moving away from the rotating wheel at a constant relative velocity (although moving toward and away from you because of your distance from the axis). He would also be traveling sideways at a constant velocity, due to the angular velocity of the wheel. So the apparent spiral motion would still be that of a constant velocity (unless you view his apparent motion toward and away with each revolution of the wheel to be a varying acceleration since his velocity will appear to change in respect to you). You've got me wondering about the centrifugal acceleration that acts on a person against the wall, however. The centripetal force that holds the wall in place is weakened by the centrifugal. But what about the person against the wall. There is no centrifugal force acting on him as far as I can tell. But wouldn't that mean that a person doesn't necessarily have to be pushed against the wall? Couldn't they just float anywhere between the axis and wall while the whole thing just rotates around them?

publius
2006-Jul-25, 02:38 AM
So the apparent spiral motion would still be that of a constant velocity (unless you view his apparent motion toward and away with each revolution of the wheel to be a varying acceleration since his velocity will appear to change in respect to you).

Grav,

That's what I mean by *coordinate acceleration*! Yes, that guy moving in an inertial staight line is accelerating like gangbusters from a rotating frame of reference. I've been trying and trying all through this thread to get you to see gravity in the same way, to see the guy freefalling toward the ground as just this same "apparent motion" that one sees in a non-inertial frame of reference!

Just for information purposes, and to exercise my brain cells a bit, let's derive Coriolis. "It can be shown", as the texts like to say when they give something without proof, that the time derivative in a rotating frame of reference, defined by a (vector) axis of rotation w is

d/dt = (d/dt)* + w x .

Now what that means is the time derivative in the inertial frame, on the left, is equal to the time derivative in the rotating frame (denoted as the "starred" coordinate system) plus the cross product of w against the thing whose derivative we are taking. Now, to get the acceleration, d^2/dt^2, we just operate the right hand side on itself twice. When we do that, we get:

d2/dt2 = d2*/dt2 + 2w x d*/dt + w x w x + dw/dt x .

For constant rotation, the dw/dt term is zero.

First term on the right is coordinate acceleration in the rotating (starred) frame. Second is the Coriolis force. Third is the centrifugal. Note the nice vector form where we can have w pointed in any direction. On the left, is just 'a', our inertial, Newtonian acceleration. Multiply by m to get the real force, F:

F = ma = ma* + 2m w x v* + m w x w x r*


So good so far?

-Richard

publius
2006-Jul-25, 03:18 AM
Actually, I am considering myself as a distant observer to the whole situation, with the same frame of reference as the axis of rotation. It seems you are seeing yourself from the same original frame of reference as the person on the wall (maybe his buddy?). But even so, the spiral you would observe for your friend has nothing to do with acceleration. He would be moving away from the rotating wheel at a constant relative velocity (although moving toward and away from you because of your distance from the axis). He would also be traveling sideways at a constant velocity, due to the angular velocity of the wheel. So the apparent spiral motion would still be that of a constant velocity (unless you view his apparent motion toward and away with each revolution of the wheel to be a varying acceleration since his velocity will appear to change in respect to you). You've got me wondering about the centrifugal acceleration that acts on a person against the wall, however. The centripetal force that holds the wall in place is weakened by the centrifugal. But what about the person against the wall. There is no centrifugal force acting on him as far as I can tell. But wouldn't that mean that a person doesn't necessarily have to be pushed against the wall? Couldn't they just float anywhere between the axis and wall while the whole thing just rotates around them?

Grav,

On second thought, we'll forget the complexities of Mr. Coriolis' frame. The above quote makes me think all my running at the mouth for pages in this thread has sort of been for naught because we aren't really on the same page about the meaning of basic terms.

You're thinking (and stuck on) the notion of "real acceleration", and that has some definite meaning (yet when thinking of gravity you switch frames in which the acceleration is real :) ) always. In Newton, it does, and that's from an inertial frame.

Let's stick with Newton, and imagine a simple, translating accelerating coordinate system. That is, a simple x-y-x cartesian coordinate system whose origin is accelerating at some constant value, which, apropos of nothing,we'll call, let's say, oh 'g'.

Now, what is the equation of motion for a body as seen from this accelerating coordinate system. Remember, nothing real is accelerating. This is just an accelerating coordinate system.

Do you see how all inertial bodies, experiencing no net real force, will appear to be accelerating? This is what I mean by "coordinate acceleration". This is a varying velocity as measured from some coordinate system.

Now, by your own protest that my spiralling buddy flung off the centrifuge is moving in straight line, and not really accelerating, do you see how an accelerating coordinate system will see coordinate accelerations when "none is really there"? And by knowing how that non inertial frame is moving, we can write expressions for the coordinate acceleration relative to an inertial frame? This is what I was doing with Mr. Coriolis' equations above.

For the simple, translating accelerating coordinate system, that equation, in full vector form looks like this:

F = ma = ma* + mg

Solving for a-star, the coordinate acceleration in the starred coordinate system, we get

a* = F/m - g

Now, F is the real force that may be acting. But this says a body that is experiencing no real force appears to accelerate at -g, opposite to the direction our coordinate system is accelerating at. If a real force F exactly balances mg, then we see no acceleration. But from an inertial, F = ma, and that m sucker is really accelerating.

See how this works? -Richard

grav
2006-Jul-25, 03:49 AM
Well, I can see what you mean about the g-sheet. If all bodies were accelerating at the same rate, none would know they were accelerating at all, becuse their relative velocities would remain the same. But in your centrifugal example, the apparent acceleration would appear to move back and forth. If we expanded this into a spacetime coordinate system, it would look like a sine wave. But then, how would this be different from the apparent epicycles of planets, for instance?

publius
2006-Jul-25, 04:57 AM
Grav,

Spin yourself around (and do it near a couch where you can fall over when you get dizzy!) and watch the room around you. Does anything appear to go back and forth? No, it goes around in circles, orbiting you.

That path of an object not moving in the inertial frame centered with the rotating frame is just a circle. Now, give that object some velocity, moving in a straight line in that inertial frame. Now, what do you see as you spin around?

With each revolution, the radius of the object gets farther and farther, but it continues to go around in a circular path at the same time. That describes a spiral.

All of that complex motion is given by the solving the equation of motion using the fictitious forces, the coriolis and centrifugal.

Let's look at the path of those objects stationary in the inertial frame, that appear to moving in exact circles from your rotating frame. The Coriolis force at any spot is -2 w x v. The centrifugal force is - w x w x r, where r is the position vector at that spot.

Now, if the object isn't moving in the inertial frame, that means it has a velocity of -w x r at some initial time in the rotating frame. So the Coriolis force to start with is -2 w x (-w x r) = 2 w x w x r. That just twice the centrifugal force (most fortunate, that factor of 2 isn't not). So our net force as viewed by the rotating frame is w x w x r, or a net *centripetal force* just necessary to keep that object going around in a circle.

So the Coriolis force minus the centrifugal force becomes a fictitious centripetal force and pushes the object around in a circle. :lol: (I know you hate me for wording it that way, a fictitious centripetal force, but I can't help it) But inertially, that object is not moving at all. That's the equation of motion of the room spinning around you.

Now add some inertial velocity to that. The Coriolis force is no longer twice the centrifugal, and it doesn't quite move in a circle, but gains a radial component of velocity.

-Richard

Squashed
2006-Jul-25, 02:02 PM
...which it will have to be eventually, ...
publius,

Why do you say, in this post (http://www.bautforum.com/showpost.php?p=791123&postcount=95), that it will have to eventually have to loop back to itself? Does it need both ends connected to a "battery" to establish the current or are you implying the "shape of the universe"?

- - - - - - - - - - - - - - - - - - - - - - -

I guess all this talk of gravity, in some ways, is in response to the question of time and whether a freefaller experiences GR time dilation or if only a stationary participant in a gravitational field experiences the time dilation of gravity.

Squashed
2006-Jul-25, 02:21 PM
Nope, the observed redshift of light can be thought of either as a loss of energy for photons escaping or as an overall change in the rate of time (i.e., the fundamental processes that produce the radiation are slowed down by an equal amount). They are the same thing. ...

Grey,

In your above post (http://www.bautforum.com/showpost.php?p=789616&postcount=62) you state that

1.) redshift of "in-transit" light (energy removed) is the same as
2.) "produced" redshift (due to slower process speeds from a slower rate of time)

but I would contend that these two are not the same because the first process affects an already existing photon whereas the second process affects the "photon factory" (the former affects the product and the latter affects the producer).

Grey
2006-Jul-25, 02:44 PM
Grey,

In your above post (http://www.bautforum.com/showpost.php?p=789616&postcount=62) you state that

1.) redshift of "in-transit" light (energy removed) is the same as
2.) "produced" redshift (due to slower process speeds from a slower rate of time)

but I would contend that these two are not the same because the first process affects an already existing photon whereas the second process affects the "photon factory" (the former affects the product and the latter affects the producer).Let me ask this question. How would you, as an observer seeing the photon, determine whether the effect was something that happened in transit or because of a difference when produced?

Squashed
2006-Jul-25, 02:53 PM
Let me ask this question. How would you, as an observer seeing the photon, determine whether the effect was something that happened in transit or because of a difference when produced?

An observer could not tell but let me state these observations:

If the redshift does not occur "in-transit" then the talk of lightwaves being redshifted into oblivion while coming out of a blackhole gravitational field is a waste of time because if the wavelength is elongated to infinity then the photon was never created in the first place because the "photon factory" has never completed the order.

Also gravitational blueshift would not occur because the receiving gravity field would have no effect on the wavelength because it all happened at the source.

Grey
2006-Jul-25, 04:47 PM
An observer could not tell but let me state these observations:If an observer cannot distinguish the two cases, how can they be considered different? That sounds like a flip question, but that's the essence of the equivalence principle, and indeed, much of the scientific approach. That is, if there is no measurable difference between two possible circumstances, there is no difference. Differences that cannot be measured are not considered meaningful. They're just alternate ways of looking at the situation.


If the redshift does not occur "in-transit" then the talk of lightwaves being redshifted into oblivion while coming out of a blackhole gravitational field is a waste of time because if the wavelength is elongated to infinity then the photon was never created in the first place because the "photon factory" has never completed the order.In either case, an external observer won't see such a photon. These are two ways of looking at that fact, but since they give the same observable result, they aren't really different cases.


Also gravitational blueshift would not occur because the receiving gravity field would have no effect on the wavelength because it all happened at the source.Sure there would. An observer deep in a gravitational field would think that all processes higher in the potential would be running faster, and so would expect light from such a source to be blueshifted relative to a local source.

Squashed
2006-Jul-25, 07:36 PM
An observer deep in a gravitational field would think that all processes higher in the potential would be running faster, and so would expect light from such a source to be blueshifted relative to a local source.

So, in your opinion, does light wavelength change enroute? I have always thought that it can via gravity and spacetime inflation/expansion.

Is the doplar redshift "source driven" or "receiver driven" or both?

I can see your point and I have been trying to figure out which happens: enroute or at the source.

Ken G
2006-Jul-26, 12:57 AM
I'll field that one Squashed. A fundamental element of relativity is that you can't say how the wavelength change occured in any unique way, because light has no medium. With sound, you can always say exactly where the wavelength shift occured, because it happens in the medium (air). But light has only the source and receiver, and the difference in speed and gravitational potential. It's all about differences from source to receiver, but you may model the effect any way you care to visualize it. You can have the change occur right when the light is emitted, or when it is absorbed, or gradually along the way (this latter choice is generally used in cosmology for the convenience of "comoving-frame coordinates"). So the choice generally occurs at the moment that you choose a reference frame from which to analyze the process. So one should generally avoid "absolute" terms when describing such physical processes, and instead be clear on the chosen reference frame, although this is not always done in practice and it causes a lot of unnecessary argument and confusion.

trinitree88
2006-Jul-26, 01:42 AM
I'll field that one Squashed. A fundamental element of relativity is that you can't say how the wavelength change occured in any unique way, because light has no medium. With sound, you can always say exactly where the wavelength shift occured, because it happens in the medium (air). But light has only the source and receiver, and the difference in speed and gravitational potential. It's all about differences from source to receiver, but you may model the effect any way you care to visualize it. You can have the change occur right when the light is emitted, or when it is absorbed, or gradually along the way (this latter choice is generally used in cosmology for the convenience of "comoving-frame coordinates"). So the choice generally occurs at the moment that you choose a reference frame from which to analyze the process. So one should generally avoid "absolute" terms when describing such physical processes, and instead be clear on the chosen reference frame, although this is not always done in practice and it causes a lot of unnecessary argument and confusion.


Ken G. I think I will respectfully disagree here, too, Ken. Light does not have only the source and the receiver....it must always traverse the intervening depths of the neutrino sea. There is no vacuum. Assumptions that the intervening neutrino sea is relatively isotropic, and of no physical consequence are unfounded., and subject of controversey. That leaves a relatively continuous train of interaction, and the light may participate anywhere along the line. ....Pete..

trinitree88
2006-Jul-26, 01:42 AM
I'll field that one Squashed. A fundamental element of relativity is that you can't say how the wavelength change occured in any unique way, because light has no medium. With sound, you can always say exactly where the wavelength shift occured, because it happens in the medium (air). But light has only the source and receiver, and the difference in speed and gravitational potential. It's all about differences from source to receiver, but you may model the effect any way you care to visualize it. You can have the change occur right when the light is emitted, or when it is absorbed, or gradually along the way (this latter choice is generally used in cosmology for the convenience of "comoving-frame coordinates"). So the choice generally occurs at the moment that you choose a reference frame from which to analyze the process. So one should generally avoid "absolute" terms when describing such physical processes, and instead be clear on the chosen reference frame, although this is not always done in practice and it causes a lot of unnecessary argument and confusion.


Ken G. I think I will respectfully disagree here, too, Ken. Light does not have only the source and the receiver....it must always traverse the intervening depths of the neutrino sea. There is no vacuum. Assumptions that the intervening neutrino sea is relatively isotropic, and of no physical consequence are unfounded., and subject of controversey. That leaves a relatively continuous train of interaction, and the light may participate anywhere along the line. ....Pete..

publius
2006-Jul-26, 03:26 AM
publius,

Why do you say, in this post (http://www.bautforum.com/showpost.php?p=791123&postcount=95), that it will have to eventually have to loop back to itself? Does it need both ends connected to a "battery" to establish the current or are you implying the "shape of the universe"?



Basically the "battery thing", that currents have to be in closed loops, unless charge is building up. And an infinite sheet of current would be building up quite a lot of charge on the ends at infinity. An infinite amount, in fact.

Building up a charge means a static electric field is going come in and oppose the current. To keep a "neutral current" going, it has to be going in a closed path.

-Richard

Ken G
2006-Jul-26, 04:01 AM
Light does not have only the source and the receiver....it must always traverse the intervening depths of the neutrino sea...
That is true, I am merely referring to the principle of relativity which is formed in the absence of any material influence by neutrinos. If that theory proves wrong due to an influence by neutrinos, then you will have been right all along.

publius
2006-Jul-26, 06:21 AM
Ken and Grey,

I don't know if you read the ATM section much, but I was reading over a thread there, http://www.bautforum.com/showthread.php?t=44142 and stumbled across the so-called "borehole anomaly" of measurements in g down in boreholes. Then I found this paper:

http://arxiv.org/PS_cache/physics/pdf/0512/0512109.pdf

This is quite interesting to me, as the fine structure constant is claimed to just fall out of this empirically. -Richard

Grey
2006-Jul-26, 01:12 PM
So, in your opinion, does light wavelength change enroute? I have always thought that it can via gravity and spacetime inflation/expansion.I'm saying that it depends on how you look at it. I usually think the slower clock explanation works better, since you'll see all processes slowed, not just redshifted light. But you can explain that by the motion of the signals through curved spacetime, too.


Is the doplar redshift "source driven" or "receiver driven" or both?A Doppler redshift is specifically related to the relative velocity of the source (at emission) and the receiver (at reception). Note that gravitational and cosmological redshifts have slightly different mechanisms, but you can often choose to see them as Doppler-like shifts, given the right choice of coordinates.


I can see your point and I have been trying to figure out which happens: enroute or at the source.I think that you can look at it either way. And since there's not a measurable difference between the two, there is no "right" answer.

Grey
2006-Jul-26, 01:18 PM
Assumptions that the intervening neutrino sea is relatively isotropic, and of no physical consequence are unfounded., and subject of controversey.Of course, assumptions that the intervening neutrino sea has any significant effect at all on electromagnetic radiation are also unfounded. And would seem to be contradicted by a lack of weird electromagnetic effects near nuclear reactors, or differences in Doppler effects based on distance to the Sun. If neutrinos are responsible for these effects, we should see differences based on local neutrino intensity.

Grey
2006-Jul-26, 01:27 PM
I don't know if you read the ATM section much, but I was reading over a thread there...Fairly often, but I had not read that particular thread closely.


This is quite interesting to me, as the fine structure constant is claimed to just fall out of this empirically. -RichardThat's fascinating! I'd be interested to see responses to the paper, if any.

publius
2006-Jul-26, 03:52 PM
Grey,

Yes, very fascinating. I hope this "Borehole Gravity Theory" gets some serious experimental attention. I would love for there to be more careful measurements of the "borehole" g distribution and see if this seems to be consistent. As it is, it is still possible that is just a coincidental trick of density variations.

And then can any predictions be wrung out of their theory that would produce unexpected results in something that can be measured under precise lab conditions? Something that comes to mind is g inside a hollow uniform density sphere. Does their theory predict the "a/2 overmass" effect there as well, or does it still cancel? If it doesn't, that might be something we could detect in a lab.

I've always believed that if there's anything different about gravity, it will be very subtle and look like we think it does most of the time. This is meeting that test.

Outside the mass distribution, we've got something indistinguishable from inverse square. But *inside*, we get small differences, and those differences make it look like more mass is there than really is.

And if this survives the Newtonian limit test, I'd love to see how it would modify full GR in the strong field regime.


-Richard

Ken G
2006-Jul-26, 04:42 PM
I hope this "Borehole Gravity Theory" gets some serious experimental attention.
You can color me spectacularly skeptical on this (I'm sure you guys are skeptical too, but see the potential in it.) My problem is, it has many of the hallmarks of pseudoscience. The mathematical rigor is better than most, but here are the hallmarks I see:
1) the paper is very long on making promises about what the theory does, but most of the support of the promises comes in the form of references to other similarly ATM papers. That doesn't make it wrong, but it is like laying out a maze for the interested follower, who ends up giving up trying to find the cheese instead of just presenting the argument again now that the "new results" strengthen the claims of a connection with alpha.
2) the paper overinterprets its own results. For example, instead of offering an alternative to the dark matter hypothesis, the paper actually states categorically that that hypothesis has led to a "misguided and fruitless search" for dark matter. Any author who could make such an ill-advised remark at this early stage of the theoretical and experimental search for dark matter has forfeited any credibility they earned with me by being able to do math.
3) the paper cherry-picks the data. For example, it is claimed that their approach works for both black holes and the dark matter in spirals, yet their figure 2 offers some cryptic explanation for why it doesn't work for spirals, saying that because they infall, they are more "contingent". Huh? Worse, there is no mention at all of the most significant contribution of dark matter, which is to the overall flattening of the comoving coordinates that has been observed in the recent cosmological data. My guess is, this is because their theory doesn't work there either.

I admit I didn't get far past these three flags, so perhaps if I had kept going in this paper, or looked up the other papers, these concerns would be alleviated. As I doubt that, I haven't invested the time. I'd be interested to hear from anyone who does.

Tensor
2006-Jul-26, 05:06 PM
And if this survives the Newtonian limit test, I'd love to see how it would modify full GR in the strong field regime.


-Richard

While interesting, you can also color me sceptical. You may want to check out this paper (http://www.arxiv.org/PS_cache/gr-qc/pdf/0407/0407059.pdf) along with a look at "Process Physics". Along with Cahill's claim that absolute motion (http://www.scieng.flinders.edu.au/cpes/people/cahill_r/CahillMM.pdf) was detected in the MM and subsequenct similar experiments.


I also question his claim that Newtonian and GR gravity models are are "flawed and disproven" simply on the basis of some possible (or not) anomolies, without providing calculation's showing how his idea's predictions match observed planetary orbits, planetary precession, and binary inspirals.

publius
2006-Jul-26, 10:11 PM
Yes, Brother Cahill's wording is strong and self-insistent, and the statement about GR and Newton "being disproven and fundamentally flawed" is way over the top. Someone trying to call attention to a very interesting gravitational anomaly and trying to get others to look at it and the alpha connection could be a lot more humble and polite. We can make some guesses about his personality from this, of course.

And Tensor's link has someone showing his "velocity field" (or some variant therefore) produces "absurd results" when trying to incorporate it (or some aspect) into GR. So that kills his strong tone.

But the borehole anomaly and the alpha/2 connection is something independent of Brother Cahill's abrasive tone. The fact that the globular cluster stuff (spherical) agrees closely with alpha/2 as well adds to my curiosity.


-Richard

grav
2006-Jul-26, 10:35 PM
Ken and Grey,

I don't know if you read the ATM section much, but I was reading over a thread there, http://www.bautforum.com/showthread.php?t=44142 and stumbled across the so-called "borehole anomaly" of measurements in g down in boreholes. Then I found this paper:

http://arxiv.org/PS_cache/physics/pdf/0512/0512109.pdf

This is quite interesting to me, as the fine structure constant is claimed to just fall out of this empirically. -Richard
Publius, even though they didn't come right out and say it, you do know that the paper for the second link you gave refers to an ether in three-dimensional space, don't you? Look up Miller and his experiments on Mt. Wilson. The paper that I present on my website also has the alpha constant falling out of equations empirically, but I considered it more of an annoyance, since I then had trouble determining just how it should be placed.

Tensor
2006-Jul-26, 11:30 PM
But the borehole anomaly and the alpha/2 connection is something independent of Brother Cahill's abrasive tone. The fact that the globular cluster stuff (spherical) agrees closely with alpha/2 as well adds to my curiosity. -Richard

The borehole anomaly is quite interesting, I'd like to see more work on it, independent of his ideas. The alpha/2 is also interesting, but could be a coincidence (remember Ashmore's paradox? Does his close match to alpha show up using other sytems of measurements?). I'd like to see his ideas a little more fleshed out and matching other observations out there before he declares 400 years of physics, that closely matches to observations, to be flawed.

That kind of tone; the speed to declare Newton and GR flawed, on some pretty flimsy evidence; and the comment that the calculations being "far from simple" (never mind he doesn't produce the equations) raises all sorts of red flags in my mind.

trinitree88
2006-Jul-26, 11:59 PM
Of course, assumptions that the intervening neutrino sea has any significant effect at all on electromagnetic radiation are also unfounded. And would seem to be contradicted by a lack of weird electromagnetic effects near nuclear reactors, or differences in Doppler effects based on distance to the Sun. If neutrinos are responsible for these effects, we should see differences based on local neutrino intensity.

Grey. If neutrinos are responsible......leads to the gravitational redshift experiment by Rebka, Pound...Harvard Towers, Lyman Lab..circa 1960. The resonance detuning and detection of the Mossbauer gamma rays as they traveleled upwards a few decades of meters.
It's true that one can apply GR here. It's also true that one can assign the photon a mass, out of SR ...E/c2=m....lift that mass through mgh...and subtract the potential energy gain from hv, to approximate the redshift. But, because the Earth's interaction with solar, and sea neutrinos has a day night oscillation at SNO.....the inverse square law for sea neutrinos, and solar neutrinos applies here. The Mossbauer photon climbs up against an ever diminishing gradient of them too. Choosing a site for this test...anti-solar zenith, and solar zenith should produce exactly the weird sensitivity you look for in Grav-redshift data. Since it's the weak interaction, and Fermi&Co produced a good bottle of Chianti when the Chicago squash courts yielded "friendly natives", from the first atomic pile...I'll bet a good bottle of Chianti on it. A first rate theory predicts.
I also bet a good bottle of Chianti at Vassar....that pulsars will always be ejected from the same magnetic pole of their progenitors. A first rate theory predicts. Pete.

Edit: It's actually simpler than that...the effect should follow the monthly high "spring" tides...near the full moon.SNO should be seeing this oscillation also.

publius
2006-Jul-27, 05:36 AM
Publius, even though they didn't come right out and say it, you do know that the paper for the second link you gave refers to an ether in three-dimensional space, don't you? Look up Miller and his experiments on Mt. Wilson. The paper that I present on my website also has the alpha constant falling out of equations empirically, but I considered it more of an annoyance, since I then had trouble determining just how it should be placed.

Grav,

No, I didn't know that, but Tensor's second link to another paper by Cahill shows he is a "non-nuller". That's another strike against Brother Cahill, and further explains the tone he takes.

If he believes there is an (a)ether, in the pre-Einstein sense of a classically behaving medium of propagation of EM, motion with respect to which produces differences in the measured speed of light, well, he's wrong.

Now, if one wishes to think of the "fabric of space-time", which behaves like SR and GR (with possible tweaks) say it does, as some sort of "aether", then have at it, but it is not the classical (a)ether.


-Richard

publius
2006-Jul-27, 05:54 AM
The borehole anomaly is quite interesting, I'd like to see more work on it, independent of his ideas. The alpha/2 is also interesting, but could be a coincidence (remember Ashmore's paradox? Does his close match to alpha show up using other sytems of measurements?). I'd like to see his ideas a little more fleshed out and matching other observations out there before he declares 400 years of physics, that closely matches to observations, to be flawed.



Tensor,

Unless I'm misreading it, he is modelling his "alpha" as a dimensionless constant -- this is not where a numerical value in some unit system equals some other numerical value.

Basically, he says the gravitational acceleration, g, is some (fluid flow looking) expression involving his "velocity field" of 3D space. That expression looks like a total (or directional) time derivative of the velocity field itself, ie
(d/dt + (v dot del) ) v (where I used d/dt as the partial here).

Now, that is his g expression, and for an irrotational v, he writes this

div g + (a/8) * (some tensor looking derivative expression on v) = -4piG(rho)


Now, this 'a' is a dimensionless constant, and the borehole data plus the globular cluster stuff show this 'a' looks close to the fine structure constant.

And outside of a spherical mass, he gets this:

g(r) = G(1 + a/2)M/r^2. Which would mean our current G is actually the "real G" times this alpha factor. But if you use the "correct" G, this can also look like an overmass of the same factor, and that actually comes out in the interior solution.

-Richard

Grey
2006-Jul-27, 06:01 AM
A first rate theory predicts.With due respect to Mr. Lomonosov, a prediction is meaningless scientifically unless it is quantitative. What are your estimates for these predicte effects, and what are the calculations that lead to them? If you're forced to make assumptions for those calculations, what are they, and how will the final estimate change if we later determine those assumptions should be modified? Are there other possible explanations for these effects?

And of course, while it's good to make predictions before they are observed, whether the theory is right or not depends on whether the observations actually match those predictions in the end. Until then, these are indeed bold predictions, but certainly not a well-supported theory.

publius
2006-Jul-28, 06:14 AM
All,

If anyone didn't see it in the "MOND" thread, I found another paper by Brother Cahill, where he purports to derive the flat rotation curve for spirals in Eq 24:

http://www.mountainman.com.au/process_physics/HPS28.pdf


His tone here is nowhere as bad as the first paper.


-Richard

grav
2007-Jan-20, 12:04 AM
publius,

It has been bugging me for a very, very long time (obviously :) ), but with my recent work on the gravitational effects of wires, cylinders, and rings, it relates directly to what we were discussing from post #80 to #88 in this thread about the the effects inside of a cylinder or ring. I figure since we both work with these types of situations quite a bit, we should get it straight.

Running through some programs for the integrations, I find that the gravitational force inside of a ring is indeed proportional to the distance from the center, similar to centrifugal force, as you said, but not precisely the same since the acceleration is not directly proportional to the distance, but only close to proportional near the center (that might be important to remember, too). So what I said about rings is incorrect. However, I must have done something differently, then, when I ran the integrations way back, and thought that I did it with rings or something but did not (probably wires around the ring to form the cylinder), because what I said about an infinitely long cylinder still stands. We will feel 0g at any point within the cylinder, in any direction, just like for a sphere. This is because the integration of infinitely long wires around the same ring will change the acceleration toward each point the wires cross to the direct inverse of the distance, instead of its square when integrating for point masses around the ring, as you know. Or, instead of wires, placing rings on top of each other to form the cylinder changes the angles at which the gravitational effects are felt from the closest and furthest points, resulting in the same thing.

Anyway, just thought I'd run this past you since it might be an important thing to know, especially since we both work with these geometries quite a bit. Looks like we were both wrong, or right, or wrong, or... Aw heck, I don't know. :p

publius
2007-Jan-20, 12:27 AM
Grav,

I'm thinking about Rindler geodesics now, not Netwonian gravity, and don't want to even think of this now....:lol:

Hold on and I'll get around to it. But, no, the g-field in the center plane inside a hollow cylinder is indeed "centifugal" all the way, not just approximately near the center.

-Richard

grav
2007-Jan-20, 02:09 AM
Grav,

I'm thinking about Rindler geodesics now, not Netwonian gravity, and don't want to even think of this now....:lol:

Hold on and I'll get around to it. But, no, the g-field in the center plane inside a hollow cylinder is indeed "centifugal" all the way, not just approximately near the center.

-RichardThat's fine. Take your time. It is actually the ring that is almost centrifugal near the center, though, not the cylinder. My integrations for d<r show a*d^2/(G*m) to be

-.5*(d/r)^3 -.5625*(d/r)^5 -.5859375*(d/r)^7 -.59814453125*(d/r)^9 -...

for which I find the formula to be

the integration of
-8 *(2n-1)!^2 *(d/r)^(2n+1)
16^n *n *(n-1)!^4

So the acceleration would be this times G*m/d^2. The first term then becomes proportional to d, the second to d^3, the third to d^5, and so on. For d<<r, very close to the center, the rest of the terms become very small in comparison to the first because the exponent of (d/r) increases, so the acceleration is almost proportional to the distance and away from the center (which is what the negative sign means here, although it is probably usually used the other way around). As d approaches r, though, the acceleration is far from directly proportional to the distance from the center, blowing up at d=r. It doesn't seem like that should be happening, though. Maybe you can show me how you get it when you have time.

I also ran the integration with d<r for a hollow cylinder. It was much simpler. The formula for that is just a*d^2/m=0. :)

grav
2007-Jan-21, 09:16 PM
As d approaches r, though, the acceleration is far from directly proportional to the distance from the center, blowing up at d=r. It doesn't seem like that should be happening, though. Actually, this is making more sense to me now. It appears that we cannot find for d=r along the z axis unless there is also thickness in both the x and y axes also, otherwise it just blows up like a singularity. So this goes for an infinitely thin wire or ring, edge of a disk, or a point, of course. It is okay finding at some angle to the surface of a hollow sphere or cylinder, disk, or infinite plane.

vieuxnez
2007-Feb-25, 08:08 AM
Ok, I get that the increasing relativistic mass will make smaller and smaller accelerations still feel like 1 G, but does anyone have the math to describe this handy? i.e. what function of acceleration f(acceleration) would describe a constant force of 1 G all the way up to approaching c? hmmm... ok, so I'm realizing that it would approach 0 as it approached c, does anyone know if it would follow the same curve as the ratio of time on the accelerating object to that of the stationary observer?

astromark
2007-Feb-25, 10:48 AM
Umm. . . Is meters per second per second not acceleration. At one G constant rate of acceleration would it just double the speed every meter theoretically to C. Looks like my understanding is faulty. where's that program download,? :) but it can not be done right.

Kaptain K
2007-Feb-25, 01:01 PM
Welcome!

You are mixing frames of reference.
To an outside observer, you're acceleration will decrease as your ship approaches the speed of light.
To you, in the ship, your acceleration remains constant as you approach, and pass, the speed of light (subjectively). At a constant one gravity, you could cross the galaxy in just a few years of subjective time. Of course, when you returned, a couple of hundred thousand years would have passed on Earth!

EvilEye
2007-Feb-25, 01:14 PM
Concepts I can figure out. even some simple equations, but now I understand why I did so poorly in college. The math still has me boggled. (not to mention that my prof. didn't speak english very well.)

Are you trying to find out what the gravity is at the center of a mass is? Or trying to find out if there is no center point?

If we had a tunnel from the north pole to the south pole through the earth, would you accelerate half way and accelerate the other half shooting out? Or would you just boing back and forth forever like a yo-yo? Or..[edit] would you get stuck "near" the center? (Of course this example doesn't take into account rotation for the purpose of easing the question)

Kaptain K
2007-Feb-25, 08:06 PM
You would accelerate from the surface to the center (at an ever decreasing rate) and decelerate from the center to the surface, arriving with zero velocity (assuming no air resistance). Interestingly, the time for the round trip back to your starting point is the same as the time it would take (theoretically) to orbit the Earth at zero altitude.

publius
2007-Feb-25, 09:40 PM
Ok, I get that the increasing relativistic mass will make smaller and smaller accelerations still feel like 1 G, but does anyone have the math to describe this handy? i.e. what function of acceleration f(acceleration) would describe a constant force of 1 G all the way up to approaching c? hmmm... ok, so I'm realizing that it would approach 0 as it approached c, does anyone know if it would follow the same curve as the ratio of time on the accelerating object to that of the stationary observer?

We go into great detail about constant proper acceleration trajectories in this thread:

http://www.bautforum.com/showthread.php?t=52372

The trajectory of "constant force" acceleration as seen from any inertial frame is hyperbolic. The interesting thing is these are the equivalent of "circles" in Minkowski space. Events on the trajectory are all constant time-like "distance", S, from some event, just like a circle or sphere are the points at constant distance from a point in Euclidean space.

A simple defintion of curvature in the plane is such that a circle has constant curvature. In Minkowsku, these hyperbolas are the analog of that -- paths of constant curvature. That constant curvature is just the proper acceleration.

-Richard

-Richard

vieuxnez
2007-Feb-26, 04:40 AM
Thank you Publius, I think that thread points me in the right direction, and it looks like I should have asked for a function of force describing the acceleration.