View Full Version : Is gravity energy?

patrick

2006-Sep-21, 07:48 PM

I was wondering: suppose a meteorite, or lightbeam, passes a heavy object (e.g. planet, blach hole) in space, it's path gets bended due to gravity.

Since changing it's path would require energy, my question thus is if mass generates energy how does that effect conservation of energy? (since energy cannot be created or destroyed)

korjik

2006-Sep-21, 08:25 PM

The energy is exchanged between the two bodies. One gains the other loses. Conservation still holds

patrick

2006-Sep-21, 08:35 PM

But their mass do not change, so how can one gain energy?

StevenCrum

2006-Sep-21, 08:50 PM

Your question about gravity and gravitation is one that the best scientists in the world have never figured out, including Einstein. And, the best of quantum mechanics and everyone else still consider it as an unknown.

In spite of all of that, the following is gravity as it exists in the universe. And yes, I factually do know.

The situation with gravitation between planets and stars, and all other objects in the universe is that all of the objects are not only completely made up of electromagnetic objects that have gravitational attraction for opposite charges of other building block objects, all of the objects themselves are fully electromagnetic and have opposite charge attraction also.

And, the standard mass thing of all mass attracts all other mass is all total slop as far as real science.

The answer to your question is entirely in the area of what causes the electromagnetic objects to be attracted gravitationally toward each other.

Describing how a proton is attracted to an electron expalins it fully because all electromagnetic objects from the very smallest component parts of the atom and all the way through planets, stars, galaxies and the universe ball-shaped structure itself operate all the same.

To make a long thing short, a proton is just like an atom in that it has an innner working structure that not only is self-generating to keep in operating continuously, but that functioning interior causes a fan-shaped radiating of photon pulses that are shot out in strings from a side about twenty degrees below where their equator would be. On earth this is about twenty dgrees below the equator and in the centarl part of South America.

The strings of photons are spaced a distance apart as they are going outward, and a magnetic charge is passed between photons in the string. For protons the charge coming from the core of the proton is positive. For electrons it is the same string formation and negative.

The situation then is that if an oppositely charged electron has its fan-shaped strings of electrons get close enough to a proton's fan of strings the two string ends bond together with a + photon magnetically attached to a - photon. The line of charge in each objects string line then pulls them together. This is the basic principle involved.

In our solar system stars have a negative string set and planets have positive. Moons have negative, and galaxies have positive. Neutron stars have negative and the universe core itself has positive.

Atoms have positive, protons positive, electrons negative, sub-protons positive, sub-electrons negative, and the next level down have the proton-like objects positive and the electron-likes negative.

The only other thing needed to be mentioned is the the sitaution of people and objects standing on earth is the situation of the core of earth being positive and its radiated photons also being in a positive string, but the shell surfcae of earth is negative. This comes from all of the electromagnetic objects having their photons being the core charge and the shell charge being opposite of the cores.

So, objects on earth are gravitationally attrcted to the negative shell of earth by the opposite positive charge that is coming form all of the aotms in bodies and everything else that sticks to earth. The positive atom charge is passed through the atom's photon strings, just the same.

So, this is above world-class physics whether anyone understands the dead-on right fact or not, and exactly how gravitation factually works in the entire universe. And, you might note, not Einstein or any other scientist in this world has ever even once gotten gravitation right. Never. And, THAT answers your question.

Doodler

2006-Sep-21, 09:02 PM

But their mass do not change, so how can one gain energy?

Momentum transfer. Take New Horizons and Jupiter as an example. NH is picking up a nice clip of velocity from its swing around Jupiter, at the same time, that same amount of energy is being robbed from Jupiter. The amount of energy transferred will be expressed in New Horizons as about 10,000 kph of additional velocity. Its a LOT of energy to a little probe. To the monster of a planet it just passed, the amount taken amounts to next to nothing compared to its total momentum, so the effects aren't even perceptable by our best instruments.

ToSeek

2006-Sep-21, 09:19 PM

StevenCrum, please note Rule 13 of this forum:

13. Alternative Concepts

If you have some idea which goes against commonly-held astronomical theory, then you are welcome to argue it here. Before you do, though READ THIS THREAD FIRST (http://www.bautforum.com/showthread.php?t=16242). This is very important. Then, if you still want to post your idea, you will do so politely, you will not call people names, and you will defend your arguments. Direct questions must be answered in a timely manner.

People will attack your arguments with glee and fervor here; that's what science and scientists do. If you cannot handle that sort of attack, then maybe you need to rethink your theory, too. Remember: you came here. It's our job to attack new theories. Those that are strong will survive, and may become part of mainstream science.

Additionally, keep promotion of your theories and ideas to only those Against the Mainstream (http://www.bautforum.com/forumdisplay.php?f=17) threads which discuss them. Hijacking other discussions to draw attention to your ideas will not be allowed.

If it appears that you are using circular reasoning, depending on long-debunked arguments, or breaking any of these other rules, you will receive one warning, and if that warning goes unheeded, you will be banned.

(Again, please note emphasized sentences.) You already have two threads to discuss your ideas; do not bring them into threads that were started for other purposes. Even though this is your second rules violation (in a matter of hours), I will cut you some slack since you're new here, and it is a different rule you've violated this time. However, I would strongly advise you to review the rules (http://www.bautforum.com/showthread.php?t=32864) before posting further. Future rules violations are unlikely to be treated with leniency.

patrick

2006-Sep-21, 09:33 PM

Momentum transfer. Take New Horizons and Jupiter as an example....

You mean that due to the mass of Jupiter energy is passed on to the smaller object giving it a swing.

So in theory, if 'zillions' of smaller objects would interact the same way on jupiter, it would imply that Jupiter would eventually run out of energy?

I found that hard to believe, since the mass of jupiter remains the same, so the gravitational pull would still be there.

ToSeek

2006-Sep-21, 09:38 PM

I was wondering: suppose a meteorite, or lightbeam, passes a heavy object (e.g. planet, blach hole) in space, it's path gets bended due to gravity.

Since changing it's path would require energy, my question thus is if mass generates energy how does that effect conservation of energy? (since energy cannot be created or destroyed)

The rule is that if you add up the total momentum of the two interacting objects, you will get a single vector pointing in one direction. That total cannot change, no matter what the interaction. If a meteorite changes its trajectory, then the planet will change its trajectory in the opposite way to compensate.

With the gravity assist, what happens is that the space probe is accelerated in its speed around the Sun, while Jupiter is slowed. However, since Jupiter is so big compared with the space probe, the change is so small that it would take a vast number of probes (billions would not be enough) to change Jupiter's trajectory significantly.

(Since this is a straightforward question and not an ATM issue, I'm going to move it to Q&A.)

pghnative

2006-Sep-21, 09:47 PM

The more global point is that energy can be exchanged without a change in mass. As I sit in the sunshine, I am absorbing radiant energy from the sun, but I am not gaining in mass. (well, except for the cheesecake I'm chowing down on...)

korjik

2006-Sep-21, 09:47 PM

You mean that due to the mass of Jupiter energy is passed on to the smaller object giving it a swing.

So in theory, if 'zillions' of smaller objects would interact the same way on jupiter, it would imply that Jupiter would eventually run out of energy?

I found that hard to believe, since the mass of jupiter remains the same, so the gravitational pull would still be there.

The energy being transfered is Jupiters gravitational potential energy with the sun, and Jupiter's kinetic energy. A small amount of this orbital energy is transfered.

Basically, in your scenario, after you zinged zillions of objects by Jupiter. Jupiter would slow down and crash into the sun.

grant hutchison

2006-Sep-21, 09:50 PM

So in theory, if 'zillions' of smaller objects would interact the same way on jupiter, it would imply that Jupiter would eventually run out of energy?Yes. Its mass would stay the same, its gravity would stay the same, but it would be slowed in its orbit by its interaction with all those smaller objects. Losing energy of motion in this way, it would drop into a lower orbit around the sun. This movement "downhill" within the sun's gravitational field would give it back some kinetic energy, so it would speed up again.

Fire enough small objects past Jupiter on slingshot trajectories, and it would gradually fall inwards until it hit the sun.

Grant Hutchison

PS: Posted at the same time as korjik

RussT

2006-Sep-21, 10:09 PM

Okay, everyone is clearly saying that the gravity field is from the 'planets energy', so...can this gravity field create 'more' gravity of its own???

ToSeek

2006-Sep-21, 10:12 PM

The more global point is that energy can be exchanged without a change in mass. As I sit in the sunshine, I am absorbing radiant energy from the sun, but I am not gaining in mass.

I bet you are, actually. Even if all the Sun is doing is exciting the electrons in the atoms in your skin, the excited electrons will be slightly heavier than they were.

Tensor

2006-Sep-21, 10:16 PM

Okay, everyone is clearly saying that the gravity field is from the 'planets energy', so...can this gravity field create 'more' gravity of its own???

Yes. (http://www.bautforum.com/showpost.php?p=830408&postcount=27)

Doodler

2006-Sep-21, 10:42 PM

Okay, everyone is clearly saying that the gravity field is from the 'planets energy', so...can this gravity field create 'more' gravity of its own???

Gravity creates momentum when it acts upon another massive (in the sense of "object having mass" moreso than "big") object, and that other object acts upon it. Mutual attraction. I don't see how it can "create" more gravity. It can only impart more momentum by a more efficient transfer. Gravity is a product of mass and density. Mass has gravity, but you can have a stronger pull by being in a higher density configuration, because the force of gravity is stronger the closer you are to the center of mass.

Compare the Sun to a 1 solar mass black hole. Same mass, same gravity, but because the black hole is denser, its possible to become more physically proximate to the center of mass, so you can have more gravitational pull than the star.

Orbiting 30 million miles from the center of the black hole would be no different than orbiting 30 million miles from the center of the Sun. You're equidistant from the center of gravity, the forces are equal. Now, orbting 30 million miles from the surface of the Sun versus 30 million miles from the surface of a black hole is a whole different ballgame. In that case, you're closer to the black hole's center of mass, and you're subject to higher gravitational attraction.

Its that reason that the black hole can impart more momentum to objects passing it, because its possible to get deeper into the gravitational field than you would be able to an equally massive star.

RussT

2006-Sep-21, 11:43 PM

Gravity creates momentum when it acts upon another massive (in the sense of "object having mass" moreso than "big") object, and that other object acts upon it. Mutual attraction. I don't see how it can "create" more gravity. It can only impart more momentum by a more efficient transfer. Gravity is a product of mass and density. Mass has gravity, but you can have a stronger pull by being in a higher density configuration, because the force of gravity is stronger the closer you are to the center of mass.

Yes, well they will say that this is pure Newtonian Physics.

[I don't see how it can "create" more gravity]

But, yes, I still have a problem with this.

Compare the Sun to a 1 solar mass black hole. Same mass, same gravity, but because the black hole is denser, its possible to become more physically proximate to the center of mass, so you can have more gravitational pull than the star.

Orbiting 30 million miles from the center of the black hole would be no different than orbiting 30 million miles from the center of the Sun. You're equidistant from the center of gravity, the forces are equal. Now, orbting 30 million miles from the surface of the Sun versus 30 million miles from the surface of a black hole is a whole different ballgame. In that case, you're closer to the black hole's center of mass, and you're subject to higher gravitational attraction.

Its that reason that the black hole can impart more momentum to objects passing it, because its possible to get deeper into the gravitational field than you would be able to an equally massive star.

Sorry, not quite Doodler.

If our sun 'suddenly became' a black hole, the mass would be the same, just in a smaller area, and so Earth's orbit would not be affected.

See my question here.

http://www.bautforum.com/showpost.php?p=830428&postcount=28

Russ: that's exactly what doodler said. He stated that 30m miles from either one's CM would result in exactly equal forces.

Tim Thompson

2006-Sep-22, 02:45 AM

Okay, everyone is clearly saying that the gravity field is from the 'planets energy', so...can this gravity field create 'more' gravity of its own???

Yes. Since the gravitational field has energy, and energy has equivalent mass, then a gravitational field must itself generate additional gravity.

Physics News Update 454, Oct 26, 1999 (http://www.aip.org/pnu/1999/split/pnu454-1.htm).

Aside from that paper, I am sure I have seen one that presents an astrophysical observation of the effect, but I can't find it.

publius

2006-Sep-22, 05:05 AM

All,

I hesitate to mention this, because it's so darn complex and may just add confusion, but the energy of the "gravitational field" in GR is a very vexing subject.

Note the paper Tim links to speaks actually of the gravity of the gravitational binding energy. As I was rambling about in those various threads about what a free fall into a black hole looks like from a far stationary reference frame, in the strict geometrical view, there is no "field energy" as is with an EM field (and I presume the other two forces as well).

In the geometrical (geodesic) view, when an object free falls, it's rest energy (as measured in our frame) gets "converted" to kinetic energy. Well, actually, that is only what it looks like to us. As space-time curves, the "direction" of time gets warped into what we think as space sort of roughyl speaking, and the objects' "motion through time" looks like motion through space. The relativistic energy of the object remains constant to us. Rest energy can sort of be thought as the "kinetic energy of motion through time".

And actually, if I'm thinking about it right, the total relativistic energy remains constant to all observers watching the fall, but they will disagree on what that constant value actually is.

In this view, there is no energy transfer between a field and the object, as there is with an EM field, where mechanical energy can be exchanged with field energy (or kinetic to potential energy), there is just geometry changing the directions of space and time, making things appear to accelerate between frames of reference.

So really, when we speak of the "gravity making gravity" part of this, it's best to not to think of the gravitational field having energy, I don't think. I've read that any attempt to define some construct purporting to be "the energy of the gravitational field" will not work, as it can be transformed away to 0 locally by a coordinate transform.

Indeed, one can see that even classically, the notion of the gravitational field energy is weird. It has to be *negative* (and even in a Heaviside Maxwell-like gravity, the energy of both g and B_g has to be negative). To see this, conside that Newton and Coulomb are the same save for the minus sign. Bring two like charges together, and the total field energy, EdotE integrated over all space increases. The g field of two similiar mass brought together increases in like fashion, yet it is attractive and produces energy (kinetic energy of the masses as they fall towards each other). The only way to make "more be less" is for g dot g to be negative. And that's the meaning of the negative sign which makes like mass charges attract rather than repel.

All that is well and good, until you consider gravitational radiation. If the gravitational field has no real energy, then how the heck can gravitational radiation carry it away from a system. It can be shown strictly that it's possible for a gravitational wave to pass through a system of test masses and increase the total energy.

This is one of the vexing questions of GR, and I would love to have it explained to me. And related to this is the question of the gravitational radiation reaction force. Do we "feel" that? Does a radiating mass follow geodesics, or does the fact that it is radiating mean that it is actually changing the geodesics as it moves, and so feels some of kind of "kick back" force.

And either way, when that gravitational radiation is carrying energy, that would seem to mean there is stress-energy "wrapped up" with it, unlike a (non-radiation) static vacuum field where the stress-energy is zero. I would think, if a quantum description is ever found, that this stress-energy wrapped up in the radiation field would be the mass and momenta of the "real gravitons" associated with the radiation.

-Richard

publius

2006-Sep-22, 05:34 AM

And I'd better make it clear before Ken G. jumps into remind me, the above "rest energy to kinetic energy" view is just that, one view of the situation based on the accounting of a particular coordinate system extended globally through curved space-time.

And this global vs. local thing is apparently one way out of the gravitational radiation energy question, if I'm remembering something I read correctly. You can say conservation of energy is "just a local" thing, and so not worry about where the energy "carried away" by radiation goes as that involves the global accounting of some global coordinate system. And when that gravitational radiation deposits energy to a distant system, you use the same out. That is not a very sastifying way to see this, IMHO. I want global accounting of energy. :)

-Richard

Ken G

2006-Sep-22, 06:03 AM

Rest energy can sort of be thought as the "kinetic energy of motion through time".

Actually, I rather like that way of saying it. :clap: Any objection I raised in the past was simply pointing out that folks around here tend to hear "ways of saying things" as "uniquely true statements about reality", and that leads to all kinds of confusion.

It can be shown strictly that it's possible for a gravitational wave to pass through a system of test masses and increase the total energy.

Furthermore, systems that would be expected to be radiating gravitationally are indeed observed to lose energy over time. Note the classical view would be that gravitational fields contain negative energy, that is, they are not a place to put energy but rather a place where energy has been taken from. From what, I couldn't say, but whatever it is, could perhaps be carrying the positive energy of gravitational waves.

Does a radiating mass follow geodesics, or does the fact that it is radiating mean that it is actually changing the geodesics as it moves, and so feels some of kind of "kick back" force. Yes I think it must, again based on observations of systems dissipating energy as they radiate gravitationally (binary pulsars, for example). There might be global accounting of energy from gravitational waves, but not gravity itself, I just don't know enough about it.

publius

2006-Sep-22, 06:36 AM

Ken,

This is just my musing, not rigorously defined, about how gravitational radiation would sort of work in the Maxwell-like GEM. But here, we would be saying the fields actually have negative energy.

Consider the difference between an EM radiation field and the "regular field" (which can be seen as a reactive energy -- the field at a point always has positive energy, but you can mathematically separate the contributions of the radiation field part and see it as "real energy" and the rest as reactive. This is just a way of accounting that gives a circuit theory like view). You've probably seen pictures of radiating systems showing how the lines of E and B "detach" from the sources -- that "detached" part would be the "real energy" part.

And so you could do that with GEM gravitational radiation. But what would happen, since the total field energy is always negative, is that the radiation part would *raise* the energy, that is, lower the magnitude of a negative number over the region as it passed. And when it interacted with another system of masses, (the "receiver" as it were), that difference would go to kinetic energy.

So you could see a little "positive lump" rippling through a sea of negative energy. The total would always remain negative, but the positive lump would reduce the magnitude of the negative number as it rolled through and eventually could transfer itself to some receiving system.

But in this picture, we are associating energy with a field, and that field is carrying it, just in the "less is more", negative way.

-Richard

Glutomoto

2006-Sep-22, 07:27 AM

I was wondering: suppose a meteorite, or lightbeam, passes a heavy object (e.g. planet, blach hole) in space, it's path gets bended due to gravity.

Since changing it's path would require energy, my question thus is if mass generates energy how does that effect conservation of energy? (since energy cannot be created or destroyed)

Maybe I'm confused. I thought the momentum transfer answer given only applies to the meteorite.

With light it's path is bent because space is curved by gravity. Therefore no energy is exchanged between the massive object and the light that passes it.

err, is that right ?

GL

:)

RussT

2006-Sep-22, 07:49 AM

Russ: that's exactly what doodler said. He stated that 30m miles from either one's CM would result in exactly equal forces.

Orbiting 30 million miles from the center of the black hole would be no different than orbiting 30 million miles from the center of the Sun. You're equidistant from the center of gravity, the forces are equal. Orbiting 30 million miles from the center of the black hole would be no different than orbiting 30 million miles from the center of the Sun. You're equidistant from the center of gravity, the forces are equal. Now, orbting 30 million miles from the surface of the Sun versus 30 million miles from the surface of a black hole is a whole different ballgame. In that case, you're closer to the black hole's center of mass, and you're subject to higher gravitational attraction.

Yes he did, in the first sentence, and then I focused on...

[Now, orbting 30 million miles from the surface of the Sun versus 30 million miles from the surface of a black hole is a whole different ballgame. In that case, you're closer to the black hole's center of mass, and you're subject to higher gravitational attraction.]

So I should have been more clear, and said that his first sentence was correct, but that since the difference in radius of the sun, to the radius of what the sun mass black hole would be, would not move the object inward enough from the 30 million miles, to make that much of a difference, at least not enough to call it a whole new ballgame, BUT, I could be wrong here.

Sorry Doodler, that I wasn't clear here, and I appologize fully if I am wrong.

Doodler

2006-Sep-22, 01:50 PM

Yes he did, in the first sentence, and then I focused on...

[Now, orbting 30 million miles from the surface of the Sun versus 30 million miles from the surface of a black hole is a whole different ballgame. In that case, you're closer to the black hole's center of mass, and you're subject to higher gravitational attraction.]

So I should have been more clear, and said that his first sentence was correct, but that since the difference in radius of the sun, to the radius of what the sun mass black hole would be, would not move the object inward enough from the 30 million miles, to make that much of a difference, at least not enough to call it a whole new ballgame, BUT, I could be wrong here.

Sorry Doodler, that I wasn't clear here, and I appologize fully if I am wrong.

It was meant as two separate examples.

The 30 million from the CM versus 30 million from their respective surfaces. No biggie, will keep in mind to use more paragraph breaks in the future. :)

As for gravity making more gravity, I'll add it to the list of things that gives me headaches trying to comprehend. :)

Ken G

2006-Sep-22, 02:27 PM

With light it's path is bent because space is curved by gravity. Therefore no energy is exchanged between the massive object and the light that passes it.

Whenever the mass-energy of one object is negligibly small compared to the other, the energy transfer to the bigger object is far smaller still. That is not a function of Newton vs. GR, it must hold either way. However, that energy transfer might not be negligible to the smaller object, as in the slingshot effect. Note that the slingshot effect would also happen to light, though would generally be small because the light is usually deflected through a very tiny angle.

Ken G

2006-Sep-22, 02:32 PM

So you could see a little "positive lump" rippling through a sea of negative energy. The total would always remain negative, but the positive lump would reduce the magnitude of the negative number as it rolled through and eventually could transfer itself to some receiving system.

That's what I had in mind as well, but note there is a sticky problem. With EM radiation, and I presume this is also true for quadrupole radiation, the fields get stronger in magnitude when the wave passes. For gravity, if we wanted there to be an energy density in the waves, and the energy is negative in the fields, then we'd have to make the field magnitude drop within the wave. That doesn't sound like what would happen, so I have no clue how you can have a field theory for gravitational radiation....!

Tim Thompson

2006-Sep-22, 03:59 PM

There is something that we call gravity. We don't know what it is, and we may never know what it is. But we do know how it behaves. The history of our attempts to scientifically understand gravity is a series of mathematical & physical models which reproduce the behavior of gravity, to match the accuracy & precision with which we can observe that behavior. Eventually that observational accuracy & precision reached the level where the Newtonian model would not work, and another model, general relativity, was created to replace it. Note that the model you choose depends on the level of accuracy & precision you need. In most cases, we can navigate our way around the solar system using unadorned Newtonian gravity, and we will get close enough to where we want to be. But if our purpose is to understand the universe at its most fundamental level, then we need all the accuracy & precision we can get, and maybe more.

All the stuff about gravitational fields & energy versus curvature depends entirely on how you choose to model gravity. The key point to remember is that all of your models need to be able to reproduce the observed behavior of gravity, one way or another. No matter what model you use, you need to get the same answer from all of them. And that's where the "gravity makes gravity" thing becomes so important.

In a strictly Newtonian model of gravity, the net gravitational field at any point is simply the vector sum of all the fields that interact at that point. Combine that with the usual conservation of energy & momentum and voila, you get gravitational sling shot tricks, and transfer of energy & momentum through a gravitational field.

But, as Publius point out, how can you do this using a curvature model, where there is no field & no energy? You do it by noting that curvature generates more curvature, or "gravity makes gravity". In a Newtonian model, the resultant field is the vector sum of all the interacting fields. So, you might think that, in the curvature model, the net (resultant) curvature measured at any given point is just the sum of the curvatures generated by all of he observed masses. You would be wrong, GR does not work that way. The resultant curvature is the sum of the curvatures generated by the observed masses, plus the extra curvature generated by the original curvature. And of course, the extra curvature generates extra curvature, and so on ad infinitum. Fortunately for us, the ad infinitum part converges fairly rapidly.

If I am not mistaken (I am forced to confess that I sometimes am), that extra curvature induces behavior that is exactly equivalent to, for instance, the loss of energy through gravitational radiation in the models that are based on a gravitational field. Consider a close binary pair of neutron stars or black holes. As they lose energy in the form of gravitational radiation, they spiral in towards each other until they merge. Now, in a curvature model, the curvature experienced by the binary pair is the resultant curvature of their masses, plus all that extra curvature. All that extra curvature will have the same effect as curvature induced from the outside, it means that the binary pair is moving in space that is more strongly curved than their own masses would indicate, so the binary pair has to change orbits to compensate for the added curvature. And so on, ad infinitum, until they merge. The gravitational radiation is nothing more complicated than curvature propagating from the binary pair.

In the end, both models thankfully predict exactly the same observable behavior, and that's the real goal.

And it should also be noted that a real understanding of general relativity will, alas, not come to anyone on a discussion board like this one. The only way to pull that off is to pull the books down & go to work. There is always the "Bible" of the subject, Gravitation (http://www.amazon.com/Gravitation-Physics-Charles-W-Misner/dp/0716703440/sr=1-2/qid=1158939497/ref=sr_1_2/104-5591355-1985532?ie=UTF8&s=books), by Misner, Thorne & Wheeler. It's about 30 years old, thick enough to replace your kevlar body armor, heavy enough to hold down the house in a tornado, and definitely intended for those not intimidated by the presence of more math than English. But it is still in print for a reason. And a few other suggestions:

Clifford Will (http://wugrav.wustl.edu/people/CMW/)'s books are pretty good.

Spacetime and Geometry: An Introduction to General Relativity (http://www.amazon.com/Spacetime-Geometry-Introduction-General-Relativity/dp/0805387323/sr=1-3/qid=1158939616/ref=sr_1_3/104-5591355-1985532?ie=UTF8&s=books), by Sean Carroll (http://preposterousuniverse.com/), 2003

Gravity: An Introduction to Einstein's General Relativity (http://www.amazon.com/Gravity-Introduction-Einsteins-General-Relativity/dp/0805386629/sr=1-5/qid=1158939616/ref=sr_1_5/104-5591355-1985532?ie=UTF8&s=books), by James Hartle (http://www.physics.ucsb.edu/~hartle/), 2002

Mister Earl

2006-Sep-22, 04:09 PM

Here is an eaasy way of explaining it all. Say you have two people rollerskating down a street side-by-side. One of them weighs seven hundred pounds (Jupiter). He has a little person next to him (unidentified small body). Suddenly, the fat guy grabs the arm of the little person and swings them down the street! The fat guy who launched the little person would slow down a tiny bit. The grabbing and launching is gravity, and the strength of the fat guy is not diminished (gravity field) and his mass didn't change, but he's skating slower down the street (orbital momentum) and the little person is leaving a smoke trail (exchange of orbital momentum).

Kind of messy, but I hope that gets the idea across somewhat.

If you send enough little people (random small bodies) by the fat man (Jupiter) and he launches every one (transfer of orbital momentum), eventually he'll slow down enough to hit a parked car (lose orbital stability and crash into the sun or other large body).

Nereid

2006-Sep-22, 04:54 PM

All,

I hesitate to mention this, because it's so darn complex and may just add confusion, but the energy of the "gravitational field" in GR is a very vexing subject.

Note the paper Tim links to speaks actually of the gravity of the gravitational binding energy. As I was rambling about in those various threads about what a free fall into a black hole looks like from a far stationary reference frame, in the strict geometrical view, there is no "field energy" as is with an EM field (and I presume the other two forces as well).

In the geometrical (geodesic) view, when an object free falls, it's rest energy (as measured in our frame) gets "converted" to kinetic energy. Well, actually, that is only what it looks like to us. As space-time curves, the "direction" of time gets warped into what we think as space sort of roughyl speaking, and the objects' "motion through time" looks like motion through space. The relativistic energy of the object remains constant to us. Rest energy can sort of be thought as the "kinetic energy of motion through time".

And actually, if I'm thinking about it right, the total relativistic energy remains constant to all observers watching the fall, but they will disagree on what that constant value actually is.

In this view, there is no energy transfer between a field and the object, as there is with an EM field, where mechanical energy can be exchanged with field energy (or kinetic to potential energy), there is just geometry changing the directions of space and time, making things appear to accelerate between frames of reference.

So really, when we speak of the "gravity making gravity" part of this, it's best to not to think of the gravitational field having energy, I don't think. I've read that any attempt to define some construct purporting to be "the energy of the gravitational field" will not work, as it can be transformed away to 0 locally by a coordinate transform.

Indeed, one can see that even classically, the notion of the gravitational field energy is weird. It has to be *negative* (and even in a Heaviside Maxwell-like gravity, the energy of both g and B_g has to be negative). To see this, conside that Newton and Coulomb are the same save for the minus sign. Bring two like charges together, and the total field energy, EdotE integrated over all space increases. The g field of two similiar mass brought together increases in like fashion, yet it is attractive and produces energy (kinetic energy of the masses as they fall towards each other). The only way to make "more be less" is for g dot g to be negative. And that's the meaning of the negative sign which makes like mass charges attract rather than repel.

All that is well and good, until you consider gravitational radiation. If the gravitational field has no real energy, then how the heck can gravitational radiation carry it away from a system. It can be shown strictly that it's possible for a gravitational wave to pass through a system of test masses and increase the total energy.

This is one of the vexing questions of GR, and I would love to have it explained to me. And related to this is the question of the gravitational radiation reaction force. Do we "feel" that? Does a radiating mass follow geodesics, or does the fact that it is radiating mean that it is actually changing the geodesics as it moves, and so feels some of kind of "kick back" force.

And either way, when that gravitational radiation is carrying energy, that would seem to mean there is stress-energy "wrapped up" with it, unlike a (non-radiation) static vacuum field where the stress-energy is zero. I would think, if a quantum description is ever found, that this stress-energy wrapped up in the radiation field would be the mass and momenta of the "real gravitons" associated with the radiation.

-RichardThis may be of some interest (I'm sure several 'regulars' here are already familiar with it): Is Energy Conserved in General Relativity? (http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html) (Michael Weiss and John Baez).

Perhaps we could go through this, or parts of it, to clear up various questions on the relationship between energy and gravity, in GR? Though, as Tim Thompson has pointed out, it'll quickly become impossible without the relevant maths under your belt.

(And, FWIW, both MTW and Will, two of Tim Thompson's suggestions, are on the Weiss and Baez reference list).

patrick

2006-Sep-22, 04:58 PM

If you send enough little people (random small bodies) by the fat man (Jupiter) and he launches every one (transfer of orbital momentum), eventually he'll slow down enough to hit a parked car (lose orbital stability and crash into the sun or other large body).

Suppose, in theory of course, the universe consists of only 1 planet (named Jupiter), and zillions of little rocks each having a little rocket attached behind them.

One by one these rocks are sent to Jupiter. Each time every rock passes Jupiter it's path is modified due to Jupiter's gravity.

Perhaps it's a flawed reasoning, but in such scenario there wouldn't be lose of orbital stability since there is no orbit. Jupiter might move little by little owing to the rocks, but the energy to change the path of passing rocks remains the same, regardless if Jupiter has moved to a different location.

Doodler

2006-Sep-22, 05:18 PM

The concept isn't that hard to follow, really. Tensor's link gave a decent nuts and bolts explanation.

Gravity is a form of energy, which must be represented by a particle (in keeping with the relativity principle of particle/wave duality), which must have its own mass, and therefore its own gravity. Therefore, gravity itself must generate some level of gravity by the nature of its very existance.

If that isn't the most brainwarpingly recursive logic I've ever seen...at the same time, its defensible grounds for gravitational pull being effectively infinite in range, if tenuous in strength.

Mister Earl

2006-Sep-22, 05:26 PM

If gravity generates gravity we'd see "graviton stars" I guess. Kind of like strangelet stars, just made solely of gravitons... A strange, high-gravity region with no mass O_o

Now THAT is an interesting thought... a massive gravity field with no mass. You could use it as kind of a slingshot to get to other places faster.

Nereid

2006-Sep-22, 05:31 PM

If gravity generates gravity we'd see "graviton stars" I guess. Kind of like strangelet stars, just made solely of gravitons... A strange, high-gravity region with no mass O_o

Now THAT is an interesting thought... a massive gravity field with no mass. You could use it as kind of a slingshot to get to other places faster.Apart from the fact that 'gravitons' are purely hypothetical (we don't have a quantum theory of gravity yet), such an object would be no more meaningful than a 'photon star' ... both travel at c.

OTOH, we used to live inside one kind of 'photon star', and, depending on how you look at it, we are living inside a 'graviton star' right now!

(the former is the universe, in the radiation-dominated era; the latter is the universe).

Mister Earl

2006-Sep-22, 05:37 PM

What would make a hypothetical object like that travel at C? Gravitons generating gravitons would slow itself down, would it not?

BigDon

2006-Sep-22, 06:17 PM

Isn't this the theory of how "hot Jupiters" get in so close to their primaries? After forming in the outer systems like our own beloved gas giants they begin to huck out several of their own masses worth of asteroids and other orbital debris until they lose enough orbital momentum and scrooch in close to their parent stars?

Mister Earl

2006-Sep-22, 06:22 PM

Gravity is a form of energy, which must be represented by a particle (in keeping with the relativity principle of particle/wave duality), which must have its own mass, and therefore its own gravity. Therefore, gravity itself must generate some level of gravity by the nature of its very existance.

But wouldn't that only count if gravitons have mass?

nokton

2006-Sep-22, 06:59 PM

Your question about gravity and gravitation is one that the best scientists in the world have never figured out, including Einstein. And, the best of quantum mechanics and everyone else still consider it as an unknown.

In spite of all of that, the following is gravity as it exists in the universe. And yes, I factually do know.

The situation with gravitation between planets and stars, and all other objects in the universe is that all of the objects are not only completely made up of electromagnetic objects that have gravitational attraction for opposite charges of other building block objects, all of the objects themselves are fully electromagnetic and have opposite charge attraction also.

And, the standard mass thing of all mass attracts all other mass is all total slop as far as real science.

The answer to your question is entirely in the area of what causes the electromagnetic objects to be attracted gravitationally toward each other.

Describing how a proton is attracted to an electron expalins it fully because all electromagnetic objects from the very smallest component parts of the atom and all the way through planets, stars, galaxies and the universe ball-shaped structure itself operate all the same.

To make a long thing short, a proton is just like an atom in that it has an innner working structure that not only is self-generating to keep in operating continuously, but that functioning interior causes a fan-shaped radiating of photon pulses that are shot out in strings from a side about twenty degrees below where their equator would be. On earth this is about twenty dgrees below the equator and in the centarl part of South America.

The strings of photons are spaced a distance apart as they are going outward, and a magnetic charge is passed between photons in the string. For protons the charge coming from the core of the proton is positive. For electrons it is the same string formation and negative.

The situation then is that if an oppositely charged electron has its fan-shaped strings of electrons get close enough to a proton's fan of strings the two string ends bond together with a + photon magnetically attached to a - photon. The line of charge in each objects string line then pulls them together. This is the basic principle involved.

In our solar system stars have a negative string set and planets have positive. Moons have negative, and galaxies have positive. Neutron stars have negative and the universe core itself has positive.

Atoms have positive, protons positive, electrons negative, sub-protons positive, sub-electrons negative, and the next level down have the proton-like objects positive and the electron-likes negative.

The only other thing needed to be mentioned is the the sitaution of people and objects standing on earth is the situation of the core of earth being positive and its radiated photons also being in a positive string, but the shell surfcae of earth is negative. This comes from all of the electromagnetic objects having their photons being the core charge and the shell charge being opposite of the cores.

So, objects on earth are gravitationally attrcted to the negative shell of earth by the opposite positive charge that is coming form all of the aotms in bodies and everything else that sticks to earth. The positive atom charge is passed through the atom's photon strings, just the same.

So, this is above world-class physics whether anyone understands the dead-on right fact or not, and exactly how gravitation factually works in the entire universe. And, you might note, not Einstein or any other scientist in this world has ever even once gotten gravitation right. Never. And, THAT answers your question.

Steven, you made your point in your first three paragraphs. Stay there.

Your rest is so contended. Oh my, foot in the water. As well as gravity,

there is energy, energy can be positive or negative. Once proposed that

stars, or Galaxies, could attract or repel each other, much in the same way

as the magnetic force operates. That did not involve antimatter.

Take your point and concept Steven, but elaboration not serve you well.

Nokton

Doodler

2006-Sep-22, 07:18 PM

But wouldn't that only count if gravitons have mass?

That was the impression I got from the page Tensor linked too.

Nereid

2006-Sep-22, 08:16 PM

What would make a hypothetical object like that travel at C? Gravitons generating gravitons would slow itself down, would it not?Well, whatever else gravitons might be, they are 'force carriers', just like gluons, photons, and the Z and W bosons (that carry the weak force).

Although the 'speed of gravity' hasn't been measured directly yet, it is c in GR, and GR has passed all the tests we've been able to subject it to, to date.

So, c is the speed they would travel at ... unless the relevant quantum theory of gravity that replaces/extends GR says otherwise ...

Tensor

2006-Sep-22, 11:38 PM

What would make a hypothetical object like that travel at C? Gravitons generating gravitons would slow itself down, would it not?

Some of the properties of the graviton can be determined by how gravitiy, as we currently know it, behaves. They would be massless. Gravity travels at c (according to GR), so the particle would have to be massless, because only massless particles travel at c. They would have a spin of 2, becuase gravity is only attractive and spin 2 bosons have the property of only attraction (as oppossed to spin 1 particles, which have attractive and repulsive properties). Real gravitons, as gravitational energy, would create virtual gravitons, the exchange particle of gravity. This is similar to real photons being EM radiation and virtual photons being the exchange particle of EM.

Tensor

2006-Sep-23, 12:01 AM

And it should also be noted that a real understanding of general relativity will, alas, not come to anyone on a discussion board like this one. The only way to pull that off is to pull the books down & go to work.

AMEN, Brother Tim. :lol:

There is always the "Bible" of the subject, Gravitation (http://www.amazon.com/Gravitation-Physics-Charles-W-Misner/dp/0716703440/sr=1-2/qid=1158939497/ref=sr_1_2/104-5591355-1985532?ie=UTF8&s=books), by Misner, Thorne & Wheeler. It's about 30 years old, thick enough to replace your kevlar body armor, heavy enough to hold down the house in a tornado, and definitely intended for those not intimidated by the presence of more math than English. But it is still in print for a reason. And a few other suggestions:

Clifford Will (http://wugrav.wustl.edu/people/CMW/)'s books are pretty good.

Spacetime and Geometry: An Introduction to General Relativity (http://www.amazon.com/Spacetime-Geometry-Introduction-General-Relativity/dp/0805387323/sr=1-3/qid=1158939616/ref=sr_1_3/104-5591355-1985532?ie=UTF8&s=books), by Sean Carroll (http://preposterousuniverse.com/), 2003

Gravity: An Introduction to Einstein's General Relativity (http://www.amazon.com/Gravity-Introduction-Einsteins-General-Relativity/dp/0805386629/sr=1-5/qid=1158939616/ref=sr_1_5/104-5591355-1985532?ie=UTF8&s=books), by James Hartle (http://www.physics.ucsb.edu/~hartle/), 2002

I fully agree with Tim that if you want an understanding of GR, the only way to do it is to get one of the above books and actually learn it. That said, and for those who are not mathematically inclined, a pretty decent understanding of GR can be found in Kip Thorne's Black Holes and Time Warps (http://www.amazon.com/Holes-Time-Warps-Einsteins-Outrageous/dp/0393312763). As a bonus, you get a neat history of GR and Black Hole research.

peteshimmon

2006-Sep-23, 12:32 AM

Now I remember reading the gravitational

field is determined by the kinetic energy of

the matter as well as the matter itself! So

there is an experiment. Will a supercooled

test mass have less gravitational attraction

than it has at room temperature? But I suppose

the difference is minute.

publius

2006-Sep-23, 01:22 AM

That's what I had in mind as well, but note there is a sticky problem. With EM radiation, and I presume this is also true for quadrupole radiation, the fields get stronger in magnitude when the wave passes. For gravity, if we wanted there to be an energy density in the waves, and the energy is negative in the fields, then we'd have to make the field magnitude drop within the wave. That doesn't sound like what would happen, so I have no clue how you can have a field theory for gravitational radiation....!

Ken,

With the GEM (gravito-electromagnetism) approximation, the equations would predict dipole radiation (and are therefore wrong, big-time, on gravitational radiation), but we can at least play around with them. You can gain some insight on how the negative sign changes things.

I *think*, think mind you, it would work out that the radiation field would indeed reduce the field intensities because of the way the minus sign changes things in the coupling. We can write GEM thus (and I'll use 'H', after Heaviside, for the gravitomagnetic force constant -- in this version the 4pi appears explicitly and isn't wrapped up in the permeabilities)

div g = - 4piG * rho

div B_g = 0

curl g = k1*(-dB_g/dt)

curl B_g = k2*-4piH*J + H/G dg/dt

and G/H = c^2

[The k1 and k2 are there to handle the factors of 4 and 2 that come out of linearizing GR -- in Heaviside's original formulation by direct analogy with EM, they would be 1. From GR, however, k2 = 4 and k1 = 1/2 if we say the gravitomagnetic acceleration is v x B_g. This is done in all sorts of different ways, some put in a 2 or 4 on the Lorentz force-like expression. This can be very confusing. And also in the GR linearization, J, the mass current, would have a factor of gamma in front as well. I mention all this for completeness, but the k's and gammas do not affect anything for our purposes here.]

Now, let's imagine making a radiator. We can just use a little ball of mass and jiggle it back and forth sinusoidally. This would be equivalent in EM not to a dipole, but to taking one bit of positive charge and wiggling it back and forth. Very different -- we start out with a monopole field and wiggle that, rather than a dipole field. And wait, this fact might make our radiation quadrapole anyway! But I'm not sure -- I'm certain GEM would predict that accelerating mass would radiate just like EM, but in GR, it's the next derivative up, "jerk".

Compare GEM with Maxwell. Note the signs of the time derivative terms are the same. But the source terms have a minus sign in front. With a positive electric charge, we'd start out with a monopole field, with E pointing out from the charge. When we started wiggling it, the radiation field would add.

But with GEM, our monopole field is pointing in the opposite direction, toward the mass. But since the time derivatives are the same sign, I believe this would have the radiation field 'g' pointing in the opposite direction, and therefore subtracting from the original monopole field, rather than adding. The B_g energy density would be negative too, so its contribution would be to make more negative energy, since we didn't have any B_g to start with, but I think the radiation g part will subtract and we can indeed see a lump of less negative energy propagating out with the radiation wave.

And note another big difference here, the monopolar nature of gravity. With EM, we can start out with a dipole, or even a magnetic dipole of neutral current that has no electric field to start with at all. The radiation E is the only E we have there.

But with gravity, we don't have negative mass, and have to start with a monopolar field. And we can't have a "neutral mass current" -- that is current with no net charge. The mass of our currents is making a monopolar 'g' to start with as well. It would be like a moving net charge density.

-Richard

Ken G

2006-Sep-23, 01:43 AM

Now, let's imagine making a radiator. We can just use a little ball of mass and jiggle it back and forth sinusoidally. This would be equivalent in EM not to a dipole, but to taking one bit of positive charge and wiggling it back and forth. Very different -- we start out with a monopole field and wiggle that, rather than a dipole field.That's true for the background field, but not the linearized far-field radiation. Wiggling a single charge is still wiggling a dipole moment, so still makes the same dipole radiation as stretching and squeezing a true dipole. Note that gravitational radiation could also be dipole if you wiggled a mass, but you can't-- the center of mass of a closed system does not wiggle, so that's why you don't have dipole gravitational waves.

But with GEM, our monopole field is pointing in the opposite direction, toward the mass. But since the time derivatives are the same sign, I believe this would have the radiation field 'g' pointing in the opposite direction, and therefore subtracting from the original monopole field, rather than adding.

I think it's trickier than this. Note that normal EM radiation is a field variation that is perpendicular to the background field, not along it. This is why there's no energy to first order in a linearized wave. But that's OK-- the energy is second order, and comes in because you increase the length of a vector a little when you add something perpendicular. But note it doesn't matter the sign-- perpendicular adds field magnitude, in any direction.

The B_g energy density would be negative too, so its contribution would be to make more negative energy, since we didn't have any B_g to start with, but I think the radiation g part will subtract and we can indeed see a lump of less negative energy propagating out with the radiation wave. Perhaps there is something that can be done here. If the energy associated with g is negative, but with B_g is positive, perhaps the increase in negative energy associated with g is more than compensated by the increase in positive energy associated with the B_g generated by the wave. I'm not sure if that could solve our problem, since as you point out, GEM is not GR.

The mass of our currents is making a monopolar 'g' to start with as well. It would be like a moving net charge density.

True, but note that current is the only thing that appears in the time derivatives. The background field may be subtracted away with no loss of generality about the radiation-- in other words, you could imagine negative masses that do nothing but cancel the real background field, and then assess the radiation anyway. The GEM equations are linear and admit the superposition principle, but I don't know what happens with the nonlinearities of GR. (Tim Thompson above claimed that they are crucial for gravitational radiation, I really just don't know.)

publius

2006-Sep-23, 04:52 AM

Ken,

Oh fiddlesticks. I had faith this should work out in the positive lump fashion, but you're shaking that faith. :)

The B_g energy has to be negative as well from similiar considerations as for g itself. Like mass currents are repulsive rather than attractive (which makes like B_g poles attractive and opposites repulsive). With EM, as two like currents are separated, the integral of B^2 increases. And likewise for two mass currents, so the B_g energy has to be negative too.

But wait a minute. What happens with a receiver vs a transmitter. We have a receiving antenna, absorbing the energy of the radiation field. Now, the radiation field induces a similiar pattern of charge separations and currents in that as the transmitting arrangement. It's just the same process reversed.

Now, that receiver would have to take energy away from the radiation field, so the induced fields would have to be subtractive, to reduce the energy, right?

What if the minus sign arrangement makes our wiggling mass transmitter behave like that instead? ............................. See, I've got faith. :)

To be sure, only thing is to find a solution for the GEM equations with a wiggling mass.................

-Richard

publius

2006-Sep-23, 05:07 AM

That's true for the background field, but not the linearized far-field radiation. Wiggling a single charge is still wiggling a dipole moment, so still makes the same dipole radiation as stretching and squeezing a true dipole. Note that gravitational radiation could also be dipole if you wiggled a mass, but you can't-- the center of mass of a closed system does not wiggle, so that's why you don't have dipole gravitational waves.

Ken,

That's a very good point, and one that is crucial to how GR works. Conserveration of mass-energy and momentum (at least for the "mechanical part" represented by the stess-energy tensor) is built right into the stree-energy tensor. If you try to put a source in that like my wiggling ball or something else, you've got to put the energy and momentum in that make it wiggle in the first place. If you don't, you get garbage, because that's not a valid stress-energy tensor.

This first sunk in when I read that paper of Steve Carlip's in answer to Van Flandern's speed of gravity stuff. Carlip showed that the GR field "extrapolated" acceleration of the source, better compensating for the propagation delay.

Carlip explained that the trouble with doing this is you can't just plug in an accelerating mass into the stress-energy tensor. You've got to put what is causing the acceleration in as well, so energy and momentum are conserved. Maxwell is only concerned with the EM part of the energy and doesn't care about mechanical energy necessary to make the sources do what they do. But GR very much cares.

So what he did was use a "photon rocket" metric, which had a mass being acclerated by throwing off EM radiation. This is nice because the total energy thrown out the tailpipe is very small compared to the rest energy of the mass.

If it weren't, say this was regular rocket throwing off significant mass, then the gravitational field wouldn't behave as you expected. Sort of like two masses flying apart -- the gravitational field remains pointed toward the center of mass. The photon rocket allowed most of the field to point toward the mass being accelerated, and show the "extrapolation of acceleration" as well as velocity.

-Richard

publius

2006-Sep-23, 05:51 AM

If I am not mistaken (I am forced to confess that I sometimes am), that extra curvature induces behavior that is exactly equivalent to, for instance, the loss of energy through gravitational radiation in the models that are based on a gravitational field. Consider a close binary pair of neutron stars or black holes. As they lose energy in the form of gravitational radiation, they spiral in towards each other until they merge. Now, in a curvature model, the curvature experienced by the binary pair is the resultant curvature of their masses, plus all that extra curvature. All that extra curvature will have the same effect as curvature induced from the outside, it means that the binary pair is moving in space that is more strongly curved than their own masses would indicate, so the binary pair has to change orbits to compensate for the added curvature. And so on, ad infinitum, until they merge. The gravitational radiation is nothing more complicated than curvature propagating from the binary pair.

Tim,

Ok that's well and fine. We could say that the curvature model only "makes it look like" energy is being lost by the "radiating" system, and the gravitational waves are just propagations of (energyless) curvature, radiating out from the system.

But, from what I understand, and this could very well be wrong, is that a passing gravitational wave can increase the energy of a mass system it pass through (a "reciever" of sorts). The way I understood it was you start out with a system of small test mass in flat space-time. Construct a gravitational wave packet of some sort and let it pass through. After the disturbance has passed through, space-time returns to flat again. But the mass system can gain energy.

I don't know if it could be this simple, but I imagined two masses sitting there at rest with respect to each other, but are moving after the wave passes through.

To my mind there is no way to say that "just looks like" the system gained energy, because space-time is flat before and after.

Now, an out might be the "small mass" assumption. To be truly flat initially, the masses would strictly have to be zero, and zero mass would still have zero energy even if it were moving. :) So it might be that if you took the actual curvature of finite mass into account, you could preserve the "only looks like it gained energy" picture.

But I don't know.

-Richard

trinitree88

2006-Sep-23, 11:33 AM

Now I remember reading the gravitational

field is determined by the kinetic energy of

the matter as well as the matter itself! So

there is an experiment. Will a supercooled

test mass have less gravitational attraction

than it has at room temperature? But I suppose

the difference is minute.

Pete. I'd say yes by E/c2=m. Otherwise, GR is trashing SR....but it already does that when a massive test particle enters the event horizon, and exceeds c. Pete

Tensor

2006-Sep-23, 01:42 PM

Pete. I'd say yes by E/c2=m. Otherwise, GR is trashing SR....but it already does that when a massive test particle enters the event horizon, and exceeds c. Pete

Pete, GR doesn't trash SR. The SR equations are simply a special case of GR, which can be used when the Minkowski metric is used in the GR field equations. The Minkowski Metric can only be used when there is no gravitational gradient (or where the gradient can be ignored). That is hardly the case near a Black Hole horizon.

Ken G

2006-Sep-23, 01:59 PM

The B_g energy has to be negative as well from similiar considerations as for g itself. Like mass currents are repulsive rather than attractive (which makes like B_g poles attractive and opposites repulsive). With EM, as two like currents are separated, the integral of B^2 increases. And likewise for two mass currents, so the B_g energy has to be negative too.

I see. Well, then we are up a tree on the field-energy approach.

Now, that receiver would have to take energy away from the radiation field, so the induced fields would have to be subtractive, to reduce the energy, right?

Yes, but this is subtracted from the radiative pertubation, not the background field (which never gets subtracted from because all the field perturbations are perpendicular to it). The way the subtraction is accomplished is via destructive interference, i.e., phase effects. Indeed, that's all that ever happens to radiation once created-- it is only affected by itself, or the coherent influences it induces.

Which suggests we may be thinking about this wrong. The radiated energy is always separate from the background field, and it can never actually exchange energy with it, whether positive or negative. Again it's the superposition principle-- we can (mathematically speaking) insert a compensating constant field to cancel the background field, with no impact on the radiation. The fact that the background field contains negative energy is irrelevant, except for the assertion that all g and B_g fields must contain negative energy, yet the Poynting vector for these fields must somehow be positive energy. If we insert a negative sign into the Poynting vector, then we lose conservation of energy, and if we don't, we are radiating negative energy. I see no solution, unless we make the energy density in fields also depend on their time derivatives in such a way as to end up with positive energy in the radiation. Correct me if I'm wrong, but isn't it E dot D, not E^2, in EM fields? Perhaps there's some way to get a "negative D" when dealing with gravitational radiation?

Atraveller

2006-Sep-23, 02:50 PM

There is something that we call gravity. We don't know what it is, and we may never know what it is. But we do know how it behaves.

Do we know how fast Gravity Acts?

Publius refers to gravitational radiation (in post #22 of this thread) which implies speed. Does Gravity propagate like light or magnetic energy? Has the speed of Gravity ever been measured?

Sorry if I'm departing from the direction of the thread.

-Dave

Ken G

2006-Sep-23, 03:08 PM

Does Gravity propagate like light or magnetic energy? Has the speed of Gravity ever been measured?

Gravity should propagate at the speed of light, but that is also true of "magnetic energy". I don't know that its speed has ever been directly measured, but there are several indirect measurements that strongly suggest GR is a successful description, and it includes gravity propagating at c. Most physicists expect that gravity is carried by vitual "gravitons" that kind of move at c but virtual particles are slippery when not carrying information, while gravitational radiation is carried by real gravitons that should clearly move at c.

gzhpcu

2006-Sep-23, 03:46 PM

As long as we don't have a quantum theory of gravity, why is gravity being discussed here as if it were a force? As long as GR reigns, it is only a spacetime distortion. Gravity waves are predicted by GR as well...

publius

2006-Sep-23, 05:36 PM

Gzhpcu,

I don't think anyone here is requiring that gravity be a "real" force. And indeed, I would think a quantum description of gravity could be interpreted as the quantization of space-time curvature.

And let me caution all lurkers reading our musings on GEM, that this is entirely an academic discussion, wondering if the GEM gravitational radiation can "make sense" in how it would carry energy from one place to another. GEM results from a linearization of GR that ignores the non-linear and high order terms. GEM predicts EM-like dipole radiation, while the full GR shows it is actually quadrapole. While it makes for interesting thought experiments, and some insight into gravitomagnetism and induction, that's about it.

When Newton is not accurate enough, the other terms of full GR are usually stronger than the additional GEM effects anyway. Case in point would be the precession of Mercury's perihelion. GEM would predict only tiny gravitomagnetic effects way too small to account for it. It is the non-linear behavior of 'g' itself that does that.

-Richard

gzhpcu

2006-Sep-23, 05:45 PM

Loop quantum gravity posits quantization of spacetime. M-theory includes a graviton, without any such quantization of spacetime. Puzzling....

publius

2006-Sep-23, 06:04 PM

Ken,

In GEM, you do indeed have to take the flux densities as pointing in the opposite direction of the field intensities. That is, the negative sign is actually wrapped up in the permeabilities. Well, that is the most logical way to handle it. For example, take the gravitational equivalent of D, the electric flux density. That would just be the mass flux density, and would point out from a positive mass. Divide that by the "permittivity" to get the field strength, and it has to be negative to make 'g' point toward the mass.

And the energy is indeed 1/2(E dot D) for a linear medium. In GEM we would write the same thing directly, and the negative sign would come from the fact the two point in opposite directions.

With this negative permeability convention, we can write GEM exactly like Maxwell (I'll use D and B for the flux densities, and g and h for the field intensities -- and use a "mu" and "epsilon" which we'll put in terms of G later just to keep from making mistakes)

div D = rho; div B = 0

curl g = -dB/dt

curl h = J + dD/dt

Now, that works directly. But let's add the negative permeabilities,

D = -e*g; B = -u*h. Now plug that in, and convert to g and B

div g = -rho/e; div B = 0;

curl g = -dB/dt

curl B = -uJ + ue dg/dt

See how neat that works out by wrapping up the negative sign in the constants?

The Poytning vector is actually E x H. So the GEM equivalent would be g x h. But since h points backwards to B(_g), we would write that

S ~ -g x B_g

The momentum density is D x B. And again we'd get a minus sign writing that in terms of g.

-Richard

Tensor

2006-Sep-23, 06:23 PM

Loop quantum gravity posits quantization of spacetime. M-theory includes a graviton, without any such quantization of spacetime. Puzzling....

gzhpcu, that may simply be due to the directions that each of them are attacking the problem. M-theory is working more from the particles up, QLG is working more from the background, down. Is it any wonder that M-theory hasn't specified background and QLG hasn't got to particles?

publius

2006-Sep-23, 06:26 PM

Ken,

This may be another one of my half-cocked guesses that doesn't pan out, but if the GEM S points in the opposite direction wouldn't that reverse the sense of transmitter and receiver?

Consider an EM system with a net S flux out. Say we're wiggling the charge back and forth. We're having to push on that charge against the radiation reaction, so we are doing work on the system.

So if in GEM, the sign of S reverses, wouldn't that analogous situation then be a receiver? Same field picture, but energy is flowing in the opposite direction? And rather than us pushing on the charge, the mass charge would be pushing against us?

And so an actual GEM transmitter would then look like an EM receiver system, with the ingoing energy being reversed?

Maybe I am placing too much stock in this "positive lump" idea. Energy does get transferred, but not in the picture I was thinking.

-Richard

Tim Thompson

2006-Sep-23, 06:29 PM

But, from what I understand, and this could very well be wrong, is that a passing gravitational wave can increase the energy of a mass system it pass through (a "reciever" of sorts). The way I understood it was you start out with a system of small test mass in flat space-time. Construct a gravitational wave packet of some sort and let it pass through. After the disturbance has passed through, space-time returns to flat again. But the mass system can gain energy.

I have never encountered this argument before, and it will take some effort to track it down. But it does not look right to me, speaking strictly intuitively. The only way for such a scenario to work would be for the gravitational wave to lose energy, transfering it to the mass system, otherwise conservation of energy is locally violated, which GR will not allow. But if that is the case, then we have a "tired light" type scenario, only this time it's "tired gravitational waves", losing energy to the universe as they propagate through it. That should pump the internal energy of the universe to unacceptably high levels, which are not observed. Hence, I think that the agrument is wrong.

publius

2006-Sep-23, 06:35 PM

Tim,

But is that really like tired light. EM radiation (photons) can certainly interact, and loose energy to charges. It's just the field interacting with its sources. The system of test masses is just a "receiver" for a gravitational wave. The wave is not loosing energy "to the universe" but transferring it to its sources, just an EM receiver.

That assumes a gravitational wave actually carries energy, of course, and this was suppose to show it must.

-Richard

publius

2006-Sep-23, 06:56 PM

Tim and all,

This is what Wiki says on the subject of energy of gravitational

radiation. Whether Wiki knows what it's talking about, I don't know:

Within parts of the scientific community there was initially some confusion as to whether gravitational waves could transmit energy as electromagnetic waves can. This confusion came from the fact that gravitational waves have no local energy density - no contribution to the stress-energy tensor. Unlike Newtonian gravity, Einstein gravity is not a force theory. Gravity is not a force in general relativity; it is geometry. Therefore the gravitational field was thought not to contain energy, as would a gravitational potential. But the field can most certainly carry energy as it can do mechanical work at a distance. For a number of years this issue was addressed by using gravitational stress-energy pseudotensors that transport energy. Among a number of candidates a frequently used one is the symmetric Landau-Lif****z pseudotensor. The analysis based on pseudotensors is subject to the general criticism that they are not proper tensors and thus physical conclusions based on them may not be coordinate independent. Since the components of a pseudotensor can vanish in a coordinate system but not in others, the gravitational energy density is not localizable. In physical terms this is simply a reflection of the fact that gravitational field is locally transformed away for a freely falling observer because of the Equivalence Principle. A major conceptual advance came when in 1962 Hermann Bondi and his coworkers analyzed gravitational radiation in Einstein gravity using a specially devised coordinate system which showed explicitly how radiation can carry energy out to infinity from an isolated source causing it to lose mass. The work by Bondi et al., Sachs and by Newman and Penrose form the basis of much of the current theoretical understanding of the structure of gravitational radiation field far from the sources

Note the part about "do mechanical work" at a distance. This is the receiver analog. Mass over here is wiggled in such a way as to radiate, and that radiation can cause other mass to wiggle. Transmitter --> receiver. :)

ETA: The profanity filter stupidly ****d out Lifsh-itz's name.

-Richard

publius

2006-Sep-23, 07:21 PM

I would express the part about the system "lossing mass" to infinity differently. I would express it as loosing gravitational *binding energy*. It would loose the mass of that energy, of course, and that's what is intended, but I would phrase it in terms of energy to make it more clear what is hapenning.

For example, we drop two masses toward each other. Now, in the geodesic view, some of their rest energy is converted to kinetic energy. They slam into each other and that kinetic energy is converted to heat. That is the "binding energy". When a system is radiating, the radiation carries (or appears to carry) some of that binding energy out of the system. EM radiation can certainly carry some of that binding energy away, and that happens all the time.

Note that to have an orbit, a bound gravitational system, some of that binding energy has to be shed, converted to heat or maybe even rotational energy from frictional interactions.

If that binding energy is carried away from the system by any available means, then the total gravity is reduced, as we have less total mass-energy to gravitate.

-Richard

Tim Thompson

2006-Sep-23, 07:25 PM

Do we know how fast Gravity Acts?

Publius refers to gravitational radiation (in post #22 of this thread) which implies speed. Does Gravity propagate like light or magnetic energy? Has the speed of Gravity ever been measured?

The "speed of gravity", which is the velocity with which gravitational waves will propagate, has never been directly measured. In order to do that, one must first directly detect gravitational waves, which has not been done, and then one must have enough detectors to confidently track the motion of the gravitational wave train. The first generation of interferometric gravitational wave detectors is still quite young (LIGO (http://www.ligo.caltech.edu/), VIRGO (http://wwwcascina.virgo.infn.it/), GEO600 (http://www.geo600.uni-hannover.de/), and others; LISA (http://lisa.jpl.nasa.gov/) is still in development, and delayed by NASA budget cuts), and has not yet produced any detections. So it will be some years I am sure, before that can be done.

However, if there are gravitational waves at all, then general relativity requires that the speed of gravity & the speed of light should be the same. The changing orbits of binary pulsars can be observed by precision timing experiments, and the results compared to the predictions of GR. The results & predictions are so far in agreement. And since the predictions are also sensitive not just to the presence of gravitational waves, but also to their speed of propagation (the "Shapiro Delay" depends on the speed of gravity), these results are taken as an indirect indication that the speed of gravity & the speed of light are the same. A few examples, just to show that it has been done.

A new test of general relativity - Gravitational radiation and the binary pulsar PSR 1913+16 (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1982ApJ...253..908T&db_key=AST&d ata_type=HTML&format=&high=4366fa465110829); Taylor & Weisberg, ApJ, Feb 15, 1982.

High-precision timing of PSR J1713+0747: Shapiro delay (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1994ApJ...437L..39C&db_key=AST&d ata_type=HTML&format=&high=4366fa465115029); Camilo, Foster & Wolszczan, ApJ, December 1994

A test of general relativity from the three-dimensional orbital geometry of a binary pulsar (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=2001Natur.412..158V&db_key=AST&d ata_type=HTML&format=); W. van Straten, et al.., Nature, July 2001.

The Relativistic Binary Pulsar B1913+16: Thirty Years of Observations and Analysis (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=2005ASPC..328...25W&db_key=AST&d ata_type=HTML&format=); Weisbeerg & Taylor, ASP Conference Series, 2005.

Shapiro Delay in the PSR J1640+2224 Binary System (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=2005ApJ...621..388L&db_key=AST&d ata_type=HTML&format=&high=4366fa465115029); Oliver Lohmer, et al., ApJ, March 2005.

Measurement of Orbital Decay in the Double Neutron Star Binary PSR B2127+11C (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=2006ApJ...644L.113J&db_key=AST&d ata_type=HTML&format=); B.A. Jacoby, et al., June 2006

There were recent claims that the Shapiro delay had been measured directly, by watching Jupiter eclipse a quasar (Fomalont & Kopeikin, 2003 (http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=2003ApJ...598..704F&db_key=AST&d ata_type=HTML&format=&high=4366fa465117595)) but that claim remains controversial and may be invalid (follow the 35 citations (http://adsabs.harvard.edu/cgi-bin/nph-ref_query?bibcode=2003ApJ...598..704F&refs=CITATIO NS&db_key=AST) to that paper, for a more or less complete history of the discussion; the exchange with Carlip, and the paper by Clifford Will are perhaps most important).

publius

2006-Sep-23, 08:04 PM

Ken,

I think the answer to this little problem is no more complex that "negative energy going to the right is positive energy going to the left". That is, I'm putting too much stock on the negative energy densities of the fields themselves, and not realizing what matters is the flow of energy.

I think a GEM transmitter receiver acts just bassackwards from EM case. We get waves of g and B_g, and yet they are increasing in magnitude, but what matters is the flow. In the EM case, we'd look at S, and say, aha, we have energy flowing out of region A and going into region B.

In the GEM case, that would be *negative energy* flowing out of region A and into region B. And that is equivalent to positive energy flowing out of region B into A.

If is any region is "receiving negative energy", then it is actually the transmitter -- the thing would only work (that is the fields and S would only behave like this) if actually there was a source of energy being converted to GEM gravitational radiation inside.

-Richard

publius

2006-Sep-24, 05:33 AM

Just for fun, let's look at the GEM Poynting vector. We'll do it for EM first, then carefully put the minus signs in the permeabilities and see what happens. It is very informative to go through the derivation of the meaning of the Poynting vector.

One generally starts out by taking E x H and seeing what the flux of that over a closed surface is. Consider the divergence of E x H:

div(E x H). From a vector calculus relation the divergence of a cross product of two vectors is:

H dot (curl E) - E dot (curl H). Now substitute the Maxwell curl relations:

H dot (-dB/dt) - E dot (J + dD/dt). Rearrange to get

-(H dot dB/dt + E dot dD/dt) - (J dot E)

Now for a linear medium, D = eE, E dot dD/dt = 1/2 d/dt(E dot D)

= d/dt(1/2 e E^2). This is just the negative time rate of change of the electric energy density. Or the rate of decrease. Now, there is no loss of generality for a non-linear medium where the relation between D and E (and B and H) is not linear. The energy is the integral of E dot dD (and H dot dB), so d(energy)/dt = E dot dD/dt (and H dot dB/dt).

So we have:

div(E x H) = -d(field-energy-density)/dt - J dot E.

Now, some derivations, those concerned with waves from some fixed source propagating through a medium, will write J = sigmaE to represent a lossy medium with some finite conductivity. So this becomes an ohmic looking term, (sigma*E^2). That is all well and good for those purposes, but it looses generality. J dot E is the rate of transfer of field energy to mechanical energy (looking like a general VI). Positive J dot E means energy going from the field to mechanical work. Leaving it in that form retains full generality, including all energy transfer processes as well as ohmic.

For example, consider free charges being accelerated by the E field, like in a plasma, gaining kinetic energy. d(KE)/dt = mv dv/dt. But dv/dt = a, is qE/m, so that is qv E, which becomes J dot E in the point function limit.

J dot E is the most general form, handling all EM-mechanical energy transfer. So,

div (E x H) = -d(field-energy-density)/dt - d(EM-mech.-energy)/dt

Now, integrate that over a volume, and convert the lefthand side to a surface integral:

Flux(E x H) = -d(field-energy-in-volume)/dt - d(EM-mech-in-volume)/dt

The (outward) flux of E x H is equal to the rate of decrease of stored field energy, less the rate at which mechanical energy is being transferred in the volume. The rate of decrease can be confusing at first blush, so we can write it thus:

Flux( -E x H) = d(field energy)/dt + d(mechanical-energy)/dt

The (inward) flux of the Poynting vector is equal to the rate at which the field energy is increasing plus the work being done by the fields.

If E x H is "going into a volume", then the energy must either go into the field inside, or be expended as mechanical work.

That is Mr. Poynting's vector. Now smoke that over, and then we'll convert to GEM by (carefully!) inserting the minus signs of the permeabilities)

-Richard

grav

2006-Sep-24, 01:46 PM

I just thought of something. Could the "gravity creates more gravity" thing be the "cause" of GR? What about the gravity dependence on mass plus kinetic energy? How does that work?

Ken G

2006-Sep-24, 02:21 PM

That is Mr. Poynting's vector. Now smoke that over, and then we'll convert to GEM by (carefully!) inserting the minus signs of the permeabilities)

That's a nice summary, but it seems that just allowing a negative permeability only gives us a negative energy density-- it's not going to let our waves propagate positive energy. Note that negative energy going to the left is not the same as positive energy going to the right. The energy flux is the same, but if you dissipate it, you don't get heating either way.

publius

2006-Sep-24, 05:01 PM

That's a nice summary, but it seems that just allowing a negative permeability only gives us a negative energy density-- it's not going to let our waves propagate positive energy. Note that negative energy going to the left is not the same as positive energy going to the right. The energy flux is the same, but if you dissipate it, you don't get heating either way.

Ken,

I ain't through yet. :) I'm still thinking about the GEM Poynting vector. It may not work out, but I'm gonna try. :lol: The summary was to get us in a Poynting frame of mind..... :)

-Richard

publius

2006-Sep-24, 05:19 PM

I just thought of something. Could the "gravity creates more gravity" thing be the "cause" of GR? What about the gravity dependence on mass plus kinetic energy? How does that work?

Grav,

To really understand "how that works", you're going to have learn some more math than you know now. Calculus/differential-eq --> vector calculus and then on to tensors, sort of a generalization of vectors. In the meantime, all the word pictures anyone tries to put together are going to be incomplete and even misleading.

In a nutshell, gravity depends on "stress-energy", which is a way of saying the source of gravity is mass-energy, mass-energy flow, momentum, and momentum flow. Well, really momentum and mass-energy flow are the same thing.

How does kinetic energy work as source? Well, just think of dividing that energy by c^2 ala E = mc^2 and getting the mass equivalence of that energy. But when a mass distribution is moving, the gravitational field is going to depend on the momentum and energy-flow as well.

-Richard

nokton

2006-Sep-24, 05:49 PM

As long as we don't have a quantum theory of gravity, why is gravity being discussed here as if it were a force? As long as GR reigns, it is only a spacetime distortion. Gravity waves are predicted by GR as well...

Thanx Phil for bringing reason to the table.

Nokton.

Ken G

2006-Sep-24, 06:43 PM

I ain't through yet.

OK, press on, and may the force be with you.

grav

2006-Sep-24, 08:25 PM

Grav,

To really understand "how that works", you're going to have learn some more math than you know now. Calculus/differential-eq --> vector calculus and then on to tensors, sort of a generalization of vectors. In the meantime, all the word pictures anyone tries to put together are going to be incomplete and even misleading.

In a nutshell, gravity depends on "stress-energy", which is a way of saying the source of gravity is mass-energy, mass-energy flow, momentum, and momentum flow. Well, really momentum and mass-energy flow are the same thing.

How does kinetic energy work as source? Well, just think of dividing that energy by c^2 ala E = mc^2 and getting the mass equivalence of that energy. But when a mass distribution is moving, the gravitational field is going to depend on the momentum and energy-flow as well.

-Richard

Thanks, publius. And you're right. I do need to learn some higher math. :o

publius

2006-Sep-24, 11:57 PM

Grav,

This may get off on a bit of a tangent for this thread here, but it may be as good a place to do it as any. I'm going to try paint a picture of how "gravity works" in GR's geometric view. In Newton, you think of an accleration field g(r), and solve the equations of motion from that. In GR, it's a bit different. You could do it that way, but it would be very complicated. It's complicated enough as it is.

Einstein's Field Equation (EFE) is a tensor differential equation for the "metric", with a source term of the stress-energy tensor (mass-energy and momentum and how it's moving). In the more familiar field picture say of Maxwell, the metric is sort of a like a potential. So the EFE is sort of like EM equations written in terms of the potentials.

But the metric has a more direct meaning, and that's the "length" of paths in space-time. You are familiar I'm sure with the pythagorean distance formula, the lenght of a line. s^2 = x^2 + y^2 + z^2. That is just the dot product of the position vector r dot r (sum of the products of the components).

Now imagine a curve in 2 dimensions, the X-Y plane. What is the length of a curve y(x) from two points x1 to x2? Well, we integrate the differential length element, ds^2 = dy^2 + dx^2. Since we have y as a function of x, we have ds^2 = (1 + (dy/dx)^2) dx^2. So the length of the curve is then

s = integral[ sqrt(1 + (dy/dx)^2) dx] from x1 to x2

If the curve is just a straight line, dy/dx = constant, then s is just a multiple of x2 - x1. But when it's a curve, that is not the case. For example, say y(x) = x^2, a parabola. dy/dx = 2x, and ds = sqrt(1 + 4x^2) dx.

Now, in thinking about this, we imagined a curve in 2D. But the above ds is basically 1D. We have a taken a straight, one dimensional line, and curved it (but we need a flat 2D "embedding space" to imagine that curvature). In general, if we want to curve a n dimensional space, we need an n + 1 dimensional space to do it in. But let's make the abstaction of forgetting about the embedding space, and just think that for a curved line, our length element has some factor which a function of the coordinate x multiplying it.

For a curved 2D surface, we we'll imagine as some z(x, y), the length element is

ds^2 = (1 + (dz/dy)^2)dy^2 + (1 + (dz/dx)^2)dx^2

where dz/dx and dz/dy are the *partials* of z. That is the length element along a path on some plane that has been curved to some z(x, y) surface in

3D space. But, again, we can forget about the embedding space and just see our length element as depending on functions of the (x,y) coordinates.

We can express this in matrix form:

ds^2 = (dx, dy) * | 1 + (dz/dx)^2, 0 | *| dx |

| 0, 1 + (dz/dy)^2 | | dy |

This is the dot product of the dr vector with this matrix sandwiched in between.

That matrix is the "metric". A rank-2 tensor can be represented as a such a matrix (a rank-3 tensor would require a 3D matrix, and a rank-4 would be a 4D array). While a tensor can be represented as 2D matrix, just like a vector can be by a 1D matrix, any matrix is not a tensor. Tensors have various properties and rules that are more restrictive. Just think of the

2D matrix as a way to represent it.

In the traditional parlance of GR, this metric tensor is denoted by 'g' (not to be confused with the gravitational acceleration, but think 'g' is for *gravity* )

I wrote the above in terms of taking 1 and 2D "planes" and warping them in 2 and 3D spaces, and so was bound to cartesian coordinates. The metric idea generalizes to all sorts of coordinate systems, and in rough matrix-like notation we can write in general

ds^2 = dr g dr,

where g is the metric tensor, and dr is the differential coordinate displacement vector in that coordinate system. There is a lot, a *LOT* that can said about this. A lot of math and theorems and rules and principles. Way more than I could even begin to try. What little I do know compared to what is known is pitiful.

But, this 'g', this metric, is what GR revolves around. In GR, our dr's are not simple 1, 2, or even 3 dimensional coordinate "planes", but 4 dimensional space-time (and to curve 4 dimensions, you need actually a 5D embedding space!). And a further complication to our Euclidean intuition is flat space-time is not Euclidean itself. dr in space-time can be 0 or negative.

Anyway, the g in GR gives us the metric for this 4D 'dr' space-time "length". The EFE determines the metric from the stress-energy tensor.

That's a lot of stuff above, and I'll stop this message here, and continue with how this 4D metric acts in a little bit.

-Richard

grav

2006-Sep-25, 12:41 AM

Wow, publius! I appreciate that. I'm sorry that I need such a math lesson, but it is very good of you and patient to post it like that. I think I actually understood much of it. It was in a language I can understand (algebra). I have looked up lessons on the internet and read a whole textbook on matrices, and another (part of one) on calculus, but they seem to always assume the reader already has some knowledge of the terminology involved. And so my brain just clicks off at the first words or symbols I don't recognize, and I don't retain any of it. I don't know why, though. I really need to learn this stuff.

So r dot r is really just the square of the total distance between points? What about v dot r? Also, the way you were talking, I thought time might be the factor that curves the third dimension. But since time is also distorted, I guess this can't be true. So what distorts time? Would other time dimensions be necessary, or is this an effect of the distortion of space-time overall (due to motion, gravity, etc)?

publius

2006-Sep-25, 01:53 AM

Grav,

A dot (or inner) product has a rigorous coordinate indepedent definition that is beyond my scope, but for rough purposes, writing a vector in components of *orthogonal coordinates*, the dot product is the sum of the products of the components. If r is the position vector, the r dot r is the square of the length. A dot B would be (Ax*Bx + Ay*By + Az*Bz), and that can be shown to be the product of the length times the cosine of the angle between the vectors. AB*cos(theta). And that means you can think of A dot B as the projection of A along B (or vice versa). If two vectors are orthogonal, their dot product is zero.

-Richard

publius

2006-Sep-25, 02:09 AM

Grav,

Now imagine a particle moving along that parabola. It's *speed*, which is ds/dt (arc length per time) is constant. Now, what about dx/dt? You can see that is not constant. If the curve were a straight line, dy/dx = constant, ds/dt would just be a constant times dx/dt. But for the curve, dx/dt would vary as the particle moved.

It would, (drum roll.......) *appear to accelerate*. Now, think about what 'x' is. That is the x coordinate we use by defining a parabola of y = x^2 in and x-y coordinate system. But what if we put the origin at another point on the parabola, and defined some other plane at an angle? That same curve would be some other y'(x'), where the primed coordinates are those other coordinates. But it would be the same curve.

Now, in view of this, our x coordinate is just the length along the tangent to that parabola at the origin. The x' coordinate would be the length along the tangent at some other point. These "tangent coordinates" are the "local coordinates" of observers. There length and time are defined by these tangents "planes" to the curved surfaces (manifold is the general term).

So in this language, as our particle moves at constant speed, ds/dt = k,

the speed of the *projection of that* along our tangent line varies. It appears to accelerate.

Now, grav, that is the how the geometric picture of gravity works. This is the key insight. The particle is not accelerating, ds/dt = constant, but it only appears to accelerate to us "stationary observers" watching it. "Real acceleration" is ds/dt not equal to a constant. And that may or may not appear to be an acceleration to us along the tangent.

Now, an exercise: Find the ds/dt along the parabola to make dx/dt be constant. In GR, this would be the similar problem to finding the force required to keep something stationary in a gravitational field.

Now, the above picture is very helpful in understanding how GR gravity works. But don't get too carried away with it, because above I just threw time in sort of arbitrarily, imagining a particle moving along a curve. In GR, time is part of the "manifold" that curves along with space. I'll get that in another message.

-Richard

grav

2006-Sep-25, 02:29 AM

So maybe it's like looking at something travelling around in a circle from the side? Although it appears to travel at a constant speed when looking straight on from the front, from the side it would appear to accelerate upward until it passed the midpoint, then decelerate to a stop, then accelerate downward again. So maybe we're looking at the acceleration of gravity from the side, and to look at it head-on would require an extra dimension? This all seems simple enough, so it must be just the terminology in the mathematics, not the mathematics itself so much, that I don't get (yet).

publius

2006-Sep-25, 03:11 AM

Grav,

Yes, it's like watching the projection, say the shadow of something moving in a circle cast on a wall. But the "extra dimension" is just the "embedding space" concept. By abstracting it in terms of the metric tensor idea, we don't need to worry about the extra dimensions to see it.

With space-time, the extra dimension would get a little weird anyway, because flat space-time is not Euclidean anyway. Space-time is like a 4D space, with time being the 4th dimension(well, the "0th" dimension, usually -- rather than letting the coordinate indices run for 1 to 4, you let it go from 0 to 3, with 0 being the time dimension/coordinate).

Without further ado, here is the "length" in flat space-time (this is the "background" of SR, and SR texts go into great detail on how flat space-time works):

ds^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2)

I've defined this with a "positive time-like signature". The time coordinate is positive, while space is negative, but it is sometimes done the opposite way. The "distance" along the time direction is just ct. That puts everything in terms of length units. You could do it in terms of time, and divide all the space coordinates by c. Or we can write that more compactly as

ds^2 = (c dt)^2 - (dr dot dr), where r is the 3D space position vector. This is "time squared" minus space squared. That is the "length" or norm of 4D space-time. It can be 0 or negative. In fact the zero length is very special. What is the relation between r and t to make the path length zero? Well,.

0 = (c dt)^2 - (dr)^2 --> (dr/dt)^2 = c --> r = ct

The length along the path light travels is always zero. :) This is a "null geodesic". That's not an accident, of course, the whole thing was designed that way because the speed of light is special, the limiting value, and this is how it works out.

Anything travelling at less than light speed will have a positive value of s along it's path. And that positive value of s, divided by c is just the "proper time" of the object travelling that path. This the length of time a clock moving along with the particle would record along the path. These are then called "time like" paths. Anything with rest mass much travel along a time-like path. Light or any other particle with zero rest mass follows the null paths (which sort of means its clock is stopped :) )

You can see it's pretty much impossible to geometrically visualize some space where the length between two different points can be zero or negative, so that's why I caution about trying to visualize it.

The metric, g, of this flat space-time is then just a diagonal matrix with values of 1 down the diagonal: (1, -1, -1, -1) (If we did it with a negative time-like signature, the space parts would be +1 and the time part would -1 ).

So roughly, like in the Euclidean metrics, flat means 1s down the diagonal (for Cartesian coordinates), but we have - there instead of all +.

Now, GR comes along and curves that, making the metric be different from all 1s down the diagonal, and depend on the coordinates, just like with my examples above. But note the time coordinate is included in this.

-Richard

publius

2006-Sep-25, 03:34 AM

Grav,

Now, the big bad EFE is the equation for determining the metric 'g' for space-time in the presence of matter and energy. That "litte equation" is very complex. In tensor notation, which wraps up a lot of complexity in very succinct symbols, it looks simple. However, choose a coordinate system, and try to solve it, and in general you have about 10 coupled partial differential equations to solve. It's a mess. Solving them numerically is no picnic either. IOW, gravity is complicated. :)

But in certain cases, exact solutions can be found. Mr. Schwarzschild found the solution for a non-moving, spherical mass distribution of mass M. If that mass is concentrated at a point, we have a classic Schwarzschild black hole. If not at a point, the metric valid outside the mass (solutions for inside the mass do exist, but all that gets complicated).

So, the Schwarzschild metric is the "vacuum solution", or the metric of the vacuum surrounding a static spherical mass. This is the simplest GR there is, like Netwon for a spherical or point mass. But there's a lot more than Netwon there.

Now, the Schwarzschild metric is done in spherical coordinates not cartesian, because that is the symmetry of the problem. In the above "tutorial", I didn't bother with non-cartesian coordinates because they complicate things. The metric there is not 1s down the diagonal, even though the space they would describe is flat -- this is coordinate curvature, not real curvature, and there is mathematical machinery to tell the difference. But don't worry with that, as it is a needless complication for our purposes.

Remember that business about the local coordinates being along the tangent to the curve? Well our Schwarzschild coordinates here are "tangent at infinity". These are the spherical coordinates of an observer at infinity.

The metric is thus:

ds^2 = (1 - R/r) (c dt)^2 - dr^2/(1 - R/r) + r^2 d(theta)^2

+ (r sin(theta))^2 d(phi)^2

Note the angular coordinate part remains "flat". And we'll forget about that for now, and concern ourselves only with radial paths. Dropping that, we have:

ds^2 = (1 - R/r) (c dt)^2 - dr^2/(1-R/r)

R is the Schwarzchild radius, 2GM/c^2, the "event horizon" of black hole.

In flat space time in this coordinates, there would be 1s there, not the

(1 - R/r) term. This factor comes from the curvature. And note the special symmetry. This factor multiplies the time coordinate and divides the radial distance coordinate.

This the metric, or "gravity" for the space-time around a spherical mass.

-Richard

publius

2006-Sep-25, 04:55 AM

Grav,

Now that is the Schwarzschild metric, describing space-time around a spherically symmetric or point mass. What can we tell from that? We can tell just about everything we want to know, but that gets complicated, but we can do some simple things. Remember how light follows null geodesics. Well it follows null geodesics in curved space time as well as flat. The radial equation of motion for light (remember we are holding the angular coordinates constant to keep it simple) fall right out by setting ds = 0:

0 = (1 - R/r) (c dt)^2 - dr^2/(1 - R/r), rearrange and solve for dr/dt

dr/dt = v(r) = +/- c*(1 - R/r); + being travelling radially out, - for "falling" radially inward. These are the equations of the null radial geodesics.

This gives us the velocity, as measured in our distant stationary spherical coordinates of a beam of light in the gravitational field. Note it "slows down", and comes to a stop at r = R, the event horizon. Ain't that interesting?

Now, remember that ds for "time like" paths is just the proper time along that path? Same holds for curved space-time here. Consider an observer stationary at some point r down in the gravity well. r is constant, and therefore dr = 0. ds is then

ds^2 = (1 - R/r) ( c dt)^2

ds/c = dT = sqrt(1 - R/r) dt.

That is the proper time, dT, of that stationary observer. His clock is slowed by a factor of sqrt(1 - R/r) relative to ours. This is gravitational time dilation. And note that for r >> R, his clock his close to ours at infinity. Thus we conclude that these Schwarzschild coordinates are close to those of any observer far enough away from the source of gravity.

Now, let's do just the opposite. Hold time constant, and consider a "path" length of some dr.

ds^2 = - dr^2/(1 - R/r).

Negative lengths are "space-like". These are just distances between points at constant time. So we throw away the negative sign and say that is the "proper length" at the point r. I'll call that proper (radial) lenght X.

dX = dr/sqrt(1 - R/r), or dr = sqrt(1 - R/r)*dX

Yardsticks "shrink" in the gravitational field. Or "space expands", however you think about it. In our frame, a small distance dr is longer down in the well. In fact, as we get close to the horizon, dX becomes infinite. The "last inch" to the horizon gets longer and longer the closer you get (for stationary observers, remember, not free-fallers).

dX is the "proper distance" (radial) for a stationary observer at some point r. The "proper speed" is the ratio of these, dX/dT.

dX/dT = (dr/dt)/(1 - R/r). We already know the coordinate speed of radial light paths is just c (1 - R/r). So the local speed, the proper speed just cancels out that factor, and the local speed of light dX/dT = c.

The slower clocks and shorter rulers just cancel out and he measures the speed of light to be c. In fact, the local speed of all null paths in any metric must always be c.

-Richard

publius

2006-Sep-25, 06:22 AM

Grav,

Now, we've looked at null geodesics and found the proper time and proper (radial) distance in the Schwarzschild gravity well. Now, the next thing is to find the non-null, time-like geodesics, which are the inertial paths that free-falling particles would follow. These are the paths that straight time-like paths are curved into.

There is a general procedure to get these, but it involves some rather complex mathematical principles. There are some cute tricks to get them here, but I'm not exactly sastified with them as they are not general. So, I'll just say "it can be shown" that the radial time like geodesics for a free fall "dropped" with zero initial velocity from infinity is this:

dr/dt = v(r) = c(1 - R/r) * sqrt(R/r)

Now, let's look at the local speed of that free-fall, measured by stationary observers as the particle flies by them. We just divide by the same (1 - R/r) factor and we have:

dX/dT = c * sqrt(R/r) = c sqrt( 2GM/rc^2) = sqrt( 2GM/r)

This is just exactly the Newtonian result, from 1/2mv^2 = GM/r, kinetic energy = potential energy.

Amazingly, in the Schwarzschild metric, the local radial free fall speed, which is the escape velocity measured locally is the exact same thing as Newtonian gravity. But that is local. We do not see that in our far away frame when r gets close to R. We see the (1 - R/r) mulitplying it.

Another fundamental result is what is the force required to keep an observer sitting stationary in the gravity well. This is just the "real" local acceleration of path r = constant. That is bit complex to get, but it turns out to be simply:

GM/r^2 * 1/sqrt(1 - R/r)

It's the Netwonian acceleration multiplied by the metric factor. Note this goes to infinity at r = R, not r = 0. It's impossible to remain stationary at the horizon, and close to it, the force gets a might bit steep.

You will also note the metric "blows up" at r = R. :) That is actually a fault of the coordinates, not the metric itself. But a discussion of that gets very involved. It does take infinite time in our coordinates for something to cross the horizon. But not in the proper time of the free-faller.

You will note that for r < R, the time-part becomes negative, while the r part becomes positive. What that "means" is time and space flip relative to our coordinates inside the horizon. The curvature changes time into space and space into time. :) Well, that's one way to say it.

-Richard

publius

2006-Sep-25, 06:47 AM

Grav,

Well, I really went on about that and I apologize to Ken because I got off on this tangent and didn't get back to the GEM Poynting. The force must've petered out on me. Maybe it'll get back.

At any rate, Grav, I wanted to show you sort of how "GR works", how it looks at things in the simplest case. You see it's all about the "metric", and how one gets geodesics which describe the "free fall" paths, the inertial paths that would be straight lines in flat space, and how they are curved. It's a bit different from the traditional "F = ma" approach that one learns in physics 101, of course, where one gets an equation of motion from the acceleration caused by the forces. You could do it in a similiar way in GR, but that get's complicated. Very complicated. The geodesics from the metric are the simplest, and that is a big enough mess in the general case where you have non-diagonal elements in the metric. BTW, "frame dragging" or gravitomagnetic effects would show up as off-diagonal elements in the time-like part of the metric, the g_0i and g_i0 components.

The g_ij for i and j from 1 to 3 are the space-parts of the metric. Note the g_00 part is roughly the Newtonian potential, with an offset. The "frame dragging" parts look sort of like a vector potential.

The GEM approximation comes from these time-like parts, actually, and ignored the spatial part of the metric. The first row of the metric tensor is then like a 4-potential, analogous to the EM case.

-Richard

grav

2006-Sep-25, 10:58 PM

Thanks, publius, for taking the time to "tutor" me in the mathematics of GR. It feels mighty strange to have it laid out this way and aimed directly at me in a thread like that, but hopefully others can learn from it, too. You have probably just quadrupled my understanding of the mathematics involved, although it is surely still miniscule compared to all that is known. I still don't know how to apply anything useful to what you have shown yet, but I have printed off what you wrote and will use it as a reference whenever I am confronted with equations such as these. I probably won't be so quick to turn away anymore. Although I will still dedicate my time to trying to incorporate GR into the "conventional" model of the universe with "real", measurable mechanisms, and without the distortion of space and time, which I deem unnecessary, but only as a supplement, even though I have yet to come up with a better one that is complete, and would probably take much time and effort, if I succeed at all, you have tried to convey to me the concepts behind the mathematics of GR, which would probably be next to impossible to incorporate without such knowledge to begin with, and so I will try to keep an open mind about it all as well. The way I see it, if the mathematics work, it works, so I am really just attempting to come up with another way of thinking about it, which is proving difficult so far. I still have a couple of ideas of how this can be done, but it would be a difficult thing indeed without knowing the mathematics of what I am attempting to incorporate in the first place. So much thanks again. :D

publius

2006-Sep-26, 03:52 AM

OK, press on, and may the force be with you.

Ken,

I was hoping to feel the force, but to no avail, then realized since this is gravity there is no force.

Let's jump in the EM Poynting vector here:

Flux(E x H) = -d(field-energy-in-volume)/dt - integral(J dot E)over volume

J dot E is the EM to mechanical power. If it's negative, then J is "bucking" E, just like current bucking voltage, and we have a general "generator" device, something putting energy into the EM fields. And the above says that work goes into increasing the energy density or Poynting flux out of the volume enclosing it.

Now, switch to GEM. -J dot g would be the rate of mechanical work (per volume) done against gravity as well. For example, you're steadily lifting a weight against gravity, you're doing work at a rate of mg*dh/dt = (mv)*g. (This shows mass current is the same as momentum density). ETA:{Note to lurkers: this may confusing. What is meant is mass current density, mass current per unit area is equivalent to volume momentum density, momentum per volume}

Now, we can leave the above field volume energy the same and not worry with putting the minus sign on the permeabilities. The field energy is the field energy. But I'm going to switch the sense of the mechanical work and say doing work against the field is positive.

Flux(g x h) = -d(field energy)/dt + d(work)/dt

Now, we'll conver the h to -B_g/u, putting the minus sign in:

Flux( -g x B_g/u) + d(field energy)/dt = d(work)/dt

What this means is doing work on the gravitional system results in the field energy increasing (which means g^2 + B^2 decrease in the region), or goes to GEM Poynting. If there is net GEM Poynting Flux, then g and B_g would not have to decrease around as much as they otherwise would.

But the GEM Poytning points bassackwards, so inward flux means energy leaving the volume. As we were doing work, if there was no GEM Poynting flux, g^2 + B_g^2 would be decreasing, as they were taking in positive energy -- ie lifting the weight is separating mass, lowering the total g field. But if there was GEM Poynting flux, the two would not have to decrease. But since GEM points backwards, that would look like EM Poynting flux coming in. And that would make sense because EM Poynting coming in would raise the energy density (make E^2 + B^2 increase, or go to mechanical work positive J dot E). So this does look like transmitter and reciever being reversed.

The only way to see how the fields would be behaving in detail would be to solve the GEM equations for some such situation. EM solutions could be used if we could find one that made sense for a meaningful GEM setup, and we would have to *carefully* handle the minus signs in those solutions.

-Richard

grav

2006-Sep-27, 03:28 AM

ds^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2)

I've defined this with a "positive time-like signature". The time coordinate is positive, while space is negative, but it is sometimes done the opposite way. The "distance" along the time direction is just ct. That puts everything in terms of length units. You could do it in terms of time, and divide all the space coordinates by c. Or we can write that more compactly as

ds^2 = (c dt)^2 - (dr dot dr), where r is the 3D space position vector. This is "time squared" minus space squared. That is the "length" or norm of 4D space-time. It can be 0 or negative. In fact the zero length is very special. What is the relation between r and t to make the path length zero? Well,.

0 = (c dt)^2 - (dr)^2 --> (dr/dt)^2 = c --> r = ct

The length along the path light travels is always zero. This is a "null geodesic". That's not an accident, of course, the whole thing was designed that way because the speed of light is special, the limiting value, and this is how it works out.

Anything travelling at less than light speed will have a positive value of s along it's path. And that positive value of s, divided by c is just the "proper time" of the object travelling that path. This the length of time a clock moving along with the particle would record along the path. These are then called "time like" paths. Anything with rest mass much travel along a time-like path. Light or any other particle with zero rest mass follows the null paths (which sort of means its clock is stopped )

You can see it's pretty much impossible to geometrically visualize some space where the length between two different points can be zero or negative, so that's why I caution about trying to visualize it.

You say it is pretty much impossible to visualize zero or negative length, but that seems to me to be the easy part. Zero length would just be a point (or null). A negative length, although impossible in reality, merely represents a negative direction, or opposite direction for motion. But the formulas you show here wouldn't define a negative length anyway. With spatial dimensions greater than ct, it would be complex. Geometrically, that would mean that the points lie outside of the parameters of our diagram. In GR, I guess that would mean outside of space-time itself.

I have always had trouble visualizing a fourth dimension, but with these formulas, I now see how it is possible. It is laid out so that ct is the hypotenuse of the spatial dimensions. This makes s the fourth spatial dimension, so that (ct)^2=x^2+y^2+z^2+s^2. And regardless of the number of dimensions s is also composed of, so that we may really have many extra dimensions which are all related by the sum of their squares, the resultant of all of those dimensions is always ct. It just keeps add ing more perpendiculars, but the resultant stays the same.

We could visualize these dimensions on a graph, with space on one side and time on the bottom. With motion, each dimension could begin at their perspective distances from an origin, and the motion of all dimensions could be followed simultaneously through time. We could also visualize our three dimensions plus one for time in a similar fashion to an engineer's 3D blueprints, where different faces are looked at, from top, front, and side, one for each dimension. For space-time, each dimension related with time would represent a time cone. An expanding sphere, for instance, beginning at some particular radius, would look like a truncated cone. Since all three spatial dimensions are the same for a sphere, this is all that is necessary, and would be its geometry in space-time. For an accelerating expanding sphere, it would look like a hat. :)

grav

2006-Sep-27, 03:32 AM

Grav,

A dot (or inner) product has a rigorous coordinate indepedent definition that is beyond my scope, but for rough purposes, writing a vector in components of *orthogonal coordinates*, the dot product is the sum of the products of the components. If r is the position vector, the r dot r is the square of the length. A dot B would be (Ax*Bx + Ay*By + Az*Bz), and that can be shown to be the product of the length times the cosine of the angle between the vectors. AB*cos(theta). And that means you can think of A dot B as the projection of A along B (or vice versa). If two vectors are orthogonal, their dot product is zero.

-Richard

So would v dot r would be vxx+vyy+vzz?

Also, is dot always spelled out this way or is there an actual dot involved?

grav

2006-Sep-27, 04:12 AM

I just realized that (ct)^2=x^2+y^2+z^2+s^2 with ct as the resultant of all spatial dimensions, means that time would not be the fourth dimension, or the zeroeth, or any dimension in itself from which space-time is composed. Instead, coupled with c, it would become the ultimate dimension. It would become the length of space-time itself.

publius

2006-Sep-27, 04:22 AM

So would v dot r would be vxx+vyy+vzz?

Also, is dot always spelled out this way or is there an actual dot involved?

Yes, that would be v dot r in cartesian coordinates. The dot is an actual dot. :) I just write it out because I'm limited (or limiting myself) to text.

-Richard

publius

2006-Sep-27, 04:30 AM

I just realized that (ct)^2=x^2+y^2+z^2+s^2 with ct as the resultant of all spatial dimensions, means that time would not be the fourth dimension, or the zeroeth, or any dimension in itself from which space-time is composed. Instead, coupled with c, it would become the ultimate dimension. It would become the length of space-time itself.

Grav,

Whooooaaaaa there pardner; hold your horses. Now, what did you do? You transposed the spatial part and see a sum of four squares with positive signs in front of them. But remember, ds^2 can be negative. :) How can the square of something "normal" be negative? Suppose space^2 came out to be 25, and (ct)^2 was 16. ds^2 would be -9. That would be a space-like "distance", outside of the light cone. And 25 + (-9) = 16. The "hypotenuese" is shorter than the one of the legs.

By "length", I mean norm, the *magnitude*. By definition, that is positive definite for "normal" Euclidean vector spaces. That is the sum of a squares and so is positive definite. Here, the hypotenuese is always longer than than any of the legs.

But in (non-Euclidean) space-time we define that norm, that magnitude as a difference of squares.

-Richard

publius

2006-Sep-27, 04:52 AM

Grav,

Let me try to paint a picture of space-time for you, and why the norm is defined as it is.

Conside a normal Euclidean space, a 2D plane. Distances between points are the same, no matter what the coordinates are. We might have an XY plane defined with the origin at a given point, and another coordinate system, X'Y', with it's origin at some other point, and with the axes rotated with respect to the first. The coordinates of a point will be different. But they will agree on the distance between points. The norm is therefore the same, or *invariant* between coordinate systems.

In space-time, it is 's' that is invariant between frames. A "point" in space-time is usually called an "event". It occurs at certain point in space at a certain point in time. We can write E = (ct, r), where r is the 3D spatial position vector. Another coordinate system, will see that event occuring at possibly a different time and a different point in space. But they agree on the 'norm', s between two such events.

The way SR works, this invariant quantity has to be defined as the difference between the time squared and space squared, not the sum.

Consider a particle moving at constant velocity 'v' in our coordinates. It starts at the origin at t = 0. At any time t, that event is then (ct, vt). The particle is at position vt (say it is moving along the x axis just to keep this simple). The "norm" between those points is just (ct)^2 - (vt)^2.

Now, in the frame of the particle, it is at rest. So in that frame, those two events occur at the same point in space. x' = 0. But the norm is invariant. So how much time passed in that frame. That is the t' coordinate. In that frame, ds^2 = (ct')^2. But these must be equal, so:

(ct')^2 = (ct)^2 - (vt)^2 --> t' = sqrt(1 - (v/c)^2) * t

His clock is running slow. :) That is just the gamma factor. You see, space-time was defined so the above relation would hold -- well, that is itself a consequence actually, a consequence of the speed of light being invariant. The space-time norm was defined so this would work out.

-Richard

grav

2006-Sep-27, 04:52 AM

Okay. I was thinking that our universe expanded with time, so that ct is that rate of expansion. If {s} were positive, we shouldn't be able to see past fifteen billion light years or so, if that is how long our universe had been in existence. But now that I think of it, I don't see how s^2 can be anything but negative (or zero). If it were positive, it would be the same as our other three dimensions, and we would know about it. So it can only be the result of a complex component, which means it would lie outside of our universe (or space). Therefore, the formula can really only be stated one way, and that is s^2=x^2+y^2+z^2-(ct)^2, since ct is always greater than the total distance that can be traversed in t (c>v). Does that sound about right?

publius

2006-Sep-27, 05:42 AM

Grav,

Again, I have to say whooaa pardner. We're talking about SR, flat space-time, not an expanding universe. An expanding universe gets into some heavy GR territory where the metric is dynamic, that is changing with time itself (whose time? Everyone's time, but in different ways for different observers). You're jumping way ahead and beyond the simple thing we're talking here. That's why I say whooaa pardner. We can't start running, but first we need to crawl, then walk, then finally run.

I prefer a positive time-like norm. That is (ct)^ - dr^2, but many use the opposite. Yes s must be positive along the trajectory of any particle with rest mass, but speaking of general points (events), s can be anything. When s is negative, then that distance/norm/interval (all terms used to describe s) is said to be "space-like", which means no causal relation could exist between those events, as it would require something faster than light. And 's' along the trajectory of any particle must be time-like. Except for light and other massless particles, where it is 0, or "light-like".

For example, sitting on earth, "now" on the moon is space-like. But 2 seconds in the future on the moon is time-like. So time-like = positive, space-like = negative, and light-like = 0.

-Richard

publius

2006-Sep-27, 06:11 AM

Grav,

I just realized you're trying to think of s as a dimension. It is not -- it's the "norm"/"length"/"separation" between events. Time and space are the dimensions, s is the norm.

And s/c is the proper time in the rest frame of a particle following that trajectory. In that frame (even accelerated ones), the distance is zero, and since s is invariant, that must (ct)^2, where t is the proper time. s/c is the time a clock moving along that trajectory would measure.

It's a bit more tricky, but when ds^2 is negative, that means ds can be intrepreted as a distance in an inertial frame where those two events are simultaneous.

-Richard

hhEb09'1

2006-Sep-27, 03:18 PM

But now that I think of it, I don't see how s^2 can be anything but negative (or zero). s2 is a short hand for ds2, the "distance" between two points in spacetime--or, another way of thinking about it, it is the distance from a point to the point (0,0,0,0). Either way, you can have positive (space-like) or negative (time-like) distances.

grav

2006-Sep-28, 12:50 AM

s2 is a short hand for ds2, the "distance" between two points in spacetime--or, another way of thinking about it, it is the distance from a point to the point (0,0,0,0). Either way, you can have positive (space-like) or negative (time-like) distances.

Yes. I figure x to be the distance between the x coordinates of two points, and so on. If we add the squares of these distances for all three dimensions, we get the total distance (in Euclidean space) between the two points. This would definitely be better expressed in calculus, where dx would be the distance between x1 and x2, but apparently I still have some learning to do in calculus, since I would have thought the square of the distance, dx, to be d2x, not dx2. dx2 looks to me like x1^2-x2^2, not (x1-x2)^2. So where I post in this thread, the x would actually be dx unless referred to as actual coordinates, and we can go from there.

<<<<<>>>>>

Okay, publius. Since I am just barely learning this stuff, I will slow down. But I still think my initial observation about that particular formula is correct. Let's look at it again. If it is s^2=(ct)^2-x^2-y^2-z^2, and if x^2+y^2+z^2=r^2, then we have s^2=(ct)^2-r^2. Let's divide both sides by t^2 and observe motion. Now we have (s/t)^2=c^2-(r/t)^2. A distance {r} travelled in time {t} is the velocity that we observe, so now we have v'^2=c^2-v^2, where v'=s/t. Maybe this way you can see what I mean. If c is always greater than the velocity we observe {v}, then c^2-v^2 is always positive. That means s^2 is always positive as well, and s can be positive or negative. But this would make it exactly the same as our three real spatial dimensions, where (ct)^2=r^2+s^2, so r and s can be interchanged since they become indistinguishable. {s} would be just as much a dimension as the other three. But this is obviously not possible because it is only three that we actually observe. That makes the formula incorrect.

So let's try it the other way around, where s^2=r^2-(ct)^2. In this case s^2 is always negative (or zero). Of course, we can't have the square root of a negative number for the value of {s} unless its value is complex. If its value is complex, then according to geometry, this would mean that {s} lies outside of the parameters of our space. It would contain one or more extra dimensions that lie outside of the observable universe. And that makes sense, at least to me, since that is exactly what any extra dimensions would have to do. The formula can also be written as (ct)^2=r^2-s^2. Since s^2 is the square of a complex number, it is negative. So if we subtract it in the formula, then we are really adding its absolute value. Since c is always greater than v, then (ct)^2 is always greater than r^2, and since s^2=r^2-(ct)^2, s^2 is always less than (ct)^2 as well, but negative (the absolute value is always less than (ct)^2). This means the total value of the sum of all vector velocities for all dimensions, real or complex, can never be greater than c. It also means that since c is a constant speed in this formula, for which all vector velocities are related, then their vector sums can never be less than c as well. c, then is a constant speed for space-time, to which all dimensions add up, and ct is always the total distance travelled through space-time over time {t}, regardless of the distance that is actually travelled through "real" space. And time, then, on its own, would really not be another dimension, but ct would be the ultimate space-time length from which all other dimensions, real or not, are composed.

hhEb09'1

2006-Sep-28, 04:07 PM

That means s^2 is always positive as well, and s can be positive or negative. That is true in your example, where you are assuming that something moved from the first point to the second point at a velocity v less than c. But s2 does not have to be positive, in general. Take any two arbitrary points of spacetime, then s2 can be negative.

The negative sign indicates that even a light beam could not have arrived at the second point from the first point. That indicates a time-like distance, rather than a space-like distance.

grav

2006-Sep-28, 08:59 PM

That is true in your example, where you are assuming that something moved from the first point to the second point at a velocity v less than c. But s2 does not have to be positive, in general. Take any two arbitrary points of spacetime, then s2 can be negative.

The negative sign indicates that even a light beam could not have arrived at the second point from the first point. That indicates a time-like distance, rather than a space-like distance.

Oh, sorry, hhEb09'1. That was meant to show that s^2 could not be positive. If it were, then s would become a spatial dimension that would be indistinguishable from our normal three. I guess I should have said, "If that were the case, then s^2 would be positive", but that cannot be the case as far as I can tell. So s^2 must always be negative (or zero) in order for it to lie outside of our other three dimensions, making s complex. But since its square is then subtracted in the formula, then we are really adding a double negative, where the subtraction of its square becomes positive, but otherwise it is complex, and lies outside of our three dimensional universe. Then ct is greater than r or s, and ct becomes the resultant of all the dimensions, so that the path that light travels would be the "null geosedic" of all of space-time. I am still having trouble imagining what this is relative to, however, since space-time is not meant to be absolute. If it is relative to an object "at rest", then that object, although at rest, would still have to be travelling at the speed of light through the extra dimensions, in which case the object, or any object, must always travel at the speed of light through space-time itself.

grav

2006-Sep-28, 09:09 PM

Publius,

A question. Light must always travel at c when travelling away from us, right? It must also travel at c when travelling toward us, right? Well, if you and I were on two separate light beams travelling parallel with each other, what would see or look like to each other if we were to "ride the waves"?

publius

2006-Sep-28, 10:14 PM

Publius,

A question. Light must always travel at c when travelling away from us, right? It must also travel at c when travelling toward us, right? Well, if you and I were on two separate light beams travelling parallel with each other, what would see or look like to each other if we were to "ride the waves"?

Grav,

Legend has it that this very thought experiment, riding a beam of light, is how Einstein developed SR. :) The answer is, we can't ride a beam of light, because we have rest mass. However, we can imagine moving very close to the speed of light relative to some observer we leave behind. You and I would look like we always do relative to each other. However that observer we left behind would look very different, very contracted and his clock running very slow.

Consider what riding a beam of light would like in the classical wave picture. We would see standing, static, spatial waves of E and B. An electric field can't do that in a source free vacuum, according to Maxwell's equations. We can't be in a frame where we'd see such a thing.

[Note: with an (a)ether, this would be possible, just as we would see a standing pressure static pressure wave in a frame moving at the speed of sound -- rather than relative velocities, B and E and the forces on charges would depend on ether-relative velocities, and partial time derivaties would become ether-velocity relative directional derivatives]

-Richard

RussT

2006-Sep-29, 12:43 AM

So, if the 'aether' was made up of physical Planck size/mass, Neutrino-like inert particles, traveling at 'C', then what?

publius

2006-Sep-29, 01:01 AM

So, if the 'aether' was made up of physical Planck size/mass, Neutrino-like inert particles, traveling at 'C', then what?

Russ,

I don't have a clue. :)

-Richard

RussT

2006-Sep-29, 01:32 AM

Russ,

I don't have a clue. :)

-Richard

I know, it's pretty weird Huh?

But this does go back to my DM/DE, "could they possibly be the same thing?" question and earlier in this thread where the discussion of 'gravity making more gravity' and 'space making more space' took place. and now I see where Ken G has said...

http://www.bautforum.com/showpost.php?p=833699&postcount=59

Quote:Ken G

[Photons generate "gravitons" (in the small-field limit of GR) all the time, by virtue of their energy. You don't need rest mass, the relativistic mass m=E/c^2 works fine for generating gravity.]

So, with gravity fields of baryonic massive bodies (I thought that they were gravity wells, not necessarily 'fields') 'creating' 'more' gravity, etc,etc as Tensor suggested, and with DE 'creating' 'more' gravity, and now photons even 'creating' (albeit very little) 'more' gravity, do you think that our picture of gravity could be OFF?

grav

2006-Sep-29, 02:45 AM

publius,

I'm still struggling to get past you're post number 79. I can see the reasoning for everything that came before it. I can see where s in s^2=x^2+y^2+z^2 would be the "norm", or the total distance in flat space geometry. But then how that transforms into s^2=(ct)^2-(x^2+y^2+z^2) or s^2=(x^2+y^2+z^2)-(ct)^2, I cannot tell. They seem to be the same variable for completely different things. For the flat space geometry, s is the norm. I can see that. But in the last formula, ct becomes the norm if I'm looking at it right. This makes s a dimension, and it can only be an extra dimension, because the other three are already accounted for.

So I can't figure out how to incorporate post 79 in with the posts that come before. For one thing the flat dimensions suddenly become negative, but work the other way around if s is considered an extra dimension. Also, it seems that the speed of light and time are now incorporated, but with an arbitrary distance. But if we write the formula as s^2=(ct)^2-(vt)^2, then this makes sense in a way, because now we are comparing velocities, and if v were greater than c, s would be complex, which ordinarily shows its impossibility. This would mean that v can never be greater than c, which is true in relativity. But then, what does that make s? If expressed this way, then s becomes real, but it is not the norm and it is not an extra dimension, nor could it be a real dimension either.

If I ignore this little mind squeeze for a moment, however, I can then follow its lead and see that s^2=[1-(v/c)^2](ct)^2, so s=(ct)*[1-(v/c)^2]^1/2, the latter half of which is fundamental in Einstein's formulas. So one can probably backtrack from the formulas to get this result, but I still don't see how it connects with flat space geometry. The only things we could really say about distance, time, velocity and the speed of light in this way (as far as I can think of), which are all present here, is how far a body will actually travel (through Euclidean space) at what "should" be {v}, relative to another, due to relavistic effects. Is that what this formula is meant to express? Although a body could potentially travel at up to what "should" be infinity (for it, according to its previous acceleration away from a point of rest), it would never travel past c relative to anything else. Maybe we can work with this. What would be the formula be for that?

Tensor

2006-Sep-29, 02:48 AM

So, with gravity fields of baryonic massive bodies (I thought that they were gravity wells, not necessarily 'fields') 'creating' 'more' gravity, etc,etc as Tensor suggested, and with DE 'creating' 'more' gravity, and now photons even 'creating' (albeit very little) 'more' gravity, do you think that our picture of gravity could be OFF?

No, becuase, except for what DM is, GR predicts all of that. Besides, there are other legitimate reasons for suspecting there are problems with GR.

publius

2006-Sep-29, 03:45 AM

Grav,

You're thinking Euclidean, which is the reason I warned you not to try to visualize space-time. :)

Think of s^2 as a *definition*. (ct)^2 - r^2, where r is the 3-space, Euclidean norm. And note I'm using the positive time-like convention, subtracting space from time. Some prefer the other way, and it will work either way.

Now, the reason this is defined that way, the difference between time and space is because that quantity is what is invariant between reference frames, indepedent of the coordinates. Again, consider a 3 space example. I have coordinate system here with my x axis pointed north. You have a coordinate system on the moon with your x axis pointed somewhere else. Our coordiantes for points in space will be different, but we will calculate the same distance between them. That is the norm.

In space-time, this thing that is invariant, the the thing that all observers calculate the same is this difference between the time and space, (ct)^2 - r^2.

And (for time-like "distances"), s/c turns out to be the proper time experienced in a frame moving between those events.

I see a particle moving at velocity 'v'. I start my clock when that particle is at some point and call that the origin. (we'll consider only one space dimension) That event is the (0, 0) in my coordinate system.

Now, some time later, that particle is at some other point, which occurs at some t and some point x. s^2 is then (ct)^2 - x^2. But since the particle is moving at constant velocity, x is just vt. And that's where I got this: s^2 = (ct)^2 - (vt)^2.

Now, in a reference frame moving with the particle, it didn't move. It that frame those events occur at the same point in space, which is the origin of his coordinate system. But they occur at different times. In that frame,

s^2 = (cT)^2 - 0^2 where T is the time on his clock.

These must be the same, so (cT)^2 = (ct)^2 - (vt)^2, and we get the gamma factor.

-Richard

RussT

2006-Sep-29, 07:50 AM

No, becuase, except for what DM is, GR predicts all of that. Besides, there are other legitimate reasons for suspecting there are problems with GR.

[No, because, except for what DM is, GR predicts all of that.]

With all due respect, Don't you mean that FWR universe(s), expanding from a point predict all of that...using GR?

[Besides, there are other legitimate reasons for suspecting there are problems with GR]

Do you mean singularities not 'appearing' to be physical; and The Unification of GR and QFT?

grav

2006-Sep-29, 05:46 PM

Grav,

You're thinking Euclidean, which is the reason I warned you not to try to visualize space-time.

Think of s^2 as a *definition*. (ct)^2 - r^2, where r is the 3-space, Euclidean norm. And note I'm using the positive time-like convention, subtracting space from time. Some prefer the other way, and it will work either way.

It's starting to look like I might not get very far in learning GR, much less incorporating it into anything, if I can't even get past that first space-time formula you posted. This is probably aggravating you, but maybe you're used to it from me by now. You know how I love to tear apart these formulas and study their "ingredients". :) How else could I really learn? But you have freed by brain a little with this last post. If I just think of {s} as a "definition", other than try to apply it to what I already know within Euclidean geometry, then we can move on. It would be the same as any other definition we might apply to something new. Look at energy, for example. It is m*v*v/2. What is that!? But it works in the formulas we use every day. Just as {s} would work for GR. The only problem I still have now, then, is determining how s^2 will work either way. The results are completely different. One way would give the opposite sign of the other, and +s^2 can be positive or negative {s} while -s^2 gives complex values. Very different indeed. So I don't see how it could work either way. It could only be one or the other. Regardless of the actual values used in the formula, one way would give a real result and the other is complex (except for when s=0, in which case they are indeed the same).

grav

2006-Sep-29, 05:54 PM

These must be the same, so (cT)^2 = (ct)^2 - (vt)^2, and we get the gamma factor.

So, if we break this down, we get T=t*[1-(v/c)^2]^1/2. Is this correct? It looks right. T would be the amount of time a "stationary" observer would see pass for the traveller, right?

grav

2006-Sep-30, 02:42 AM

Okay, publius. Help me out here. I just saw in another recent thread (http://www.bautforum.com/showthread.php?t=47342) (pg 2) that last formula written as t=T*[1-(v/c)^2]^1/2. So now I'm confused (again). But I think by the definitions you gave for t and T that yours is correct.

publius

2006-Sep-30, 04:23 AM

Okay, publius. Help me out here. I just saw in another recent thread (http://www.bautforum.com/showthread.php?t=47342) (pg 2) that last formula written as t=T*[1-(v/c)^2]^1/2. So now I'm confused (again). But I think by the definitions you gave for t and T that yours is correct.

Grav,

Both are correct..........................:whistle:

It has to do with simultaneity, which varies between frames. That moving observer could claim he was stationary, write X = vT, and come up with the same formula with t and T reversed, as above.

His clock ticking T = t/gamma is simultaneous with my clock ticking t *in my frame*. In *his* frame my clock ticking T/gamma is simultaneous with his clock ticking T. Say gamma is 2. When my clock ticks 2 hours, his clock has just ticked 1 hour in my frame. But in his frame, when his clock has ticked 1 hour, my clock has only ticked off 30 minutes.

To see how this works, and how it is indeed mathematically consistent, although it will drive you nuts getting your head wrapped around it at first, we need to look at the full Lorentz transforms. The relativity of simultaneity is something that will trip you up in subtle ways.

-Richard

hhEb09'1

2006-Sep-30, 04:54 AM

This makes s a dimension, and it can only be an extra dimension, because the other three are already accounted for. Usually it is "t" that is considered the extra dimension, with a twist. It is a time-like dimension, and the others are space-like. That's why when s2 turns out to be negative, it is called time-like.

grav

2006-Sep-30, 02:05 PM

Yes. I think I can see all of that. Ohhhh, this is going to be a loooonnngg thread. Especially if we are to eventually usher things back on topic. :rolleyes:

Tensor

2006-Sep-30, 07:44 PM

[No, because, except for what DM is, GR predicts all of that.]

With all due respect, Don't you mean that FWR universe(s), expanding from a point predict all of that...using GR?

Nope, you better go through it again.

[Besides, there are other legitimate reasons for suspecting there are problems with GR]

Do you mean singularities not 'appearing' to be physical; and The Unification of GR and QFT?

Not sure what you mean by this. Just the fact that at a distance smaller than the Planck Length, the equations fail. This indicates there are problems with GR at that distance. We suspect that the GR/QM unification will solve this, but we don't know.

hhEb09'1

2006-Oct-01, 03:52 AM

Oh, sorry, hhEb09'1. That was meant to show that s^2 could not be positive. If it were, then s would become a spatial dimension that would be indistinguishable from our normal three. I guess I should have said, "If that were the case, then s^2 would be positive", but that cannot be the case as far as I can tell. Sorry I missed this post, but of course it can be positive. I have no idea why you are saying it cannot be positive. There are all sorts of examples where it is positive, I think some of them appeared earlier. What in the world are you thinking of here? :)

publius

2006-Oct-01, 03:55 AM

Sorry I missed this post, but of course it can be positive. I have no idea why you are saying it cannot be positive. There are all sorts of examples where it is positive, I think some of them appeared earlier. What in the world are you thinking of here? :)

I started this out with a positive time-like convention, and you are using a positive space-like convention, so we may be confusing grav on the meaning of positive vs negative. I prefer positive time-like, but that's just me.

-Richard

hhEb09'1

2006-Oct-01, 04:30 AM

I started this out with a positive time-like convention, and you are using a positive space-like convention, so we may be confusing grav on the meaning of positive vs negative. I prefer positive time-like, but that's just me.No, I know it's not just you, but grav is not saying it should be negative when it is positive, or vice versa. He's saying it's one and not the other.

Besides, my first post was in response to grav's post (http://www.bautforum.com/showthread.php?p=833547#post833547) where he clearly writes "s^2=x^2+y^2+z^2-(ct)^2" so that confusion occurred long before I got involved. Going back to your post (http://www.bautforum.com/showthread.php?p=831994#post831994) where you define ds^2, I notice something odd. You mention "The 'distance' along the time direction is just ct" but you don't make a similar remark about space. Why not--what is the "distance" along a space direction? :)

publius

2006-Oct-01, 05:08 AM

You mention "The 'distance' along the time direction is just ct" but you don't make a similar remark about space. Why not--what is the "distance" along a space direction? :)

What I meant was actually converting units of time to units of spatial distance (well, I should dimension of time to dimension of distance, but that's a dimension of a slightly differenc color). And I also mentioned we could do it in time units by dividing all the spatial coordinates by c.

-Richard

hhEb09'1

2006-Oct-01, 05:13 AM

What I meant was actually converting units of time to units of spatial distance (well, I should dimension of time to dimension of distance, but that's a dimension of a slightly differenc color). And I also mentioned we could do it in time units by dividing all the spatial coordinates by c.

Yes, but what is the "distance" along a space direction, using your convention? :)

hhEb09'1

2006-Oct-01, 05:25 AM

So really, when we speak of the "gravity making gravity" part of this, it's best to not to think of the gravitational field having energy, I don't think. I've read that any attempt to define some construct purporting to be "the energy of the gravitational field" will not work, as it can be transformed away to 0 locally by a coordinate transform.I'm not sure why that would be an objection. After all, even kinetic energy of an object can be transformed away to 0 by a coordinate transform, right?

publius

2006-Oct-01, 05:25 AM

Yes, but what is the "distance" along a space direction, using your convention? :)

Oh, I see what you're getting at. For a space-like interval, I'll get a negative ds^2? Well, I just say, aha, that's negative so it's space-like, throw away the minus sign and take the square root, and say that's a spatial distance. :lol:

-Richard

hhEb09'1

2006-Oct-01, 05:30 AM

Oh, I see what you're getting at. For a space-like interval, I'll get a negative ds^2? Well, I just say, aha, that's negative so it's space-like, throw away the minus sign and take the square root, and say that's a spatial distance. Instead of saying ix? Why so much trouble? :)

hhEb09'1

2006-Oct-01, 05:35 AM

Oh, I see what you're getting at. For a space-like interval, I'll get a negative ds^2? Well, I just say, aha, that's negative so it's space-like, throw away the minus sign and take the square root, and say that's a spatial distance."Throw away the minus sign"? :) is that some new-fangled mathematical operation?

publius

2006-Oct-01, 05:53 AM

"Throw away the minus sign"? :) is that some new-fangled mathematical operation?

Yes, it's called "positation". I'm surprised you haven't heard of it.

Seriously, and this is beyond my mathematical pay grade by a few notches, so don't this as anything but my prejudices, I don't like using a complex formulation of this. In AC circuits, I use it all the time of course, but that's a little different.

For example, IIRC, if we're going to complex vector spaces, where each of our dimensions can be complex (2 dimensions per dimension? ....), we'd normally define the norm as the dot product of the vector with its conjugate. But in space-time, we couldn't do that, because we'd loose the minus sign.

In AC "phasors", the norm is indeed the product of the phasor and its conjugate. And VI, the power product works out rather cutely if you define that as V^*I. The real part of that is the "dot product" of V and I, and the imaginary part is sort of a cross product (if we pretend we've got a z/k direction there along which it lies). The real part is the real power and the imaginary part is the reactive power.

There is more fun like this to be had with quaternions -- a quaternion product gives you a vector part involving the cross product and a "scalar part" involving the dot products.

-Richard

hhEb09'1

2006-Oct-01, 02:21 PM

Yes, it's called "positation". I'm surprised you haven't heard of it. That's why physicists think they don't need mathematicians, and why they always will :)

Seriously, and this is beyond my mathematical pay grade by a few notches, so don't this as anything but my prejudices, I don't like using a complex formulation of this.But you already are, you just avoid it the mention of it. There's nothing wrong with it--or else the equations would be wrong. Perhaps that's true, but we don't know that yet.

grav

2006-Oct-01, 02:42 PM

Well, hhEb09'1, I take it from the last few posts that you see what I have been getting at by now. I suppose a positive s^2 might work for a time-like interval, just not spatial, although I still have know idea what that means exactly.

Oh, I see what you're getting at. For a space-like interval, I'll get a negative ds^2? Well, I just say, aha, that's negative so it's space-like, throw away the minus sign and take the square root, and say that's a spatial distance. :lol:

-Richard

I guess that's about the jest of it, though I would still say that one would have to keep the negative sign for a complex value in the square root for the space-like interval. This is because if we don't, s becomes a real spatial dimension, which it can't be, since the other three are already accounted for. In other words, it can be rearranged into r^2+s^2=(ct)^2, and s can have any number of spatial dimensions itself, say, a and b, so that we get (a^2+b^2)+(x^2+y^2+z^2)=a^2+b^2+x^2+y^2+z^2=(ct)^2 . As one can see, any of these spatial dimensions can be easily interchanged. Who's to say which is which when none are absolute? Of course, we only have three dimensions, so this can't be the case for space-like intervals. It cannot be real, only complex. It is like identifying the points that lie on a circle in real space. For this, x^2+y^2=R^2, where -R<x<R, and -R<y<R. If we try to plot a point for x>R or x<-R, then y is complex. Ordinarily, this means that it is impossible to plot for that particular x coordinate within Euclidean geometry. But we could also think of it as a coordinate for y that lies in another dimension other than our own three, outside of ordinary Euclidean geometry, where the coordinate becomes complex.

Interestingly, I did that many years ago when I came up with a graph with complex coordinates to the left of the origin, and positive and negative values on the right for the x axis, and the same for the y. In other words, instead of negative to positive x, I did it for negative to positive square roots ( (-x^2)^1/2=ix to (x^2)^1/2=(+/-)x ). The interesting part was that each geometric figure seemed to "step up". The real quandrant presented the figure as normal, but quadruply folded. On the complex side, however, lines became circles, circles became parabolas, and parabolas became hyperbolas, or something like that (it's been awhile). I'm wondering if such a graph would help with all of this, to lay out what is really happening. I'm beginning to suspect now more than ever that GR is really a sort of a reduced "shortcut" result for something that is really much more complex, occuring in "real" space-time. These sort of formulas are useful in that respect. But as I said before, if it works, it works.

hhEb09'1

2006-Oct-01, 02:57 PM

Well, hhEb09'1, I take it from the last few posts that you see what I have been getting at by now. I suppose a positive s^2 might work for a time-like interval, just not spatial, although I still have know idea what that means exactly.That's right. Positive (in publius's convention) is time-like, negative is space-like.

I guess that's about the jest of it, though I would still say that one would have to keep the negative sign for a complex value in the square root for the space-like interval. This is because if we don't, s becomes a real spatial dimension, which it can't be, since the other three are already accounted for.Heh? (gist, BTW) It's not another dimension, it's a measurement in spacetime. The diagonal of a box is not a fourth dimension, added to its length, width, and depth.

grav

2006-Oct-01, 03:00 PM

Grav,

Both are correct..........................:whistle:

It has to do with simultaneity, which varies between frames. That moving observer could claim he was stationary, write X = vT, and come up with the same formula with t and T reversed, as above.

His clock ticking T = t/gamma is simultaneous with my clock ticking t *in my frame*. In *his* frame my clock ticking T/gamma is simultaneous with his clock ticking T. Say gamma is 2. When my clock ticks 2 hours, his clock has just ticked 1 hour in my frame. But in his frame, when his clock has ticked 1 hour, my clock has only ticked off 30 minutes.

To see how this works, and how it is indeed mathematically consistent, although it will drive you nuts getting your head wrapped around it at first, we need to look at the full Lorentz transforms. The relativity of simultaneity is something that will trip you up in subtle ways.

-Richard

Yes. I see how that could trip someone up very easily. Defining the formula as t=T*gamma for one and T=t*gamma for the other is very confusing and quite unworkable. If one attempts to cancel out for t, for instance, one gets t=T*gamma and t=T/gamma, so T*gamma=T/gamma, so gamma=1/gamma, and gamma^2=1. Of course, this is only true for v=0. So yes. It can get very confusing. If it were me, however, I would define it one way and one way only. Instead of t=tA and T=tB, which is the way this looks to be defined to me, so that the formula changes for each frame of reference, I would define it according to that frame of reference, which would become ttraveller, as seen by observer=tobserver*gamma. Now it works either way. But of course, I'm not the one making up the rules. So that's just me. :)

grav

2006-Oct-01, 03:12 PM

Heh? (gist, BTW) It's not another dimension, it's a measurement in spacetime. The diagonal of a box is not a fourth dimension, added to its length, width, and depth.

This is the other point I was trying to get at. Publius originally described {s} as the "norm". But the norm is the square root of the sum of the squares of the perpendicular dimensions. If we have s^2=(ct)^2-r^2, then that becomes r^s+s^2=(ct)^2. The other way around, we would have s^2=r^2-(ct)^2, or r^2-s^2=(ct)^2, but in this case, {s} is complex, so we still get r^2+s^2=(ct)^2 for the positive (absolute) values. Either way, then, the resultant is ct, not s, and ct becomes the norm (ct is the diagonal), making s another "real" dimension if it is real (not complex). This means that ct is the furthest distance one can travel through space-time. And since c is a constant, this is the distance one always travels through space-time for a given time, even when at "rest", and so regardless of the actual value of v (r=vt).

hhEb09'1

2006-Oct-01, 03:28 PM

This is the other point I was trying to get at. Publius originally described {s} as the "norm". But the norm is the square root of the sum of the squares of the perpendicular dimensions. If we have s^2=(ct)^2-r^2, then that becomes r^s+s^2=(ct)^2. The other way around, we would have s^2=r^2-(ct)^2, or r^2-s^2=(ct)^2, but in this case, {s} is complex, so we still get r^2+s^2=(ct)^2 for the positive (absolute) values. Either way, then, the resultant is ct, not s, and ct becomes the norm (ct is the diagonal), making s another "real" dimension if it is real (not complex). This means that ct is the furthest distance one can travel through space-time. And since c is a constant, this is the distance one always travels through space-time for a given time, even when at "rest", and so regardless of the actual value of v (r=vt).You're confusing the measurement with the concept of dimension, as I said. You cannot twist the definition of s^2 to suit your own concept of what a norm "should be", it is defined that way and not "your" way. Whether s^2 is on one side of the equal side or the other, s^2 can still be negative--the sign in front of it does not change things.

grav

2006-Oct-01, 04:08 PM

You're confusing the measurement with the concept of dimension, as I said. You cannot twist the definition of s^2 to suit your own concept of what a norm "should be", it is defined that way and not "your" way. Whether s^2 is on one side of the equal side or the other, s^2 can still be negative--the sign in front of it does not change things.

Well, if it is defined as the "norm", that's fine. But I took it to mean that it is the resultant like you used for the diagonal of a box. Can you think of any way that the resultant of a set of vectors can be less than one of its sides? Of course, you might say by using complex measurements, but in this case, even if we do that for {s}, the subtraction of its square becomes positive. So the resultant can only be that which is found by the sum of the squares, which becomes ct either way. One can also think of the resultant as being the one with the largest absolute value for its square, which is always ct in this formula. This means that light will always travel a null geosedic through space-time, and r and s are its "components".

hhEb09'1

2006-Oct-01, 04:16 PM

Well, if it is defined as the "norm", that's fine. But I took it to mean that it is the resultant like you used for the diagonal of a box.Big mistake

Can you think of any way that the resultant of a set of vectors can be less than one of its sides?In what context? In the context that we are talking about, it makes sense. The usual way of calculating a distance (the diagonal of a box), or norm, was found to be not useful, whereas this one is very useful.

This means that light will always travel a null geosedic through space-time, and r and s are its "components".The "null" in "null geodesic" means that s=0. There is no component there.

grav

2006-Oct-01, 05:14 PM

In what context? In the context that we are talking about, it makes sense. The usual way of calculating a distance (the diagonal of a box), or norm, was found to be not useful, whereas this one is very useful.It makes sense to me this way, too. I just wouldn't consider {s} to be the norm, but ct. But then, my way of thinking is probably a little different anyway. A little backwards, even, some might say. :p But that's usually how I do my best work. ;) Not always, though. I guess it doesn't matter what we call it. I'm not going to try to go against convention here. If it is easier for everyone else to think of it that way, then that's fine by me, just as long as I know how it's defined.

The "null" in "null geodesic" means that s=0. There is no component there.

Yes. Come to think of it, that would be the case. If by null geosedic, we are simply talking about the distance light will travel over time, then v=c, so s^2=(ct)^2-r^2=(ct)^2-(vt)^2=(ct)^2-(ct)^2=0. So it's not the null geosedic, but just the path that matter travels through space-time that I am referring to.

hhEb09'1

2006-Oct-01, 05:33 PM

It makes sense to me this way, too. I just wouldn't consider {s} to be the norm, but ct. But then, my way of thinking is probably a little different anyway. A little backwards, even, some might say. :p But that's usually how I do my best work. ;) Not always, though. I guess it doesn't matter what we call it. I'm not going to try to go against convention here. If it is easier for everyone else to think of it that way, then that's fine by me, just as long as I know how it's defined.However, if you take two points (x1, y1, z1, t1) and (x2, y2, z2, t2) of spacetime, you're saying that the appropriate calculation is just ct? c times (t2-t1), in other words? That's not right.

So it's not the null geosedic, but just the path that matter travels through space-time that I am referring to.The null geodesic is the path that light takes through spacetime, s=0.

grav

2006-Oct-01, 06:22 PM

However, if you take two points (x1, y1, z1, t1) and (x2, y2, z2, t2) of spacetime, you're saying that the appropriate calculation is just ct? c times (t2-t1), in other words? That's not right.

No. I'm saying that the resultant (it would seem to me, anyway) would be ct, where (ct)^2=r^2-s^2. In this case, ct is the path that matter (or light) always follows through space-time.

The null geodesic is the path that light takes through spacetime, s=0.

Right. The null geosedic applies only to light (or the speed of light), where r=ct, so s=0. I meant to say before that ct would also be the path that matter travels through spacetime as well, where v<c, so s<>0.

hhEb09'1

2006-Oct-01, 06:27 PM

No. I'm saying that the resultant (it would seem to me, anyway) would be ct, where (ct)^2=r^2-s^2. In this case, ct is the path that matter (or light) always follows through space-time.

Right. The null geosedic applies only to light, where r=ct, so s=0. I meant to say before that ct is also the path that matter travels through spacetime as well, where v<c, so s<>0.I'm not sure how you are interpreting that. I can't see how you can make sense of it. You can calculate a value for two points that are sufficiently separated that nothing--light or matter--can travel from one point to the other.

So, how do you mean that?

grav

2006-Oct-01, 06:50 PM

I'm not sure how you are interpreting that. I can't see how you can make sense of it. You can calculate a value for two points that are sufficiently separated that nothing--light or matter--can travel from one point to the other.

So, how do you mean that?

That would be true if the points and time were arbitrary, in which case the formula would be meaningless. It would only be useful (as far as I can tell), as a relativity of velocities. That means that the points (x1, y1, z1, t1) and (x2, y2, z2, t2) would really be points that are separated by a distance of vt=r, where t=t2-t1 and r^2=(x2-x1)^2+(y2-y1)^2+(z2-z1)^2. So we would have v*(t2-t1)=[(x2-x1)^2+(y2-y1)^2+(z2-z1)^2]^1/2. Also, vx=(x2-x1)/(t2-t1), vy=(y2-y1)/(t2-t1), and vz=(z2-z1)/(t2-t1), where v^2=vx^2+vy^2+vz^2.

publius

2006-Oct-01, 07:19 PM

That's why physicists think they don't need mathematicians, and why they always will :)But you already are, you just avoid it the mention of it. There's nothing wrong with it--or else the equations would be wrong. Perhaps that's true, but we don't know that yet.

Hee-hee-heeeee. Physicists don't need mathematicians. Until they do need them. When they need 'em, they *NEED* 'em.

I was wrong about "positation". The correct term (I just looked it up...in my own mind) is "square-rootabus-minus1icus resultus interruptus".

But wait a minute, why I can't say I'm just dividing out the metric factor of -1. Ie

s^2 = c(t2 - t1) g_tt c(t2 - t1) + (r2 - r1) g_rr (r2 - r1), where g_rr = -1.

So, I get s^2 = -a^2. I say, in a frame where those events were simultaneous t2 = t1, I have

s^2 = -a^2 = (r2 - r1) (-1) (r2 - r1)

(r2 - r1)^2 = s^2/-1 = -a^2/-1 = a^2.

-Richard

hhEb09'1

2006-Oct-01, 08:59 PM

That would be true if the points and time were arbitrary, in which case the formula would be meaningless. It would only be useful (as far as I can tell), as a relativity of velocities. publius used it a few posts back (http://www.bautforum.com/showthread.php?p=833559#post833559) for two examples, the first had two spacetime points: (here and now) and (the moon and now); the second had (here and now) and (the moon and now). The calculation was far from meaningless.

But wait a minute, why I can't say I'm just dividing out the metric factor of -1. That's easy for you to say. :)

I'd just like to see your version of what s equals, from the formula, before prostration or whatever it's called

grav

2006-Oct-01, 11:13 PM

Originally Posted by grav

That would be true if the points and time were arbitrary, in which case the formula would be meaningless. It would only be useful (as far as I can tell), as a relativity of velocities.

publius used it a few posts back for two examples, the first had two spacetime points: (here and now) and (the moon and now); the second had (here and now) and (the moon and now). The calculation was far from meaningless.

Yes, but those would be just the special cases for when t=0 or r=0. And I think publius just meant them as a simple demonstration of the difference between space-like and time-like. For t=0, s^2=r^2, so s=r (space-like), or r=0, s^2=(ct)^2, so s/c=t (time-like). But s^2 would indeed be positive for t=0, or r>ct, in which case even light hasn't had a chance to transfer information from one location to another yet, which would indeed take time. I suppose one could use this as a loose translation of the formula if they desire, to determine a description of the value of {s} for the difference in the space-time coordinates of particular events, but it would seem to me that those events themselves, then, would still be arbitrary, and I'm not sure exactly what practical application this would have.

hhEb09'1

2006-Oct-01, 11:41 PM

Yes, but those would be just the special cases for when t=0 or r=0. No, better check the examples again. The second one does not have either t nor r equal to zero.

grav

2006-Oct-02, 12:05 AM

No, better check the examples again. The second one does not have either t nor r equal to zero.

Okay. Yes, I guess so. But all I'm really saying is that a positive s^2 (for space-like) is possible only for random, arbitrary points in space-time. In order for it to have any real meaning (for the formula to be workable), the events should be related in some way. And that something would be a transfer of information (energy or a body of matter or some such), which would always travel at or below c, and therefore take at least t=r/c to occur (or be transferred).

hhEb09'1

2006-Oct-02, 01:08 AM

Okay. Yes, I guess so. But all I'm really saying is that a positive s^2 (for space-like) is possible only for random, arbitrary points in space-time. In order for it to have any real meaning (for the formula to be workable), the events should be related in some way.The first example is the earth and moon and the distance between them. They're related, not random or arbitrary. What's wrong with that?

PS: remember, publius and his ilk considers positive s^2 to be time-like

grav

2006-Oct-02, 02:03 AM

The first example is the earth and moon and the distance between them. They're related, not random or arbitrary. What's wrong with that?

We can pick any two sets of coordinates and apply a time between separate events that occur at those coordinates. I consider that arbitrary. You might not, I don't know. We seem to have a miscommunication in definitions here, but I think you know what I mean. Don't make me have to break out the dictionary for this. :p

PS: remember, publius and his ilk considers positive s^2 to be time-like

Yes. That's what I was saying. A space-like interval could only be negative. Otherwise, it would become indistinguishable from our three spatial dimensions. A positive s^2, then, would have to be time-like. But it also depends on which way we use the formula. :)

hhEb09'1

2006-Oct-02, 02:52 AM

We can pick any two sets of coordinates and apply a time between separate events that occur at those coordinates. I consider that arbitrary. You might not, I don't know. We seem to have a miscommunication in definitions here, but I think you know what I mean. Don't make me have to break out the dictionary for this. You said "In order for it to have any real meaning (for the formula to be workable), the events should be related in some way." Why does it not have any real meaning? The distance to the moon from the earth doesn't mean anything?

Yes. That's what I was saying. A space-like interval could only be negative. Otherwise, it would become indistinguishable from our three spatial dimensions.?? That's why it's called space-like.

A positive s^2, then, would have to be time-like. But it also depends on which way we use the formula. :)I was responding to what you had said: "a positive s^2 (for space-like)"

grav

2006-Oct-02, 01:31 PM

Originally Posted by grav

We can pick any two sets of coordinates and apply a time between separate events that occur at those coordinates. I consider that arbitrary. You might not, I don't know. We seem to have a miscommunication in definitions here, but I think you know what I mean. Don't make me have to break out the dictionary for this.

You said "In order for it to have any real meaning (for the formula to be workable), the events should be related in some way." Why does it not have any real meaning? The distance to the moon from the earth doesn't mean anything?

I mean real meaning as far as working with the formula to determine something in physics such as a transfer of energy or the motion of a body from point A to point B or some such. Anything like this would occur in time {t} where t=>r/c.

Yes. That's what I was saying. A space-like interval could only be negative. Otherwise, it would become indistinguishable from our three spatial dimensions.

?? That's why it's called space-like.

To me, it would make no sense to call it space-like if it were positive unless v>c, so r=>ct (t<=r/c, where the distance is greater than the time for ct), as with "arbitrary" coordinates. If that is how it is really referred to, I don't know. I'm new to all of this. But that is what it sounds like, and so I agree that's the way it should be. For instance, s^2=r^2-(ct)^2, where t=0 would give us s=r for pure distance. This would be positive and a pure distance, but I consider this arbitrary because as far as the time units are concerned, there is no way something could travel any distance (except zero) in zero time through space-time.

A positive s^2, then, would have to be time-like. But it also depends on which way we use the formula.

I was responding to what you had said: "a positive s^2 (for space-like)"

A distance could be positive and space-like for real distance where the time is less than r/c. See above. Since nothing can travel faster than c, this could never happen, except when the coordinates are purely arbitrary, or to put it another way, when we are talking pure coordinates in space-time, without any regard to something being transferred between those points in time {t}.

I'm thinking of this as actually defining time as the fourth dimension. All that really takes is to imagine a graph where one side is time and the bottom is for distance. We do that distance to time thing all the time this way. We could also imagine it on a plane if we include the z axis (in and out of the graph). We could follow how points move through time this way. A line on the graph would just be the velocity, r/t (I'm using r here instead of d because that's really what r is and that's what we've been using in this thread). An orbit would become a spiral in the x,z plane that moves upward (or downward depending on the positive and negative directions as they are laid on the graph) with time. It would look like a spring. Expansion in all directions of a body would look like a cone. And so on.

Now let's make the z direction the s axis. So we have distance in real space for x, time for y, and s for z. Now we can plot how objects move (or energy is transferred) through space-time. The s coordinates are complex, however, since they don't lie within three dimensional space. And the resultant for these dimensions is always ct. Now, if t=0, we would have s^2=r^2-(ct)^2, so s=r. But for a real "transfer" of an object through space-time, this would mean r=vt, so v=infinity. Even for some t where ct<r, then v>c. This cannot even be plotted in the graph for s, because the complex coordinates for s (s^2 is negative) become positive again. This would be like coming up with a complex coordinate in our own three dimensions. For instance, for a circle, R^2=x^2+y^2, where x and y are real. These coordinates can only be real if -R=<x=<R and -R=<y=<R. If x or y lie outside that range, the other becomes complex. Now, this can't be graphed in real space since one of the coordinates is complex, which means that geometry is impossible in real space. But it can be graphed for s. In other words, for R^2=x^2+y^2 for a circle, x^2 and y^2 must be positive for real space. But we could indeed come up with a coordinate for x^2 or y^2 that is negative if either of the values lies outside of the given parameters. Then the other becomes complex, and lies outside of real space. The same goes for s^2. If we determine a value for r that lies outside of the range for r=vt, where -c=<v<=c, then s^2 becomes positive. That means it lies outside of its parameters. Interestingly enough, however, for v>c, it makes it real again, which would mean, though, in this case, that not only does it lie outside of real space, but it would also lie outside of space-time itself. This is because nothing can travel faster than c, through real space or space-time. So if we come up with a positive value for s^2, then we are talking pure coordinates, and we are taking it out of the context of the formula (or at least applying it in a completely different way than what I am discussing here).

hhEb09'1

2006-Oct-02, 04:08 PM

I mean real meaning as far as working with the formula to determine something in physics such as a transfer of energy or the motion of a body from point A to point B or some such. Like computing the strength of the earth's gravity on the moon, using Newton's formula? Surely, we've all done that sort of thing, right? The formula uses the inverse of the distance squared. That hasn't real meaning?

grav

2006-Oct-02, 05:50 PM

Like computing the strength of the earth's gravity on the moon, using Newton's formula? Surely, we've all done that sort of thing, right? The formula uses the inverse of the distance squared. That hasn't real meaning?

You're trying to confuse me here, aren't you? :naughty: :) I don't know why you want to debate it with me, really. I'm still just starting to learn it all myself, and only in the best ways I can conceive of it, which would be through a real and complex coordinate system, which I rarely work with either, but would definitely be useful in this case. So any discussions we have from this point, we are learning about together, unless of course, you already know more about it than I do, which you probably do. Maybe you're trying to make a point about it that I can't see yet, I don't know. It did take me a while to see what you meant about the rotation thing, and it was very enlightening to say the least, and so I will continue to engage you with this as well.

I said before that there may be ways that the purely coordinate part of it might be useful, but that I just couldn't think of any at the moment. But as far as the gravity between the Earth and moon goes, I don't see how this would be purely coordinate anyway. It would be a transfer of energy, so yes, I guess that would have real meaning as far as the way this formula seems to be meant to be applied (real meaning within the context of the formula, so that it can actually be applied to further calculations). But then, Newtonian gravity considers the speed of gravity to be instantaneous, which enevitably would make it purely a spatial coordinate system, where s=r, so s^2 is positive and s is real, but in this case, it would really be just throwing the coordinates back in real space, since t<r/c. If gravity instead travels at the speed of light, though, then s=0, and any speed less than this would then define s as complex. So in as far as it appears (to me) that this formula is meant to be applied, nothing travels faster than c, and so s is always zero or complex.

I am saying that according to the foundations of relativity, nothing should travel faster than the speed of light, so nothing can be transferred from point A to point B faster than this. Of course, we can always speculate on two points and give a time between them that is less than the time it would take for light to travel that distance, and then apply them to the formula, but to what end I don't know. That doesn't mean it wouldn't be applicable in some way, but I just wouldn't know in what way that is in keeping with the implied intent for the context of the formula to begin with. That's just how I see it. You may think differently.

bRainDance

2006-Oct-02, 06:03 PM

If gravity, or whatever is holding/pushing us unto this planets surface WASN'T a large amount of energy ....

Then why are our thighs so big?

Too fight the power( of gravity)! lol

peace

bRainDance

2006-Oct-02, 06:06 PM

p.s. im sure what your all banging on about makes perfect sense to you in your own little weird way

but at all times that you are trying to appear like a techno ******** remembre this ...

Genius lies in simplicity :D

punkrockbong151

2006-Oct-02, 08:12 PM

I was wondering: suppose a meteorite, or lightbeam, passes a heavy object (e.g. planet, blach hole) in space, it's path gets bended due to gravity.

Since changing it's path would require energy, my question thus is if mass generates energy how does that effect conservation of energy? (since energy cannot be created or destroyed)

this is a good question, but from wat i have read i haven't come across any answers. but first of all if anything came close at all to a black hole, it's dimensions would be stretched like spahgetti (or however u spell it) and sucked in, and ther's no point, how would we harness something using the gravity, it would have to go miles high and come back down and repeat that process, but a good thought tho

publius

2006-Oct-03, 04:45 AM

All,

I think this was thread where we discussed GEM in relation to the energy transfer picture. Something in that stationary earth thread came up that reminded me of something I meant to say about the GEM linearization.

g and B_g do NOT transform like E and B do between inertial frames. It's a bit different, primarily because of the gamma factor on the mass current, I think. I think if the velocity of that mass current (ie rho*v) is small, it would be about like EM, but there may be other differences. Anyway, that's something to keep in mind in constructing GEM analogies based on EM.

-Richard

hhEb09'1

2006-Oct-03, 04:50 AM

You're trying to confuse me here, aren't you? :naughty: :) I don't know why you want to debate it with me, really.No, I'm not trying to confuse you. :)

I'm not really debating either--I'm just offering examples that seem to contradict the notions that you've been professing. I give you credit for trying though.

I'm still just starting to learn it all myself, and only in the best ways I can conceive of it, which would be through a real and complex coordinate system, which I rarely work with either, but would definitely be useful in this case. So any discussions we have from this point, we are learning about together, unless of course, you already know more about it than I do, which you probably do. Maybe you're trying to make a point about it that I can't see yet, I don't know. It did take me a while to see what you meant about the rotation thing, and it was very enlightening to say the least, and so I will continue to engage you with this as well. Thanks. I think. :)

I said before that there may be ways that the purely coordinate part of it might be useful, but that I just couldn't think of any at the moment. But as far as the gravity between the Earth and moon goes, I don't see how this would be purely coordinate anyway. It would be a transfer of energy, so yes, I guess that would have real meaning as far as the way this formula seems to be meant to be applied (real meaning within the context of the formula, so that it can actually be applied to further calculations). But then, Newtonian gravity considers the speed of gravity to be instantaneous, which enevitably would make it purely a spatial coordinate system, where s=r, so s^2 is positive and s is real, but in this case, it would really be just throwing the coordinates back in real space, since t<r/c. If gravity instead travels at the speed of light, though, then s=0, and any speed less than this would then define s as complex. So in as far as it appears (to me) that this formula is meant to be applied, nothing travels faster than c, and so s is always zero or complex.Newton's equations are approximations to the Einsteinian ones, but the Einsteinian ones produce the Newtonian ones not just in the limit but also at the level that we typically use them--in other words, the Einsteinian ones behave like Newtonian ones with infinite speed of gravity. So, I think your objection is unfounded, unless you are trying to insist that we can only use the full set of Einsteinian equations, all the time--and I definitely do not get that impression from your posts so far. :)

I am saying that according to the foundations of relativity, nothing should travel faster than the speed of light, so nothing can be transferred from point A to point B faster than this. Of course, we can always speculate on two points and give a time between them that is less than the time it would take for light to travel that distance, and then apply them to the formula, but to what end I don't know. That doesn't mean it wouldn't be applicable in some way, but I just wouldn't know in what way that is in keeping with the implied intent for the context of the formula to begin with. That's just how I see it. You may think differently.Regardless of the justification that I gave above, a computation is just a computation. Whether the result is physically realizable or not does not make the attempt at calculation a fool's game--sometimes it's what you do to see what can be done, and you have to know how to interpret the results.

p.s. im sure what your all banging on about makes perfect sense to you in your own little weird waySo, you haven't reached 170 yet? Keep at it

publius

2006-Oct-03, 04:59 AM

Grav,

GR has some mighty clever tricks up its sleeve about the propagation delay of gravity. We've discussed this before, but heck, I couldn't find it. :)

GR hides the propagation delay much better than EM forces. You can see radiation as a consequence of propagation delay, the forces "missing" from equal and opposite. Momentum and energy will be lost, and the radiation is what carries it away.

Same in GR -- radiation comes about from the propagation delay, but the question of "real energy and momentum" in those waves is a very complex and vexing question, as you saw in this thread. But since GR does a better job of "hiding" the propagation delay, the radiation is much weaker, coming in at quadrapole order rather than dipole.

And this is what allows Newton to work so well over the solar system, even though the light travel delays are significant. As Tom Van Flandern notes, if we took the GEM approximation here, and used that for the solar system, rather than instantaneous Newton alone, the solar system would "blow up" in short order on astronomical time scales. It wouldn't be stable.

And what is amazing is this better cancellation of propagation delay just fell right out of the EFE, Einstein did not start out to put it in.

ETA: Van Flandern may not be taking into account a GEM like gravity, but putting the propagation delay right in Netwon's inverse square. This would make the "missing" much worse, as it is the velocity dependent behavior that modifies things beyond that. At any rate, GEM would have the solar system "blowing up", but it might last longer than Van Flandern says. He is very stubborn about refusing to believe GR cancels the propagation delay much better.

-Richard

grav

2006-Oct-03, 02:08 PM

hhEb09'1,

Since I am just learning this, maybe I was a little quick to say that nothing useful could come from a purely coordinate approach. I still can't think of anything yet, but that doesn't mean it it's not possible. Maybe you could give me an example of how {s} could be used this way to further calculations. I'm not sure if the apparent infinite speed of Newtonian gravity counts, especially since the formula is for GR, but I could be wrong. The way I'm looking at it so far, I can easily visualize what the formula entails. But I would have to see how {s} is actually applied in other formulas to know for sure.

The way I see it, the coordinates (x,y,z,t) can be applied to this formula in any way desired for a positive or negative s^2. But to me it would appear that a positive s^2 is purely coordinate and a negative s^2 is a transfer of energy where v=<c (or coordinate as well). Because of my way of looking at it, however, the is a definite relationship between the distance travelled and the time to get there, where v=<c. In this case, the formula would better serve if written as s^2=(vt)^2-(ct)^2. We could still extrapolate the distance after the fact with r=vt=(vx^2+vy^2+vz^2)^1/2*t=(x^2+y^2+z^2)^1/2. Since v=<c for this, s^2 is always zero or negative. Also, ct is always the largest absolute term, and so it is always the resultant, where (ct)^2=(vt)^2-s^2. We could further break this down into s^2=[(v/c)^2-1](ct)^2, so s=i[1-(v/c)^2]^1/2*ct=ict*gamma. For small v, this breaks down further still into s=ict*[1-Ekin/Erest]. If we use the formula the other way around like publius does, we could drop the pesky {i}. But I still don't see how that could be the case. If v can be greater than c, as in the coordinate system, then we would have s=i[1-(v/c)^2]^1/2*ct=i*i*[(v/c)^2-1]^1/2*ct=-[(v/c)^2-1]^1/2*ct=-[r^2-(ct)^2], which, now that I think about it, if I did this correctly, would make s=-r for t=0, not s=r, although (r^2)^1/2 is +/-r, so I guess we could also have s=-(r^2)=-(-r)=r after all. Anyway, we need to move on so I can see how the value of s is actually used. In what other formula(s) is {s} expressed?

grav

2006-Oct-03, 02:13 PM

Grav,

GR has some mighty clever tricks up its sleeve about the propagation delay of gravity. We've discussed this before, but heck, I couldn't find it. :)

GR hides the propagation delay much better than EM forces. You can see radiation as a consequence of propagation delay, the forces "missing" from equal and opposite. Momentum and energy will be lost, and the radiation is what carries it away.

Same in GR -- radiation comes about from the propagation delay, but the question of "real energy and momentum" in those waves is a very complex and vexing question, as you saw in this thread. But since GR does a better job of "hiding" the propagation delay, the radiation is much weaker, coming in at quadrapole order rather than dipole.

And this is what allows Newton to work so well over the solar system, even though the light travel delays are significant. As Tom Van Flandern notes, if we took the GEM approximation here, and used that for the solar system, rather than instantaneous Newton alone, the solar system would "blow up" in short order on astronomical time scales. It wouldn't be stable.

And what is amazing is this better cancellation of propagation delay just fell right out of the EFE, Einstein did not start out to put it in.

-Richard

Yes! In looking at this post, I'm thinking that the formulation for the propagation delay is what I really need to learn before I can incorporate GR into what I have been working on. It would appear that this would be the first and most obvious step in determining the difference between Newtonian gravity and GR. Everything else is probably just details.

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