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grav
2009-Jun-20, 12:23 AM
Let's say we have two observers, Alice and Carol, that are stationary to each other with a distance of 100 meters between them and their clocks are synchronized. Another observer, Bob is circling around Alice at a distance of 100 meters, constantly properly accelerating toward her, and so orbits Alice and passes Carol once per orbit. Bob sets his clock to read the same as Carol's when they coincide, then Bob continues travelling around until he reaches Carol again, where they compare times on their clocks. According to Alice and Carol's frame, Bob is constantly accelerating, so his clock ticks sqrt[1 - g R / c^2] slower than theirs. Since g = v^2 / R, that becomes sqrt[1 - 2 (v/c)^2]. Also, observers in each frame sees clocks in the other ticking sqrt[1 - (v/c)^2] slower because of the time dilation due to the relative speed.

Okay, so Alice and Carol see Bob's clock ticking sqrt[(1 - 2 (v/c)^2) (1 - (v/c)^2)] slower than theirs and Bob sees their clocks ticking sqrt[(1 - (v/c)^2) / (1 - 2 (v/c)^2)] faster than his own. The problem here is that when Bob and Carol meet back up and compare times on their clocks, each should say the other is the inverse of their own, so if Carol's says her clock ticked twice as fast as Bob's, then Bob should say his clock ticked at half the rate of Carol's, but that is not how the ratios work out. It does for the time dilation of the fictitious gravity but not when that for the relative speed is included. I have tried using simultaneity effects to resolve this, but so far haven't been able to make it work. So what's going on here? How do we make the times between SR and GR work out inversely for the times on the clocks between the frames of the observers?

tusenfem
2009-Jun-20, 09:17 AM

grav
2009-Jun-20, 12:25 PM
Goodness Grieve Charlie Brown, ANOTHER twin paradox thread?Yep. :) Just coincidence, though, really. I was trying to work through some math for this recent thread (http://www.bautforum.com/science-technology/89397-new-version-twin-paradox-accelerated-twin-older.html), but found I had trouble doing so.

StupendousMan
2009-Jun-20, 01:14 PM
Let's say we have two observers, Alice and Carol, that are stationary to each other with a distance of 100 meters between them and their clocks are synchronized. Another observer, Bob is circling around Alice at a distance of 100 meters, constantly properly accelerating toward her, and so orbits Alice and passes Carol once per orbit. Bob sets his clock to read the same as Carol's when they coincide, then Bob continues travelling around until he reaches Carol again, where they compare times on their clocks. According to Alice and Carol's frame, Bob is constantly accelerating, so his clock ticks sqrt[1 - g R / c^2] slower than theirs. Since g = v^2 / R, that becomes sqrt[1 - 2 (v/c)^2]. Also, observers in each frame sees clocks in the other ticking sqrt[1 - (v/c)^2] slower because of the time dilation due to the relative speed.

Nope, you're not correct here. Because the direction of Bob's motion is constantly changing, the apparent speed of Bob's clock, as seen by Carol and Alice, is not constant. In the same way, the apparent speed of Carol's clock and Alice's clock, as seen by Bob, is not constant.

You've taken the ordinary twin paradox -- Bob flies straight away from Alice at high speed, then flies straight back -- and simply added some curvature to the motion. Until you understand the ordinary twin paradox, you won't understand this situation, either.

Go work on the ordinary twin paradox until you understand it thoroughly.

grav
2009-Jun-20, 01:31 PM
Nope, you're not correct here. Because the direction of Bob's motion is constantly changing, the apparent speed of Bob's clock, as seen by Carol and Alice, is not constant. In the same way, the apparent speed of Carol's clock and Alice's clock, as seen by Bob, is not constant.

You've taken the ordinary twin paradox -- Bob flies straight away from Alice at high speed, then flies straight back -- and simply added some curvature to the motion. Until you understand the ordinary twin paradox, you won't understand this situation, either.

Go work on the ordinary twin paradox until you understand it thoroughly.Oh, I have. :) I've worked it out using SR in a straight line and instant acceleration at the point of turnaround at least. There is no problem with that scenario. I am currently trying to find it for constant acceleration along a straight line as well, though. I am having trouble applying simultaneity shifts. I integrated for the orbit, but found that as the orbitting observer conceptualizes Carol moving past-back during the first part of the orbit, he conceptualizes her moving future-forward during the last part, so it evens out, but I may have done it incorrectly. I am also trying to think about it in terms of many observers stationary to Alice that are placed around the circumference and Bob passing each of them while coinciding in the places they happen to be.

grav
2009-Jun-20, 01:37 PM
Nope, you're not correct here. Because the direction of Bob's motion is constantly changing, the apparent speed of Bob's clock, as seen by Carol and Alice, is not constant. In the same way, the apparent speed of Carol's clock and Alice's clock, as seen by Bob, is not constant.I can see why Bob would see Alice's clock as constant and Carol's clock as moving past-back when receding and then future-forward upon approach, but why wouldn't Alice and Carol see Bob's clock as constant? Are you including time of flight of light effects for over a distance? I am not for this since it is not necessary to do so, but only using the pure conceptualized effects.

grav
2009-Jun-20, 02:31 PM
You've taken the ordinary twin paradox -- Bob flies straight away from Alice at high speed, then flies straight back -- and simply added some curvature to the motion. Until you understand the ordinary twin paradox, you won't understand this situation, either.

Go work on the ordinary twin paradox until you understand it thoroughly.Here is a quick example of the ordinary twin paradox as I see it. Alice and Bob are originally stationary to each other in the same place. A space station is located 3 light-years away. Bob instantly accelerates to .6 c toward the space station. When Bob reaches the space station, he instantly accelerates and turns around to travel back to Alice at .6 c, where he instantly decelerates back to rest with Alice.

Now, Alice just sees him travel .6 c both ways over a distance of 3 light-years, so she sees the journey take 10 years according to her clock. With a Lorentz contraction of .8, Bob now measures the distance to the space station as (.8)(3) = 2.4 light-years away, so Bob clocks the journey as only taking 8 years. Alice sees time dilation for Bob's clock during his journey, so reads it as (.8)(10) = 8 years passing on his clock, which is the same as he reads. Bob sees time dilation also occurring on Alice's clock, so would read (.8)(8) = 6.4 years passing for hers, but there is also a simultaneity shift when Bob accelerates at the turnaround of tl = L Z d (2v) / [c^2 (1 - (v / c)^2)] = (.8) (.8) (3) (1.2) / (1 - (.6)^2) = 3.6 years that Bob additionally sees pass on Alice's clock, making 10 years in all that he sees for her clock, and both observers agree with the times the other's clock reads. In the last equation, L is the Lorentz contraction, Z is the time dilation, d is the distance as measured in the original frame, and 2v is the total change in relative speed from one direction to the other.

Okay, so here's the thing. I've been trying to find for the simultaneity shift with a constant acceleration around an orbit or in a straight line. The problem is I keep coming up with no overall simultaneity shift. I may be doing something incorrectly, because it would seem like there should be some, but so far it's looking like the simultaneity shift only figures in when there is originally some distance between the observers before the acceleration occurs, but not for constant acceleration the entire way.

mugaliens
2009-Jun-21, 11:56 PM
The answer is the paradox is only apparent, depending upon observational perspective.

It is not real.

If you cannot accept that, you do not yet understand the four-dimensional manifold of Minkowski spacetime and Poincare's symmetry group, and are probably still trying to wrap your noggin around it from different three-dimensional perspectives.

The key is understanding that time and distance are synomomous at the rate of c. One that's grasped, one can visual the four-dimensional interellations and the apparent paradox disappears.

For the math, see this (http://en.wikipedia.org/wiki/Basic_introduction_to_the_mathematics_of_curved_sp acetime).

grav
2009-Jun-22, 12:44 AM
The answer is the paradox is only apparent, depending upon observational perspective.

It is not real.

If you cannot accept that, you do not yet understand the four-dimensional manifold of Minkowski spacetime and Poincare's symmetry group, and are probably still trying to wrap your noggin around it from different three-dimensional perspectives.

The key is understanding that time and distance are synomomous at the rate of c. One that's grasped, one can visual the four-dimensional interellations and the apparent paradox disappears.

For the math, see this (http://en.wikipedia.org/wiki/Basic_introduction_to_the_mathematics_of_curved_sp acetime).Thanks Mugaliens. I'm not really sure what you're saying though. I understand the original twin paradox is actually resolved, so now I'm just trying to do it for acceleration, but I'm having trouble lining up the times observed.

mugaliens
2009-Jun-22, 03:27 AM
Thanks Mugaliens. I'm not really sure what you're saying though. I understand the original twin paradox is actually resolved, so now I'm just trying to do it for acceleration, but I'm having trouble lining up the times observed.

Well, here's another resource (http://en.wikipedia.org/wiki/Special_relativity#The_geometry_of_space-time) covering the basic geometry (sans GR). Do you understand partial differentials, matrix calculus, and world lines? If so, you should be able to follow the field equations, there, in particular, this for the velocity and acceleration in 4D (http://en.wikipedia.org/wiki/Special_relativity#Velocity_and_acceleration_in_4D ).

This may help, too (http://en.wikipedia.org/wiki/Four-acceleration).

That's the math approach, anyway. If you'd like to state your current problem, please do so and I'll do my best to translate the math into English.

- Mugs

mugaliens
2009-Jun-22, 03:39 AM
Lemme take a stab at it with what you have in your OP:

Also, observers in each frame sees clocks in the other ticking sqrt[1 - (v/c)^2] slower because of the time dilation due to the relative speed.

Here's your error. The 4D problem involves one of acceleration, not of relative speed. Thus, this statement is not correct. Alice and Carol see one another's clocks as ticking at a synchronous rate with each other. Due to their distance, however, each sees the other's clock as slight behind their own.

Meanwhile, Alice sees Bob's clock as ticking at a slower rate than her own, but sees that slower rate as a constant due to Bobs constant 100 m distance.

Carol, on the other hand, see's Bob's clock as ticking at a slightly slower rate when the distance between Bob and Carol is decreasing, and at a significantly slower rate when the distance between Bob and Carol is increasing. Over one of Bob's orbits, the mean of the variance in rate that Carol sees matches the rate that Alice sees.

That's it - there is no paradox, except the apparent one created by the error in your statement, quoted above.

grav
2009-Jun-22, 04:09 AM
Lemme take a stab at it with what you have in your OP:

Here's your error. The 4D problem involves one of acceleration, not of relative speed. Thus, this statement is not correct. Alice and Carol see one another's clocks as ticking at a synchronous rate with each other. Due to their distance, however, each sees the other's clock as slight behind their own.

Meanwhile, Alice sees Bob's clock as ticking at a slower rate than her own, but sees that slower rate as a constant due to Bobs constant 100 m distance.

Carol, on the other hand, see's Bob's clock as ticking at a slightly slower rate when the distance between Bob and Carol is decreasing, and at a significantly slower rate when the distance between Bob and Carol is increasing. Over one of Bob's orbits, the mean of the variance in rate that Carol sees matches the rate that Alice sees.

That's it - there is no paradox, except the apparent one created by the error in your statement, quoted above.Yes, that's exactly how I see it also :), but here's the thing. There is also proper acceleration taking place for Bob, so in addition to the time dilation due to relative speed, he should also be experiencing a gravitational time dilation due to proper acceleration with the fictional gravity of the constant acceleration inward. I have now broken it up into four steps to accomplish. The first is trying to find the time dilation observed for Bob by Alice when he accelerates in a straight line away from her. That is what I'm currently asking about and trying to work out in this thread (http://www.bautforum.com/space-astronomy-questions-answers/89762-dilation-fictional-gravity.html). The next step is to find the time dilation Bob observes for Alice as he accelerates away from her. Then I need to find the overall simultaneity shift and add that to the time dilation Bob observes for Alice and it should match what time has passed for Alice. After I have worked it out for acceleration in a straight line, then I'll apply it again to orbit.

mugaliens
2009-Jun-22, 04:26 AM
Not so, as the sum of his acceleration over 1 orbit amounts to zilch! The only difference for Bob is his relative velocity around Alice.

Minkowski introduced the concept of proper time (http://en.wikipedia.org/wiki/Proper_time)in 1908. I think one of the things that's getting you is the difference between an accelerated clock and an inertial clock. In fact, Example 2: The rotating disk (http://en.wikipedia.org/wiki/Proper_time#Example_2:_The_rotating_disk)is right up your ally, as it shows the difference as: dT*sqrt(1-(rw/c)2)

Since g = v^2 / R, that becomes sqrt[1 - 2 (v/c)^2]. Also, observers in each frame sees clocks in the other ticking sqrt[1 - (v/c)^2] slower because of the time dilation due to the relative speed.

Since v=rw (radius times angular velocity)

...but that is not how the ratios work out. It does for the time dilation of the fictitious gravity but not when that for the relative speed is included.

Again, as the summation of the acceleration over each orbit is zero, the only effect is the relative velocity of a moving Bob with respect to a stationary Alice and Carol.

Bob's orbital velocity is what matters. Nothing else.

grav
2009-Jun-22, 05:54 AM
I'm not sure what you mean when you say the summation of the acceleration over each orbit equals zero. That might occur for the simultaneity shifts, but the proper acceleration is a constant, as is the relative speed. The part about the rotating disks in the link did look similar to what I'm working on and it made me hopeful, but it still only seems to refer to relative speed without mention of proper acceleration, although it does mention acceleration in general so I can't be sure. I also clicked on "proper acceleration" at the bottom of the link, which I've looked at before but apparently not carefully enough, because in the "acceleration in (1 + 1)D (http://en.wikipedia.org/wiki/Proper_acceleration#Acceleration_in_.281.2B1.29D)" section, it contains a scenario including both proper and coordinate acceleration along a straight line, which is what I've been trying to find for that. I don't understand how they are saying it quite yet, but I'll keep reviewing it.

grav
2009-Jun-22, 07:34 PM
Well, I worked it out using the formulas in that section of Wiki's proper acceleration I linked to. Let's say Bob travels a distance of 2 d = 20 light-seconds from Alice to Carol as the distance is measured in the stationary frame, accelerating and then decelerating for half the way each. So d / c = 10 seconds and we'll say he properly accelerates such that a / c = .1 sec-1. So according to that section, y_mid = 1 + a d / c^2 = 2. Then the time that passes according to Bob is t' = 2 (c / a) acosh(y_mid) = 2 (c / a) log[x + sqrt(x^2 - 1)] = 2 (10) (1.316957897) = 26.33915794 seconds. The time that passes in the stationary frame is t = 2 (c / a) sinh[acos(y_mid)] = 2 (c / a) sinh[t' / (2 (c / a))] = 2 (10) sqrt(3) = 34.64101615 seconds.

Those equations are formulated a little different than usual, so now let's compare that to what we get when using just the ordinary integrations with SR only. The distance the stationary frame sees Bob travel is d = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1]. Rearranging that to find the time in terms of the distance travelled, we get

d = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1]
a d / c^2 = sqrt(1 + (a t / c)^2) - 1
1 + a d / c^2 = sqrt(1 + (a t / c)^2)
(1 + a d / c^2)^2 = 1 + (a t / c)^2
(a t / c)^2 = (1 + a d / c^2)^2 - 1
a t / c = sqrt[(1 + a d / c^2)^2 - 1]
t = (c / a) sqrt[(1 + a d / c^2)^2 - 1]

Finding that for acceleration and then again for deceleration gives t = 2 (10) sqrt[2^2 - 1] = 2 (10) sqrt(3) = 34.64101615 seconds, exactly what we had before. Then figuring it for the time of the accelerating observer's clock with t' = Int dt sqrt[1 - (v/c)^2], where the relative speed in terms of the stationary frame's time is (v/c) = (a t / c) / sqrt(1 + (a t / c)^2), giving t' = Int dt / sqrt(1 + a t / c)^2) = 2 (c / a) log[a (t/2) / c + sqrt(1 + (a (t/2) / c)^2)], we get t' = 2 (10) (1.316957897) = 26.33915794 seconds, same as we had before also. So only the time dilation in terms of the relative speed was used in the previous calculations, and proper acceleration is only considered in terms of the time dilation of the coordinate acceleration it produces, but not the time dilation of the proper acceleration itself. The Wiki page for the twin paradox also does the same thing. But if we do it in this way, then besides time dilation for proper acceleration not being accounted for, the time that Bob reads for Alice's clock while accelerating, which is Int dt' sqrt[1 - (v/c)^2] = Int 2 e^(a t' / c) / (e^(2 a t' / c) + 1) = 2 (c / a) [atan(e^(a t' / c)) - pi / 4] for SR only in terms of the relative speed Bob measures over his own time, plus the simultaneity shifts when they are added in are always larger than what Alice's clock actually reads, so something's not right. I know I'm not the first person to consider this, so I'm slightly disappointed, but I guess I'll just have to find a way to determine it on my own, although of course, that's kinda the way I like it :), and I have a couple of ideas about how to go about doing that.

mugaliens
2009-Jun-22, 08:49 PM
Well, I gave you the answer, Wiki did, too, and you even restated it in your post, grav - right before you made the same mistake you did earlier, the one I pointed out, and where you notice "something's not right."

There's the water, but only you can take the drink! Good luck, grav.

grav
2009-Jun-22, 09:15 PM
Well, I gave you the answer, Wiki did, too, and you even restated it in your post, grav - right before you made the same mistake you did earlier, the one I pointed out, and where you notice "something's not right."

There's the water, but only you can take the drink! Good luck, grav.If the answer is that the time dilation from proper acceleration cancels out, I didn't see that in Wiki, and I don't see how that could be. Proper acceleration in this context is a scalar, not a vector, just like the relative speed is a scalar, even though the vectors of relative velocity could cancel out, but the time dilation is experienced according to the scalar, so the time dilation due to proper acceleration itself shouldn't cancel out, any more than gravitational time dilation would cancel out between one side of the planet and the other, unless it just happens to work out the same as if it did in the end mathematically or something, which I'll eventually find out.

mugaliens
2009-Jun-23, 07:14 AM
Well the answer is that there is no paradox, so start eliminating each of your assumptions until the answers jibe both between each other and with what you know the answer to be, and when that happens, you'll have eliminated the error.

grav
2009-Jun-23, 03:18 PM
Well the answer is that there is no paradox, so start eliminating each of your assumptions until the answers jibe both between each other and with what you know the answer to be, and when that happens, you'll have eliminated the error.Yes, I try to make as few assumptions as possible, and if I'm unsure about something, I'll run the mathematics separately for each possiblility if I can and compare results to see which possibilities might be eliminated. For example, one assumption I must run on now is that, although time and space can be measured differently by different frames, the relative speed directly measured between two frames is an invariant. That is, if Alice sees Bob travelling at some relative speed to her, then Bob also measures the same relative speed to Alice, although at different times and distances, but at some point it must occur. Relativity seems to be built upon this one invariant and the speed of light being constant to all frames, and all comparisons can be taken between frames for where the relative speed is measured to be the same by each.

If that is the case, then let's say Alice is a faraway observer or stationed in the center of a planet with very strong gravity on its surface and experiences no gravitational time dilation. Bob is on the surface of the planet and experiences a very strong gravitational time dilation, so that his clock runs one tenth as fast as Alice's. The planet is very small, and so is its rate of rotation, say one rotation per ten thousand years. This way, time dilation due to relative speed is insignificant and only gravitational time dilation persists strongly. Now, if Alice and Bob measure their relative speeds to each other as the same, then since Alice measures Bob's speed as v = 2 pi R / P, where R is the radius of the planet Alice measures and P is its period of rotation, then Bob will also measure his own relative speed to Alice as v = 2 pi R' / P', and if his clock is running ten times slower than Alice's, then he will measure one tenth the time for the period of rotation, since his clock will only read one tenth the passage of time for one complete rotation, and if v = 2 pi R / P = 2 pi R' / P', then he should measure one tenth the radius for the planet as well than Alice measures.

So far its looking like length contraction occurs in this way, at least along the direction the proper acceleration due to gravity acts. If that also occurs with proper acceleration while firing thrusters on a ship, then the accelerating observer would see distances behind him shorten as he accelerates. This would be in addition to the shortening due to coordinate acceleration and elongation of the ship and its rulers during acceleration. If some of these factors produce a Rindler horizon, where the distance to an observer originally stationary at a certain distance behind the accelerating observer never changes during constant acceleration, then that produces a problem with this assumption, however, since the accelerating observer always measures the relative speed of that stationary observer at the horizon to be zero, while the stationary observer observes that the accelerating observer is always moving away, so has some relative speed. If the relative speed is not considered an invariant, nor time nor space, and even the speed of light changes during acceleration, then I don't see what else there is to run on.

grav
2009-Jun-25, 02:36 AM
The simultaneity shift for how Bob observes Carol during orbit moves past-back when originally moving away, does so for a quarter of the orbit, then moves future-forward again by the same amount in reverse until there is no overall simultaneity shift when Bob is moving opposite Carol, then future-forward again for another quarter and past-back until they meet, for no overall simultaneity shift observed again. I've calculated it a few different ways to make sure, and it always varies with (sine x) R, so cancels out twice per orbit.

The contraction seen of the distance to Carol will vary also with the simultaneity shift, however, so while he measures R to Alice directly in front of him, he measures the contracted distance to Carol as the simultaneity shift changes, so always sees himself at the center of the short end of an ellipse type oval relative to her, R measured directly forward and contracted to sqrt[1 - (v/c)^2) on each side. If he were orbitting the Earth, he would see the Earth shaped this way, but the shape revolving with him as the Earth remains stationary.

As I see it so far, however, the time dilation due to proper acceleration while orbitting will act perpendicularly to this direction and always have him view himself at the center of the long side of an ellipse type oval, contracted toward Alice, and proportional to (cos x) R for the simultaneity shift. If the time dilation due to the relative speed and that due to the proper acceleration are equal when orbitting, then Bob will end up seeing the entire two dimensional plane of his orbit contracted by sqrt[1 - (v/c)^2], and he will see himself orbitting Alice in an overall circular orbit after all, and Carol always remaining the same distance from Alice also, but that distance and the radius he observes for his orbit will be contracted. In that case, if he measures a lesser time per revolution according to his own clock and a lesser radius in the same proportion, then he will measure the same relative speed to Alice as she measures for him. I am still attempting to work out how those times are measured differently with no overall simultaneity shifts observed, though.