# Time Dilation in Relativity

Discussion in 'Physics & Math' started by RJBeery, Feb 1, 2017.

1. ### Confused2Registered Senior Member

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Quickie (at work)... can I suggest a diversion to the muon experiment where we have actual results to back up our thoughts.

3. ### Q-reeusValued Senior Member

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What about it though? It well confirms SR time dilation as a relative effect. A far slower decay rate measured in the earth frame relative to when a muon is at rest there. From the muon's rest frame the distance covered between locations A and B is far less than that distance interval measured in the earth rest frame. Thus the probability of decay between locations A and B is correspondingly reduced in the muon frame. There is no overall paradox - just different pov's.
And being a one-way constant relative speed velocity case, cannot form the basis for a travelling twin 'paradox'.

5. ### RJBeeryNatural PhilosopherValued Senior Member

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Nice! The problem here is in our definition of "accelerated". In the paper I wrote:
One consequence of this altered definition is that an object in free fall can still be under acceleration. A gravitational field devoid of tidal forces is not possible, and any spatially-extended object would feel acceleration even in a geodesic path. Your mention of the slingshot trajectory and the paper referencing the "accelerated twin" both are both considering the free-fall orbit path to be unaccelerated. GR predicts the same instantaneous time dilation at a point in a gravitational field whether we are passing through that point in free-fall or we are remaining at that point because we are standing on a surface. Because of this, an orbital path is simply subjected to the expected gravitational time dilation at that altitude plus the additional effects of torque as the body spins through its orbit.

7. ### RJBeeryNatural PhilosopherValued Senior Member

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Good catch, thanks Confused2. I'll correct this later today!

8. ### Confused2Registered Senior Member

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On reflection the difference between an 'out' journey and a 'return' journey (Twin Paradox) is best left to another thread.

9. ### Q-reeusValued Senior Member

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I never bothered to try and untangle that passage. It's vague. Given your below para, in retrospect the last bit above seems to be describing tidal forces on an object that net cancel to zero. How could that possibly define 'acceleration' of an object? All the tidal stresses do is stress the object internally. And given they decline as ~ r^-3 as against 'g forces' declining as ~ r^-2, relative effect of tidal stresses become negligible as radius r from a central mass M becomes large. And, once again, tidal stresses have no impact on the body's overall trajectory, except a possible small shift in the effective centre of mass.
Let's ignore the last part referring to 'torque'. No an object will not feel any *net* acceleration when moving along a geodesic. By definition in fact. And see my comment above. Set out a clear definition of exactly what you consider acceleration to be. It must specify whether it is based on a proper or coordinate basis. This you have not done. I can tell you right now, whichever unambiguous definition is settled on, there is no chance it will allow a fully consistent explanation for time dilation. But please do try.

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Fine by me.

11. ### RJBeeryNatural PhilosopherValued Senior Member

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Modern gravitometers can measure the tidal forces between your head and your feet. Taken to the extreme an object undergoing spaghettification on its geodesic path toward a small black hole would certainly "feel" acceleration. I'm not talking about overall trajectory, I'm talking about the stresses imparted to an object in a gravitational field (even in free fall) and the absence of those stresses outside of such field. I think we are actually in agreement on this point and you are rejecting my attempt at a revised definition of acceleration, which is fine, because it's just a label. If we can model time dilation in terms of "foobar and distance" rather than gravitational potential and there are aesthetic reasons to do so including a test to discern which method is correct then I'd like to explore it.

12. ### SchmelzerValued Senior Member

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The problem with your definition of acceleration from a GR point of view is that it is not well-defined what would be the meaning of no acceleration. Your "the second derivative of the function of position of an object" depends on the choice of the function of position, in other words, on the choice of non-accelerated coordinates.

13. ### RJBeeryNatural PhilosopherValued Senior Member

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An object floating within a massive shell would be experiencing no acceleration.

(Also, not that it's relevant, but acceleration as measured from any non-accelerated coordinate system is absolute.)

14. ### Q-reeusValued Senior Member

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Again, you seem to be trying to link tidal stresses to 'acceleration' but without making any clear statement as to what defines 'acceleration', nor how it could be precisely linked to tidal stresses. There can be no such consistent linkage. And btw it IS possible to produce an arbitrarily uniform g-field rather simply. Any spherical cavity within and offset from the centre of a larger uniformly dense spherical mass will have an uniform g-field existing within. Zero tidal forces.

15. ### Q-reeusValued Senior Member

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And the object floating within the shell has a reduced clock-rate wrt far outside, as determined by it's depressed potential. Which case was referenced to early on this thread. The simple definition of your last line is proper acceleration. What an on-board accelerometer would detect.

16. ### RJBeeryNatural PhilosopherValued Senior Member

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As I mentioned I'm willing to substitute the word 'acceleration' for something else, as long as it describes 'nonhomogeneous or asymmetric forces experienced by an object'. That phrase may be vague but we both agree that tidal forces would cause such a condition.
From page 7:
This is fantastic, I'll have to explore.

17. ### QuarkHeadRemedial Math StudentValued Senior Member

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Just read up on Newton's Shell Theorem. You will find that, for a massive spherical shell, the gravitational potential sums (or rather integrates) to zero inside, hence the metric field is constant throughout the interior of the shell, and hence the spacetime curvature is identically zero there

18. ### RJBeeryNatural PhilosopherValued Senior Member

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Hi QuarkHead, I'm familiar with Shell Theorem, I was not familiar with the arrangement that Q-reeus described. I did not think it was possible to construct a uniform g-field without an infinite wall. What this means is that my model would predict no gravitational time dilation even in a g-field if that field contains no gradient.

19. ### QuarkHeadRemedial Math StudentValued Senior Member

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Yes, well it depends what you mean by a "g-field". If you mean the metric field, then this field must be constant throughout the interior of the shell - that is the Shell Theorem in this application.

If by "g-field" you mean the gravitational field, then with a little work you can show they it is the same thing (use the field equations!).

Either way, with a constant metric field in a region of spacetime, and hence vanishing curvature, there can be no gravitational time dilation.

Einstein's theory says exactly this

20. ### RJBeeryNatural PhilosopherValued Senior Member

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While I agree with you on this, I don't think Q-reeus will. The potential is greatest at the core of a ball even when there is no curvature. This is exactly the issue I'm proposing to test.

21. ### QuarkHeadRemedial Math StudentValued Senior Member

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But we are not discussing balls - I and Q were referring to massive spherical shells. Specifically to their interior

22. ### RJBeeryNatural PhilosopherValued Senior Member

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Same issue. The Earth is a ball, but if there were a cavity at its core the potential would be maximized and the gravitational acceleration would be zero. GR's current interpretation says that time dilation would continue to increase from the Earth's surface to its core even when we are within said cavity, devoid of acceleration of any sort.

This is why I want to perform a Pound-Rebka experiment under ground. I predict that time dilation is maximized at the surface where tidal forces are greatest and tapers to zero as we move down.

Last edited: Feb 6, 2017
23. ### Q-reeusValued Senior Member

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We can agree on defining tidal forces or rather tidal stresses or rather tidal fields, but to what useful end? How does mostly negigible internal stresses on a body have any appreciable bearing on your program to define time dilation in terms of acceleration (that is yet to be defined)?
Incorrect. EP merely states it's impossible to locally distinguish between a uniform gravitational field or uniform acceleration. Time dilation otoh is never a locally measurable thing but relational and requires a globally based comparison - 'here' vs 'there'. It's what gravitational redshift is about. Your initial faulty assumption then negates what follows in that passage.
Took me much longer than expected to find, but explore away courtesy of: https://irodovsolutionsmechanics.blogspot.com/2008/09/irodov-problem-1215.html
Note: the result is exact only for strictly 1/r^2 fields, thus Electrostatics and Newtonian gravity. However departures from that in GR will ordinarily be exceedingly small, and anyway likely to be made arbitrarily even smaller by a suitable slight redistribution of matter density in the surrounding mass.