Rest length in general relativity

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In general relativity, the concept of rest length is not absolute and is dependent on the observer's chosen reference frame. The rest length is defined as the length of an object measured in its rest frame, but selecting this frame can be complex, especially for large or non-stationary objects. Observers can agree on the rest length if they are in the same frame, yet in non-stationary spacetimes, the rest length may vary over time due to spatial contraction. The proper length is determined through integrals that are frame-dependent, leading to ambiguity in defining a unique rest length. Ultimately, while the rest length can be invariant, the lack of a unique frame complicates its measurement in general relativity.
  • #121
Hurkyl said:
Parallel transport respecting a metric is supposed to preserve local geometry -- i.e. it's supposed to preserve the metric. I.e. if \tau denotes parallel transport along some curve, we're supposed to have \langle \tau(v), \tau(w) \rangle = \langle v, w \rangle.

For the circle, any metric can be described by an everywhere nonzero periodic scalar function g, and the inner product by

\langle X, Y \rangle = g \cdot X \cdot Y

(again, ordinary function multiplication)

The metric compatability condition can be expressed locally by

\nabla_X g = 0

for any vector field X, which translates here to

X g' + X g = 0.

This implies g has the form g(t) = K e^{-t} -- but this cannot be both periodic and nonzero. Therefore, there does not exist any metric compatable with this connection.


If you're uncomfortable with my assertion that the derivative of the metric is expressed by the derivative of g, we have the alternative differential formulation of metric compatability:

X \langle Y, Z \rangle = \langle \nabla_X Y, Z\rangle + \langle Y, \nabla_X Z \rangle

which you can check again leads to a contradiction.

Please be patient with me since I'm not well versed in the finer details of differential geometry and tensor calculus. Especially the notation that you're using. Then again I'm here to learn! :)

Pete

ps - Why is my smiley face function, as well as all the format functions, not working for me?
 
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  • #122
Hurkyl said:
Conclusion: this geometry cannot be expressed by a metric.
Okay. I've had some time to think about this. I don't know what you mean when you say that the "geometry" cannot be expressed by a metric. Recall the question
Can you think of a case where one can define a geodesic but for which a metric cannot be defined?
A metric can certainly be defined for a unit circle.

Whether " this geometry cannot be expressed by a metric" (if that actually has a meaning) is true is another matter. Metrics are defined on manifolds, not on "geometries".

Pete
 
  • #123
pmb_phy said:
A metric can certainly be defined for a unit circle.
But that metric has absolutely nothing to do with these geodesics -- if you go back a little further into the thread, you made the assertion:

There are two equivalent definitions of a geodesic. One is, as you've said, a curve which parallel transports its tangent, the other is a curve which has a stationary value for its "length". Each is, equivalently, a more general definition.​

But in this case, we have a class of geodesics which can be defined by "a curve which parallel transports its tangent" (i.e. geodesics for an affine connection), but cannot be defined by "a stationary value for length" (i.e. geodesics for a metric tensor).


Consider the two notions:
(1) geodesics for a (pseudo)Riemannian metric
(2) geodesics for an affine connection

As I understand it, the point under contention was gel's statement
a geodesic is defined more generally as a curve which parallelly transports its own tangent vector.​
Specifically, you were questioning whether (2) really is more general than (1).

That (1) is a special case of (2) is clear: the geodesics for any metric tensor are the same as the geodesics for its corresponding Levi-Civita connection. That (2) really is more general can be seen from my example: it's an example of (2) that cannot be described in terms of (1).
 
  • #124
Hurkyl said:
But that metric has absolutely nothing to do with these geodesics -
The question was "Can you think of a case where one can define a geodesic but for which a metric cannot be defined?" In this case a metric can be defined.

I'm not going to get into this more since its taking this thread off topic.

Pete
 
  • #125


In the presence of torsion, the two definitions of "geodesic" differ
according to these notes "General Relativity with Torsion: Extending Wald’s Chapter on Curvature" by Steuard Jensen:
http://web.archive.org/web/20070316...com/~steuard/teaching/tutorials/GRtorsion.pdf (retrieved through archive.org). See section 3.3.
There are terms that distinguish the two characterizations... but I don't remember them or references to them at this time.

Geodesics are determined by the "Projective Structure" (which can be defined without a metric), as emphasized by Weyl.
See:
"Geometry in a manifold with projective structure" by J. Ehlers and A. Schild (Comm. Math. Phys. 32, no. 2 (1973), 119–146)
http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.cmp/1103859104
which has a summary of their construction of the pseudo-Riemannian metric structure from the Projective (geodesic, i.e. free-fall) and Conformal (light-cone, i.e., light-propagation) structures.
and:
"Classical General Relativity" by David Malament
http://arxiv.org/abs/gr-qc/0506065 (see section 2.1, in particular page 8)

I haven't fully absorbed these articles... but they have been on my to-read list.
 

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