Rest length in general relativity

[gel]

However I could likewise say that geodesics require only the concept of a metric to be defined. This can be defined by an affine connection but only requires the existence of a metric, which makes it the more general definition. :)
Pete

no, a metric defines a connection, but not the converse.

 

[pmb_phy]

no, a metric defines a connection, but not the converse.

That the metric determines the connection is of no relevance in determining whether the metric or the connection provides a more general definition of geodesic. If it were the the metric would be more general since the connection canbe obtained from it.

I'm also not certain that the metric can't be obtained from the connection either (apart from a constant/conformal factor or something similar).

Pete

 

[gel]

by more general, I mean it applies in more situations, even those where there isn't a metric.

 

[pmb_phy]

by more general, I mean it applies in more situations, even those where there isn't a metric.

And by more general I could say that it applies in more situations, even those where there isn't a connection.

What are the "more cases" that you're referring to?

Pete

 

[gel]

And by more general I could say that it applies in more situations, even those where there isn't a connection.

Could you? How would it do that?

You can define connections on Lie groups without any need for a metric. I think some approaches to quantum gravity use non-metric connections.

 

[pmb_phy]

Could you? How would it do that?

I say I could say it. I didn't say I could prove it. :) That's why I asked you what are the "more cases" that you're referring to? From your response it seems that they are unrelated to geodesics.

You can define connections on Lie groups without any need for a metric. I think some approaches to quantum gravity use non-metric connections.

Sorry but I'm not familiar with Lie groups. Would a geodesic even have a meaning in that case?

Can you think of a case where one can define a geodesic but for which a metric cannot be defined?

Pete

 

[Hurkyl]

I don't think "spacelike geodesic" makes sense.

It just struck me what's going on.

You were thinking local optima of paths -- which doesn't work here because the metric is not positive definite. For a spacelike path, a 'spatial perturbation' increases length, and a 'temporal perturbation' decreases length. Therefore, you saw a big problem with the notion of geodesic.

I was thinking unit vectors and parallel transport -- which still works here. Therefore, I didn't see any problem at all!

(edit: Oh, haha, now I see everyone else figured that out last page. :blush:)
(edit: also fixed my double post)

 

[Fredrik]

It just stuick me what's going on.

You were thinking local optima of paths -- which doesn't work here because the metric is not positive definite. For a spacelike path, a 'spatial perturbation' increases length, and a 'temporal perturbation' decreases length. Therefore, you saw a big problem with the notion of geodesic.

I was thinking unit vectors and parallel transport -- which still works here. Therefore, I didn't see any problem at all!

That's right. That was a mistake by me. What my argument shows (I think) is that the alternative definition of a geodesic fails under certain conditions, but my argument isn't a problem for the standard definition.

I still don't think we can define the rest length of an arbitrary object, for several reasons. First of all, the geodesic we're considering is going to intersect the world lines of some other parts of the object and there's no reason to expect those parts to be "stationary" at those events. It's not even clear what stationary means here. And even if we can make sense of those things, I don't see a natural way to choose the endpoints of the path.

 

[Fredrik]

About metrics vs. connections...

If you have a metric, you can always construct a connection, but you can't construct a metric from a connection. So there are definitely spaces that have a connection but no metric. (I don't have an example, but I remember that's how it was presented in the books I've read).

A connection defines parallel transport, and that's all we need to define a geodesic. If we're given a metric and we would like to use it to define a geodesic, the standard way to do it is to first use the metric to define a connection, and then use the connection to define a geodesic. The alternative, which is to use the metric directly to define a geodesic as the shortest/longest path doesn't seem to work for arbitrary paths in spaces of 3 or more dimensions with a metric that isn't Riemannian. (It works just fine for time-like paths in a space with a -+++ metric, but it doesn't seem to work for space-like paths, for the reasons I mentioned in #92).

 

[Fredrik]

Very minor point here, the proper length along constant t' is 0.8. I think you missed a square root. I only mention it because I was careful to pick numbers for which the square root worked out nicely :/

D'oh. I noticed that I got a different result than you, but I thought the mistake was on your end. You're right though. I calculated $ds^2$, not $\sqrt{ds^2}$.

 

[Hurkyl]

Can you think of a case where one can define a geodesic but for which a metric cannot be defined?

Yes. Let's consider the Euclidean unit circle.

I will label the points of the circle by angular position -- i.e. by real numbers, with the condition that x and x + 2 pi denote the same point.

I can represent scalar fields as real functions satisfying f(x) = f(x + 2 pi)
I can also represent vector fields as real functions satisfying f(x) = f(x + 2 pi)

The tangent vector to a curve y at y(t) is simply y'(t).
The exterior derivative is given by (df)(X) = f' X

Now, consider the following connection:
$$\nabla_X Y = X \cdot Y' + X \cdot Y$$
where multiplication here is ordinary multiplication of real-valued functions.

(check that it satisfies the axioms of a connection!)

Geodesics for this connection are curves of the form $y(t) = A + B e^{-t}$.

Let v be a tangent vector at 0. Parallel transporting it around the circle gives the vector $v e^{-t}$, where t measures angular distance.

In particular, parallel transporting once clockwise about the circle rescales any tangent vector by $e^{-1}$, and thus cannot be an isometry under any metric.

Conclusion: this geometry cannot be expressed by a metric.

 

[pmb_phy]

Conclusion: this geometry cannot be expressed by a metric.

Let me get back to you on this at a later date.

Pete

 

[Hurkyl]

Why?

Because I have proven that parallel transport is not an isometry.

 

[pmb_phy]

Because I have proven that parallel transport is not an isometry.

What does it mean for parallel transport is not an isometry?

 

[Hurkyl]

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.

 

[pmb_phy]

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?

 

[pmb_phy]

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

 

[Hurkyl]

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).

 

[pmb_phy]

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

 

[robphy]

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.