Rest length in general relativity

[Hurkyl]

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

What's wrong with the obvious definition?

 

[Fredrik]

What's wrong with the obvious definition?

I don't know. Maybe nothing. What's the obvious definition?

This is what I'm thinking right now:

In 1+1 dimensional Minkowski space, a geodesic between space-like separated events A and B can be defined as the curve from A to B that maximizes the integral of $\sqrt{dx^2-dt^2}$ along the path. This is a straight line, because every other path gets a bigger contribution from -dt², and that makes the result smaller.

This definition seems to make sense in 1+1 dimensions, but if we add more spatial dimensions, the integral can be made as large as we want by having the path take a long detour into the other dimensions. So it doesn't seem to make sense in 3+1 dimensions. Maybe there's a way around it, but I don't see one right now.

 

[Hurkyl]

The definition you quoted actually states that a geodesic is a local optimum: it isn't required to be a global optimum of the arclength function. For a nice example embedded in Euclidean space, consider the sphere: there are exactly two geodesics between any two (non-antipodal) points on the sphere. Their union is the great circle passing through those two points. One of them is the shortest path between the two points, the other is not. (But it is a local minimum: any local perturbation gives a longer path)

Incidentally, I was thinking about parallel transport: a geodesic is a curve whose tangent vector has length 1, and remains a tangent vector under parallel transport.

Either way, my point is that nothing breaks down in the definition of geodesic here.

Edit: ignore this post, read this one instead.[/color]

 

[JesseM]

You're using Euclidean intuition in a Minkowski geometry. You can't determine the length of a worldine merely by looking at how long it is. E.g. for a time like worldline a squiggly path connecting two events has a smaller value of proper time than a straight worldline between the two.

But I wasn't considering the length of just any path by looking at how long it was, I was specifically considering a spacelike path which lies in a single SR surface of simultaneity (the same surface that the straight-line path lies in). Is ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 for this path not just equal to the spatial length of the path in the frame that uses this definition of simultaneity? After all, in the coordinate system where the path lies on a single surface of simultaneity, dt will always be zero. If that's right, that would be an argument to suggest why a geodesic doesn't maximize ds for spacelike paths, but that's OK because you said it minimized it.

Regarding your example of null geodesics. Its possible to connect any two events on a timelike geodesic by two straight null geodesics.

Yes, but for timelike geodesics, a geodesic is not supposed to minimize $$\sqrt{-ds^2}$$ integrated along it, it's supposed to maximize it. My argument about connecting two events by a path consisting of two null geodesics was meant to show why I didn't think ds integrated along a spacelike path would be minimized. But maybe the answer has something to do with Hurkyl's distinction between being a "local optimum" and a "global optimum of the arclength function".

In any case, it occurs to me that even if you can find a unique geodesic between any two spacelike separated events, there is a problem with defining the length of the object at the time of event A on the worldline of one end by finding the event B on the other end such that the length of the geodesic between them is maximized. Although this definition would correspond to the rest length for a rigid object in SR, I don't think it'd make much sense for an object in SR whose length is changing; so in curved spacetime this definition would at best only be useful in the case where the object was rigid (the internal stresses at any point inside the object don't change over time), and where you could pick any event on the worldline for A and the "length" would always be the same. This might work in a static spacetime but I'd guess it wouldn't work otherwise.

 

[jostpuur]

In general relativity, we cannot even define what it means for things at two different locations in space-time to be at rest with each other, so "observer at rest with the mesaured object" is, strictly speaking, nonsense.

What justification do you have for this assertion?

I suppose it's the usual: You need to do parallel transporting to compare velocities at different space time points. The relative velocity depends on the chosen path of parallel transport.

This is altogether different. Parallel transport has nothing to do with proper distance.

We need the rest frame in order to choose the spatial path along which the proper distance is calculated. We need to know relative velocities to solve the rest frame, and then we need parallel transport to solve relative velocities.

 

[jostpuur]

general relativity question about inertial frames

This isn't like SR where there is a preferred way to construct an observer's "rest frame" based on the fact that there's a special set of coordinate systems where the laws of physics take the same form, in GR all coordinate systems are on equal footing as far as physics is concerned, no?

I've understood that the theory of gravitation is formulated so that all, also non-inertial, coordinate sets are equal, but I wouldn't be so sure about all physics. If we want to do stuff with electromagnetism, shouldn't we use something like inertial frames? I'm guessing: Wouldn't frames where geodesic paths are straight lines be the inertial frames?

 

[MeJennifer]

In a Minkowski spacetime a geodesic is defined as the longest path between two events not the shortest path. In a Lorentzian spacetime the principle is similar but in some cases of curvature the longest path is only locally the longest path. Lightlike geodesics, also called null geodesics, have a zero length in spacetime.

Note that spacetime is more a chronometric than a geometric description of reality. Distances are derived in GR, they are not primary.

 

[JesseM]

In a Minkowski spacetime a geodesic is defined as the longest path between two events not the shortest path.

This is true for timelike paths, but pmb_phy seems to say it's the opposite for spacelike paths. And as I said, it seems to me that if you look at two events with a spacelike separation and draw a squiggly path between them which lies entirely in the surface of simultaneity which contains both, this will have a greater spatial length (in the inertial coordinate system which defines simultaneity this way) than a straight-line path (which also lies within this surface of simultaneity). And in the inertial coordinate system where this surface of simultaneity has constant t, then dt is going to be 0 for every line element on the path, so ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 reduces to ds^2 = dx^2 + dy^2 + dz^2, so the spatial length of the path in this coordinate system is the same as ds integrated along the path.

 

[pmb_phy]

But I wasn't considering the length of just any path by looking at how long it was, ...

I was referring to the statement you made, i.e.

a squiggly path between two points has a greater length than a straight-line path.

What did you mean by "longer"?

I was specifically considering a spacelike path which lies in a single SR surface of simultaneity (the same surface that the straight-line path lies in). Is ds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 for this path not just equal to the spatial length of the path in the frame that uses this definition of simultaneity?

The proper distance is defined for such a path, yes.

After all, in the coordinate system where the path lies on a single surface of simultaneity, dt will always be zero. If that's right, that would be an argument to suggest why a geodesic doesn't maximize ds for spacelike paths, but that's OK because you said it minimized it.

To be precise, a geodesic is a worldline for which "s" has a stationary value.

Yes, but for timelike geodesics, a geodesic is not supposed to minimize $$\sqrt{-ds^2}$$ integrated along it, it's supposed to maximize it.

I was merely giving you an example of a timelike geodesic for which events can be connected by two null worldlines.

Pete

 

[gel]

So the proper length along a path of constant t' is 0.64 (if we take v to be 0.6).

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 :/

Things get a lot more complicated in GR, so I'm not at all convinced that there's a way to get around the problem there. I would have to see a proof to believe that there is.

For a one dimensional object like a long thin rod or a piece of string, then you can come up with a definition of its length.
Each point on the object has a 4-velocity (tangent to its world line). Given any space-time point in the world-sheet of the object, you can pass a space-like curve through it such that its tangent vector remains orthogonal to the local 4-velocity of the object.
As long as the object is smoothly embedded in space, the existence and uniqueness of the curve will follow from uniqueness of solutions to first order diff eqns.

Just use the proper length to such a curve to define the length of your rod/string. Of course, it can still vary in time but that's unavoidable.

 

[gel]

This is true for timelike paths, but pmb_phy seems to say it's the opposite for spacelike paths. And as I said, it seems to me that if you look at two events with a spacelike separation and draw a squiggly path between them which lies entirely in the surface of simultaneity which contains both, this will have a greater spatial length...

A spacelike geodesic neither minimizes nor maximizes the length, even locally. Which you can see by perturbing it in either a timelike or spacelike direction. However its length will be stationary (to first order under small perturbations). In any case, a geodesic is defined more generally as a curve which parallelly transports its own tangent vector.

 

[pmb_phy]

A spacelike geodesic neither minimizes nor maximizes the length, even locally. Which you can see by perturbing it in either a timelike or spacelike direction. However its length will be stationary (to first order under small perturbations). In any case, a geodesic is defined more generally as a curve which parallelly transports its own tangent vector.

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.

Pete

 

[gel]

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.

when I said "more generally" I was referring to the fact that geodesics only require the concept of parallel transport to be defined. This can be defined by a metric, but only requires the existence of a connection, which makes it the more general definition.

 

[pmb_phy]

when I said "more generally" I was referring to the fact that geodesics only require the concept of parallel transport to be defined. This can be defined by a metric, but only requires the existence of a connection, which makes it the more general definition.

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

 

[JesseM]

I was referring to the statement you made, i.e.

a squiggly path between two points has a greater length than a straight-line path.

What did you mean by "longer"?

I think if you looked at the context of that quote, you can see I was talking about a squiggly path through the surface of simultaneity which contained both events:

On the other hand, the straight-line path in flat spacetime also doesn't seem to maximize the value of ds integrated along it, since the value of ds integrated along a spacelike path is just the length of the path in the surface of simultaneity that contains it, and obviously in a given surface of simultaneity, a squiggly path between two points has a greater length than a straight-line path.

In that context, I just meant having a longer spatial length in the coordinate system which used that definition of simultaneity.

To be precise, a geodesic is a worldline for which "s" has a stationary value.

Does this mean that all small perturbations to the path change s in the same way, i.e. for a given path, either all small perturbations increase s, or else all small perturbations decrease s? (as suggested by Chris Hillman's post here) If so, is it possible to come up with examples of timelike paths where all small perturbations increase s (increase the proper time), or examples of spacelike paths where all small perturbations decrease s? Or do all timelike geodesics maximize the proper time with respect to small perturbations, and all spacelike geodesics minimize the length with respect to small perturbations? This review paper does seem to say that spacelike geodesics minimize s in some sense, in section 2.2, if I'm interpreting the language correctly:

2.2. Special properties of geodesics in spacetimes depending
on their causal character. We will mean by co–spacelike any geodesic such
that the orthogonal of its velocity is a spacelike subspace at each point, that is: all
the geodesics in the Riemannian case and timelike geodesics in the Lorentzian one.

...

Timelike and co–spacelike geodesics. It is well known that conjugate points
along a timelike (resp. Riemannian) geodesic in a Lorentzian (resp. Riemannian)
manifold cannot have points of accumulation. Even more:

(1) Any timelike geodesic maximizes locally d in a similar way as any Riemannian
geodesic minimizes locally its corresponding d. Nevertheless, there are two
important differences:

– Riemannian geodesics minimize locally length among all the smooth
curves connecting two fixed points p, q. Nevertheless, the timelike ones
maximize only among the causal curves connecting p, q;

Yes, but for timelike geodesics, a geodesic is not supposed to minimize $$\sqrt{-ds^2}$$ integrated along it, it's supposed to maximize it.

I was merely giving you an example of a timelike geodesic for which events can be connected by two null worldlines.

You didn't give an example, just stated that it would be possible to do so--but anyway, I agree (you just need an event C between A and B such that A is on the past light cone of C and B is on the future light cone of C). Still, as I said, my argument was trying to show that there could always be a path with smaller s, whereas for timelike geodesics my understanding was that the geodesic maximizes s, at least compared to small perturbations. But perhaps the answer here is that this 0-s path is not itself a spacelike path, and it's possible to find a separate spacelike path which is minimal with respect to small perturbations?

 

[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.