Rest length in general relativity

[MeJennifer]

I think that when they say that the velocity vector (of the dust - or body in this thread) is orthogonal to the t=const surfaces, this gives a reasonable definition of length.

We are talking in this topic about non-stationary spacetimes right?

 

[gel]

We are talking in this topic about non-stationary spacetimes right?

yes - at least, I am.

 

[pmb_phy]

We are talking in this topic about non-stationary spacetimes right?

MJ - Perhaps you didn't see the question that I had asked you above.

I had posted the following comment

MeJennifer was referring to non-stationary spacetimes. A body in such a spacetime would, in general, have its rest length a function of time because the spatial contraction would be a function of time.

But then I realized that I was trying to speak for you when I really didn't know for sure what your concerns were. So I asked you

Ummmm ... at least this is what I think she was referring to. Is this correct MJ?

But you never responded. Can you explain to me why you were referring to non-stationary spacetimes and why you have concerns about defining rest length? I would very much appreciate it. It would help me in responding to your posts. Thank you.

Pete

 

[MeJennifer]

why you have concerns about defining rest length?

In non stationary spacetimes there is no notion of rest not even as a linear function of time because, as like you said, time becomes an integral part of the non-linear equations in non-stationary spacetimes.

 

[pmb_phy]

I never said the new coordinate system was a new "inertial frame", in fact I specifically said it wasn't in post #38 where I suggested I could provide a flat spacetime example:..

Then I'm confused as to what you meant when you wrote original inertial frame. Why did you use the term "original"? Some coordinate transformations merely change the way that events are labeled. E.g. a switch in spatial coordinates from Cartesian coordinates to spherical coordinates is an example of that. Then there are coordinate transformations which correspond to a change in the relative motion of the observer's frame of reference. A Lorentz transformation from one Lorentz frame (i.e. one inertial frame) to another is just such a transformation. What kind of coordinate transformation do you think that your example represents. To me it's the relabeling of events.

Pete

 

[pmb_phy]

In non stationary spacetimes there is no notion of rest not even as a linear function of time because, as like you said, time becomes an integral part of the non-linear equations in non-stationary spacetimes.

I don't understand what you mean by "no notion of rest". If I had a spring in my hand whose length was oscillating but for which the center was at rest then, at least to me, it can be said to be at rest. It will just have a changing length.

So I take it that you view this that if somethings size is changing then it can't be said to be at rest, even if the center of mass was at rest or even if the center of the spring example above was nailed to a desk top. Is that correct?

Pete

 

[JesseM]

Then I'm confused as to what you meant when you wrote original inertial frame. Why did you use the term "original"?

I meant the original frame, which was an inertial one, as opposed to the second frame, which was non-inertial. As an analogy, if I contrasted an adult chicken with the "original egg it hatched from", this wouldn't imply the chicken was an egg.

Some coordinate transformations merely change the way that events are labeled. E.g. a switch in spatial coordinates from Cartesian coordinates to spherical coordinates is an example of that. Then there are coordinate transformations which correspond to a change in the relative motion of the observer's frame of reference. A Lorentz transformation from one Lorentz frame (i.e. one inertial frame) to another is just such a transformation. What kind of coordinate transformation do you think that your example represents. To me it's the relabeling of events.

You are still ignoring my point, which is about coordinate systems in curved spacetime rather than flat spacetime. Do you still think the distinction between "coordinate transformations which correspond to a change in the relative motion of the observer's frame of reference" and "coordinate transformations which merely change the way that events are labeled" makes sense in this context? 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? If we do a coordinate transformation on Schwarzschild coordinates based on some arbitrary function, then aside from aesthetic tastes there is no reason to consider the new coordinate system (or its definition of simultaneity) any less physical than the Schwarzschild coordinate system.

 

[pmb_phy]

You are still ignoring my point, which is about coordinate systems in curved spacetime rather than flat spacetime.

I'm not ignoring your point intentionally. I think what happened is a miscommunication. I had assumed that you started with a an inertial frame of reference in flat spacetime. If so then its impossible to use a coordinate transformation to obtain a curved spacetime. So what is this curved spacetime that you started out in?

Let us straighten this out before moving on.

By the way, did you check to see if the Jacobian of the transformation was non-zero? This is a requirement if it is to be a valid coordinate transformation. I suspect that its zero, i.e. that its not a valid coordinate transformation. Don't quote me on that just yet!

Pete

 

[pmb_phy]

ok, let me have a go instead.
Suppose we have flat space and (t,x,y,z) are standard inertial coordinates.
Suppose that there is a long and thin rod at rest with respect to these coords, at y=z=0 and 0 <= x <= 1, so it has length 1.
Change to coords (t',x',y',z') with x'=x, y'=y, z'=z and t' = t + 0.6 x/c.
Note, the rod is still at rest in the new coord frame, and occupies the same region of space. However, proper length (s) along a curve with y=z=0 is,
<br /> ds^2 = dx^2-c^2dt^2 = (dx&#039;)^2 - (c dt&#039;-0.6 dx&#039;)^2<br />
Along a slice of constant t', use dt'=0 to get
<br /> ds^2 = (dx&#039;)^2 - 0.36 (dx&#039;)^2=(0.8 dx&#039;)^2<br />
So, the length of the rod measured using the (t',x',y',z') coord system is 0.8, not 1.

You do realize that you must also transform the components of the metric when you do a coordinate transformation, right? This may be the reason that you got the wrong result. However I'm not sure of this because I have a sneaking suspicion that this transformation leaves the metric unchanged. Not sure. Check this out if you have the time.

Pete

 

[JesseM]

I'm not ignoring your point intentionally. I think what happened is a miscommunication. I had assumed that you started with a an inertial frame of reference in flat spacetime. If so then its impossible to use a coordinate transformation to obtain a curved spacetime. So what is this curved spacetime that you started out in?

My example did not use any curved spacetime, because as I said before, I don't have the knowledge of the math of GR to come up with a numerical example involving curved spacetime. The example was just illustrating how you could have two different coordinate systems such that a given object was at rest in both (i.e. every part of it was maintaining a constant position coordinate as the time coordinate varies), but they had different surfaces of simultaneity (i.e. different surfaces of constant t); the idea was that you could do the same thing in GR, and that in the GR case there would be no physical reason to prefer either of the two coordinate systems. Also, I imagine that whatever formula you're using to calculate "rest length" given a coordinate system with accompanying metric in curved spacetime (to be clear, are you agreeing with jostpuur's formula or are you using a different one?) could also be applied to the two coordinate systems in flat spacetime I offered; if it turned out that the formula yielded different answers for the "rest length" in these two coordinate systems, then that would make it plausible that the same would be true if you applied the same formula to two different coordinate systems in a curved spacetime.

 

[pmb_phy]

JesseM - Quick question; Is it your contention that the quantity ds², whose square root is the proper distance between two points, is coordinate dependent? I.e. that a coordinate transformation will change its value?

Pete

 

[Fredrik]

You do realize that you must also transform the components of the metric when you do a coordinate transformation, right? This may be the reason that you got the wrong result. However I'm not sure of this because I have a sneaking suspicion that this transformation leaves the metric unchanged. Not sure. Check this out if you have the time.

You're right. The coordinate change t'=t+vx, x'=x changes the components of the metric to

\begin{pmatrix}-1 &amp;&amp; -2v\\0 &amp;&amp; 1\end{pmatrix}

so

ds^2=-dt^2+dx^2=-dt&#039;^2-2v dt&#039; dx&#039;+dx&#039;^2.

Since dt'=dt=0 along the relevant path, the proper length remains the same. A coordinate change that doesn't change which slices of space-time we consider space can't possibly change the proper length. To change the proper length, you have to change what path to use in the integral.

Proper length is only ambiguous when it's not clear which slices of space-time we should think of as space at different times.

 

[JesseM]

JesseM - Quick question; Is it your contention that the quantity ds², whose square root is the proper distance between two points, is coordinate dependent? I.e. that a coordinate transformation will change its value?

Pete

That's a very good question, and thinking about it helps me clarify the argument I'm making. In curved spacetime ds would normally refer to an infinitesimal line element, but if you take a specific spacelike curve between two events, you can integrate ds and get a coordinate-independent length, yes? So certainly if you pick a particular spacelike curve which is bounded on one end by an event on the worldline of the left edge of an object, and bounded on the other end by an event on the worldline of the right edge of an object, then you will get a unique answer for the length along that particular curve. The problem is that in curved spacetime you can pick an infinite variety of spacelike curves whose ends correspond to events on opposite sides of the object, and for each spacelike curve, you can find a coordinate system in which the object is at rest and where every event on the spacelike curve has the same time-coordinate. So if "length" is understood as the spatial distance from one end to another at a particular instant in time, in GR this doesn't pick out a unique curve precisely because there's an infinite variety of coordinate systems which define "instant in time" (i.e. simultaneity) differently.

 

[Fredrik]

What I said in #72 can't be right. The metric must be symmetric. I don't have time to figure out what I did wrong right now, but I will in a couple of hours.

 

[Fredrik]

I forgot to transpose a matrix in the calculations. Now I get

g=\begin{pmatrix}-1+v^2 &amp;&amp; -v\\-v &amp;&amp; 1\end{pmatrix}

so that

ds^2=-dt^2+dx^2=-(1-v^2)dt&#039;^2-2v dt&#039;dx&#039;+dx&#039;^2.

 

[Mentz114]

Jessem,

The problem is that in curved spacetime you can pick an infinite variety of spacelike curves whose ends correspond to events on opposite sides of the object, ...

If the length is defined as the time taken for light to travel the path, then you integrate along the null geodesic, and there is no ambiguity in the choice of integral.

M

 

[JesseM]

Jessem,


If the length is defined as the time taken for light to travel the path, then you integrate along the null geodesic, and there is no ambiguity in the choice of integral.

M

If you integrate ds along a null geodesic isn't it always zero?

 

[Mentz114]

If you integrate ds along a null geodesic isn't it always zero?

We won't be integrating the proper interval, only the spatial or temporal part, they being equal in the case of light. Either will give us the time and distance. Since we measure times, we will integrate the temporal part between our readings, t_0 and t_1.

Am I making sense ?

M

 

[JesseM]

We won't be integrating the proper interval, only the spatial or temporal part, they being equal in the case of light. Either will give us the time and distance. Since we measure times, we will integrate the temporal part between our readings, t_0 and t_1.

But unlike ds, integrating only the spatial part would be coordinate-dependent, no?

 

[Mentz114]

But unlike ds, integrating only the spatial part would be coordinate-dependent, no?

I don't know, but yes I suspect it. I'll work on it when my stock of old envelopes is restored.

M

 

[gel]

You do realize that you must also transform the components of the metric when you do a coordinate transformation, right? This may be the reason that you got the wrong result.

The result was right. The metric components are different expressed in terms of x',t' than in x,t, as I wrote. The length is different because the path with t=0 as different from the path with t'=0.
...I think this is what Frederick was saying in post #72

 

[snoopies622]

Suppose we have flat space and (t,x,y,z) are standard inertial coordinates...Change to coords (t',x',y',z') with x'=x, y'=y, z'=z and t' = t + 0.6 x/c.

Is this legitimate? It looks like if \frac{dx}{dt}=c then \frac{dx&#039;}{dt&#039;}=\frac{c}{1.6}

Isn't the speed of light invariant in a flat space?

 

[pmb_phy]

The result was right. The metric components are different expressed in terms of x',t' than in x,t, as I wrote. The length is different because the path with t=0 as different from the path with t'=0.
...I think this is what Frederick was saying in post #72

How did the path change? I had the chance to sit down with pen and paper and do the math yet but I will get around to it. Perhaps tonight!

Pete

 

[pmb_phy]

ok, let me have a go instead.
Suppose we have flat space and (t,x,y,z) are standard inertial coordinates.
Suppose that there is a long and thin rod at rest with respect to these coords, at y=z=0 and 0 <= x <= 1, so it has length 1.
Change to coords (t',x',y',z') with x'=x, y'=y, z'=z and t' = t + 0.6 x/c.
Note, the rod is still at rest in the new coord frame, and occupies the same region of space. However, proper length (s) along a curve with y=z=0 is,
<br /> ds^2 = dx^2-c^2dt^2 = (dx&#039;)^2 - (c dt&#039;-0.6 dx&#039;)^2<br />
Along a slice of constant t', use dt'=0 to get
<br /> ds^2 = (dx&#039;)^2 - 0.36 (dx&#039;)^2=(0.8 dx&#039;)^2<br />

So, the length of the rod measured using the (t',x',y',z') coord system is 0.8, not 1.

Okay. I'm working on this problem tonight.

The proper length of the rod, d\sigma, is related to the spacetime interval ds² by

d\sigma = \sqrt{-ds^2}

Recall that the proper length of the rod is 10, not 1.

I figured out what the problem is. And I'm sooo embarrassed! The problem here is in the false assumption that the value of spacetime interval will remain invariant under all coordinate transformations. That is incorrect. The spacetime interval is a Lorentz invariant. This means that its value remains unchanged upon a Lorentz transformation. "diffeomorphism invariance" refers to the fact that tensor equations retain their form under an arbitrary coordinate transformation. It does not mean that scalars remain unchanged under arbitrary coordinate transformations. The spacetime interval remains invariant under Lorentz transformations. That doesn't mean that it remains invariant under an arbitrary coordinate transformation. E.g. if you employed a Galilean transformation then there's no reason to expect that the spacetime interval will remain unchanged. (sigh) I should have know that! I guess I'm getting old! :)

Pete

 

[Fredrik]

Is this legitimate? It looks like if \frac{dx}{dt}=c then \frac{dx&#039;}{dt&#039;}=\frac{c}{1.6}

Isn't the speed of light invariant in a flat space?

It's equal to c in all inertial frames, but this isn't a Lorentz transformation, so the new coordinate system isn't an inertial frame. (Yes, it's legitimate).

 

[Fredrik]

The result was right. The metric components are different expressed in terms of x',t' than in x,t, as I wrote. The length is different because the path with t=0 as different from the path with t'=0.
...I think this is what Frederick was saying in post #72

Actually when I looked at your transformation and saw that it only changes the time coordinate, I assumed (incorrectly) that it would tilt the t axis and leave the spatial axes unchanged. That's why I expressed myself the way I did. I thought it was obvious that the path doesn't change here (again, incorrectly), so I thought that by calculating the line element to be "something*dt'+something*dt'dx'+1[/color]*dx'", I was proving that the length doesn't change. Of course the calculation was wrong too. Not just the one in #72, but also my "correction" in #75.

I found the mistake today, and corrected it. Now I'm pretty sure I have the correct answer. I haven't had anyone else look at it, but at least the result comes out the same regardless of whether I'm using the primed or unprimed coordinates.

The coordinate change you suggested is

\begin{pmatrix}t&#039;\\x&#039;\end{pmatrix}=\Lambda\begin{pmatrix}t\\ x\end{pmatrix}

with

\Lambda=\begin{pmatrix}1 &amp;&amp; v\\0 &amp;&amp; 1\end{pmatrix}

and v=0.6. (I'm setting c=1). Assume that the world lines of the endpoints of the rod in the unprimed system are just the lines x=0 and x=1. When we say that the length of the rod is 1, we really mean that the proper length along a line of constant t from x=0 to x=1 is 1. The new coordinate system treats a different set of "slices" of space-time as "space". A line of constant t' is a line of constant t+vx. So put t+vx=k and find out the k of the line of constant t' that goes through t=0,x=1 (an arbitrary point on the world line of the right end of the rod). It's obviously v. So the line is t=-vx+v and it intersects the world line of the left end of the rod (x=0, remember) at t=v,x=0.

So the new path is a straight line from t=v,x=0 to t=0,x=1. Since it's a straight line, we don't have to do an integral. We can just write dt=0-v=-v, dx=1-0=1 and ds^2=-dt^2+dx^2=-v^2+1[/tex]. So the proper length along a path of constant t&#039; is 0.64 (if we take v to be 0.6).<br /> <br /> We can do this in the primed coordinates too. The components of the metric in the new frame are<br /> <br /> g&amp;#039;=(\Lambda^{-1})^T\eta\Lambda^{-1}=\begin{pmatrix}-1 &amp;amp;&amp;amp; v\\v &amp;amp;&amp;amp; 1-v^2\end{pmatrix}<br /> <br /> so the line element can be expressed as<br /> <br /> ds^2=-dt^2+dx^2=-dt&amp;#039;^2+2vdt&amp;#039;dx&amp;#039;+(1-v^2)dx&amp;#039;^2.<br /> <br /> The coordinates of the endpoints of the path are just t&#039;=v,x&#039;=0 and t&#039;=v,x&#039;=1, so dt&#039;=0 and dx&#039;=1, and that gives us the same result as before.

 

[Fredrik]

I think the coordinate change that gel came up with shows very effectively that proper length is a problematic concept even in Minkowski space. We don't have to consider shape-shifting objects in non-stationary curved space-times to see that there's a problem. The problem is present even in special relativity, even when every part of the object is stationary at all times in some inertial frame.

The question is, is there a way around the problem? In this particular case there is. It makes sense to define the proper length to be the integral of \sqrt{-ds^2} along a path of constant time in the inertial frame where every part of the object is stationary.

If the different parts of the objects are moving so that there's no such inertial frame, then choose one where the center of mass is stationary.

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.

 

[JesseM]

I have a question which might be relevant to the issue of whether there's any meaningful way to define "rest length" in general relativity. If we have two points with a spacelike separation, is it meaningful to talk about a spacelike "geodesic" between them? It's not obvious to me that the concept of spacelike geodesics would make sense. Thinking about flat spacetime, if you draw a straight line between two events with a spacelike separation, which would presumably be the geodesic if one exists at all, I don't think this path would minimize the value of ds integrated along it--consider that if you have two points A and B with a spacelike separation, you can find a third point C such that A and B both lie on the surface of C's past light cone, and a path which went along the light cone from A to C and then back from C to B should have 0 length when you integrate ds along it, and it's possible to find spacelike paths which are arbitrarily close to this path. 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.

But I don't know, I may be thinking about this wrong. If it is possible to find a unique spacelike geodesic path between two events with a spacelike separation, then maybe we could define the "rest length" at a given moment by picking an event A on the worldline of one end of the object, and then finding the event B on the worldline of the other end such that the length of the geodesic path between A and B is maximized (in analogy with the flat spacetime case where the length of an object is maximized in its own rest frame).

 

[Fredrik]

(I haven't tried to find a rigorous argument, but...) I don't think "spacelike geodesic" makes sense. You probably have to slice up a region of spacetime into spacelike hypersurfaces first, and then you can talk about geodesics in each slice.

That "slicing" is of course exactly what a coordinate system does, and there are always many possible choices. I would be surprised if there's a "natural" choice even for a steel rod, and very surprised if there's a natural choice for a spinning soft rubber bag that's shaped like the statue of liberty and is half filled with water.

 

[pmb_phy]

I have a question which might be relevant to the issue of whether there's any meaningful way to define "rest length" in general relativity. If we have two points with a spacelike separation, is it meaningful to talk about a spacelike "geodesic" between them?

Absolutely. A geodesic in flat spacetime, in Lorentz coordinates, is a straight line. Likewise, any straight line is a geodesic. This is readily seen if one considers the requirement of being a geodesic; that the "length" of the worldline be stationary. When this is the case the geodesic equation holds along that worldine. Consider the geodesic equation; in flat spacetime in Lorentz coordinates all of the Christoffel symbols will be zero. This will yield the equation of a line.

It's not obvious to me that the concept of spacelike geodesics would make sense. Thinking about flat spacetime, if you draw a straight line between two events with a spacelike separation, which would presumably be the geodesic if one exists at all, I don't think this path would minimize the value of ds integrated along it--...

It actually does minimize the length of the path.

consider that if you have two points A and B with a spacelike separation, you can find a third point C such that A and B both lie on the surface of C's past light cone, and a path which went along the light cone from A to C and then back from C to B should have 0 length when you integrate ds along it, and it's possible to find spacelike paths which are arbitrarily close to this path. 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.

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.

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

Pete