Is coordinate speed affected by gravitational waves?

[Haorong Wu]

TL;DR Summary

Why a coordinate speed for light which is not equal to one does not contradict relativity?

As in Bernard Schutz's A first course in general relativity, page 220, we suppose a gravitational wave travels in the z-direction with pure "+" polarization, so that the metric in the TT coordinate system is given by$$ds^2=-dt^2+[1+h_{+}(z-t)]dx^2+[1-h_{+}(t-z)]dy^2+dz^2 .$$ Suppose that two objects lie on the x-axis, one of them at the origin $x=0$ and the other at coordinate location $x=L$. Now a photon from the origin traveling along the x-axis moves along a null world line so $$ds^2=0, dy=dz=0.$$That means the effective speed of the photon is $$(\frac {dx} {dt})^2=\frac 1 {1+h_{+}}$$.
Now, the book reads that, although this is not equal to one, this is just a coordinate speed, so it does not contradict relativity.

I can not understand this statement.

First, what is coordinate speed? According to Wikipedia, coordinate speed is the coordinate distance measured by the observer divided by the coordinate time of the observer. In my opinion, in the present of gravitational wave, the distance between the two objects will change as the wave passes them. So the effective speed will not be one. But why this does not contradict relativity? Does this means that this TT coordinate system is not a inertial frame, so the speed of the light could be not equal to one?

Second, I am still confused by coordinate transform. Suppose if there is not wave, the distance between the two objects is 2m, so the light would need time of 2m to travel from one object to the other one. And suppose when the wave passes the objects, the distance between them are contracted to 1m, then the light would need only time of 1m to travel. Then let the wave disappear so that the distance between them is restored to 2m. But does that means the light travel in a speed higher than 1? I think I used a wrong coordinate system to describe the situation so the contradiction appears.

 

[Nugatory]

First, what is coordinate speed? According to Wikipedia, coordinate speed is the coordinate distance measured by the observer divided by the coordinate time of the observer.

Yes, that is exactly what it is. You make it come out to be anything you want just by choosing coordinates that give you the answer you're looking for. For example...

Let's use the non-inertial coordinates in which I am at rest (coordinate speed zero) on the surface of the rotating earth. Using these coordinates, Alpha Centauri goes around me in a circle with a radius of 4.3 light-years once every 24 hours. That's a distance of about 27 lightyears every day and 9850 lightyears every year - the coordinate velocity using these coordinates is 9850 times the speed of light.

Or I could use coordinates in which the center of mass of the sun is at rest. Using these coordinates, the position of Alpha Centauri changes very little with time so the coordinate velocity is close to zero.

 

[Dale]

Summary:: Why a coordinate speed for light which is not equal to one does not contradict relativity?

But why this does not contradict relativity? Does this means that this TT coordinate system is not a inertial frame, so the speed of the light could be not equal to one?

That is right. The second postulate only claims that the one way speed of light is c in an inertial frame. It can be any arbitrary value in non inertial coordinates.

 

[pervect]

The coordinate speed is just dx/dt, the derivative of the position x with respect to time.

I am not sure if you've already observed that "x" is not a distance, because the distance is $\sqrt{\left( 1+h_+ \right)\,\Delta x}$?

Thus $dx/dt$ is the rate of change of a coordinate with respect to time, not the rate of change of a distance with respect to time.

 

[Haorong Wu]

Thanks, guys.

@ Nugatory So the coordinate system at rest on the Earth is not a inertial frame, and the speed measured in this system is not the proper speed. But is the second coordinate system an inertial frame? I am not sure about this. the sun is moving around the center of our galaxy and the Alpha Centauri is not within our solar system. So could this frame still be regarded as an inertial frame?@pervect I am sorry I have not noticed the difference between distance and coordinate. So suppose there is a rod. If the spacetime around the rod is curved, then the coordinate I read in the new system changes, but the real length of the rod does not alter? That sounds like distance is the proper length? I think I may catch something here.

 

[PeroK]

Thanks, guys.

@ Nugatory So the coordinate system at rest on the Earth is not a inertial frame, and the speed measured in this system is not the proper speed. But is the second coordinate system an inertial frame? I am not sure about this. the sun is moving around the center of our galaxy and the Alpha Centauri is not within our solar system. So could this frame still be regarded as an inertial frame?@pervect I am sorry I have not noticed the difference between distance and coordinate. So suppose there is a rod. If the spacetime around the rod is curved, then the coordinate I read in the new system changes, but the real length of the rod does not alter? That sounds like distance is the proper length? I think I may catch something here.

You're on page 220 of Schutz and studying gravitational waves, but you seem to have skipped the basics of GR.

 

[Ibix]

Imagine walking across a piece of graph paper. Your coordinates can be read off the graph paper at any time, and your coordinate speed is the rate at which you cross lines on the paper.

Now walk across logarithmic graph paper. If you move in the positive direction then the number of lines you cross per second is proportional to $10^{vt}$.

Using non-homogeneous coordinates like that in flat spacetime would be a bit odd. But curved spacetime forces you to use coordinate systems like that all the time, which is why you must use the metric to relate coordinate changes to distances.

 

[PeterDonis]

So suppose there is a rod. If the spacetime around the rod is curved, then the coordinate I read in the new system changes, but the real length of the rod does not alter?

"changes" compared to what? "does not alter" compared to what?

You seem to have a mental picture where we start with the rod and spacetime being flat, and record some numbers, and then we make spacetime curved and see if the numbers change. That's not how it works. In the presence of matter, spacetime is curved, period. It doesn't "start out flat and then curve". It just is curved. There is no way to compare "how things are when spacetime is curved" with "how things are when spacetime is not curved".

 

[Haorong Wu]

Thanks, again.

@PeroK Well, I am sad about that. I have gone through Schutz's book. I can solve most of the problems. But I always feel I do not grasp something important so that the theory of GR does not come as a whole to me. However, at least now I know that the distance is the proper length and how the gravitational wave induces a system of spring and particles to oscillate.

@Ibix Great example. I understand the differences and relations between coordinates and distances better now.

@PeterDonis Thanks! Maybe I misunderstood the example in Schutz's book. It suppose that there is a system of two particles connected by a spring. Then a gravitational wave passes it. The metric is changing due to the wave so the proper length of the spring changes as well. Then the system starts to oscillate. I thought this is a situation where spacetime is changed from flat to slightly curved. Where did I make mistake?

 

[PeroK]

@PeroK Well, I am sad about that. I have gone through Schutz's book. I can solve most of the problems. But I always feel I do not grasp something important so that the theory of GR does not come as a whole to me. However, at least now I know that the distance is the proper length and how the gravitational wave induces a system of spring and particles to oscillate.

One of the most fundamental things is that light travels on null paths. And that, in general relativity, is the generalisation of the invariant speed of light (across IRF's) from special relativity.

As a simple example we can take a null radial path in Schwarzschild coodinates: $$ds^2 = -(1 - \frac {2GM}{r})dt^2 + (1 - \frac {2GM}{r})^{-1}dr^2$$ and $$ds^2 = 0 \ \Rightarrow \ \frac{dr}{dt} = (1 - \frac{2GM}{r})$$ And you can see that $\frac{dr}{dt} \ne 1$.

You need to understand that - it's absolutely fundamental.

Also $ds^2 = 0$ is coordinate-independent; whereas something like $\frac{dr}{dt} = 1$ or $\frac{dx^1}{dx^0} = 1$ cannot be a law of physics, as it is not covariant.

 

[PeroK]

PS to labour this point. If we take a local observer, stationary at radius $r$, then they measure radial distance according to: $$ds = (1 - \frac{2GM}{r})^{-1/2}dr$$ and the local proper time is: $$d\tau = (1 - \frac{2GM}{r})^{1/2}dt$$ Hence the locally measured speed of light on its null radial path is: $$\frac{ds}{d\tau} = (1 - \frac{2GM}{r})^{-1}\frac{dr}{dt} = 1$$ And there's your locally invariant measurement of the speed of light.

 

[vanhees71]

@PeterDonis Thanks! Maybe I misunderstood the example in Schutz's book. It suppose that there is a system of two particles connected by a spring. Then a gravitational wave passes it. The metric is changing due to the wave so the proper length of the spring changes as well. Then the system starts to oscillate. I thought this is a situation where spacetime is changed from flat to slightly curved. Where did I make mistake?

Don't worry. That's a trap due to reading textbooks overemphasizing the geometric interpretation of GR. One can also understand GR as the relativistic theory to describe the gravitational interaction. As elegant as the "geometry-only approach" is, you loose the feeling for gravity as being an interaction as the other fundamental interactions are. For the more physical approach, see

S. Weinberg, Gravitation and Kosmologie, Wiley&Sons, Inc.,
New York, London, Sydney, Toronto (1972).

 

[PeterDonis]

It suppose that there is a system of two particles connected by a spring. Then a gravitational wave passes it. The metric is changing due to the wave so the proper length of the spring changes as well. Then the system starts to oscillate. I thought this is a situation where spacetime is changed from flat to slightly curved.

Spacetime doesn't "change"; it's a 4-dimensional geometry that already includes all of the "changes" that happen along some particular worldline.

In the particular example you describe, the spring and the particles are assumed to have negligible mass, so whatever spacetime curvature exists is not due to them. It's due to the gravitational wave. So the local metric changes along the worldline of the spring/particle system due to the gravitational wave passing, yes.

In the post of yours that I responded to, however, you were not talking about this example. You were talking about a different scenario, with a rod, and you said nothing about any gravitational wave, so the only possible source of spacetime curvature was the rod itself. That was the scenario I was responding to.

 

[Nugatory]

So the coordinate system at rest on the Earth is not a inertial frame, and the speed measured in this system is not the proper speed.

There is no such thing as a “proper speed”. All speeds are necessarily relative to something else and a speed $\frac{\mathrm{d}x}{\mathrm{d}t}$ calculated using the coordinates of an inertial frame isn't any more physically meaningful than any other coordinate velocity.

But is the second coordinate system an inertial frame? I am not sure about this. the sun is moving around the center of our galaxy and the Alpha Centauri is not within our solar system. So could this frame still be regarded as an inertial frame?

It is a very very very good approximation to one. It would be exact if the sun were not being pulled into a non-inertial circular path by the gravitational attraction of the galaxy, an effect that is well and thoroughly negligible for all practical purposes.

(Note that I am using the language of special relativity: gravity is treated as a force, spacetime is flat, and it is possible to construct global inertial frames. You do not want to take on curved spacetime and general relativity until you understand coordinate velocity using non-rectilinear and non-inertial coordinates in special relativity.)

 

[pervect]

Thanks, guys.

@ Nugatory So the coordinate system at rest on the Earth is not a inertial frame, and the speed measured in this system is not the proper speed. But is the second coordinate system an inertial frame? I am not sure about this. the sun is moving around the center of our galaxy and the Alpha Centauri is not within our solar system. So could this frame still be regarded as an inertial frame?@pervect I am sorry I have not noticed the difference between distance and coordinate. So suppose there is a rod. If the spacetime around the rod is curved, then the coordinate I read in the new system changes, but the real length of the rod does not alter? That sounds like distance is the proper length? I think I may catch something here.

The rod needs to be rigid of course, but then yes. Of course rigid rods are an idealization, actual physical rods are not rigid. There are some subtle issues regarding the idealization of rigidity as well, but it'd probably be better to avoid them for the moment.

 

[PAllen]

Don't worry. That's a trap due to reading textbooks overemphasizing the geometric interpretation of GR. One can also understand GR as the relativistic theory to describe the gravitational interaction. As elegant as the "geometry-only approach" is, you loose the feeling for gravity as being an interaction as the other fundamental interactions are. For the more physical approach, see

S. Weinberg, Gravitation and Kosmologie, Wiley&Sons, Inc.,
New York, London, Sydney, Toronto (1972).

I hope you are aware that Weinberg later regretted the degree to which he de-emphasized the geometric approach in this book.

 

[Haorong Wu]

Thanks, friends! I learned much from your posts. I will need more time to digest them.

But first of all, I can see my mistake is that I forgot we should only consider covariant quantities in GR since they are frame invariant, and the partial derivatives are clearly not covariant. So as to $dr/dt$, I should use their covariant counterpart, i.e., the proper length and proper time and I now get the correct speed for the photon. Cheers!

 

[vanhees71]

I hope you are aware that Weinberg later regretted the degree to which he de-emphasized the geometric approach in this book.

Maybe, but it's good that there exist textbooks with different points of view. Of course, geometry in the sense of Klein's Erlanger Programm underlies all of physics, but that doesn't mean that one should forget about the physical context too.

Weinberg's book is written in 1971. I think at this time the hope was to find a quantum theory of gravitation along the same lines as the gauge theories of the 3 other interactions were successfully quantized. Today we know that this is not the case. Maybe that's why Weinberg regretted the "de-emphasis" of geometry later.