# Meaning of Schwarzschild solution in isotropic/anisotropic coordinates

1. Apr 4, 2013

### Agerhell

According to the Schwarzschild solution in the most common anisotropic (Schwarzschild?) coordinates the proper time and the coordinate time are related as:

$$\frac{dt}{d\tau}=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{|\bar{v}\times\hat{r}|^2}{c^2}-\frac{1}{1-\frac{2GM}{rc^2}}\frac{(\bar{v}\cdot\hat{r})^2}{c^2}}}$$

Now as i read on wikipedia, quote:

I interpret this as that the velocity of light (in coordinate time)according to the Schwarzschild solution in Schwarzschild coordinates is different in different directions and that the velocity of light is: $$c_{coordinate}=c(1-\frac{2GM}{rc^2})$$ vertically and $$c_{coordinate}=c\sqrt{1-\frac{2GM}{rc^2}}$$ horizontally. Correct?

Now, according to wikipedia you can transform to isotropic coordinates to get the speed of light to be the same in all directions. What set of coordinates corresponds to the real physical situation? Is the speed of light isotropic in a gravitational field or is it not?

I am trying to set up an expression for $\frac{d\bar{v}}{dt}$ that is supposed to produce planetary orbits similar to what is exptected from the Schwarzschild solution. To check my expression I must compare with the variant of the Schwarzschild solution with coordinates that correspond to the real life situation. Which set of coordinates is that?

2. Apr 4, 2013

### Staff: Mentor

Where are you getting this from? It has at least one major issue that I can see: you have v and r as 3-vectors with dot and cross products. Relativistic equations should be written in terms of 4-vectors. I'm not sure the 3-vectors you are using (particularly the r one) are even well-defined in a curved spacetime.

Only the coordinate velocity, and coordinate velocities have no direct physical meaning. That's true for both charts; see further comments below.

(Note: Wikipedia is not a good source for this kind of stuff. At best you can sometimes use it as a source of links to references that may be better. It's not that what it says is wrong, exactly--though sometimes it is--but it is often not put in a way that aids understanding.)

Neither one.

Since the only notion of "speed" that has any invariant meaning in a curved spacetime is speed measured locally--in this case, a measurement of the speed of a beam of light that is just passing you--and since such local measurements will give isotropic results in a gravitational field, I think the answer to your question is yes.

3. Apr 4, 2013

### PAllen

You really need to ask how you are going to measure this. Let's say we directly measure the round trip light time over some distance laid out; and we do this in two orientations. If distance is defined as currently, in terms of light properties (and you independently apply the definition in each direction), then you will get isotropic c. If you imagine you have a physical ruler that matches proper length computed using SC simultaneity (which matches radar simultaneity), then over some distance you can get different values than c for the round trip time. I do not know whether the result would be isotropic or not, but some time ago here, Passionflower produced nice graphs of the radial light speed measured this way (and how it diverged from c when measured over 'significant' radial distance).

Since the effect would only be detectable for significant gravitational change between radii, a key question is becomes how real physical rulers behave. Do they match proper distance using radar simultaneity as is typically assumed for 'ideal rulers'? This is a non-trivial question, which I also cannot answer.

Another variant would be to apply e.g. the current light definition to construct e.g. a 10 meter rigid ruler, then turn the ruler to different orientations and measure 2 way light speed with it (in a scenario where gravitational acceleration was detectably different over 10 meters). The ruler would be under different stresses in different orientations. You could define rigidity such that it maintains proper length, but whether that matches a plausible theory of matter in GR is not so simple.

Last edited: Apr 4, 2013
4. Apr 4, 2013

### Staff: Mentor

They do if we assume that the rulers are not significantly affected by proper acceleration: i.e., that the forces induced in the rulers by holding them at rest in a gravitational field, instead of letting them free fall, do not significantly change their length.

We could test this as follows: take two rulers of identical construction. Hold one at rest in the gravitational field. Take the second and toss it upward in free fall from some location well below the first, in such a way that it momentarily comes to rest with its center of mass exactly coincident with the center of mass of the first ruler. If the two rulers exactly overlap at this moment, then the first ruler is not significantly affected by proper acceleration.

If you assume ideal rulers as above, then the speed of radially moving light will be measured differently by observers at different altitudes. This is obvious if we set up a light beam to bounce radially back and forth between two such observers: the round-trip travel time of the same light beam, covering the same radial distance, will be shorter as measured by the observer who is at a lower altitude.

The part I would have to think about would be how to compare this kind of measurement with a measurement taken tangentially. For any proper distance that is significant compared to the size of the central mass, a "tangential" light beam will not be at a constant altitude while it travels. Also its spatial path will bend, which has to be taken into account.

5. Apr 4, 2013

### PAllen

Which is why, on that old thread, nobody dared tackle the tangential case ...

6. Apr 4, 2013

### Agerhell

Well, fellow Recognized science advisor Bill K kind of told me so. He said: (se old thread)

By defining $\bar{v}$ as one usually do the expression in the first post of this thread is what you get. Please inform me if there is a more suitable way of defining v and still use the threedimensional world I prefer to live in...

Well, the coordinate velocity of light sure has physical significance. If you are going to send a light signal from point A (somewhere on earth for instance) to point B (a spacecraft in orbit around one of Jupithers Moons for instance) you want to know the coordinate velocity of light in order to calculate when the light signal will arrive. You cannot just say that the speed of light is always locally c and expect to get the right result.

Last edited: Apr 4, 2013
7. Apr 4, 2013

### Staff: Mentor

He gave you a way to define v as a scalar; that's not the same as defining it as a 3-vector. Also he didn't say anything about r as a 3-vector. Also he pointed out in that same thread that the formula is only valid for circular motion.

I don't see how you get this. I see an expression with just v^2/c^2 under the square root in the other thread (where v here is just a scalar), but I don't see how you are getting from that to the 3-vector expression you wrote. One key thing I'm not sure you have considered is, as I noted before, how to define r as a 3-vector. You can't do it the way I suspect you are thinking you can, because the spacetime is curved, and the definition I suspect you are implicitly using assumes flat spacetime.

Not really. In cases like the one you describe, the effects of spacetime curvature are very small, so everybody's time is very close to coordinate time, and expressions like "when the light signal will arrive" can be interpreted as "the coordinate time at which the light signal will arrive" without too much of a problem. But that's an artifact of the particular scenario; it's not a general principle.

You don't need to ask what the speed of light is at all to solve this problem; you just need to integrate a null geodesic between two points. It's similar to what Bill_K showed you in the other thread, except that the length of a null vector is zero instead of -1 (it's -1 for a timelike object like a planet or an orbiting satellite, which is the case you were discussing in the other thread). You can rewrite the integrand so it has "v" in it if you want, but there's no need to; what allows you to eliminate all the coordinates except one is the fact that there are constants of the motion (E and L), not the fact that you know what "v" is, since you don't anyway--you would have to solve for it in terms of the same constants of the motion. But that's just adding an intermediate step that doesn't need to be added.

8. Apr 4, 2013

### Agerhell

Note that all I did here is to identify $\frac{dr}{dt}=\bar{v}\cdot\hat{r}$ and $|\bar{v}\times\hat{r}|^2=r^2(\frac{d\phi}{dt})^2$. Are these identifications illegal?

What Bill K did whas to say that in the special case of a circular orbit $\bar{v}\cdot\hat{r}=0$ in which case you get a simplified expression.

That is not what I am trying to do. I have an expression for $\frac{d\bar{v}}{dt}$ in ordinary three space that I use in a simple Runge-Kutta step by step procedure to produce orbits. The orbits become very similar to those expected by the Schwarzschild solution in Schwarzschild coordinates but they are not 100 percent identical. When deriving my expression I assumed that $c_{coordinate}=c\sqrt{1-\frac{2GM}{c^2}}$ isotropically in all directions which I assume might be the reason for the slight mismatch. I could try to compensate for an anisotropic speed of light, but it might be difficult.

Therefore I would like to know if the anisotropy of the speed of light is a reality or not. If the Schwarzschild solution in Schwarzschild coordinates produces the correct anomalous perihelion shift but the wrong speed of light, it seems like there is something fishy about it.

9. Apr 4, 2013

### pervect

Staff Emeritus
They don't really make a lot of sense. The biggest problem is that a change in one unit of r does not represent a change in distance of one unit, because of the metric coefficient.

This means that dr/dt isn't really a velocity, because dr doesn't represent a distance. It only represents a coordinate change.

There are some other issues as well. If we had a better idea of what you actually intended to do with your "v", or if we had a better idea of how you proposed to actually measure your v, we might comment further on your proposal.

You might want to ask Bill K if he stands behind your final expression. I suspect there's a bit of a disconnect between what he thinks of defining v appropriately and what you think of as defining v appropriately. (If you read the fine print, he did mention that v did need to be defined appropriately).

I can add that the way *I* would define v would be from the viewpoint of a co-located static observer. This is something that corresponds to a simple actual measurement.

One can sketch out the apparatus needed to make such a local measurement easily.

If, as I suspect, you are focussed on some "gods-eye view at infinity", I think you'll find it difficult to even describe the apparatus and processing needed to measure "v" from what you can observe from at infinity (which would be a bunch of radar signals).

10. Apr 4, 2013

### PAllen

One thing about my example of measuring (two way) light speed with an ideal ruler and a clock is that, even though it is likely anisotropic (over significant distances) and does not match c, it has nothing to do with the coordinate isotropy of SC coordinates. The anisotropy in these coordinates exists for light speed at a single point. Meanwhile, the measurement I describe converges to isotropic c over small distances, and diverges (probably anisotropically) from c the greater the distance over which you measure. Also, the measurement could be computed, and come out the same, in any coordinates; and the proper distance involved does not match coordinate quantities in any of the common coordinates for SC geometry.

Thus: there is a physical sense in which light in SC geometry can be measured to be different from c and anisotropic, but this has no relation to SC coordinate anisotropy.

11. Apr 4, 2013

### Staff: Mentor

That already indicates an issue, because the 3-spaces of constant time in Schwarzschild coordinates are not Euclidean 3-spaces, which is what I think you mean by "ordinary three space". So your procedure for computing orbits already makes at least one false assumption. It may be approximately valid (depending on how accurate you want your results to be), but you appear to be treating it as though it were, or ought to be, exact.

Of course not, since, as above, you are assuming Euclidean 3-space, incorrectly.

It depends on what you are trying to use the "speed of light" for. As I said before, for this type of problem I think you would be better off not even trying to frame it in terms of a single "speed". You can derive separate equations for the rate of change of each spatial coordinate with respect to time; you don't need to have a single "speed" at all.

No, there is something fishy about your assumptions about space, specifically, as I said above, that you are assuming it's Euclidean when it isn't. The speed of light given in Schwarzschild coordinates is not "wrong"; it just doesn't mean what you think it means.

12. Apr 5, 2013

### Passionflower

I would like to point out it is possible to setup a physical experiment in determining the velocity of light but the interpretation will perhaps remain inconclusive.

In a Schwarzschild chart:

One can define two static observers A and B at two different r-coordinate values, each observer can send a light signal to the other observer which is reflected immediately. When one calculates the round trip times it turns out that ABA is not equal to BAB. One might conclude that this means that the velocity of light changes in the field but on the other hand A and B undergo different proper accelerations thus one could argue that this influences the results.

One can do a similar setup for two free falling (from infinity) observers and have them send light signals in a similar way but the problem here is how to interpret 'distance between A and B' while they are moving in the field and with respect to each other as there is no single interpretation for this.

13. Apr 5, 2013

### Agerhell

Gentlemen, you seem to be totally unfamiliar with the post-Newtonian expansion as used by the Jet Propulsion Laboratory to compute orbits for spacecrafts and ephemeride production for computing future positions of celestial objects in the solar system. The post-Newtonian expansion is used in ordinary three-space where "r" and "v" are supposed to have their usual three-space interpretation. Sure, the procedure of getting to the post-Newtonian expansion from the analytic Schwarzschild solution in Schwarzschild coordinates (JPL documentations says from isotropic coordinates) might introduce some errors, I am not a big fan of the post-Newtonian expansion myself, but the procedure of using relativity in ordinary three-space to produce relativity-corrected orbits has a history of some forty years I guess, and is not something new. The post-Newtonian expansion for the spherically symmetric case taken to the first post-Newtonian 1PN level looks like:
$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2})\hat{r}+\frac{4GM}{r^2}(\hat{r}\cdot \hat{v})\frac{v^2}{c^2}\hat{v}$$
My alternative expression for computing orbits under Schwarzschild conditions looks like:
$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}(\hat{r}\cdot\hat{v})(1-3\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}+\frac{v^4}{c^4(1-\frac{2GM}{rc^2})^2})\hat{v}+\frac{GM}{r^2}(\hat{r}\times\hat{v})\times\hat{v}$$

Now I want to test how well my expression (that is derived assuming an isotropic velocity of light) corresponds with what the Schwarzschild solution says. That is what I am looking to do. I read in the JPL documentation by Theodore Moyer that the post-Newtonian expression above is derived from "the Schwarzschild isotropic one-body point mass metric", so it seems that JPL is betting for an isotropic velocity of light...