# What is a metric for uniformly moving frame?

1. Aug 26, 2014

### wil

The Schwarzschild Metric has a form:

$ds^2 = Kdt^2 - 1/K dr^2 - r^2dO^2$

where: K = 1 - a/r;

There is a time scaled by K, but a space radially by 1/K.

This is a typical time dilation and a space contraction, which is known from SR,
but the Schwarzschild metrics is spherically symmetric, and in second case a space is axially symmetric only - contraction is along x axis, not radially.

Thus shouldn't be the metrics in SR, for Minkowski space, similar to the Schwarzschild metrics?

$K = 1 - v^2/c^2$
and the metrics for the moving frame should be:

$ds^2 = K dt^2 - 1/K\cdot dx^2 - dy^2 - dz^2$

Is this correct - legal, and why not?

2. Aug 26, 2014

### stevendaryl

Staff Emeritus
No, in inertial coordinates, the metric looks the same for any uniformly moving frame:

$ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$

The time-dilation factor $\sqrt{1-\frac{v^2}{c^2}}$ doesn't have to be inserted by hand, it is derivable from the metric. To see this, figure out $ds^2$ for an observer moving at a constant velocity $v$ in the x-direction:

In that case, $dx = v dt$, $dy=0$, $dz=0$. So we have:

$ds^2 = c^2 dt^2 - v^2 dt^2 = c^2(1-\frac{v^2}{c^2}) dt^2$

The proper time, $\tau$ is related to $s$ by $\tau = s/c$. So $\tau$ obeys:

$d\tau^2 = (1-\frac{v^2}{c^2}) dt^2$

So

$d\tau = \sqrt{1-\frac{v^2}{c^2}} dt$

So the time dilation factor is not something that you put into the metric, it comes out of the metric.

3. Aug 26, 2014

### WannabeNewton

These are gravitational effects i.e. they are due to the space-time geometry of the central mass. They are not kinematic i.e. they are not due to SR as applied to an observer moving or at rest in the gravitational field. The canonical Schwarzschild metric is written in terms of the asymptotic Lorentz frame of an observer at infinity. Using the gravitational time dilation factor you can scale the frame to get the Schwarzschild metric in the local frame of a static observer in the gravitational field. Then you can apply a Lorentz transformation to this frame in the usual way to get the Scwharzschild metric in terms of e.g. an observer moving in circular orbit around the central mass or an observer freely falling radially.

Doing this (which is very easy to do) you will find that the Schwarzschild metric in the local frame contains both the gravitational time dilation effects and the kinematic SR time dilation and length contraction effects. Don't confuse the two.

4. Aug 26, 2014

### wil

I suppose this reasoning can be applied to the GR metric and the result will be the identity matrix also.

There is: K * 1/K = 1, and this means the light speed is preserved, because the space contraction is compensated by a time dilation at any place.

5. Aug 26, 2014

### wil

I know that, but in a mathematical sense we have rather the same situation
in both cases.

There is a subtle geometric difference only:
1. In GR there is spherical space, and we can identify a special point - the centre.
2. and in SR we can identify a special direction only - there is no center, just a line.

The moving frame is different, distorted someway from the point of view of stationary frame.
Locally the situation is the same in both cases again: the identity metrics.

Full analogy.
I don't know where is the reason to treat these two geometries in completely different ways: in GR space is deformed, but in SR it isn't, despite the explicit transformation, which is not the identity transform!

6. Aug 26, 2014

### Staff: Mentor

No, there is a spherical space with the center a special point only in a few specific situations, such as when we have a spherically symmetric distribution of mass in an an otherwise empty universe, as in the Schwarzchild spacetime. General relativity works just as well for other distributions of matter, including the easiest case of a completely empty and hence flat spacetime.

Indeed, the reason we call them "general" relativity and "special" relativity is that GR is the general theory that works for all spacetimes whether flat or not and no matter how mass is distributed within the spacetime, whereas SR works only for the special case of a flat spacetime, which is to say one in which there are no gravitational effects. If you take the Einstein field equations of GR, and solve them for that particular case, SR is what pops out.

7. Aug 26, 2014

### pervect

Staff Emeritus
No, the metric for Minkowskii space is:

$ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$

If you do a Lorentz boost in the x direction

$t' = \gamma(t - vx)$
$x' = \gamma(x - (v/c^2) t)$

the resulting metric is:
$ds^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2$

I don't entirely follow your argument, but since it's coming to incorrect conclusions you can be sure it's wrong. I believe part of your problem may be thinking of the Schwarzschild metric as a "spherical space", when actually it's curved. For instance if you embed the (r, phi) plane on a 2d surface, you don't get a flat disk, you get Flamm's paraboloid, http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid

8. Aug 27, 2014

### Staff: Mentor

No, the Minkowski metric gives a space which is symmetric under 4 directions of translation, 3 axes of spatial rotations, and 3 directions of boosts.

This is certainly a legal metric. One common approach of solving problems in GR is exactly what you are attempting: guess the form of the metric and then see what properties your guessed metric has.

In this case, your metric is flat because the Riemann curvature tensor is all 0. It also is inertial since the Christoffel symbols are all zero also.

Basically, this metric describes a situation where you use different units in x than in y and z. For example, if y and z are in meters then x could be in miles.

Last edited: Aug 27, 2014
9. Aug 27, 2014

### wil

These metrices are identical - the same, because the naming of variables in equations has no meaning.

I'm talking about the metrics of moving frame but with the coordinates of the stationary frame.
|--------------> v the moving system

O - a stationary

The stationary observer at O describes the moving system,
so, he uses naturally his own coordinates, and You suggest he should to use the local coordinates of the moving observer!

Yes. The spherical space is curved, in the sense of Gaussian curvature, but the Minkowski is flat - probably like a cilinder: $k_g = k_1\cdot k_2 = 0$ thus in this case: k1 = 0 or k2 = 0.

10. Aug 27, 2014

### Staff: Mentor

I don't know how to interpret this sentence in terms of the mathematical framework of relativity. The metric, usually denoted g or equivalently ds^2, is a single geometric object. It can be expressed in arbitrary coordinates, but the metric itself belongs to the manifold, not the frames. So there is no sense in which the moving frame has a different metric from a stationary frame. They both have the same metric, expressed in different coordinates.

11. Aug 27, 2014

### wil

But when we place x axis along the radial r direction of the Schwarzschild metrics,
then in any fixed point: r0, we obtain just the metric I wrote!
Maybe approximately only, but quite good.

Thus the metrics isn't an identity, because it is for a stationary and remote observer, not for a local one.

12. Aug 27, 2014

### wil

Does this means that the metric for an observer, falling into a gravitational field is always the same, regardless of its speed?

I think the observer should detect some differences...

13. Aug 27, 2014

### Staff: Mentor

Yes. The metric is the same for any observer in any spacetime regardless of their motion. Only the coordinates that they use to label the metric is different.

Thus the "metrics of moving frame but with the coordinates of the stationary frame" is the same thing as the "metric of the stationary frame".

14. Aug 27, 2014

### Staff: Mentor

Yes, this is true. At any point in the Schwarzschild you can construct a local inertial frame of this form.

This shows that the standard Schwarzschild coordinates are anisotropic. The r coordinate has a different length than the theta or phi coordinates.

There are also isotropic Schwarzschild coordinates which get rid of this problem:
https://en.wikipedia.org/wiki/Isotropic_coordinates
https://en.wikipedia.org/wiki/Schwarzschild_metric#Alternative_coordinates

Not sure what you mean here.

15. Aug 27, 2014

### Staff: Mentor

You are confusing the metric (which is a tensor) with the values of its components in a given coordinate system (which are real-valued functions of the position). The same metric will be written in different ways in different coordinate systems - for example the Schwarzchild metric which describes the space-time around a spherically symmetric mass has very different components in Schwarzchild coordinates and Kruskal coordinates but it's the same metric tensor describing the same spacetime.

16. Aug 27, 2014

### wil

By the way, I found a problem with the falling frame, or rather a clock.

What time measures the falling radially clock with speed equal to the escape velocity: v_i^2 = 2GM/r ?
Let's assume the falling distance is very small, so the r coordinate changes a little only: dr/r -> 0
M--------r0 v<--- C - a clock
The time will be shorter than the measured by a fixed clock at the point r0, or rather longer?

17. Aug 27, 2014

### wil

I noticed this.

I'm talking about much more general problem, because about the observed, measured properties, not only the static/fixed geometric objects in a math space.

18. Aug 27, 2014

### Staff: Mentor

That depends on whose frame you use to make the comparison.

19. Aug 27, 2014

### wil

This is frame independent, because we compare two clocks only.
the space is almost flat due to a small distance:

r0
|------dr------| to ~= dr/v, for the local clock, fixed at r0.
|<---v --------| tf = ?

and the falling clock measures: tf, then we just compare: to > tf ?

20. Aug 27, 2014

### Staff: Mentor

No, it is frame dependent. The number of clocks, the distance, and the curvature are irrelevant. The only circumstance where clock comparisons are frame independent is if they start and end co-located.