What is a metric for uniformly moving frame?

[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?

 

[stevendaryl]

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?

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.

 

[WannabeNewton]

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

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.

 

[wil]

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.

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.

 

[wil]

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.

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!

 

[Nugatory]

1. In GR there is spherical space, and we can identify a special point - the centre.

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 Schwarzschild 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.

 

[pervect]

The Schwarzschild Metric has a form:

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?

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

 

[Dale]

in second case a space is axially symmetric only - contraction is along x axis, not radially.

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.

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?

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.

 

[wil]

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

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!

I believe part of your problem may be thinking of the Schwarzschild metric as a "spherical space", when actually it's curved.

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.

 

[Dale]

I'm talking about the metrics of moving frame but with the coordinates of the stationary frame.

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.

 

[wil]

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.

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.

 

[wil]

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.

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...

 

[Dale]

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...

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".

 

[Dale]

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.

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

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

Not sure what you mean here.

 

[Nugatory]

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.

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 Schwarzschild metric which describes the space-time around a spherically symmetric mass has very different components in Schwarzschild coordinates and Kruskal coordinates but it's the same metric tensor describing the same spacetime.

 

[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?

 

[wil]

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 Schwarzschild metric which describes the space-time around a spherically symmetric mass has very different components in Schwarzschild coordinates and Kruskal coordinates but it's the same metric tensor describing the same spacetime.

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.

 

[Dale]

The time will be shorter than the measured by a fixed clock at the point r0, or rather longer?

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

 

[wil]

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

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 ?

 

[Dale]

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

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.

 

[wil]

This is a simple stationary setup.
Experiment very similar to the Pound-Rebka on a tower,
but with one difference: the second clock is falling and it measures the flying time between two points.

A speed must be big, and in the optimal case: equal to the escape velocity, which is about 11 km/s on the Earth surface, but a several km/s should be good enough.
The big speed is important, because in this way we eliminate the typical - local gravitational redshift.

In this way we can check a falling clock time dilation.

 

[Dale]

Again, the outcome depends on which reference frame you use. This is unavoidable unless the clocks begin and end together. This isn't something you can avoid by judicious choice of details. If the clocks start and end colocated then the elapsed time is invariant, otherwise it is not.

 

[wil]

You probably don't understand yet what is a difference of two numbers.
This is equivalent to the two angles difference, phases, ect.

Further discussion is pointless.

 

[Dale]

I understand the math. I also understand the physics. The physical quantity you want to calculate is frame variant. Do you understand why?

Don't get mad at me, I didn't make the rules of physics. You are not going to be able to learn physics if every time something challenges your mindset you get angry and quit. Further discussion is only pointless if you refuse to learn.

 

[wil]

You missed something.

Quick test:
do the moon phases depend on the observer-frame also?

 

[Dale]

do the moon phases depend on the observer-frame also?

Yes. This is due to the relativity of simultaneity (which is usually the most difficult concept for students to learn in SR).

 

[wil]

Wrong.
The moon phase is completely a local phenomenon: the sun light reflects off the moon's surface to the direction of the observer on the earth, or not.

 

[Dale]

Sure, but simultaneity is not local.

This is just the Lorentz transform. To make it concrete, suppose I am at rest wrt the moon at a distance of 1 light-month from the moon. Suppose that you pass next to me heading towards the moon at 0.6 c. In my frame the moon is a new moon (beginning of cycle) when you pass me, and in your frame the moon is slightly past full (0.6 of the way through its cycle) when you pass me.

This is just the Lorentz transform of the event (t,x)=(0,0.8) from your frame into mine.

 

[wil]

This example is absolutely irrelevant for the lunar phases problem, because the lunar phase is unambiguously defined phenomena for the local observer on the earth, not for any other, especially on the Jupiter or on the Andromeda galaxy.

But your 'calculations' are still wrong:
the observed state of the distant moon in this example will, and must be the same for both observers,
because the same light reaches both observers: the moving and the stationary.
There is only one stream of light for all observers, not many - infinite many independent streams - one for any individual frame, separately generated at the moon surface.

 

[Dale]

The question "what is the phase of the moon now" is a question that can be asked by any observer anywhere at rest in any reference frame and the answer will depend on the reference frame.

Even restricting it to observers on earth, the answers will differ in different reference frames, although the differences will be relatively small since the distance is only ~1.3 light-seconds. Specifically, if the moon is now a new moon for an observer at rest on the Earth then for an observer on the Earth moving at .6 c towards the moon the phase will be just past new (300 ppb of a cycle).

My calculations are correct for answering the question "what is the phase of the moon now". Not "what image of the moon am I receiving now". Do you understand the distinction? If we visually see a star go supernova today and if the star is 1000 ly distant do you believe that the star went supernova today or do you understand that it went supernova 1000 years ago in our frame?

 

[Nugatory]

the observed state of the distant moon in this example will, and must be the same for both observers

But you have asked for something different: The observed state of the moon now, and thanks to the relativity of simultaneity, "now" for one observer is not the same time as "now" for another observer. They will of course agree about the angle that a particular light signal makes as it strikes the moon's surface and is reflected, but they will find that event happens at different times.

And of course this is after backing out the effects of light travel time. The observer ten light-years distant will correctly conclude that the light left the moon ten years before it reached his eyes, and the observer twenty light-years distant will correctly conclude that the light left the moon twenty years before it reached his eyes... but those will not in general be the same time in their respective frames if they are moving relative to one another.

 

[wil]

This will be exactly the same moment of time - before and after the transformation, and any transformation.

You can apply any convention of time measurement, for example:
t = 80000000 or t' = -6777777 - it is the same moment of time, but in two different time keeping methods.

Any convention of simultaneity can't give insight into future, nor to the past.

 

[Dale]

This will be exactly the same moment of time - before and after the transformation, and any transformation.

You can apply any convention of time measurement, for example:
t = 80000000 or t' = -6777777 - it is the same moment of time, but in two different time keeping methods.

You are missing the point of relativity of simultnaety. Not only are the numbers different for the time assigned to a given event, but also the set of events that occur simultaneously with the given event.

 

[Ken G]

Yes, or put differently, a key lesson of relativity is that it is not "the same moment" because there is no unambiguous global meaning of "the same moment" in the first place. That is the physical basis of the "spacetime" concept-- there is only "here and now", and they come together-- so there is no unambiguous meaning to "there and now."

 

[wil]

No. This is the change of a time coordinate only, means: the numbers which observer uses to label the events, nothing more.
The events, facts alone are always fixed in its original place.

This is probably an ideal analogy to the tensors in GR:
change of coordinates, changes nothing in a geometry, and in physics.