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.

 

[Dale]

Yes, simultaneity is just a labeling and changes nothing in the physics.

The quantities you have been asking about depend on the labeling. That is what it means when I said that they depend on the reference frame.

 

[Ken G]

Right, we are all agreeing that the coordinate choices can't matter-- as long as we choose coordinates that locally correspond to things we can measure (so our coordinates mean something), we can extend them elsewhere any way we like, and equip that extension with instructions for how to convert to the measurements that someone else in that elsewhere place can do (so they still mean something). The concept of simultaneity is thus a purely coordinate construct, we only thought simultaneity meant something more than that when we didn't realize it was all just coordinates.

The problem is, it sounds like wil wants to say, unless I misread, that "things are really simultaneous, and we just can't tell because we have different coordinates". But actually we should say "whether or not we choose to regard events as simultaneous is an arbitrary function of our chosen coordinates, all that physically matters are the invariant causality relationships, so if event A cannot effect B and B cannot effect A, then there is some perfectly valid coordinate system that can regard them as simultaneous without any changes to physics as we know it."

Framed like this, it means that pre-relativity, we used to think simulataneity was a very narrow class of events because we used to think signals could travel at any speed, so either A could affect B or B could affect A for essentially any events A and B except that very narrow class of "truly simultaneous" events. But when signal speeds were found to be limited by c, it means the class of events that cannot have any causal connection between them got much wider, and all of them can equally be regarded as simultaneous, simultaneity is in effect everything outside the light cone.

 

[wil]

The quantities you have been asking about depend on the labeling. That is what it means when I said that they depend on the reference frame.

The problem I asked is really frame independent, like the lunar phases - exactly!

This is a comparison of the speed of local processes in the two particular systems,
this means we have two reference frames only, not more.

You can represent a time measured by my wrist watch using other time unit, but what that changes?

I measure, in some process, 1200000000 the cesium transitions, and thus this is a global fact - in any other frame the number of the transition (of my cesium) must be perfectly the same!

On the Earth lives now, say: 7 billions of humans;
the 'now' means an event defined wrt some objective fact, for example the precise position - coordinates of the solar system in the Galaxy.

The question is: how many humans lives on the Earth, in the same moment of time, wrt other reference frame?

 

[Ken G]

The point is, those 1200000000 cycles are measured by you between two events, A and B, that are happening to you, and so your clock is present at A and B. Anyone else can agree that you did indeed measure 1200000000 cycles between those events, but the problem is, what do they measure as the number of cycles between those events? There is no unambiguous way to answer that question, because events A and B did not happen to them, they happened to you, and their clock was not present at both A and B. So anyone else that wants to say how many cycles they would reckon occurred between A and B has to first find events C and D that do happen to them, and they have to claim that C is simultaneous with A and D is simultaneous with B, and only then can they associate their cycles between C and D with your cycles between A and B. Since there is considerable freedom to match C and D against A and B, while still honoring all the causality connections that physics needs to be able to explain, there is this thing known as the "relativity of simultaneity."

There is no way around this-- there simply is not an unambiguous way to "correctly" match C and D to A and B, such that comparison of numbers of cycles means anything more than just an arbitrary coordinate system. This is also the reason that if one uses the Einstein simultaneity convention to match events C and D to A and B, then both people think the other one will measure fewer cycles between the events in question, as they will not be the same events. You will think the other person has chosen the wrong two events C and D to match up with your A and B-- you will think they should have used events C' and D', and so they should have gotten fewer cycles than 1200000000 between C' and D' when in fact they found more than 1200000000 between C and D.

 

[Dale]

This is a comparison of the speed of local processes in the two particular systems

The clocks were only co-located at one moment. For any other moment they are separated, not local, and therefore any comparison requires a choice of simultaneity convention. Therefore it is frame variant.

As I said several times above, the only frame invariant way to compare the clocks is for them to be co-located at two events. Otherwise they cannot be local for both.

 

[wil]

The problem is, it sounds like wil wants to say, unless I misread, that "things are really simultaneous, and we just can't tell because we have different coordinates". But actually we should say "whether or not we choose to regard events as simultaneous is an arbitrary function of our chosen coordinates, all that physically matters are the invariant causality relationships, so if event A cannot effect B and B cannot effect A, then there is some perfectly valid coordinate system that can regard them as simultaneous without any changes to physics as we know it."

We can use any convention of time measurement, but I'm afraid the convention will be always limited to some subset of reality.

The claim of type: 'if event A cannot effect B and B cannot effect A'
is highly dependent on our knowledge.

We can discovery in the future same C event, which joins indirectly and causally these events: A with B: A -> C -> B, thus the two events will be no longer arbitrarily ordered, means our assumed time convention can lost its applicability.

 

[Nugatory]

I measure, in some process, 1,200,000,000 the cesium transitions, and thus this is a global fact - in any other frame the number of the transition (of my cesium) must be perfectly the same!

You are correct that that fact is frame-independent.

You and your cesium clock are traveling through space-time on some timelike worldline. At some point on that worldline you zero your clock and start it counting. At some later point on that worldline the counter reads 1200000000, you correctly say that 130 milliseconds has passed. Everyone, regardless of frame and relative velocity, will agree that your clock read zero at the first event (you started it running) and read 1200000000 at the second event (you stopped it running). They will also agree that the interval along your worldline between these two events is 130 milliseconds (check my division, please), that's how much you aged between the two events, it's how much time you experienced between them. That's your proper time and it is also frame invariant.

Now consider me, equipped with a similar cesium clock but traveling along a different worldline than you, and moving relative to you. Which event on my worldline is "at the same time" as the event when you started your clock? Which event on my worldline is "at the same time" as the event when you stopped your clock because it had reached 1200000000 cycles? I'll say that two things happened at the same time if they have the same $t$ coordinate, and so will you - but where did these $t$ coordinates come from? They are frame-dependent, and as a result any method for associating points on my world line with yours is also frame-dependent.

You can see this effect even with special relativity in flat spacetime: It's the way that two observers in inertial motion relative to one another can both correctly find that the other one's clock is running slow. Using your coordinates, the endpoints of the measurement interval along your worldline are separated by 1200000000 cycles, but when we identify the two points on my worldline that have the same $t$ coordinates as your endpoints, the interval between them is less than 1200000000. And the exact same thing happens when we start with the endpoints of my 1200000000 cycle interval and find the corresponding points on your worldline using my $t$ coordinates - they correspond to an interval that is different from and shorter than your 1200000000-cycle interval.

 

[wil]

The clocks were only co-located at one moment. For any other moment they are separated, not local, and therefore any comparison requires a choice of simultaneity convention. Therefore it is frame variant.

As I said several times above, the only frame invariant way to compare the clocks is for them to be co-located at two events. Otherwise they cannot be local for both.

The setup is frame invariant:

O----O'
A-----------B <--- v C,

One observer with a clock is at a point O, which is fixed at r = r0 - in the metric.
He measures a distance AB = h, which is small h/r0 -> 0, this is to simplify the problem only, because now we can assume the v ~ const between A and B.

Alternatively the O can be in the middle position of O' - for a better approximation.

Second observer with a clock C falls freely in the gravity, and notices two events:
he is in position of the point A, and next at B - these both are frame independent,
that means these are the facts, like a bomb explosion, or a breaking of glass, due to direct contact with hammer in our hand.

Both clocks measure his own proper times, and we can compare these times;
the result is unambiguously and frame independent.

Try to compute this using GR, and the result will be easily verifiable in the described setup -
we have quite good clocks already, thus there is no problem.
This is an example of a real experiment to verify the GR theory.

 

[Dale]

O is not colocated with B and O' is not colocated with either A or B. Therefore you will have to use some frame dependent simultaneity convention to map the time of A and B to events on the world line of O or O'.

Are you familiar with spacetime diagrams?

 

[wil]

You can see this effect even with special relativity in flat spacetime: It's the way that two observers in inertial motion relative to one another can both correctly find that the other one's clock is running slow. Using your coordinates, the endpoints of the measurement interval along your worldline are separated by 1200000000 cycles, but when we identify the two points on my worldline that have the same $t$ coordinates as your endpoints, the interval between them is less than 1200000000. And the exact same thing happens when we start with the endpoints of my 1200000000 cycle interval and find the corresponding points on your worldline using my $t$ coordinates - they correspond to an interval that is different from and shorter than your 1200000000-cycle interval.

Do not assume the result in advance - just check it experimentally.
In addition: unverifiable theories are useless.

And the last sentence should be the motto of the PF.

Goodbye, for all the pupils.

 

[Dale]

Do not assume the result in advance - just check it experimentally.
In addition: unverifiable theories are useless..

i agree completely.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[stevendaryl]

i agree completely.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Well, there are two kinds of physics discussions: One is about what a theory predicts, and the second is about experimental results. Very often, when there is a more-or-less standard theory that is relevant to a problem, people say: "It's impossible for such-and-such to happen", when they really mean "According to quantum mechanics (or Special Relativity, or the laws of themodynamics, etc.), it's impossible for such-and-such to happen".

Physics newbies often don't realize when people are talking about what the theory predicts, and when they are talking about what experiments show. A basic rule of thumb is that if anyone claims that something is impossible, it's almost always from the standpoint of some standard theory.

 

[Dale]

The setup is frame invariant:

O----O'
A-----------B <--- v C

Here is a spacetime diagram to show why the measurement you are trying to make is not frame invariant and why it depends on the choice of simultaneity convention. As you specified there is a clock C (red worldline) which moves between event A and event B. Meanwhile, observer O (blue worldline) is colocated with C at event A, but because C is moving relative to O they are not colocated at B.

The proper time from A to B on C's worldline is frame invariant. However, B does not occur on O's worldline. To compare the time on C's clock at B to the time at O's clock at B we must somehow map B to some event on O's worldline (black lines), which is called a simultaneity convention.

If we use O's simultaneity convention then we will map B to B? and conclude that O is running faster than C. If we use C's simultaneity convention then we will map B to B?? and conclude that O is running slower than C. If we use some 3rd party's simultaneity convention then we may map B to B? and conclude that O is running at the same rate as C.

 

[wil]

Here is a spacetime diagram to show why the measurement you are trying to make is not frame invariant and why it depends on the choice of simultaneity convention. As you specified there is a clock C (red worldline) which moves between event A and event B. Meanwhile, observer O (blue worldline) is colocated with C at event A, but because C is moving relative to O they are not colocated at B.

The proper time from A to B on C's worldline is frame invariant. However, B does not occur on O's worldline. To compare the time on C's clock at B to the time at O's clock at B we must somehow map B to some event on O's worldline (black lines), which is called a simultaneity convention.

If we use O's simultaneity convention then we will map B to B? and conclude that O is running faster than C. If we use C's simultaneity convention then we will map B to B?? and conclude that O is running slower than C. If we use some 3rd party's simultaneity convention then we may map B to B? and conclude that O is running at the same rate as C.

This is obvious, because this setup is frame invariant, thus it must be simultaneity convention invariant also.

Because this fact the measurement is perfect - the ideal to veryfy any theory.

In that setup we measure a local time duration between two real events, not between two abstract coordinates, which are not a real entities.

 

[Ken G]

That was the point of it-- since as far as we now know, all setups are "frame invariant," relativity asserts that you cannot have a set up that is not frame invariant. Also, all measurements are regarded as "ideal" for verifying theories, why would we intentionally build in flaws into our measurements if we want to test something? Finally, all meaningful time durations are a "local time duration between two real events", since physics is an empirical mode of inquiry so we no not regard abstract durations between unreal events as "measurements."

I think all that is happening here is that you are adopting a stance that one particular frame is the one in which "true simultaneity" is established, and you are basically saying "well, if you just do measurements between real events, you can't tell that this is actually true." It is certainly true that nothing in relativity tells us that you have to be wrong-- relativity does not say that there cannot be one frame where the simultaneity is the true one and all other frames really have time slowed down, indeed that was the original interpretation supplied by Lorentz and Poincare. Einstein's interpretation did not change any of the equations, but it is regarded as a more streamlined interpretation, and is preferred expressly because Einstein removed the extraneous elements and just pointed out that what is "actually true" is only what we can demonstrate to be actually true with real measurements between real events.

Thus, it seems to me that all you are saying is that you don't agree with an interpretation that eliminates "abstract" elements and sticks to what can actually be demonstrated as true. This is the nature of interpretations-- if you are using all the same equations, you are using the same theory, but you are supplying a different justification for the equations, a different language for talking about them. Interpretations are not right and wrong, but they can be the preferred mainstream language as in Einstein's approach, or they can be regarded as clunky, arbitrary, and possibly even missing the key lessons of the theory, as many would say your interpretation is doing. All the same, tomorrow new observations could be made that require correcting relativity in a way that rules out Einstein's interpretation but allows yours, so we always have to bear this in mind. Still, in the absence of any such observations, you really cannot argue that your interpretation is the correct one, for you have no evidence for that, but there is clear evidence that your interpretation involves an arbitrary and unnecessary choice to designate one frame as the one with "true simultaneity", and arbitrary and unnecessary elements are generally disfavored in a good interpretation.

An analogy would be claims that the Earth is at the center of the Big Bang expansion. Any observer is free to make that claim, since that is exactly how things appear when any observer looks outward at an expansion that obeys the Hubble law. But we recognize it is just an arbitrary and unnecessary interpretation, because the nature of the expansion is such that any observer anywhere could equally make that claim, so why would we favor our own point of view? Instead, we should take it as a lesson that any observer could make that claim, and put the claim in perspective as a result. But that doesn't mean the Earth really isn't the center of the expansion, and an observation (say, of primordial gravitational waves or something) could conceivably be done that demonstrates it really is. So even though it could be the only surviving interpretation of some new theory that replaces our current one, it is not a favored interpretation of the theory we have now.