Space time curvature caused by fast electron

[matthias31415]

Hi everybody!

what happens if an electron passes by with a speed of, say, 99.999999999...% of the speed of light (relative to me). Its mass will then be enormous. Will this electron cause a relevant curvature of spacetime? Can it be so fast that it acts like a black hole?

I guess not. But why?

 

[harrylin]

Hi everybody!

what happens if an electron passes by with a speed of, say, 99.999999999...% of the speed of light (relative to me). Its mass will then be enormous. Will this electron cause a relevant curvature of spacetime? Can it be so fast that it acts like a black hole?

I guess not. But why?

Hi, although I think that you are right that a very high energy electron will cause a lot of spacetime curvature, it cannot turn into a black hole. For that you need a high enough rest energy density.
That's easy to see if you transform to a frame in which the electron is in rest: in that frame its kinetic energy is zero, and so it can never become a black hole.

 

[matthias31415]

Hi, although I think that you are right that a very high energy electron will cause a lot of spacetime curvature, it cannot turn into a black hole. For that you need a high enough rest energy density.
That's easy to see if you transform to a frame in which the electron is in rest: in that frame its kinetic energy is zero, and so it can never become a black hole.

thanks harrylin,

what I don't understand is when to use rest energy and when to use the energy with respect to the actual reference frame (e.g., me).

Let's assume the electron moves horizontally. I understand that its horizontal inertia is really high given that it cannot further be accelerated. But what about vertical inertia/gravity?

If it is only rest energy that decides if an object turns into a black hole, why isn't rest energy all that counts for spacetime curvature?

 

[PeterDonis]

Hi, although I think that you are right that a very high energy electron will cause a lot of spacetime curvature

No, it won't. The spacetime curvature caused by any object is a frame-independent, invariant quantity. Since the object's velocity is not a frame-independent, invariant quantity, the curvature the object causes can't depend on the velocity. (Same argument for the kinetic energy.)

what I don't understand is when to use rest energy and when to use the energy with respect to the actual reference frame (e.g., me).

You don't use either. You use the stress-energy tensor, which takes into account both the object's rest energy and the object's motion (and pressure and internal stresses in the object) in such a way that the curvature caused by the object is frame-invariant. The equation that expresses the relationship is the Einstein Field Equation (EFE).

By "frame-invariant" I mean that you can write down the EFE in any frame you like to get answers to questions about actual physical observables, like whether or not an object can become a black hole. So if you write down the EFE in a frame in which the object is at rest, the stress-energy tensor in that frame (for an object like an electron that has no internal pressure or stresses) does contain *only* the object's rest energy density, and as harrylin says, you can then predict whether the object will become a black hole just by looking at its rest energy density.

(Note: For a more realistic object that could collapse into a black hole, such as a collapsing star, the stress-energy tensor in the object's rest frame will also contain components representing the pressure inside the object, which can significantly affect whether the object will become a black hole. That's why I put in the qualifier above about the object having no internal pressure or stresses.)

 

[Naty1]

Matthias: Several years ago Dr Greg was kind enough to explain two types of spacetime curvature in response to a question I posted...

The key is that gravitational curvature IS observer independent (as already noted) and is reflected as curvature of the spacetime manifold ("graph paper" as described below). Frame dependent curvature (observer dependency) is a variable overlay on top of this fixed background curvature,

From Dr Greg:
What we call the "curvature of spacetime" has a technical meaning; the equations that describe it are very similar to the equations that describe, say, the curvature of the Earth's surface in terms of latitude and longitude coordinates, or any other pair of coordinates you might choose. This "curvature" need not manifest itself as a physical curve "in space".

For the rest of this post let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. …
In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame ([an accelerated observer] but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarize, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer.

 

[harrylin]

No, it won't. The spacetime curvature caused by any object is a frame-independent, invariant quantity. Since the object's velocity is not a frame-independent, invariant quantity, the curvature the object causes can't depend on the velocity. (Same argument for the kinetic energy.)

When we speak of a high energy electron, what we mean is the kinetic energy; thus I don't understand what you mean with "same argument for the kinetic energy". And now I have a similar question as the OP, for the rest energy and mass of a plasma with high speed electrons is in principle higher than that of a plasma with slow speed electrons. How can that not affect spacetime curvature? Here below you discuss how to calculate the amount of curvature in a frame independent way and you seem to agree that the electron's kinetic energy ("motion") contributes to curvature.

Thanks,
Harald

PS this seems to be related to the citation by Naty, but some more elaboration could be helpful.

You don't use either. You use the stress-energy tensor, which takes into account both the object's rest energy and the object's motion (and pressure and internal stresses in the object) in such a way that the curvature caused by the object is frame-invariant. The equation that expresses the relationship is the Einstein Field Equation (EFE). [..]

 

[Passionflower]

No, it won't. The spacetime curvature caused by any object is a frame-independent, invariant quantity. Since the object's velocity is not a frame-independent, invariant quantity, the curvature the object causes can't depend on the velocity. (Same argument for the kinetic energy.)

I understand why you say this but is the matter really so black and white?

For instance take the rest mass of a gold atom, a lot of that is contributed by electrons moving at relativistic speeds.

 

[matthias31415]

Matthias: Several years ago Dr Greg was kind enough to explain two types of spacetime curvature in response to a question I posted...

From Dr Greg:
...If we switch to a non-inertial frame ([an accelerated observer] but still in the absence of gravitation),...

an "accelerated observer ... in the absence of gravitation"?

isn't that a key assumption of general relativity that what an accelerated observer experiences cannot be distinguished from gravitation?

Anyway, thanks for the citation, it looks very instructive, I'll just have to read it a few more times ;-)

 

[PeterDonis]

When we speak of a high energy electron, what we mean is the kinetic energy; thus I don't understand what you mean with "same argument for the kinetic energy".

The OP was talking about a single electron; for that case the rest energy density of the electron is the only thing that causes spacetime curvature. The kinetic energy is frame-dependent, just as the velocity is; in the electron's rest frame it is zero, and we can predict all physical observables, like whether the electron forms a black hole, by solving the EFE in the electron's rest frame.

And now I have a similar question as the OP, for the rest energy and mass of a plasma with high speed electrons is in principle higher than that of a plasma with slow speed electrons. How can that not affect spacetime curvature?

Now you are talking about a different case, where we have a bunch of electrons all moving in different ways, but there is some average "rest frame" for the system as a whole. To model this in GR, you need to assign a stress-energy tensor to the system as a whole, and that stress-energy tensor will contain components describing fluid pressure in the system's rest frame. That pressure is due to the internal motions of the parts of the system, in this case electrons. The pressure makes an additional contribution to the spacetime curvature caused by the system, and it increases with the energy (more precisely, with the temperature) of the plasma, so you are correct that a plasma with higher temperature (and hence higher-speed electrons in it) will cause more spacetime curvature than a low-temperature one. But that's a different case than a single electron.

Here below you discuss how to calculate the amount of curvature in a frame independent way and you seem to agree that the electron's kinetic energy ("motion") contributes to curvature.

Not sure what you're referring to, but if you mean my description of the stress-energy tensor for a *single* electron, in a frame in which the electron is not at rest, that tensor will contain components describing the electron's momentum, which are not there in the electron's rest frame, but the energy component will also be different than it is in the electron's rest frame, so that the final result, the predicted spacetime curvature, is the same.

If you are referring to a stress-energy tensor for a system with many particles, in the "average" rest frame of the system, see above for how it incorporates the motion of the individual particles.

 

[A.T.]

The spacetime curvature caused by any object is a frame-independent, invariant quantity

Wouldn't the spacetime curvature caused by a fast moving spherical mass be described by Lorentz contracted version of the Schwarzschild metric? Or do you mean that the curvature measures would not be affected by the uniform contraction, because they depend on the second derivates of the metric?

 

[PeterDonis]

I understand why you say this but is the matter really so black and white?

For instance take the rest mass of a gold atom, a lot of that is contributed by electrons moving at relativistic speeds.

See my post in response to harrylin just now. I'm not sure if anyone has ever written down a stress-energy tensor for an atom that tries to actually model the internal motions of the electrons as pressure, but the SET for a white dwarf does the same thing with the "electron fluid" in the white dwarf (and the SET for a neutron star does the same thing for the "neutron fluid" in the neutron star). So a gold atom at rest does cause more spacetime curvature due to its relativistic electrons than it would if its electrons were at rest inside it. But a gold atom moving as a whole does not cause any more spacetime curvature than a gold atom at rest.

 

[PeterDonis]

Or do you mean that the curvature measures would not be affected by the uniform contraction, because they depend on the second derivates of the metric?

Yes.

 

[DrGreg]

an "accelerated observer ... in the absence of gravitation"?

isn't that a key assumption of general relativity that what an accelerated observer experiences cannot be distinguished from gravitation?

Locally, yes. If you look at a small piece of spacetime near you, it's very hard to tell whether it's curved or flat. The curvature, or lack of curvature, won't become noticeable until you look at a larger region. (In the same way as it's difficult to tell whether we live on a flat Earth or a spherical Earth if you only look around your immediate neighbourhood.)

 

[pervect]

The numerical values of the components of the curvature (for instance the tidal forces) that you measure would definitely be affected by the velocity of the moving mass for a transverse boost, i.e. if the mass is whizzing by you and not approaching head on. Of course, in an abstract sense it is the "same tensor", just seen from a different viewpoint.

In the limit of an ultra-relativistic transverse velocity, you get the Aichelburg-Sexl solution. http://arxiv.org/abs/gr-qc/0110032 actually writes down the Riemann tensor for this solution.

Note that, for a transverse boost, said tensor contains delta functions, i.e. values that approach infinity (the integrals are finite, though). So this is considerably different than the finite non-delta function values for the Riemann curvature tensor of a stationary mass.

This is the behavior for a transverse boost. For a parallel boost, the outcome is completely different and pretty much as Peter originally stated - the components of the curvature tensor don't change with velocity. This is discussed in the paper I quoted above, and in MTW as well. I believe there was some good reasons for this, but I don't recall what they were offhand. But one of the consequences of this interesting fact is that if you are falling directly into a black hole (with no transverse component of your velocity), the tidal forces don't depend on the velocity of your approach.

If one is familiar with how the electric field of an electron transforms according to SR, one can gain a lot of insight as to how its gravitational field transforms. For instance, the electric field of a relativistically moving electron is not spherically symmetric, and one can't apply Coulomb's law to derive it.

YOu can compare the statements about the boost of the Riemann tensor to behavior of the electric field - they're pretty similar. The transverse component of the E-field gets boosted by gamma, the parallel component of the E-field is not affected by the boost.

 

[PeterDonis]

The numerical values of the components of the curvature (for instance the tidal forces) that you measure would definitely be affected by the velocity of the moving mass for a transverse boost, i.e. if the mass is whizzing by you and not approaching head on. Of course, in an abstract sense it is the "same tensor", just seen from a different viewpoint.

I should clarify that I didn't mean to imply that the *components* of the curvature can't change with relative motion. I only meant that the physical invariants calculated from those components don't change. For example, the tidal force experienced by an observer following a given worldline can be expressed as a contraction of the observer's 4-velocity with the curvature tensor, i.e., as an invariant.

 

[harrylin]

The numerical values of the components of the curvature (for instance the tidal forces) that you measure would definitely be affected by the velocity of the moving mass for a transverse boost, i.e. if the mass is whizzing by you and not approaching head on. Of course, in an abstract sense it is the "same tensor", just seen from a different viewpoint. [..] This is the behavior for a transverse boost. For a parallel boost, the outcome is completely different and pretty much as Peter originally stated - the components of the curvature tensor don't change with velocity. [..]

Thanks for the clarifications.
Indeed, it would not make much sense to me if total space time curvature would not be the equal to the sum of the contributions of all the relevant elements - such the OP's electron that passes by at high speed.

Harald

 

[PAllen]

Thanks for the clarifications.
Indeed, it would not make much sense to me if total space time curvature would not be the equal to the sum of the contributions of all the relevant elements - such the OP's electron that passes by at high speed.

Harald

Well, of course it isn't the sum. GR is non-linear. Further, any situation where you ask about a rapidly moving massive body's effect on a stationary test body can be transformed to a question about the interaction between a rapidly moving test body and a stationary massive body. The results must be identical, as to any invariant or measured quantity. Thus all observables relating to a rapidly moving massive body can be answered as if the body is stationary. Period.

Thus, I think Peter is perfectly correct here:

- The expression of curvature tensor (like any tensor) is coordinate dependent.
- The 'amount' of curvature as measured by an invariant (e.g. the Kretschman invariant) or by an observable, is not dependent on the speed of the body. Observations may well depend on the relative motion of an instrument and a source of gravity.

The amount of curvature produced by system of interacting particles is another thing altogether - motion of the parts definite affects the invariant curvature. However, the invariant effect can be analyzed in coordinates where there is no overall motion of the system.

Another way to look at this: if you have a coherently moving packet of dust (all particles moving in the same direction at the same speed), all observables may be treated assuming you have a stationary packet of dust (including the small contribution of self-gravity causing the dust to coalesce over time). On the other hand, if you have a dust packet with random internal motions, then the KE of the particles contributes intrinsically to the curvature produced. More curvature will be produced than the coherent packet.

 

[harrylin]

Well, of course it isn't the sum. GR is non-linear.

Thanks for the precision, and sorry if my formulation was sloppy: what I meant is that it would not make sense if in GR the total contribution of zero effects is unequal to zero.

Thus I appreciated your clarification that indeed the numerical values of the components of the curvature that you measure would definitely be affected by the velocity of the moving mass, if the mass is whizzing by you: that is exactly how I understood the OP.

[...] all observables relating to a rapidly moving massive body can be answered as if the body is stationary. Period. [..]

Indeed, that's also what I stressed in my first reply.

Harald

 

[zonde]

Further, any situation where you ask about a rapidly moving massive body's effect on a stationary test body can be transformed to a question about the interaction between a rapidly moving test body and a stationary massive body. The results must be identical, as to any invariant or measured quantity. Thus all observables relating to a rapidly moving massive body can be answered as if the body is stationary. Period.

Idea that result depends only from relative velocities follows from principle of relativity. But how can we check that GR respects principle of relativity? Or maybe we know already that it doesn't?

 

[TrickyDicky]

Idea that result depends only from relative velocities follows from principle of relativity. But how can we check that GR respects principle of relativity? Or maybe we know already that it doesn't?

Interesting question, it has been the source of some debates from science historians and Einstein texts specialists. The conclusion seems to be that although Einstein was convinced at first that GR not only respected but generalized to arbitrary motion the principle of relativity, later he changed his mind about this and had to admit GR doesn't generalize the principle of relativity, and it only respects it in a local way (by the equivalence principle), deviating from it as soon as curvature makes an entrance.
Reference:
"The Twins and the Bucket: How Einstein Made Gravity rather than Motion Relative in General Relativity." Michel Janssen
http://www.tc.umn.edu/~janss011/

 

[PAllen]

Idea that result depends only from relative velocities follows from principle of relativity. But how can we check that GR respects principle of relativity? Or maybe we know already that it doesn't?

It follows axiomatically from the differential geometry GR is expressed in. All observables in GR are defined in terms of coordinate independent quantities (not just covariant quantities). A contraction formed from the world line of a test instrument and tensor is invariant. That is sufficient to justify my statement.

 

[PeterDonis]

The conclusion seems to be that although Einstein was convinced at first that GR not only respected but generalized to arbitrary motion the principle of relativity, later he changed his mind about this and had to admit GR doesn't generalize the principle of relativity, and it only respects it in a local way (by the equivalence principle), deviating from it as soon as curvature makes an entrance.

Interesting paper. From the abstract:

Two different relativity principles play a role in these accounts: (a) the relativity of non-uniform motion, in the weak sense that the laws of physics are the same in the two space-time coordinate systems involved; (b) what Einstein in 1920 called the relativity of the gravitational field, the notion that there is a unified inertio-gravitational field that splits differently into inertial and gravitational components in different coordinate systems. I provide a detailed reconstruction of Einstein’s rather sketchy accounts of the twins and the bucket and examine the role of these two relativity principles. I argue that we can hold on to (b) but that (a) is either false or trivial.

In hindsight, since non-uniform motion, in the sense of proper acceleration, is a direct physical observable (accelerometer reading), it's not surprising that Einstein's attempt to "relativize" it didn't hold up. But that doesn't affect what PAllen was saying; locally, a body with non-zero proper acceleration, and a body with zero proper acceleration but which is momentarily at rest with respect to the accelerating body, will "look the same" as far as the principle of relativity is concerned. It's only by looking at the two bodies' worldlines over a large enough distance that we can see the difference between them.

 

[PAllen]

In hindsight, since non-uniform motion, in the sense of proper acceleration, is a direct physical observable (accelerometer reading), it's not surprising that Einstein's attempt to "relativize" it didn't hold up. But that doesn't affect what PAllen was saying; locally, a body with non-zero proper acceleration, and a body with zero proper acceleration but which is momentarily at rest with respect to the accelerating body, will "look the same" as far as the principle of relativity is concerned. It's only by looking at the two bodies' worldlines over a large enough distance that we can see the difference between them.

Also, I was implicitly assuming (should have stated) that the massive body is inertial; this is an invariant feature.

 

[zonde]

It follows axiomatically from the differential geometry GR is expressed in. All observables in GR are defined in terms of coordinate independent quantities (not just covariant quantities). A contraction formed from the world line of a test instrument and tensor is invariant. That is sufficient to justify my statement.

This is too vague. And your sentence about invariant contraction does not make any sense to me.

Observables are invariants axiomatically. That's good.
Now what happens with curvature under transformation?

EDIT: Thought that it is not very meaningful to ask what happens with curvature. Rather I should ask what happens with description of curvature i.e. tensor field.

 

[PeterDonis]

And your sentence about invariant contraction does not make any sense to me.

He's using the term "contraction" the way I used it in an earlier post, to mean an expression that is formed by multiplying vectors or tensors together component by component and then summing the results over all indexes, to obtain an expression that is frame-invariant. For example, the invariant "length" of a 4-vector is found by contracting it with itself; thus, if an electron has 4-momentum p^{a}, its rest mass is given by

m = \sqrt{p^{a} p_{a}} = \sqrt{g_{ab} p^{a} p^{b}}

where we use the metric to lower the index of one "copy" of p^{a}. This expression will give the same number regardless of the coordinate system used, which matches our observation that electrons all have identical rest mass, regardless of their state of motion.

Now what happens with curvature under transformation?

EDIT: Thought that it is not very meaningful to ask what happens with curvature. Rather I should ask what happens with description of curvature i.e. tensor field.

Um, it transforms like a tensor? That's the whole point: that as long as you describe everything in terms of vectors and tensors, the way they transform under a change of coordinates automatically guarantees that all observables, i.e., all scalar invariants that can be formed by contracting vectors and tensors together, remain invariant.

 

[zonde]

Um, it transforms like a tensor? That's the whole point: that as long as you describe everything in terms of vectors and tensors, the way they transform under a change of coordinates automatically guarantees that all observables, i.e., all scalar invariants that can be formed by contracting vectors and tensors together, remain invariant.

Let's say that I describe spacetime around non-rotating gravitating body using coordinate time and flat geometry. In that case transforming it to other state of motion makes different parts of gravitating body move at different speeds proportionally to local (coordinate) speed of light.

So if we do the same in GR it rises the question if we can apply the same transformation across the whole spacetime region containing gravitating body and get consistent result? Or do we have to apply different transformations at different points in order for them to stick together?

 

[pervect]

Let's say that I describe spacetime around non-rotating gravitating body using coordinate time and flat geometry. In that case transforming it to other state of motion makes different parts of gravitating body move at different speeds proportionally to local (coordinate) speed of light.

So if we do the same in GR it rises the question if we can apply the same transformation across the whole spacetime region containing gravitating body and get consistent result? Or do we have to apply different transformations at different points in order for them to stick together?

Lets back up a bit.

Imagine you have the 2-d surface of a sphere. At any point on the surface of the sphere there will be a flat plane tangent to the sphere, the tangent plane,.

This result can be generalized - if you have a n-dimensional curved manifold, such as a 4 dimensional space-time manifold, at any point on the surface there will be a flat manifold tangent to it at that point. This is called the "tangent manifold", or perhaps confusingly "tangent space". It would be logical to call it a tangent space-time if your manifold is a space-time manifold, but I'm not sure I've ever seen anyone do this.

Tensors live in the tangent manifold at a point, and every point in the manifold has a different tangent manifold just like every point on the surface of the sphere does.

If you think of a small enough section of space time, you can blur the distinction between the tangent manifold, and the actual manifold, in the same way that you can ignore the curvature of the Earth if you're only concerned with a small portion of its surface. This doesn't actually make them the same, so it's not necessarily a good idea, but people do it all the time anyway.

The lowest order tensors are basically vectors that span the tangent "space", and their duals. If you're not familar with linear algebra, or familiar but rusty, it's not a bad idea to read up the topic. One way of looking at dual vectors is a linear map from a vector to a scalar.

Suppose you have some coordinate system in the tangent space, which is conveniently flat. And you make the transformation x' = 2x.

Under that transformation, some tensor components will double, and some will be cut in half, it depends on what sort of tensor they are, if they are components in the x direction. Components in the y or z direction won't be affected.

If you have a vector and a dual vector, which as I mentioned is a linear map to a scalar, then there is some natural number associated with the pair, when you apply the dual vector to the vector.

This number has to be independent of the coordinates you use to have any physical meaning. Physical quantites won't change if you change the coordinates, by definition.

So you can immediately see that dual vectors and vectors can't transform the same way if you want this natural "product" to be preserved when you perform a coordinate transformation.

If you focus on the transformation properties, you can define a tensor formally by how it transforms when you change coordinates. The example I chose of a coordinate transformation, x'=2x, was just a simple example to illustrate the basic idea, you can find a formal definition of the tensor transformation rules in the textbooks. However, this simple specific example is good enough to get the basic idea of how and why tensors transform the way they do, and the textbooks will fill in all the details that follow from this broad overview.

 

[PeterDonis]

Let's say that I describe spacetime around non-rotating gravitating body using coordinate time and flat geometry. In that case transforming it to other state of motion makes different parts of gravitating body move at different speeds proportionally to local (coordinate) speed of light.

If it's a "gravitating body", then the spacetime around it is not flat and can't be described by flat geometry. So it's not clear what "transforming it to other state of motion" means. See below.

So if we do the same in GR it rises the question if we can apply the same transformation across the whole spacetime region containing gravitating body and get consistent result? Or do we have to apply different transformations at different points in order for them to stick together?

In GR you can use any coordinates you want, and therefore any transformation between coordinates is "valid", as long as you transform everything accordingly, including the metric. (There are some technical points about continuity, differentiability, etc. but we can restrict discussion here to cases where those points aren't important.) This includes transformations that depend on the coordinates, which can be thought of as "applying different transformations at different points". As long as you transform everything, including the metric, according to the proper rules for the type of object it is (pervect's post discusses some of those rules), then any quantities formed by contracting vectors and tensors (and their duals, as pervect points out) will be the same in the new coordinates as in the old.

As far as the physical "meaning" of a coordinate transformation, strictly speaking, there isn't one. Coordinates don't have direct physical meaning, so neither do transformations between them. In particular, in a curved spacetime, you can't really find a global transformation that uniquely corresponds to "putting the gravitating body in a different state of motion" the way you could with a test body in flat spacetime by making a global Lorentz transformation.

 

[atyy]

Idea that result depends only from relative velocities follows from principle of relativity. But how can we check that GR respects principle of relativity? Or maybe we know already that it doesn't?

GR does not respect the Principle of Relativity. The Principle of Relativity says that there is a preferred class of reference frames called inertial frames.

In GR, inertial frames can only be used locally, and don't cover all of spacetime.

Also, the relative velocity of distant objects (in the SR sense) is not defined.

GR respects the principle of general covariance, but so does SR.

The distinguishing feature of GR is "no prior geometry", not general covariance.

 

[PAllen]

GR does not respect the Principle of Relativity. The Principle of Relativity says that there is a preferred class of reference frames called inertial frames.

I would say instead, GR observes a less simple form of Principle of Relativity. First, general covariance per se gives a sort of trivial formulation: any coordinates can be used, so you can always pick coordinates where some chosen body is at rest. More substantively, you can start with locally a locally Minkowski frame built around a chosen timelike geodesic. This will locally give expression to physics which is locally like SR. Such a local frame can be extended over a large region (though you may not be able to cover the whole manifold), in natural (but non-unique) ways. In the sense I used, where you have a test body and a gravitational source, you can have an exact statement:

If you build coordinate system from the massive body at rest, in which a test body moves near c; and a different coordinate system where the test body is at rest and the massive body moves relativistically; then any computed observable or geometric invariant must be that same in both coordinate systems. In particular, all curvature invariants (rather than component representations of the curvature tensor) are the same.