I Inertial & non-inertial frames & the principle of equivalence

[Frank Castle]

One particular form of the equivalence principle states that

The laws of physics for freely falling particles in a gravitational field are locally indistinguishable from those in a uniformly accelerating frame in Minkowski spacetime

My question is, does one arrive at this conclusion from a Newtonian perspective, i.e. before positing that gravity is the manifestation of spacetime curvature?

 

[Frank Castle]

If this is the case, then this form of the EP makes sense, since according to an observer on the surface of a planet, within the gravitational field (which we assume to be uniform), particles in free-fall are in a uniformly accelerating reference frame. Viewing gravity as a real force in this context, the observer on the planet would see that according to the freely falling reference frame, the physical laws are those of special relativity. Equivalently, an observer at rest in a uniformly accelerating rocket in deep space (far from an gravitational fields) would observe that freely falling particles are accelerating uniformly (in exactly the same manner as in the gravitational field), and again, the rocket observer would conclude that the laws of physics in this free fall frame are those of special relativity. The two cases are indistinguishable. Thus, the laws of physics of freely falling particles in a gravitational field and a uniformly accelerating reference frame in Minkowski spacetime are locally indistinguishable.

 

[PeterDonis]

One particular form of the equivalence principle states

Where are you getting this from? Please give a reference.

 

[Frank Castle]

Where are you getting this from? Please give a reference.

Apologies, I’ve been reading this: http://www.nottingham.ac.uk/~ppzap4/AGR.pdf(pp 5-6)

And this:
http://www.rug.nl/research/portal/files/34926446/Complete_thesis.pdf (page 23)

 

[PeterDonis]

(page 23)

This appears to contain more or less the formulation you quoted in the OP. I would add some clarification as follows:

Locally in spacetime, the laws of physics for freely falling particles in a gravitational field in a frame at rest with respect to the source of the field are the same as those in a uniformly accelerating frame in Minkowski spacetime.

(Note that the first clarification is actually in the source, but you abbreviated it to "locally" in your formulation in the OP; I think the longer form is better. The second clarification is critical, since without it your statement is false.)

does one arrive at this conclusion from a Newtonian perspective

How can one, since Newtonian physics does not even contain the concept of "spacetime", let alone "Minkowski spacetime"?

 

[Frank Castle]

How can one, since Newtonian physics does not even contain the concept of "spacetime", let alone "Minkowski spacetime"?

Sorry, this is a bad phrasing on my part. What I meant by this, is does this argument originate from initially considering gravity as a real force? I’m a little confused by this formulation of the equivalence principle since I thought that free-fall frames were locally inertial frames, since objects in free-fall follow the laws of physics. However, in this formulation seems to state that the laws of physics in a freely falling laboratory in a gravitational field are locally indistinguishable from those in a uniformly accelerating reference frame in Minkowski spacetime. Especially the part on page 23 of the second reference that say “locally an observer can always eliminate gravity by moving to an accelerated frame of reference” (I’ve paraphrased a bit here). Maybe I’m misinterpreting this?!

Is the point that, locally in spacetime, the laws of physics for an observer at rest in an arbitrary gravitational field are indistinguishable from those in a uniformly accelerating reference frame in Minkowski spacetime. Furthermore, locally in spacetime, the laws of physics in a freely falling frame of reference are those of special relativity.

 

[PAllen]

No, you misunderstand the source as described by Peter. A frame at rest with respect to a gravitation source is NOT a free fall frame. Free falling particles in this frame will be accelerating, just like inertial particles in an SR accelerating frame.

A different aspect of the equivalence principle is local equivalence of SR inertial frames an free fall frames near a gravity source.

 

[PeterDonis]

does this argument originate from initially considering gravity as a real force?

No. That would be inconsistent, since the conclusion of the argument is that gravity is not a real force.

I thought that free-fall frames were locally inertial frames

They are. But the source you quoted from is not considering that aspect of the equivalence principle. It's considering a different aspect: the local equivalence of uniformly accelerated frames in Minkowski spacetime with frames at rest in a gravitational field.

 

[Frank Castle]

No, you misunderstand the source as described by Peter. A frame at rest with respect to a gravitation source is NOT a free fall frame. Free falling particles in this frame will be accelerating, just like inertial particles in an SR accelerating frame.

A different aspect of the equivalence principle is local equivalence of SR inertial frames an free fall frames near a gravity source.

No. That would be inconsistent, since the conclusion of the argument is that gravity is not a real force.
They are. But the source you quoted from is not considering that aspect of the equivalence principle. It's considering a different aspect: the local equivalence of uniformly accelerated frames in Minkowski spacetime with frames at rest in a gravitational field.

You’re right, I’ve got myself completely mixed up on this one.

So is it correct to say that the equivalence principle states that locally, it is impossible for an observer to distinguish whether they are at rest in an arbitrary gravitational field, or in a uniformly accelerating frame of reference in Minkowski spacetime, by carrying out any kind of experiment. That is, locally the laws of physics in a frame at rest in an arbitrary gravitational field are identical to those in a uniformly accelerating reference frame in Minkowski spacetime.

Furthermore, locally a freely falling reference frame in an arbitrary gravitational field constitutes a local inertial reference frame and the laws of physics will reduce to those of special relativity in such frames.

 

[PeterDonis]

is it correct to say that the equivalence principle states that locally, it is impossible for an observer to distinguish whether they are at rest in an arbitrary gravitational field, or in a uniformly accelerating frame of reference in Minkowski spacetime, by carrying out any kind of experiment

Yes.

locally the laws of physics in a frame at rest in an arbitrary gravitational field are identical to those in a uniformly accelerating reference frame in Minkowski spacetime

Yes. And these laws of physics are just the laws of special relativity, expressed in a uniformly accelerating frame.

locally a freely falling reference frame in an arbitrary gravitational field constitutes a local inertial reference frame

Yes.

and the laws of physics will reduce to those of special relativity in such frames

More precisely, they will reduce to the laws of SR, expressed in an inertial frame. These are the same laws as above, just expressed in a different frame.

 

[Frank Castle]

Yes. And these laws of physics are just the laws of special relativity, expressed in a uniformly accelerating frame.

Is this because an observer cannot locally distinguish between an arbitrary gravitational field and a uniformly accelerating reference frame in Minkowksi spacetime by performing any kind of experiment. The laws of physics in Minkowski spacetime are those of SR and so the laws of physics for the observer should be SR, but expressed in a uniformly accelerating reference frame?

Would it be correct then to say that locally within an arbitrary gravitational field, the laws of physics reduce to those of SR for all reference frames (inertial and non-inertial)?

 

[PeterDonis]

Is this because an observer cannot locally distinguish between an arbitrary gravitational field and a uniformly accelerating reference frame in Minkowksi spacetime by performing any kind of experiment.

Isn't that what you already said in post #9, and I agreed to in post #10?

The laws of physics in Minkowski spacetime are those of SR and so the laws of physics for the observer should be SR, but expressed in a uniformly accelerating reference frame?

Isn't that what I already said?

Would it be correct then to say that locally within an arbitrary gravitational field, the laws of physics reduce to those of SR for all reference frames (inertial and non-inertial)?

Statements about the laws of physics are independent of any choice of reference frame. At least, that's the proper way to consider such statements in relativity. You can deduce particular consequences of that for how the laws will appear when expressed in a particular frame, but the laws themselves don't depend on which frame you choose.

 

[Frank Castle]

Isn't that what you already said in post #9, and I agreed to in post #10?

Isn't that what I already said?

Sorry, yes I'm repeating things :/

Statements about the laws of physics are independent of any choice of reference frame. At least, that's the proper way to consider such statements in relativity. You can deduce particular consequences of that for how the laws will appear when expressed in a particular frame, but the laws themselves don't depend on which frame you choose.

If one is starting from a basic heuristic argument though, would the chain of arguments go like this:

1. The laws of physics reduce to those of SR in a local free-fall frame;

2. One then recasts the mathematical form of these laws into tensorial form;

3. The equations hold in one (set of) reference frame (a local free-fall frame), and hence, since they are in tensorial form, they hold in all reference frames.

 

[PeterDonis]

would the chain of arguments go like this

No. Logically speaking (though not historically), putting the laws in tensorial form comes first. You don't have to pick a frame at all to do that.

 

[Frank Castle]

No. Logically speaking (though not historically), putting the laws in tensorial form comes first. You don't have to pick a frame at all to do that.

Ok. I was trying to understand how it was done historically, and how Einstein arrived at these conclusions.

Given the requirement that the laws of physics should reduce to those of SR in local reference frames, are the generalisations of these equations to curved spacetime unique?

 

[PeterDonis]

I was trying to understand how it was done historically, and how Einstein arrived at these conclusions.

Historically, Einstein's recognition of the equivalence principle as crucial for extending relativity to cover gravity predates the expression of the laws of physics in tensorial form. So Einstein's original understanding of the EP didn't involve the "tensorial form" stuff at all.

In fact, Einstein's recognition of the EP as crucial predates even the Minkowski spacetime formulation of SR (Einstein's recognition of the EP as crucial was in 1907; Minkowski came up with his formulation in 1908). So if you want to understand Einstein's original thought process about the EP, you have to go back to his 1905 formulation of SR.

Given the requirement that the laws of physics should reduce to those of SR in local reference frames, are the generalisations of these equations to curved spacetime unique?

If you mean, are the expressions of the laws in tensorial form unique, yes.

 

[Frank Castle]

Historically, Einstein's recognition of the equivalence principle as crucial for extending relativity to cover gravity predates the expression of the laws of physics in tensorial form. So Einstein's original understanding of the EP didn't involve the "tensorial form" stuff at all.

In fact, Einstein's recognition of the EP as crucial predates even the Minkowski spacetime formulation of SR (Einstein's recognition of the EP as crucial was in 1907; Minkowski came up with his formulation in 1908). So if you want to understand Einstein's original thought process about the EP, you have to go back to his 1905 formulation of SR.

Ah ok. Maybe I'll stick with the modern interpretation then

If you mean, are the expressions of the laws in tensorial form unique, yes.

Ok cool, that's what I thought.

Thanks for your time and patience.

 

[Frank Castle]

Historically, Einstein's recognition of the equivalence principle as crucial for extending relativity to cover gravity predates the expression of the laws of physics in tensorial form. So Einstein's original understanding of the EP didn't involve the "tensorial form" stuff at all.

In fact, Einstein's recognition of the EP as crucial predates even the Minkowski spacetime formulation of SR (Einstein's recognition of the EP as crucial was in 1907; Minkowski came up with his formulation in 1908). So if you want to understand Einstein's original thought process about the EP, you have to go back to his 1905 formulation of SR.
If you mean, are the expressions of the laws in tensorial form unique, yes.

Sorry to resurrect this post, but I have a further question that I'm hoping you can help me with.

By the equivalence principle, within a sufficiently small neighbourhood of a given spacetime point the laws of physics are those of SR in both inertial and non-inertial reference frames, so what is the difference when one considers a larger region of a given coordinate chart?

The reason I ask is because in a non-inertial frame, Christoffel symbols will naturally be present in the equations due to fact of being in an accelerated reference frame. The same is true in a coordinate system in the presence of a gravitational field, however, if one considers a large enough region within a given coordinate chart one should be able to detect the effects of curvature. Shouldn't one be able to distinguish between the laws of SR expressed in non-inertial coordinates, and the case in which curvature has to be taken into account?

Is the point that, if we are in a non-inertial reference frame in which the laws of SR apply we can always find a coordinate transformation that reduces the metric to the Minkowksi metric. However, for larger regions around a spacetime point it is impossible to choose a coordinate system in which the metric reduces to the the Minkowski metric and curvature effects have to be taken into account.

 

[PeterDonis]

what is the difference when one considers a larger region of a given coordinate chart?

That you can no longer ignore the effects of spacetime curvature, if it is present (see below).

Shouldn't one be able to distinguish between the laws of SR expressed in non-inertial coordinates, and the case in which curvature has to be taken into account?

Yes, by computing the Riemann tensor and seeing whether it is zero or nonzero. If it's zero, you are in flat spacetime and any stuff like Christoffel symbols, etc. is just due to choosing non-inertial coordinates. If the Riemann tensor is nonzero, spacetime curvature is present and there is no such thing as global inertial coordinates--any inertial chart only works in a small enough patch of spacetime that curvature can be ignored.

 

[Frank Castle]

Yes, by computing the Riemann tensor and seeing whether it is zero or nonzero. If it's zero, you are in flat spacetime and any stuff like Christoffel symbols, etc. is just due to choosing non-inertial coordinates. If the Riemann tensor is nonzero, spacetime curvature is present and there is no such thing as global inertial coordinates--any inertial chart only works in a small enough patch of spacetime that curvature can be ignored.

So when one says that within a small enough patch of spacetime the laws of physics reduce to SR, is it the case that one could start in non-inertial coordinates within this sufficiently small patch, such that the Christoffel symbols don't vanish, but it would be possible to find a coordinate transformation such that they vanish and one can explicitly see that the laws are those of SR? If the patch is larger, then such a transformation does not exist as a consequence of spacetime curvature.

Is it correct to say that the equivalence principle implies that within a sufficiently small neighbourhood of each spacetime point the laws of physics are those of SR, regardless of ones choice of coordinates (i.e. in both inertial and non-inertial coordinates)?

If this is true though, then I'm confused about the fact that non-inertial frames are included since the Riemann tensor will not vanish (since the metric will only be Minkowski to second order).

 

[PAllen]

So when one says that within a small enough patch of spacetime the laws of physics reduce to SR, is it the case that one could start in non-inertial coordinates within this sufficiently small patch, such that the Christoffel symbols don't vanish, but it would be possible to find a coordinate transformation such that they vanish and one can explicitly see that the laws are those of SR? If the patch is larger, then such a transformation does not exist as a consequence of spacetime curvature.

You don’t need a coordinate transform to verify flatness. The curvature tensor is an expression based on the Christoffel symbols. If it is zero, this fact is true in all coordinates. For example, polar coordinates for a plane have nonzero christoffel symbols, but the curvature tensor constructed from them vanishes. Similarly, Rindler coordinates in SR have non vanishing christoffel symbols, but vanishing curvature tensor constructed from. You can never make the curvature tensor vanish at a point unless there is no curvature there.

 

[Frank Castle]

You don’t need a coordinate transform to verify flatness. The curvature tensor is an expression based on the Christoffel symbols. If it is zero, this fact is true in all coordinates. For example, polar coordinates for a plane have nonzero christoffel symbols, but the curvature tensor constructed from them vanishes. Similarly, Rindler coordinates in SR have non vanishing christoffel symbols, but vanishing curvature tensor constructed from. You can never make the curvature tensor vanish at a point unless there is no curvature there.

This is what I'm confused about though. The equivalence principle (EP) states that in a local free-fall frame the laws of physics should reduce to those of SR, which mathematically, means that within a sufficiently small region around a spacetime point, a local coordinate transformation exists such that the metric is Minkowksi (up to second order), with the Christoffel symbols vanishing within this region. However, the EP also states that locally, the effects of gravitation should be indistinguishable from that of a uniformly accelerating reference frame in Minkowski spacetime. So the laws of physics should reduce to those of SR in non-inertial reference frames within a sufficiently small patch of spacetime too, right?

 

[PAllen]

This is what I'm confused about though. The equivalence principle (EP) states that in a local free-fall frame the laws of physics should reduce to those of SR, which mathematically, means that within a sufficiently small region around a spacetime point, a local coordinate transformation exists such that the metric is Minkowksi (up to second order), with the Christoffel symbols vanishing within this region. However, the EP also states that locally, the effects of gravitation should be indistinguishable from that of a uniformly accelerating reference frame in Minkowski spacetime. So the laws of physics should reduce to those of SR in non-inertial reference frames within a sufficiently small patch of spacetime too, right?

The EP is effectively a first order statement. The curvature tensor measures second and higher order deviations, and physics attributable to curvature generally vanishes in the limit at a point, while curvature does not. Consider a trivial calculus analog. To first order, the tangent to a curve approximates the curve near the point of tangency. But the second derivative does not vanish at a point.

 

[PeterDonis]

is it the case that one could start in non-inertial coordinates within this sufficiently small patch, such that the Christoffel symbols don't vanish, but it would be possible to find a coordinate transformation such that they vanish and one can explicitly see that the laws are those of SR?

Yes. More precisely, you can explicitly see that the laws are those of SR as expressed in an inertial frame. But you don't have to express the laws of SR in an inertial frame to see that they're the laws of SR. You can write those laws in tensorial form, independent of any choice of coordinates.

Is it correct to say that the equivalence principle implies that within a sufficiently small neighbourhood of each spacetime point the laws of physics are those of SR, regardless of ones choice of coordinates (i.e. in both inertial and non-inertial coordinates)?

Go back and read what I said in post #12 (the last part).

I'm confused about the fact that non-inertial frames are included since the Riemann tensor will not vanish

Yes, it will. Don't make assumptions. Try it. For example, try computing the Riemann tensor for Rindler coordinates on Minkowski spacetime.

https://en.wikipedia.org/wiki/Rindler_coordinates

 

[Frank Castle]

Yes. More precisely, you can explicitly see that the laws are those of SR as expressed in an inertial frame. But you don't have to express the laws of SR in an inertial frame to see that they're the laws of SR. You can write those laws in tensorial form, independent of any choice of coordinates.

Does this mean that in an arbitrary frame of reference the laws of physics are those of SR, but in coordinate form include terms taking into account curvature?

I think what I've got myself confused about is what happens when one isn't considering a sufficiently small patch of spacetime. Are the non-gravitational laws of physics still those of SR (since they are in tensorial form and thus coordinate independent) when curvature cannot be neglected?

Is the point that the EP principle requires that within a sufficiently small neighbourhood of each spacetime point the laws of physics are those of SR, and this amounts to being able to find a local coordinate transformation in which the metric reduces to the Minkowski metric (to second order), where one can explicitly see that the laws of physics take on their SR form. Thus, any observer (for example a Rindler coordinate frame) can find such local coordinates. However, this coordinate transformation will only hold for a sufficiently small region, after which it breaks down and the effects of curvature become apparent and one must take into account the Christoffel symbols. One can furthermore calculate the Riemann tensor and find that it is non-vanishing and hence the presence of the Christoffel symbols is due to the curvature of spacetime and not the effects of an accelerated reference frame.

This is distinguished from SR, in which spacetime is globally Minkowski. An observer who starts of in Rindler coordinates will be able to find a global coordinate transformation that transforms the metric in its Minkowski form. Furthermore, if they compute the Riemann tensor in their (initial) frame of reference they will find that it vanishes and hence they can conclude that they are in flat spacetime.

 

[PeterDonis]

Does this mean that in an arbitrary frame of reference the laws of physics are those of SR, but in coordinate form include terms taking into account curvature?

There is no curvature in SR. In SR spacetime is flat.

Are the non-gravitational laws of physics still those of SR (since they are in tensorial form and thus coordinate independent) when curvature cannot be neglected?

If spacetime curvature cannot be neglected, there are no "non-gravitational laws of physics". "Spacetime curvature cannot be neglected" is the same as "gravity cannot be ignored".

after which it breaks down and the effects of curvature become apparent and one must take into account the Christoffel symbols

No. When the effects of curvature become apparent, one must take into account curvature. "Christoffel symbols" are not the same as "curvature". Curvature means a nonzero Riemann tensor, not nonzero Christoffel symbols.

You really need to spend some time looking at specific examples. Both @PAllen and myself have already suggested Rindler coordinates in Minkowski spacetime. In those coordinates, the Christoffel symbols are nonzero; but spacetime is flat, so the Riemann tensor is zero.

 

[Frank Castle]

No. When the effects of curvature become apparent, one must take into account curvature. "Christoffel symbols" are not the same as "curvature". Curvature means a nonzero Riemann tensor, not nonzero Christoffel symbols.

Apologies, this was very bad wording on my part.

You really need to spend some time looking at specific examples. Both @PAllen and myself have already suggested Rindler coordinates in Minkowski spacetime. In those coordinates, the Christoffel symbols are nonzero; but spacetime is flat, so the Riemann tensor is zero.

I have done the example of Rindler coordinates in Minkowksi spacetime, and I get that one can be in a non-inertial reference frame such that Christoffel symbols are non-zero, however, the Riemann tensor is zero indicating that spacetime is flat.

What I've confused myself over is that in tensorial form the laws of physics are valid in an arbitrary coordinate system, and so whether or not one is in a sufficiently small patch of spacetime they will be valid right (they will just take into account gravity, i.e. curvature for larger (local) regions? Now, if I consider a small enough region of spacetime, but happen to be in a non-inertial reference frame, how do I know that the laws of physics are those of special relativity within this patch, since if I compute the Riemann tensor it will not vanish, indicating that I'm in a curved spacetime? Is the point that, within this patch, I will be able to find a coordinate transformation such that the non-gravitational laws of physics are explicitly in SR form?

 

[PeterDonis]

if I consider a small enough region of spacetime, but happen to be in a non-inertial reference frame, how do I know that the laws of physics are those of special relativity within this patch, since if I compute the Riemann tensor it will not vanish, indicating that I'm in a curved spacetime?

I strongly suggest that you actually try this exercise instead of just waving your hands. The answer depends on what coordinates you use on the small patch of spacetime.

If you use the natural coordinates for the small region of spacetime--coordinates that assume that that small patch is flat, and ignore curvature--then if you compute the Riemann tensor in those coordinates, it will be zero. But of course those coordinates are only an approximation; they don't exactly describe the actual geometry of the spacetime, and if you go outside the small patch of spacetime, the approximation becomes unusable because the error it introduces is too large. (Note that the error is not zero except at one single point, the one you picked as the spacetime origin of the coordinates, i.e., the point at the "center" of the small patch.)

If you use global coordinates in a curved spacetime, and compute the Riemann tensor in those coordinates, then you will find that it is nonzero. But those coordinates will not look the same as the coordinates described above, even in the small patch of spacetime referred to above. The differences will be small within the small patch of spacetime (if you've done things correcty, they will be too small to actually measure), but they will not be zero (except at the single point at the "center" of the small patch, if you've chosen coordinates appropriately).

Is the point that, within this patch, I will be able to find a coordinate transformation such that the non-gravitational laws of physics are explicitly in SR form?

There won't be a coordinate transformation, in the usual sense, that takes you from global coordinates on a curved spacetime, to local coordinates in a small patch of that spacetime in which the metric is assumed to be flat. The local coordinates are an approximation; as above, they don't describe the geometry of the actual spacetime. (To be precise, they describe the flat geometry of the tangent space at the point you picked as the spacetime origin--the point at the "center" of the small patch.)

 

[PAllen]

@PeterDonis, can you explain more what you mean by natural coordinates, and your claim around this? What I know is that using either Riemann normal coordinates or Fermi Normal coordinates, you can make the connection vanish at a point, as well as the metric be Minkowski. However, due to the Riemann tensor depending on derivatives of the connection, this all has no effect on the curvature tensor at than point. Any contraction of it will be the same at this point as for any other coordinates.

 

[PeterDonis]

can you explain more what you mean by natural coordinates

I mean something like Minkowski coordinates or Rindler coordinates--coordinates that describe a flat metric. In a small patch of spacetime, if we are working in a local inertial frame, we are using Minkowski coordinates on the small patch.

using either Riemann normal coordinates or Fermi Normal coordinates, you can make the connection vanish at a point, as well as the metric be Minkowski

Yes, but only at that single point. Over a finite region, the metric is not Minkowski in these coordinates. So over a finite region, even a small one, there will be differences between these coordinates and the coordinates of a local inertial frame. The differences will be second order, so over a small enough patch of spacetime they can be ignored--but the way they are ignored is to just use the Minkowski metric as an approximation on the small patch.

 

[PeterDonis]

due to the Riemann tensor depending on derivatives of the connection, this all has no effect on the curvature tensor at than point. Any contraction of it will be the same at this point as for any other coordinates.

For any other coordinates that describe the actual curved metric of the spacetime, yes. But not for approximate coordinates that treat spacetime as flat within a small patch (such as Minkowski coordinates in a local inertial frame). In the approximate coordinates, the Riemann tensor will be zero (because the nonzero Riemann tensor components in the actual curved metric are second order, so in the approximation being used they are ignored).

 

[PAllen]

For any other coordinates that describe the actual curved metric of the spacetime, yes. But not for approximate coordinates that treat spacetime as flat within a small patch (such as Minkowski coordinates in a local inertial frame). In the approximate coordinates, the Riemann tensor will be zero (because the nonzero Riemann tensor components in the actual curved metric are second order, so in the approximation being used they are ignored).

Still a bit confused by this. Coordinates don’t determine a metric, only it’s expression on a coordinate basis. Would it be correct to describe what you are proposing as approximating a small region of the manifold by its tangent plane, and that for most measurements over a small region you won’t detect the inconsistencies from doing so? But not for all expressions of measurements, e.g. the ratio of the deviation angles in triangle from 180 to the area of the triangle approaches a fixed nonzero limit in a given plane of measurement (this is a particular curvature scalar). However, if below a certain size you cannot detect the angular deviation, you lose the ability to detect curvature.

I’ve always seen this approached in terms of perturbation of the Minkowski metric. That a general metric can be expresed near any point as Minkowski plus secon order terms in appropriate coordinates. Then you argue that that for sufficiently local measurements the effect of the second order terms is below the reachable precision of whatever you are measuring, compared to just using the Minkowski metric. This approach doesn’t pretend that curvature has vanished.

 

[PeterDonis]

Would it be correct to describe what you are proposing as approximating a small region of the manifold by its tangent plane

Yes. The reason we do that is that the laws of SR, strictly speaking, only hold in flat spacetime, so if we are going to say that the laws of SR hold in a sufficiently small patch of spacetime, we have to approximate that small patch by its flat tangent space.

This approach doesn’t pretend that curvature has vanished.

No, it doesn't. But if curvature is nonzero, the laws of SR are no longer the complete laws of physics, because the laws of SR don't include any effects due to spacetime curvature. So another way of viewing the approximation I've been describing is that it ignores any laws of physics that involve gravity (i.e., spacetime curvature), and only deals with non-gravitational laws of physics. For many purposes, this is sufficient. In fact, even some effects that are usually thought to involve gravity, such as gravitational time dilation over a small height range, can be analyzed within these limitations (this is how Einstein originally derived gravitational time dilation as a prediction). A more precise statement would be that the approximation I've been describing ignores any laws of physics that involve tidal gravity.

 

[Frank Castle]

I strongly suggest that you actually try this exercise instead of just waving your hands. The answer depends on what coordinates you use on the small patch of spacetime.

Using a Riemann normal coordinate system, one has that $g_{\mu\nu}=\eta_{\mu\nu}+\mathcal{O}(\varepsilon^{2})$, where $\varepsilon<<1$ parametrises the higher order curvature corrections to the metric, and $\Gamma^{\mu}_{\;\;\alpha\beta}=0$. Given this, it follows that the Riemann tensor vanishes to second order, i.e. $R^{\mu}_{\;\;\alpha\nu\beta}= 0 +\mathcal{O}(\varepsilon^{2})$.

Would this be correct at all? Is this what you mean by neglecting curvature such that the Riemann tensor is zero and one can use the laws of SR?

I mean something like Minkowski coordinates or Rindler coordinates--coordinates that describe a flat metric. In a small patch of spacetime, if we are working in a local inertial frame, we are using Minkowski coordinates on the small patch.

By Minkowski coordinates do you mean locally Cartesian coordinates?

 

[vanhees71]

@PeterDonis, can you explain more what you mean by natural coordinates, and your claim around this? What I know is that using either Riemann normal coordinates or Fermi Normal coordinates, you can make the connection vanish at a point, as well as the metric be Minkowski. However, due to the Riemann tensor depending on derivatives of the connection, this all has no effect on the curvature tensor at than point. Any contraction of it will be the same at this point as for any other coordinates.

I guess one good example are a tetrad of the tangent space in one given point of spacetime. The equivalence principle is telling us that you can always find such a tretrad (and thus infinitely many) at any (regular) point of a spacetime.

 

[PeterDonis]

and $\Gamma^{\mu}_{\;\;\alpha\beta}=0$.

Riemann normal coordinates do not give $\Gamma^{\mu}{}_{\alpha\beta}=0$ everywhere in a small patch of spacetime, only at one single point. At other points within the patch, the Christoffel symbols do not vanish.

By Minkowski coordinates do you mean locally Cartesian coordinates?

I mean coordinates in which the metric is $\eta_{\mu \nu}$.

 

[PeterDonis]

Given this, it follows that the Riemann tensor vanishes to second order, i.e. $R^{\mu}_{\;\;\alpha\nu\beta}= 0 +\mathcal{O}(\varepsilon^{2})$.

No, it doesn't. Even in Riemann normal coordinates, the Riemann tensor does not vanish at the chosen point. Only the Christoffel symbols do (and the metric is exactly $\eta_{\mu \nu}$ at that point).

 

[Frank Castle]

No, it doesn't. Even in Riemann normal coordinates, the Riemann tensor does not vanish at the chosen point. Only the Christoffel symbols do (and the metric is exactly ημν\eta_{\mu \nu} at that point).

That's what I thought, which is why I was saying that this is only to second order, i.e. if we neglect terms that are second order in derivatives of the metric.

Like you say, the metric only equals $\eta_{\mu\nu}$ at the single point (such that the tangent space to that point is Minkowksi, right?), but for a small region around that point, the metric should be $\eta_{\mu\nu}$ up to second-order, and these corrections will be small for a small enough neighbourhood of said point.

I was trying to understand your comments about the Riemann tensor vanishing. I was under the impression that it is impossible to set up a Minkowski coordinate system on a curved manifold and the best we can do is to use Riemann normal coordinates (RNC), in which the metric is only exactly Minkowski at a single point?!

 

[PeterDonis]

which is why I was saying that this is only to second order

Which is not correct. At the chosen point, $\epsilon$ is zero, so what you were claiming was that the Riemann tensor is zero at the chosen point, and its components at other points in the small patch go like $\epsilon^2$. That's wrong. The Riemann tensor is nonzero even at the chosen point.

for a small region around that point, the metric should be $\eta_{\mu\nu}$ up to second-order

The corrections to the metric are second order in $\epsilon$. The metric is not the same as the Riemann tensor.

Have you actually tried to do this calculation yourself? Or looked at a detailed treatment in a textbook? Sean Carroll's online lecture notes, chapter 2 (pp. 50-51 in the version I have) discuss this precise topic.

I was trying to understand your comments about the Riemann tensor vanishing.

The Riemann tensor vanishes in flat spacetime. The tangent space at any point in any spacetime is flat. So the tangent space has a vanishing Riemann tensor. So when we use Minkowski coordinates in a local inertial frame centered on some point in a curved spacetime, we are, as I responded to @PAllen , approximating the actual spacetime in a small patch around the chosen point by the tangent space at that point.

I was under the impression that it is impossible to set up a Minkowski coordinate system on a curved manifold

It is. But the tangent space at any point is not curved.

the best we can do is to use Riemann normal coordinates (RNC)

Riemann normal coordinates describe the actual small patch of spacetime, not the tangent space. So heuristically, Riemann normal coordinates tell you the error involved in approximating a small patch of spacetime around a chosen point by the tangent space at that point.

 

[Frank Castle]

Which is not correct. At the chosen point, ϵϵ\epsilon is zero, so what you were claiming was that the Riemann tensor is zero at the chosen point, and its components at other points in the small patch go like ϵ2ϵ2\epsilon^2. That's wrong. The Riemann tensor is nonzero even at the chosen point.

Sorry, you’re right. I didn’t think that one through properly. I realize I’ve been being a bit cavalier with certain notions. I’ll try to be more careful.

The corrections to the metric are second order in ϵϵ\epsilon. The metric is not the same as the Riemann tensor.

Have you actually tried to do this calculation yourself? Or looked at a detailed treatment in a textbook? Sean Carroll's online lecture notes, chapter 2 (pp. 50-51 in the version I have) discuss this precise topic.

I haven’t gone through the complete calculation myself, but I have read through Carroll’s notes and Schutz’s book. Carroll talks about the equivalence principle as corresponding to our ability to choose a RNC system locally around any given point, such that the laws of physics take on their SR form. Surely, more precisely, this is strictly only true at a single point, and the laws are only approximately SR for nearby points?

The Riemann tensor vanishes in flat spacetime. The tangent space at any point in any spacetime is flat. So the tangent space has a vanishing Riemann tensor. So when we use Minkowski coordinates in a local inertial frame centered on some point in a curved spacetime, we are, as I responded to @PAllen , approximating the actual spacetime in a small patch around the chosen point by the tangent space at that point.

I haven’t come across any GR notes that discuss using Minkowski coordinates in curved spacetime, unless this corresponds to choosing a tetrad frame?

 

[PeterDonis]

Surely, more precisely, this is strictly only true at a single point, and the laws are only approximately SR for nearby points?

I don't think this correctly describes what is going on.

The laws are tensor laws, which, as I said, relate quantities at the same spacetime point. So as long as you are talking about non-gravitational laws (i.e., laws that don't involve the Riemann tensor or any tensor derived from it), you can always make them take exactly their SR form at any point. If you are concerned that they might not take their SR form at some other point nearby, you just switch to coordinates centered on the new point.

It is true that, since the laws are differential equations, you can't just consider a single point if you want to give them physical meaning; properly defining derivatives requires considering an open neighborhood about the chosen point. And since using treating that open neighborhood as flat is an approximation, it would be true that the SR forms of the laws, which must treat the open neighborhood as flat, are only approximate. But that would be true at the chosen point itself, not just at nearby points, because the approximation basically means you are approximating derivatives at the chosen point in the actual curved spacetime (which would be covariant derivatives) as derivatives in the flat tangent space (which are just partial derivatives).

Also, as I noted above (and as I noted in post #33 in response to @PAllen ), if you look at laws that involve gravity (i.e., the Riemann tensor or any tensor derived from it--the Einstein Field Equation is such a law), then there is no such thing as "SR form" for these laws, even considered as tensor equations at a single spacetime point, because in SR the Riemann tensor and all tensors derived from it vanish, so any "laws" involving them are vacuous. So, for example, you can't write the Einstein Field Equation in any "SR form", because the Einstein tensor in flat spacetime vanishes, but the stress-energy tensor does not, so they can't possibly be equal to one another.

I haven’t come across any GR notes that discuss using Minkowski coordinates in curved spacetime

That's because you don't. You use Minkowski coordinates on the tangent space at a point, which is flat.

 

[Frank Castle]

The laws are tensor laws, which, as I said, relate quantities at the same spacetime point. So as long as you are talking about non-gravitational laws (i.e., laws that don't involve the Riemann tensor or any tensor derived from it), you can always make them take exactly their SR form at any point. If you are concerned that they might not take their SR form at some other point nearby, you just switch to coordinates centered on the new point.

Ah ok, so one just shifts to a new RNC frame centred around the new point that you’re considering.

Also, as I noted above (and as I noted in post #33 in response to @PAllen ), if you look at laws that involve gravity (i.e., the Riemann tensor or any tensor derived from it--the Einstein Field Equation is such a law), then there is no such thing as "SR form" for these laws, even considered as tensor equations at a single spacetime point, because in SR the Riemann tensor and all tensors derived from it vanish, so any "laws" involving them are vacuous. So, for example, you can't write the Einstein Field Equation in any "SR form", because the Einstein tensor in flat spacetime vanishes, but the stress-energy tensor does not, so they can't possibly be equal to one another.

Good point. Would it be correct to say that the equations describing non-gravitational physics reduce to their SR form in RNCs as they are minimally coupled to the metric (as required by the EP)?

That's because you don't. You use Minkowski coordinates on the tangent space at a point, which is flat.

Is this essentially what one does when choosing to neglect gravity, for example, if one is interested in standard model physics in situations where gravity is negligible? What this correspond to working in a tetrad frame?

Are there any GR books that discuss this at all?

 

[PeterDonis]

Would it be correct to say that the equations describing non-gravitational physics reduce to their SR form in RNCs

Not really, because in RNCs the metric is not flat. RNCs are coordinates on the actual curved spacetime, not the flat tangent space. Strictly speaking, the equations describing non-gravitational physics can only take their SR form in the flat tangent space, because SR assumes that spacetime is flat.

Is this essentially what one does when choosing to neglect gravity, for example, if one is interested in standard model physics in situations where gravity is negligible?

The standard model is quantum field theory in flat spacetime, so yes.

 

[PeterDonis]

Are there any GR books that discuss this at all?

I'm not sure GR books discuss the precise issue we have been discussing. Carroll's online lecture notes, for example, talk about the tangent space at a point p, but then segue directly into Riemann normal coordinates without really making it clear that those are coordinates on the actual curved manifold, not the tangent space at a point p.

 

[Frank Castle]

I'm not sure GR books discuss the precise issue we have been discussing. Carroll's online lecture notes, for example, talk about the tangent space at a point p, but then segue directly into Riemann normal coordinates without really making it clear that those are coordinates on the actual curved manifold, not the tangent space at a point p.

This is the problem I find. I’ve read several set of notes that make it sound as if the laws of physics reduce to SR in normal coordinates, but this isn’t really true, as you’ve pointed out.

From what I understand, from our discussions in this thread, the laws of physics will reduce to exactly those of SR at only one point in RNCs, i.e. the point $p$ that the coordinate system is centred around. At nearby points they will be approximately SR, but this approximation will break down as one moves further away from $p$.

What I’m unsure of now, is how the equivalence principle is satisfied. Is the point that one can always the laws of physics to exactly SR at each point in spacetime, and that the laws of physics will stay SR to a good enough approximation at sufficiently nearby points such that one can not detect curvature by carrying out non-gravitational experiments.

Or, is it that at each point $p$ there is a tangent space equipped with a Minkowski metric and so is flat. In this frame the laws are those of SR but once one moves away from the point $p$ one is strictly no longer on the manifold in this coordinate system, but for a sufficiently small neighbourhood the approximation holds? In practice how does one do this? Does this tangent space correspond to the tangent vector space that exists at each point on the manifold? How does one construct a coordinate system here?

 

[PeterDonis]

the laws of physics will reduce to exactly those of SR at only one point in RNCs, i.e. the point $p$ that the coordinate system is centred around

No, that's not correct. Remember what I said about derivatives. The laws of SR assume spacetime is flat; if the actual spacetime (which is what RNCs are coordinates on) is not flat, then the derivatives that appear in the laws of SR, which assume spacetime is flat, will not be exactly equal to actual derivatives in the actual curved spacetime. This is true even for derivatives at the point $p$.

The only exact statement that can be made about the point $p$ in RNCs is that the metric is exactly Minkowski at that point.

What I’m unsure of now, is how the equivalence principle is satisfied.

It's satisfied by not expressing it in vague ordinary language, but in precise math. It turns out that there are a number of different ways of translating the vague ordinary language into precise math. These include, at least, the two ways we have been discussing: in terms of the approximation that uses the tangent space at a point (where we say the laws of SR hold exactly in the tangent space, but the tangent space is only an approximation to the actual curved spacetime), or in terms of the approximation that uses Riemann normal coordinates on the actual curved spacetime (where we say the laws of SR only hold approximately, and comparing the RNC with Minkowski coordinates tells us how good the approximation is).

 

[PAllen]

Actually, for Fermi Normal coordinates for a geodesic world line, the metric is exactly Minkowski all along the world line, and the connection (based on metric first derivatives) vanishes all along the world line. It is only curvature, based on connection derivatives (metric second derivatives), which is non vanishing on the world line defining the FNC.

 

[PeterDonis]

for Fermin Normal coordinates for a geodesic world line, the metric is exactly Minkowski all along the world line, and the connection (based on metric first derivatives) vanishes all along the world line.

Yes, this is true, but once you move off the worldline, the "corrections" to the Minkowski metric and the connection behave differently, as functions of the coordinates, than they do in Riemann normal coordinates. Heuristically, RNCs are "optimized" to make things as close as possible to the Minkowski metric and zero first derivatives in a small patch of spacetime around a single event. FNCs are "optimized" to make things as close as possible to the Minkowski metric and zero first derivatives in a small "world tube" around a single geodesic worldline. Different tools for different purposes.

 

[Frank Castle]

No, that's not correct. Remember what I said about derivatives. The laws of SR assume spacetime is flat; if the actual spacetime (which is what RNCs are coordinates on) is not flat, then the derivatives that appear in the laws of SR, which assume spacetime is flat, will not be exactly equal to actual derivatives in the actual curved spacetime. This is true even for derivatives at the point ppp.

The only exact statement that can be made about the point ppp in RNCs is that the metric is exactly Minkowski at that point.

Ah, I missed that subtlety. I naively assumed that because the metric is exactly Minkowski, and the Christoffel symbols are zero, at a point that everything would reduce to SR form.

So is it acceptable then to state that the equivalence principle corresponds mathematically to the ability to be able to construct RNCs in a sufficiently small neighbourhood of each spacetime point, and within this region the laws of physics will be those of SR to a good approximation (so long as we are sufficiently close to that point)?

Mathematically, does the tangent space, in which we construct a Minkowski coordinate system, correspond to the tangent space of vectors $T_{p}M$ at each point $p$? Is there some kind of "tangent map" that one can use to construct said Minkowski coordinates?

 

[PAllen]

An observation worth making is the scale at which deviations from SR are detectable near earth. If one restricts oneself to indications of spatial curvature, for a long time there was no prospect of detecting deviations from Euclidean geometry. On scales of 10 meters, we are still 5 to 10 orders of magnitude away from detecting anything due to earth. However, LIGO has effectively measured spatial Euclidean curvature deviations due to GW over kilometer scales. On the other hand, curvature of spacetime is currently detectable in volumes of order 6 inches by 1 second. Thus deviations from SR near Earth can now be detected over truly remarkably small regions. Note, I am not talking about just clocks detecting ‘gravitational red shift’, as this does not really detect curvature over small regions. Instead, I am referring to SQUID accelerometers which detect deviations from Rindler behavior over 6 inch distance scales.