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

[PeterDonis]

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)?

That's one way of translating the ordinary language into math, yes, as I said.

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 $p$?

Yes.

Is there some kind of "tangent map" that one can use to construct said Minkowski coordinates?

The origin of the coordinates corresponds to the zero vector at the point $p$. Points with nonzero coordinates correspond to nonzero vectors with appropriate lengths and directions. Basically we are identifying vectors with the points at which the tips of their arrows are if the tails are at the origin; or, to put it another way, we are identifying vectors with the displacements to which they give rise once we pick a standard unit of distance.

 

[Frank Castle]

The origin of the coordinates corresponds to the zero vector at the point ppp. Points with nonzero coordinates correspond to nonzero vectors with appropriate lengths and directions. Basically we are identifying vectors with the points at which the tips of their arrows are if the tails are at the origin; or, to put it another way, we are identifying vectors with the displacements to which they give rise once we pick a standard unit of distance.

So would the coordinates be of the form $x^{\mu}(p)=x^{\mu}(p_{0})+tX^{\mu}(p)$?

By the way, I think I’ve finally found a useful book that discusses this, de Felice’s “Relativity on curved manifolds”.

One thing I’m still a little confused about is why the laws of physics won’t take on their SR form at the centre of a RNC system? Surely as the metric reduces to the Minkowski metric at that point and the Christoffel symbols vanish, such that covariant derivatives reduce to partial derivatives, equations will reduce to the form they have in SR at that point?

 

[PeterDonis]

would the coordinates be of the form $x^{\mu}(p)=x^{\mu}(p_{0})+tX^{\mu}(p)$?

No, they're just Minkowski coordinates.

 

[PeterDonis]

Surely as the metric reduces to the Minkowski metric at that point and the Christoffel symbols vanish, such that covariant derivatives reduce to partial derivatives, equations will reduce to the form they have in SR at that point?

"Equations will reduce to the form they have in SR at that point" is not quite the same as "the laws of SR hold exactly at that point", which is the statement I was responding to earlier. The laws of SR assume that spacetime is flat. There is no getting around that. So if spacetime is not in fact flat, the laws of SR cannot hold exactly. But yes, in RNCs centered on a chosen point, the equations (more precisely, the non-gravitational equations, the ones that don't involve the Riemann tensor or any tensor derived from it) will have the same form they have in SR, since, as you say, the metric is exactly Minkowski at that point and the Christoffel symbols are exactly zero at that point, so at that point covariant derivatives in RNCs are equivalent to partial derivatives.

 

[Frank Castle]

"Equations will reduce to the form they have in SR at that point" is not quite the same as "the laws of SR hold exactly at that point", which is the statement I was responding to earlier. The laws of SR assume that spacetime is flat. There is no getting around that. So if spacetime is not in fact flat, the laws of SR cannot hold exactly. But yes, in RNCs centered on a chosen point, the equations (more precisely, the non-gravitational equations, the ones that don't involve the Riemann tensor or any tensor derived from it) will have the same form they have in SR, since, as you say, the metric is exactly Minkowski at that point and the Christoffel symbols are exactly zero at that point, so at that point covariant derivatives in RNCs are equivalent to partial derivatives.

Thanks for the clarification.

No, they're just Minkowski coordinates.

Does the spacetime metric induce a Minkowski metric on the tangent space at each point? Does this correspond to using vielbeins? Is the reason why a RNC system is possible because it corresponds to an exponential map from the tangent space at a given point, and the tangent space is Minkowski?

 

[PeterDonis]

Does the spacetime metric induce a Minkowski metric on the tangent space at each point?

No, because the tangent space is not a subspace of the actual spacetime. We know the tangent space has a Minkowski metric because it's flat by definition.

Does this correspond to using vielbeins?

Not really, no.

Is the reason why a RNC system is possible because it corresponds to an exponential map from the tangent space at a given point, and the tangent space is Minkowski?

Not really, no.

Where are you going with these questions? It doesn't seem to me like we are getting anywhere.

 

[Frank Castle]

Where are you going with these questions? It doesn't seem to me like we are getting anywhere.

I'm just still a bit confused about the tangent space stuff to be honest. I get RNCs I guess because they are coordinates on the actual spacetime manifold, although I they corresponded to an exponential map from the tangent space at a point to an infinitesimal neighbourhood about that point.

Re. the tangent space. Is the point that tangent vectors in the tangent space to a given can be treated as infinitesimal displacements around that point and so can reasonably approximate the manifold in an infinitesimal neighbourhood at that point? Does one then simply choose Minkowski coordinates on the tangent space?

Sorry, I have taken in what you've previously said, it's just I seem to be stuck on the understanding of these points. I've read a couple of texts that have very briefly mentioned that the equivalence principle can be translated mathematically to the ability to approximate the laws of physics to those of SR in an infinitesimal neighbourhood of a point by using the tangent space to that point, but neither of them went into any detail at all about this.

Maybe I'm getting too bogged down in the minutiae of it all, but it feels like one should be really precise and careful when it comes to GR.

We know the tangent space has a Minkowski metric because it's flat by definition.

This is because the tangent space has a vector space structure so must be flat, right?

 

[Frank Castle]

I think I might understand things a little better...

Is it correct to say that in SR, because Minkowski space is a vector space we can naturally identify it with it’s tangent space at each point, and then “glue” these together to define global inertial frames. However, when we transition to general relativity in which spacetime is curved, it no longer has a vector space structure. Hence there is no way to naturally connect neighbouring tangent spaces at each point on the manifold and no global inertial frames can exist. The best we can do is identify the manifold and it’s tangent space at each given point such that it well approximates the manifold within an infinitesimal neighbourhood each point. The tangent space has a vector space structure and so is, by definition, flat Minkowski space. Thus, as you said, the laws of physics are exactly SR on each tangent space - global coordinate frames exist and are related by the usual Lorentz transformations. This translates to the manifold in such a way that the non-gravitational laws of physics take on their SR form for an infinitesimal neighbourhood around a given point, with the manifold being approximately Minkowski within this neighbourhood.

 

[PeterDonis]

they corresponded to an exponential map from the tangent space at a point to an infinitesimal neighbourhood about that point.

Why do you think that?

Is the point that tangent vectors in the tangent space to a given can be treated as infinitesimal displacements around that point and so can reasonably approximate the manifold in an infinitesimal neighbourhood at that point?

That's useful, but it's not the main point. The main point of the tangent space is that it is a vector space, which the curved spacetime itself is not. And in order to write the laws of physics in tensorial form you need to have a vector space to work with (because tensors and all the operations we want to do on tensors require a vector space).

This is because the tangent space has a vector space structure so must be flat, right?

No. The metric is additional structure on the tangent space, over and above the vector space structure. A vector space does not have to have a metric at all.

However, the reverse of what you're saying is valid: if a metric space (a space with a metric, which does not have to be a vector space at all) happens to be flat, then you can always derive a vector space structure from the metric. That is how we know that the tangent space is a vector space: we define it to be flat, so we can use the vector space structure derived from the flat metric.

 

[Frank Castle]

Why do you think that?

I thought that the Riemann normal coordinate system was an exponential mapping from the tangent space at a point to the manifold at that point?

And in order to write the laws of physics in tensorial form you need to have a vector space to work with (because tensors and all the operations we want to do on tensors require a vector space).

So is it the case that when one uses a coordinate basis induced by a coordinate system on the spacetime manifold the component form of the mathematical representations of the laws don't take on their special relativistic form exactly because the manifold is curved?

The metric is additional structure on the tangent space

If the metric is defined on the tangent spaces of the spacetime manifold, how does it describe the curved geometry on the manifold? Is it that by representing it in a coordinate basis (induced by some coordinate system on the manifold) one gains a description of the manifold geometry in that coordinate patch, because one can use it to measure the lengths of the coordinate basis vectors and hence the distances along the curves that they are tangent to?

Naively, I've always thought of $g_{\mu\nu}(x)$ as describing the geometry within a given coordinate patch, but I'm confused as to how this can be so when the metric tensor is defined at each particular point acting on the tangent space to that point? I get that the metric tensor is a tensor field that acts on vectors in the tangent space to each point on the manifold and intuitvitely describes the dot product between them. In this way one can use it to determine lengths of vectors and also angles between to given tagent curves that intersect at a point. I know I must be missing something here though, as if it were only ever defined within an infinitesimal neighbourhood of each spacetime point, then one could always use coordinates such that it reduces to the Minkowski metric.

 

[PeterDonis]

I thought that the Riemann normal coordinate system was an exponential mapping from the tangent space at a point to the manifold at that point?

And I asked why you think that. Just repeating it doesn't answer the question. Where are you getting it from? What textbook? What peer-reviewed paper?

is it the case that when one uses a coordinate basis induced by a coordinate system on the spacetime manifold the component form of the mathematical representations of the laws don't take on their special relativistic form exactly because the manifold is curved?

You shouldn't be thinking about coordinates until you understand things in coordinate-independent terms. SR assumes spacetime is flat, so obviously whatever the laws of physics are in a curved spacetime, they can't be the exact laws of SR. That is true regardless of what coordinates you choose.

If the metric is defined on the tangent spaces of the spacetime manifold, how does it describe the curved geometry on the manifold?

The metric on the tangent space is additional structure on the tangent space. The metric on the tangent space is not the same as the metrc on the actual manifold. That should be obvious since the tangent space metric is flat and the metric on the actual manifold is curved.

 

[Frank Castle]

And I asked why you think that. Just repeating it doesn't answer the question. Where are you getting it from? What textbook? What peer-reviewed paper?

Apologies, I misunderstood what you were getting at here. I’ve been reading Nakahara’s book “Geometry, Topology and Physics” . In section 7.4.4 he discusses normal coordinate systems in terms of a map $EXP: T_{p}M\rightarrow M$, $X_{q}\mapsto q$ where $X_{q}\in T_{p}M$. I realize now that the notation might be misleading and he’s not referring to an exponential map given how he’s defined the actual mapping of tangent vectors to points.

The metric on the tangent space is additional structure on the tangent space. The metric on the tangent space is not the same as the metrc on the actual manifold. That should be obvious since the tangent space metric is flat and the metric on the actual manifold is curved.

But I thought the metric tensor was defined as $g:T_{p}M\times T_{p}M\rightarrow\mathbb{R}$, i.e. it acts on tangent vectors? In Nakahara’s book he makes no mention of there being to different metrics defined, one on the manifold and one on the tangent spaces.

 

[PeterDonis]

I thought the metric tensor was defined as $g:T_{p}M\times T_{p}M\rightarrow\mathbb{R}$, i.e. it acts on tangent vectors?

All tensors act on tangent vectors. But there are still two metrics and you have to avoid confusing them. The tangent space itself is flat, and has its own flat metric, distinct from the metric of the actual curved manifold. In other words, in addition to the tensor $g$, the metric of the curved manifold, there is another tensor, usually called $\eta$, which is the metric of the flat tangent space. Both of these are tensors and so they act on tangent vectors; but they are different tensors.

 

[Frank Castle]

All tensors act on tangent vectors. But there are still two metrics and you have to avoid confusing them. The tangent space itself is flat, and has its own flat metric, distinct from the metric of the actual curved manifold. In other words, in addition to the tensor $g$, the metric of the curved manifold, there is another tensor, usually called $\eta$, which is the metric of the flat tangent space. Both of these are tensors and so they act on tangent vectors; but they are different tensors.

Ah ok. I’ve never seen this distinction being explicitly made before. Do you know of any good textbooks that discuss this?

In practice, how does one make the distinction between them? Is it simply that one chooses to evaluate $g$ on coordinate basis vectors that are tangent to coordinate curves on the actual manifold, thus describing the geometry on the manifold

 

[PeterDonis]

I’ve never seen this distinction being explicitly made before.

Probably because textbooks generally have no need to discuss the actual metric of the tangent space, as distinct from the metric of the manifold, in any detail; just saying that the tangent space is flat is enough.

In practice, how does one make the distinction between them?

If you see the metric written in the form $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$, or something similar, that's how. But that form is usually used only in the weak field approximation, where it is assumed that $h_{\mu \nu} \ll 1$.

Is it simply that one chooses to evaluate $g$ on coordinate basis vectors that are tangent to coordinate curves on the actual manifold

No; the distinction between any two tensors is independent of any choice of coordinates.

 

[Frank Castle]

Probably because textbooks generally have no need to discuss the actual metric of the tangent space, as distinct from the metric of the manifold, in any detail; just saying that the tangent space is flat is enough.

Ah ok, so it’s implicitly assumed then.

If you see the metric written in the form gμν=ημν+hμνgμν=ημν+hμνg_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}, or something similar, that's how. But that form is usually used only in the weak field approximation, where it is assumed that hμν≪1hμν≪1h_{\mu \nu} \ll 1.

So in this case is $g_{\mu\nu}$ the pullback metric from the tangent space?

No; the distinction between any two tensors is independent of any choice of coordinates.

Sorry, what I meant by this is, if I represent both metrics in a coordinate basis (induced by a coordinate chart on the manifold) how do I differentiate between the two? Does one simply calculate the Riemann tensor for both?

 

[PeterDonis]

if I represent both metrics in a coordinate basis (induced by a coordinate chart on the manifold) how do I differentiate between the two?

Um, by their components? [Edit: actually, strictly speaking, this won't work--see my post #69.]

Does one simply calculate the Riemann tensor for both?

To do that you need their components anyway, so you need to know those first.

 

[PeterDonis]

in this case is $g_{\mu\nu}$ the pullback metric from the tangent space?

No. Why would you think that?

 

[PeterDonis]

if I represent both metrics in a coordinate basis (induced by a coordinate chart on the manifold)

Actually, strictly speaking, you can't even do this, because the actual spacetime and its tangent space, considered as manifolds, are different manifolds, so you would need two different coordinate charts (one for the spacetime and one for the manifold), and each chart can only represent one of the two metrics.

 

[PeterDonis]

I’ve been reading Nakahara’s book “Geometry, Topology and Physics” .

I should note that I am not an expert on this topic, or on this textbook. Also, it is a very advanced textbook, and I'm not sure you have the background for it. We seem to be increasing your confusion in this thread instead of reducing it.

It might be better at this point to go back to your original question about the EP. Has it been answered? If not, what has not been answered? To be clear, answering your original question about the EP should not require all of this advanced differential geometry and topology. The fact that we are getting deeper into those topics indicates, to me, that we have gotten off the track.

 

[Frank Castle]

Um, by their components?

I was meaning in terms of which one is the flat metric and which one corresponds to the curved manifold, but I guess this question has been answered now.

I've been re-reading Sean Carroll's notes and he talks about the exponential map as a local mapping of the tangent space to the manifold via $exp_{p}:T_{p}M\rightarrow M$, $exp_{p}(k^{\mu})=x^{\mu}(\lambda =1)$ where $x^{\mu}(\lambda)$ is a solution to the geodesic equation subject to $\frac{dx^{\mu}(0)}{d\lambda}=k^{\mu}$. Is this what you were referring to on being able to approximate the manifold locally with the tangent space near a given point?

No. Why would you think that?

Sorry, ignore me on this one. I was mis-remembering a section I'd read in Sean Carroll's notes.

 

[Frank Castle]

I should note that I am not an expert on this topic, or on this textbook. Also, it is a very advanced textbook, and I'm not sure you have the background for it. We seem to be increasing your confusion in this thread instead of reducing it.

It might be better at this point to go back to your original question about the EP. Has it been answered? If not, what has not been answered? To be clear, answering your original question about the EP should not require all of this advanced differential geometry and topology. The fact that we are getting deeper into those topics indicates, to me, that we have gotten off the track.

The original question has been answered, and I think I understood it. Apologies for getting so side-tracked, I managed to get myself muddled with some concepts that I thought I understood. I guess I'm going to have to go back and read/re-read some things. Are there any particularly good textbooks or notes that you've read and found helpful?

 

[PeterDonis]

he talks about the exponential map as a local mapping of the tangent space to the manifold via $exp_{p}:T_{p}M\rightarrow M$, $exp_{p}(k^{\mu})=x^{\mu}(\lambda =1)$ where $x^{\mu}(\lambda)$ is a solution to the geodesic equation

Note that last qualifier: a solution to the geodesic equation. That's crucial. What Carroll is saying here is that a point in spacetime and a particular tangent vector at that point, taken together, determine a unique geodesic throughout the spacetime (or at least throughout some open connected region of the spacetime). We can then construct a map from the tangent space at the chosen point to the spacetime by mapping each tangent vector to the point in the spacetime that lies exactly one unit (of affine parameter) along the unique geodesic in the spacetime determined by that tangent vector.

Is this what you were referring to on being able to approximate the manifold locally with the tangent space near a given point?

No. The mapping I described above, heuristically, is a mapping from the tangent space at a point to a "unit circle" in the spacetime centered on that point. It is not a mapping from the tangent space, considered as a flat manifold, to an open neighborhood of the chosen point in the actual curved manifold.

Once again, we seem to be getting further away from the actual topic of this thread; this discussion is not supposed to be a general discussion about differential geometry. Is there anything specifically about the equivalence principle that still needs to be clarified?

 

[PeterDonis]

Are there any particularly good textbooks or notes that you've read and found helpful?

I personally think Carroll's lecture notes are enough of a treatment of differential geometry for GR unless you are actually doing active research in the field. MTW give a more detailed treatment, but it can be hard to follow. Wald also gives a more detailed treatment, but it's more abstract and I'm not sure the physical meaning comes through as clearly.

 

[Frank Castle]

Is there anything specifically about the equivalence principle that still needs to be clarified?

Just a (hopefully) quick clarification. By the way, thanks for answering my further questions despite going of on a massive tangent (pardon the pun), sorry it ended up being so long.

So, if I've understood correctly, the equivalence principle corresponds to our ability to construct a RNC system on the spacetime manifold, within which the laws of physics take their SR form (mathematically), however, the metric is only Minkowski to first-order (apart from at the origin of the coordinate system). Since this is a coordinate system on the actually manifold, physics is only approximately that of SR within the local neighbourhood of the origin of this coordinate system. Alternatively, one can work in the tangent space to a given point in which the laws of physics are exactly those of SR - this will approximate an infinitesimal neighbourhood of the manifold around a given point precisely because the manifold is (pseudo-) Riemannian.

The will (probably) be the last thing I wanted to ask related to this is, does a RNC system generally cover a smaller patch of the manifold than a more general coordinate system, in which the laws of physics do not reduce to their SR form?

I personally think Carroll's lecture notes are enough of a treatment of differential geometry for GR unless you are actually doing active research in the field. MTW give a more detailed treatment, but it can be hard to follow. Wald also gives a more detailed treatment, but it's more abstract and I'm not sure the physical meaning comes through as clearly.

Thanks very much for the recommendations.

 

[PeterDonis]

if I've understood correctly, the equivalence principle corresponds to our ability to construct a RNC system on the spacetime manifold, within which the laws of physics take their SR form

The EP is independent of any choice of coordinates. And the ability to construct RNC centered on a point is a property of any manifold, as a matter of mathematics, independent of any physical interpretation. So I don't know if what you say here is a useful way of looking at it.

 

[Frank Castle]

The EP is independent of any choice of coordinates. And the ability to construct RNC centered on a point is a property of any manifold, as a matter of mathematics, independent of any physical interpretation. So I don't know if what you say here is a useful way of looking at it.

I think this is still a bit of a sticking point for me. I get that the EP also requires that the laws of physics are those of SR for a sufficiently small patch of spacetime in uniformly accelerating reference frames as well as free-fall frames, but I’m unsure how this is realized in practice? I mean, if one is in a non-inertial reference frame, how does one know how local a region around a given point one has to be for the EP to hold?
In RNCs these is more explicitly obvious, since the derivative of the Christoffel symbols, $\frac{\partial\Gamma^{\mu}_{\;\alpha\beta}}{\partial x^{\nu}}=-\frac{1}{3}\left(R^{\mu}_{\;\alpha\beta\nu}+R^{\mu}_{\;\beta\alpha\nu}\right)$, determines how far one can move from the origin of the coordinate system before curvature becomes non-negligible (explicitly one uses the Jacobi equation to calculate the geodesic deviation).

 

[PeterDonis]

I get that the EP also requires that the laws of physics are those of SR for a sufficiently small patch of spacetime in uniformly accelerating reference frames as well as free-fall frames

That is because the EP is independent of your choice of coordinates, and all you are doing when you use an accelerating frame instead of a free-fall frame is to choose different coordinates (Rindler coordinates vs. Minkowski coordinates).

how does one know how local a region around a given point one has to be for the EP to hold?

This has nothing to do with your choice of coordinates. It has to do with how curved the spacetime is as compared to how accurate your measurements are.

In RNCs these is more explicitly obvious

Yes, but that's a calculational convenience, not a necessity.

 

[Frank Castle]

That is because the EP is independent of your choice of coordinates, and all you are doing when you use an accelerating frame instead of a free-fall frame is to choose different coordinates (Rindler coordinates vs. Minkowski coordinates).

So does one simply exploit the EP by noting that one can calculate a quantity using SR in either an inertial or non-inertial frame and this calculation will be valid for a sufficiently small region in curved spacetime?

This has nothing to do with your choice of coordinates. It has to do with how curved the spacetime is as compared to how accurate your measurements are.

Can one not calculate the geodesic deviation of test particles to determine the range of validity of ones chosen local inertial coordinates (i.e. the range at which curvature causes the geodesics to intersect)?

 

[PeterDonis]

does one simply exploit the EP by noting that one can calculate a quantity using SR in either an inertial or non-inertial frame and this calculation will be valid for a sufficiently small region in curved spacetime?

You can do that, yes, and the EP says it will work.

Can one not calculate the geodesic deviation of test particles to determine the range of validity of ones chosen local inertial coordinates (i.e. the range at which curvature causes the geodesics to intersect)?

You can do that for any coordinates. Geodesic deviation is independent of coordinates.

 

[Frank Castle]

You can do that for any coordinates. Geodesic deviation is independent of coordinates.

By this do you mean that one can use any coordinate system you want (inertial or non-inertial) such that the laws of physics are those of SR for a sufficiently small neighbourhood - one can calculate the geodesic deviation in any of these coordinate systems to determine how small this neighbourhood has to be in order for the approximation to of SR to hold?

 

[PeterDonis]

By this do you mean that one can use any coordinate system you want (inertial or non-inertial) such that the laws of physics are those of SR for a sufficiently small neighbourhood - one can calculate the geodesic deviation in any of these coordinate systems to determine how small this neighbourhood has to be in order for the approximation to of SR to hold?

Yes.

 

[Frank Castle]

Yes.

Ok great, I think I'm getting it now. So is the point that if one considers larger regions of a given coordinate system the approximation breaks down and one has to take into account the effects of the gravitational field? The equations (for the non-gravitational laws of physics) will look the same as they do in a non-inertial reference frame (i.e. including connection terms), however, the Riemann tensor will be non-zero indicating that the spacetime is curved (this is true in an infinitesimal neighbourhood of a point too, but the point is the tidal effects are too small to be observable for small enough regions). Furthermore, the geodesic deviation for finite patches of the coordinate system will be non-negligible meaning that full GR is required in order to correctly describe physical experiments.

 

[PeterDonis]

is the point that if one considers larger regions of a given coordinate system the approximation breaks down and one has to take into account the effects of the gravitational field?

Not larger regions of a given coordinate system. Larger regions of the spacetime. All of this is independent of any choice of coordinates. As I said, it depends on how curved the spacetime is and how accurate your measurements are. Those are independent of coordinates.

 

[Frank Castle]

Not larger regions of a given coordinate system. Larger regions of the spacetime. All of this is independent of any choice of coordinates. As I said, it depends on how curved the spacetime is and how accurate your measurements are. Those are independent of coordinates.

Sorry, by larger region I was assuming this corresponded to covering a larger region of spacetime.

So depending on how accurate one's measurements are and how curved spacetime actually is will determine the size of the region of spacetime around each point in which the laws of SR (approximately) hold, and this will be true for any coordinate system?

If one can always choose a RNC system, and furthermore, because the laws of physics are in tensorial form, one can choose any coordinate system in which the laws of physics are those of SR for sufficiently small neighbourhoods of each point, is it the case that the only real point where GR comes in is determining the geodesics of spacetime such that a RNC can be constucted, and working out the geodesic deviation such that one can determine how small the region around each point has to be in order for curvature to be negligible?

 

[PeterDonis]

depending on how accurate one's measurements are and how curved spacetime actually is will determine the size of the region of spacetime around each point in which the laws of SR (approximately) hold

Yes.

and this will be true for any coordinate system?

It is true independently of coordinates. You seem to have the logic backwards. You don't first choose coordinates and then figure out the size of the region. You first figure out the size of the region, using coordinate-independent facts (the accuracy of your measurements and the curvature of spacetime are both coordinate-independent), and then, if you must, you choose coordinates and calculate what the coordinate-independent facts translate to in those coordinates.

is it the case that the only real point where GR comes in is determining the geodesics of spacetime such that a RNC can be constucted, and working out the geodesic deviation such that one can determine how small the region around each point has to be in order for curvature to be negligible?

You make it sound like this isn't very much. In fact it's everything. "GR comes in" in determining the actual curved geometry of the spacetime. That is everything. It's not just a small thing added on.

 

[Frank Castle]

You make it sound like this isn't very much. In fact it's everything. "GR comes in" in determining the actual curved geometry of the spacetime. That is everything. It's not just a small thing added on.

Sorry, I realize it's a much bigger deal than I make it sound. I was just wondering how this enters the non-gravitational laws of physics - since they are in tensorial form they "look" the same whether or not spacetime is curved, it's just in coordinate form that they differ, i.e. partial derivatives becoming covariant derivatives and the metric becoming non-Minkowski, however, this would be true in a non-inertial frame in flat spacetime too. Can differences be seen, for example, from the EM wave equation, in which a term proportional to curvature appears in curved spacetime?

 

[PeterDonis]

partial derivatives becoming covariant derivatives and the metric becoming non-Minkowski

Both of these statements are independent of coordinates.

 

[Frank Castle]

Both of these statements are independent of coordinates.

Ah ok. So this is the key point - the fact that the derivatives become covariant derivatives and the metric non-Minkowski is a due to the manifold being curved, which is a coordinate independent statement.

 

[PeterDonis]

the fact that the derivatives become covariant derivatives and the metric non-Minkowski is a due to the manifold being curved, which is a coordinate independent statement

Yes.

 

[Frank Castle]

Yes.

Ok great, I think it’s finally sinking in. Thanks for your time and patience, it’s much appreciated.

 

[jeremyfiennes]

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

What does "in Minkowski spacetime" mean in this context? Does it apply to both frames, or only the second?

 

[Ibix]

What does "in Minkowski spacetime" mean in this context? Does it apply to both frames, or only the second?

This is analogous to saying that the surface of the Earth is indistinguishable from a Euclidean plane, over a small enough region.

 

[jeremyfiennes]

This is analogous to saying that the surface of the Earth is indistinguishable from a Euclidean plane, over a small enough region.

So "Minkowski" spacetime is simply "flat" spacetime, a region small enough for dx²+dy²+dz²-c²dt² to be effectively zero?

 

[Ibix]

So "Minkowski" spacetime is simply "flat" spacetime, a region small enough for dx²+dy²+dz²-c²dt² to be effectively zero?

I think you have the right idea, but your maths is badly wrong. First of all, $dx^2+dy^2+dz^2-c^2dt^2\simeq 0$ is (in actual Minkowski spacetime) the region near the surface of the light cone in some coordinate system, and is of infinite extent. In curved spacetime I don't think it's necessarily meaningful. I think you probably meant $dx^2+dy^2+dz^2+c^2dt^2\simeq 0$, which is still coordinate dependent but at least defines a small volume.

The next problem is that spacetime may be "flat enough" over a region that isn't the same scale in different directions in spacetime. For example, at the Lagrange points of a reasonably distant binary black hole system, you could sit there for a long time - but you can't move far before curvature becomes apparent.

One way to approach this is to pick coordinates such that the metric is diagonal at your location. Spacetime is approximately Minkowski in the region where the second derivatives of the metric are zero to whatever precision your measures will tolerate.

 

[haushofer]

So "Minkowski" spacetime is simply "flat" spacetime, a region small enough for dx²+dy²+dz²-c²dt² to be effectively zero?

My cents: you should be carefull by identifying Minkowski spacetime via its line-element (metric), because a coordinate transformation can seriously mangle it up without changing the spacetime itself! "Flat" spacetime means that the Riemann tensor is zero,

<br /> R^{\rho}{}_{\mu\nu\sigma} = 0 \ \ \rightarrow \ \ \ g_{\mu\nu} = \eta_{\mu\nu} \,.<br />

I denoted the (components of the) Minkowski solution by $\eta_{\mu\nu}$. And since this is a tensor equation, it will be zero in any coordinate system.

 

[jeremyfiennes]

Spacetime is approximately Minkowski

This resumes my query, which is basically one of terminology and definition. 1) what does this statement mean, exactly? 2) What is the criterion for saying whether a spacetime is "Minkowski"?

 

[sweet springs]

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).

Rieman tensor, second order differential of metric, comes from difference of second order from the flat space.

 

[Ibix]

This resumes my query, which is basically one of terminology and definition. 1) what does this statement mean, exactly? 2) What is the criterion for saying whether a spacetime is "Minkowski"?

Minkowski spacetime has a metric $ds^2=dt^2-dx^2-dy^2-dz^2$ everywhere. Any metric can be made into this form at any point, but cannot be made to have the form globally. So you can always change coordinates so that special relativity applies at the place you are, and the smoothness of the metric means that SR will be a decent approximation in a small region around you (e.g. inertial objects initially at rest with respect to each other will remain near enough at rest). But if you go far away from your chosen point you'll find SR gets to be a worse and worse prediction.

 

[PeterDonis]

Minkowski spacetime has a metric $ds^2=dt^2-dx^2-dy^2-dz^2$ everywhere.

As @haushofer cautioned in post #96, you really shouldn't identify Minkowski spacetime this way because the form of the line element can change if you change coordinates. (Consider, for example, Rindler coordinates on Minkowski spacetime). It's safer to say that Minkowski spacetime is globally flat--the Riemann tensor vanishes everywhere. That statement is independent of coordinates (because if a tensor vanishes in one coordinate system it must vanish in all).