Inertial & non-inertial frames & the principle of equivalence

[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.

 

[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.