Inertial & non-inertial frames & the principle of equivalence

[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,

$$R^{\rho}{}_{\mu\nu\sigma} = 0 \ \ \rightarrow \ \ \ g_{\mu\nu} = \eta_{\mu\nu} \,.$$

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).

 

[jeremyfiennes]

Sorry, I'm not there yet. In standard 3-d space dt²–dx²–dy²–dz²=0. Is ds² as defined here then a measure of deviation from flatness, i.e. of curvature? But that seems to go against your "Minkowski spacetime is globally flat". And for that statement to make any sense to me, I have to know what 'Minkowski spacetime' is, i.e. how it is defined, which I still don't.

 

[PeterDonis]

Is ds² as defined here then a measure of deviation from flatness, i.e. of curvature?

No. It's the [Edit: squared] spacetime distance between two points. Curvature is described by the Riemann curvature tensor.

 

[Ibix]

Is ds2 as defined here then a measure of deviation from flatness, i.e. of curvature?

No. $ds^2$ is the distance-squared between two events, $(x,y,z,t)$ and $(x+dx,y+dy,z+dz,t+dt)$.

Think of a flat plane. It's flat, it has simple rules like "initially parallel lines remain parallel".

Now think of the surface of the earth. It has a much more complex geometry. For example, parallel lines don't stay parallel, they cross, and eventually meet themselves coming the other way. However, in any small region, you can pretend that the Earth is flat - this is why you can tile your floor using square tiles without bothering about the curvature. The Earth is approximately Euclidean over a small region.

The same is true of curved spacetimes. They are approximately flat over a small region. Just as initially parallel lines on the floor remain close enough to parallel for you to tile your kitchen, but not enough to plan an intercontinental flight, inertial objects initially at rest will stay at rest close enough for bouncing around the inside of the ISS but not for its whole orbit. The only difference is that a flat plane is Euclidean. A flat spacetime is Minkowski.

Measures of curvature are measures of how bad an approximation Minkowski (or Euclidean) geometry is. The Riemann tensor is the full deal for this.

 

[PAllen]

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).

But the question being answered was what approximately Minkowski means. Part of that is that the metric has given signature, which does mean it can be diagonalized to a certain form at any point. This is one common definition of a pseudo-Riemannian manifold.

 

[jeremyfiennes]

Thanks all. I realize I have to go back and really get behind the Lorentz transformation. In <http://www.ulb.ac.be/sciences/ptm/pmif/ProceedingsHP/Reignier.pdf>, for instance, I find that Lorentz "without any further justification" defined local time as t' = γ(t - vx/c²). And that Poincaré used the same relation, "but didn’t give any detail about his calculation". And so on. My question is: exactly how did Poincaré and Lorentz arrive at this formula? Based on what reasoning? I can't find it anywhere.