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

[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).

 

[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.

 

[Nugatory]

My question is: exactly how did Poincaré and Lorentz arrive at this formula?

By working backwards, asking themselves which formula would lead to consistent and usable results.

Einstein's contribution was to work forwards, showing that the transformations follow from previously underappreciated facts (the two postulates of the 1905 paper) about how the universe works.

 

[Ibix]

As I understand it, it's a mathematical patch that makes the Maxwell equations transform consistently. That's all the justification there is for Lorentz and Poincare, and it's specifically used in electromagnetism.

Einstein was able to derive the transforms from a clear basis - his postulates.

 

[jeremyfiennes]

As I understand it, it's a mathematical patch that makes the Maxwell equations transform consistently. That's all the justification there is for Lorentz and Poincare, and it's specifically used in electromagnetism

Thanks.

 

[Ibix]

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.

Agreed, but that wasn't quite what I said, as @PeterDonis pointed out. I said that the metric of Minkowski spacetime is diag(1,-1,-1,-1), which isn't quite the same as what you are (correctly) saying.

I think (?) that the correct statement in coordinate terms is that one can always choose coordinates such that the metric in those coordinates is diag(1,-1,-1,-1) at some chosen point and "nearly" that in some neighbourhood. Hence, expressed in those coordinates, physics equations in that neighbourhood can't be very different from physics equations in inertial frames in SR. In a genuinely flat spacetime one can pick coordinates such that the metric takes that simple form everywhere, and the "neighbourhood" is infinite. But this is not possible in curved spacetime.

I think that's right. But as Peter pointed out in #100, working in terms of tensors is safer because the coordinate-invariance is built in.

 

[PeterDonis]

I think (?) that the correct statement in coordinate terms is that one can always choose coordinates such that the metric in those coordinates is diag(1,-1,-1,-1) at some chosen point and "nearly" that in some neighbourhood.

That's correct. And you can even define what "nearly" means more precisely: it means you can choose coordinates such that all of the first derivatives of the metric are zero at the chosen point, but you cannot choose coordinates such that all of the second derivatives of the metric are zero at the chosen point. The second derivatives contain information about the curvature; heuristically, you can set all but twenty of them to zero, and the twenty remaining will be the components of the Riemann curvature tensor, which cannot be made to vanish in any coordinate system unless the spacetime is globally flat (i.e., Minkowski).

 

[DAH]

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

Yes.

Hi Peter

I'm no expert, far from it really, but I was thinking about the Equivalence Principle and if it would be possible by experiment to distinguish between gravity and an accelerating frame such as a rocket, even when the tidal effects are negligible.

The Pound-Rebka experiment successfully detected a blueshift in light when photons were sent towards the surface of the Earth from a height of 22.6 m. The formula for the blueshift is given by: $$
z=-\frac{gh}{c^{2}}$$

According to EP if the same experiment was carried out on a rocket accelerating at 1g then they would detect the exact same blueshift in light, is that correct?

However, I was thinking about how the blueshift would be observed over time in the accelerating frame, and i came across this paper: https://arxiv.org/abs/1907.06332

According to the paper above the blueshift would drift with time and if photons were sent in the other direction then the redshift will also drift with time. The formula for the blueshift drift is given by:
$$z=-\frac{aL}{\left ( c+at \right )^{2}}$$

So, I would like to know, if this is a flaw in EP and could this experiment be used to distinguish between gravity and an accelerating frame?

Thanks
DAH

 

[PeterDonis]

if it would be possible by experiment to distinguish between gravity and an accelerating frame such as a rocket, even when the tidal effects are negligible.

It isn't.

According to EP if the same experiment was carried out on a rocket accelerating at 1g then they would detect the exact same blueshift in light, is that correct?

If we assume that tidal effects are negligible, yes.

According to the paper above the blueshift would drift with time

I don't think this paper is a reliable source. It does not appear to have been peer reviewed, and the "drift" effect it claims does not seem consistent with what is known about Rindler observers in flat spacetime.

 

[DrGreg]

I don't think this paper is a reliable source. It does not appear to have been peer reviewed, and the "drift" effect it claims does not seem consistent with what is known about Rindler observers in flat spacetime.

I just had a quick skim through that paper and it states it's not about Rindler observers, but about observers who all experience identical constant proper acceleration. (So I think it might be what are sometimes called "Bell's spaceship observers".)

It's not a fair comparison with the Pound-Rebka experiment because the transmitter and receiver aren't a fixed distance apart (as measured by themeselves).

 

[Ibix]

So, in short, if we set up a Pound-Rebka type experiment with both ends at rest in the sense meant by this paper, the ends of the experiment would move apart as measured by (e.g.) a radar set next to it. I think that could be interpreted as tidal gravity (local accelerometers read the same, the relative velocity is initially zero, yet the distance changes), so we would not expect the equivalence principle to apply on any timescale over which the distance change would be detectable.

 

[PeterDonis]

I just had a quick skim through that paper and it states it's not about Rindler observers, but about observers who all experience identical constant proper acceleration. (So I think it might be what are sometimes called "Bell's spaceship observers".)

If that is the case, then I'm confused by the statement in the caption to Fig. 1 of the paper, that the detector will observe blueshift from source A. For the Bell congruence, there will be redshift observed in both directions. Only the Rindler congruence will have blueshift observed in the "downward" direction.

 

[DrGreg]

If that is the case, then I'm confused by the statement in the caption to Fig. 1 of the paper, that the detector will observe blueshift from source A. For the Bell congruence, there will be redshift observed in both directions. Only the Rindler congruence will have blueshift observed in the "downward" direction.

Fig 1 appears in a section discussing the problem within Newtonian physics.

By the way, the paper seems to have been published this year in Chinese Physics C, Volume 44, Number 7 (see abstract at https://iopscience.iop.org/article/10.1088/1674-1137/44/7/075103/pdf), which is on the Web of Science list.

 

[PeterDonis]

Fig 1 appears in a section discussing the problem within Newtonian physics.

Ah, I missed that.

 

[PeterDonis]

if we set up a Pound-Rebka type experiment with both ends at rest in the sense meant by this paper, the ends of the experiment would move apart as measured by (e.g.) a radar set next to it. I think that could be interpreted as tidal gravity (local accelerometers read the same, the relative velocity is initially zero, yet the distance changes)

No, it can't, because we are talking about flat spacetime, in which tidal gravity is zero.

You can't look at deviation of non-geodesic worldlines to assess tidal gravity. You have to look at geodesic deviation. In other words, it's not enough to have "local accelerometers read the same" as part of the initial condition; it has to be "local accelerometers read zero".

 

[Sagittarius A-Star]

For the Bell congruence, there will be redshift observed in both directions. Only the Rindler congruence will have blueshift observed in the "downward" direction.

The discussed paper says, that there is a blueshift in down-direction in the Bell spaceships scenario. This text is not in the "Newton" chapter:

A. Hyperbolic metric and redshift drift
...
Similarly, we can consider light source A on the right of the detector in Figure 1. The result shows that a blueshift, namely, z <0, is observed by the detector

Source:
https://arxiv.org/pdf/1907.06332.pdf

Consistent to that is the result for a "ping" scenario between Bell spaceships in another paper, see Fig.3:

As noted in the text, the leading spaceship loses sight of the trailing ship, with the observed energy tending to zero. The trailing spaceship initially sees an increase in the blueshifting of the leading spaceship, before it decreases back towards unity.

Source:
https://arxiv.org/pdf/1712.05276.pdf

 

[Sagittarius A-Star]

For the Bell congruence, there will be redshift observed in both directions.

In the inertial reference frame, in which both Bell spaceships started and keep distance, it can be shown, that the "downwards" direction must be blueshifted, not redshifted. When the leading spaceship sends a short light puls, both spaceships have the same momentarily velocity $v_1$. Later, when the trailing spaceship receives that ligth pulse, it moves with $v_2> v_1$ into the light. From that follows a Doppler blueshift.