Metric tensor for a uniformly accelerated observer

[Pyter]

TL;DR Summary

Computing the metric tensor and the Riemannian manifold topology for a simple accelerating observer

Hello all,

let's suppose we have, in a flat spacetime, two observers O and O', the latter speeding away from O, with an uniform acceleration $a$.
In the Minkowski spacetime chart of O, the world-line of O' can be drawn as a parable.
We know that the Lorentz boost at every point of the world-line can be approximated with that of an O' moving with constant speed $v(t)=at$ (as $v$ increases, the $x'$ and $t'$ axes draw closer scissors-style, as it were).
If we consider this scenario from a GR point of view, the Minkowski chart is the Euclidean projection of a 2D Riemannian manifold (we might consider 1 space dimension only for the sake of simplicity), with $x^0=t$ and $x^1=x$,
My questions are:

1. What are the values of the $g^{\mu\nu}$ as a function of $(x,t)$, i.e. $g^{\mu\nu}(x^0, x^1)$?
2. What's the shape of the manifold embedded in the 3D space?.
3. What are the functions mapping the manifold to the Euclidean 2D plane?

Of course the situation of O' should be physically equivalent to that of a still observer in a gravitational field.

 

[PeterDonis]

My questions are

Look up "Rindler coordinates".

 

[Pyter]

Look up "Rindler coordinates".

Thanks, I'm checking those out and they seem to refer to "constant proper acceleration" and "hyperbolic motion" in which the path appears as a hyperbola in a Minkowski chart: I guess this it the real equivalent of a still observer in a gravitational field.
In my proposed scenario the acceleration is constant as measured by O, and O' world-line is a parabola and not a hyperbola. So I guess the acceleration measured by O' is not constant and she would equivalently feel immersed in a varying gravitational field.
Anyway the latter is the example I find more often in textbooks about GR, and I was curious on how it translates to GR.

 

[PAllen]

Constant acceleration of some body in an inertial frame can only continue for a short period of time before reaching the c limit. The proper acceleration (or pseudo gravity experienced) of the body rapidly approaches infinite.

 

[PeterDonis]

the latter is the example I find more often in textbooks about GR

It is? All of the examples I know of in GR textbooks have constant proper acceleration, not constant coordinate acceleration. What textbooks are you referring to?

 

[Pyter]

I've just re-read my 1st question and the trivial answer came to me: O is in a flat space, so its $g^{\mu\nu}==\eta^{\mu\nu}$ (diagonal).
But O' doesn't experience flat space, so $g^{\mu\nu}$ in her frame is generally different from $\eta^{\mu\nu}$.
If she transform her $g^{\mu\nu}$ with a Lorentz boost at every point of her world-line, she gets a $\eta^{\mu\nu}$ too. If she doesn't, she has the same metric tensor of a still observer in a gravitational field.
So my questions can be re-phrased as:

1. what's O' metric tensor -> answer: in the Newtonian approximation, it's very close to $\eta^{\mu\nu}$ with a very small extra term added to $g^{00}$ representing the gravitational potential. But outside the Newtonian approximation?
2. what's the shape of the Riemannian manifold of an (1-space dimension) observer immersed in a gravitational field?
3. (unchanged)

 

[Pyter]

It is? All of the examples I know of in GR textbooks have constant proper acceleration, not constant coordinate acceleration. What textbooks are you referring to?

I've studied for instance a collection of lectures called Intro to Differential Geometry and General Relativity by S. Waner.

 

[Dale]

In my proposed scenario the acceleration is constant as measured by O, and O' world-line is a parabola and not a hyperbola.

This is not physically possible for an extended time.

Anyway the latter is the example I find more often in textbooks about GR, and I was curious on how it translates to GR.

No. The Rindler coordinates are by far more common. I have never even heard of anyone trying what you suggest.

 

[pervect]

Constant coordinate acceleration in an inertial frame is impossible, because the accelerating object would exceed the speed of light.

I'm not sure what other coordinates would be interesting - constant coordinate acceleration doesn't have a meaning until one specifries the coordinates.

 

[PeterDonis]

I've studied for instance a collection of lectures called Intro to Differential Geometry and General Relativity by S. Waner.

You mean these?

https://www.zweigmedia.com/diff_geom/tc.html

If so, which chapter, section, exercise, etc?

 

[PeterDonis]

I guess the acceleration measured by O' is not constant and she would equivalently feel immersed in a varying gravitational field.

That is correct for the case of constant coordinate acceleration, yes. And, as others have already noted, that case can only continue for a finite time, since it is impossible in relativity for O' to reach the speed of light.

 

[Pyter]

You mean these?

https://www.zweigmedia.com/diff_geom/tc.html

If so, which chapter, section, exercise, etc?

In my edition, between Corollary 9.5 and Proposition 9.6 there's a Minkowski chart in which the accelerating ball world-line is parabola-shaped, or so it seems.

 

[PeterDonis]

In my edition, between Corollary 9.5 and Proposition 9.6 there's a Minkowski chart in which the accelerating ball world-line is parabola-shaped, or so it seems.

On what basis do you think it's a parabola? There is no equation given, just a diagram. (I'm looking at a PDF of the book.)

Further, the text after the diagram says, in describing the event labeled as "R" in the diagram:

"R: Your velocity (in the original frame) is approaching light speed, so your world line becomes almost parallel to the photon's world line."

This implies, to me, that the curve is intended to be a hyperbola. However, the acceleration is not specified; it is not even specified that it is constant (either constant proper acceleration, or constant coordinate acceleration in the inertial coordinates in which the diagram is drawn), so I don't think any particular conclusion is justified about the magnitude of the acceleration. Nor is the diagram intended to support any such conclusion: it is only intended, as far as I can see, to show that any coordinates in which an observer with nonzero acceleration is at rest cannot be inertial.

Also, you said "textbooks", multiple. Can you give an example from any other textbook which you are interpreting as describing a parabolic worldline for an accelerated observer?

 

[pervect]

Constant coordinate acceleration in an inertial frame is impossible, because the accelerating object would exceed the speed of light.

[add]
That was a bit imprecise - constant coordinate acceleration in an inertial frame is only possible for a limited time. At some point, one runs into the speed of light barrier, and it become physically7 impossible, even with infinite proper acceleration.

I'm not sure what other coordinates would be interesting - constant coordinate acceleration doesn't have a meaning until one specifries the coordinates.

 

[Nugatory]

I've just re-read my 1st question and the trivial answer came to me: O is in a flat space, so its so its $g^{\mu\nu}==\eta^{\mu\nu}$ (diagonal). gμν==ημν
But O' doesn't experience flat space, so gμν in her frame is generally different from ημν.

O' most certainly does experience flat space - their motion has nothing to do with whether the spacetime is flat or not. However, when we choose coordinates in which O' is at rest (that is, Rindler coordinates) we will find that the individual metric components are not equal to $\eta^{\mu\nu}$ - those are the values of the metric components written in Minkowski coordinate, and in general the components of a tensor change when we change coordinate systems. (This is similar to how vectors behave - the vector $(1,1)$ Cartesian coordinates might be the vector $(\sqrt{2},\pi/4)$ in $r,\theta$ polar coordinates, but it's the same vector either way).

We can use the tensor transformation rule to calculate the components of the metric tensor using Rindler coordinates from $\eta^{\mu\nu}$ and we'll end up with something that looks different. When we use these to calculate the curvature (Riemann tensor, Weyl tensor, Ricci tensor and constant) we'll get exactly the same zero curvature flat spacetime result that we get when we use Minkowski coordinates.

 

[Pyter]

Also, you said "textbooks", multiple. Can you give an example from any other textbook which you are interpreting as describing a parabolic worldline for an accelerated observer?

Basically just that one, the others were short tutorials I've read somewhere, or maybe that's just the way I remember it. So far I'd never heard of Rindler, or Born, coordinates: I'm looking into it.

When we use these to calculate the curvature (Riemann tensor, Weyl tensor, Ricci tensor and constant) we'll get exactly the same zero curvature flat spacetime result that we get when we use Minkowski coordinates.

That's interesting, I thought that, as per the equivalence principle, the situation of an accelerating observer in a flat space was virtually indistinguishable from that of a still observer in a gravitational field caused by the stress-energy tensor.
You say that's possible for O' to tell the two cases apart by computing the curvature tensor/scalars? Of course in this case the gravitational field is homogeneous as opposed to a non-homogeneous one generated by matter, but for a rotating observer the situation should look more like a "true" gravitational field. Can she also tell them apart?

 

[Ibix]

You say that's possible for O' to tell the two cases apart by computing the curvature tensor/scalars?

Of course. The equivalence principle is only true locally, and the definition of "not local" is "a region large enough such that curvature is non-negligible". Anything you might do such as looking out of the window and seeing if the stars have a changing redshift or not is a way of detecting curvature.

 

[Pyter]

The equivalence principle is only true locally

An accelerating observer can also locally approximate her frame with a Lorentz one. So can she locally see both an inertial frame, and a frame subject to gravity, at the same time?

 

[PeroK]

An accelerating observer can also locally approximate her frame with a Lorentz one. So can she locally see both an inertial frame, and a frame subject to gravity, at the same time?

An accelerating observer's rest frame instantaneously coincides with an inertial frame. That doesn't mean that they are approximately an inertial observer.

 

[Ibix]

An accelerating observer can also locally approximate her frame with a Lorentz one. So can she locally see both an inertial frame, and a frame subject to gravity, at the same time?

Both an observer in a rocket and one on Earth can immediately detect that they are accelerating with local measurements - e.g. a weighing scale. They can both become inertial by jumping, and can choose to use a local accelerating frame by regarding the floor as at rest, or a local inertial frame by regarding a free-falling object as at rest. So they cannot distinguish between the planet and the rocket by local experiments.

The difference between the two situations is the presence or absence of spacetime curvature. Looking at the redshift of distant stars (looking for kinematic versus gravitational redshift) or very precisely comparing the hang angle of two pendulums (looking for inhomogeneity) are two ways to detect the curvature. In the first approach we are doing it by waiting a long time, extending our experiment in time so that it covers a large region of spacetime and any curvature becomes apparent. In the second approach we are improving our measurement accuracy so that our experimental space isn't "local enough".

In summary, any observer may choose a local inertial or local non-inertial frame. The difference in the presence of gravity is that you cannot extend those frames in the same way you can in flat spacetime, either in timelike or spacelike directions, and that failure is detectable.

 

[pervect]

To expand a bit on what others have said, a local observer can determine if they are accelerating or not by consulting an accelerometer.

There is a different instrument the local observer can use to determine if a nearby mass is present. This would be some version of the "Forward mass detector", https://en.wikipedia.org/w/index.php?title=Robert_L._Forward&oldid=980634130#Forward_Mass_Detector. This sort of instrument can potentially do more than detect the presence of mass - with some analysis of the results, it can do some amount of imaging of the density of the source. Bell areospace and perhaps a few other companies, offer commercial services which do this - not to detect the obvious presence of gravity of the Earth, but to survey for the presence of oil fields through the effect the density changes associated with the oil cause. Specifically, the effects these density changes cause in the gravitational field of the Earth.

Google for "Bell Geosspace: FTG gravity gradiometery", or more generically for "gravity gradiometer" or "full tensor gravity gradiometer" for more information. The Forward mass detector is a specific implementation of the more general concept of the gravity gradiometer.

The tensor in question that the full tensor instruments measure is the Riemann curvature tensor, or rather, some components of it. In spite of the name, don't believe that they actually measure all the components in the sense a physicist would use. What they measure is the gravity gradient, in three dimensions. This turns out to be closely related to the Riemann tensor, through the Bel decomposition of said tensor, not to be confused with Bell areospace. This is rather technical, but see for instance wiki https://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=982635460 , what the "full tensor" gravity gradiometers measure is referred to as the "electrogravitc part of the Riemann tensor" in this article stub. It's also known by the name "tidal tensor".

It's essentially a precision measurement of tidal forces, the same forces that cause tides on the Earth, in three dimensions. You could regard a measurement of the tides on the Earth due to the sun and moon as another specific example of detecting the presence of mass via observation. The presence of the tides indicates the presence of the nearby masses. If there were no nearby masses, there would be no tides.

 

[Pyter]

The equivalence principle is only true locally,

It's pretty useless then, isn't it?
In one of his main papers, Einstein computes the time dilation due to gravity by equating the observer subject to the gravitational field to another one in a flat space and rotating with constant angular velocity. So in his view they were equivalent for an indeterminate $ds^2$?

They can both become inertial by jumping,

Do they really need to change their state of motion to switch frames?

To expand a bit on what others have said, a local observer can determine if they are accelerating or not by consulting an accelerometer.

That is a weighing scale. And the result would be the same for both.

It's essentially a precision measurement of tidal forces, the same forces that cause tides on the Earth, in three dimensions.

Yes, that would definitely tell a still observer in a gravity field apart from an accelerating observer in flat space, since the latter would experience a homogeneous acceleration field with no gradient whatsoever.

 

[PeterDonis]

It's pretty useless then, isn't it?

Not at all. I strongly suggest that you take a step back and take some time to learn more relativity before forming an opinion.

In one of his main papers, Einstein computes the time dilation due to gravity by equating the observer subject to the gravitational field to another one in a flat space and rotating with constant angular velocity.

He does? I thought he did it by equating an observer at rest in a gravitational field to an observer at rest inside a rocket accelerating in a straight line in flat spacetime. Please give a specific reference to the paper you are talking about.

in his view they were equivalent for an indeterminate $ds^2$?

Of course. The whole point of the equivalence principle is that spacetime geometries that are different globally, i.e., have different $ds^2$, still look the same locally. That's how we figure out what the laws of physics look like in curved spacetime, with gravity present: by taking the known laws of physics in flat spacetime, with gravity absent, and applying the equivalence principle in each local "patch" of a curved spacetime, and then fitting all the local "patches" together. The latter process is where all the complications of GR come in. But without the starting point given by the former process--being able to carry over all of our knowledge of the flat spacetime laws of physics into a local patch of a curved spacetime--we wouldn't even be able to get GR off the ground. And Einstein of course knew that quite well: his recognition of the basic point of the equivalence principle--what he called the happiest thought of his life--was an essential step towards GR.

Do they really need to change their state of motion to switch frames?

He didn't say they were switching frames by jumping. He said they were becoming inertial by jumping; that's a physical change in their state of motion (accelerometer goes from reading nonzero to reading zero), not an abstract change in the mathematical description they are using, which is what switching frames would be.

 

[PeterDonis]

that would definitely tell a still observer in a gravity field apart from an accelerating observer in flat space, since the latter would experience a homogeneous acceleration field with no gradient whatsoever.

No, that's not correct. There is still a gradient in the "acceleration field" for an observer inside an accelerating rocket in flat spacetime; the acceleration is slightly smaller at the top of the rocket than at the bottom.

What is absent in the accelerating rocket is geodesic deviation: freely falling objects that are at rest relative to each other at one time, stay at rest relative to each other in the rocket, whereas they do not in the presence of a gravitating mass like the Earth. The gravity gradiometer is actually measuring geodesic deviation, not "gradient of the acceleration field". That is what @pervect meant by "tidal forces".

 

[Ibix]

It's pretty useless then, isn't it?

Depends what you use it for. It's a statement that we will never find an object with gravitational mass different from its inertial mass - if we did find such a thing, we'd have to throw out GR (and any other metric theory of gravity). It's also got a rigorous mathematical form, which is that the first derivatives of the metric can be made to vanish at any point. It also means that you don't need to worry too much about the effects of gravity in reasonably small spaces.

So in his view they were equivalent for an indeterminate $ds^2$?

I'm not quite sure what you are asking here. If PeterDonis' reply didn't help, ask again in a different way.

Do they really need to change their state of motion to switch frames?

Of course not. Changing frames is little more than a choice of what to call "at rest". But the point was that to attempt to distinguish between an Earth lab and a rocket lab you need to do experiments. One such experiment might be to investigate free-fall - changing their state of motion to inertial. The experimenter may use inertial or non-inertial coordinates at any time during the experiment, whichever is more convenient.

 

[A.T.]

The equivalence principle is only true locally,

It's pretty useless then, isn't it?

It's a limiting case of General Relativity, just like Newtonian Gravity is. Newtonian Gravity is not useless, just because it applies only in the limit of weak gravitation. And the Equivalence Principle is not useless, just because it applies only in the limit of weak tidal forces.

 

[Pyter]

Please give a specific reference to the paper you are talking about.

A. Einstein - The Special and General Theory of Relativity

In the first place, we see from this expression that two clocks of identical construction will go at different rates when situated at different distances from the centre of the disc. This result is also valid from the standpoint of an observer who is rotating with the disc.

Now, as judged from the disc, the latter is in a gravitational field of potential $\phi$, hence the result we have obtained will hold quite generally for gravitational fields.

It sounds to me like the EP is valid for as long as it's needed, and not only for infinitesimal space-time regions.

The whole point of the equivalence principle is that spacetime geometries that are different globally, i.e., have different ds2, still look the same locally

It came out wrong, by indeterminate $ds^2$ I meant arbitrary space-time regions, not only infinitesimal, if that's what is meant by locally.

 

[PeterDonis]

A. Einstein - The Special and General Theory of Relativity

This is not an actual paper. It is a popular book written for laymen, and the presentation in Appendix III (c) is not Einstein's original argument for gravitational time dilation; that original argument, as I said, used a rocket accelerating in a straight line in flat spacetime.

It is of course true that the time dilation effect Einstein describes in your reference does exist, and can, if we use "gravitational field" to mean "any situation in which time dilation exists", be described as due to a "gravitational field". However, Einstein was not using this argument as an argument for the equivalence principle, or as an application of it; he was using it, as the context makes clear, as a general argument that we should expect to see gravitational time dilation in any gravitational field, and with any measurement we like of time dilation, not just a measurement restricted to a small patch of spacetime.

It sounds to me like the EP is valid for as long as it's needed, and not only for infinitesimal space-time regions.

Nope. Einstein was not using the equivalence principle here. See above.

by indeterminate I meant arbitrary space-time regions, not only infinitesimal, if that's what is meant by locally.

"Infinitesimal" is what is meant by "locally". So if you weren't using "indeterminate" to just mean "infinitesimal", then your use of "indeterminate" is not correct if we are talking about the equivalence principle.

 

[A.T.]

It sounds to me like the EP is valid for as long as it's needed, and not only for infinitesimal space-time regions.

Something that holds locally can be integrated to derive its global consequences.

 

[Pyter]

So the takeaway is: a rotating observer in a flat space and a still observer in a gravity field are not equivalent (in fact the latter can detect space curvature with some experiment), but "equivalent enough" that their clock dilation obey the exact same rules.

 

[PeroK]

So the takeaway is: a rotating observer in a flat space and a still observer in a gravity field are not equivalent (in fact the latter can detect space curvature with some experiment), but "equivalent enough" that their clock dilation obey the exactly same rules.

There is no local experiment to measure time dilation. Any local experiment, by definition, will use its own local proper time.

 

[Pyter]

There is no local experiment to measure time dilation. Any local experiment, by definition, will use its own local proper time.

I mean the time dilation with respect to an inertial observer in flat space.

Cit.:

[ 157 ]
If we represent the difference of potential of the centrifugal force between the position of the clock and the centre of the disc by ϕ i.e. the work,considered negatively, which must be performed on the unit of mass against the centrifugal force in order to transport it from the position of the clock on the rotating disc to the centre of the disc, then we have

From this it follows that

In the first place, we see from this expression that two clocks of identical construction will go at different rates when situated at different distances from the centre of the disc. This result is also valid from the standpoint of an observer who is rotating with the disc.

Now, as judged from the disc, the latter is in a gravitational field of potential ϕ, hence the result we have obtained will hold quite generally for gravitational fields. Furthermore, we can regard an atom which is emitting spectral lines as a clock, so that the following statement will hold:

An atom absorbs or emits light of a frequency which is dependent on the potential of the gravitational field in which it is situated.

 

[pervect]

Yes, that would definitely tell a still observer in a gravity field apart from an accelerating observer in flat space, since the latter would experience a homogeneous acceleration field with no gradient whatsoever.

It's not that different for an accelerating observer than a stationary one, though it's harder to implement because the common-mode accelerations need to be compensated for if you are interested in the tidal components parallel to the direction of the acceleration. If you're only interested in the traverse tidal components, the acceleration doesn't matter at all.

The airplane mounted gravity gradiometers used in the oil surveys certainly have a non-zero proper acceleration, due to both gravity and turbulence. But the gravity gradiometer instruments can operate anyway, they are designed to reject the common-mode signal, basially by using matched accleration sensors attached to a rotating wheel.

There are some minor theoretical relativistic effects that at practical accelerations are not measurable. At 1 g (10 m/s^2) acceleration, I make these effects out of the order of $10^{-15}$ (meters/second^2) / meter, i.e. for a 1 meter spaceship accelerating with a proper acceleration of 10 $m/s^2$ at the nose, the proper acceleration at the tail would be very slightly higher, 10 m^s^2 plus $10^{-15}$ m/s^2. This is well beyond the resolution of practical instruments, so rejecting the common mode signal is the only practical problem for using a gravity gradiometer on an accelerating spaceship if it's accelerating at reasonable accelerations.

If the hypothetical spaceship is trying to hold station near a black hole and pulling a few billion or trillion g's by doing so, things might be different.

It's also different if the spaceship is very long - for instance, if you had 1 light year (born rigid) spaceship accelerating at 1g at the nose, the proper acceleration at the tail would be infinite.

 

[Pyter]

No, that's not correct. There is still a gradient in the "acceleration field" for an observer inside an accelerating rocket in flat spacetime; the acceleration is slightly smaller at the top of the rocket than at the bottom.

Don't they experience the same proper acceleration at the top and the bottom?

 

[PeroK]

I mean the time dilation with respect to an inertial observer in flat space.

You need to be careful with that argument above. If we are considering a case where there is curved spacetime, then there is no "inertial observer in flat space". There are no global inertial frames.

With a centrifuge in flat spacetime, then there can be an inertial observer.