I Non-Inertial Relativistic Dynamics

[Halc]

TL;DR Summary

Exploring why some coordinate effects become physical effects and v-v relative non-inertial frames

This topic is not about the pinned pages, but I went to the dilation page on the FAQ and it seemed incomplete. https://www.physicsforums.com/threads/time-dilation-definition-what-is-time-dilation.763074/
A quick review since the page is locked
"Time dilation is the factor by which an inertial observer measures another observer's clock as going slow"
I would have said 'computes' since 'measures' implies a direct observation, and an incoming clock is measured to go fast. Some astronomers forget this and mistake a sufficiently fast incoming object as actually moving FTL instead of just appearing to.

Second comment is that part way in the font suddenly changes needlessly to something I cannot read below 200% magnification.

"Third: Gravitational time dilation is greater (the clock is slower) where gravity is stronger (and gravitational potential is higher)."
The bold part is wrong. Gravity is pretty weak on say Mercury, but clocks run slower there than on Earth because the potential is higher on Earth.
Even Sabine Hossenfelder seems to make this mistake when dilation is a function of gravitational strength and of acceleration in one of her you-tubes. She should know better.
I don't want to link it here, but ZdrZf4lQTSg gets to it.
If somebody cares, I can copy or link to my critique of that video on another forum, which is a lot quicker than watching the whole thing.

--- Feature presentation: ---

Time dilation is described as a coordinate effect (at least under SR, not necessarily with gravity), not a physical effect since it is frame dependent. Linear speed is relative, so there's no local experiment that can determine an absolute rest inertial frame.

But rotation changes everything. Absolute rotation can be locally determined, and the dilation and length contraction and such suddenly seem absolute, suggesting that these things are more than just coordinate effects.

One example is (relative to an inertial frame) a stationary circular track stuffed bumper-to-bumper with objects that travel along it. As the objects pick up speed, they contract and more objects can be fit in. This is length contraction due to motion and since more objects can be fit into the same space, there's at least something physical about it.
Now consider the same scenario from a rotating frame where the objects are stationary and it is the track spinning. In this case, it is the stationary objects that are contracted and the moving track that is not. Things seem backwards in this non-inertial frame.
Clocks (all stationary) run faster at the center and slower the further from the center you go. There is a limited distance beyond which it is not possible to be stationary. This is similar to gravitational time dilation. There is not gravity here, nor is there spacetime curvature, but there very much is a difference in potential between stationary clocks running at different rates.

Locally, we put an observer in one of the objects on the track. It seems to be at least an accelerating frame. Clocks objectively up high run faster than the ones near the floor. Tidal forces can be measured. Rotation is more difficult to detect, but a gyro constitutes a local test that will show it. Notice that I say local, but relativistic effects like the clock and the tide tests, while being within a box, are not technically local. An accelerometer is local, but is not a relativistic effect.


I also often see cited in twin paradox topics (great reference, eh?) that Einstein, in some lecture, used gravitational equations and EP to explain simultaneity changes with Earth as computed in the frame of an accelerating ship going to some distant star & back, and this seems wrong since the equivalence principle is a local principle and an accelerating object isn't local. A gravitational field is curved spacetime and the accelerating object/frame is not. That's an objective difference rendering the equations for one irrelevant to the other.

Am I wrong there? The proponents of the equivalence cite that it only works for a uniform gravitational field, but there cannot be such a thing under GR. There can be no infinite plane of matter, and while the field inside an off-center hollow in a planet may be uniform under Newtonian gravity, it isn't uniform under GR. Contradictions result if it was uniform.

 

[PeterDonis]

I would have said 'computes'

Yes, that's true; "time dilation" as the term is normally used in SR does not include the relativistic Doppler effect, so it does not reflect what is directly observed.

The bold part is wrong.

It's not wrong with the correct interpretation of "gravity is stronger". The thread makes clear that "gravity" means the total gravitational potential. That means you have to take into account all gravitating masses.

Gravity is pretty weak on say Mercury, but clocks run slower there than on Earth because the potential is higher on Earth.

The potential due to Mercury is weaker on Mercury, than the potential due to Earth is on Earth.

But both of those potentials are negligible compared to the potential of the Sun. The Sun's potential is stronger on Mercury than on Earth, and that effect dominates the relative clock rates. So taking the Sun into account, it is not true that "gravity is pretty weak" on Mercury; gravity (the Sun's gravity) is stronger on Mercury than on Earth.

rotation changes everything

In the sense that, for example, an observer at the center of a rotating disk will agree with an observer at the edge of the disk (who is rotating around with the disk) that the latter observer's clock is running slower, yes--that form of "time dilation" is not symmetric the way standard SR "time dilation" between inertial observers is.

But that form of "time dilation" is not "described as a coordinate effect". Only the symmetric, standard SR "time dilation" between inertial observers is.

I also often see cited in twin paradox topics (great reference, eh?) that Einstein, in some lecture, used gravitational equations and EP to explain simultaneity changes with Earth

Please read this:

https://www.physicsforums.com/threads/when-discussing-the-twin-paradox-read-this-first.1048697/

It's a sticky at the top of this forum for a reason.

 

[PeterDonis]

a uniform gravitational field ... there cannot be such a thing under GR

In the sense of a field whose "acceleration due to gravity" is the same everywhere, yes, that's true.

However, the term "uniform gravitational field" usually means the "gravitational field" experienced by, for example, a family of Rindler observers in Minkowski spacetime. The "acceleration due to gravity" of this field varies inversely with altitude. I agree that "uniform" isn't the best term to describe this, but it's the one we're stuck with for historical reasons.

 

[PeterDonis]

A gravitational field is curved spacetime

Not the way that term is usually used, no. For example, the term is used, as I noted in my last post, to refer to what is experienced by a family of Rindler observers in flat Minkowski spacetime.

One of the issues with this topic, even in the literature, is that the term "gravity" has multiple meanings (to give the three most common: "acceleration due to gravity", "gravitational potential", and "spacetime curvature"), and it's not always made clear which one is meant. "Gravitational field" usually means the first of the three, sometimes the second, and very rarely if ever the third.

 

[Halc]

Thanks Perter for the replies.

It's not wrong with the correct interpretation of "gravity is stronger". The thread makes clear that "gravity" means the total gravitational potential.

I was unaware that 'gravitational strength' and 'gravitional potential' meant the same thing. The topic did not make it that clear. I took 'strength' to mean how much the potential changes locally over distance, the derivative of the potential.

I could not find any verification of this equivalence on the web nor was it clear in the FAQ thread.
Perhaps if
"where gravity is stronger (and gravitational potential is higher)."
was changed to
"where gravity is stronger (where gravitational potential is higher)."
if would be far more clear that the former wording actually meant the latter.


I don't mean to be argumentative, but I didn't think the thread made that equivalence clear at all.

Please read this:
https://www.physicsforums.com/threads/when-discussing-the-twin-paradox-read-this-first.1048697/
It's a sticky at the top of this forum for a reason.

I'm not new to the twin thingy. Nothing in there relevant except the link to the usenet article, which in turn has a link to something dealing with applying EP to the situation.
https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

It does say that GR does not apply since there is only pseudo-gravity, not real gravity. Good.

Parts there that upset me:
"Uniform "gravitational" time dilation: Say you have two identically constructed clocks. One is deep down in a uniform "gravitational" potential well (or "pseudo-potential", if you prefer); the other is higher up. "

You address this here:

However, the term "uniform gravitational field" usually means the "gravitational field" experienced by, for example, a family of Rindler observers in Minkowski spacetime. The "acceleration due to gravity" of this field varies inversely with altitude.

Agree with this, except calling it gravity since no gravitational field can have this property. OK, the the quote I posted puts 'gravitational' consistently in scare quotes, indicating that it is talking about pseudo-gravity, not actual gravity.

At best, gravity can approach the geometry of an accelerating field as a mass grows without limit, and altitude is the distance from the event horizon. This implies that no matter how massive a black hole might be, a 'stationary' point a light year from the event horizon would experience about 1g (potential per meter) and no more. It's not intuitive that it must drop off that fast no matter how massive the black hole.

Back to the linked article.
There is a disclaimer that this GR treatment is not applicable. It is being used for the accelerating frame of 'Stella' as she turns around. 'Terrance' is not stationary in this accelerating frame, but is rather in freefall, and thus (in Stella's accelerating frame) takes a sort of parabolic trajectory as he arcs from 1 LY all the way up to 7 LY and back down to 1 LY over the course of almost ~13.7 years on his clock and a day on Stella's clock.

In the end, no numbers are presented. EP is referenced only in the sense that in both cases, clocks at higher potentials log more proper time than clocks at low potentials. Nowhere in the article is a gravitational (or any) dilation equation mentioned. Nowhere is it suggested that the EP implies that the equations for the aging of Terrance would be the same in an accelerating scenario as it would in a gravitational scenario.

Bottom line: My concern was not addressed. I don't have the text of Einstein's lecture. Perhaps all he did was the above: show that Terrance aging more than Stella happens in both cases, and not going to far as to suggest that the computation would be identical in both cases.

As a moderator elsewhere, I occasionally need to reply correctly to somebody using Einstein's lecture to assert that gravity and acceleration are non-locally equivalent.

 

[PeterDonis]

I was unaware that 'gravitational strength' and 'gravitional potential' meant the same thing.

They don't necessarily.

The topic did not make it that clear.

The topic directly talks about gravitational potential. But I agree the wording could be improved.

except calling it gravity since no gravitational field can have this property.

More precisely, no curved spacetime will admit a family of accelerated observers whose acceleration has this property.

At best, gravity can approach the geometry of an accelerating field as a mass grows without limit, and altitude is the distance from the event horizon. This implies that no matter how massive a black hole might be, a 'stationary' point a light year from the event horizon would experience about 1g (potential per meter) and no more.

To put this another way: in a small enough patch of spacetime close enough to the horizon of a black hole, the "acceleration field" of hovering observers will look similar to that of Rindler observers in Minkowski spacetime. Yes, this is true.

Back to the linked article.
There is a disclaimer that this GR treatment is not applicable.

Are you talking about the Insights article? Or the Usenet Physics FAQ article?

And specific quotes would be really helpful. I can't really respond to vague comments that don't reference specific statements that you are concerned about.

Nowhere is it suggested that the EP implies that the equations for the aging of Terrance would be the same in an accelerating scenario as it would in a gravitational scenario.

That's good, because any such implication would be false. The equations for a curved spacetime are not the same as the equations for a flat spacetime.

In fact, in many curved spacetime scenarios the EP is not applicable at all, because the scenario cannot be contained within a single local inertial frame, i.e., within a single patch of the spacetime that is small enough that the spacetime curvature can be ignored and the patch can be approximated as flat.

My concern was not addressed. I don't have the text of Einstein's lecture

I'm not sure how your concern could be addressed. We would need some kind of specific quote from Einstein that you are concerned about.

 

[PeterDonis]

Perhaps if
"where gravity is stronger (and gravitational potential is higher)."
was changed to
"where gravity is stronger (where gravitational potential is higher)."
if would be far more clear that the former wording actually meant the latter.

I've made a change to the post you referenced along these lines.

 

[Halc]

Are you talking about the Insights article? Or the Usenet Physics FAQ article?

The one I linked, indirect through usenet, on math.ucr,edu

And specific quotes would be really helpful. I can't really respond to vague comments that don't reference specific statements that you are concerned about.

Best one is this "Real (not pseudo) gravitational time dilation (i.e., fields due to matter) is a different story. These fields are never uniform, and the derivations just mentioned don't work.". This explicitly states that since the derivations are different between the two cases, the mathematics of one cannot be used to compute the other. Now the question is, did Einstein say as much in his lecture?

I'm not sure how your concern could be addressed. We would need some kind of specific quote from Einstein that you are concerned about.

Yea, I have to see if I can find a transcript of it. For starters, it was done in German AFAIK.

I've made a change to the post you referenced along these lines.

Thanks for that. The word 'potential' went away, replaced by 'position'.

The tiny font at the bottom half is still there.

 

[PeterDonis]

The word 'potential' went away, replaced by 'position'

Not in the part I changed. I just took out the word "and" before "the gravitational potential is higher".

 

[PeterDonis]

Best one is this "Real (not pseudo) gravitational time dilation (i.e., fields due to matter) is a different story. These fields are never uniform, and the derivations just mentioned don't work.". This explicitly states that since the derivations are different between the two cases, the mathematics of one cannot be used to compute the other.

All the article means here is that curved spacetime is not the same as flat spacetime. You can use the same general framework to do both computations, but the specifics will be different because the spacetime geometry is different, and therefore the expressions for the metric, and so for time dilation, will be different.

Now the question is, did Einstein say as much in his lecture?

The only way to tell will be to look at a transcript.

 

[PeterDonis]

why some coordinate effects become physical effects and v-v relative non-inertial frames

Just to be clear: this does not happen. Anything that is a "physical effect" (i.e., an invariant) is that in any frame. For example, you can derive the results discussed for observers on a rotating disk using an inertial frame whose spatial origin is the center of the disk. You don't need to use a non-inertial frame in which the disk is at rest.

 

[Halc]

Just to be clear: this does not happen. Anything that is a "physical effect" (i.e., an invariant) is that in any frame.

It does happen in any frame: The fitting of more objects on the circular track. OK, the size/shape of the track and the size of any specific object moving along it is all frame dependent. It's how many more you can pack in that is invariant.

Likewise, consider two identical thin rings (thin to avoid Ehrenfest issues). Spin one up and now it fits through the other. That's a physical effect. It goes through in any frame. Things that seem to be coordinate effects in linear motion are often physical effects with rotating motion.

You don't need to use a non-inertial frame in which the disk is at rest.

No, of course not. I just found it to be an interesting property of such a non-inertial frame.
Another one is that a light pulse can slow, halt, and return back the way it came. This goes on somewhere near the orbit of Neptune relative to the frame of Earth's surface.

 

[PeterDonis]

It does happen in any frame

Then it isn't a coordinate effect.

Things that seem to be coordinate effects in linear motion are often physical effects with rotating motion.

No, that's not correct. The physical effects you describe are not the same as the coordinate effects that usually get described as "length contraction". The fact that the same term sometimes gets used to describe both does not mean they're the same thing, or that one is the "rotating motion version" of the other.

Note, btw, that even in linear motion you can have actual physical effects due to "length contraction"--for example, a pole in fast motion fitting inside a barn that it would not fit in if it were at rest relative to the barn.

I just found it to be an interesting property of such a non-inertial frame.

It's not. It's not a "property of a non-inertial frame". An actual physical effect can't be a property of any frame. It's an invariant.

a light pulse can slow, halt, and return back the way it came. This goes on somewhere near the orbit of Neptune relative to the frame of Earth's surface.

I don't know what you're referring to here.

 

[Halc]

The physical effects you describe are not the same as the coordinate effects that usually get described as "length contraction".

That is perhaps our disconnect, using the same language to describe both linear and rotational effects.

Note, btw, that even in linear motion you can have actual physical effects due to "length contraction"--for example, a pole in fast motion fitting inside a barn that it would not fit in if it were at rest relative to the barn.

Come on now, that's a coordinate effect, not a physical one, since the pole does not fit in the barn in frames where the barn moves faster. The whole scenario is just an exercise in relativity of simultaneity in disguise.

I don't know what you're referring to here.

In a rotating frame there's a distance beyond which nothing can be stationary due to c restrictions. That distance is nearly the distance to Neptune for a 1-day angular rate. So picture a light pulse in an inertial frame that moves eastward tangentially to the circle at that radius. It touches the circle for a moment, and then continues straight, never coming inside.
Relative to the rotating frame of Earth, that light pulse spirals in from the east, curving ever inward and slowing until it is coming straight at us as its speed hits 0. Once it stops for that brief moment, it turns around and exits in a spiral mirroring the path in to the west. We see none of this since the light never gets to Earth. It's just a coordinate curiosity.

 

[PeterDonis]

that's a coordinate effect, not a physical one, since the pole does not fit in the barn in frames where the barn moves faster.

That's true, but one can easily add an element to the scenario that gives an invariant representing "the pole fits in the barn in the barn's rest frame". For example, one could have a light signal emitted from the center of the barn in such a way that it will only cause the doors to close if it first passes each end of the pole before it hits the doors. Note that the events in question along each light beam's worldline (passing the end of the pole and then hitting the corresponding door) are null separated, so their ordering is invariant. Therefore, whether the doors close or not in this version of the scenario is also an invariant, and they will only close if the pole fits in the barn in the barn's rest frame.

This illustrates, btw, that one has to be very careful in thinking about what's a coordinate effect and what isn't.

 

[PeterDonis]

In a rotating frame there's a distance beyond which nothing can be stationary due to c restrictions.

If "stationary" means "having constant spatial coordinates in some frame", yes.

But that meaning of "stationary" has no physical meaning. So your description of "how the light moves" also has no physical meaning.

Which illustrates why one should not use such language to describe coordinate effects, such as the coordinate behavior of light in the rotating frame. For example, the proper way to describe "how light moves" in relativity is to pick out the light cones, which are invariant. (This becomes much more important in curved spacetime, where descriptions in terms of coordinate effects become even more problematic.)

 

[Halc]

one could have a light signal emitted from the center of the barn in such a way that it will only cause the doors to close if it first passes each end of the pole before it hits the doors.

It would indeed then be an invariant that the barn doors both close. It just wouldn't be an invariant that those doors are closed simultaneously.

 

[Histspec]

I don't have the text of Einstein's lecture.

As far as I know from reading Einstein's collected papers, there is only one published paper of Einstein in which he used the EP to discuss the twin paradox (besides some letters from 1917-1918 in which he also sketched that solution). It's in his popular article (without equations) from 1918:
Dialog about Objections against the Theory of Relativity (English translation)
For a discussion of that paper, see Janssen: The twins and the bucket: How Einstein made gravity rather than motion relative in general relativity.
The corresponding approximate equations were also explicitly stated in the contemporary relativity textbooks of Kopff (1921) p. 117f. or Born (1922) p. 256f.

We have two clocks U1 ($t_{1}$) and U2 ($t_{2}$), with U2 making the round-trip away from U1 and back. Assuming the time of acceleration of U2 is negligible small compared to the inertial phases, we have in the inertial frame of U1:
$$\Delta t_{1}=\frac{\Delta t_{2}}{\sqrt{1-v^{2}/c^{2}}}=\Delta t_{2}\left(1+\frac{1}{2}\frac{v^{2}}{c^{2}}+\dots\right)$$
So U2 is retarded at reunion with respect to U1 in first approximation by
$$(e1)\quad \frac{1}{2}\Delta t_{2}\frac{v^{2}}{c^{2}}$$
From the viewpoint of U2, however, symmetry of time dilation during the inertial phases requires
$$\Delta t_{1}=\Delta t_{2}\sqrt{1-v^{2}/c^{2}}=\Delta t_{2}\left(1-\frac{1}{2}\frac{v^{2}}{c^{2}}+\dots\right)$$
so U1 is retarded by with respect to U2 in first approximation
$$ (e2)\quad \frac{1}{2}\Delta t_{2}\frac{v^{2}}{c^{2}}$$
Yet during turnaround that lasted for $\Delta\tau_{2}$, a uniform gravity field pops up leading to gravitational time dilation:
$$\Delta\tau_{1}=\Delta\tau_{2}\left(1+\frac{hg}{c^{2}}\right)$$
with $hg$ as gravitational potential, $g$ as constant acceleration, $h$ as distance between U1 and U2. As U1 is located at a higher potential during turnaround, it is advancing with respect to U2 by
$$\frac{\Delta\tau_{2}hg}{c^{2}}$$
Since we evidently have $\Delta\tau_{2}=2v/g$ and $h=\tfrac{1}{2}\Delta t_{2}v$, this advance can be written as
$$(e3)\quad \frac{\Delta\tau_{2}hg}{c^{2}}=\Delta t_{2}\frac{v^{2}}{c^{2}}$$
So the advance of U1 (e3) during turnaround is the double of its retardation (e2) in the inertial phases, which gives the total advance of U1 with respect to U2 by
$$(e3)-(e2)=\Delta t_{2}\frac{v^{2}}{c^{2}}-\frac{1}{2}\Delta t_{2}\frac{v^{2}}{c^{2}}=\frac{1}{2}\Delta t_{2}\frac{v^{2}}{c^{2}}=(e1)$$
in agreement with (e1).

 

[PeterDonis]

It would indeed then be an invariant that the barn doors both close. It just wouldn't be an invariant that those doors are closed simultaneously.

Yes, that's correct.