Einstein's unidimensional elevator and Rindler coordinates

  • #1
343
29
Consider an "unidimensional elevator" of size L accelerating w.r.t. a given inertial reference frame. Suppose each elevator's point accelerates with a constant proper acceleration ##g## according Rindler acceleration profile. In the given inertial frame with coordinates ##(x,t)## the elevator points's worldlines are described by:
$$x = R\cosh\eta, \qquad t = R\sinh\eta$$
AFAIK based on Equivalence Principle (EP) all experiments executed inside the "unidimensional elevator" can be explained introducing an uniform gravitational field inside it. The Equivalence here is not just local but we can extend it for the entire region of flat spacetime involved.

Now suppose you leave a particle at rest at quote ##R_{0}## in the elevator. At that moment I believe it shares the proper acceleration of the elevator point in which it is at rest and then, during its free-fall, it follows a geodesic described as just a straight line in ##(x,t)## coordinates. Basically its initial proper acceleration has been "converted" into the elevator ##(R,\eta)## coordinate acceleration.

If what said is correct, now each elevator point (with spatial ##R_{x}## coordinate inside the elevator itself) has a different proper acceleration (according Rindler acceleration profile) thus the coordinate acceleration for objects released from rest at different elevator quotas is not a constant; in other words my conclusion is that uniform gravitational field inside the elevator is not the case !

Can you help me in understanding if the argument is right or where is wrong ? Thanks.
 

Answers and Replies

  • #2
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Hi @cianfa72 and welcome to PF!

The Equivalence here is not just local but we can extend it for the entire region of flat spacetime involved.
There are limits to this extension. There is a Rindler horizon below the elevator that limits how far it can extend in the "downward" direction. And the proper acceleration varies with height in the elevator so the gravitational field can only be considered "uniform" over small enough ranges of height that the change in proper acceleration is not significant. See further comments below.

suppose you leave a particle at rest at quote ##R_{0}## in the elevator. At that moment I believe it shares the proper acceleration of the elevator point in which it is at rest and then, during its free-fall, it follows a geodesic described as just a straight line in ##(x,t)## coordinates.
Yes.

Basically its initial proper acceleration has been "converted" into the elevator ##(R,\eta)## coordinate acceleration.
By this you mean, I take it, that the proper acceleration of the elevator (which was only shared by the particle before it was released) is numerically equal to the coordinate acceleration of the particle in the ##(R, \eta)## coordinates after it is released? This is correct only when the particle is first released. As it "falls", its coordinate acceleration in the ##(R, \eta)## coordinates changes.

If what said is correct, now each elevator point (with spatial ##R_{x}## coordinate inside the elevator itself) has a different proper acceleration (according Rindler acceleration profile)
It is true that the proper acceleration of objects at rest inside the elevator varies with height (your ##R_x## coordinate) in the elevator. But I don't see how you are deducing that from your thought experiment about a particle being released from rest into free fall. The particle's coordinate acceleration in ##(R, \eta)## coordinates changes because those coordinates are non-inertial; the same would be true (although the numerical details would be different) for the motion of any freely falling particle in any non-inertial coordinates. That fact, in itself, does not tell you anything useful about the proper acceleration of objects at rest in the non-inertial coordinates. You have to compute that from the actual metric in those coordinates.

my conclusion is that uniform gravitational field inside the elevator is not the case
It is true that the gravitational field is not "uniform" in the sense of the proper acceleration of objects at rest being the same everywhere--or in the sense of the coordinate acceleration of objects in free fall at the instant they are released from rest being the same everywhere. See my statement near the top of this post. This is a well-known and well-understood feature of the Rindler congruence--meaning the family of worldlines in flat spacetime that describe objects at rest in the elevator at various heights.

One way of understanding it in geometric terms is to view this set of worldlines as the hyperbolic equivalent of a set of concentric circles. Just as the curvature of the circles varies with their radius (i.e., their distance from the common center), the curvature of the hyperbolas in spacetime that are the various Rindler worldlines varies with their distance from a common center (which is the intersection of the common asymptotes of the hyperbolas). And the curvature of the hyperbolas is the same as the proper acceleration of the objects that have those hyperbolas as worldlines.
 
  • #3
1,478
122
An astronaut on an accelerating rocket drops a ruler. The rest frames of the ruler and the astronaut start becoming increasingly different. The astronaut observes ruler's length changing, just as special relativity says.

Then I guess the astronaut should say something about the coordinate accelerations of the different parts of the ruler, because that is what we are interested about ... right?


Well that is explained in great detail here in post #42:
https://www.physicsforums.com/threads/bells-spaceship-paradox-length-contraction.951719/page-3

(It's about a rocket that is length-contracting, and Bob and Alice are strapped on that rocket, Bob is seated closer to the rear of the rocket)




Oh, that does not quite work, because the astronaut is non-inertial, which makes his observations quite different. Oh dear.

But special relativity should be able to handle this ... Oh yes, I forgot the relativity of simultaneity. That's why there was no "gravitational time dilation".
 
Last edited:
  • #4
343
29
There are limits to this extension. There is a Rindler horizon below the elevator that limits how far it can extend in the "downward" direction. And the proper acceleration varies with height in the elevator so the gravitational field can only be considered "uniform" over small enough ranges of height that the change in proper acceleration is not significant. See further comments below.
Sure, in my thought experiment the elevator size is L (as measured at ##t=0## in the "fixed" inertial frame with ##(x,t)## coordinates) so the EP extension applies actually to a limited spacetime region

By this you mean, I take it, that the proper acceleration of the elevator (which was only shared by the particle before it was released) is numerically equal to the coordinate acceleration of the particle in the ##(R, \eta)## coordinates after it is released? This is correct only when the particle is first released. As it "falls", its coordinate acceleration in the ##(R, \eta)## coordinates changes.
sure, I was meaning that

It is true that the proper acceleration of objects at rest inside the elevator varies with height (your ##R_x## coordinate) in the elevator. But I don't see how you are deducing that from your thought experiment about a particle being released from rest into free fall. The particle's coordinate acceleration in ##(R, \eta)## coordinates changes because those coordinates are non-inertial; the same would be true (although the numerical details would be different) for the motion of any freely falling particle in any non-inertial coordinates. That fact, in itself, does not tell you anything useful about the proper acceleration of objects at rest in the non-inertial coordinates. You have to compute that from the actual metric in those coordinates.
ok, sorry for my English I expressed myself incorrectly (I was not deducting that from my thought experiment)

Thus my new question is: does it exist any "rigid" accelerating frame inside it we can assume an "uniform" gravitation field non just locally but for a limited region ? Note: as you said I mean uniform gravitational field in the sense of the proper acceleration of objects at rest being the same everywhere--or in the sense of the coordinate acceleration of objects in free fall at the instant they are released from rest being the same everywhere
 
  • #5
pervect
Staff Emeritus
Science Advisor
Insights Author
9,954
1,136
Consider an "unidimensional elevator" of size L accelerating w.r.t. a given inertial reference frame. Suppose each elevator's point accelerates with a constant proper acceleration ##g## according Rindler acceleration profile. In the given inertial frame with coordinates ##(x,t)## the elevator points's worldlines are described by:
$$x = R\cosh\eta, \qquad t = R\sinh\eta$$
Are you imagining this elevator (I'd call it a congruence of worldlines) represents a rigid (Born rigid) body? Bell's spaceship paradox indicates that it would not be.

The usual congruence of worldines for a rigid elevator is given in, for instance, https://en.wikipedia.org/wiki/Rindler_coordinates#/media/File:Rindler_chart.svg

The hyperbolae in this diagram the worldlines of points on the elevator. Every different point has it's own worldline.
 
  • #6
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
does it exist any "rigid" accelerating frame inside it we can assume an "uniform" gravitation field non just locally but for a limited region ? Note: as you said I mean uniform gravitational field in the sense of the proper acceleration of objects at rest being the same everywhere
'
No. In relativity, it is impossible to have a set of observers that satisfy both of the following properties:

(1) The observers are all at rest relative to each other;

(2) The observers all have the same nonzero proper acceleration.

You have to pick one or the other. If you pick (1), you get the Rindler congruence. If you pick (2), you get the Bell congruence (the set of worldlines that shows up when you analyze the Bell spaceship paradox).
 
  • #7
1,478
122
Thus my new question is: does it exist any "rigid" accelerating frame inside it we can assume an "uniform" gravitation field non just locally but for a limited region ? Note: as you said I mean uniform gravitational field in the sense of the proper acceleration of objects at rest being the same everywhere--or in the sense of the coordinate acceleration of objects in free fall at the instant they are released from rest being the same everywhere

Put some electric charge on a tip of a long rod, then take the rod in an uniform electric field and test how the force felt at the other end of the rod changes when you move the charged end to different positions. You will notice that the force does not change.

Tie a test mass on a tip of a long rod, then take the rod in an uniform gravity field and test how the force felt at the other end of the rod changes when you move the test mass end to different positions. You will notice that the force does not change.

This far things seem to work the same way in both fields. But in gravity field there is always the gravitational time dilation, which causes acceleration of test mass to be different at different positions. And that is why there is no gravity field that is uniform in the same way as an uniform electric field is uniform.
 
  • #8
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Tie a test mass on a tip of a long rod, then take the rod in an uniform gravity field and test how the force felt at the other end of the rod changes when you move the test mass end to different positions. You will notice that the force does not change.
This experiment cannot be done; there is no such thing as a "uniform gravity field" in the sense you are using the term (and in the sense that you can have a "uniform electric field", for example between two plates of an idealized capacitor). Go read my post #6.
 
  • #9
1,478
122
This experiment cannot be done; there is no such thing as a "uniform gravity field" in the sense you are using the term (and in the sense that you can have a "uniform electric field", for example between two plates of an idealized capacitor). Go read my post #6.

Yes there are that kind of "uniform" gravity fields. Like for example 1000 km above the event horizon of a super massive black hole. If you conduct my experiment there, the force you feel at your end of the rod will be very close to constant, as you move the rod around.

I mean the force is only very slightly larger when the test mass is 1 cm above the event horizon, compared to the case when the test mass is 1000 km above the event horizon.

It is important that the person conducting this experiment does not move up or down in the gravity field, that's why it's important to use a long rod.

Idealized capacitor has infinitely large plates ... so to be fair I must be allowed to use an infinitely large black hole, in which case the force is exactly constant.


(I don't have any sources for any of this, but I remember that we once agreed that when a mountain climber descends along a rope, the force on the anchor stays constant)
 
  • #10
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Yes there are that kind of "uniform" gravity fields.
No, there are not. See below.

Like for example 1000 km above the event horizon of a super massive black hole. If you conduct my experiment there, the force you feel at your end of the rod will be very close to constant, as you move the rod around.
"Very close to constant" is not the same as "exactly constant". But in any case, you are missing the point of this thread. Go read my post #6, which addresses the question the OP of the thread is actually asking. If you agree with it, then please stop posting irrelevancies in this thread. If you disagree with it, then please explain why.
 
  • #11
343
29
'
You have to pick one or the other. If you pick (1), you get the Rindler congruence. If you pick (2), you get the Bell congruence (the set of worldlines that shows up when you analyze the Bell spaceship paradox).
IMO Einstein's elevator argument as described at GR introductory level is not clear about this point: it seems EP extends over the entire elevator while, as discussed here, the equivalence with an uniform gravitational field applies with different values at different points being at rest in accelerating elevator (because, as said before, each different point at rest in the elevator has actually a different (constant) proper acceleration)
 
Last edited:
  • #12
pervect
Staff Emeritus
Science Advisor
Insights Author
9,954
1,136
IMO Einstein's elevator argument as described at GR introductory level is not clear about this point: it seems EP extends over the entire elevator while, as discussed here, the equivalence with an uniform gravitational field applies with different values at different points being at rest in accelerating elevator (because, as said before, each different point at rest in the elevator has actually a different (constant) proper acceleration)
I believe Einstein's elevator argument is purely a local one. Consequently, the equivalence principle can be and frequently is misunderstood by improperly interpreting this local equivalence improperly as something more global.

I also agree that each different point in the elevator has a different proper acceleration, assuming the elevator is Born rigid.

It sounds like you might be trying to say something additional that I wouldn't necessarily agree with, but I think we may agree on these two points.

The modern statement of the equivalence principle breaks it down into three forms. WIki has some discussion of this. It's not the most authoritative source, but it has some discussion of the three major forms of the modern statement of the EP, the weak EP, the Einstein EP, and the strong EP. There is also some discussion in wiki of a more exact meaning of "local" in this context.

<<wiki link>>
 
  • #13
1,478
122
IMO Einstein's elevator argument as described at GR introductory level is not clear about this point: it seems EP extends over the entire elevator while, as discussed here, the equivalence with an uniform gravitational field applies with different values at different points being at rest in accelerating elevator (because, as said before, each different point at rest in the elevator has actually a different (constant) proper acceleration)
Do you think real gravity fields don't have that effect of different proper accelerations at different points?
 
  • #14
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,499
2,635
Do you think real gravity fields don't have that effect of different proper accelerations at different points?
Well, the variation of ##g## with height is different for the "real" gravity near a massive planet and the "artificial" gravity inside an accelerating elevator. In the first case, ##g## decreases according to an inverse-square law, while in the second case, ##g## decreases according to 1/r law.
 
  • #15
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
the equivalence with an uniform gravitational field applies with different values at different points being at rest in accelerating elevator
No. If you are talking about a scenario where the difference in proper acceleration is detectable, you are already outside the purview of the equivalence principle. The EP, strictly speaking, applies to a single observer at a single point inside the elevator. The elevator is really there to emphasize the fact that the observer can only make local observations, by providing an easily understandable barrier that prevents the observer from, for example, seeing light from distant objects.

In practice, as long as the "elevator" is small enough that differences in the acceleration are not detectable within it, the EP can be applied inside the entire elevator; but what counts as "small enough" depends on the situation.
 
  • Like
Likes cianfa72 and SiennaTheGr8
  • #16
343
29
No. If you are talking about a scenario where the difference in proper acceleration is detectable, you are already outside the purview of the equivalence principle.
Here with detectable do you refer basically to any possible experiment done inside the elevator (e.g. by measurements of accelerometers at rest at different elevator points) ?

 
  • #17
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Here with detectable do you refer basically to any possible experiment done inside the elevator (e.g. by measurements of accelerometers at rest at different elevator points) ?
Yes. If you have accelerometers at different points inside the elevator that give detectably different readings, you are outside the purview of the equivalence principle. Any real accelerometer has a finite accuracy, so it is possible to have a finite elevator with accelerometers everywhere giving the same readings to within measurement error.
 
  • #18
1,478
122
Well, the variation of ##g## with height is different for the "real" gravity near a massive planet and the "artificial" gravity inside an accelerating elevator. In the first case, ##g## decreases according to an inverse-square law, while in the second case, ##g## decreases according to 1/r law.

Here's a very simple argument that I came up with. I can make it more complete if needed, but maybe you get the idea:

Inside an accelerating spaceship a small rocket can hover using one of its ten rocket motors at position A. At position B all ten rocket motors are needed to hover.

There is time dilation effect inside the spaceship according to the crew. The crew says rocket motors burn fuel at 1/10th rate at position B, compared to position A.

Proper acceleration is ten times larger when all ten motors run, right?


Okay now this is quite interesting : There is a time dilation effect in a "real" gravity field too.
 
  • #19
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Inside an accelerating spaceship a small rocket can hover using one of its ten rocket motors at position A. At position B all ten rocket motors are needed to hover.

There is time dilation effect inside the spaceship according to the crew. The crew says rocket motors burn fuel at 1/10th rate at position B, compared to position A.

Proper acceleration is ten times larger when all ten motors run, right?
You can't independently specify the proper acceleration and the time dilation.
 
  • #20
1,478
122
You can't independently specify the proper acceleration and the time dilation.
Well it happens to be so that the correct time dilation ratio for a proper acceleration ratio of ten is ten, like I said.

I even checked that.

(Luckily I found a simple formula for the proper acceleration ratio in Wikipedia and a simple formula for time dilation ratio in Physics Forums)
 
  • #21
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Well it happens to be so that the correct time dilation ratio for a proper acceleration ratio of ten is ten, like I said.

I even checked that.

(Luckily I found a simple formula for the proper acceleration ratio in Wikipedia and a simple formula for time dilation ratio in Physics Forums)
Please show your work.
 
  • #22
1,478
122
Please show your work.



Proper times are related like this when the larger of the two proper accelerations is a:
$$\tau_1 = (1 + \frac{aL}{c^2}) \tau_0$$


Proper accelerations are related like this when the larger of the two proper accelerations is a:

$$a_2=\frac{a}{1+\frac{aL}{c^2}}$$





Now we have ##1+\frac{aL}{c^2}## in both formulas, let's substitute that with k:


Proper times:
$$\tau_1 = k \tau_0$$

Proper accelerations:

$$a_2=\frac{a}{k}$$

The same factor k is in both formulas.

At the end of this page is the formula for proper accelerations
https://en.wikipedia.org/wiki/Bell's_spaceship_paradox

Post #20 here has the formula for proper times:
https://www.physicsforums.com/threa...dpoints-of-a-rocket-post-acceleration.906461/




(I hope it was correct to say that "a" is the larger one of the two proper accelerations. Let's see ... a guy at the rear of an accelerating rocket might feel a very large acceleration, and then the upper guy's time would be running very fast for the lower guy, so the lower guy needs to multiply his time with a large "k" to get the time that the upper guy's clock advanced ... yes it seems correct)
 
  • #23
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
I hope it was correct to say that "a" is the larger one of the two proper accelerations.
Yes, that's right. Your formulas are OK.

However, a factor of 10 in acceleration and time dilation is going to be too large for the EP to be valid; with that much of a change you will be able to distinguish an elevator in flat spacetime from one in curved spacetime, because tidal gravity (i.e., the difference in how proper acceleration and time dilation change with height in the elevator) will be detectable.
 
  • #24
1,478
122
Yes, that's right. Your formulas are OK.

However, a factor of 10 in acceleration and time dilation is going to be too large for the EP to be valid; with that much of a change you will be able to distinguish an elevator in flat spacetime from one in curved spacetime, because tidal gravity (i.e., the difference in how proper acceleration and time dilation change with height in the elevator) will be detectable.

Take a black hole large enough, get close enough to it, now you have factor of 10 proper acceleration difference between your head and your feet. Tidal forces are not the reason for that, because the black hole was large enough.


A longer version of that:


A spaceship with a powerful rocket motor starts slowly descending towards a large black hole. The rocket motor is at the rear of the ship, and the ship descends rear first. As gravitational time dilation starts slowing down the motor, the pilot increases the thrust to compensate for that. Passengers at the rear of the ship do not observe any time dilation effect on the motor, they only observe that the pilot is increasing the thrust. The passengers at the rear might say that gravity is getting stronger as the ship descends.

Also the passengers at the front of the ship feel that gravity inside the ship is getting stronger, but as the motor is some distance below them, they observe that as the gravity is getting stronger, time dilation is having an increasingly large effect on the motor.

After enough time has passed, the passengers at the rear feel ten times larger proper acceleration compared to the passengers at the front. (At that point the ship may be quite close to the event horizon, but it's still above it.)
 
Last edited:
  • #25
PeterDonis
Mentor
Insights Author
2020 Award
32,978
11,450
Take a black hole large enough, get close enough to it, now you have factor of 10 proper acceleration difference between your head and your feet. Tidal forces are not the reason for that, because the black hole was large enough.
Again, please show your work. And remember, the issue is not the ratio of proper accelerations by itself; it's that the ratio of proper accelerations has to always be the same as the ratio of time dilations (as you showed it is for Rindler observers in flat spacetime). If those two ratios are different, you know you're in a curved spacetime.
 

Related Threads on Einstein's unidimensional elevator and Rindler coordinates

  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
6
Views
6K
  • Last Post
Replies
2
Views
1K
  • Last Post
2
Replies
33
Views
5K
Replies
2
Views
4K
Replies
4
Views
2K
  • Last Post
Replies
9
Views
909
Replies
5
Views
647
Replies
15
Views
4K
Top