# I Relativistic centrifugal force

#### Alan McIntire

Summary
Without considering relativity, centrifugal force is mv²/r.
I assume there should be a correction for relativity in measuring centrifugal force from the point of view of a particle on the rim of a rotating disc. Is the correction
1/√(1-v²/c²) , or is there an additional correction for acceleration in a perpendicular direction?
I first thought of this problem when I came across the "Ehrenfest Paradox", and realized that as velocity approaches c, the measured force must diverge to infinity as the velocity approaches c.

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#### Dale

Mentor
Unfortunately, this question does not have a unique answer. There are several different rotating reference frames available, each with their own answer to the question. Which would you prefer?

Here is a list that describes several different ones you might consider.

The one defined by 2.17 is probably the best “traditional” one to use, but 2.19 has some nice mathematical properties.

#### Alan McIntire

Unfortunately, this question does not have a unique answer. There are several different rotating reference frames available, each with their own answer to the question. Which would you prefer?

Here is a list that describes several different ones you might consider.

The one defined by 2.17 is probably the best “traditional” one to use, but 2.19 has some nice mathematical properties.
Thanks for that reference. It's just what I was looking for.

I realize that clocks on a rotating disc cannot be synchronized, but the force at a point on the rim should have a unique answer. Either the force of a given weight tied to a string with a given tensile strength on the rotating rim is great enough to break or it isn't.
The whole point of the Ehrenfest Paradox was that “rigid bodies” cannot actually be completely rigid, but stretch and shrink under various forces. If the forces are great enough, as in high acceleration or angular acceleration, the disc cannot hold its shape- it’s going to break apart no mater HOW "indi
structable" the materal.

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#### jartsa

Summary: Without considering relativity, centrifugal force is mv²/r.
I assume there should be a correction for relativity in measuring centrifugal force from the point of view of a particle on the rim of a rotating disc. Is the correction
1/√(1-v²/c²) , or is there an additional correction for acceleration in a perpendicular direction?

I first thought of this problem when I came across the "Ehrenfest Paradox", and realized that as velocity approaches c, the measured force must diverge to infinity as the velocity approaches c.

The relativistic force needed to cause acceleration transverse to the motion of a particle is gamma times the Newtonian force. So that would be the coordinate centripetal force $F=\gamma mv²/r$.

When such coordinate force is applied to a particle that is moving to the perpendicular direction to the force, it feels like a gamma times larger force for the particle.

So therefore the particle feels a force $F'=\gamma^2 mv²/r$

I admit that that was not a calculation done in a rotating frame.

We know that a particle on the rim observes gamma times larger angular velocity of fixed stars than a particle at the center observes. So it seems we could quite easily derive an equation that uses that angular velocity of fixed stars observed by a particle. That equation we could then maybe call "centrifugal force in a rotating frame".

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#### Ibix

So therefore the particle feels a force $F′=\gamma^2mv^2/r$
Just to be clear (I'm not sure if your next line was intended to convey this), $\gamma$, $v$ and $r$ in this formula are those measured in the rest frame of the center of rotation.
So it seems we could quite easily derive an equation that uses that angular velocity of fixed stars observed by a particle. That equation we could then maybe call "centrifugal force in a rotating frame".
This is where I think you need to look at the PhD thesis Dale linked - I suspect "the angular velocity of the distant stars" depends on your simultaneity convention.

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#### PeterDonis

Mentor
The whole point of the Ehrenfest Paradox was that “rigid bodies” cannot actually be completely rigid, but stretch and shrink under various forces.
More precisely, there can't be any such thing as "rigid motion" of a body except in a few very special cases. Spinning up a non-rotating disc to make it rotating is not one of those special cases, so it is impossible for such a disc to remain rigid and retain its exact non-rotating shape after it is rotating.

The more precise version of the above goes by the name of "Born rigidity", and the special cases where it can be realized are classified by the Herglotz-Noether theorem. The Wikipedia article on Born rigidity gives a brief overview:

#### PeterDonis

Mentor
Here is a list that describes several different ones you might consider
I'm a little leery of the procedure undertaken in section 2 of this thesis, because it is deriving "spacetime metrics" without verifying that the different coordinate charts being used are all valid. In particular, metric 2.19 is using a discontinuous definition of "time", which I suspect might lead to a discontinuity in the coordinates that is not being taken into account.

(The other two metrics, 2.15 and 2.17, are using Born coordinates as they are usually defined, and Born coordinates with the time coordinate rescaled. So 2.15 is using a known valid chart, and 2.17's chart is probably valid.)

#### PeterDonis

Mentor
The one defined by 2.17 is probably the best “traditional” one to use
Actually, 2.15 is the more common one, since, as I noted before, it is just Born coordinates:

#### PeterDonis

Mentor
I'm a little leery of the procedure undertaken in section 2 of this thesis
Also, the "problem" stated in section 1.7, that the metric tensor of the "space" of the rotating disk appears to be non-Euclidean, whereas spacelike slices of constant time in the lab frame are Euclidean, does not have a "solution", if by that is meant an actual set of non-Euclidean spacelike slices "cut" out of the spacetime in some way. There aren't any. The spatial geometry derived in section 1 for the disk does not correspond to any spacelike slice.

#### PeterDonis

Mentor
I realize that clocks on a rotating disc cannot be synchronized, but the force at a point on the rim should have a unique answer.
Yes, it does. More precisely, the proper acceleration of an observer rotating with the disk at any point is an invariant (since proper acceleration always is). The Wikipedia page on Born coordinates that I linked to computes the proper acceleration in both lab frame coordinates (the "Langevin observers" section) and Born coordinates (corresponding to metric 2.15 in the thesis @Dale linked to), and it can be seen that they are the same.

#### Dale

Mentor
In particular, metric 2.19 is using a discontinuous definition of "time", which I suspect might lead to a discontinuity in the coordinates that is not being taken into account.
It is definitely discontinuous, but I think it is pretty clearly identified and taken into account.

However, it is not problematic to remove that discontinuity from the chart and cover the manifold in two such charts if needed.

That metric is actually the one I find most interesting here.

#### PeterDonis

Mentor
it is not problematic to remove that discontinuity from the chart and cover the manifold in two such charts if needed
But doing that eliminates the possibility of describing the "space" of the disk as a whole at some instant of time using a spacelike slice of this metric, since the discontinuity means there is no way of covering the entire "space" with a single chart. This is a consequence of the fact that the "space" in question is a quotient space and does not correspond to any spacelike slice in the spacetime.

#### Dale

Mentor
But doing that eliminates the possibility of describing the "space" of the disk as a whole at some instant of time using a spacelike slice of this metric, since the discontinuity means there is no way of covering the entire "space" with a single chart. This is a consequence of the fact that the "space" in question is a quotient space and does not correspond to any spacelike slice in the spacetime.
Yes, completely agreed. I usually don’t care about that, so the simplicity of the metric is appealing.

#### Dale

Mentor
I realize that clocks on a rotating disc cannot be synchronized, but the force at a point on the rim should have a unique answer. Either the force of a given weight tied to a string with a given tensile strength on the rotating rim is great enough to break or it isn't.
The reason that I said there was no unique answer to your question is because your question specifically asked about the centrifugal force. The centrifugal force is a fictitious force, in the language of tensors it is one of the Christoffel symbols. The fictitious forces / Christoffel symbols depend on your choice of coordinates.

Now, the measured tension in a given string attached to a given weight is indeed unique. As you say, the observable fact of whether or not it breaks cannot be frame dependent, but that is not the same as the centrifugal force. The difference is mainly due to different definitions of time in the different coordinate systems.

#### PAllen

So therefore the particle feels a force $F'=\gamma^2 mv²/r$
There are better (more rigorous) derivations, but this is, indeed, the magnitude of 4-force felt by a 'particle' of the disc with constant angular velocity per the lab frame. As @Ibix noted, all quantities as measured in the lab frame, but this resulting magnitude is invariant.

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#### DrGreg

Gold Member
There are better (more rigorous) derivations, but this is, indeed, the magnitude of 4-force felt by a 'particle' of the disc with constant angular velocity per the lab frame. As @Ibix noted, all quantities as mentioned in the lab frame, but this resulting magnitude is invariant.
Agreed, or, to put it another way, $\gamma^2 v²/r$ is the proper acceleration of a point on the disk, and is the value measured by an accelerometer attached to that point.

#### m4r35n357

I find it intriguing that a simple spinning disk can appear to defy analysis whereas a black hole (even a spinning one) is described fairly simply!

As I understand it, we cannot use SR to analyze the spinning disk because it is not globally flat, but what is it precisely that prevents a solution to einstein's equations (I guess in 2+1D, there is plenty of symmetry)?

Perhaps we do not need the full mechanism of GR (stress-energy tensor), or perhaps it is inescapable owing to the presence of curvature. Is there any concept of GR kinematics (I hesitated to even ask that last part!)?

#### Dale

Mentor
I find it intriguing that a simple spinning disk can appear to defy analysis whereas a black hole (even a spinning one) is described fairly simply!

As I understand it, we cannot use SR to analyze the spinning disk because it is not globally flat, but what is it precisely that prevents a solution to einstein's equations (I guess in 2+1D, there is plenty of symmetry)?

Perhaps we do not need the full mechanism of GR (stress-energy tensor), or perhaps it is inescapable owing to the presence of curvature. Is there any concept of GR kinematics (I hesitated to even ask that last part!)?
A spinning disk does not defy analysis. There are in fact multiple ways to analyze a spinning disk. The difference is that by the time people get to GR they know that the coordinates have no physical significance and they are used to clearly specifying the coordinates without expecting clocks and rulers to measure coordinate times and distances. People analyzing the spinning disk often are trying to do so using Einstein synchronization, but nobody analyzing a Kerr black hole tries that.

The spacetime is globally flat so SR is all that is needed. There is no need to find any solution to the Einstein field equations other than the trivial flat spacetime solution of SR.

What you do want to use is tensors and differential geometry. That is not GR, but it is a mathematical tool that is heavily used by GR.

#### Ibix

find it intriguing that a simple spinning disk can appear to defy analysis whereas a black hole (even a spinning one) is described fairly simply!
It's not enormously problematic if you use tensor tools. It's just that there isn't just one obvious answer for how to set up a coordinate system, and naive approaches based on adapting standard inertial frame SR lead directly to elephant traps.
we cannot use SR to analyze the spinning disk because it is not globally flat
Spacetime remains flat. But it's very easy to produce non-flat spatial slices, fail to realise you've done it, and end up in a mess - as has been demonstrated fairly comprehensively over the last month or so.

#### m4r35n357

A spinning disk does not defy analysis.
I was careful to say "appear" ;)
The spacetime is globally flat so SR is all that is needed.
So the global metric for the spinning disk is Minkowski? I need to get my head round that!
I think I'm too used to hearing that the circumference "contracts" whilst the radius does not. To me that points to a "conical" (ish) spacetime. [Thinks:] But then that is flat apart from the point . . . ow my head.

#### A.T.

Summary: Without considering relativity, centrifugal force is mv²/r.
I assume there should be a correction for relativity in measuring centrifugal force from the point of view of a particle on the rim of a rotating disc. Is the correction
1/√(1-v²/c²) , or is there an additional correction for acceleration in a perpendicular direction?

I first thought of this problem when I came across the "Ehrenfest Paradox", and realized that as velocity approaches c, the measured force must diverge to infinity as the velocity approaches c.
Centrifugal forces and the "Ehrenfest Paradox" are different issues. Even if you perfectly balance the centrifugal forces with some externally applied centripetal forces, a rigid disc will still break, when you try to spin it up.

#### jbriggs444

Homework Helper
So the global metric for the spinning disk is Minkowski?
The metric is independent of any coordinate system(*). It is an invariant feature of the space-time. The space-time is flat. The motion of the disk and the motion of any coordinate system that one might want to attach to the disk are irrelevant.

(*) If you want to write down the metric you will likely want to do so using a coordinate system. The representation of a metric using a particular coordinate system will depend on the coordinate system. [And here we reach the limit of my knowledge of the subject].

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#### m4r35n357

The metric is independent of any coordinate system(*). It is an invariant feature of the space-time. The space-time is flat. The motion of the disk and the motion of any coordinate system that one might want to attach to the disk are irrelevant.
So it can be Minkowski because it is flat, but you would need to do coordinate transormations in order to represent various observers, eg. someone riding on the disk at a certain radius (we are after all discussing centrifugal forces). Is that any closer; I don't recall encountering any such metric, so perhaps it is not even useful?
I suppose even defining rotational speed is a problem where there is no rigidity!

#### Nugatory

Mentor
So the global metric for the spinning disk is Minkowski?
The spacetime containing the spinning disk is Minkowski; that manifold uses the Minkwoski metric. That is, every event on the worldlines of every point that makes up the disk is an event in Minkowski spacetime; collectively these worldlines form a worldtube in that spacetime. However, we can choose subsets of these events to form submanifolds that are not Minkowski (rather as the non-Euclidean surface of a two-sphere can be embedded in Euclidean three-space). So what's going on here is that there is no spacelike slice through that worldtube that is both Euclidean and corresponds to our intuitive notion of "the surface of the disk".

#### PAllen

The spacetime containing the spinning disk is Minkowski; that manifold uses the Minkwoski metric. That is, every event on the worldlines of every point that makes up the disk is an event in Minkowski spacetime; collectively these worldlines form a worldtube in that spacetime. However, we can choose subsets of these events to form submanifolds that are not Minkowski (rather as the non-Euclidean surface of a two-sphere can be embedded in Euclidean three-space). So what's going on here is that there is no spacelike slice through that worldtube that is both Euclidean and corresponds to our intuitive notion of "the surface of the disk".
Actually, forget the Euclidean part. There is no spacelike slice (at all) through the world tube that corresponds to intuitive notions of "surface of the disk". Specifically, no spacelike slice that is a local time slice for every particle of the disk surface. Mathematically, the stationary rotating congruence has no hypersuface orthogonal spatial slices at all.

"Relativistic centrifugal force"

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