Physics Help Forum Am I doomed to never understand the relativistic rotating disk?

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 Aug 18th 2017, 03:56 AM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 Am I doomed to never understand the relativistic rotating disk? Ok, I am really out of my depth here so I am wondering whether I can get any mileage with this with just flat minkowski space with uniformly rotating polar co-ordinates and some vector calculus and linear algebra? (I'm really not good with tensors). Is it worth me trying to understand some basic concepts about this now or do I really need "to climb the tensor mountain" to get there? Or can I do some "good enough" analysis with just some vector calculus and linear algebra? I am basically trying to understand why, in rotating polar co-ordinates, that the circumference undergoes fitzgerald contraction but the radius does not and whether it annihilates the minkowski flat space-time assumption.
Aug 18th 2017, 08:22 AM   #2
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 Originally Posted by kiwiheretic Ok, I am really out of my depth here so I am wondering whether I can get any mileage with this with just flat minkowski space with uniformly rotating polar co-ordinates and some vector calculus and linear algebra? (I'm really not good with tensors). Is it worth me trying to understand some basic concepts about this now or do I really need "to climb the tensor mountain" to get there? Or can I do some "good enough" analysis with just some vector calculus and linear algebra? I am basically trying to understand why, in rotating polar co-ordinates, that the circumference undergoes fitzgerald contraction but the radius does not and whether it annihilates the minkowski flat space-time assumption.
I understand your dilemma since I too had a problem with it for a very long time until I realized the nature of the geometry.

First of all, the spacetime is flat regardless of the geometry. It's not possible to introduce spacetime curvature simply by changing from an inertial frame to a non-inertial frame.

This problem is known as the Ehrenfest Paradox. It's described here with a possible resolution to the paradox given by Gron.

I have a solution that I'm considering at the moment myself. Its possible that the geometry changes from a surface of a flat disk to the surface of a cone. A cone, while having a different spatial geometry than the disk still has zero intrinsic curvature so its still flat in that sense.

Last edited by Pmb; Aug 18th 2017 at 08:31 AM.

 Aug 18th 2017, 03:52 PM #3 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 I notice they use a different metric tensor for rotating polar/spherical co-ordinates as opposed to straight Euclidean co-ordinates. Is this just "cosmetic" or does changing to a different metric tensor signify a warping of the manifold (if that's the correct term)?
 Aug 19th 2017, 08:01 AM #4 Member   Join Date: Dec 2012 Location: Boulder, Colorado Posts: 63 You're making it a lot harder than it needs to be. When viewed from the inertial frame with spatial origin at the center of the disk (but not rotating with it), basic SR says that there will be length contraction along the direction of the motion, but no length contraction in the direction perpendicular to the motion. So in the direction parallel to the circumference of the circle, there will be length contraction, but along the radius, there will be no length contraction. The length contraction along the circumference means that it takes more yardsticks laid out (and stationary wrt the disk) along the complete circumference that there would be if the disk weren't rotating. But the number of yardsticks laid out (and stationary wrt the disk) along a radius will be the same with or without rotation of the disk. If the disk weren't rotating, the ratio of the number of circumference yardsticks to the number of radius yardsticks would be 2pi, but when rotating, the ratio is greater than 2pi. That's when Einstein realized that, according to the creatures at rest on the disk (and who believe that the disk is stationary in a very weird gravitational field), their geometry isn't Euclidean. And Einstein realized then that his general relativity theory would have to utilize non-Euclidean geometry. __________________ Mike Fontenot
 Aug 19th 2017, 02:36 PM #5 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 That's kind of a mind bender though. What would an observer see while looking at that given that we can only see 3 dimensions?
 Aug 19th 2017, 05:45 PM #7 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 534 So, if I understand you correctly, we can just add in minkowski space and set it to a rotating polar co-ordinate system so all we need is the standard polar co-ordinate vector calculus expression $\displaystyle \mathbf{\ddot{r}}= (\ddot{r} - r \dot{\phi}^2)\mathbf{\hat{r}} + (r \ddot{\phi}+2 \dot{r}\dot{\phi})\hat{\mathbf{\phi}}$ to make the disk appear to be stationary with respect to an observer ??
Aug 20th 2017, 08:09 AM   #8
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 Originally Posted by kiwiheretic So, if I understand you correctly, we can just add in minkowski space and set it to a rotating polar co-ordinate system so all we need is the standard polar co-ordinate vector calculus expression $\displaystyle \mathbf{\ddot{r}}= (\ddot{r} - r \dot{\phi}^2)\mathbf{\hat{r}} + (r \ddot{\phi}+2 \dot{r}\dot{\phi})\hat{\mathbf{\phi}}$ to make the disk appear to be stationary with respect to an observer ??
You don't need to do ANYTHING beyond what I've already specified. The stationary observer at the center of the rotating disk (who is not rotating wrt the disk) just computes the centripetal acceleration v*v/r acting on a creature fixed at a point on the disk. The creature perceives a force of that same magnitude that is pulling him radially outward (called the centrifugal force). Since v = omega*r, v*v/r = omega*omega*r, so the perceived centrifugal force increases linearly with the distance from the center of the disk. The creatures conclude that they are under the influence of a strange gravitational field that causes that force.

The creatures on the disk don't NEED any mathematical equations of motion, because they aren't (in their opinion) moving. They are entitled to insist that they are stationary. If they DO decide to move on the surface of the disk, it is a very slow and irregular movement that requires no equations or formalism. For example, they can cooperate among themselves to slowly lay out the yardsticks along the circumference of the disk and along some radius extending out to that circumference. Or they can slowly lay out yardsticks along any circle of any radius on the disk ... they need no equations or mathematical formalism to do those things. And once they have done those things, they can conclude (without any further help) that in their world, the ratio of a circumference to the corresponding radius ISN'T 2pi (it's greater than 2pi), and therefore that the geometry of their world ISN'T Euclidean. THAT is the critically important result that this example gave to Einstein. It established a constraint that the (at that time) unknown general theory of relativity HAD to satisfy, and it pointed Einstein in the right direction to (eventually) formulate that theory.
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Mike Fontenot

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