Relativistic Rotation: Does v=rω?

[Nabeshin]

I know that for normal Newtonian motion, we can write v=rω. Does this hold for relativistic motion as well? My gut tells me no, but I can't find (in my GR book or online) a comparable equation or discussion of relativistic rotation. It should be noted that while the case of relativistic motion in flat space time is interesting, ultimately I am interested in this equation in a curved space time.

 

[atyy]

Yes, you can have a "rigidly" rotating disc with v=rw, which is the protagonist in the Ehrenfest paradox.

For GR, maybe http://relativity.livingreviews.org/Articles/lrr-2003-3/

 

[Nabeshin]

Hrm. Well, the case I'm interested in simply involves a particle moving with some angular speed(dϕ/dτ) at a radius of r, and I want to find the linear velocity. So for this case, there isn't really any rigid body in question...

 

[JesseM]

In any given coordinate system I think all the usual kinematical laws have to apply--coordinate velocity is coordinate distance over coordinate time, the coordinate circumference of a path which is circular (relative to the coordinates) is 2pi times the coordinate radius, etc. Only when you're comparing multiple frames, or looking at coordinate-independent quantities like proper distance and proper time, would you have to worry about ways that relativity is different from Newtonian mechanics.

 

[kista]

Lets say you are in an inertial frame watching a particle moving in circle of radius r with velocity w. Then v=rw is valid. Only thing is that w would be different from w as measured in the instantaneous rest frame of particle because of time dilation. So value of w would be different but relation would be valid i think.

 

[atyy]

Hrm. Well, the case I'm interested in simply involves a particle moving with some angular speed(dϕ/dτ) at a radius of r, and I want to find the linear velocity. So for this case, there isn't really any rigid body in question...

Try defining ϕ in terms of x,y,z,t, the cartesian coordinates of an inertial frame, then use v=|(dx/dt, dy,dt, dz,dt)|.

 

[yuiop]

Post 8 of this thread https://www.physicsforums.com/showthread.php?p=2446850&highlight=tangential#post2446850 might be of interest to you.

 

[Nabeshin]

Thanks everyone who responded! That post is very similar to what I was doing that I needed this for, so it helped a lot. Thanks, kev!

 

[yuiop]

Cheers Nabeshin,

I had started this post before you posted your response, so I thought I might as well finish it and you or others might still find something useful in it.

..., ultimately I am interested in this equation in a curved space time.

Consider a stationary observer located on a tower a distance r from the centre of a non rotating gravitational body. If he were to measure the linear velocity of a passing satellite with a circular orbit of radius r, then yes, he could calculate the angular velocity ω of the satellite using the Newtonian equation ω = v/r. An observer at infinity would measure the linear velocity to be:

$$v \sqrt{1-\frac{r_s}{r}} =\ r\omega \sqrt{1-\frac{r_s}{r}}$$

and the angular velocity as :

$$\frac{v}{r} \sqrt{1-\frac{r_s}{r}} =\ \omega\sqrt{1-\frac{r_s}{r}}$$

An observer on board the satellite (considering himself to be stationary), would measure the instantaneous velocity of the observer on the tower to be v (or rω) as he passed by. When he multiplies the velocity by the period it takes the tower to complete one full rotation, he works out the circumference of the path taken by the tower observer to be:

$$2\pi r \sqrt{1-v^2/c^2}$$

so effectively he measures the circumference to be length contracted and the geometry does not appear Euclidean to the observer on board the satellite. Note that both the observer on the tower and the observer at infinity measure the circumference to be simply 2pi*r.

The Newtonian formula for the velocity of an body of mass m orbiting a gravitational body of mass M is:

$$v = \sqrt{\frac{GM^2}{(M-m)r}$$

which for very small m compared to M approximates to:

$$v = \sqrt{\frac{GM}{r}}$$

This is the linear orbital velocity as measured by an observer at infinity. (It is slightly odd that the Newtonian equation for angular velocity is equivalent to the local angular velocity in the Schwarzschild metric while the Newtonian equation for orbital velocity is equivalent to the measurement made by the observer at infinity in the same metric.)

Relative to the speed of light the last equation can be restated as:

$$\frac{v}{c} = \sqrt{\frac{GM}{rc^2}} = \sqrt{\frac{r_s}{2r}}$$


The relativistic linear orbital velocity $v_r/c$ as measured by a stationary observer at radius r is time dilated so:

$$\frac{v_r}{c} \sqrt{1-\frac{r_s}{r}} = \sqrt{\frac{r_s}{2r}}$$

For an orbiting photon, $v_r/c =1$ so the orbital radius $r_p$ of a photon can be determined by:

$$\sqrt{1-\frac{r_s}{r_p}} = \sqrt{\frac{r_s}{2r_p}}$$

$$1-\frac{r_s}{r_p} = \frac{r_s}{2r_p}$$

$$r_p = \frac{3r_s}{2}$$

 

[Demystifier]

Nabeshin, the following papers may also be helpful:
http://xxx.lanl.gov/abs/gr-qc/9904078 [Phys.Rev. A61 (2000) 032109]
http://xxx.lanl.gov/abs/gr-qc/0307011 [a free chapter of the book "Relativity in Rotating Frames"]