# Rotating rigid sphere stress-energy tensor

1. Mar 21, 2009

### aatw

Hy.

Can somebody please show me the way, how to transform stress-energy tensor for sphere in rest frame to stress-energy tensor in rotating frame using Lorentz transformations?

2. Mar 21, 2009

### tiny-tim

Welcome to PF!

Hy aatw! Welcome to PF!

Whyever would you want to use a rotating frame?

Anyway, the rotations in the group of Lorentz transformations are exactly the same as in ordinary Euclidean space.

3. Mar 21, 2009

### aatw

Maybe I didn't express myself so well.
Let's start from stress-energy tensor for sphere, which can be written as $$T^{ab} = (\rho + p) u^a u^b + p g^{ab}$$, but in rest frame where sphere doesn't rotate, it's only $$T^{00}$$ element that's nonzero (I think so!).
But now sphere starts to rotate with angular speed $$\omega$$ around z axis and I want to use Lorentz transformation to determine $$T^{ab}$$ for rotating sphere.

4. Mar 21, 2009

### Mentz114

Have you tried plugging in this velocity 4-vector ?
$$\left[ \begin{array}{c} -\sqrt{\omega^2r^2-c^2}\\\ -\omega r\sin(\omega t) \\\ \omega r\cos(\omega t) \\\ 0 \end{array} \right]$$
where $r^2=x^2+y^2$ into the EMT ? This seems to be in cartesian coords but easy enough to transform to polar.

I suppose another approach would be to boost in the x and y directions with velocities as given above. I'll try that later if I have time.

Last edited: Mar 21, 2009
5. Apr 2, 2009

### Mmmm

No, if $g = \eta$ then $T^{ij} = P \delta^i_j$ where P is the pressure between the elements (just use the formula you quoted with $u^a = (1,0,0,0)$ -which is the 4-velocity in the rest frame).

ok, so just use an ordinary boost in all 3 spacial directions with $v=\omega r$

Last edited: Apr 2, 2009