Special Relativistic Time Dilation -- A computer in a very fast centrifuge Ok, I've stumbled onto what I think is a bit of a paradox. 
First off, say you had some computer in a very fast(near light speed) centrifuge. You provide power to this computer via a metal plate on the "wall" of the centrifuge's container, so it works similar to how subways and streetcars are powered.
If the computer normally would consume 200 watts, how much power would it consume at say 1/2 of light speed? Would it consume 400 watts from our still viewpoint?
Also, what if you were to be capable of communicating with this computer? Would the centrifuge-computer receive AND transmit messages faster from our still viewpoint? I'm a bit lost in even thinking about it.
 A: One thing to be aware of is that the principle of relativity would not apply to this computer--rotating reference frames are not inertial, and therefore, will not be related to 'stationary' reference frames by simple Lorentz transformations.  Also note that if there are any capacitors or anything along those lines in the computer, then they would be accelerating with respect to a rest frame, and this acceleration would produce electromagnetic waves (and even failing this, accelerating current-carrying wires would have a similar effect), which would create a potentially measurable difference between the frame of the computer and the rest frame.  
My instinct for this problem, though, is that the power consumption of the computer would go down, not up--the computer uses energy in its rest frame, which has a slower clock than the lab frame.  So, it would go on using 200 J every second of ITS proper time, which would correspond to more time in the lab frame.  
A: If you want your computer to run fastest, leave it at rest ;-).
A: A rotating frame has some similarities with special relativity, but since it is an accelerated frame it is not SR.  To describe rotating coordinates consider the transformation between frame $F$ and $F^\prime$
$$
x^{\mu\prime}~=~{\Lambda^{\mu}}_\nu x^\nu,
$$
In cylindrical coordinates the radial and azimuthal coordinates are $r^\prime~=~r$ and $z^\prime~=~z$ and
$$
t^\prime~=~\gamma(t~-~\Omega\chi^t),~\theta^\prime~=~\gamma(\theta~-~\Omega\chi^\theta),
$$
where $\gamma$ is the Lorentz factor.  The angular velocity $\Omega$ gives the tangential velocity $v~=~\Omega r$ as measured by a nonrotating observer.  The rotation in these coordinates is
$$
\eta~=~{{d\theta}\over{dt}},~\eta^\prime~=~{{d\theta^\prime}\over{dt^\prime}},
$$
where in the rotating frame is given by
$$
\eta^\prime~=~\frac{d\theta~-~\Omega\Big(\frac{\partial\chi^\theta}{\partial t}dt~+~\frac{\partial\chi^\theta}{\partial\theta}d\theta\Big)}
{dt~+~\Omega\Big(\frac{\partial\chi^t}{\partial t}dt~-~\frac{\partial\chi^t}{\partial\theta}d\theta\Big)}.
$$
Define the velocities $v~=~r\eta$ and $v^\prime~=~r\eta^\prime$,
for $\chi^t~=~\theta$ and $\chi^\theta~=~t$.  This gives a formula
for the addition of rotation
$$
\eta^\prime~=~{{\eta~-~\Omega}\over{1~-~r^2\Omega\eta/c^2}},
$$
which is analogous to the linear equation for velocity additions.
The metric for a rotating frame is then
$$
ds^2~=~dt^2~-~r^2\Big(d\theta~+~{\Omega\over c} dt\Big)^2~-~dr^2
$$
or
$$
ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2.
$$
The horizon for any observer on the rotating frame exists at $r~=~c/\Omega$. A solution to these coordinates with $dr~=~0$ according to the proper time $s$ is found to be
$$
t~=~ g^{-1}\sinh(gs),~\theta~=~{{g^{-1}}\over r}\cosh(gs)~-~g^{-1}{\Omega\over c} \sinh(gs),
$$
where $g~=~{{\Omega^2r}\over\sqrt{1~-~\Omega^2r^2/c^2}}$ is the acceleration parameter. 
A rotating frame has a sort of event horizon, a Rindler wedge horizon $r~=~c/\Omega$.  However, this is on the frame, so if you send photons along a radially directed fiber optic they get sort of trapped there a bit like a black hole.  If you send them along a free path then there is no horizon.  This is also related to the Sagnac effect, where sending photon pulses along a circular fiber optic in both directions results in one group returning before the other, or in an actual measurement a phase shift. 
A: The clock speed is reduced by the amount of the Lorentz factor.
The waste heat is more energetic by the amount of the Lorentz factor.
So the power that the computer draws is unchanged. 
When speed is 0.8 c, waste heat photons that have two times the normal energy are produced at rate that is half of the normal rate.
There is an important point missing in this answer:
Let us consider a laptop. When we accelerate the laptop to relativistic speed, it takes some energy to accelerate the energy in the laptop battery. When this relativistic laptop computes, the kinetic energy of the chemical energy in the battery turns into kinetic energy of waste heat. 
