Can the total energy-momentum of a rigidly accelerated extended body be expressed in terms of its mass, acceleration and velocity? Suppose an extended body is rigidly accelerated, hence maintaining a constant mass and distance between its points in its rest frame. Is it possible to express its total energy in any frame in terms of its mass, acceleration and velocity?
The main problem with this is that the various parts will have different velocities and accelerations; however, it might work if a particular point on the extended body is chosen to represent the mass and velocity of the body. I don't know how to do this or if it's already a standard result in relativistic mechanics.
 A: I can't imagine this is an actual homework problem, and I thought it was interesting so I spent some time working out an answer.
A hyperbolically accelerated point particle has constant acceleration $a$ with respect to the momentarily comoving frame. In terms of the proper time $\tau$ the coordinates $(t,x)$ in an inertial frame are,
$$t(\tau)=t_0+\frac{1}{a}\sinh (a\tau),\quad x(\tau)=x_0+\frac{1}{a}\left(\cosh (a\tau)-1\right),$$
where I'm setting $c=1$ for now.
Now suppose there is another accelelerated particle a distance $\xi$ away in the momentarily comoving frame. The four-vector pointing to this particle in this frame is just $(0,\xi).$
Transforming this displacement to the $(t,x)$ frame, and adding to the trajectory of the original particle we get the displaced trajectory,
$$t(\tau,\xi)=t_0+\frac{1}{a}\left(1+a\xi\right)\sinh (a\tau),$$
$$x(\tau,\xi)=x_0-\frac{1}{a}+\frac{1}{a}\left(1+a\xi\right)\cosh (a\tau).$$
These accelerated coordinates $(\tau,\xi)$ are well known and are called Rindler coordinates, so I haven't gone into too much detail in the steps.

Now suppose your rigid body has uniform line density $\lambda$ in the comoving frame. Then an infinitessimal bit of mass $dm$ is $dm=\lambda d\xi,$ and if we find the energy for that piece we can integrate over $\xi$ to find the answer. 
The energy of a piece $dm$ is equal to the time component of its four-momentum.
$$dE=dm \frac{dt}{d\tau}=dm(1+a\xi)\cosh(a\tau)$$
Now the problem is different $dm$ at the same proper time $\tau$ correspond to different times $t$. I didn't notice this in my first answer. It is not too hard to solve in terms of $t$ using
$$\tau=a^{-1}\sinh^{-1}\left(\frac{a(t-t_0)}{1+a\xi}\right)$$
then the energy becomes
$$dE=\lambda \sqrt{(1+a\xi)^2+a^2(t-t_0)^2}d\xi=\lambda \gamma \sqrt{1+\frac{(1+a\xi)^2-1}{1+a^2(t-t_0)^2}}d\xi$$
where I have pulled out the scale factor of $\gamma$ for the $\xi=0$ piece since in the limit that the size is small we expect that to be the answer for energy.
If we integrate we can get a closed form answer, but it is rather ugly, so let's make a harmless approximation. If we restore a factors of $c$, $a\xi$ is really $a\xi/c^2$, so let's take this as a very small number. It is small even as the velocity becomes relativistic, since it doesn't depend on time. I'll expand to first order in $a\xi$.
$$dE\approx\lambda \gamma \left(1+\frac{a\xi}{1+a^2(t-t_0)^2}\right)d\xi$$
Now if we integrate, defining the total mass $M$ and the left and right endpoints of the rigid body $\xi_L,\xi_R$, the total energy is
$$E=\gamma M\left(1+\frac{a}{2}\frac{\xi_L+\xi_R}{1+a^2(t-t_0)^2}\right)=\gamma M\left(1+\frac{a\bar{\xi}}{\gamma^2}\right)$$
where $\bar\xi$ is an average of the endpoints. We can see the correction term becomes even smaller as $\gamma$ increases, so when $a\xi$ is small taking the rigid body as a point particle is a good approximation.
