T(p)=\sqrt{p^2c^2+m^2c^4}
I checked that this returned the velocity when the derivative with respect to p was taken and it did, correctly (I believe), return the momentum divided by m\gamma with \gamma being given as a function of p.
I then thought I'd have a look at a relativistic harmonic oscillator using V(x)=\frac{1}{2}kx^2
I was able to reduce this to the equation (for some reason, the equations seemed much easier to solve for the momentum than the position):
\ddot p^2 + \frac{p^2\ddot p^2}{m^2c^2} - \frac{k^2p^2}{m^2}=0
Which bears an obvious resemblance to the classical form with a relativistic correction of \frac{p^2 \ddot p^2}{m^2c^2} which will, as expected, produce large deviations for large momenta as well as high jerk situations and, according to wolfram alpha, the solutions appear to be approximately sinusoidal (which is good as it should approximate the classical sinusoidal harmonic oscillator at low momenta).
That said, I am unable to reduce this to a closed form (and, as the equation is non-linear, it appears that in all likelihood, no such form exists) and, not being familiar with the techniques of differential equations at all, am unable to plot the solutions (I am particularly interested in what happens to the family of \ddot p(0)=0 curves as p(0) becomes close to m\gamma c.
All of this led to me to wonder the following:
1. Do hamilton's equations still hold for relativistic physics (with an appropriate hamiltonian)?
2. If anyone could give me a clue as to how to plot the solutions to the differential equation (thinking about it physically, I think that it should be triangle wave in the limit of large p(0) and a sine wave in the limit of small p(0)).
3. If anyone had any idea why it might be (or at least seem) easier to solve for momentum than position.
