It helps to write the full action:
$$S = \int \frac{-mc^2}{\gamma}dt - \int U dt $$
The first term can be put in a much better form by noting that $d\tau = \frac{dt}{\gamma}$ represents the proper time for the particle. The action is then:
$$S = -mc^2\int d\tau - \int U dt$$
The first term is Lorentz invariant, being only the distance between two points given by the Minkowski metric, and is good in relativity. The second term however, isn't (assuming that $U$ is a scalar); there is no way it can be a relativistic action.
There are two easy ways out:
- The first is simply to change the term to $\frac{U}{\gamma}$. This gives the action:
$$S = -\int (mc^2+U)d\tau$$
- The second is to "promote" the term (a terminology used in Zee's Einstein Gravity in a Nutshell) to a relativistic dot product, giving the action:
$$S = -mc^2\int d\tau - \int U_\mu dx^{\mu}$$
The former has no real world classical analog (that I know of), and the latter is more or less the interaction of a particle with a static electromagnetic field. But the original form is recovered from the latter when the spatial components of $U_\mu$ vanish, leaving only $U_0$.
The kinetic energy is obtained by transforming the Lagrangian to the Hamiltonian (see here).
The expressions are not true in general. The first one should be $E^2 = m^2c^4 + p^2c^2$, and the momentum is in general $p = \gamma m v$. The rest energy is $E_0 = m c^2$ and it doesn't depend on the frame (by definition), and the kinetic energy is always $T=E-mc^2 = (\gamma - 1) mc^2$.
You are (understandably) confused because the question is not telling you that momentum is $mc$. You are being told that in a specific situation and in a specific frame, it just so happens that the momentum is equal to $mc$. You should be able to find the velocity from this, and then the kinetic energy.
Alright, since you're having trouble let's get our equations straight. First we define $\gamma$, which is a function of velocity $v$, as $1/\sqrt{1-v^2/c^2}$. The momentum $p$ of a particle with mass $m$ moving with velocity $v$ is given by $p = mv/\sqrt{1-v^2/c^2} = \gamma m v$. The expression $\gamma mv$ looks simpler but don't forget that $v$ is hidden inside $\gamma$.
There are two expressions for the energy. Obviously both are true and can be proved to be equal to each other; the only difference is whether you want the energy in terms of $p$ or $v$. So we have $E^2 = p^2c^2 + m^2c^4$ and $E = mc^2/\sqrt{1-v^2/c^2} = \gamma m c^2$. Kinetic energy $T$ is defined as $T = E-mc^2 = (\gamma-1)mc^2$. As always, this depends on $v$ through $\gamma$.
All these equations are true in any frame. The quantities themselves (such as $v$, $p$ or $E$) change when you change frames, but they change in such a way that all the equations remain correct.
Now, you have been told that in some particular frame, a particle is moving with $p=mc$. This will not be true in general, since the formula for $p$ is $mv/\sqrt{1-v^2/c^2}$; it just so happens that in our situation, $v$ is such that $p = mc$. This is an equation you can use to find $v$; knowing $v$, you can use the formula for kinetic energy $T$ (which, don't forget, depends on $v$) and find what you are being asked for.
Best Answer
If you go all the way back to the derivation of the Lagrangian, I think you can see why the relativistic Lagrangian is what it is. Usually, one starts with d'Alembert's principle $$ \sum_i \left(F_i - \dot{p}_i\right) \delta x_i = 0, $$
where $F_i$ are the forces acting on particle $i$, $p_i$ is its momentum, $\delta x_i$ is a virtual displacement, and the dot designates differentiation with respect to coordinate time $t$. For the relativistic case, we use the proper momentum $p = \gamma m v$.
Focusing on the relevant term only:
$$-\dot{p}\, \delta x = -\frac{d(\gamma m v)}{dt}\, \delta x $$
We use the chain rule (or integrate by parts):
$$ = -\frac{d}{dt}\left(\gamma m v\, \delta x\right) + \gamma m v \frac{d(\delta x)}{dt}$$
When we do the full variation the first term vanishes because $\delta x=0$ on the boundary. Under typical assumptions the ordinary and variational derivatives commute:
$$ = \gamma m v\, \delta v$$
We want to end up with a statement like $\delta(\mathrm{stuff})$. If you are smarter than me, you might see that:
\begin{align} \delta\left(\frac{1}{\gamma}\right) &= \frac{1}{2} \left(1-\frac{v^2}{c^2}\right)^{-1/2} \frac{-2v}{c^2}\,\delta v = -\frac{1}{c^2}\,\gamma v \, \delta v \\ \delta\left(-\frac{mc^2}{\gamma}\right)&= \gamma m v\, \delta v \end{align}
or you could know the answer and `astutely guess' like I did.
When you go through the whole variational rigmarole you'll end up with something like:
$$ \int_{t_i}^{t_f}\sum_i \left(F_i - \dot{p}_i\right) \delta x_i \,\mathrm{d}t = \delta \int_{t_i}^{t_f}\left(-\sum_i\frac{m_i c^2}{\gamma_i} - V \right)\mathrm{d}t = 0 $$
So for a single particle $$ L = - \frac{m c^2}{\gamma} - V.$$