Question
Consider the following Lagrangian for a particle of mass $m$ ($c$ is a constant)
$L = −mc \sqrt{c^2 - r' .r'} $
Show that it is invariant under translations, rotations and compute the conserved angular momentum $L$, linear momentum $p$ and energy.
What I've Done:
To show the translation $r$ -> $r + a$ so $(r + a)' = r'$ since $a$ is a constant.
For the rotation put $O$ which represents an orthogonal rotation matrix so $r$ -> $Or$ so $(Or)' = Or'$then $Or'.Or'$ = $r'^T O^T . Or' = r'^T.r' = r'.r'$
Im not sure if what I did is right but I think it is so if you could just check that it would be helpful and help me with the computing the momentum and energy thank you.
What you've done to show invariance under translations and rotations is correct.
Then you can calculate the momentum, angular momentum and energy as follows. I've used a dot instead of a prime for the time derivative. We have $$\mathcal{L} = - mc^2 \sqrt{1-\frac{\dot{r}^2}{c^2}}.$$ Then the momentum is given by $$p = \frac{\partial \mathcal{L}}{\partial \dot{r}} = \frac{m \dot{r}}{\sqrt{1-\frac{\dot{r}^2}{c^2}}}$$
The Euler Lagrange equation is then$$\dot{p} = \frac{\partial \mathcal{L}}{\partial r}.$$ This Lagrangian doesn't depend on $r$, so $\dot{p} = 0$ and $p$ is conserved.
Then the angular momentum is given by $$L = r \times p = \frac{m }{\sqrt{1-\frac{\dot{r}^2}{c^2}}}r\times \dot{r}.$$To see that $L$ is conserved, note that $$\dot{L}=\dot{r}\times p + r\times \dot{p}.$$ We already know $p$ is conserved so the second term is zero, and the first term is zero because $p$ is proportional to $\dot{r}$.
The energy is given by the Hamiltonian; $$E = \mathcal{H} =\dot{r}\cdot p - \mathcal{L} = \frac{m c^2}{\sqrt{1-\frac{\dot{r}^2}{c^2}}}.$$
To show that the energy is conserved, we'll rewrite in terms of $p$ instead of $\dot{r}$. To do this, first define the Lorentz factor $\gamma := \frac{1}{\sqrt{1-\frac{\dot{r}^2}{c^2}}}$. Then rearrange to see that $ \dot{r}^2 = c^2\frac{\gamma^2-1}{\gamma^2} $.Now square the relation between $p$ and $\dot{r}$, and substitute for $\dot{r}^2$ in terms of $\gamma$ to give that $p^2 = m^2c^2(\gamma^2-1).$ Solve this equation for $\gamma$ to see that$$\frac{p^2}{m^2c^2} + 1 = \gamma^2.$$ Finally, substitute this into the expression for the energy to arrive at Einstein's famous mass-shell relationship,$$E^2 = m^2c^4 \gamma^2 = m^2c^4 + p^2c^2.$$ In the particle's reference frame $p=0$ and hence $E=mc^2$. We can also see now that the energy is manifestly conserved, because it's a function only of $p$ and $p$ is conserved.