Lagrangian equation compute conserved angular and linear momentum. 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.
 A: 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.
