energy momentum relation in SR and GR
There is the famous energy-momentum relation
$$E^2 - p^2 c^2 = m^2 c^4.$$
I thought it is always valid. "Always" means it is valid in general, so in GR and SR. It is something like an elemantary construction. But now I am not sure and if I understand right it is valid only in SR. Right?
Below I write some more calculations and thoughts for a better understanding of my question.
In SR it is very simple to derive the formula. We start with the metric
$g_{\mu\nu} = \eta_{\mu\nu} = {\rm diag}(1,-1,-1,-1)$
and the product
$g_{\mu\nu} p^\mu p^\nu = m^2 c^2$
where $p^\mu = m u^\mu = m \frac{{\rm d} x^\mu}{{\rm d} \tau}$ is the four-momentum. This gives
$g_{\mu\nu} p^\mu p^\nu = (p^0)^2 + \eta_{ij} p^i p^j = (E/c)^2 - p^2 = m^2 c^2$
with the definitions $p^0 = E/c$ and $\eta_{ij} p^i p^j = {\mathbf{p}}^2 = p^2$. 394
The last equation is equivialent to the above energy-momentum relation.
Now, when we turn to GR and consider the general metric in spherical coordinates
$g_{\mu\nu} = {\rm diag}(g_{00}, g_{11}, -r^2, -r^2 \sin^2 \vartheta)$
gives
$g_{\mu\nu} p^\mu p^\nu = g_{00} (p^0)^2 + g_{ij} p^i p^j = g_{00} (p^0)^2 - p^2 = m^2 c^4$
Here we have to be carefully now with the definition of $p^0$. In nearly all books about GR we find for the above spherical metric the following relation (or equivalent)
$g_{00} m \frac{{\rm d} x^\mu}{{\rm d} \tau} = E/c = \rm const$
which is interpreted as the conservation of energy. In SR we have $g_{00} = \eta_{00} = 1$ and therefore $p^0 = E/c$. But in GR we have the $g_{00}$ coefficent, e.g.
$g_{00} p^0 = E/c$
This would give the following energy-momentum relation
$\frac{E^2}{g_{00}} - p^2 c^2 = m^2 c^4$
Am I missing or do I mix something, e.g the definition of the time-component $p^0$ through an energy?
Or are the calcutions all right and in GR the energy-momentum relation is indeed different compared to the above famous and well known, which seems to be valid only in SR.
I am heavily confused.
My only idea how to resolve this is to redefine the energy in the general case, e.g.
$\sqrt{g_{00}} p^0 = \mathcal{E}/c$
Note the square root compared to the above definition. This would give an epxression analogue to the famous relation
$\mathcal{E}^2 - p^2 c^2 = m^2 c^4$
But $\mathcal{E}$ is not conserved here any more, $\mathcal{E} \neq \rm const$. Instead we have $\sqrt{g_{00}} \mathcal{E} = E = \rm const$.
Is this a problem?
In SR we have simply $E = \mathcal{E}$ and there is no need to worry. So, SR automatically implies that $p^0$ is conserved. But it is not necessary the case in GR, right? Or is this all just a question of definition/convention? 394