# Kinetic, potential and total orbital energy in General Relativity

In Schwarzschild geodesics the total orbital energy $E$ is

$$E = \dot{t} \left( 1 - \frac{r_{\rm s}}{r} \right) m \, c^2$$

with the time dilation factor $\dot{t}$ in dependence of the local velcity $v$

$$\dot{t} = \frac{1}{\sqrt{ \left( 1-\frac{r_{\rm s}}{r} \right) \left( 1-\frac{v^2}{c^2} \right)}}$$

so plugged into the equation for $E$ we get

$$E = \frac{m \ c^2 \ (r-r_{\rm s})}{\sqrt{r \ (r_{\rm s}-r)(v^2/c^2-1)}}$$

which seems to be

$$E = m \ c^2 + E_{\rm \ kin} + E_{\rm \ pot}$$

But how would one factor out the kinetic and the potential component of the total Energy in terms of the coordinate derivatives $\dot{r}, \dot{\phi}, \dot{t}$ or in terms of $v^2=v_{\perp}^2+v_{\parallel}^2$ (radial and transverse components)?

The other constant of motion, the angular momentum, is easy to get because with

$$\dot{r} = v_{\parallel} \sqrt{\frac{1-2 M/r}{1-v^2}} \ , \ \dot{\phi} = \frac{ v_{\perp}}{r \sqrt{1-v^2}}$$

we get

$$L = m \ \dot{\phi} \ r^2 =\frac{m \ v_{\perp} \ r}{\sqrt{1-v^2}}$$

but what about $E_{\rm \ kin}$ and $E_{\rm \ pot}$? Those seem to be very different than with Newton or Special Relativity, at least one of them since the sum does not match up. I only managed to calculate to total energy but failed to split it into it's components.

Your expression for the total energy is

$$E=\frac{mc^2(r-r_s)}{\sqrt{r(r-r_s)(1-v^2/c^2)}}=mc^2\gamma\sqrt{1-\frac{r_s}{r}}$$

If you wish to split this up into kinetic and potential energy, we recall that the kinetic energy in Special relativity is $E_{\text{kin}}=mc^2(\gamma-1)$, and so we have

$$E=mc^2+E_{\text{kin}}+E_{\text{pot}}$$

Where

$$E_{\text{pot}}=-mc^2\gamma\left(1-\sqrt{1-\frac{r_s}{r}}\right)=-mc^2\gamma\left(1-\sqrt{1-\frac{2GM}{rc^2}}\right)$$

Let's do a sanity check. In the nonrelativistic limit, $mc^2(\gamma-1)\sim mv^2/2$ and

$$E_{\text{pot}}\sim-mc^2\left(1-\left(1-\frac{GM}{rc^2}\right)\right)=-\frac{GMm}{r}$$

Which agree with the nonrelativistic expressions!

Your expression for the total energy can be written as:

$$E=mc^2\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2}}}$$

This is slightly wrong because in general relativity under spherical symmetry you have:

$$v_{light}=c(1-\frac{2GM}{rc^2})$$

$$v_{light}=c\sqrt{1-\frac{2GM}{rc^2}}$$

$$E=mc^2\left(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})^2(\hat{r}\cdot\hat{v})^2+(1-\frac{2GM}{rc^2})|\hat{r}\times\hat{v}|^2\right)}}}\right).$$
$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\right).$$