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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.

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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!

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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:

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

in the radial direction and:

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

in the pure non-radial direction.

You really need to compensate the Lorentz factor for this, and get:

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

This can be re-written as:

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

Written in this form, the Schwarzshild radius, the innermost stable circular orbit and the photon radius is not so hard to find. You can still find the classical kinetic and potential energy in the weak field limit.

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