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In general relativity the energy of a test-body moving in a spherically symmetric gravitational field can be written as:

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

The strange part is just because the speed of light is not the same in the radial direction as it is transverse to the radial direction. To be precise I mean velocity in coordinate time and treat the r-parameter of the Schwarzschild solution as a real distance. For a pure non-radial motion the expression above becomes a little bit simpler:

$$E=mc^2\left(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2\left(1-\frac{2GM}{rc^2}\right)}}}\right). $$

This can be rewritten as:

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}\right).$$

For a circular motion you have $v=\sqrt{GM/r}$ in general relativity as well as classically, and by replacing $v^2$ with $GM/r$ above you get an expression where you find that you need infinite energy to stay in a circular orbit at the photon sphere radius and by taking the derivative of that expression with respect to "r" you find the innermost stable radius.

If you define $c(r)=c\sqrt{1-\frac{2GM}{rc^2}}$ and $m(r,v)=\frac{m}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left(1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2)+|\hat{r}\times\hat{v}|^2\right)}}}$

which for pure non-radial motion reduces to: $$m(r,v)=\frac{m}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}.$$

you can write: $E=m(r,v)(c(r))^2$

Question: Can you write, is it correct to write, the energy of a test-particle moving under Schwarzschild conditions in the form $E=m(r,v)(c(r))^2$ like I have outlined above?


There is a specific reason why I am asking and that is that I want to generalize Newtons gravitational law by writing:

$$\frac{d(m\gamma\bar{v})}{dt}=-\frac{GMm\gamma}{r^2}\hat{r}$$ and when I use $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ I can only explain one third of the "anomalous perihelion shift". In order to explain all of it I can use $\gamma=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}{\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}}$. If you do this you get an expression that looks like:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(\hat{r}-3\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})} +\frac{v^4(\hat{r}\cdot\hat{v})\hat{v}}{c^4(1-\frac{2GM}{rc^2})^2}\right)$$.

(Note: here $\frac{\bar{r}}{r}=\hat{r}$ and $\frac{\bar{v}}{v}=\hat{v}$)

(Edit: Maybe it possible to find the acceleration by the condition that the time derivative of the energy expression must be zero, given that the expression for the energy is correct.)

The resulting orbits from using this expression is close to what is expected from GR but not quite right, I think it could be improved. In the plot of numerical integrations using the last expression below the green circle represents the Schwarzschild radius and the red circle represents the innermost stable circular orbit.

Orbits closing in and passing the ISCO

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    $\begingroup$ Please try using "\left(" and "\right)" to get nested larger brackets in Mathjax/LaTex. This will try and size the brackets to match the enclosed object and look significantly better. $\endgroup$ – StephenG May 13 at 5:05
  • $\begingroup$ Thanks for the new edit, however in that last equation I think you need $\frac{GM}{r^\color{red}{3}}$ and not $\frac {GM}{r^\color{green}{2}}$ $\endgroup$ – StephenG May 13 at 13:13
  • $\begingroup$ I meant for $\bar{r}$ to mean radial distance and $\hat{r}$ to mean a unit vector in the radial direction but this might be non-obvious. $\endgroup$ – Agerhell May 13 at 13:35
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    $\begingroup$ I see what you mean, and it's non-obvious to me : I'd expect $\hat{r}$ to be the position vector (and I'd assume $\hat{v}=\frac{d\hat{r}}{dt}$ which would not be the case in your use ). I would not normally expect to see them as unit vectors unless stated. Thanks. $\endgroup$ – StephenG May 13 at 15:22
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In general relativity, if a spacetime has a timelike Killing vector $\xi$ (i.e., is stationary), test particles have a conserved quantity $v^a\xi_a$. For a massive particle, with the appropriate normalization, this is interpreted as the energy. In the Schwarzschild spacetime, this is $(1-2m/r)(dt/d\tau)$, where $\tau$ is the proper time, $m$ is the mass of the black hole, and the units are such that $G=c=1$.

In general relativity the energy of a test-body moving in a spherically symmetric gravitational field can be written as:[...]

I'm not sure what you mean by this. By Birkhoff's theorem, a vacuum spacetime that is spherically symmetric must be Schwarzschild. If that's what you mean, then I'll take your word for it that the extremely complicated expression you give is equivalent to the standard way of expressing it, for $r>2m$. It looks wrong for $r\le 2m$, because your square roots become imaginary, although I suppose it's possible that analytic continuation makes it equivalent to the correct expression. (BTW, you could do yourself a favor by working in a system of units where you don't have to write all the factors of $G$ and $c$.)

To be precise I mean velocity in coordinate time[...]

Meaning your $v=dr/dt$? This sounds unlikely to make sense for $r<2m$, where $r$ is the timelike coordinate and $t$ is the spacelike one.

The strange part is just because the speed of light is not the same in the radial direction as it is transverse to the radial direction.

Not true. The speed of light is the same in all directions. The coordinate velocity differs, but coordinate velocities are not physically meaningful.

Can you write, is it correct to write, the energy of a test-particle moving under Schwarzschild conditions in the form E=m(r,v)(c(r))2 like I have outlined above?

I haven't checked your algebra, but if your algebra is correct, then there is nothing to stop you from writing it this way. But the notion that $c$ depends on $r$ is not likely to lead you anywhere of interest. In a sensible system of units, $c=1$.

There is a specific reason why I am asking and that is that I want to generalize Newtons gravitational law by writing:

You can't get general relativity by fiddling with the form of Newton's law of gravity. Newton's law of gravity describes instantaneous action at a distance, which is impossible in relativity. What is your goal here? Are you looking for a post-Newtonian approximation that works in the weak-field limit, for the Schwarzschild spacetime?

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  • $\begingroup$ I am looking for an expression comparable to the post-Newtonian approximation but from Schwarzschild coordinates instead of isotropic coordinates yes. However I would preferably like for the expression to function also in the strong field limit. I really do not want anything to be able to cross the event Horizon, is that a must? $\endgroup$ – Agerhell May 13 at 14:38
  • $\begingroup$ In the post-Newtonian Schwarzshild approximation you have a very strange negative inverse cube term,see this post, that I do not like. I have a question worth 50 point on the sister site that might be easy for you. It is here. $\endgroup$ – Agerhell May 13 at 15:00
  • $\begingroup$ I really do not want anything to be able to cross the event Horizon, is that a must? The fact that it seems completely wrong at or inside the horizon suggests that it's at best only approximately right outside the horizon. $\endgroup$ – Ben Crowell May 13 at 17:01
  • $\begingroup$ At least you get reasonable orbits down to the Schwarzshild radius. Here is an old example of an orbit computed using a slightly different expression: Spinning around and falling to Schwarzschild radius You have a kind of a 0/0 situation at the event horizon. $\endgroup$ – Agerhell May 13 at 17:58

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