Can Schwarzschild acceleration be written in exact form as a function of position and velocity vector?

Question

For the special case of only one massive point-shaped or spherically symmetric non-rotating body and a small "test mass" body under general relativity the movement of that small body is given by the Schwarzschild solution of the Einstein field equations. Ignoring relativity and staying with Newton we have:

$${d(m\bar{v})\over{mdt}}={d\bar{v}\over{dt}}=f(\bar{r},GM)=-{GM\over{r^2}}\hat{r}.$$

Now is there a found function g such that ${d\bar{v}\over{dt}}=g(\bar{r},\bar{v},GM)$ creates a motion of the test body such that it exactly replicates what the motion would be according to the Schwarzschild solution?

If I do not want to use the analytical solution but create a numerical step by step integration using the acceleration at all times to recreate the Schwarzschild orbits not only by shape but also by orbital velocity, what would this function g look like?

I found a paper by Rosswog/Tejeda from 2013 trying to find a as good "function g" as possible, so I guess at least 10 years ago there was no perfect solution. I also know there is the "post-Newtonian expansion" as described to the 3PN-level in this thread here on stack exchange.

Questions:

1. Can you write the gravitational acceleration of a test body under Schwarzschild conditions as a function of both the position vector and velocity vector in such a way that the orbits given using stepwise integration exactly replicates the orbits given from the analytical Schwarzschild solution?

2. What is the expression for the acceleration of a test body under Schwarzschild conditions as a function of position and velocity that when used in step wise integration best replicates the Schwarzschild orbits?

Answer 1

The well known post-Newtonian expansion to the 1PN "first post-Newtonian" order reduces under stated conditions of only one massive spherical symmetric body to:

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

This is expression 4-26 on page 4-19 in Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation, by Theodore Moyer.

This expression quite perfectly reproduces the correct GR perihelion shift for the weak fields of our solar system. It does not reproduce classical values for the orbital velocity or for the initial acceleration of an object at rest. I do not no if that matters. At stronger fields you get really strange orbits if you try it. The post-Newtonian expansion is available to at least the 3PN order and to that order it looks like this under stated conditions:

\begin{align} \frac{d\bar{v}}{dt} &=-\frac{GM}{r^2}\left(1-4\frac{GM}{rc^2} + 9\left(\frac{GM}{rc^2}\right)^2 - 16\left(\frac{GM}{rc^2}\right)^3\right)\hat{r} \\ &\qquad-\frac{GM}{r^2}\left(\frac{v^2}{c^2}-\frac{2GM}{rc^4}\left(\bar{v}\cdot\hat{r}\right)^2 +\frac{(GM)^2}{r^2c^6}(\bar{v}\cdot\hat{r})^2\right)\hat{r}\\ &\qquad-\frac{GM}{r^2}\left(-4\frac{(\bar{v}\cdot\hat{r})}{c^2}+2\frac{GM}{rc^4}(\bar{v}\cdot{\hat{r}})-4\frac{(GM)^2}{r^2c^6}(\bar{v}\cdot\hat{r})\right)\bar{v} \end{align}

I do not know how much "better" orbits you get with this expanded version or if there is a "5PN" version calculated by someone that gives even better results.

(As there are no answers to the inital questions as of yet I throw this in as an approximation to be replaced with better approximations if possible.)

Answer 2

In 2013 one approximation to get the Schwarzschild acceleration as a function of velocity and position was published in MNRAS (Monthly Notices of the Royal Astronomical Society) by Rosswog/Tejeda. It seems to me that at least until then there were no exact solution, otherwise coming up with an approximation would make no sense. Unfortunately they are not using cartesian coordinates which makes the paper somewhat hard for me to grasp. In the abstract they write:

This potential reproduces several relativistic features with higher accuracy than commonly used pseudo-Newtonian approaches. The new potential reproduces the exact location of the marginally stable, marginally bound, and photon circular orbits, as well as the exact radial dependence of the binding energy and the angular momentum of these orbits. Moreover, it reproduces the orbital and epicyclic angular frequencies to better than 6%.

In cartesian coordinates this expression for the acceleration (expression A.5 in the paper) looks like:

$$-{GM\over{r^2}}\left(1-{2GM\over{rc^2}}\right)^2\hat{r}+{2GM\over{r^2(1-{2GM\over{rc^2}})}}{v^2\over{c^2}}(\hat{v}\cdot\hat{r})\hat{v} \\ -{3GM\over{r^2}}{v^2\over{c^2}}|\hat{v}\times\hat{r}|^2\hat{r}$$

($\hat{r},\hat{v}$ unit vectors)

I have not really analyzed this expression more than what is in the paper. Obviously they are introducing a repulsive $1/r^3$ and an attractive $1/r^4$ term besides the classical Newtonian $1/r^2$-term. They also have one velocity-dependent term that vanishes for a circular orbit and one that vanishes for an object falling straight down to the center of mass.

Perhaps this is still the best approximation of an exact solution to the quest for producing an expression for gravitational acceleration as a function of position and velocity that reproduces Schwarzschild orbits?

Answer 3

So i put in my own alternative: $$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(\hat{r}-2\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})} -\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})^2}\left((\hat{v}\cdot\hat{r})^2+(1-\frac{2GM}{rc^2}-\frac{v^2}{c^2})|\hat{v}\times\hat{r}|^2\right)\right)\tag{1}$$.

($\hat{r},\hat{v}$ unit vectors)

You get this from inserting:

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

into: $$\frac{d(m\gamma\bar{v})}{dt}=-\frac{GMm\gamma}{r^2}\tag{3}$$

The term (2) is a resistance to acceleration term. The physical meaning of the part under the "root sign" to the left is that the resistance to acceleration becomes infinite in the direction you are traveling when you reach the speed of light. Note that in coordinate time the speed of light is not constant in a gravitational field but varies and is different in the radial direction versus in the plane tangential to the radial direction. The physical meaning of the part under the "root sign" to the right is that the resistance to acceleration becomes infinite at the Schwarzschild radius.

The somewhat strange appearance of (1) is because I took (2) for the case of pure radial motion and put into (3). Then I took (2) for the case of pure non-radial motion and put into (3). Expression (1) is just a superposition of these two special cases. I hope this is a correct way of doing it.

The expression for acceleration (1) gives the correct anomalous precession in the weak field limit. It gives the correct value for the orbital velocity of a circular orbit: $$v=\sqrt{GM/r}\tag{4}$$ which is the same as the classical value and it also gives the same value as classically for the acceleration of a body at rest, which I think is correct according to GR.

I think from this discussion that (1) gives the correct expression for the acceleration of a body in pure radial infall.

In the strong field limit you can see that for circular orbits using identity (4) the expression (2) becomes infinite at the photon radius ($r=3GM/c^2$) and taking the derivative of that new expression with respect to r I think reveals the innermost stable circular orbit at $r=6GM/c^2$.When you get really close to the black hole you also get the expected result of the orbiting body falling down and stopping at the Schwarzshild radius. I ran a lot of simulations a few years ago using an expression very similar to (1).

In many ways I think (1) does a good job at replicating Schwarzshild orbits.

Answer 4

−GMr2(1−2GMrc2)2r^+2GMr2(1−2GMrc2)v2c2(v^⋅r^)v^−3GMr2v2c2|v^×r^|2r