3
$\begingroup$

From the documentation ”Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation ” expression 4-61 on page 4-42 it can be seen that JPL uses the following expression to account for the effects of relativity under Schwarzschild conditions:

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

When I use this expression in an integrator it replicates the ”anomalous precession of perihelion” correctly. However, in the Schwarzschild solution in Schwarzschild coordinates you should get the same orbital velocity in a circular orbit as classically. Also, the initial acceleration as you drop an object from rest should be the same as classically (I believe). This JPL expression fails in accomplishing that. Someone told me that JPL uses isotropic coordinates instead of Schwarzschild coordinates and that this could be an effect of that, but that seems strange to me.

If you use the ”relativistic mass” concept, that works quite well to calculate the relativistic acceleration of a charged particle under influence of the Lorentz force, on gravity you end up with:

$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}(\hat{r}-\frac{v^2}{c^2}(\hat{r}\cdot\hat{v})\hat{v}) $

This can only generate one third of the perihelion shift, but the expression is better than the JPL expression in the sense that it reproduces correct values for the orbital velocity and the initial acceleration of an object at rest. By cheating and inserting a factor of three:

$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}(\hat{r}-3\frac{v^2}{c^2}(\hat{r}\cdot\hat{v})\hat{v}) $

you get an expression that reproduces the correct perihelion shift but also the correct orbital velocity of an object in circular orbit and the initial acceleration of an object at rest.

The condition for circular motion is $\mathbf{v} \cdot \mathbf{r}=0$, there is no radial part of the motion. Then you set the acceleration terms that do not vanish for $\mathbf{v} \cdot \mathbf{r}=0$ equal to $v^2/r$, the centrifugal acceleration and solve. You see that in the case of no motion,$v=0$, and the case of no radial motion the second and the third expression above reduces to the classical Newtonian gravitational acceleration, which is expected also from the Schwarzschild solution in Schwarzschild coordinates, but the "JPL expression" do not. I would be very happy if someone from JPL could tell me why you are using the first expression above. There is a rudimentary derivation of the expression in the documentation but it is rather high level and not so easy to comprehend.

Note that according to JPL $v = \sqrt{GM/R}$ no longer holds true for a circular orbit but instead you have:

$v=\sqrt{\frac{GM}{r}\frac{(1-4GM/(rc^2))}{(1-GM/(rc^2))}}$

Also when dropping an object from rest, according to JPL, the acceleration goes as:

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

From this last expression we actually see that JPL, in all of their ephemeris calculations, actually use a small "negative inverse r cube" gravitational term, which is a bit odd.

Questions:

1.Why is JPL using the first expression above and not something similar to the third?

2.What is the correct expression for the orbital velocity for a body in circular motion according to JPL?

3.What is the correct initial acceleration of an object at rest according to JPL?

I would be very happy to get some answers.


Strong field orbits

I did spend a lot of time, see old messy paper, trying to come up with some physical explanation for why, at least in the weak field limit, the third expression above should hold true by experimenting with a "general relativistic relativistic mass" of the type $\gamma(r,v)$ instead of just $\gamma(v)$ but I did not quite succeed. If you insert $\gamma=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$ into $\frac{d(m\gamma\bar{v})}{dt}=-\frac{GMm\gamma}{r^2}\hat{r}$ you end up with $\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(\hat{r}-3\frac{v^2(\hat{r}\cdot\hat{v})\hat{v}}{c^2(1-\frac{2GM}{rc^2})} +\frac{v^4(\hat{r}\cdot\hat{v})\hat{v}}{c^4(1-\frac{2GM}{rc^2})^2}\right)$.

In the strong field limits this expression results in orbits as shown below where the green circle represents the Schwarzschild radius and the red circle represents the radius of the "innermost stable circular orbit" located at a distance of three Schwarzschild radiuses. The result is similar to what is expected from GR. Boundary cases where the objects spins around in a circle close to the radius of the innermost stable circular orbit and then falls down into the black hole

If you use the JPL-formula in the strong field limit you can get very strange "bouncing" effects as shown below, this is because of the repulsive inverse r-cube term:

1PN post-Newtonian expansion strong field orbits

This is not at all what is expected from GR. I realized there is a higher order version of the JPL formula that includes an attractive inverse $r^4$ term as well as a repulsive inverse $r^5$ term. Still I think it is very strange to simulate GR by using a repulsive $r^3$ term and I do not really know the reason for why it is common practice to do just that.

$\endgroup$
  • $\begingroup$ The first equation is extensively referenced in answers to How to calculate the planets and moons beyond Newtons's gravitational force? It's not just people who work at JPL who use it, use seems pretty widespread. $\endgroup$ – uhoh Mar 24 at 13:27
  • $\begingroup$ I just thought most people can say "JPL uses this and it seems to work even though I do not exactly know why", but JPL is somehow "responsible" for their own expression. Since I could not find an email adress I wrote a paper mail to Theodore Moyer who wrote the official JPL documentation several years ago asking basically the same questions as above. I got one polite response before he got tired of me but he could not really tell me if he thought the differences I pointed out above should be there or if they are there because of approximation errors in deriving his expression. $\endgroup$ – Agerhell Mar 24 at 14:40
  • 1
    $\begingroup$ For circular motion you start with $\bar{v}\cdot\bar{r}=0$, there is no radial part of the motion. Then you set the force equal to v^2/r, the centrifugal force and solve. You see that in the case of no motion and the case of no radial motion the second equation in the question reduces to the classical Newtonian gravitational accelerations. I did some comparizons in page 8 to 10 in this paper: vixra.org/pdf/1303.0004v1.pdf $\endgroup$ – Agerhell Mar 25 at 5:33
  • 1
    $\begingroup$ @uhoh Do you have access to a reasonably good integrator? It would be interesting to know how much each of the four terms in the jpl formula contributes to the perihelion precession. It seems to me that they have complemented the classical inverse square gravitational acceleration with a small inverse cube part that causes most of the precession. That is a little strange since relativity is not suppose to be about adding an inverse cube part to the classical Newtonian acceleration. $\endgroup$ – Agerhell Mar 27 at 15:12
  • 1
    $\begingroup$ I'm not really familiar with GR, but judging by eq. 4-60, they do indeed use isotropic coordinates, since the term depending on spatial coordinates in the expression for $ds^2$ is of the form $f(\bar{r})(dx^2+dy^2+dz^2)$. I guess that the approximation formula for $d\bar{v}/dt$ in a weak field is a result of this choice. $\endgroup$ – Litho May 27 at 8:13
4
$\begingroup$

To understand the meaning of the formula, one needs to face its origins in general relativity. GR is not just about corrections to Newtonian expressions for gravitational forces and accelerations. It is more profound: The key point is that there is no longer a simple relationship between space and time coordinates and physical measurements of space and time.

All physical processes, including the length of rulers and the ticking of clocks, are affected by the gravitational field (metric), in such a way that the laws of physics have the same form in every coordinate system. (The Schwarzschild solution is not a law of physics, but the field equation that it satisfies is.) It takes some work to figure out what is still a well-defined physical observable, given that the choice of coordinates has so much more freedom than in Newtonian physics (or even special relativity).

This complication applies even in the post-Newtonian approximation of GR. The use of isotropic coordinates is a convention that affects all coordinate-based expressions but cannot affect the physics. In particular, we cannot take for granted what coordinate-based expressions for position, velocity, and acceleration mean unless we explicitly relate them to something observable (operationally defined).

The perihelion shift is observable because it is defined relative to asymptotically flat spacetime at large distances (the "fixed stars"). As another example, by integrating an equation for the propagation of light rays in any given coordinates, we could predict well-known physical measurements of light bending and time delay.

But the "speed" of a body in circular orbit has no unique or natural definition once we go beyond the Newtonian limit. Speed "should be" the circumference divided by the period. Is the circumference defined by placing a measuring tape around the orbit, or placing it radially to the sun and multiplying by $2\pi$? Is the period defined by clocks riding on the body, clocks held at rest on the orbit, or clocks at infinity? Whatever physical measurement we are interested in, GR can give us the prediction (independent of what coordinates we use), but they are all different.

"Relativistic mass" reasoning is not valid in GR. Your hypothesized formula (with the factor of 3) might conceivably result from the Schwarzschild solution in some choice of coordinates, and that is the only way it would be justified. Absent this, it would be impossible to make physical predictions because we don't know how clocks, rulers, light, etc., behave relative to the coordinates in which the formula is written.

$\endgroup$
  • $\begingroup$ The derivation of the "JPL Schwarzschild approximation formula" is outlined on pages 4-22 to 4-24 in the mentioned documentation. It is a bit beyond me. In ephemeris simulations I think you basically start with the model of the accelerations then you fit your model to the revolution time. So "r" is not measured but fitted into the model. You could basically have ten different models, all prescribing different orbital velocity but the same perihelion shift. Whatever model you use to fit the inital "r" will give the the orbital velocity that fits the measurements. $\endgroup$ – Agerhell Mar 27 at 13:19
  • $\begingroup$ If you use the "JPL Schwarzschild approximation formula" instead of the classical Newtonian expression to fit the radial distance between the Earth and the Sun you end up with the Earth being a little more then 2 kilometers closer to the sun. The difference is so small that I think it is hard to experimentally find out which model is more correct in the weak field solar system environment. Anyhow, if you have any information on the physical interpretation of the four terms in the JPL formula, please share it. It seems to me that they get the perihelion shift right by using inverse cube gravity $\endgroup$ – Agerhell Mar 27 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.