# Planetary Orbits calculated from the Metric

1. Jul 10, 2011

### Philosophaie

I am learning about General Relativity. The planetary orbits can be calculated with more precision especially Mercury. I am stuck on how to get from the Schwarzschild Metric:

a four variable Differential Equation

to a radius(r,theta,phi,t) and velocity(r,theta,phi,t) of a single planet in an ecliptical orbit around a single sun in a somewhat inertial space.

Any hints or suggestions would be helpful!

2. Jul 10, 2011

### bcrowell

Staff Emeritus
If all you want to do is get the precession of Mercury's perihelion, it's fairly easy: http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken]

The general problem of orbital motion in the Schwarzschild metric is actually kind of fun, but complicated for the same reasons that it's fun. There is a nice treatment in Exploring Black Holes by Taylor and Wheeler, using only simple calculus.

Last edited by a moderator: May 5, 2017
3. Jul 10, 2011

### pervect

Staff Emeritus
I'm not quite sure what your background in relativity is. Usually one looks for the differential equations relating four functions of proper time

r(tau), theta(tau), phi(tau), and t(tau)

This is not precisely what you wrote above, however. What you did write doesn't seem very promising.

The equation relating the four functions above is known as the geodesic equation. Directly solving for the geodesic equation is possible, but not recommended. Considerable simplification can be achieved by noting that some of the equations are of the form (d/dtau) (some conserved quantity) = 0, which is equivalent to saying that the conserved quantity is a constant of motion.

By velocity, I assume you mean the velocity relative to a static observer. You can calculate this, by means of frame fields, but it's not needed to write down the differential equations of motion, though it helps in interpreting them.

The radius is pretty much not needed, and is problematical to define in any event.

If you are just interested in the problem of Mercury's orbit, Goldstein has a section on it in classical mechanics

4. Jul 11, 2011

### Philosophaie

I do know College Calculus and Differential Equations.

I am shooting for the Geodesy of a Satellite orbiting a large planet (or even black hole) with an observer on the Planet (or in the black hole?). With the:

velocity(r(tau),theta(tau),phi(tau),t(tau))

I want to know the stability of the orbit and what corrections need the be made to keep a stable orbit. Would Newonian Mechanics suffice or would Relativistic Mechanics more accurate results?

5. Jul 11, 2011

### pervect

Staff Emeritus
I would guess you'd be best off doing the Newtonian analysis, around a planet (rather than a black hole) first. I still don't have a clear idea of what you're trying to calculate and why. IT might be clearer if you did a detailed Newtonian analysis.

6. Jul 11, 2011

### bcrowell

Staff Emeritus
Given your math background and what you want to figure out, Exploring Black Holes by Taylor and Wheeler is exactly the book you need to read.

7. Jul 11, 2011

### pervect

Staff Emeritus
8. Jul 12, 2011

### Samshorn

The referenced link presents the usual formula relating (dr/ds)^2 to r and the constants of motion, and then it groups together some of the terms in that formula and calls that quantity U^2, and then it says "A planet in a nearly circular orbit oscillates between perihelion and aphelion with a period that depends on the curvature of U^2 at its minimum." Apparently the word "curvature" here means [second derivative with respect to r], because the link defines a parameter k as this second derivative, and then it says the period is 2pi * sqrt(2/k).

Why does the period from apihelion to perihelion equal this quantity?

Is there a more detailed explanation of the reasoning available?

Last edited by a moderator: May 5, 2017
9. Jul 12, 2011

### Philosophaie

Would not the Potential Energy in the curvature be at a maximum at the Apihelion or at least shortly there after and minimum shortly after the Perihelion. The speed and curvature both decrease after the closest point and the uphill trek to the point farthest form the sun.

10. Jul 12, 2011

### pervect

Staff Emeritus
That would be the Newtonian approach, but GR does things differently. You're not particularly likely to guess how GR handles this on your own from a knowledge of only Newtonian theory, you'll need to actually track down a book (like the one recommended) and read it...

Most of the material on the WWW that I can think of is either graduate level, or glosses over too much, unfortunately. Which means a visit to the library, or the bookstore.

11. Jul 12, 2011

### bcrowell

Staff Emeritus
Conservation of energy in simple harmonic motion, with m=1, reads like this:
$$\frac{1}{2}v^2+U=E$$
and the resulting period is $2\pi\sqrt{1/k}$. The corresponding equation in the present case is
$$\dot{r}^2+U^2=E^2,$$
which is an equation of the same form with the substitutions $U\rightarrow U^2/2$ and $E\rightarrow E^2/2$. The factor of 2 in the substitution involving the U's effectively cuts k in half, which is what leads to the 2 inside the square root in the result, $2\pi\sqrt{2/k}$.

12. Jul 13, 2011

### Samshorn

That equation represents simple harmonic motion only if the derivative of U is a linear function of r, which it isn't for the orbital equation.

Okay, I see what you're doing. You're basically saying that, over the small range of r values for a nearly circular orbit we can approximate the function U(r)^2 by an expression of the form (1/2)kr^2 + Br + C for some suitable constants k, B, and C. So the orbital equation is (r')^2 + (1/2)kr^2 + Br + C = E^2. Differentiating twice with respect to r and multiplying through by r'/2 gives r"' + (k/2)r' = 0, which is a simple harmonic equation with frequency sqrt(k/2), and the estimated value of k is the second derivative of U(r)^2.

One could question whether this method is accurate for Mercury, which has a fairly large eccentricity, so its radial distance from the Sun varies significantly. Fortunately, we know from more rigorous derivations that this doesn't matter if m/r is sufficiently small.

Last edited: Jul 13, 2011
13. Jul 13, 2011

### yuiop

This might the sort of thing you are looking for http://www.fourmilab.ch/gravitation/orbits/

I will try and simplify the equations given on that website to hopefully make things a bit clearer. That website uses units such that G=c=1

The first equation is for effective potential (V):

$$V = \sqrt{\left(1-\frac{2M}{r}\right) \left(1+\frac{L^2}{r^2}\right) }$$

where L is the angular momentum per unit mass of the test particle and M is the mass of the central massive gravitational body. V is a function of L and r. M and L are constants for a given test particle in free fall following a geodesic. The instantaneous radius of the particle from the central body is r.

The radius of the particle from the centre evolves in proper time according to:

$$\frac{dr}{d\tau} = \sqrt{E^2 - V^2}$$

where E is the potential energy of the particle at infinity which is also a constant for a given particle following a geodesic. It can be seen from the above equation that if dr/dtau is zero at infinity then V is equal to E at infinity.

Angular motion about the centre of attraction in terms of proper time is then:

$$\frac{d\theta}{d\tau} = \frac{L}{r^2}$$

Again it can be seen that the value of V at infinity is independent of L because $d\theta/d\tau$ always goes to zero at infinity.

Now if you want the equations in terms of coordinate time rather than proper time the website gives the conversion factor as:

$$\frac{dt}{d\tau} = \frac{E}{(1-2M/r)}$$

Now the above indicates that the ratio of coordinate time to proper time is only a function of radius and independent of the velocity of the particle, but I suspect this is just an approximation that uses the assumption that the particle is moving at sub relativistic speeds such that velocity time dilation effects are negligible. Maybe one of the more knowledgeable people here could confirm that?

You indicate an interest in stable and unstable orbits. These are found from differentiating the effective potential and then solving for dV/dr = 0 to find the maxima and minima to obtain the radii of the possible circular orbits:

$$r_{circular} = \frac{L \left(L \pm \sqrt{L^2 - 12M^2}\right)}{2M}$$

where the smaller value is the unstable circular orbit on a knife edge effective potential peak and the larger value is a stable orbit in an effective potential valley. If the angular momentum L^2 is less than 12M^2 there is no stable orbit and the particle will reach the surface of the gravitational body.

Hope that helps some.