Schwarzschild conservation of angular momentum

1. Apr 3, 2013

Agerhell

What is the definition of angular momentum that is to be conserved in the Schwarzschild solution of general relativity, expressed in coordinate time?

I am trying to put together an expression for gravitational acceleration that is to emulate the Schwarzschild solution. I need to test whether the angular momentum is conserved in the same manner that the Schwarzschild solution predicts.

After some googling I see that the quantity $$r^2d{\varphi}/d{τ}$$
is supposed to be conserved according to the Schwarzschild solution, correct?

What is the correct expression for angular momentum that is to be conserved for a test-body in the Schwarzschild solution, expressed in coordinate time and not proper time?

2. Apr 3, 2013

WannabeNewton

Yes that is the conserved angular momentum. It is hard for me to gauge what exactly it is you want because there isn't just a single way to do these things and one way can have greater utility than another even if they have the same end goal. Having said that, the most obvious way to do it would be to just use the chain rule $$L = r^{2}\frac{\mathrm{d} \phi}{\mathrm{d} \tau} = r^{2}\frac{\mathrm{d} \phi}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} \tau} = r^{2}(1 - \frac{2GM}{r})^{-1}E\frac{\mathrm{d} \phi}{\mathrm{d} t}$$

3. Apr 3, 2013

Agerhell

Thank you. "E" in this context is just a constant?

4. Apr 3, 2013

WannabeNewton

It's the conserved energy of the orbit, so technically yes it is a constant of motion: $E = (1 - \frac{2GM}{r})\frac{\mathrm{d} t}{\mathrm{d} \tau}$.

5. Apr 3, 2013

Agerhell

Thank you. Now since I do not know if E is always conserved in my numerical simulation, the simulation might have errors. I would like to use $$dt/d{\tau}$$ directly. I assume the correct expression for this quantity is:

$$dt/d{\tau} = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}$$

Is this correct?

So when I check whether the angular momentum is conserved in the sense expected by the Schwarzschild solution I want:

$$L=r^2\frac{d\phi}{dt}\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}= constant$$

And to check if the energy, as expected by the Schwarzschild solution, is a constant of motion in my simulation I have to keep track of:
$$E= \frac{1-\frac{2GM}{rc^2}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}= constant$$

6. Apr 3, 2013

WannabeNewton

How did you get this?

7. Apr 3, 2013

Agerhell

Someone told me before that the correct expression for energy according to the Schwarzschild solution is as I wrote above. By taking that expression for energy and divide it by the factor $$(1-\frac{2GM}{rc^2})$$

you get the expression for $$\frac{dt}{d\tau}$$ with the help of the formula you provided.

Hopefully the expression for the energy is the correct one. At least it reduces to the classical potential and kinetic energy in the weak field slow velocity limit.

8. Apr 3, 2013

WannabeNewton

I'm assuming you mean the energy expression you wrote in the very last line of your preceding post. Can you point to a place where it is shown to be that (website or textbook etc.)? I can't find such an expression in any of my texts.

9. Apr 3, 2013

Bill_K

Lookit, Agerhell, you're getting sidetracked. You have two equations

L = r2(dφ/dτ)
E = (1 - 2GM/c2r)(dt/dτ)

why don't you just divide one by the other and check that

L/E = r2/(1 - 2GM/c2r)(dφ/dτ)

is constant.

10. Apr 3, 2013

Bill_K

No. This is only for a particle in circular motion, where you've defined v to be r(dφ/dt).

11. Apr 3, 2013

Agerhell

Yes I mean that expression. I asked someone through email and he said that the quantity:

E=mc^2*sqrt(1-2GM/(c^2 r))/sqrt(1-v^2/c^2)

could be shown to be conserved in the Schwarzschild case. However, he stated that v in the formula above was intended to mean:

v=sqrt(-g_ij dx^i dx^j/dt^2/(1-2GM/(c^2r))

In the fomula I wrot above I just used "normal v" and divided it by the factor sqrt(1-2GM/(c^2r)). I figured this makes sense as the velocity of light in a gravitational field varies as sqrt(1-2GM/(c^2r)) if taken in coordinate time. This way you get the formula i wrote above. If you divide both the divisor and the dividend by sqrt(1-2GM/(c^2r)) you get a perhaps more common-looking expression:
$$E=\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}}}$$

Is this familiar?

12. Apr 3, 2013

Agerhell

So if you have a gps-satellite in a perfectly circular orbit, you will have:

$$\frac{dt}{d\tau}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}-\frac{2GM}{rc^2}}}$$

but if you have a gps-satellite in the very same place moving with the same speed, compared to the frame of a non-spinning earth, but in a different direction (with an r-component that is not zero) you will get $$\frac{dt}{d\tau}=something else$$

Is that what you are saying?

Last edited: Apr 3, 2013
13. Apr 3, 2013

Bill_K

The length of the 4-velocity vector is a constant. This gives

1 = - (1 - 2GM/c2r)-1/c2(dr/dτ)2 - r2/c2(dφ/dτ)2 + (1 - 2GM/c2r)(dt/dτ)2

Solve this for dt/dτ and define v appropriately (!), and you'll get your expression.

14. Apr 3, 2013

pervect

Staff Emeritus
Take a look at http://www.fourmilab.ch/gravitation/orbits/

Using coordinate time is going to make the expressions very messy, but as others have said, all you need to do is start with the basic relationships in terms of proper time, and do some algebra.

The basic expressions you need are:

Just divide the two expressiosn and you'll get $d\phi / dt$

If you want a complete set of differential equations. you'll need to add in the following two equations, which I'll call Set B.

where

You can derive these equations from the previous 2 and the length of a 4-vector, as Bill K suggests. Or you can just use them and not worry about where they came from.

The point is that you can solve for $dr / d\tau$ knowing E, L, and r by using the equations in Set B, and then you can find dr/dt by dividing $$dr / d\tau[/itex] by $dt / d\tau$ You'll then have the differential equations for dr/dt and $d\phi / dt$. You can probably manipulate these to find L and E given (dr/dt) and $d\phi / dt$ at some point, if that's what your'e trying to do. (That seems to be what you asked initially, but I don't quite get the feeling that that's what you're actually looking for). If you're trying to do anything with velocities, you need to realize that dr/dt isn't a velocity, because r doesn't measure distance, you need to correct for the metric. Also t won't necessairly measure the value of time you need to measure a velocity. I"m not sure if you are actually using velocities in the sense of something you'd measure (with for example a doppler radar). If not, you don't need to worry about this. Last edited: Apr 3, 2013 15. Apr 4, 2013 Agerhell So, you are saying: [tex]\frac{dt}{d\tau}=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}-\frac{|\bar{v}\times\hat{r}|^2}{c^2}-\frac{1}{1-\frac{2GM}{rc^2}}\frac{(\bar{v}\cdot\hat{r})^2}{c^2}}}.$$

So if you travel horizontally on or just above a spherically symmetric non-spinning planet with a certain speed as measured with coordinate time your clock will tick slightly faster then if you travel vertically with the same speed? The expressions:

$$E=(1-\frac{2GM}{rc^2})\frac{dt}{d\tau}$$

and:

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

are correct if you define $dt/d\tau$ as I did now?