# Urgent Help required- schwarzschild solution

1. May 10, 2005

### Romeo

Consider a model in which we let a body fall (radially) into a star, being the simplest example of the schwarzschild solution, in which the angular parts of the solution may be ignored, so that we consider:
$$ds^2 = -(1- GM/r) dt^2 + (1- GM/r)^{-1}dr^2.$$

I have been told that this may be adapted to a Lagrangian of form:

$$L(r, \dot{r}) = -(1- GM/r) (\frac{dt}{d\lambda})^2 + (1- GM/r)^{-1}(\frac{dr}{d\lambda})^2,$$

and then solved to find a general time for a body to 'fall' from rest, from a general distance R. How, I don't know (...I love supervisors).

I now have 13 books out upon general relativity, and am royally stuck. It is important that a solution avoids tensor calculus- i am aware that this is possible.

Any help with this would be incredibly appreciated.

Regards

Romeo.

2. May 10, 2005

### pervect

Staff Emeritus
I think most of what you need should be in the Lagrangian - you shouldn't need any GR since you have the Lagrangian. But I think perhaps your intepretation of what variables the Lagrangain is a funciton of is a bit off, and after looking at the problem even more I think that maybe the Lagrangian itself is a bit off. So you should ask where this Lagrangian came from.

You have a system where $$\lambda$$ is the "time" variable, and you have two functions

$$r(\lambda)$$, $$t(\lambda)$$

Letting $$\dot{r} = \frac{dr}{d\lambda}$$ and $$\dot{t}=\frac{dt}{d\lambda}$$

we can write

$$L(r,\dot{r},t,\dot{t}) = -(1-GM/r)\dot{t}^2+\frac{\dot{r}^2}{1-GM/r}$$

which gives by Lagrange's equations a couple of equations, one of which is

$$\frac{d}{dt} (2\frac{\dot{r}}{1-GM/r}) =0$$

This implies that

$$\frac{\dot{r}}{1-GM/r} = K$$ where K is some constant.

Unfortunately, if we compare this to MTW pg 656, eq 25.16a, or some other source, we cannot veryify this expression, which therfore seems suspect. It's possible that the expression for the Lagrangian is correct, but not if $$\lambda$$ is supposed to be "proper time". But my guess is that the Lagrangain isn't complete. It really doesn't look like it has the right form.

From MTW, for an infalling body, with $$\lambda$$ being the proper time, we should have

$$(\frac{dr}{d\lambda})^2 = E^2 - (1-2GM/r)$$

[I've taken various liberties in converting MTW's equations into a comparable form]

for the correct expression, here E is a constant which depends on the energy of the infalling particle, and E=1 when the particle falls from infinity.

In the "fall from infinity" case, $$(\frac{dr}{d\lambda})^2 = 2GM/r$$

Last edited: May 10, 2005
3. May 10, 2005

### Zanket

Romeo, the book Exploring Black Holes covers all the major scenarios for a Schwarzschild black hole, including the one you mention (on pg. 3-31). It doesn't always give the full derivation, but gives lots of hints. (There's an answer book for it, that's mentioned in the acknowledgements but I haven't seen it.) No tensor calculus in the book. You might want to add this one to your library.

4. May 11, 2005

### Romeo

Pervect- very kind, thank you. It's annoying that MTW is the only text i couldn't get my hands on (on-loan), so have a plethora of others (although Weinberg and Schultz have been useful).

Zanket- Who authors that text? (I'm going to feel very silly if that's the MTW title... )

Best

Romeo

5. May 11, 2005

### pervect

Staff Emeritus
Anything that talkes about the "effective potential" should enable you to find dr/dtau as a function of r.

http://www.fourmilab.ch/gravitation/orbits/

has the effective potential equations written down, and a neat Java applet. In your case, L is zero (no angular momentum). It also has the auxillary equations that you need to find t and theta (but since L=0, theta is a constant in your case).

6. May 11, 2005

### Zanket

The W from MTW and Edwin F. Taylor. For more info look here.

7. May 26, 2005

### James

Zanket's choice is excellent. Taylor & Wheeler's book is very insightful. Easy to follow [as evidenced by my completing all the exercises].