# Swartzchild metric and free fall

1. Jul 15, 2007

### daniel_i_l

I just finished reading Black Holes by E. Taylor and J. Wheeler. Throughout the book they use the SC metric for the the metric near a massive object(for radial motion only:
$$d \tau^2 = \left( 1 - \frac{ 2M }{r} \right) dt^2 - \frac{dr^2}{\left( 1 - \frac{ 2M }{r} \right)}$$
Where dt is measured very far away from the massive body, r is measured as the circumference of a sphere whose center is in the middle and whose outer shell reaches the point divided by two times pi. M,t and r are mearured in meters.
Then, in order to calculate to path of a particle near a massive object in between two events they use the metric to find the path that give the maximal proper time (the Principal Of Maximal Aging).

Now my question is, to use POMA you need to have initial and final events and then calculate the path between them via the POMA. But I want to prove that the SC metric also predicts that an object at rest next to a massive object will start falling in. To do that shouldn't I start at t0 and r0 and then calculate which r1 will give the maximal proper time for a giver t1? Shouldn't this r1 be smaller than r0 - meaning that in order to maximize proper time the object should start falling towards the massive object? I tried to do this but I got that for any r1=/=r0 the proper time is smaller for a giver t1 then if r1=r0. Is that right? If so, how does the SC metric predict that an object that starts at rest next to a massive body will move towards it?
Thanks.

Last edited: Jul 15, 2007
2. Jul 15, 2007

### StatusX

That principle applies to the shortest path between two points, so you have to specify both the starting and ending locations completely, one coordinate of the final position isn't enough. More specifically, by leaving out this coordinate, you're not imposing the condition that the particle is initially at rest.

Last edited: Jul 15, 2007
3. Jul 15, 2007

### daniel_i_l

Then how can I use the metric to show that a particle released from rest near a massive object starts falling?
Thanks.

4. Jul 15, 2007

### StatusX

You can derive the geodesic equation, which determines the local conditions on a path so that it maximizes proper time globally. All the solutions can be verified to satisfy the maximal time conditions between their endpoints, but being local, they allow you to calculate the path of a particle knowing only its initial conditions, ie, position and velocity.

More visually, if you were to draw the paths maximizing time between (t0,r0) and (t1,r) for all r, these would all have different slopes through (t0,r0), and the condition that the particle is initial at rest picks out one of these, and so one final r.

Last edited: Jul 15, 2007
5. Jul 15, 2007

### daniel_i_l

Can you explain what you mean by:
"determines the local conditions on a path so that it maximizes proper time globally"?
How would I go about deriving this equation? Can you give me some starting points?
Thanks.

6. Jul 15, 2007

### StatusX

Did you see my edit? I think that's a more satisfying explanation. With that picture in mind (ie, a drawing of all possible geodesics passing though each point), note that if a given path maximizes proper time between its endpoints, it also maximizes proper time between any two points lying on it, since otherwise we could replace this section with a longer path and make the overall proper time bigger.

So being a geodesic is not just a global property, it requires that a certain condition must be satisfied between any two points, no matter how close they are. In particular, as you move the endpoint closer and closer to the starting point, you determine how any geodesic passing through a point must evolve given only its initial direction. It turns out that the condition specifices its second derivative with respect to proper time.

The derivation uses calculus of variations. It's very similar to the derivation of Lagrange's equations, and if you've seen these, it's not that hard to derive it yourself. I can get you started if you want, but it should be in any GR textbook.

Last edited: Jul 15, 2007
7. Jul 15, 2007

### daniel_i_l

I've never seen the derivation of the Lagrange's equations (the only quantitive use of the calculus of variations I've seen was in Feynman's Lectures on physics where he discussed the principal of least action - which is awfully similar to the POMA now that I think about it). But can you get me started please?
(Is it basically like this: http://www.eftaylor.com/pub/newton_mechanics.html ?)
And is this the general idea: Get an expression for the path from (t0,r0) to
(t1,r) as a variable of r and find the one whose slope at r0 is zero?
Thanks!

Last edited: Jul 15, 2007
8. Jul 15, 2007

### StatusX

That site presents a derivation of Lagrange's laws that's a little different than the usual one. If you take the time to go through it, it might be a little more intuitive than the standard derivation, but it takes a while to get there, and I think it would be more trouble than it's worth for the geodesic equation.

I've started trying to do the derivation a few times here, but I can't see a way to do without being confusing in less than 2 or 3 pages. I'd really suggest finding a book on GR, there are plenty of good ones. I was planning on writing up something on calculus of variations for the tutorial section here pretty soon, and if I do I'll be sure to include a derivation of the geodesic equation.

It's not a bad picture to have, but when deriving the equations this isn't what's being done directly. Although the way they do it on that site seems to be doing quantitatively what I described above, of imposing the condition that the path must be a maximum over every subpath, and using this condition on infinitesimal sections to derive differential equations governing the path. (ie, doing what you suggested for t1=t0+dt, r=r0+dr)

Last edited: Jul 15, 2007
9. Jul 15, 2007

### pervect

Staff Emeritus
If you just want to work out orbits (which will be geodesics) in the Schwarzschild space-time, the easy approach is to take advantage of the existence of a conserved energy and angular momentum.

Any orbit will occupy a plane, so you can eliminate one coordinate right off. Usually one sets $\theta=0$ for the equatorial plane.

Then you have two coordinates, r and $\phi$. So you need two equations.

The geodesic equations will be equivalent to

d/dt (E) = 0
d/dt (L) = 0

For massive particles one usually uses $\tilde{E}$, which is the energy per unit mass of the orbiting particle, rather than E, and also $\tilde{L}$

I would expect that "Exploring black holes" would cover some of the same material, you can also look at http://www.fourmilab.ch/gravitation/orbits/

Last edited: Jul 16, 2007
10. Jul 16, 2007

### daniel_i_l

Thanks everyone! I'll read those links and see if I can prove it. I'll post my conclusions if anything interesting comes out:)