# Schwarzschild Geometry

1. Aug 11, 2009

### 23571113

To make a long story short I'm suppose to be learning how to "obtain non-zero curvature components of Schwarzschild geometry". However, I'm not sure what all that entails (tensors? differential geometry?). So any advice on what level of math/physics will be needed would be great!

2. Aug 11, 2009

### atyy

There is a nice introduction to the curvature tensor in 2D in Chapter 7 "Riemann's counting argument" of Richard Koch's differential geometry notes (you can generalise it to 4D yourself): http://www.uoregon.edu/~koch/

3. Aug 11, 2009

### 23571113

So obtaining those non-zero curvature components does in fact require tensors and differential geometry?

4. Aug 11, 2009

### atyy

No, just elementary differentiation and formulas (components of the Schwarzschild metric and the components of the Riemann tensor in terms of the components of an arbitrary metric). But it will probably seem meaningless without basic differential geometry.

5. Aug 11, 2009

### 23571113

Okay - here is my situation:

I'm in 11th grade and I'm only now starting Single Variable Calculus, so I don't have the mathematical background to handle differential geometry. I'm trying to learn how to obtain the curvature components of Schwarzschild geometry in order to work on a research project with a professor. So I don't need rigorous maths just enough to get me by. Is there a link or book which I could use to learn about obtaining the curvature components?

6. Aug 11, 2009

### atyy

7. Aug 11, 2009

### atyy

Quick rough explanation: mass-energy curves spacetime. In curved spacetime you cannot setup spacetime coordinate axes that are at "right angles" to each other everywhere. But of course, maybe spacetime is not curved and you just happened to use crooked coordinates. The Riemann curvature tensor tells you whether spacetime is really curved, or whether youu just happened to use strange coordinates.

8. Aug 11, 2009

### atyy

So the Riemann tensor is a geometric invariant.

An analogy of an invariant in 2D flat space Euclidean geometry: depending on how you rotate your coordinates a vector could be (x,y) or (xb,yb) - but the length x^2+y^2=xb^2+yb^2 is a geometric invariant.

A caution - in flat space there are position vectors and velocity vectors. In curved space there are only velocity vectors.

9. Aug 11, 2009

### 23571113

The link seems to be very informative: however right now the matrices, vectors, and tensors are completely greek to me, including equation 7.15. I have some limited exposure to the Schwarzschild metric, but nothing like this =D

So is the process by which you obtain the curvature components lengthy? Or is it a plug and chug sort of thing?

Sorry, if I'm missing something you already explained...

10. Aug 11, 2009

### atyy

If you have the Schwarzchild metric components, you basically just differentiate them twice to get the Riemann curvature components - it's lengthy but plug and chug.

But c'mon - are you telling me you don't know what a velocity vector is?

11. Aug 11, 2009

### 23571113

I know what a velocity vector is (like plane old 2-D; the ones you learn about in pre-cal)... but I haven't taken vector calculus yet, and I have no idea about manifolds.

12. Aug 11, 2009

### Phrak

The bare bones is supplied in Equations 7.13-7.15 in the link atty supplied. What is missing is how to obtain the Christoffel connection, gamma from g and how to obtain the Riemann tensor, R from gamma, and lastely how to contract the Riemann tensor to the Ricci tensor.

So if you find the missing equations (4.86 and 3.67 in the same paper), that would be halfway there, but you would need to apply index notation to get from one to the other.

...or there's eq. 3.77 which defines the Riemann tensor in terms of the metric for you. Now you still have the mysterious elements that look like subscipted partial derivative operators--and in fact, they are.

$$\partial_{\mu} = \frac{\partial}{\partial x^{\mu}}$$

However, once you have the Reimann tensor, or the Einsten tensor derived from it, it might be nice to look at, but I don't think it would give you much sense of what curvature is like in the Schwazchild metric.

Last edited: Aug 11, 2009
13. Aug 11, 2009

### atyy

Ok, then in principle you know everything. I don't think there's anything markedly different about vector calculus. A manifold is just something you can put smooth coordinates on. A matrix transforms one vector into another one. Take your time and read Martin's book. Or maybe Phrak can help you out more.

14. Aug 11, 2009

### Phrak

Ha! Or maybe you can, and I should escape this thread. I'm up to your tricks

15. Aug 27, 2009

### 23571113

I realize that it it will probably seem meaningless to obtain the components of Schwarzschild geometry without some differential geometry. However, as mentioned above I'm trying to learn this for a research project - I'm not endeavoring to self-study the material in order to gain complete knowledge or understanding of the subject. So my professor has told me that the next "step" for me (before I can start the actual research) is getting the ability to obtain the curvature components - whether or not I understand them is another story. I'm currently taking single variable calculus at my high school, and I would like to learn the rest of the necessary (to obtain the curvature components) math during next summer. Would vector calculus be enough? Or do I need to take differential equations and or linear algebra? Also do I need to learn those maths completely or could I pick out a smaller portion?

I know this isn't the ideal way to learn physics - but right now I'm just trying to become competent enough to do a research project. I definitely plan on learning these subjects correctly at university.

Thanks!