# Schwarzschild coordinate r

1. Mar 27, 2014

### Vrbic

Schwarzschild coordinate "r"

Hello, Im a newguy here, so if my question dont belong to this section, please let me know.
My question:
In spherical symmetric spacetime discrabed by Schwarzschild coordinate ds2=-a(r)dt2+b(r)dr2+r2(dΘ2+Sin2(Θ)d$\varphi$2), "r" is defined as r=$\sqrt{A/(4\pi)}$ where "A" is an area of sphere dΘ2+Sin2(Θ)d$\varphi$2. What is relation between "r" and real distance from the center of coordinate?

Thank you all.

2. Mar 27, 2014

### PAllen

In both SR and GR, there is no such thing as 'real distance' (in SR, it is frame dependent, in GR there is commonly no unique definition of it; there are distance conventions that can be adopted, and this r coordinate is an example). In this case, there may no center, even mathematically: you could have and event horizon, with a singularity inside, in which case there is timelike world line that in any way represents the history of a center.

3. Mar 27, 2014

### Vrbic

So if I want to know exact radial distance between to evens only I can do is ∫grrdr with limits of my r1,r2. Is it right?

4. Mar 27, 2014

### Bill_K

Should be ∫√grrdr, but yes, that's basically the right idea.

5. Mar 28, 2014

### pervect

Staff Emeritus
Some remarks in addition to the missing square root:

It might be helpful to recall from special relativity that "real distance" depends on the frame of reference, it's not an observer independent property in special relativity. Being observer dependent, the adjective "real" may be confusing when applied to distance, since different observers measure different distances.

GR doesn't have exactly the same notion of "frame of reference" as SR. But in this case you can consider that what you are computing with your formula is the distance as measured by a static observer. Static observers are observers that have constant Schwarzschild coordinates, they are basically "at rest" in the Schwarzschild coordinate system.

Last edited: Mar 28, 2014
6. Mar 28, 2014

### bcrowell

Staff Emeritus
Note that in the case where there's an event horizon, $\sqrt{g_{rr}}$ becomes imaginary inside the horizon. Physically, there are no static observers inside the horizon.