# B Schwarzschild Metric

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1. Jul 17, 2016

### AleksanderPhy

Hello I have been reading about Schwarzschild metric and scources what I read said that Schwarzschild metric is used to describe a non-rotating black holes. And what I can not understand is what can you calculate with it? It would be good if you give some examples where you can use it.

2. Jul 18, 2016

### Simon Bridge

The metric contains everything you need to know about the geometry of spacetime in the region for which it is valid. Thus, you use it for every calculation involving space and time ... say if you want to find the trajectory of an object flying by a black hole...
Historically it has been used to predict properties of black holes that aid astronomers and astrophycisists in finding likely candidates in Nature.

3. Jul 18, 2016

### AleksanderPhy

but can I use Scharzschild metric to calculate mass of a black hole

4. Jul 18, 2016

### haushofer

The mass is the only parameter you have to use for this metric, just like the gravitational field of a point source in Newtonian gravity. A difference with Newtonian gravity is that a rotating or charged source will change the gravitational field in GR.

You can use the Schwarzschild metric for any spherical symmetric source. If you approach the earth as being spherically symmetric and non-rotating, you can also use it for outside the earth (outside, because it is a vacuum solution!)

5. Jul 18, 2016

### Simon Bridge

You can use it to work out the mass of a black hole from other observations much as you may work out the mass of gravitating bodies in Newtonian mechanics.
The observations need to involve space and time.

6. Jul 18, 2016

### Staff: Mentor

The Schwarzschild metric is not just for black holes; it describes the gravitational effects around any non-rotating (or slowly rotating) spherically symmetric mass. It has been used to
- Calculate the orbits of the planets with greater accuracy (google for "Mercury precession relativity")
- Work out the formulas for gravitational time dilation.
- Just about anything else involving gravity around a planet (including Earth) or a star (including the sun).

7. Jul 18, 2016

### Markus Hanke

What a metric does, in physical terms and in the context of General Relativity, is tell you how neighbouring events are related to one another. Since events are points in space at given instances in time, this means that a metric enables you to calculate ( among other things ) everything there is to know about the motion of test particles. For example, if you let a clock freely fall towards a central ( uncharged, non-rotating ) body in an otherwise empty region, then the Schwarzschild metric allows you to calculate the trajectory the falling clock will follow, what the clock reads as seen from the perspective of any arbitrary observer at any moment of ( his own ) time, what distance the clock moves in a given period of an observer's time etc etc etc. In essence, the metric allows you to quantify all relevant dynamics of test particles in a Schwarzschild spacetime. Furthermore, because the connection is known, you can also use it to compute certain more abstract geometric properties of the spacetime itself, such as curvature, horizon surfaces within it, symmetries and constants of motion, and so on.
In a way ( and this is meant only as an analogy, not an exact correspondence ), the metric in GR plays a role somewhat similar to the gravitational potential in Newtonian gravity, in terms of being the fundamental entity from which you can compute all your relevant quantities.

8. Jul 18, 2016

### pixel

If the mass is spherically symmetric, then why the restriction to non-rotating or slowly rotating? From symmetry, nothing would change to an outside observer if the mass rotated.

9. Jul 18, 2016

### Yukterez

If that were the case, why use the complicated Kerr metric instead of the easy Schwazschild metric? Of course there would be a change in symmetry from spherically symmetric to axially symmetric.

10. Jul 18, 2016

### Staff: Mentor

I've asked myself the same question and took a brief look on Wiki. As far as I understood it, it goes:
The Schwarzschild metric is a solution to the Einstein field equations which couple spacetime to the energy-stress tensor.
So if you change charge or rotation, you change this tensor and it seems evident, that this also changes the Schwarzschild solution.
Thus the short answer is: because you changed the system's energy which is coupled to spacetime.

But this is only my layman's view. I might be wrong.

11. Jul 18, 2016

### Yukterez

Compare the Schwarzschild case with the Kerr case; in the upper part you have a non rotating Schwarzschildschild black hole, and in the lower part a rotating Kerr black hole. The mass of the black hole and the initial conditions for the test particle are the same, the difference is only in the spin-parameter, which leads to a completely different trajectory:

http://org.yukterez.net/schwarz.vs.kerr.gif

As you can see the Schwarzschild scenario can easily be displayed in 2D because, to quote Wikipedia:

Since the Schwarzschild metric is symmetrical about θ = π/2, any geodesic that begins moving in that plane will remain in that plane indefinitely (the plane is totally geodesic).

while for the same scenario with Kerr you need 3D (except if you limit your motion to the equatorial plane).

Last edited: Jul 18, 2016
12. Jul 18, 2016

### A.T.

It's not just the energy, which is a scalar and would be the same regardless which way it rotates. There are directional effects too:

https://en.wikipedia.org/wiki/Frame-dragging

13. Jul 18, 2016

### AleksanderPhy

Is it too much to ask but can somebody give me link about calculating the trajectory of object falling into a black hole or give me an example how to calculate a trajectory of a object that falls into a black hole?

14. Jul 18, 2016

### Staff: Mentor

You'll find a bunch of stuff if you google for "Scwarzschild geodesic equation"; you're looking for solutions to the geodesic equation for the Schwarzschild metric. The wikipedia page at https://en.wikipedia.org/wiki/Schwarzschild_geodesics will give you a feeling for how the calculations work.

Unfortunately, nothing you find is likely to be an easy read - there's a reason why you don't usually encounter general relativity until after you've completed a serious four-year undergraduate physics program. If this stuff were easy it wouldn't have taken two and a half centuries and Einstein to figure out this next step after Newtonian gravity.

15. Jul 18, 2016

### Yukterez

If you know how to calculate the trajectory of an object in Newtonian mechanics you should have no problems in doing so. The only difference between the Newton and the Schwarzschild equation of motion is the red term:

$\color{black}{\ddot{r}(t) = -\frac{G\cdot M}{r(t)^2} + r(t)\cdot \dot{\theta}(t)^2} \color{red}{ - \frac{3\cdot G\cdot M\cdot \dot{\theta}(t)^2}{c^2}} \color{black}{ \text{ , } \ddot{\theta}(t) = \frac{-2\cdot \dot{r}(t)\cdot \dot{\theta}(t)}{r(t)}}$

and the line element which gives you the time dilation of the test particle with respect to an observer at infinity. Examples can be found here.