# Frames in GR

1. Apr 6, 2013

### HomogenousCow

Hello I am having problems in GR because I do not understand how observations frames work in this theory. I have a few more specific questions.

-In the schwarzschild metric, which frame is the metric written in? (The one with swarzschild coordinates)
-In the non-vaccum solutions, how do we actually prescribe the components for the stress energy tensor, more specifically from which frame do we take the observed values from?

All other tips and pieces of wisdom are welcome

Last edited: Apr 6, 2013
2. Apr 6, 2013

### bcrowell

Staff Emeritus
This is SR, not GR. The metric has the same form in any inertial frame.

At a given point in spacetime, they can be expressed in any frame you like. Their values in different frames are related by the tensor transformation law. You don't even need to have a frame of reference; you can just have some coordinate system that covers some open subset of spacetime.

3. Apr 6, 2013

4. Apr 6, 2013

### pervect

Staff Emeritus
I would suggest that you not worry about frames at all, and simply understand GR as a theory about generalized coordinates. You can worry about frame-fields later.

The metric is written in coordinates, not in a frame.

Lets assume you have some coordinates for some point A = (t,p,q,r), t being your time coordinate.

Then the metric gives you the Lorentz interval from A to B, where B is a nearby point with coordinates t+dt, p+dp, q+dq, r+dr

Because the Lorentz interval is independent of frame, you don't need to specify what "frame" you use, as long as A and B are sufficiently close.

So you forget about the "frames" entirely, and concentrate on the Lorentz Interval, as something that you can measure that's *INDEPENENT* of your choice of frame.

Again, you don't need a frame. It's easier to visualize with one less dimension, so imagine one time coordinate t and two space coordinates p and q.

Then planes of constant t will slice your box up into slabs. Adding in planes of constant p and q dice your box into small cubes.

The stress-energy tensor gives the energy and momentum density in one of these small cubes.

5. Apr 6, 2013

### bcrowell

Staff Emeritus
OK. The Schwarzschild coordinates do not form a frame of reference. In GR, frames of reference are local, not global.

6. Apr 7, 2013

### HomogenousCow

But then what does the coordinate time in the metric actually mean?

7. Apr 7, 2013

### pervect

Staff Emeritus
It's a fairly "natural" choice of coordinate for a stationary observer. But like most coordinates, it doesn't have any great physical significance. If you realize this early on, you can save yourself a lot of misunderstandings.

If you want to leverage your intuition about flat space-time as much as possible, the isotropic form of Schwarschild coordinates is useful:

http://ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/IsotropicCoordinates.htm

Isotropic coordinates share the same time coordinates as the non-isotropic Schwarschild coordinates, but have a different definition for the radial coordinate.

Trying to assign coordinates to a curved space-time is rather similar to the task of assigning coordiantes to a curved Earth. Any representation of the surface of the Earth on a flat sheet of paper will be distorted. Similarly, any representation of curved space-time via global coordinates will be distorted.

The metric coefficients when properly interpreted tell you in what manner your choice of coordinates is distorted.

If your metric coordiates are all nearly a unit diagonal, the distortions due to curvature are minimal, and you can use your intuition with good results.

If you want to use your intuition in some region where the metric is not nearly a unit diagonal, it can be handy to choose a different set of coordinates that make the metric nearly a unit diagonal. Unfortunately, this can only be done locally, you can't make the metric a unit diagional everywhere, though you can make it a unit diagional at any particular point of interest.

8. Apr 7, 2013

### Agerhell

The Schwarzschild radius becomes different using isotropic coordinates (according to wikipedia).
What is the orbital velocity of a body in circular orbit using isotropic coordinates, is it still $v=\sqrt{GM/r}$?
If you drop something from standstill using isotropic coordinates (all velocities equal to zero) what is the initial acceleration, $\frac{d^2r}{dt^2}$?

The proper time will be given by the same expression even if isotropic coordinates are used?

9. Apr 7, 2013

### Agerhell

In the Schwarzschild solution in Schwarzschild coordinates the coordinate time can be said to be the time as measured by a clock located infinitely far away from the center of the gravitational field and with no veloctiy relative to it. In the solar system we could think of a clock infinitely far away from the center of mass of the solarsystem as the "coordinate time".

In general, clocks will be effected by nearby masses and there motion, they always measure "proper time".

Coordinate time is useful as a concept if you want to compare, for instance a clock somewhere on earth with a clock in a spacecraft encirculing one of the moons of saturn. You try to calculate how fast each clock ticks in coordinate time and then you compare them with each other...