Schwarzschild solution and velocity of stationary observer

[dianaj]

Hey all,

I suddenly find myself very confused about velocity and coordinate systems. I have a feeling this is very simple, but sometimes the mind just curls up, you know? ;)

When you ask what an observer observe, you need to see things from his point of view - his reference frame. And his reference frame must be a frame moving with him. But this reference frame can look very different depending on what coordinates he choose, right? So this frame can e.g. be an inertial frame, but it can also be some weird twisted frame, right?

Consider the Schwarzschild solution and a stationary observer at a distance r from the center (beyond 2GM). He must of course be using some sort of thrust to stay still.

Now, my book says: "Work in inertial coordinates such that the observer is in the rest frame. Then the velocity of the observer is U = (1,0,0,0)." So the observer, using inertial coordinates, sees himself as having velocity U=(1,0,0,0)?

Another place in my book, the velocity of this stationary observer is described as

$$U=((1-\frac{2GM}{r})^{-1/2},0,0,0)$$

This is obviously in a coordinate system where the metric is the Schwarzschild metric itself. So this is another kind of reference frame. But is this what the observer himself measures? I think I have a tendency to consider a omnipresent observer looking down at the whole S.T., and I think I have considered

$$U=((1-\frac{2GM}{r})^{-1/2},0,0,0)$$

as the velocity that this omnipresent observer sees the stationary observer having. But this is wrong, right?

So basically, when you want to know your velocity, you choose a coordinate system, find the corresponding metric and then calculate the velocity from the relation

$$U^\mu U^\nu g_{\mu \nu} = -1.$$

And what you find is the velocity that you see yourself having. Right?

 

[Jonathan Scott]

An observer's own four-velocity is always (1,0,0,0), isn't it?

 

[atyy]

So basically, when you want to know your velocity, you choose a coordinate system, find the corresponding metric and then calculate the velocity from the relation

$$U^\mu U^\nu g_{\mu \nu} = -1.$$

And what you find is the velocity that you see yourself having. Right?

Yes.

Edit: Ooops, I meant just the LHS, not sure if the RHS is right - does it work out in Schwarzschild coordinates?

 

[Mentz114]

Have a look at this page.

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

See 'Static observers in Schwarzschild vacuum'.

 

[dianaj]

Yes.

Edit: Ooops, I meant just the LHS, not sure if the RHS is right - does it work out in Schwarzschild coordinates?

The equation holds in any coordinate system, so it works in Schwarzschild coordinates as well. This is the neat thing about tensors equations and scalars - they don't care much about coordinates. :)

 

[dianaj]

Have a look at this page.

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

See 'Static observers in Schwarzschild vacuum'.

Thanks - just the sort of thing I was looking for! :)

 

[dianaj]

An observer's own four-velocity is always (1,0,0,0), isn't it?

No, only in special relativity.

 

[Jonathan Scott]

No, only in special relativity.

OK - I guess the way I learned GR it had never occurred to me to use "own velocity" in a different way from SR.

 

[atyy]

The equation holds in any coordinate system, so it works in Schwarzschild coordinates as well. This is the neat thing about tensors equations and scalars - they don't care much about coordinates. :)

OK, good, that's what I thought from Fermi-Walker transport, but sometimes I need to do the computation in another coordinate system for a sanity check :)