Can I express stationary spacetime and static spacetime in this way?

[yicong2011]

Can I express "stationary spacetime" and "static spacetime" in this way?

Can I express "stationary spacetime" and "static spacetime" in this way?

stationary spacetime: if we arrange a set of observers locating in any places, then each of them should observe nothing change at the place where s/he is located.

static spacetime: if our observers can contact with each other (let us neglect the technical problems of the "contact" amongst them), they may find out that what they observed is the same and same over time.

 

[yicong2011]

Actually, I am a bit unclear about the physical meaning of the "timelike killing vector field orthogonal to the hypersurface"...

The first note in this thread is my speculative "understanding" of the two concept...

 

[WannabeNewton]

Stationary means, roughly, that given any hypersurface of time the surface itself will always look the same or in other words the space - time admits a time - like killing vector field. A time - like killing vector field means that any infinitesimal diffeomorphisms generated by a time - like vector field (so any "flows" in time on the space - time such as time reversals) will keep the metric and therefore the space - time invariant. Also roughly, the condition that the time - like killing vector field be orthogonal to a family of hypersurfaces translates to there be no cross terms between space - like coordinates and the time - like coordinate in the metric itself and this is a sufficient condition for a static space - time when added with the stationary condition.

Just out of curiosity is there a text you are reading at the moment where you came upon this particular section and statement?

 

[yicong2011]

Stationary means, roughly, that given any hypersurface of time the surface itself will always look the same or in other words the space - time admits a time - like killing vector field. A time - like killing vector field means that any infinitesimal diffeomorphisms generated by a time - like vector field (so any "flows" in time on the space - time such as time reversals) will keep the metric and therefore the space - time invariant. Also roughly, the condition that the time - like killing vector field be orthogonal to a family of hypersurfaces translates to there be no cross terms between space - like coordinates and the time - like coordinate in the metric itself and this is a sufficient condition for a static space - time when added with the stationary condition.

Just out of curiosity is there a text you are reading at the moment where you came upon this particular section and statement?

Thanks.

I have read the definition of "static" and "stationary" spacetime in my book (d'Inverno's).

I am just pondering that "what the difference of the two spacetime that can observers at each point see". I just want to find a more physical understanding.

 

[WannabeNewton]

Yeah that is what I figured you were using =D. I'm sorry but I'm not good at physical explanations so I'll let someone else take a crack at it. In the meantime if you have access to Spacetime and Geometry - Carroll it has a relatively detailed section on Birkhoff's Theorem that also goes more deeply into this topic that might make the physical aspect of it all more apparent I would assume. If you don't I think the free online lecture notes by the same author also talks about it (less detailed but useful nonetheless) http://preposterousuniverse.com/grnotes/

 

[yicong2011]

Thanks.

 

[pervect]

If a set of coordinates exist so that none of the metric coefficients is a function of time, that space-time is either static or stationary.

Note that such coordinates might exist and you might not happen to choose them. The fact that you choose coordinates to make this untrue doesn't really change the actual nature of the space-time.

The fact that none of the metric coefficients are functions of time means that the distances between objects at constant coordinates don't change. If all the observers can synchronize their watches, the space-time is static, otherwise it's stationary.

One thing I'm not sure about. If you have a rotating disk, you can't synchronize your watches on it. But you can synchronize your watches if you use coordinates based off a non-rotating disk. So the underlying space-time is "really" static, but I'm not sure everyone talks about it this pure coordinate independent sense.

 

[atyy]

stationary spacetime: if we arrange a set of observers locating in any places, then each of them should observe nothing change at the place where s/he is located.

Yes. A stationary spacetime has a preferred time. If we use the preferred time to define space and time, then we can have a rigid lattice of observers for whom the gravitational field doesn't change with time. If observer A sends a light signal to B, who sends it to C, who sends it back to A, the light signal will always take the same time to complete the loop.

In a static spacetime, the additional property is that the time taken for light to complete the loop is the same whether it goes ABCA or ACBA (ie. the lattice isn't rotating).

Rindler's text talks about this. Also, there may be more than one rigid lattice. A stationary spacetime means that there is at least one such lattice, but not necessarily only one.

 

[DrGreg]

One thing I'm not sure about. If you have a rotating disk, you can't synchronize your watches on it. But you can synchronize your watches if you use coordinates based off a non-rotating disk. So the underlying space-time is "really" static, but I'm not sure everyone talks about it this pure coordinate independent sense.

I think in that case you could say that spacetime is static, but your rotating coordinates are stationary but not static.