# Why can't we determine our motion relative to spacetime?

Tags:
1. Aug 2, 2014

### 21joanna12

Hi all,

I was thinking about special relativity and spacetime and had the thought of measuring our velocity with respect to spacetime. I was wondering why this would't work and what makes it different from the ether, which we would be able to measure our velocity with repsect to to determine absolute motion. Especially when considering gravitational waves- if they pass through you when you are 'stationary' and when you are moving away from their source at a constant velocity, would you not be able to tell the difference?

I actually had this thought when considering background independence for general relativity- how is it background independent if it relies on the warping of spacetime... although I'm sure that's a question for another thread!

Sorry if this is a really stupid question, and thank you for any replies!

2. Aug 2, 2014

### Staff: Mentor

One of the main ideas of special relativity is that there is no such thing as a preferred frame of reference from which to measure motion against. All motion is relative to another object and thus it is always equally valid to say one object is motionless while the other is moving or vice versa. Spacetime is not something that has a preferred frame of reference. Indeed there is no such thing as a frame of reference for spacetime since all motion takes place within spacetime. Think of spacetime in SR more like a way of labeling the position of objects in space and time and not as a a strange substance like the aether.

As for gravity waves, they behave similar to EM waves in that they travel at c in all inertial frames. So it doesn't matter what direction you are moving relative to their source, they always travel at the same speed.

3. Aug 2, 2014

### Staff: Mentor

Spacetime isn't a physical thing that you can somehow place a fixed mark in and use that to measure your position and velocity relative to the mark.

In both cases, the waves travel at a speed c relative to you, same as with light.

4. Aug 2, 2014

### WannabeNewton

If a spacetime is static then it contains a privileged family of observers who can be thought of as being at rest in the spacetime. These are the observers who follow orbits of the twist-free timelike Killing field. They define a rigid global rest frame relative to which you can define the velocities of particles moving through the frame. This is as close as you can get to a notion of velocity with respect to spacetime and is the GR version of the velocity of objects falling in gravitational fields in Newtonian gravity.

In stationary spacetimes things get a bit more complicated due to source rotation but suffice it to say the idea is basically the same.

In a radiating spacetime e.g. Vaidya, you cannot define such a notion of rest with respect to the spacetime geometry.

5. Aug 3, 2014

### Matterwave

But thinking about this physically, this family of observers can not necessarily be "at rest" w.r.t. each other, since in Minkowski spacetime obviously you can not choose between any Lorentz frame, and so this "family of observers" for Minkowski spacetime must be those observers traveling at constant velocity. In that sense, different observers give you different velocities, and so do not necessarily define a global rigid rest frame.

6. Aug 3, 2014

### WannabeNewton

The observers all have to be following orbits of the same time-like Killing field $\xi^{\mu}$. Clearly these observers will always be at rest with respect to one another since $\nabla_{(\mu}\xi_{\nu)} = 0$. In Minkowski space-time there exists no privileged state of rest since any time-like Killing field corresponding to inertial observers can be boosted to give another time-like Killing field of inertial observers. But the observers following orbits of any one such field will always be at rest with respect to one another in the Lie-transported global rest frame adapted to the field. This is easy to see as Killing fields are necessarily Born-Rigid. In stationary space-times with gravitational fields from a source there is a privileged state of rest and it is the one I have spelled out above. This is also known as the Copernican frame since it is the frame fixed to the distant stars. It is the stationary gravitational field that singles out this privileged rest state.

7. Aug 3, 2014

### Matterwave

Then I must revise my statement to be only that this notion of observers on time like Killing fields as defined in your previous post, need not necessarily give rise to a privileged notion of rest, as can be seen explicitly by looking at the Minkowski case.

In other words, in some cases, you won't be able to choose which family of observers to define your state of rest from, as there may be many families of such observers.

8. Aug 3, 2014

### WannabeNewton

Minkowski space-time is the only exception and for obvious reasons as there is no gravitational field. The same thing happens in Newtonian mechanics.

9. Aug 3, 2014

### Matterwave

But it is an important exception. ;)