Particle in Minkowski Space: Motion in a Flat Universe

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Suppose I take 2d Minkowski space $$ds^2=-dt^2+dx^2$$ and put a test particle in there. I would expect that since we have a flat space with no matter inside that it should just "sit still" so to speak i.e. not move anywhere.

However, there will be an integral of motion (since we have a timelike Killing vector) given by $$E=\dot{t}$$ where dot denotes differentiation with respect to proper time.

Then I can use the fact that $$ds^2=-1$$ for timelike geodesics and rearrange to get $$-1=-E^2+\dot{x}^2 \Rightarrow \dot{x} = \sqrt{E^2-1} \Rightarrow x(\tau)=\sqrt{E^2-1} \tau$$

In other words the particle will move to infinity along a timelike geodesic. Why is it moving?

Thanks.

 

[Orodruin]

There is no such thing as absolute motion in special relativity (nor in classical mechanics). There will exist inertial frames where it is moving and there will exist inertial frames where it is standing still.

 

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There is no such thing as absolute motion in special relativity (nor in classical mechanics). There will exist inertial frames where it is moving and there will exist inertial frames where it is standing still.

Thanks. I thought the reason would be something like this. What is wrong with my analysis though?

This is how I am used to obtaining geodesic motion in GR and for most spacetimes I've considered so far e.g. AdS, I can tell whether the particle will move or not. For example in AdS, I can do a similar analysis and find that particles have periodic motion i.e. we get closed timelike curves which presumably can't be cured by changing frames. So why is it ok to do this analysis for AdS (also a vacuum) but not for Minkowski?

 

[Orodruin]

What is wrong with my analysis though?

Nothing. Depending on which inertial frame you use, the value of $E$ will be different. In the inertial frame where the test particle is at rest you will have $E = 1$.

 

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Nothing. Depending on which inertial frame you use, the value of $E$ will be different. In the inertial frame where the test particle is at rest you will have $E = 1$.

Thanks very much. This makes sense. I do have a follow up (related) question regarding the situation in AdS. In this case, timelike geodesics are given (in global coords) by $$\sin{\rho}=\sqrt{1-\frac{L^2}{E^2}} \sin{\tau}$$ where L is the AdS radius.

By analogy with your post, I see that the rest energy of such a particle would be E=L. Now I can draw various trajectories on a Penrose diagram (sin curves of differing amplitudes depending on the value of E) and my question is do these different sin curves represent:

A) a single particle from the perspective of different inertial frames
B) multiple particles with different energies
C) both A and B are true

Thanks again.