# Particle in Minkowski

1. Nov 3, 2015

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.

2. Nov 3, 2015

### Orodruin

Staff Emeritus
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.

3. Nov 3, 2015

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?

4. Nov 3, 2015

### Orodruin

Staff Emeritus
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 $E = 1$.

5. Nov 3, 2015

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.