Killing vectors of minkowski space

[WannabeNewton]

How does one know from the general form of the killing vectors in minkowski space:
X$$^{a}$$ = $$\omega$$_a_b(x$$^{a}$$) + t$$^{a}$$

that there are 3 rotational isometries, 3 boosts, 3 spatial translations, and 1 time translation from that general form? It has me very confused >.<

 

[Mentz114]

I can't read your notation so I could be wrong. The four translations are obviously from t^a and if $\omega$ is traceless then it has 3 symmetric degrees of freedom ( boosts) and 3 anti-symmetric df (rotation) .

 

[WannabeNewton]

I can't read your notation so I could be wrong. The four translations are obviously from t^a and if $\omega$ is traceless then it has 3 symmetric degrees of freedom ( boosts) and 3 anti-symmetric df (rotation) .

Omega is a rank two anti symmetric tensor sorry I couldn't get latex to work properly. I understand the part with the translation but how can you show that there are 6 components of omega with 3 being boosts and the other 3 being rotations? Also a lot of my books represent the translations in terms of $$\partial$$ / $$\partial$$x$$^{a}$$ - are they simply representing translations as displacements or is there a way in which you can extract the differentials from t$$^{a}$$?

 

[George Jones]

The thread

https://www.physicsforums.com/showthread.php?t=454237

might help.

 

[Mentz114]

Omega is a rank two anti symmetric tensor ...

In that case I'm baffled. I'll have to read George Jones' thread again because I didn't see how spatial rotations and boosts are accommodated.

 

[WannabeNewton]

The thread

https://www.physicsforums.com/showthread.php?t=454237

might help.

Thanks I think I get it now. If minkowski space killing vectors are this confusing to me I can't imagine schwarzschild or robertson metrics. Sucks for me considering how important these isometries seem to be.

 

[K^2]

Any vector in Minkowski space is a Killing vector, because any directional derivative of the metric with respect to displacement is zero. You can think of it as degenerate eigen vectors. Any complete set spanning the space works as well as any other.

As a consequence, the energy and momentum in Minkowski space are absolutely conserved.