Are all Killing Vectors in 2D Spacetime Hypersurface Orthogonal?

[cristo]

Just a quick question. Suppose we are considering a (1+1) spacetime. Are all vector fields hypersurface orthogonal? I think the answer is yes, since in the formula \xi_{[a}\partial_b \xi_{c]}, two indices will always be the same, which I think automatically makes the expression equal to zero, but am just seeking clarification!

 

[Dick]

I think you are correct. In 2d, a vector field has a 'pretty unique' orthogonal direction (up to direction reversal). So use that orthogonal direction to integrate a curve from a starting point. That curve is the orthogonal 'hypersurface'. So you can visualize it geometrically as well.

 

[cristo]

Thanks, Dick. It's just this question I'm doing on an old exam paper. I've found a killing vector, and transformed into new coordinates so that it is of the form d/dt. The question then asks 'is this spacetime static'. Since the 2d metric is lorentzian, d/dt is clearly timelike => stationary. Then I calculated the hypersurface orthogaonality equation and saw that it was clearly satisfied, thus the spacetime is stationary, then thought about the situation and conjectured that all KV's in 2d are hypersurface orthogonal.

I just wanted to know I was right really, since this question will be on the exam, and saw this is a nice shortcut.

Anyway, thanks again for the help!