Proving Surface Gravity with Killing's, Hypersurface, Geodesic Equation

[LAHLH]

Could someone point me in the right direction to prove that the surface gravity is given by $$\kappa^2=-\frac{1}{2}(\nabla_{\mu}\chi_{\nu})(\nabla^{\mu}\chi^{\nu})$$

I know it involves Killings equation, the hypersurface othog equation and the geodesic equation somehow but I'm not entirely sure where to go from there as I've tried expanding each of these and not really got anywhere.

Thanks

 

[George Jones]

See section 5.2.4 of Eric Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

which evolved into the excellent book, A Relativist's Toolkit: The Mathematics of Black Hole Mechanics.

 

[LAHLH]

Thanks a lot, just what I was looking for!

While I'm here, I'm also trying to figure out (a few pages later in Carroll ch6 where I was reading about surface gravity), why the four acceleration $$a^{\mu}=U^{\mu}\nabla_{\sigma}U^{\mu}$$ can be written as $$a_{\mu}=\nabla_{\mu}\ln{(V)}$$

Here V is the so called "redshift factor", i.e. if you have a static observer with four velocity proportional to the time translation Killing Vector K then V(x) is the proportionality constant. $$V=\sqrt{-K_{\mu}K^{\mu}}$$

He says "as you can easily verify", so I tried to do so and didn't seem to reproduce what he has, I'm not sure if I'm treating the covariant derivative correctly, I'm just assuming the usual chain rules etc apply, because I was pretty sure that I remember they do, but hmm

 

[jarlostensen]

Just out of curiosity; what did you get when you calculated it..?

 

[jarlostensen]

Btw, I got;

$$\frac{1}{V^{2}}g_{\mu \nu} K^{\mu} K^{\nu}_{;\alpha}}$$

which sort of looks right..?

 

[LAHLH]

Btw, I got;

$$\frac{1}{V^{2}}g_{\mu \nu} K^{\mu} K^{\nu}_{;\alpha}}$$

which sort of looks right..?

I got exactly this too. Given that $$U^{\mu}=K^{\mu}/V(x)$$ it looks really close, but I don't know how to massage that into $$\frac{K^{\sigma}}{V} \nabla_{\sigma}\frac{K^{\mu}}{V}$$

We have

$$\frac{1}{V^2}K^{\rho} \nabla_{\mu}K_{\rho}$$. The K's are currently dotted with each other rather than with the cov deriv, and whilst you can combine the external K and one of the V's to give a U, you can't just take the other V inside the covariant deriv to create another U, so I'm kinda stuck...

 

[LAHLH]

Anyone know how to do this? or know any books with the same equation so I can double check it is actually given by this formula?

 

[George Jones]

Anyone know how to do this?

I think that I have done it.

The K's are currently dotted with each other rather than with the cov deriv

Use Killing's equation to change this around. I also used the orthogonality of 4-velocity and 4-acceleration,

[tex]0 = U^\alpha U^\beta \nabla_\beta U_\alpha .[/itex][/tex]

 

[jarlostensen]

Ok, that gets me to;

$$0 = \frac{K^{\alpha} K^{\beta}}{V^4} ( K_{\alpha;\beta} V - K_{\alpha} V_{;\beta})$$

Can we have just one more hint please?

 

[George Jones]

Ok, that gets me to;

$$0 = \frac{K^{\alpha} K^{\beta}}{V^4} ( K_{\alpha;\beta} V - K_{\alpha} V_{;\beta})$$

Can we have just one more hint please?

To kill the last term, expand

[tex]0 = U^\alpha U^\beta \nabla_\beta U_\alpha[/itex]<br /> <br /> in terms of $K^\mu$ and $V$, explicitly calculate the derivative involved, and use the fact that something symmetric contracted with something antisymmetric gives zero.[/tex]

 

[LAHLH]

This site has been intermittently down for me the last couple of days and I had actually found a proof coming at this from the $$a^{\mu}=U^{\mu}\nabla_{\sigma}U^{\mu}$$ direction, but it sounds like exactly the same techniques used here, I'd be happy to write it out if anyone wants to see it explicitly.

Thanks a lot for the replies George, very helpful.