Question about Killing vectors in the Kerr Metric

[edwiddy]

Hi, I'm a physics undergrad working through Carroll at the moment. In the section on the Kerr black hole, he states that K= \partial_t is a Killing vector because the coefficients of the metric are independent of t. He then states in eq. 6.83 that K^\mu is normalized by:

K^\mu K_\mu = - \frac{1}{\rho^2} (\Delta - a^2 \sin^2{\theta})

where \Delta = r^2 - 2GMr + a^2 and \rho^2 = r^2 + a^2 \cos^2{\theta}.

Now I can't seem to for the life of me duplicate this from the metric. We take K^\mu = (\partial_t)^\mu = \delta ^\mu_t right? Then:

K^\mu K_\mu = g^{\mu\nu}K_\nu K_\mu

which is only non zero for \mu=\nu=t...but that doesn't match up. The crossterms in the metric need to come into play, but it seems that if anyone of the indices is \phi then it goes to zero...

Thanks.

 

[Ayfel]

It would be interesting if someone could clarify this

 

[Ayfel]

Nada you esta xD

P.D: Sorry I already got it :P

 

[Ayfel]

Now I am struggling with the following K^\mu = (\partial_t)^\mu = \delta ^\mu_t . Could anyone explain how to get the last equality.

i'm reading this from GRAVITATION(Thorne, Wheeler...) and I am quite confused as to how this is true, it is probably a silly thing but I am very new to this notation

Thank you.

 

[Ben Niehoff]

Now I am struggling with the following K^\mu = (\partial_t)^\mu = \delta ^\mu_t . Could anyone explain how to get the last equality.

This is a coordinate dependent statement. It assumes that 't' is one of your coordinates; in particular, that the manifold has a translational symmetry along the 't' direction. In other words, take a manifold with a translational symmetry, and call the axis along that symmetry 't'.

Second, I agree the notation is a bit confusing. I would simply write

K = \partial_t

If your coordinates are (t, x^1, x^2, x^3), then you could write K = (1,0,0,0). Or in other words,

K^\mu = \delta_0^\mu

The notation in your OP is simply using the label 't' instead of '0'.

 

[Ayfel]

The confusing part for me is to go from K = \partial_t to K = (1,0,0,0)

The first expression I suppose is independent of your reference system, but then I do not know how by setting them to t,r,etc you make the transition from the first to the second expression. I am thinking of K = \partial_t as the partial derivative and I do not see how this becomes K = (1,0,0,0)

 

[Ben Niehoff]

Hmm, maybe it will be more clear if we write it out completely:

K = K^\mu \partial_\mu = K^t \partial_t + K^r \partial_r + K^\theta \partial_\theta + K^\phi \partial_\phi

Now since we know K = \partial_t and the basis vectors \partial_t, \; \partial_r, \; \partial_\theta, \; \partial_\phi are all linearly independent, we are left with

K^t = 1, \qquad K^r = K^\theta = K^\phi = 0

Does that make sense?

 

[Ayfel]

Ok now I get it, I was just being confused with the basis vectors and the components of the killing vector. I feel silly, but thank you now I get it.