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.