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I'm having issues with computation of Killing equation. I'm using Mathematica to check if the given vectors are Killing vectors or not, and by hand for simple vector like $\xi=\partial_t$ I get the correct result (0=0 when plugging it in the equation), but when I try to check it wiht Mathematica I get weird results.

I don't know if I made my code right, but it should be working, since everything else is working. So I am starting to doubt my by hand method, even tho I'm getting good results :\

The equation is

$$\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0$$

And say my vector is $\xi=\partial_t$, if it is a vector, that means that $\xi^{\mu}=\delta^{\mu}_0$, right? In that case, do I need to raise index in my Killing equation?

And if so is this the correct form:

I'm raising index with metric tensor:

$$\nabla_\mu(g_{\nu\alpha}\xi^\alpha)+\nabla_\nu(g_{\mu\beta}\xi^\beta)$$

Since it applies that my metric is covariantly constant, and I can put my metric components out, and calculate the covariant derivative normally.

$$g_{\nu\alpha}\nabla_\mu\xi^\alpha+g_{\mu\beta}\nabla_\nu\xi^\beta$$

First things first: Is this correct? I mean, when calculating if the expression is true, do I need to raise index?

And do I need to put different indices in metric ($\alpha, \beta$)? Or could I write

$$g_{\nu\alpha}\nabla_\mu\xi^\alpha+g_{\mu\alpha}\nabla_\nu\xi^\alpha?$$

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And say my vector is $\xi = \partial_t$, if it is a vector, that means that $\xi^\mu = \delta^\mu_0$, right? In that case, do I need to raise index in my Killing equation?

Yes $\xi^\mu = \delta_0^\mu$. Remember that $\xi$ is the vector which is invariant under any coordinate transformations. This vector can be expanded out in terms of linear combination of basis vectors: \begin{align} \xi(p) = \xi^\mu(p) \partial^{(p)}_\mu, \end{align} where I have picked the partials of the coordinate system at point $p$ as one particular basis to work in.

In physics we usually refer to $\xi^\mu$ as the 'vector' and says it varies contravariantly with a coordinate transformation, but to be rigorous, $\xi^\mu$ are actually the components of the vector which vary contravariantly, while $\xi$ is actually the vector.

So we see that in the case you are looking at $\xi = 1 \times \partial_t$ so $\xi^\mu = \delta^\mu_0$.

Now to check the identity you can either raise the index in the Killing equation or simply work with the covariant vector $\xi_\mu$, which involves lowering $\xi^\mu$.

First things first: Is this correct? I mean, when calculating if the expression is true, do I need to raise index?

Yes that expression is correct.

And do I need to put different indices in metric ($\alpha,\beta$)?

No, they are just dummy indices for each term which are summed over, and so you can use either $\alpha$ and $\beta$ and just about anything except for indices that are used in the expressions already, e.g. $\mu$ or $\nu$. What you have written is fine.

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  • $\begingroup$ In the last expression, when I'm summing over the repeated indices, I need to have $\alpha$ and $\beta$, even though they are dummy indices, right?, because I'm summing over repeated indices, if I put $g_{\nu\alpha}\nabla_\mu\xi^\alpha+g_{\mu\alpha}\nabla_\nu\xi^\alpha?$, I'd have 4 $\alpha$, so that would be wrong, right? $\endgroup$ – dingo_d Sep 28 '13 at 13:07
  • $\begingroup$ no, that's fine. each term carries its own implicit sum, which uses a dummy index which is independent of whatever other dummy index you use for another term. in particular, you could use the same dummy index. $\endgroup$ – nervxxx Sep 28 '13 at 13:12
  • $\begingroup$ what i mean is, $g_{\nu \alpha} \nabla_\mu \xi^\alpha := \sum_{\alpha} g_{\nu \alpha} \nabla_\mu \xi^\alpha$, so in writing each term, the $\alpha$s are 'used up'/'evaluated' straightaway, so $g_{\nu \alpha} \nabla_\mu \xi^\alpha + g_{\mu \alpha} \nabla_\nu \xi^\alpha = (\sum_{\alpha} g_{\nu \alpha} \nabla_\mu \xi^\alpha) + (\sum_{\alpha} g_{\mu \alpha} \nabla_\nu \xi^\alpha)$. $\endgroup$ – nervxxx Sep 28 '13 at 13:16
  • $\begingroup$ Remember that $\xi$ is the vector which is invariant under any coordinate transformations. [...] In physics we usually refer to $\xi^\mu$ as the 'vector' and says it varies contravariantly with a coordinate transformation, but to be rigorous, $\xi^\mu$ are actually the components of the vector which vary contravariantly, while $\xi$ is actually the vector. This is one way of looking at it, but it's not the only way of looking at it, or the only rigorous way. Many people, including me, usually think of vectors as not being invariant. [...] $\endgroup$ – Ben Crowell Sep 28 '13 at 13:19
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    $\begingroup$ @dingo_d $\mu$ and $\nu$ are not repeated (there's no downstairs $\mu$ and upstairs $\mu$) so there's no sum. You're free to pick whatever values of $\mu$ and $\nu$ to evaluate at. $\endgroup$ – nervxxx Sep 28 '13 at 13:27
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I'd also note that, in the case where you pretty much already know that the vector is a Killing vector, and you're just using Killing's equation to verify this, the form that does not use the covariant derivative is almost always easier:

$$£_{\xi}g_{ab} = \xi^{c}\partial_{c}g_{ab} + g_{cb}\partial_{a}\xi^{c} + g_{ac}\partial_{b}\xi^{c}$$

In your case, you already know that the second and third terms are zero. If there is no explicit $t$ dependance in $g_{ab}$, you're done.

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  • $\begingroup$ I thought about using the Lie derivative. Now that I checked in Mathematica I get the same result. But I still don't get the good result, or I dk if I'm interpreting this right. For a Killing vector, Killing eq should hold, or in other words, Lie derivative of the metric, along the Killing vector should be 0, right? But now, for 2 vectors in 2 different papers, I don't have that condition satisfied :\ I have NHEK metric, and $\partial_t$, and $\partial_\phi$ when I put them in Killing eq. give 0, but other two don't :\ $\endgroup$ – dingo_d Sep 28 '13 at 16:04
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    $\begingroup$ @dingo_d: Killing's equation is just $£_{\xi}g_{ab}=0$. And don't use mathematica if you don't understand what it's output is. Compute it by hand, simplifying your starting metric at first, if the calculation doesn't immediately make sense. $\endgroup$ – Jerry Schirmer Sep 28 '13 at 16:22
  • $\begingroup$ I did what you suggested (oh, and I have a predefined package in Mathematica, just in case my code is wrong), and I computed by hand, and I still got that $\mathcal{L}_\xi g_{\tau\tau}\neq 0$. :S Should I made another question about this on PHY.SE? $\endgroup$ – dingo_d Sep 28 '13 at 18:13
  • $\begingroup$ @dingo_d: without knowing your metric, it's hard to say whether or not this is a problem. It might just be that $\partial_{t}$ isn't a killing vector of your spacetime. $\endgroup$ – Jerry Schirmer Sep 28 '13 at 18:29
  • $\begingroup$ It is, since it's near horizon extreme Kerr metric with $SL(2,R)\times U(1)$ isometry. $U(1)$ gives one Killing vector: $\sim\partial_\phi$, and $SL(2,R)$ gives the other three, one of which is $\sim\partial_t$, and when I calculate Lie derivative and Killing equation of those two I get 0, that's what's so confusing :\ If the other two are also Killing vectors, shouldn't their Lie derivative give 0 too? And Killing equation with those vectors should give 0? Unless I need to look at the limit because of the specifics of the metric itself (near horizon and extreme)? $\endgroup$ – dingo_d Sep 28 '13 at 18:32

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