I don't quite know where to start this question. I'm essentially not understanding how to compute the Lie derivative of a given metric and vector.

So I have the following definition:

$$ \left(\mathcal{L}_{\xi}g\right)_{ab}=\xi^{c}\partial_{c}g_{ab}+g_{cb}\partial_{a}\xi^{c}+g_{ac}\partial_{b}\xi^{c} $$

And say, as an example, I have the following vector (as an example):

$$ \xi^{a}=(\alpha(t,r),\beta(t,r),0,0) $$ So the vector is superscript $a$? This has instantly confused me! My metric is of the form $g_{ab}$, which is consistent with the expression given.

So assuming that I can re-write the above expression as $\xi^{c}$, I'm still lost as to how I proceed with this calculation. What does $\xi^{c}$ correspond to when we have multiple components?

Could anyone give a generalised introduction to how this is computed, where to go with it, and perhaps an example to get me going. (I understand much better with a few examples under my belt).

Maybe if someone could give me the example using the basic FRLW metric $$g_{ab}=diag(1,-a^{2},-a^{2},-a^{2})$$ with the same vector? My actual metric is the following: $$ g_{ab}=diag(-e^{\Phi(r)},(1-b(r)/r)^{-1},r^{2},r^{2}\sin^{2}\theta) $$ But hopefully with the example given I should be able to make a start on this metric.

I am currently at this stage:


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    $\begingroup$ ...what has the appearance of the index $a$ in the line before to do with the $a$ on the $\xi^a$? Also, it appears you are either using abstract index notation - because writing $\xi^a = (\xi^0,\xi^1,\xi^2,\xi^3)$ does not make sense if $\xi^a$ is supposed to be a component as in ordinary index notation - or you are misusing index notation. Your problem seems to be less with any specific computation and more with understanding index notation. $\endgroup$
    – ACuriousMind
    Commented Mar 14, 2016 at 11:08
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    $\begingroup$ unfortunately, the notation $x^\mu=(t,\boldsymbol x)$ is used all the time... $\endgroup$ Commented Mar 14, 2016 at 11:11
  • $\begingroup$ I'm not misusing it as far as I am aware, this is directly from a question which is being asked. The notation is as written. The vector that is defined as $\xi^{a}$ should that be $\xi^{c}$? That is what I am asking. $\endgroup$ Commented Mar 14, 2016 at 11:11
  • $\begingroup$ Let's just assume this is true, and proceed. This is really possibly just a typo by my lecturer. Would it be possible to run my through then how to calculate an example Lie derivative with the basic FRLW metric given? Hopefully then I can make paths into understand how this is done! $\endgroup$ Commented Mar 14, 2016 at 11:15
  • $\begingroup$ I don't see a difference between $\xi^a$ and $\xi^c$. $\endgroup$
    – Kyle Kanos
    Commented Mar 14, 2016 at 11:43

1 Answer 1


Writing out all the contracted indices often helps and is a good starting point... The only non-zero components of $\xi^c$ is $0$ and $1$, so

$ \xi^c\partial_c g_{ab}+g_{cb}\partial_a\xi^c+g_{ac}\partial_b\xi^c =\xi^0\partial_0 g_{ab}+\xi^1\partial_1 g_{ab}+g_{0b}\partial_a\xi^0+g_{1b}\partial_a\xi^1++g_{a0}\partial_b\xi^0++g_{a1}\partial_b\xi^1 $

Then use that the first two components of $\xi$ are functions of $t$ and $r$, and that your metric is diagonal (only $g_{00}=-e^{\Phi(r)}$, $g_{11}=(1-b(r)/r)^{-1}$ etc. contributes) and its components are functions of $r$. By computing the different derivatives properly and plugging them back into the equation above you should obtain the correct result.

  • $\begingroup$ I have a few questions. Does the 0 indice represent time, $t$ and 1 the radial extent, $r$? So would $\partial_{0}g_{ab}=0$, and then how would I evaluate $\partial_{a}\xi^{1}$ or $\partial_{b}\xi^{1}$? Could you clarify this in your answer if possible...many thanks! :) $\endgroup$ Commented Mar 14, 2016 at 13:28
  • $\begingroup$ Yeah, 0 is time, 1 is radial and so on. You need to be careful when evaluating derivatives of the metric, for example $\partial_0 g_{ab}$ would be nonzero in the FLRW metric, since $g_{11}$ is a function of time.. $\endgroup$
    – Rexbye
    Commented Mar 14, 2016 at 13:39
  • $\begingroup$ Remember the metric is diagonal, so any of diagonal components are zero. Use $a=b$ and write it out when $a=0$, $a=1$, etc. $\endgroup$
    – Rexbye
    Commented Mar 14, 2016 at 13:46
  • $\begingroup$ So, one more quick question. Apologies, but would $\partial_0g_{ab}$ be zero? Since there is no time-dependence? $\endgroup$ Commented Mar 14, 2016 at 13:48
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    $\begingroup$ I have some stuff i need to attend to through out the day, but i'll give it a quick look when i'm back home tonight. Hopefully someone else can help you as well... $\endgroup$
    – Rexbye
    Commented Mar 14, 2016 at 13:59

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