I am trying to work out $\square=\nabla^{\mu}\nabla_{\mu}$ in the metric

$ ds^{2}=-A(r)dt^{2}+B(r)^{-1}dr^{2}+r^{2}d\Omega^{2} $$

My work:

when applying $\square$ to a scalar $\phi$, then

$ \square\phi=\nabla^{\mu}\nabla_{\nu}\phi=\nabla^{\mu}\partial_{\mu}\phi=g^{\mu\nu}\nabla_{\nu}\partial_{\mu}\phi=g^{\mu\nu}(\partial_{\nu}\partial_{\mu}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda})\phi $

Christoffel symbol

\left( \begin{array}{ccc} \left\{0,\frac{A'(r)}{2 A(r)},0\right\} & \left\{\frac{A'(r)}{2 A(r)},0,0\right\} & \{0,0,0\} \\ \left\{\frac{1}{2} B(r) A'(r),0,0\right\} & \left\{0,-\frac{B'(r)}{2 B(r)},0\right\} & \{0,0,-r B(r)\} \\ \{0,0,0\} & \left\{0,0,\frac{1}{r}\right\} & \left\{0,\frac{1}{r},0\right\} \end{array} \right)

substituting the metric and affine values in the equation above, my answer came to be

$$ \square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{1}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr} $$ However, the answer happens to be

$$ \square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{2}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr} $$

Could someone please show me where the third comes from in the second term?


closed as off-topic by user10851, ACuriousMind, Kyle Kanos, JamalS, Jim Mar 31 '15 at 16:01

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  • $\begingroup$ Could you include the work you have done on it? Show us your steps? But be forewarned, "check my work" questions (if this turns out to be one of them) are off-topic here $\endgroup$ – Jim Mar 30 '15 at 22:25
  • $\begingroup$ Jimnosperm, Its actually from a research paper in arxiv, not a homewor problem arxiv.org/pdf/1001.4157v3 $\endgroup$ – MrDi Mar 30 '15 at 22:27
  • 2
    $\begingroup$ @435145 If you read the homework policy, it doesn't matter if it is homework or not, just if it is homework like $\endgroup$ – Jimmy360 Mar 30 '15 at 22:35
  • $\begingroup$ (I'm guessing $\phi$ has no angular dependence...) Ignore the time part, and pretend $B(r)=1$. Then, you've got the standard spatial metric in spherical coordinates. So you should get something like the standard Laplacian in spherical coordinates, which has that $2$. Looks like you screwed something up in calculating the Christoffel symbols. $\endgroup$ – Mike Mar 30 '15 at 22:46
  • $\begingroup$ where the factor of two come from? $\endgroup$ – MrDi Mar 30 '15 at 23:21

Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat $d\Omega^2$ more carefully:
$r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$
so you get one more r factor when using the formula @Prahar gives.
and btw the Christoffel symbol is for 2+1 and certainly wrong for 3+1.


I'll prove a formula that is probably easier to use for this. \begin{equation} \begin{split} \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} \right) g^{\mu\nu} \partial_\nu \phi + \partial_\mu \left( g^{\mu\nu} \partial_\nu \phi \right) \\ &= \frac{1}{2g} \partial_\mu g g^{\mu\nu} \partial_\nu \phi + \partial_\mu g^{\mu\nu} \partial_\nu \phi + g^{\mu\nu} \partial_\mu \partial_\nu \phi \\ &= - \frac{1}{2} g^{\alpha\mu} g^{\nu\beta} \left[ \partial_\mu g_{\alpha\beta} + \partial_\alpha g_{\mu\beta} - \partial_\beta g_{\alpha\mu} \right] \partial_\nu \phi + g^{\mu\nu} \partial_\mu \partial_\nu \phi \\ &= g^{\mu\nu} \left( \partial_\mu \partial_\nu - \Gamma^\lambda_{\mu\nu} \partial_\lambda \phi \right) \\ &= \nabla^\mu \nabla_\mu \phi \end{split} \end{equation}


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