I am stuck on exercise 20.5 part a) from Misner, Thorne, and Wheeler's Gravitation chapter 20. The Einstein summation convention is used throughout this post.

Problem Statement

Calculate $t^{\alpha\beta}_{\text{L-L}}$ for the nearly Newtonian metric \begin{equation}ds^2=-(1+2\Phi)dt^2+(1-2\Phi)\delta_{jk}dx^jdx^k\end{equation}(see $\S18.4$). Assume the source is slowly changing, so that time derivatives of $\Phi$ can be neglected compared to space derivatives.



Work Towards Solution

Equation 20.22 states that \begin{align}(-g)t^{\alpha\beta}_{\text{L-L}}&=\frac{1}{16\pi}\left\{\mathfrak{g}^{\alpha\beta}_{\,\,\,,\lambda}\mathfrak{g}^{\lambda\mu}_{\,\,\,,\mu}-\mathfrak{g}^{\alpha\lambda}_{\,\,\,,\lambda}\mathfrak{g}^{\beta\mu}_{\,\,\,,\mu}+\frac{1}{2}g^{\alpha\beta}g_{\lambda\mu}\mathfrak{g}^{\lambda\nu}_{\,\,\,,\rho}\mathfrak{g}^{\rho\mu}_{\,\,\,,\nu}-\left(g^{\alpha\lambda}g_{\mu\nu}\mathfrak{g}^{\beta\nu}_{\,\,\,,\rho}\mathfrak{g}^{\mu\rho}_{\,\,\,,\lambda}+g^{\beta\lambda}g_{\mu\nu}\mathfrak{g}^{\alpha\nu}_{\,\,\,,\rho}\mathfrak{g}^{\mu\rho}_{\,\,\,,\lambda}\right)\\+g_{\lambda\mu}g^{\nu\rho}\mathfrak{g}^{\alpha\lambda}_{\,\,\,,\nu}\mathfrak{g}^{\beta\mu}_{\,\,\,,\rho}+\frac{1}{8}\left(2g^{\alpha\lambda}g^{\beta\mu}-g^{\alpha\beta}g^{\lambda\mu}\right)\left(2g_{\nu\rho}g_{\sigma\tau}-g_{\rho\sigma}g_{\nu\tau}\right)\mathfrak{g}^{\nu\tau}_{\,\,\,,\lambda}\mathfrak{g}^{\rho\sigma}_{\,\,\,,\mu}\right\}\end{align}. Where $\mathfrak{g}^{\mu\nu}=(-g)^{1/2}g^{\mu\nu}$, and $g$ is the determinant of the contravariant metric tensor.

I succeeded in finding $t^{00}_{\text{L-L}}$ and $t^{0j}_{\text{L-L}}$, but not $t^{jk}_{\text{L-L}}$, using the following approximations: \begin{equation}-g\approx1\qquad\mathfrak{g}^{\mu\nu}_{\,\,\,,\lambda}\approx\begin{cases}4\Phi_{,\lambda}\quad\text{if all indices are spatial}\\0\quad\text{otherwise}\end{cases}\qquad g^{\mu\nu}\approx\eta^{\mu\nu}\end{equation}. Instead, I keep getting \begin{equation}t^{jk}_{\text{L-L}}\approx-60\Phi_{,j}\Phi_{,k}+46\delta_{jk}\Phi_{,l}\Phi_{,l}\end{equation}

What am I doing wrong?

  • $\begingroup$ $g$ is the determinant of the contravariant metric tensor It’s usually the determinant of the covariant metric tensor. $\endgroup$
    – G. Smith
    Sep 7 '19 at 0:37

I verified that your result (divided by $16\pi$, which you omitted) is what you get when you incorrectly take $g$ to be the determinant of the contravariant metric tensor. And I verified that, with the correct definition of $g$ as the determinant of the covariant metric tensor, you get the MTW result.

So your problem was a small conceptual error about a definition, not a calculational one.

Also, with the way you did it, $\mathfrak{g}^{\mu\nu}{}_{,\lambda}$ isn't $4\Phi_{,\lambda}$ when all indices are spatial; that's the result when all indices are spatial and $\mu=\nu$.

In the correct calculation, it is $\mathfrak{g}^{00}{}_{,i}$ that is nonzero.

  • $\begingroup$ So my friend and I tried to recalculate the problem today and we are still getting the same answer as before. Anyway you can provide more detail to what you're doing right? $\endgroup$ Sep 7 '19 at 17:53
  • $\begingroup$ What did you find for $\mathfrak{g}^{00}{}_{,i}$? $\endgroup$
    – G. Smith
    Sep 7 '19 at 17:55
  • $\begingroup$ Since this is what this site considers a homework-style question, I’m not going to provide a complete solution. $\endgroup$
    – G. Smith
    Sep 7 '19 at 17:57
  • $\begingroup$ we are still getting $\mathfrak{g}^{00}{}_{,i}$ to be 0. I don't need a step-by-step solution. Just trying to figure out how you got that term to be nonzero. I believe we still have -g = 1 $\endgroup$ Sep 7 '19 at 18:02
  • $\begingroup$ No. To get $\mathfrak{g}^{\mu\nu}{}_{,\lambda}$, you have to calculate $(-g)^{1/2}$ and $g^{\mu\nu}$ both through terms of order $\Phi$. $\endgroup$
    – G. Smith
    Sep 7 '19 at 18:07

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