# Stuck Solving MTW Gravitation Problem 20.5

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 $$$$ds^2=-(1+2\Phi)dt^2+(1-2\Phi)\delta_{jk}dx^jdx^k$$$$(see $$\S18.4$$). Assume the source is slowly changing, so that time derivatives of $$\Phi$$ can be neglected compared to space derivatives.

\begin{align}t^{00}_{\text{L-L}}&=-\frac{7}{8\pi}\Phi_{,j}\Phi_{,j},\\t^{0j}_{\text{L-L}}&=0,\\t^{jk}_{\text{L-L}}&=\frac{1}{4\pi}\left(\Phi_{,j}\Phi_{,k}-\frac{1}{2}\delta_{jk}\Phi_{,l}\Phi_{,l}\right)\end{align}

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: $$$$-g\approx1\qquad\mathfrak{g}^{\mu\nu}_{\,\,\,,\lambda}\approx\begin{cases}4\Phi_{,\lambda}\quad\text{if all indices are spatial}\\0\quad\text{otherwise}\end{cases}\qquad g^{\mu\nu}\approx\eta^{\mu\nu}$$$$. Instead, I keep getting $$$$t^{jk}_{\text{L-L}}\approx-60\Phi_{,j}\Phi_{,k}+46\delta_{jk}\Phi_{,l}\Phi_{,l}$$$$

What am I doing wrong?

• $g$ is the determinant of the contravariant metric tensor It’s usually the determinant of the covariant metric tensor. Commented Sep 7, 2019 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.

• 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? Commented Sep 7, 2019 at 17:53
• What did you find for $\mathfrak{g}^{00}{}_{,i}$? Commented Sep 7, 2019 at 17:55
• Since this is what this site considers a homework-style question, I’m not going to provide a complete solution. Commented Sep 7, 2019 at 17:57
• 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 Commented Sep 7, 2019 at 18:02
• No. To get $\mathfrak{g}^{\mu\nu}{}_{,\lambda}$, you have to calculate $(-g)^{1/2}$ and $g^{\mu\nu}$ both through terms of order $\Phi$. Commented Sep 7, 2019 at 18:07