# Ngô Quốc Anh

## January 12, 2010

### R-G: Stress energy tensor of a scalar field satisfies some semilinear wave equation

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Tags: — Ngô Quốc Anh @ 21:17

Today, we study a very interesting property of stress energy tensor of a scalar field. Recall from this topic that Einstein tensor is divergence free, that is

$\displaystyle {E^{\mu\nu}}_{;\mu }=0$.

We still know from the Einstein equation that

$\displaystyle E^{\mu\nu}=\frac{8 \pi G}{c^4}T^{\mu\nu}$

where $T$ the stress energy tensor. As a consequence, $T$ is also divergence free, that is

$\displaystyle\nabla_\beta T^{\alpha\beta}=0$

where $\nabla$ is covariant derivative. We all know from the topic where we define Einstein tensor by contracting the Bianchi identity.

We know consider a special case of the stress energy tensor, that is, stress energy tensor of a scalar field $\psi$ with potential $V(\psi)$, a function of $\psi$. In general relativity, in a local frame $T$ is given as follows

$\displaystyle T_{\alpha\beta}=\nabla_\alpha\psi\nabla_\beta\psi-\frac{1}{2}g_{\alpha\beta}\nabla_\mu\psi\nabla^\mu\psi - g_{\alpha\beta}V(\psi)$.

We firstly need to raise indexes $\alpha\beta$. Obviously, one gets

$\displaystyle\begin{gathered}{T^{ab}} = {g^{a\alpha }}{g^{b\beta }}{T_{\alpha \beta }} \hfill \\ \qquad= {g^{a\alpha }}{g^{b\beta }}\left( {{\nabla _\alpha }\psi {\nabla _\beta }\psi- \frac{1}{2}{g_{\alpha \beta }}{\nabla _\mu }\psi {\nabla ^\mu }\psi- {g_{\alpha \beta }}V(\psi )} \right) \hfill \\ \qquad= {g^{a\alpha }}{g^{b\beta }}{\nabla _\alpha }\psi {\nabla _\beta }\psi- \frac{1}{2}{g^{a\alpha }}{g^{b\beta }}{g_{\alpha \beta }}{\nabla _\mu }\psi {\nabla ^\mu }\psi- {g^{a\alpha }}{g^{b\beta }}{g_{\alpha \beta }}V(\psi ) \hfill \\ \qquad= {g^{a\alpha }}{g^{b\beta }}{\nabla _\alpha }\psi {\nabla _\beta }\psi- \frac{1}{2}{g^{ab}}{\nabla _\mu }\psi {\nabla ^\mu }\psi- {g^{ab}}V(\psi ). \hfill \\ \end{gathered}$

Thus

$\displaystyle {T^{\alpha \beta }} = {g^{a\alpha }}{g^{b\beta }}{\nabla _a}\psi {\nabla _b}\psi- \frac{1}{2}{g^{\alpha \beta }}{\nabla _\mu }\psi {\nabla ^\mu }\psi- {g^{\alpha \beta }}V(\psi )$.

Now we compute $\nabla_\beta T^{\alpha\beta}$. Indeed, we have

$\displaystyle 0={\nabla _\beta }{T^{\alpha \beta }} = {\nabla _\beta }\left( {{g^{a\alpha }}{g^{b\beta }}{\nabla _a}\psi {\nabla _b}\psi- \frac{1}{2}{g^{\alpha \beta }}{\nabla _\mu }\psi {\nabla ^\mu }\psi- {g^{\alpha \beta }}V(\psi )} \right)$.

A simple calculation with the fact that $\nabla_\alpha g^{\beta\gamma}=0$ shows that

$\displaystyle 0={\nabla _\beta }{T^{\alpha \beta }} = \big( {\nabla ^\beta}{\nabla _\beta}\psi-V'(\psi)\big)\nabla^\alpha\psi$.

Thus

$\displaystyle V'(\psi ) = {\nabla ^\alpha }{\nabla _\alpha }\psi$.

This is why we always assume that the field $\psi$ is supposed to satisfy some semilinear wave equation.