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.

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