- #1
Markus Kahn
- 110
- 14
- Homework Statement
- Let ##\phi## be a scalar field that obeys the wave equation, i.e. ##\nabla_a\nabla^a\phi = 0##. How do you need to chose the constant ##C## for the Stress energy tensor
$$T_{a b}=\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi$$
to be conserved, i.e. satisfy ##\nabla_a T^{ab}=0##?
- Relevant Equations
- All given in the question.
My try:
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla^a \left(\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi\right)\\
&\overset{(1)}{=} \underbrace{(\nabla^a\nabla_{a} \phi)}_{=0} \nabla_{b} \phi + \nabla_{a} \phi (\nabla^a\nabla_{b} \phi)-\frac{C}{2} \underbrace{(\nabla^ag_{a b})}_{=0} \nabla_{c} \phi \nabla^{c} \phi~ \underbrace{ - \frac{C}{2} g_{a b} (\nabla^a\nabla_{c} \phi) \nabla^{c} \phi -\frac{C}{2} g_{a b} \nabla_{c} \phi (\nabla^a \nabla^{c} \phi)}_{= Cg_{ab}(\nabla^a\nabla_{c} \phi) \nabla^{c} \phi }\\
&= \nabla_{a} \phi (\nabla^a\nabla_{b} \phi) - Cg_{ab}(\nabla^a\nabla_{c} \phi) \nabla^{c} \phi\\
&\overset{(2)}{=} \nabla_{a} \phi (\nabla^a\nabla_{b} \phi) - C\nabla_{a} \phi(\nabla_b\nabla^a \phi)\\
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi],
\end{align*}
$$
where in ##(1)## I assumed we are working with the Levi-Civita connection (i.e. ##\nabla_a g_{bc}=0##) and the fact that ##\phi## satisfies the wave equation. In ##(2)## I just relabeled the indices and contracted the metric with the connection.
This is where I'm stuck. I'm not really sure how to proceed from here. I thought that one could maybe make use of the fact that the covariant derivative acts especially nicely on scalar fields, i.e.
$$\nabla _a \phi = \partial_a \phi\quad \Longrightarrow \quad\nabla_b \nabla_a \phi = \partial_b\partial_a\phi - \Gamma^{k}_{ab}\partial_k \phi,$$
we modify the statement slightly to our case, i.e.
$$\nabla_a\nabla^b\phi = \nabla_a (g^{bc}\nabla_c\phi) = g^{cb} \nabla_a\nabla_c\phi = g^{cb}(\partial_c\partial_a\phi - \Gamma^{k}_{ac}\partial_k \phi).$$
This then results in
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi]\\
&=\nabla_{a} \phi [g^{ab}(\partial_b\partial_c\phi - \Gamma^{k}_{cb}\partial_k \phi)- C g^{ab}(\partial_c\partial_b\phi - \Gamma^{k}_{bc}\partial_k \phi) ]\\
&\overset{C=1}{=} \nabla_a \phi g^{ab}[ \underbrace{[\partial_b,\partial_c]}_{=0}\phi + \underbrace{(\Gamma^k_{bc}-\Gamma^k_{cb})}_{=0}\phi ]\\
&=0,
\end{align*}
$$
where the Christoffel-symbols cancel since we chose the Levi-Civita connection, i.e. a torsion-free connection.
So the result would be ##C=1##... I'm really skeptical about this, so I would appreciate if someone could take a look at it and confirm that I'm not doing nonsense or give me a hint where I went wrong.
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla^a \left(\nabla_{a} \phi \nabla_{b} \phi-\frac{C}{2} g_{a b} \nabla_{c} \phi \nabla^{c} \phi\right)\\
&\overset{(1)}{=} \underbrace{(\nabla^a\nabla_{a} \phi)}_{=0} \nabla_{b} \phi + \nabla_{a} \phi (\nabla^a\nabla_{b} \phi)-\frac{C}{2} \underbrace{(\nabla^ag_{a b})}_{=0} \nabla_{c} \phi \nabla^{c} \phi~ \underbrace{ - \frac{C}{2} g_{a b} (\nabla^a\nabla_{c} \phi) \nabla^{c} \phi -\frac{C}{2} g_{a b} \nabla_{c} \phi (\nabla^a \nabla^{c} \phi)}_{= Cg_{ab}(\nabla^a\nabla_{c} \phi) \nabla^{c} \phi }\\
&= \nabla_{a} \phi (\nabla^a\nabla_{b} \phi) - Cg_{ab}(\nabla^a\nabla_{c} \phi) \nabla^{c} \phi\\
&\overset{(2)}{=} \nabla_{a} \phi (\nabla^a\nabla_{b} \phi) - C\nabla_{a} \phi(\nabla_b\nabla^a \phi)\\
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi],
\end{align*}
$$
where in ##(1)## I assumed we are working with the Levi-Civita connection (i.e. ##\nabla_a g_{bc}=0##) and the fact that ##\phi## satisfies the wave equation. In ##(2)## I just relabeled the indices and contracted the metric with the connection.
This is where I'm stuck. I'm not really sure how to proceed from here. I thought that one could maybe make use of the fact that the covariant derivative acts especially nicely on scalar fields, i.e.
$$\nabla _a \phi = \partial_a \phi\quad \Longrightarrow \quad\nabla_b \nabla_a \phi = \partial_b\partial_a\phi - \Gamma^{k}_{ab}\partial_k \phi,$$
we modify the statement slightly to our case, i.e.
$$\nabla_a\nabla^b\phi = \nabla_a (g^{bc}\nabla_c\phi) = g^{cb} \nabla_a\nabla_c\phi = g^{cb}(\partial_c\partial_a\phi - \Gamma^{k}_{ac}\partial_k \phi).$$
This then results in
$$
\begin{align*}
\nabla^a T_{ab}
&= \nabla_{a} \phi [\nabla^a\nabla_{b} \phi - C\nabla_b\nabla^a \phi]\\
&=\nabla_{a} \phi [g^{ab}(\partial_b\partial_c\phi - \Gamma^{k}_{cb}\partial_k \phi)- C g^{ab}(\partial_c\partial_b\phi - \Gamma^{k}_{bc}\partial_k \phi) ]\\
&\overset{C=1}{=} \nabla_a \phi g^{ab}[ \underbrace{[\partial_b,\partial_c]}_{=0}\phi + \underbrace{(\Gamma^k_{bc}-\Gamma^k_{cb})}_{=0}\phi ]\\
&=0,
\end{align*}
$$
where the Christoffel-symbols cancel since we chose the Levi-Civita connection, i.e. a torsion-free connection.
So the result would be ##C=1##... I'm really skeptical about this, so I would appreciate if someone could take a look at it and confirm that I'm not doing nonsense or give me a hint where I went wrong.
