# Covariant Derivatives (1st, 2nd) of a Scalar Field

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1. Nov 27, 2016

### rezkyputra

1. The problem statement, all variables and given/known data
Suppose we have a covariant derivative of covariant derivative of a scalar field. My lecturer said that it should be equal to zero. but I seem to not get it

2. Relevant equations
Suppose we have
$$X^{AB} = \nabla^A \phi \nabla^B \phi - \frac{1}{2} g^{AB} \nabla_C \phi \nabla^C \phi$$
it should be proven that
$$\nabla_A \: X^{AB} =0$$
with $\phi$ is a scalar field

3. The attempt at a solution

Naturally, we would expand the equations.
\begin{align}\nabla_A \: X^{AB} &= \nabla_A \left[ \nabla^A \phi \nabla^B \phi - \frac{1}{2} g^{AB} \nabla_C \phi \nabla^C \phi \right] \\
&= \nabla^B \phi \left[ \nabla_A \nabla^A \phi \right] + \nabla^A \phi \left[ \nabla_A \nabla^B \phi \right] - \frac{1}{2} g^{AB} \nabla_C \phi \left[\nabla_A \nabla^C \phi\right] - \frac{1}{2} g^{AB} \nabla^C \phi \left[\nabla_A \nabla_C \phi\right]
\end{align}

We know that the covariant derivative of a scalar is its partial derivative $\nabla_A \phi = \partial_A \phi$

\begin{align}\nabla_A \: X^{AB} &= \partial^B \phi \left[ \nabla_A \partial^A \phi \right] + \partial^A \phi \left[ \nabla_A \partial^B \phi \right] - \frac{1}{2} g^{AB} \partial_C \phi \left[\nabla_A \partial^C \phi\right] - \frac{1}{2} g^{AB} \partial^C \phi \left[\nabla_A \partial_C \phi\right]
\end{align}

Now the 2nd Cov. Der. would depends on the christoffel symbol where
\begin{align}
\nabla_A \partial_B \phi &= \partial_A \partial_B \phi - \partial_C \phi \Gamma_{AB}^C \\
\nabla_A \partial^B \phi &= \partial_A \partial^B \phi + \partial^C \phi \Gamma_{AC}^B
\end{align}

so that
\begin{align}\nabla_A \: X^{AB} &= \partial^B \phi \left[ \partial_A \partial^A \phi + \partial^C \phi \Gamma_{AC}^A \right] + \partial^A \phi \left[ \partial_A \partial^B \phi + \partial^C \phi \Gamma_{AC}^B \right] \\
\nonumber
& - \frac{1}{2} g^{AB} \partial_C \phi \left[\partial_A \partial^C \phi + \partial^D \phi \Gamma_{AD}^C\right] - \frac{1}{2} g^{AB} \partial^C \phi \left[\partial_A \partial_C \phi - \partial_D \phi \Gamma_{AC}^D\right]
\end{align}

Now I'm stuck as to where should I go? How could t be proven that last (and quite long) equation be equal to zero?
Any help would be much appreciated

2. Nov 27, 2016

### andrewkirk

That doesn't sound correct to me. IIRC the covariant derivative of the covariant derivative of scalar field $\phi$ is a $\pmatrix{0\\2}$ tensor whose $(\alpha,\beta)$ element is
$$\phi_{;\alpha\beta}=\partial_\beta\partial_\alpha\phi-\Gamma^\mu_{\alpha\beta}(\partial_\mu\phi)$$
which is not in general zero.

3. Nov 27, 2016

### rezkyputra

All in all, said tensor $X^{AB}$ in my post was actually an Energy momentum tensor, which should have a covariant derivative of zero $\nabla_A X^{AB} = \nabla_A T^{AB} = 0$

or is there a special case where that would be zero?

4. Nov 29, 2016

### nrqed

You have forgotten to apply the covariant derivative to the metric $g^{AB}$.