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

[rezkyputra]

Homework Statement

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

Homework 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

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
Thanks in advance

 

[andrewkirk]

Suppose we have a covariant derivative of covariant derivative of a scalar field. My lecturer said that it should be equal to zero.

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.

 

[rezkyputra]

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.

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?

 

[nrqed]

Homework Statement

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

Homework 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

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
Thanks in advance

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