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

• rezkyputra
In summary: Also, in the second line you have mixed up partial and covariant derivatives. Here's the correct calculation:\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_A \nabla^A \phi \nabla^B \phi + \nabla^A \phi \nabla_A \nabla^B \phi - \frac{1}{2} \n 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 haveX^{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} =0with ##\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 rezkyputra said: 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)$$which is not in general zero. andrewkirk said: That doesn't sound correct to me. IIRC the covariant derivative of the covariant derivative of scalar field ##\phi## is a ##\pmatrix{0\\2}## tensor whose ##(\alpha,\beta)## element is$$\phi_{;\alpha\beta}=\partial_\beta\partial_\alpha\phi-\Gamma^\mu_{\alpha\beta}(\partial_\mu\phi)$$which is not in general zero. All in all, said tensor ##X^{AB}## in my post was actually an Energy momentum tensor, which should have a covariant derivative of zero ##\nabla_A X^{AB} = \nabla_A T^{AB} = 0 ## or is there a special case where that would be zero? rezkyputra said: ## Homework Statement Suppose we have a covariant derivative of covariant derivative of a scalar field. My lecturer said that it should be equal to zero. but I seem to not get it ## Homework Equations Suppose we have$$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
You have forgotten to apply the covariant derivative to the metric $g^{AB}$.

## 1. What is a covariant derivative of a scalar field?

A covariant derivative of a scalar field is a mathematical operation that allows us to calculate the rate of change of a scalar field along a given direction in curved space. It takes into account the curvature of the space and allows us to differentiate between vectors and tensors in a curved space.

## 2. How is the first covariant derivative of a scalar field calculated?

The first covariant derivative of a scalar field is calculated by taking the partial derivative of the scalar field with respect to each coordinate, and then subtracting the connection coefficients (also known as Christoffel symbols) multiplied by the components of the vector along each coordinate.

## 3. What does the first covariant derivative represent physically?

The first covariant derivative represents the change in the scalar field due to the curvature of space. It tells us how the scalar field changes as we move along a given direction in curved space.

## 4. How is the second covariant derivative of a scalar field different from the first?

The second covariant derivative of a scalar field is calculated by taking the first covariant derivative and applying the same process again. This means that the second covariant derivative takes into account the curvature not only in the direction of differentiation, but also in the perpendicular directions. Essentially, it measures the change in the first covariant derivative along different directions in curved space.

## 5. What is the significance of the second covariant derivative in physics?

The second covariant derivative is significant in physics because it plays a crucial role in the Einstein field equations of general relativity. These equations describe the curvature of space-time and the relationship between matter and energy. By calculating the second covariant derivative, we can understand the behavior of matter and energy in a more accurate and comprehensive way in curved space.

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