Checking derivation of the curvature tensor

In summary, the conversation involves someone trying to derive the curvature tensor by finding the commutator of two covariant derivatives and seeking help to check their work. The attempt involves taking the covariant derivatives and subtracting them from each other, resulting in the Riemann tensor. It is a common topic in differential geometry.
  • #1
peter46464
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Homework Statement


I am trying to derive the curvature tensor by finding the commutator of two covariant derivatives. I think I've got it, but my head is spinning with Nablas and indices. Would anyone be willing to check my work? Thanks

Homework Equations


I am trying to derive the curvature tensor by finding the commutator of two covariant derivatives: $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}.$$
The covariant derivative of ##V_{a}## is given by$$\nabla_{b}V_{a}=\partial_{b}V_{a}-\Gamma_{ab}^{d}V_{d}.$$

3. The Attempt at a Solution

Start by taking another covariant derivative: $$\nabla_{c}\nabla_{b}V_{a}=\partial_{c}\left(\nabla_{b}V_{a}\right)-\Gamma_{ac}^{e}\nabla_{b}V_{e}-\Gamma_{bc}^{e}\nabla_{e}V_{a}.$$
For the first right-hand side term $$\partial_{c}\left(\nabla_{b}V_{a}\right)=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}V_{d}\right).$$
Then using the product rule I get $$\partial_{c}\left(\nabla_{b}V_{a}\right)=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}.$$
For the second right-hand side term
$$-\Gamma_{ac}^{e}\nabla_{b}V_{e}=-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right).$$
And for the third right-hand side term $$-\Gamma_{bc}^{e}\nabla_{e}V_{a}=-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
Put all these together to get $$\nabla_{c}\nabla_{b}V_{a}=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right)-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
Interchanging ##b## and ##c##
gives $$\nabla_{b}\nabla_{c}V_{a}=\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\left(\partial_{c}V_{e}-\Gamma_{ec}^{d}V_{d}\right)-\Gamma_{cb}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
Subtract one from the other: $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right)-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right)-
\left(\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\left(\partial_{c}V_{e}-\Gamma_{ec}^{d}V_{d}\right)-\Gamma_{cb}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right)\right).$$
Multiplying out the brackets in the second line gives $$\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\partial_{b}V_{e}+\Gamma_{ac}^{e}\Gamma_{eb}^{d}V_{d}-\Gamma_{bc}^{e}\partial_{e}V_{a}+\Gamma_{bc}^{e}\Gamma_{ae}^{d}V_{d}-
\left(\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\partial_{c}V_{e}+\Gamma_{ab}^{e}\Gamma_{ec}^{d}V_{d}-\Gamma_{cb}^{e}\partial_{e}V_{a}+\Gamma_{cb}^{e}\Gamma_{ae}^{d}V_{d}\right).$$
Note that ##\Gamma_{ac}^{e}\partial_{b}V_{e}=\Gamma_{ab}^{e}\partial_{c}V_{e}## and ##\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}=\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}##
(because of the definition of ##\frac{\partial e_{x}}{\partial x^{z}}=\Gamma_{xz}^{y}e_{y}).##
I then end up with $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=\left(\partial_{b}\left(\Gamma_{ac}^{d}\right)-\partial_{c}\left(\Gamma_{ab}^{d}\right)+\Gamma_{ac}^{e}\Gamma_{eb}^{d}-\Gamma_{ab}^{e}\Gamma_{ec}^{d}\right)V_{d}.$$
The expression inside the brackets on the rhs is the Riemann tensor, meaning $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=R_{\phantom{\mu}abc}^{d}V_{d}.$$
 
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  • #2
It looks like more or less the right thing. I'm not going to go through the details. This is "textbook" stuff that you can find in lots of different differential geometry texts.
 

Related to Checking derivation of the curvature tensor

1. What is the curvature tensor?

The curvature tensor, also known as the Riemann curvature tensor, is a mathematical object that describes the curvature of a manifold in terms of its metric. It is used in Einstein's theory of general relativity to describe the curvature of spacetime.

2. How is the curvature tensor calculated?

The curvature tensor is calculated by taking the partial derivatives of the Christoffel symbols, which are the connection coefficients of the metric. These derivatives are then combined in a specific way to form the components of the curvature tensor.

3. Why is it important to check the derivation of the curvature tensor?

It is important to check the derivation of the curvature tensor because it is a crucial step in the development of Einstein's theory of general relativity. Any errors in the derivation can lead to incorrect predictions about the behavior of spacetime and the universe.

4. What are some common mistakes made in deriving the curvature tensor?

Some common mistakes made in deriving the curvature tensor include incorrect calculations of the Christoffel symbols, forgetting to take into account the symmetry of the tensor, and not properly applying the rules of tensor calculus.

5. How is the curvature tensor used in physics?

The curvature tensor is used in physics, specifically in Einstein's theory of general relativity, to describe the curvature of spacetime caused by the presence of matter and energy. It is also used in other areas of physics, such as in the study of black holes and gravitational waves.

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