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 Exercise [14.04] 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Exercise [14.04]
In short, once we have defined the covariant derivative on scalar fields and vector fields, we can work out the action on more general tensor fields T by induction using the identity: (\nabla T)\cdot U = \nabla (T\cdot U) - T\cdot\nabla U, where U is either an arbitrary vector field or an arbitrary covector field, the choice being such that the convariant derivative of the contracted quantity T\cdot U is known. Since U is arbirtrary we can then recover \nabla T.

Because it will be useful for later questions, I will work this out explicitly.

For each coordinate vector field \partial_c := \partial/\partial x^c, define the coefficients: \Gamma^a_{cb} := \left[\nabla {\partial_c}}\right]^a_b
(Note that \Gamma^a_{cb} is tensorial in the a and b indices by construction, but in general it will not be tensorial in the c index).

Now for an arbitrary vector field \boldsymbol{\xi}=\xi^c \partial_c} we use the Leibniz rule and the rule for the action of a covariant derivative on a scalar field:
\nabla_b \xi^a = \left[\nabla\boldsymbol{\xi}\right]^a_b = \left[\nabla\xi^c \partial_c\right]^a_b\\ <br />= \left[(\nabla\xi^c)\otimes\partial_c\right]^a_b + \left[\xi^c\nabla\partial_c\right]^a_b\\<br />= \left[(d\xi^c)\otimes\partial_c\right]^a_b + \xi^c\Gamma^a_{cb}\\<br />= \left[{\partial\xi^c\over{\partial x^d}}dx^d\otimes\partial_c\right]^a_b + \xi^c\Gamma^a_{cb}\\<br />= {\partial\xi^a\over{\partial x^b}} + \xi^c\Gamma^a_{cb}

We now calculate the covariant derivative of a covector field \boldsymbol{\alpha}=\alpha_c dx^c}. The Leibniz rule gives:
(\nabla\boldsymbol{\alpha})\cdot\boldsymbol{\xi} = \nabla(\boldsymbol{\alpha}\cdot\boldsymbol{\xi}) - \boldsymbol{\alpha}\cdot \nabla\boldsymbol{\xi}
In coordinates this becomes:
(\nabla_b \alpha_a)\xi^a = \nabla_b(\alpha_a\xi^a) - \alpha_a \nabla_b\xi^a\\<br />= {\partial (\alpha_a\xi^a)\over{\partial x^b}} - \alpha_a\left( {\partial\xi^a\over{\partial x^b}} +\xi^c\Gamma^a_{cb}\right)\\<br />= \left( {\partial\alpha_a\over{\partial x^b}} - \alpha_a\Gamma^a_{cb}\right)\xi^a
(n.b. I changed dummy indices on the \xi term in the last step)
Since this is true for any vector field \boldsymbol{\xi} we have:
\nabla_b \alpha_a = {\partial\alpha_a\over{\partial x^b}} - \alpha_a\Gamma^a_{cb}

Using induction, it is then straightforward to show that for an arbitrary tensor field T we get the following:
\nabla_b T^{i_1\cdots i_p}_{j_1\cdots j_q} = {\partial T^{i_1\cdots i_p}_{j_1\cdots j_q}\over{\partial x^b}} +  T^{c i_2\cdots i_p}_{j_1\cdots j_q}\Gamma^{i_1}_{cb} + \cdots + T^{i_1\cdots i_{p-1}c}_{j_1\cdots j_q}\Gamma^{i_p}_{cb}\\ \text{\hspace{4cm} } -  T^{i_1\cdots i_p}_{a j_2\cdots j_q}\Gamma^a_{j_1 b} - \cdots - T^{i_1\cdots i_p}_{j_1\cdots j_{q-1}a}\Gamma^a_{j_q b}

I'll prove the inductive step in the case of an arbitrary covariant tensor. The general case is similar (except that you need to contract contravariant indices with a covector field instead of a vector field).
Suppose we have proved the identity for covariant tensors with up to q indices. The case for a covariant tensor with q+1 indices is handled by contracting with a vector field \boldsymbol{\xi}
(\nabla_b T_{j_1\cdots j_q k})\xi^k = \nabla_b (T_{j_1\cdots j_q k}\xi^k) -  T_{j_1\cdots j_q k}\nabla_b \xi^k\\<br />= {\partial (T_{j_1\cdots j_q k}\xi^k)\over{\partial x^b}} -  T_{a j_2\cdots j_q k}\xi^k\Gamma^a_{j_1 b} - \cdots - T_{j_1\cdots j_{q-1}a k}\xi^k\Gamma^a_{j_q b} - T_{j_1\cdots j_q k}\left({\partial\xi^k\over{\partial x^b}} +\xi^c\Gamma^k_{cb}\right)
=\bigg( {\partial T_{j_1\cdots j_{q+1}}\over{\partial x^b}}-T_{a j_2\cdots j_{q+1}}\Gamma^a_{j_1 b} -<br />-  \cdots - T_{j_1\cdots j_q a}\Gamma^a_{j_{q+1}b} \bigg) \xi^{j_{q+1}}
(where we have renamed dummy indices in the final step). This is true for any vector field and so the induction step is proven.

04 May 2010, 04:28

Joined: 23 Aug 2010, 13:12
Posts: 33
Post Re: Exercise [14.04]
The equations immediately preceding and following

"Since this is true for any vector field we have:"

need to have some indices fixed in the gamma term (a->c,c->a)

23 Aug 2010, 13:31

Joined: 26 Mar 2010, 04:39
Posts: 109
Post Re: Exercise [14.04]
Yes, I noticed that a little while back and I've been meaning to fix it. Also in the final equation, to be consistent I should really have used the dummy index j_{q+1} throughout rather than k.

When I first saw this I was planning to fix it when Latex was working on the forum again but now it seems to be permanently broken.

It's nice to know that someone is actually checking this stuff!

24 Aug 2010, 06:28
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