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I understand these parts up there, but enter image description here

I cannot understand how the second formula of the last equality leads to the third formula. Can anyone show me what relabeling indices rules are used to lead this, and if possible, proof of these relabeling indices rules?

Thanks very much.

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The third line comes from literally plugging the second line into the first one, and the last equality is achieved by relabeling indices. What exactly don't you understand? –  Henry T. Horton Apr 27 '12 at 3:08
    
@HenryT.Horton the last equality - how does the second formula of the last equality leads to the third formula? Can you tell me about relabeling indices rules, because I seem to have forgotten them :) –  user27515 Apr 27 '12 at 3:11
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This sort of calculation is appropriately called index gymnastics. Here is the gory detail for calculating the covariant derivative of the vector field $\bar F$, $$\begin{array}{rcll} \displaystyle\frac{\partial \bar F}{\partial u^j} &=& \displaystyle\frac{\partial (f^i\bar a_i)}{\partial u^j} & \textrm{(expand in coordinate basis)} \\ &=& \displaystyle\frac{\partial f^i}{\partial u^j} \bar a_i + f^i \frac{\partial \bar a_i}{\partial u^j} & \textrm{(chain-rule)} \\ &=& \displaystyle\frac{\partial f^i}{\partial u^j} \bar a_i + f^i \Gamma^k_{ij} \bar a_k & \textrm{(definition of }\Gamma)\\ &=& \displaystyle\frac{\partial f^i}{\partial u^j} \bar a_i + f^k \Gamma^i_{kj} \bar a_i & \textrm{(switch dummy indices)}\\ &=& \displaystyle\frac{\partial f^i}{\partial u^j} \bar a_i + f^k \Gamma^i_{jk} \bar a_i & \textrm{(torsion-free)}\\ &=& \displaystyle\left(\frac{\partial f^i}{\partial u^j} \bar a_i + f^k \Gamma^i_{jk}\right) \bar a_i & \textrm{(factor out }\bar a_i). \end{array}$$ We have used your implicit assumption that the connection be torsion-free, $\Gamma^i_{jk} = \Gamma^i_{kj}$. Of course, when you get used to doing such calculations you need not write so many steps.

Note that $i$ and $j$ in $a_{ij}b^{ij} = \sum_{i=1}^n \sum_{j=1}^n a_{ij}b^{ij}$ are dummy indices, they are being summed over. (This is Einstein summation notation, a discovery Einstein was quite proud of, and rightfully so.) We can change dummy indices so long as the index we change to is free. For example, $a_{ij}b^{ij} = a_{ik}b^{ik}$, but $a_{ij}b^{ij} \ne a_{ii}b^{ii}$, since $a_{ii}b^{ii} = \sum_{i=1}^d a_{ii}b^{ii}$.

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What you are trying to do with these equations is to define the covariant derivative of a vector field $\bar{F} = f^i \bar{a}_i$ (here we use the Einstein summation notation). The first equation comes from the chain rule: $$\frac{\partial \bar{F}}{\partial u^j} = \frac{\partial (f^i \bar{a}_i)}{\partial u^j} = \frac{\partial f^i}{\partial u^j} \bar{a}_i + f^i \frac{\partial \bar{a}_i}{\partial u^j}. \tag{$\ast$}$$ Now the Christoffel symbols of the second kind are defined by the equation $$\frac{\partial \bar{a}_i}{\partial u^j} = \Gamma^k_{\phantom{k}ij} \bar{a}_k, \tag{$\ast\ast$}$$ assuming that the $\bar{a}_i$'s form a coordinate basis with respect to the local coordinates $u^i$. Plugging $(\ast\ast)$ into $(\ast)$ gives $$\begin{align} \frac{\partial \bar{F}}{\partial u^j} & = \frac{\partial f^i}{\partial u^j} \bar{a}_i + f^i \left(\Gamma^k_{\phantom{k}ij} \bar{a}_k\right) \\ & = \frac{\partial f^i}{\partial u^j} \bar{a}_i + f^k \left(\Gamma^i_{\phantom{i}kj} \bar{a}_i\right) \\ & = \left(\frac{\partial f^i}{\partial u^j} + f^k \Gamma^i_{\phantom{i}kj} \right)\bar{a}_i . \end{align}$$

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I have to catch the bus... I can put more detail tomorrow morning. Sorry. –  Henry T. Horton Apr 27 '12 at 3:36
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Also, $\Gamma$ is symmetric in its lower indices. –  anon Apr 27 '12 at 3:37
    
$\Gamma$ is symmetric in its lower indices if and only if it comes from a torsion-free connection... Which of course seems to be the case in whatever source the original equations came from. –  Henry T. Horton Apr 27 '12 at 13:27
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