# Index notation and differentiation

Let $x_i$ such that $i=1,2,\ldots,n$, and $\vec{x}=(x_1,\ldots,x_n)$

Define $$A:= M_{ij}(\vec{x})\dot{x}^i\dot{x}^j$$ where Einstein summation applies.

Also, $M$ is symmetric and invertible -- a metric.

What then is ${\partial \over \partial \vec{x}}A$ and ${\partial \over \partial \dot{\vec{x}}}A$?

Also, if anyone has any good references on the subject, and would not mind sharing, that would be greatly appreciated.

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Valentin's answer is quite correct, but for a neophyte it can still be daunting. So I would like to remind to you that dummy indices don't matter when we sum over them, so: $$A = M_{ij}(x) \dot{x}^{i}\dot{x}^{j} = M_{kl}(x)\dot{x}^{k}\dot{x}^{l}.$$ Remember, Einstein's summation convention says when an index is repeated, and one is "upstairs" while another is "downstairs", then we sum over it...thus its a dummy index.

(Contrast this with what Misner, Thorne, and Wheeler's Gravitation calls the "Euclidean summation convention" where we just sum over any repeated index, regardless whether it's repeated downstairs or upstairs. So Euclidean has $M_{ii}={M^{i}}_{i}=M^{ii}=\sum_{i}M_{ii}$ whereas Einstein says this is just the diagonal entry of $M$)

This is important because when we consider the vector $\vec{v}=\partial A/\partial\dot{\vec{x}}$ it has components \begin{aligned} v_{k}&=\frac{\partial A}{\partial \dot{x}^{k}}\\ &=\frac{\partial M_{ij}(x)}{\partial\dot{x}^{k}}\dot{x}^{i}\dot{x}^{j} +M_{ij}(x)\frac{\partial\dot{x}^{i}}{\partial\dot{x}^{k}}\dot{x}^{j} +M_{ij}(x)\dot{x}^{i}\frac{\partial\dot{x}^{j}}{\partial\dot{x}^{k}}\end{aligned} (where we use the product rule), and we have $$\frac{\partial\dot{x}^{j}}{\partial\dot{x}^{k}}={\delta^{j}}_{k}$$ be the Kronecker delta, in abstract index notation. Thus we have our components be \begin{aligned} v_{k}&=\frac{\partial M_{ij}(x)}{\partial\dot{x}^{k}}\dot{x}^{i}\dot{x}^{j} +M_{ij}(x)\frac{\partial\dot{x}^{i}}{\partial\dot{x}^{k}}\dot{x}^{j} +M_{ij}(x)\dot{x}^{i}\frac{\partial\dot{x}^{j}}{\partial\dot{x}^{k}}\\ &=0+M_{ij}(x){\delta^{i}}_{k}\dot{x}^{j}+M_{ij}(x)\dot{x}^{i}{\delta^{j}}_{k}\\ &=M_{kj}(x)\dot{x}^{j}+M_{ik}(x)\dot{x}^{i}. \end{aligned} Now we use the fact that $M$ is a symmetric matrix to rewrite this as $$M_{kj}(x)\dot{x}^{j}+M_{ik}(x)\dot{x}^{i}=M_{jk}(x)\dot{x}^{j}+M_{ik}(x)\dot{x}^{i}.$$ Then we use the fact that we can rewrite dummy indices as we please, $$M_{jk}(x)\dot{x}^{j}=M_{ik}(x)\dot{x}^{i}.$$ So we get Valentin's result $$v_{k} = 2M_{ik}x^{i}$$ where we sum over $i$, using Einstein's convention.

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Thank you very much for the detailed explanation! – Simon Bird Jul 9 '12 at 8:32
One last warning: mathematicians often use "Einstein's summation convention" sloppily and ambiguously, as strange as it sounds they abuse language and sometimes call "Euclidean summation convention" incorrectly Einstein summation. So be careful! I know representation theory commits this crime a lot. But it's a misdemeanour, not a felony... – Alex Nelson Jul 9 '12 at 17:42

$\displaystyle \frac{\partial}{\partial \vec{v}} F$ is a shorthand notation for a vector whose components are $\displaystyle \frac{\partial}{\partial v^j}F$. It is extensively used in calculus of variations in the multivariable case for writing out Euler-Lagrange equations, or in the proof of Noether's theorem, for example. In your case, apparently $$\frac{\partial A}{\partial \vec{x}}=\frac{\partial M_{ij}(\vec{x})}{\partial \vec{x}}\dot{x}^i\dot{x}^j$$ $$\frac{\partial A}{\partial \dot{\vec{x}}}=2M_{ij}\dot{x}^i$$

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Thank you, Valentin! – Simon Bird Jul 9 '12 at 8:40