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Assume a point's position is given by the coordinates $x_i$. Introducing a new set of coordinates $\Theta_i$, one can relate the differentials $d\mathbf{x}=(dx_1, dx_2, dx_3)$ and $d\mathbf{\Theta}=(d\Theta_1, d\Theta_2, d\Theta_3)$ via the Jacobian $J$:

$d\mathbf{x} = J d\mathbf{\Theta}$

or, the inverse relation

$d\mathbf{\Theta} =J' d\mathbf{x}$

where $J_{ij} = \partial x_i/\partial \Theta_j$ and $J'_{jk} = \partial \Theta_j/\partial x_k$. Due to symmetry, I would assume that $J'$ is the inverse of $J$. When I compute their matrix product, however, this is what I get:

$(JJ')_{ik} = J_{ij}J'_{jk} =\frac{\partial x_i}{\partial \Theta_j} \frac{\partial \Theta_j}{\partial x_k} = 3 \frac{\partial x_i}{\partial x_k} = 3 \delta_{ik}$

which does not seem like the identity matrix since $I_{ik} = \delta_{ik}$. Am I missing something obvious here or making some algebra mistakes?

Edit: I think my question is related to this one. There, in the first answer the $k$-sum disappears and the result is $\delta_{ij}$, but I can't see how.

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    $\begingroup$ This might be better for Math SE. $\endgroup$
    – Charlie
    Apr 15, 2020 at 21:43

2 Answers 2

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You've gotten a little confused with the index notation. I find that writing it out in full always helps when it gets too obscure to handle. In your case, you have that $$ \frac{\partial x_i}{\partial \Theta_j}\frac{\partial \Theta_j}{\partial x_k}=\sum_{j=1}^3 \frac{\partial x_i}{\partial \Theta_j}\frac{\partial \Theta_j}{\partial x_k} = \frac{\partial x_i}{\partial \Theta_1}\frac{\partial \Theta_1}{\partial x_k} +\frac{\partial x_i}{\partial \Theta_2}\frac{\partial \Theta_2}{\partial x_k}+\frac{\partial x_i}{\partial \Theta_3}\frac{\partial \Theta_3}{\partial x_k} $$ but this last expression is simply the chain rule! That is, differentiating $x_i$ with respect to $x_k$ gives $$ \frac{\partial x_i}{\partial x_k} = \frac{\partial x_i}{\partial \Theta_1}\frac{\partial \Theta_1}{\partial x_k} +\frac{\partial x_i}{\partial \Theta_2}\frac{\partial \Theta_2}{\partial x_k}+\frac{\partial x_i}{\partial \Theta_3}\frac{\partial \Theta_3}{\partial x_k} $$ so $$ \frac{\partial x_i}{\partial \Theta_j}\frac{\partial \Theta_j}{\partial x_k}=\frac{\partial x_i}{\partial x_k}. $$

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    $\begingroup$ how embarrassing, messing up the chain rule :) thanks a lot! $\endgroup$
    – Botond
    Apr 15, 2020 at 22:25
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Using the Einstein summation convention (leaving out the $\Sigma$s, but summing over repeated indices)

We have $$dx^i=\frac{\partial x^i}{\partial \Theta^j} d\Theta^j$$

But $$d\Theta^j=\frac{\partial \Theta^j}{\partial x^k} dx^k$$

So $$dx^i=\frac{\partial x^i}{\partial \Theta^j} \frac{\partial \Theta^j}{\partial x^k} dx^k$$ But $x^i$ and $x^k$ are independent variable unless $i=k$ so $$dx^i=\delta^i_k dx^k$$ and therefore $$\frac{\partial x^i}{\partial \Theta^j} \frac{\partial \Theta^j}{\partial x^k}=\delta^i_k$$ Thus the Jacobian matrices are indeed inverses.

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