# Inverse of the coordinate transform Jacobian

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.

• This might be better for Math SE. Apr 15, 2020 at 21:43

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 $$\frac{\partial x_i}{\partial \Theta_j}\frac{\partial \Theta_j}{\partial x_k}=\frac{\partial x_i}{\partial x_k}.$$
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.