# Inverse Jacobi Matrix in Spherical Coordinates

1. Feb 23, 2013

### Coelum

Dear all,
I am reading R.A. Sharipov's Quick Introduction to Tensor Analysis, and I am stuck on the following issue, on pages 38-39. The text is freely available here: http://arxiv.org/abs/math/0403252. If my understanding is correct, then the Jacobi matrices for the direct and inverse coordinates transformation are inverse of each other (when computed in the same point using the same frame of reference, of course). I want to apply the concept to spherical coordinates.
The spherical coordinates transformation can be defined as follows:
$$\begin{cases} x^1 & = x & = \rho \cos \theta \sin \phi \\ x^2 & = y & = \rho \sin \theta \sin \phi \\ x^3 & = z & = \rho \cos \phi \end{cases}$$
and its inverse is:
$$\begin{cases} y^1 & = \rho & = √x^2+y^2+z^2 \\ y^2 & = \theta & = \arctan(y/x) \\ y^3 & = \phi & = \arccos(z/r) \end{cases}$$
The Jacobi matrices for the two transformations are defined respectively as:
$$\begin{cases} S^i_j & =∂x^i/∂y^j \\ T^i_j & =∂y^i/∂x^j. \end{cases}$$
Switching to matrix notation: if those matrices are inverse to each other, then I should get $ST=I$ where $I$ is the identity matrix.
By applying the definitions, I get the following matrices:
$$S= \begin{pmatrix} \cos\theta\sin\phi & \sin\theta\sin\phi & \cos\phi \\ -\rho\sin\theta\sin\phi & \rho\cos\theta\sin\phi & 0 \\ \rho\cos\theta\cos\phi & \rho\sin\theta\cos\phi & -\rho\sin\phi \end{pmatrix}$$
and
$$T= \begin{pmatrix} x/\rho & y/\rho & z/\rho \\ -y/(r^2-z^2) & x/(r^2-z^2) & 0 \\ -xz/r^2\sqrt{r^2-z^2} & -yz/r^2\sqrt{r^2-z^2} & \sqrt{r^2-z^2}/\rho^2 \end{pmatrix} = \begin{pmatrix} \cos\theta\sin\phi & \sin\theta\sin\phi & \cos\phi \\ -\sin\theta/\rho\sin\phi & \cos\theta/\rho\sin\phi & 0 \\ -\cos\theta\cos\phi/\rho & -\sin\theta\cos\phi/\rho & \sin\phi/\rho \end{pmatrix}.$$
Each row is the gradient of a component of the vector transformation. Clearly, $ST\neq I$. However, $ST^t=I$. Why? What's wrong with my reasoning above?

2. Feb 23, 2013

### xaos

maybe this will help,

consider just the identity X=X(P(X,Y,Z),Q(X,Y,Z),R(X,Y,Z)).

differentiating with respect to X and applying the chain rule we get:

dX/dX=1
=(dX/dP)*(dP/dX)+(dX/dQ)*(dQ/dX)+dX/dR)*(dR/dX)
=row[dX/dP, dX/dQ, dX/dR] x col[dP/dX, dQ/dX, dR/dX]

this is the term first row first column of the product matrix. continue with dX/dY and dX/dZ for the first row, now collect the row terms into the rows of a matrix, and the column terms into the columns of a matrix to form the matrix product.

but by your construction you have for the first row first column without transpose:

row[dX/dP, dY/dP, dZ/dP] x col[dP/dX, dQ/dX, dR/dX]

which doesn't satisfy the chain rule. so you really need the transpose.

i hope this helps.

3. Feb 24, 2013

### Coelum

Xaos,
thanks - I agree with your analysis.

Hence, I wonder if, given the above definitions for $T$ and $S$, the statement $T=S^{-1}$ shouldn't be replaced by $T^T=S^{-1}$.

I ask since I am new to tensor calculus and I want to be sure that I am not missing some important point.

4. Feb 25, 2013

### xaos

perhaps the inverse jacobian is acting on the dual tangent space, which is a row vector representation. so transposing converts it back to acting on a column vector?

5. Feb 27, 2013

### Coelum

I understand your remark from an algebraic point of view but not from a geometric point of view: why should the spherical coordinates define covectors? The base associated to the spherical coordinates is not the dual of the original cartesian base (which is the dual of itself).
Any idea?

6. Feb 27, 2013

### xaos

since the jacobian is generally defined locally, you can certainly attach a cotangent space to the points of the submanifold in place of the tangent space. in this case, the submanifold is an inverse spherical coordinate system, which is just a spherical coordinate system in reverse (within a region which makes them 1-1). it's weird, you're in R3, and then you attach all of R3 to a point in R3 for the intersection of three surfaces. now define three linear functions of R3 into R at that point and you have a cotangent space.

that the inverse jacobian acts on the dual tangent vectors makes sense by the transformation identity into primed coordinates:

(d/dx'_i) *(dx'_j) = (d/dx_n) * (dx_m) * dx'_n/dx_i * dx_m/dx'_j =δ_ij

but the transpose is not easily seen.

this calculation makes more sense in terms of an inner product on a sphere of radius r:
<X',X'> = <JX,SX> = <X,X>=r^2 and J^t *S =I, and the transpose is made explicit. this is what you need for SO(3) geometry.

7. Aug 18, 2013

### berra

It seems to me you have written your jacobi matrix wrong. The i'th row <=> coordinate function x_i, the j'th column <=> partial derivative with respect to argument j.