# Flux in different coordinate systems

1. Nov 12, 2012

### Wox

I have an electromagnetic field with a Poynting vector that has the following form in spherical coordinates:
$$\bar{P}(R,\phi,\theta)=\frac{f(\phi,\theta)}{R^2}\bar{e}_{r}$$
The exact nature of $f(\phi,\theta)$ is not known. Suppose I measure the flux of this vector field by a flat area detector. The pixel values of an image acquired by such a detector are given by
$$I_{\text{pixel}}(x,y)=\iint_{\text{pixel}} \bar{P} \cdot \bar{\text{d}S}=\iint_{\text{pixel}} \left\|\bar{P}(x,y)\right\|\cos(\alpha(x,y))\ \text{d}x\text{d}y$$
where $\alpha$ the (known) angle between detector surface and Poynting vector and $XY$ the detector plane in the detector reference frame $XYZ$ for which we know the relation to the $\bar{P}$ coordinate system. From this I want to know the pixel values as would have been measured when the detector had another orientation and position with respect to the origin of $\bar{P}$
$$\begin{bmatrix}x'\\y'\\z'\\1 \end{bmatrix} =L\cdot\begin{bmatrix}x\\y\\z\\1\end{bmatrix}$$
where $L$ a know composition of rotations and translations or even for a spherical detector. How can I do this? The Jacobian can be used when the transformation only involved the detector plane $XY$. However this is not the case and I can also not use the divergence theorem, since we don't have a closed surface. Any ideas on how to approach this?

2. Nov 15, 2012

### chiro

Hey Wox.

Have you encountered unfixed co-ordinate system mathematics and calculus through tensors?

3. Nov 16, 2012

### Wox

I'm not sure what you mean by "unfixed co-ordinate system mathematics". You mean vector spaces without choosing a basis?

As for tensors, this is what I know: a tensor product of two vector spaces $V$ and $W$ both over field $K$ is a pair $(T,\otimes)$ where $T$ a vector space over $K$ and $\otimes\colon V\times W\rightarrow T$ a bilinear map with the property that for any bilinear map $B_{L}\colon V\times W\rightarrow X$ with $X$ a vector space over $K$, there exists a unique linear map $F_{\otimes}\colon T\rightarrow X$ so that $B_{L}=F_{\otimes}\circ\otimes$. Furthermore if $(T,\otimes)$ and $(T',\otimes')$ are two tensor products of $V$ and $W$ then there exists a unique isomorphism $F\colon T\rightarrow T'$ such that $\otimes'=F\circ\otimes$. Although I understand what all this says (I know vector spaces, bilinear maps, linear maps, bijective linear maps = vector space isomorphisms), I don't really grasp the idea or the practical implications. But any suggestions you have are welcome, if I don't understand I'll try to learn it.

4. Nov 16, 2012