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Flux in different coordinate systems

  1. Nov 12, 2012 #1

    Wox

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    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 [itex]f(\phi,\theta)[/itex] 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 [itex]\alpha[/itex] the (known) angle between detector surface and Poynting vector and [itex]XY[/itex] the detector plane in the detector reference frame [itex]XYZ[/itex] for which we know the relation to the [itex]\bar{P}[/itex] 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 [itex]\bar{P}[/itex]
    $$\begin{bmatrix}x'\\y'\\z'\\1 \end{bmatrix} =L\cdot\begin{bmatrix}x\\y\\z\\1\end{bmatrix}$$
    where [itex]L[/itex] 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 [itex]XY[/itex]. 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?
     
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  3. Nov 15, 2012 #2

    chiro

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    Hey Wox.

    Have you encountered unfixed co-ordinate system mathematics and calculus through tensors?
     
  4. Nov 16, 2012 #3

    Wox

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    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 [itex]V[/itex] and [itex]W[/itex] both over field [itex]K[/itex] is a pair [itex](T,\otimes)[/itex] where [itex]T[/itex] a vector space over [itex]K[/itex] and [itex]\otimes\colon V\times W\rightarrow T[/itex] a bilinear map with the property that for any bilinear map [itex]B_{L}\colon V\times W\rightarrow X[/itex] with [itex]X[/itex] a vector space over [itex]K[/itex], there exists a unique linear map [itex]F_{\otimes}\colon T\rightarrow X[/itex] so that [itex]B_{L}=F_{\otimes}\circ\otimes[/itex]. Furthermore if [itex](T,\otimes)[/itex] and [itex](T',\otimes')[/itex] are two tensor products of [itex]V[/itex] and [itex]W[/itex] then there exists a unique isomorphism [itex]F\colon T\rightarrow T'[/itex] such that [itex]\otimes'=F\circ\otimes[/itex]. 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.
     
  5. Nov 16, 2012 #4

    chiro

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    The application physically is widespread:

    http://en.wikipedia.org/wiki/Application_of_tensor_theory_in_physics

    It has to do with what you are saying, but basically you have identities that deal with things like del, grad, differentiation and integration in arbitrary co-ordinate systems by relating them back to Euclidean since we have a developed theory in R^n.

    The isomorphism is the best attribute that captures the description but instead of just thinking about vector spaces, add the results of calculus to that and you have a good idea of how its used.
     
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