Problem: Cartesian tensor to Spherical Tensor

  • Thread starter gaffareee
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  • #1
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Dear Frnds,

my problem is very simple. please help me.

I have a dielectric tensor like:

epsilon_T=
|1 0 0|
|0 1 0| %% note that it is in cartesian co-ordinate system
|0 0 a|

now the potential equation of dipole should be as in spherical system

V=1/(4*pi*epsilon_T (?) *epsilon_0)*(qdcos(theta)/R)

now should i change the epsilon_T value cartesian to spherical system ?

or how i can write down the code ?

should it be V_xx, V_yy and V_zz?

please hurry.
 

Answers and Replies

  • #2
fzero
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now the potential equation of dipole should be as in spherical system

V=1/(4*pi*epsilon_T (?) *epsilon_0)*(qdcos(theta)/R)
This equation doesn't make much sense. On the LHS you have a scalar, while on the RHS you seem to have an uncontracted tensor in the denominator.

I believe that you should have an expression for the potential (electric or energy, it doesn't matter) that is proportional to

[tex]\sum_{ij} (\epsilon_T)_{ij} E_i r_j.[/tex]

As a scalar, this quantity can be evaluated in Cartesian coordinates and then written in terms of spherical coordinates, which would avoid having to convert vectors to spherical coordinates.
 
  • #3
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You haven't said what type tensor it is.
 
  • #4
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First write down the D field. For example, for a single point charge the D field is
[tex]\vec{D} = q/(4\pi r^2) \hat{r}[/tex]

Then use the relationship [tex]\vec{D} = \epsilon \vec{E}[/tex] to get the E field (you have to invert your dielectric tensor).

Then perform the integral to calculate the potential from the E field. The potential will still be a scalar but I think you will find that instead of 1/r you will have something like 1/sqrt(x^2+y^2+(az)^2). But you have to check.
 

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