Fourier Tranform of electric dipole charge density

  • Thread starter naftali
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  • #1
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Homework Statement


Hi,

This is supposed to be simple, so I guess I miss something..
We have charge q at x1=d*cos (w*t), y=0, z=0. and charge -q at
x2=-d*cos (w*t), y=0, z=0. I need to do fourier transform to the charge density.

Homework Equations


The fourier transform is : [itex]\rho_{\omega} = \int \rho (r,t)e^{i\omega t} dt [/itex]


The Attempt at a Solution


I first try only for the first charge. We have [itex]\rho (r,t) = \delta(x-d\cdot cos (\omega t))\cdot\delta(y)\cdot\delta(z) [/itex]
But I don't know how to do the intergral :
[itex]\rho_{\omega} = \int \delta(x-d\cdot cos (\omega t))\cdot\delta(y)\cdot\delta(z)e^{i\omega t} dt [/itex]
Any ideas? Do I really need to do this integral?
 

Answers and Replies

  • #2
Mute
Homework Helper
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Generally, if you have an integral over a delta function of another function, you can use the relation

[tex]\delta(f(t)) = \sum_{j}\frac{\delta(t-t^\ast_j)}{|f'(t^\ast_j)|},[/tex]
where the [itex]t^\ast_j[/itex] are the solutions of [itex]f(t^\ast) = 0 [/itex]. Note that this relation doesn't work if the derivative is zero at a solution.
 
  • #3
31
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Hi,

Thanks for your answer.
Actually the question is to find the potential of this charges far from the origin,so I guess I should transform the dipole moment vector only..
Using the identity for the delta function I get weird things like exp(arccos (x/d))...
It is just interesting if the transform of the charge density could be useful.
 

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