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Electric field and Legendre Polynomials

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    I want to varify that the components of a homogenous electric field in spherical coordinates [itex]\vec{E} = E_r \vec{e}_r + E_{\theta} \vec{e}_{\theta} + E_{\varphi} \vec{e}_{\varphi}[/itex] are given via:

    [itex]E_r = - \sum\limits_{l=0}^\infty (l+1) [a_{l+1}r^l P_{l+1}(cos \theta) - b_l r^{-(l+2)} P_l cos(\theta)][/itex]

    [itex]E_{\theta} = \sum\limits_{l=0}^\infty [a_{l+1}r^l + b_{l+1} r^{-(l+3)}]sin(\theta)P'_{l+1}(cos \theta)[/itex]

    [itex]E_{\varphi} = 0[/itex]

    I have rotational symmetry about the z-axis (azimuthal symmetry).



    2. Relevant equations

    I know that the potential in charge-free space and with azimuthal symmetry can be given via the Legendre Polynomials:

    [itex]\Phi(r, \theta) = \sum\limits_{l=0}^\infty (a_l r^l + b_l r^{-(l+1)}) P_l(cos \theta)[/itex]



    3. The attempt at a solution

    Let's begin with [itex]E_r[/itex].

    [itex]\vec{E} \vec{e}_r = E_r[/itex]

    And:

    [itex]\vec{E} = - \nabla \Phi[/itex]

    So basically what I have to do is apply the gradient (in spherical coordinates) and multiply with [itex]\vec{e}_r[/itex]. In other words: Apply the [itex]\vec{e}_r[/itex] component of the gradient to the potential. Is this correct? If so: How exactly do I apply the gradient to a sum like (2)?
     
  2. jcsd
  3. Jan 18, 2014 #2

    strangerep

    User Avatar
    Science Advisor

    Not quite. First write out the representation of the gradient operator in spherical coordinates and apply it to your ##\Phi##.

    Does that get you any further?
     
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