I'm trying to follow a text I found online. The author decomposes EM fields such

$$ \mathbf{E} = \sum_{lm}\left(f_l(r) \mathbf{Y}_{lm} - i \frac{l(l+1)}{r} g_l(r) \mathbf{\Psi}_{lm} - i\left(\frac{d g_l(r)}{dr} + \frac{g_l(r)}{r} \right) \mathbf{\Phi}_{lm}\right) e^{-i\omega t} $$ $$ \mathbf{B} = \sum_{lm}\left(g_l(r) \mathbf{Y}_{lm} + i \frac{l(l+1)}{r} f_l(r) \mathbf{\Psi}_{lm} + i\left(\frac{d f_l(r)}{dr} + \frac{f_l(r)}{r} \right) \mathbf{\Phi}_{lm}\right)e^{-i\omega t} $$ Where the functions are defined as $$ \mathbf{\Phi}_{lm} = r \mathbf{\hat r} \times \mathbf{\nabla} Y_{lm} $$ $$ \mathbf{\Psi}_{lm} = r \mathbf{\nabla} Y_{lm} $$ $$ \mathbf{Y}_{lm} = Y_{lm} \mathbf{\hat r} $$ where $\mathbf{\hat r}$ is the position unit vector $\mathbf{\hat r} = \mathbf{r}/r$, $Y_{lm}$ are the vector spherical harmonics (the arguments $\theta, \phi$ are omnitted), $f_l(r),g_l(r)$ are "good behaving" radial functions and $\mathbf{\nabla}$ is the gradient.

Now the author states, that using the equations $$ \mathbf{\hat r} \times \mathbf{E} = 0 $$ $$ \mathbf{\hat r} \cdot \mathbf{B} = 0 $$ and by utilising the Spherical harmonic orthogonality we get $$ f_l(r) = 0 $$ $$ \frac{d g_l(r)}{dr} = 0 $$ But this doesn't work for me , I always get $$ \frac{dg_l(r)}{dr} + r g_l(r) = 0 $$ Could someone check the answer with me or prove me wrong? For further reading on the Functions $(\mathbf{Y},\mathbf{\Psi},\mathbf{\Phi})$ please refer to this text.

  • $\begingroup$ Perhaps further elaboration will help motivate someone to answer. The problem is a boundary one, if we consider $f_l(r),g_l(r)$ to be Spherical Bessel/Hankel functions (althought with different coeficients) the equations for E and B represent EM waves. If we denote, now the equations $ \mathbf{n} \times \mathbf{E}= 0$ and $\mathbf{n} \times B$ are the boundary conditions, if we solve the equations we should be able to find the coefficient for transmision/reflection of the waves. (note that we are missing some equations for the boundary, I have not included these as to not confuse readers. $\endgroup$
    – Nitaa a
    Apr 27, 2022 at 13:21
  • 1
    $\begingroup$ You have expressions of the form $\frac{df}{dr}+f $ in your equations, which look dimensionally inconsistent. Are you sure these equations are correct? $\endgroup$
    – Hossein
    Apr 30, 2022 at 16:08
  • $\begingroup$ Thank you, there should be, of course, $\frac{f}{r}$ $\endgroup$
    – Nitaa a
    May 1, 2022 at 13:25

1 Answer 1


Using: $$ \mathbf{\hat r}\times \mathbf \Phi_{lm} = -\mathbf \Psi_{lm} $$ $$ \mathbf{\hat r}\times \mathbf \Psi_{lm} = \mathbf \Phi_{lm} $$ $$ \mathbf{\hat r}\times \mathbf Y_{lm} = 0 $$ I get from $\mathbf{\hat r}\times \mathbf E=0$ at the boundary and the independence of the harmonic functions: $$ \frac{g_l}{r}=\frac{dg_l}{dr}+\frac{g_l}{r}=0 $$ which gives $\frac{dg_l}{dr}=0$.

Note that you second boundary condition, $\mathbf{\hat r}\cdot \mathbf B = 0$ using: $$ \mathbf{\hat r}\cdot \mathbf \Phi_{lm} =\mathbf{\hat r}\cdot \mathbf \Psi_{lm} = 0 $$ $$ \mathbf{\hat r}\cdot \mathbf Y_{lm} = Y_{lm} $$ gives only $g_l=0$ at the boundary.

There is no condition on $f_l$ based on what you've given us. Are you sure you gathered all the information, and you don't have a typo? What is your reference?

Hope this helps and tell me if something's not clear.


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