In the course of studying Thomson scattering, I obtained a current density described by the equation \begin{align} \mathbf{j}(\mathbf{r},t) =& \frac{ e^2\,E_\text{inc}}{ m\,\omega} \, \sin\left(k_o\,z -\omega \,t\right) \\ & \delta\left( x - \frac{ e\,E_\text{inc}} { m\,\omega ^2} \left[ \cos\left(k_o\,z -\omega \,t\right) - \cos\left(k_o\,z \right) \right] \right) \, \delta\left( y \right) \, \delta( z ) \, \hat{\mathbf{x}} . \end{align} Further say that I want to calculate the magnetic vector potential in the radiation zone, $\mathbf{A}_\text{rad}(\mathbf{r},t)$.

It is well known [1] $$ \mathbf{A}_\text{rad} = \frac{\mu_0}{4\,\pi\,r} \int \mathbf{j}\left(\mathbf{r}^\prime, t - \frac{r}{c} + \frac{\hat{\mathbf{r}}\cdot\mathbf{r}^\prime}{c}\right) \,d^3 \mathbf{r}^\prime . $$

So, \begin{equation} \mathbf{j}(\mathbf{r}^\prime,t) = \hat{\mathbf{x}} \, \frac{ e^2\,E_\text{inc}}{ m\,\omega} \, \sin\left(k_o\,z^\prime -\omega \,t\right) \delta\left( x - \frac{ e\,E_\text{inc}} { m\,\omega ^2} \left[ \cos\left(k_o\,z^\prime -\omega \,t\right) - \cos\left(k_o\,z^\prime \right) \right] \right) \, \delta\left( y^\prime \right) \, \delta( z^\prime ) \end{equation} and \begin{equation} \mathbf{j}\left(\mathbf{r}^\prime, t - \frac{r}{c} + \frac{\hat{\mathbf{r}}\cdot\mathbf{r}^\prime}{c} \right) = \hat{\mathbf{x}} \, \frac{ e^2\,E_\text{inc}}{ m\,\omega} \, \sin\left(k_o\,z^\prime -\omega \,\left[t - \frac{r}{c} + \frac{\hat{\mathbf{r}}\cdot\mathbf{r}^\prime}{c}\right] \right) \delta\left( x - \frac{ e\,E_\text{inc}} { m\,\omega ^2} \left[ \cos\left(k_o\,z^\prime -\omega \,\left[t - \frac{r}{c} + \frac{\hat{\mathbf{r}}\cdot\mathbf{r}^\prime}{c}\right] \right) - \cos\left(k_o\,z^\prime \right) \right] \right) \, \delta\left( y^\prime \right) \, \delta( z^\prime ) \end{equation}

I now return to the computation of the magnetic vector potential. I have that $$ \mathbf{A}_\text{rad} (\mathbf{r},t) = \hat{\mathbf{x}} \, \frac{\mu_0}{4\,\pi\,r} \, \frac{ e^2\,E_\text{inc}}{ m\,\omega} \int \sin\left(k_o\,z^\prime -\omega \,\left[t - \frac{r}{c} + \frac{x\,x^\prime+ y\,y^\prime+ z\,z^\prime }{r\,c}\right] \right) \delta\left( x^\prime - \frac{ e\,E_\text{inc}} { m\,\omega ^2} \left[ \cos\left(k_o\,z^\prime -\omega \,\left[t - \frac{r}{c} + \frac{x\,x^\prime+ y\,y^\prime+ z\,z^\prime }{r\,c}\right] \right) - \cos\left(k_o\,z^\prime \right) \right] \right) \, \delta\left( y^\prime \right) \, \delta( z^\prime ) \,d^3 \mathbf{r}^\prime . $$

Owing to the $x^\prime$, $y^\prime$, and $z^\prime$ in the arguments of the delta function, I am not clear on how to proceed.


(1) Have I misunderstood how to compute the magnetic vector potential?

(2) Baring in mind the $x^\prime$, $y^\prime$, and $z^\prime$ in the arguments of the delta function, how do I proceed?

My first attempt

$$ \mathbf{A}_\text{rad} (\mathbf{r},t) = - \frac{\mu_0}{4\,\pi\,r} \, \frac{ e^2\,E_\text{inc}}{ m\,\omega} \int \sin\left( \omega \,\left[t - \frac{r}{c} + \frac{x\,x^\prime }{r\,c}\right] \right) \delta\left( x^\prime - \frac{ e\,E_\text{inc}} { m\,\omega ^2} \left[ \cos\left( \omega \,\left[t - \frac{r}{c} + \frac{x\,x^\prime }{r\,c}\right] \right) - 1\right] \right) \,dx^\prime \,\hat{\mathbf{x}}. $$ Even when I take it this far, I am still stumped. Now, I need to determine when $$ x^\prime - \frac{ e\,E_\text{inc}} { m\,\omega ^2} \cos\left( \omega \,\left[t - \frac{r}{c} + \frac{x\,x^\prime }{r\,c}\right] \right) = -\frac{ e\,E_\text{inc}} { m\,\omega ^2}. $$ This appears to be a transcendental equation. Guidance for how to solve these problems might be obtained by studying the techniques and approaches used in solving the Lienard-Wiechert problem [1,2]

My second attempt

Drop the problematic term all together and settle for $$ \mathbf{A}_\text{rad} (\mathbf{r},t) = \frac{\mu_0\, e^2\,E_\text{inc}}{ 4\,\pi\,r\,m\,\omega} \sin\left( \omega \,t - \frac{\omega \,r}{c} \right) \,\hat{\mathbf{x}}. $$


[1] Zangwill, Modern Electrodynamics, pp 734, 871.

[2] https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential


1 Answer 1


The magnetic vector potential for pure Thomson scattering is just the same as the magnetic vector potential for an ideal oscillating electric dipole.

If you can assume that the wavelength of the incident radiation is much larger than the amplitude of the induced electron oscillation and you are considering a position that is many wavelengths away from the oscillating electron, then your second equation seems correct to me (for incident radiation linearly polarised along the x-axis).

I don't think you have misunderstood anything, it's just that if you aren't making the "electric dipole approximation" (that the retarded time is the same for all points on the source and your current density simplifies accordingly by setting ${\bf r'}=0$), then everything gets significantly more messy.


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