The equation numbers below are the same as they appear in Jackson's Classical Electrodynamics.

If a plane monochromatic electromagnetic radiation is incident on a free particle of charge $e$ and mass $m$, the particle will accelerate and emit radiation. For non-relativistic motion, the instantaneous power radiated per unit solid angle into a polarization state ${\vec\varepsilon}$ is given by $$ \frac{dP}{d\Omega} =\frac{e^2}{4\pi c^3}|\hat{{{\vec\varepsilon^*}}}\cdot\dot{\vec v}|^2 \tag{14.120} $$ where $\dot{\vec v}$ is the acceleration of the charged particle.

  • The first question is why is the polarization vector ${\vec\varepsilon}$ complex conjugated in the equation 14.120?

In the paragraph below equation 14.121, the time average of $|\dot{v}|^2$ is equated to $\frac{1}{2}{\rm Re}[\dot{\vec v}\cdot \dot{\vec v}^*]$ where the acceleration $\dot{\vec v}$ due to the incident electric field $\vec E(\vec x, t)=\vec\varepsilon_0E_0e^{i(\vec k_0\cdot{\vec x}-\omega t)}$ is $$\dot{\vec v}=\vec\varepsilon_0\frac{eE_0}{m}e^{i(\vec k_0\cdot\vec x-\omega t)}.\tag{14.121}$$ Therefore, $$\frac{dP}{d\Omega}=\frac{e^2}{4\pi c^3}|{{{\vec\varepsilon^*}}}\cdot{{\vec\varepsilon_0}}|^2|\dot{v}|^2.$$

Let us calculate $$|\dot{v}|^2=|{\dot{\vec v}}\cdot{\dot{\vec v}}|=\left(\frac{eE_0}{m}\right)^2|e^{2i(\vec k_0\cdot\vec x-\omega t)}|=\left(\frac{eE_0}{m}\right)^2$$ which is time-independent.

  • This brings me to the second question. As per the calculation above, $|\dot{v}|^2$ is time-independent and it will not make much sense to take its time average. What is wrong with my calculation of $|\dot{v}|^2$?
  • $\begingroup$ Why do you claim it doesn't make sense to time average a constant? $\endgroup$ Jun 17, 2021 at 15:50
  • 1
    $\begingroup$ I meant that it is trivial. Sorry about that. What I wanted to say is that if we do it in the way I have presented, we cannot produce Jackson's result. $\endgroup$ Jun 17, 2021 at 15:57

2 Answers 2


The formula (14.120) comes from equation (14.20), wherein Jackson has a footnote that explains the complex conjugation. To quote Jackson:

We always exhibit the polarization explicitly by writing the absolute square of a vector that is proportional to the electric field.

In this case, you are correct in assuming $\left|\vec{\epsilon}^*\cdot\vec{v}\right|=\left|\vec{\epsilon}\cdot\vec{v}\right|$ to not depend on whether or not we choose the complex conjugate of the polarization vector.

The crux is that Jackson sometimes wants to write $\vec{v}$ in a form like (14.121), where it is proportional to the electric field $$\vec{v}=\vec{\epsilon}_0\frac{e E_0}{-i\omega}e^{i(\vec{k}_0\cdot\vec{x}-\omega t)},$$ which is complex! Then, physically, we need to remember that the true field and the true velocity are both given by the real parts of this "analytic signal." Performing linear calculations with the analytic signal is straightforward, but care must be taken when taking squares etc., because $\mathrm{Re}(z^2)\neq \left[\mathrm{Re}(z)\right]^2$ in general.

You yourself have stumbled upon the importance of taking the real part before taking the absolute square. If we take the absolute square first and then look for the real part, as you do, we find no need to take a time average. If we first take the real part of $\dot{\vec{v}}$, then take the absolute square and the time average, we find an answer different by a factor of 2: $$\frac{1}{2}\dot{\vec{v}}^*\cdot \dot{\vec{v}}=\int_{\mathrm{full\,period}}dt \left[\mathrm{Re}(\dot{\vec{v}})\right]^2=\frac{1}{2}\mathrm{Re}(\dot{\vec{v}}^*\cdot \dot{\vec{v}}).$$

The above answers your questions, but leaves the question: why didn't we take the real part of $\vec{\epsilon}$ in the above computation? Alas, the complex notation is again conspiring against us. We know that the overlap of the vector $\vec{\epsilon}$ with itself should be unity, so we need to reinstate the property inner product $\langle \vec{a},\vec{b}\rangle=\vec{a}^* \cdot\vec{b}$ in order to achieve that result. Otherwise, with $\vec{\epsilon}=\vec{\epsilon}_0=\mathbf{x}+i\mathbf{y}$, we could incorrectly find $P=0$. These are all the perils of working with the analytic signal, which is most useful for understanding superpositions, relative phases, and other linear properties of electric fields.


Imaginary ("complex") representation of oscillating quantity (velocity or acceleration) works straightforwadly only in linear equations. Formulae containing other than constant or linear terms require that the actual physical value of the quantity (acceleration, electric field) is put in.

In your case, the formula contains square of the actual acceleration, $|\dot{v}|^2$ is supposed to be square of this actual acceleration. This acceleration $\dot{v}$ is a real number (or real vector), not a square of absolute value of its complex representation (phasor) $\tilde{\dot{v}}$, which is not a real number in general.

So if acceleration is $\dot{v} = a_0 \cos \Omega t$, the correct way to calculate the square of this acceleration is

$$ |\dot{v}|^2 = \dot{v} \cdot \dot{v} = a_0^2 \cos^2\Omega t. $$ Time average of this is

$$ \frac{1}{2}a_0^2 $$ hence the invention and occasional suggestion to use the strange-looking formula for time average

$$ \text{time average}\left(|\dot{v}|^2\right) = \frac{1}{2}\text{Re}(\tilde{\dot{v}}\tilde{\cdot \dot{v}}^*) $$ which gives the same result.

  • $\begingroup$ Thanks, there is also a first part of the question. Any idea about that? $\endgroup$ Jun 17, 2021 at 16:03

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