Is the intensity of electromagnetic radiation emitted by an oscillating dipole proportional to the modulus of the second derivative of dipole moment squared, that is: $$I\propto |\ddot P(\omega)|^2$$

my professor in class wrote this equation but I can't find it, is it right? How do you arrive at this formula from an oscillating electric dipole? I think I have to use retarded potentials but I can't find that equation.


2 Answers 2


Here's a sideways look at it (since the derivation of the power from an oscillating electic dipole is in almost every appropriate electromagnetism textbook). A conventional solution is provided at the foot of my answer.

If we assume that electromagnetic radiation is produced by the dynamics of an electric dipole, we can use physical intuition and dimensional analysis to get that proportionality.

Let us assume that the power radiated is proportional to the relevant physical constants to some power and is also proportional to $r^{-2}$, in order to conserve energy in spheres of different radius, and is proportional to the $m$th time derivative of the electric dipole moment raised to the power $n$. $$I \propto c^{\alpha} (4\pi \epsilon_0)^{\beta} r^{-2} \left(\frac{d^m p}{dt^m}\right)^n$$ But we can guess that $n=2$, because the intensity must be real and positiv, but let's leave it free for the moment.

From there we can do a dimensional analysis by balancing the dependencies on charge, mass, length and time on the left and right hand sides of the proportionality.

The left hand side is a power per unit area. This has dimensions of $M L^2 T^{-3} \times L^{-2} = M^1 L^0 T^{-3}$, where $M$ represents mass, $L$ length and $T$ time.

On the right hand side we can do the same thing, but we also need to introduce the dimension of charge, $Q$, so that dipole moment has dimensions of $Q^1 L^1$.

Going through the other individual terms. $c$ has dimensions $L^1T^{-1}$; from Coulomb's law we can get that $4\pi \epsilon_0$ has dimensions $Q^2/MLT^{-2} L^2 = Q^2 M^{-1} L^{-3} T^2$; $r^{-2}$ has dimensions of $L^{-2}$; and the $m$th time derivative introduces dimensions of $T^{-m}$. Putting this together, the proportionality becomes $$ M^1 L^0 T^{-3} Q^0 = (LT^{-1})^{\alpha} (Q^2M^{-1} L^{-3} T^2)^{\beta} (L^{-2}) (Q^1L^1T^{-m})^n\ ,$$ where we have explicitly assuled the left hand side does not depend on charge at all - i.e. $Q^0$.

Now we equate the powers of the various dimension on each side of the proportionality, since they must balance. Equating powers of $M$ first : $$ 1 = -\beta \ \ \ \rightarrow\ \beta = -1$$ Using $\beta = -1$, equate powers of $Q$ : $$ 0 = -2 + n\ \ \ \rightarrow\ n=2\ ,$$ which is as we suspected earlier. Now using $\beta=-1, n=2$, equate powers of $L$ : $$0 = \alpha +3 -2 +2 \ \ \ \rightarrow \alpha = -3$$ and finally, using $\alpha=-3, \beta=-1, n=2$, equate powers of $T$ : $$ -3 = 3 -2 -2m \ \ \ \rightarrow m = 2\ .$$ Thus $$I \propto c^{-3} (4\pi\epsilon_0)^{-1} r^{-2} \left(\frac{d^2 p}{dt^2}\right)^2 = \frac{ \ddot{p}^2}{4\pi \epsilon_0 c^3 r^2}$$

Conventional solution

Intensity radiated by a sinusoidally oscillating electric dipole of the form $$ p = p_0 \sin \omega t$$ is (and this really is in every textbook on electrodynamics, or see here): $$I = \frac{\omega^4 p_0^2}{32 \pi^2 \epsilon_0 c^3}\frac{\sin^2\theta}{r^2}$$ But from the definition of the dipole we see that $$\ddot{p} = - \omega^2 p_0 \sin\omega t$$ and so $$\ddot{p}^2 = \omega^4 p_0^2 \sin^2 \omega t$$ For high frequency radiaton we would normally take the time average of this ($1/2$) and substitute this into the intensity equation $$ I = \frac{2\langle \ddot{p}^2\rangle}{32 \pi^2 \epsilon_0 c^3}\frac{\sin^2\theta}{r^2} = \frac{\sin^2\theta}{4\pi} \left( \frac{\langle \ddot{p}^2\rangle}{4\pi \epsilon_0 c^3 r^2}\right)\ .$$

  • $\begingroup$ Thank you for the answer, just one question: is $I$ the Poynting vector? $\endgroup$
    – Salmon
    Commented Mar 19, 2022 at 14:09
  • 1
    $\begingroup$ It's the magnitude of the Pointing vector @Salmone $\endgroup$
    – ProfRob
    Commented Mar 19, 2022 at 15:05
  • $\begingroup$ so it is wrong to say that the modulus of $I$ is proportional to the square module of the second derivative of the dipole moment $|\ddot P(\omega)|^2$, it is proportional to $\langle \ddot{p}^2\rangle$ $\endgroup$
    – Salmon
    Commented Mar 19, 2022 at 16:32
  • 1
    $\begingroup$ @Salome what's the difference? And what is the modulus for? When you square a number it becomes positive. The time-average is a simple multiplier (a factor of 0.5 for sinusoidal waves) so doesn;t affect the proportionality. $\endgroup$
    – ProfRob
    Commented Mar 19, 2022 at 16:59
  • 1
    $\begingroup$ @Salome yes by a factor of 0.5 for sinusoidal waves. $\endgroup$
    – ProfRob
    Commented Mar 19, 2022 at 17:34

Look at the Ignatowsky equations, these are "almost" equivalent to that of Maxwell; if you studied antennas then you must have met them in some form or the other, see 1:

$$ \textbf{E} = \int \frac{[\rho]\hat{\textbf{n}}}{R^2}d\textbf{x}' +\frac{1}{c}\int \frac{([\textbf {J}]\cdot \hat{\textbf{n}})\hat{\textbf{n}}+([\textbf {J}] \times \hat{\textbf{n}}) \times \hat{\textbf{n}}}{R^2}d\textbf{x}' + \frac{1}{c^2}\int \frac{([\dot {\textbf {J}}] \times \hat{\textbf{n}}) \times \hat{\textbf{n}}}{R}d\textbf{x}' \tag{1}\label{1}$$

$$\textbf{B} = \frac{1}{c}\int \frac{[\textbf {J}]\times \hat{\textbf{n}}}{R^2}d\textbf{x}' + \frac{1}{c^2}\int \frac{[\dot {\textbf {J}}] \times \hat{\textbf{n}}}{R}d\textbf{x}'\tag{2}\label{2}$$

In the radiation field where $R\to \infty$ you can write the asymptotic field as :

$$ \textbf{E} \approx \frac{1}{Rc^2}\int [([\dot {\textbf {J}}] \times \hat{\textbf{n}}) \times \hat{\textbf{n}}]d\textbf{x}'\tag{3}\label{3}$$

$$\textbf{B} \approx \frac{1}{Rc^2}\int [\dot{\textbf {J}}] \times \hat{\textbf{n}}] \textbf{x}'\tag{4}\label{4}$$

In both $\eqref{3}$ and $\eqref{4}$ the far field depends on the time derivative of the current $\dot{\textbf {J}}$. In an oscillating dipole whose moment is $p(t)=\mathrm{e_0} \ell(t)$ the length of the dipole is $\ell(t)$ and there are charges $+e_0$ and $-e_0$ at its ends. For a harmonically oscillating dipole $\ell = \ell_0 sin (\omega t)$. The corresponding electric current of such a dipole is the time derivative of its moment, $J=\frac{dp}{dt}=\dot p=e_0\ell_0 \omega cos(\omega t)$. Therefore, the second derivative of $p$ that is the 1st derivative of the current that shows up in Ignatowsky's equations: $\dot J = \ddot p$ and for a harmonic oscillation $\dot J = -e_0\ell_0 \omega^2 sin (\omega t)$.


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