How can we motivate that electromagnetic fields carry momentum and angular momentum?

Question

When I was an undergraduate student, I was comfortable with the concepts of momentum and angular momentum of a particle or of a system of particles. However, the concept of electromagnetic fields carrying momentum and angular momentum occurred to me as somewhat odd to digest. I don't know if everyone feels the same. What could be a compelling way of introducing the concept of momentum and angular momentum of fields, both at undergraduate and graduate-level courses on classical electrodynamics, so that it becomes natural? Of course, without introducing the particle picture of light i.e. photons.

Answer 1

I found C. E. Mungan’s explanation of radiation pressure convincing (presented as an exercise in Griffiths’ Introduction to Electrodynamics). You can extend it by looking at the effect of a circularly polarized EM mode on a test charge. It shows that there is injected energy, momentum and angular momentum, which must come from the field.

To make things concrete, let $r$ be the position of the particle, $m$ its mass, $q$ its charge, $\gamma$ the damping factor, $\omega_0$ the proper frequency of oscillator, $E,B$ the EM fields. In the long wavelength limit (which justifies that the fields are not applied at $r$ and that you can neglect the magnetic force): $$ \ddot r+\gamma \dot r+\omega_0^2r=\frac{q}{m}E(x=0) $$ You can now solve the linear equation and in the stationary regime, you’ll have momentum, angular momentum and energy balance.

Let the angular frequency be $\omega$ and wave vector be $k$ so that: $$ B=\frac{1}{\omega}k\times E $$ and choose cartesian coordinates $x,y,z$ such that $k$ is along $z$ so $E$ and $r$ are in the $xy$ plane. Let $\pm \pi/2$ be the phase difference between $E_x,E_y$ so that RCP corresponds to $-\pi/2$ and LCP to $+\pi/2$. Let $E_0$ be the amplitude of the mode so that $\langle E_x^2+E_y^2\rangle=E_0^2$.

Taking the average over a period of the oscillation of the field of angular velocity $\omega$, you have the power (injected energy): $$ \begin{align} P &=\langle qE \cdot v\rangle \\ &= \omega\frac{q^2}{m}\frac{\gamma\omega}{(\omega^2-\omega_0^2)^2+(\gamma\omega)^2}E_0^2 \end{align} $$ For the force (injected momentum), the electric force vanishes, so only the magnetic one contributes: $$ \begin{align} F &=\langle qv \times B\rangle \\ &= k\frac{q^2}{m}\frac{\gamma\omega}{(\omega^2-\omega_0^2)^2+(\gamma\omega)^2}E_0^2 \end{align} $$ and the torque (injected angular momentum): $$ \begin{align} T &=\langle r \times qE\rangle \\ &=\pm e_z \frac{q^2}{m}\frac{\gamma\omega}{(\omega^2-\omega_0^2)^2+(\gamma\omega)^2}E_0^2 \end{align} $$ You can intepret the common factor: $$ \frac{q^2}{m}\frac{\gamma\omega}{(\omega^2-\omega_0^2)^2+(\gamma\omega)^2}E_0^2=\hbar \dot N $$ as the rate of absorption of photons. Note that a non vanishing $\gamma$ is crucial for it to be non zero. There is no quantum mechanics, the discussion has been purely classical, but you need to introduce the constant $\hbar$ to match the dimensions. You recover the expression of energy (sorry for the conflict), momentum and angular momentum of the mode: $$ \begin{align} E &= \omega \hbar N \\ P &= k \hbar N \\ L &= \pm e_z \hbar N \end{align} $$ which are the expected relations from field theory (note that this shows that the field has spin $1$).

The benefit of this approach is that the calculations are accessible at an undergraduate level.

Hope this helps.

Answer 2

My answer is rather throwaway.

Light can carry energy. You can invoke photoelectric effect, or just feel it with heat lamps (the Sun is one huge one).

By the Special Theory of Relativity, if light can carry energy, it must also be able to carry momentum, and thus angular momentum.

You can also invoke experiments like Compton scattering and Raman spectroscopy to directly motivate that they carry momentum, this time with the expectation that the wavelength has something to do with it.