Can someone explain the classical angular momentum in electromagnetic theory of light? If I shine elliptically polarised em wave on a black disc it rotates. I would like to know how to calculate torque in classical picture.


2 Answers 2


The Result You Seek

The Wikipedia page on angular momentum of light gives the classical angular momentum as:

$$\frac{\epsilon_0}{2i\omega}\int_{\mathbb{R}^3} \left(\mathbf{E}^\ast\times\mathbf{E}\right)d^{3}\mathbf{r} +\frac{\epsilon_0}{2i\omega}\sum_{i=x,y,z}\int_{\mathbb{R}^3} \left({E^i}^{\ast}\left(\mathbf{r}\times\mathbf{\nabla}\right)E^{i}\right)d^{3}\mathbf{r}\tag{1}$$

when the positive frequency part alone of the fields is kept (hence the complex conjugates).

Angular momentum is a great deal easier to think about when the field vectors are written as the Riemann-Silberstein vectors, which I discuss in my answer here. These are the rank-2, skew-symmetric Faraday tensor in disguise. The first term in (1) is the spin angular momentum, and, rewritten in positive frequency Riemann-Silberstein vectors when everything is roughly paraxial (i.e. near to a plane wave) it reads:

$$\hat{\mathbf{z}}\frac{1}{\omega}\int \left(|\mathbf{F}_+|^2-|\mathbf{F}_-|^2\right)d^{3}\mathbf{r}$$

i.e. $\frac{1}{\omega}$ times the right polarized energy density less the left polarized energy density in the direction of propagation of the light. Orbital angular momentum vanishes in the paraxial limit and so the last equation is the total angular momentum in this case.

What This Means and Sketch of How It Is Derived

It is important to recall how this equation is derived: one imagines an electromagnetic field crossing the boundary into a conductive medium and being absorbed there. In general, the currents induced by the incident field feel a torque. One then calculates the moment of the Lorentz force on the currents in the medium to calculate the angular impulse exerted on the medium, exactly analogously with method 3 of the linear momentum / impulse calculation in my answer here. The angular momentum density $\left(|\mathbf{F}_+|^2-|\mathbf{F}_-|^2\right)/\omega$ calculated from this most basic (in the sense of fundamental) Newtonian-Maxwell physics is the difference between the intensities of the circularly polarized base states. This calculation says that the right and left circularly polarized components transfer angular momentum $\pm E/\omega$ in the direction of light propagation, respectively, whenever energy $E$ is absorbed. So now we see that, if the photon has energy $h\nu$, then if a high number of them are to transfer the same angular momentum as classical physics reckons, the photon's angular momentum has to be $\pm h\nu/\omega$ or $\pm \hbar$ in the direction of its propagation for right and left circularly polarized photons, respectively.


A classic paper here, which actually measures the torque exerted on a quarter wave plate (which takes linear polarised, i.e. zero angular momentum, light in and outputs circularly polarised light). The paper, as mentioned in the comments, is:

R. Beth, "Mechanical Detection and Measurement of the Angular Momentum of Light", Phys. Rev. 50 1936 pp115-127

The torque exerted on a quarter wave plate when polarised light (any polarisation, as long as we have a pure polarisation state) is shone through it is $\frac{P}{\omega}$, where $P$ is the power of the light beam.

  • $\begingroup$ Technically Beth measured the angular momentum transferred to a half-wave plate, but that's just two quarter-wave plates stuck together. $\endgroup$
    – rob
    Oct 26, 2014 at 15:15
  • $\begingroup$ Note however that a circularly polarised plane wave does not carry angular momentum along its propagation direction if the Poynting vector is used. As $\bf P = \epsilon_0 \bf E \times \bf B$ is parallel to the propagation direction, $\bf J = \bf r \times \bf J$ is perpendicular to it or vanishes. This is a paradox. $\endgroup$
    – my2cts
    Jun 9, 2021 at 13:12
  • $\begingroup$ @my2cts I believe you're talking about the orbital angular momentum here, which is the $\bf{r}\times\bf{p}$, equivalent to the second term in my equation (1). A circularly polarized wave indeed bears no orbital angular momentum, but does bear spin (intrinsic) angular momentum. $\endgroup$ Jun 10, 2021 at 13:20
  • $\begingroup$ @SeleneRoutley Actually when the gauge invariant formalism is used, there is no spin angular momentum. All of the angular momentum is expressed by $\bf J = \bf r \times \bf P$. $\endgroup$
    – my2cts
    Jun 10, 2021 at 13:23
  • $\begingroup$ Edited comment: Note however that a circularly polarised plane wave does not carry angular momentum along its propagation direction if the Poynting vector is used. As $$\bf P = \epsilon_0 \bf E \times \bf B$$ is parallel to the propagation direction, $$\bf J = \bf r\times \bf P$$ is perpendicular to it or vanishes. This is a paradox. $\endgroup$
    – my2cts
    Jun 10, 2021 at 13:26

Electromagnetic field of light has two kinds of angular momentum: spin angular momentum (SAM) and orbital angular momentum (OAM). The former represents the dynamical rotation of the electric (or magnetic) field around the propagation direction and indicates the polarization of the beam. The latter represents the rotation of light around the beam axis. The verification of these internal angular momentum can be see in light matter interaction, where the total angular momentum of systems remain conserved.


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