- #1
Kostik
- 274
- 32
- TL;DR
- An infinite circularly-polarized plane wave transports no angular momentum in the direction of propagation, but a wave of finite extent does do edge effects. How to calculate the electric and magnetic fields in such a wave packet?
In an infinite plane wave propagating in the ##z## direction, the momentum density is ##\mathbf{p}=(4π)^{-1}(\mathbf{E} × \mathbf{B})## which points in the ##z## direction; therefore, the angular momentum density about the ##z##-axis ##\mathbf{L} = \mathbf{r} × \mathbf{p}## has no ##z##-component. (Naturally, the total angular momentum in any volume which is axially symmetric about the ##z##-axis is zero. In a volume element ##dV## located a distance ##\mathbf{r}## from the ##z##-axis, the angular momentum density ##\mathbf{L} = \mathbf{r} × \mathbf{p}## is orbital angular momentum, as it arises purely from the linear momentum ##\mathbf{p}##.) Thus, even a circularly polarized wave, being a sum of linearly polarized waves, has no ##z##-angular momentum density, and therefore does not exert a torque on charged matter.
Now consider a plane wave such as $$\mathbf{E} = E_0 \cos(\omega t - kz) \hat{\bf{x}}, \quad\quad
\mathbf{B} = E_0 \cos(\omega t - kz) \hat{\bf{y}} $$ of finite transverse extent, occupying a cylinder of radius ##R## around the ##z##-axis. In the plane ##z=0##, the electric and magnetic fields point in the ##\hat{\bf{x}}## and ##\hat{\bf{y}}## directions, but approaching the wave border at ##r=(x^2+y^2 )^2=R## the field strength rapidly drops to zero.
Suppose the fields have constant magnitude when ##r≤R-ε##, attenuate in the annulus ##R-ε<r<R##, and vanish when ##r>R##. In the plane ##z=2\pi/\omega## the electric and magnetic fields point in the ##-\hat{\bf{x}}## and ##-\hat{\bf{y}}## directions. Therefore, the field lines in the planes ##z=0## and ##z=2\pi/\omega## must connect with each other in the region ##0<z<2\pi/\omega## to form closed loops. (The same is true of all pairs of planes ##z=z_0## and ##z=z_0+2\pi/\omega##.) The exact shape of the field lines depends upon how the field strength is attenuated near the wave boundary, but the ##\mathbf{E}## and ##\mathbf{B}## fields will have a ##z##-component in the region ##R-ε<r<R##.
Thus, there will be some transverse energy flux. In a circularly polarized wave, the field vectors have constant magnitude and rotation, so the transverse energy flux in the annular region will be constant in magnitude and orientation (clockwise or counterclockwise) for a given ##r## in the annulus. As a result, a circularly polarized plane wave of finite extent transports angular momentum in the direction of propagation, owing to the transverse energy flow near its edge. Such a wave will possess spin angular momentum.
Jackson's "Electrodynamics" (2nd edition), p. 333, problem 7.20, says that the electric field in such a circularly polarized wave is $$\mathbf{E}(x,y,z,t) = \left[ E_0(x,y) ( \hat{\bf{x}} \pm i\hat{\bf{y}} ) + \frac{i}{k}\left( \frac{\partial E_0}{\partial x} \pm i\frac{\partial E_0}{\partial y} \right) \hat{\bf{z}} \right] e^{-i(\omega t - kz)}, \quad\quad \mathbf{B} = \mp i\mathbf{E} .$$ Setting ##E_0(x,y)=## constant, the ##\hat{\bf{z}}## term vanishes, and we are left with $$\mathbf{E}(x,y,z,t) = E_0(x,y) ( \hat{\bf{x}} \pm i\hat{\bf{y}} ) e^{-i(\omega t - kz)} $$ which is the correct expression for right- and left- circularly polarized plane waves. (We always take the real part and discard the imaginary part to obtain the physical wave.)
How does one obtain Jackson's fields? We have no charges or currents, so only Maxwell's homogeneous equations apply. The issue, of course, is the border area of the wave with finite transverse width, where ##E_0(x,y)## tapers off from some constant value throughout the "middle" of the wave quickly to zero.
This is an important idea, because the finite wave gives meaning to the "spin" of a circularly polarized wave, with a direct analogy to the spin of the photons which comprise the wave.
Thanks for an assist here.
Edit: I think maybe it's as simple as calculating ##F_{\mu\nu} = A_{\mu,\nu}-A_{\nu,\mu}## and picking off the ##E## and ##B## field components.
Now consider a plane wave such as $$\mathbf{E} = E_0 \cos(\omega t - kz) \hat{\bf{x}}, \quad\quad
\mathbf{B} = E_0 \cos(\omega t - kz) \hat{\bf{y}} $$ of finite transverse extent, occupying a cylinder of radius ##R## around the ##z##-axis. In the plane ##z=0##, the electric and magnetic fields point in the ##\hat{\bf{x}}## and ##\hat{\bf{y}}## directions, but approaching the wave border at ##r=(x^2+y^2 )^2=R## the field strength rapidly drops to zero.
Suppose the fields have constant magnitude when ##r≤R-ε##, attenuate in the annulus ##R-ε<r<R##, and vanish when ##r>R##. In the plane ##z=2\pi/\omega## the electric and magnetic fields point in the ##-\hat{\bf{x}}## and ##-\hat{\bf{y}}## directions. Therefore, the field lines in the planes ##z=0## and ##z=2\pi/\omega## must connect with each other in the region ##0<z<2\pi/\omega## to form closed loops. (The same is true of all pairs of planes ##z=z_0## and ##z=z_0+2\pi/\omega##.) The exact shape of the field lines depends upon how the field strength is attenuated near the wave boundary, but the ##\mathbf{E}## and ##\mathbf{B}## fields will have a ##z##-component in the region ##R-ε<r<R##.
Thus, there will be some transverse energy flux. In a circularly polarized wave, the field vectors have constant magnitude and rotation, so the transverse energy flux in the annular region will be constant in magnitude and orientation (clockwise or counterclockwise) for a given ##r## in the annulus. As a result, a circularly polarized plane wave of finite extent transports angular momentum in the direction of propagation, owing to the transverse energy flow near its edge. Such a wave will possess spin angular momentum.
Jackson's "Electrodynamics" (2nd edition), p. 333, problem 7.20, says that the electric field in such a circularly polarized wave is $$\mathbf{E}(x,y,z,t) = \left[ E_0(x,y) ( \hat{\bf{x}} \pm i\hat{\bf{y}} ) + \frac{i}{k}\left( \frac{\partial E_0}{\partial x} \pm i\frac{\partial E_0}{\partial y} \right) \hat{\bf{z}} \right] e^{-i(\omega t - kz)}, \quad\quad \mathbf{B} = \mp i\mathbf{E} .$$ Setting ##E_0(x,y)=## constant, the ##\hat{\bf{z}}## term vanishes, and we are left with $$\mathbf{E}(x,y,z,t) = E_0(x,y) ( \hat{\bf{x}} \pm i\hat{\bf{y}} ) e^{-i(\omega t - kz)} $$ which is the correct expression for right- and left- circularly polarized plane waves. (We always take the real part and discard the imaginary part to obtain the physical wave.)
How does one obtain Jackson's fields? We have no charges or currents, so only Maxwell's homogeneous equations apply. The issue, of course, is the border area of the wave with finite transverse width, where ##E_0(x,y)## tapers off from some constant value throughout the "middle" of the wave quickly to zero.
This is an important idea, because the finite wave gives meaning to the "spin" of a circularly polarized wave, with a direct analogy to the spin of the photons which comprise the wave.
Thanks for an assist here.
Edit: I think maybe it's as simple as calculating ##F_{\mu\nu} = A_{\mu,\nu}-A_{\nu,\mu}## and picking off the ##E## and ##B## field components.
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