Momentum in Electromagnetic Wave and Radiation Pressure

Here we investigate on the case that the electromagnetic waves carry energy. We can express the energy carried by the electromagnetic waves in terms of the momentum; in other words, the electromagnetic waves carry momentum. With the rate of change of momentum we can determine the radiation pressure; you may have experienced this when you're out to feel the sun in winter.

Figure 1 A charge (an electron) under the influence of electromagnetic wave.

We introduced Lorentz's force in Maxwell's equations which determines the total force on a charge due to both magnetic and electric fields; both electric and magnetic forces act on the charge and therefore their sum is the total force. For example, both forces act on a charge in Figure 1 above. The total force (Lorentz's force) is

\[F = q\vec E + q \,\vec v \times \vec B \]

In Figure 1 you can see that the case of an electromagnetic wave travelling in positive x-direction strikes a material surface of electrons bound to the ions. The electric field exerts a force on the charge once it encounters a charge in its path. You know that the magnetic field does not work on stationary charges but as soon as the electric field does its job (set charge into motion if it was not already in motion), the magnetic field hops in.

If you consider an electron, the force due to the electric field in one half cycle is downwards (towards -y-direction) as shown in Figure 1. In another half cycle, the electric field is downwards and the force is upwards. Therefore, the net force due to the electric field in the y-direction is zero.

We emphasized the fact that an electromagnetic wave carries momentum and hence the radiation pressure (momentum results radiation pressure). But you just saw that the net force in y-direction is zero and therefore the electric field is not responsible for providing the momentum to the electron. Then what does?

Let's look at the magnetic field. Since the magnetic force is always perpendicular to the direction of velocity of the moving charge, and the magnetic force the electron is always in +x-direction, therefore the magnetic field provides the net force on the electron responsible for the momentum or the momentum flow (force is the rate of change of momentum).

Since the magnetic force is always perpendicular to the velocity $\vec v$ of the electron, it never does any work on it. Let's look at the magnetic force $F_B$ on the electron here.

\[F_{Bx} = |q \vec v_y \times \vec B_z| = |q| v_y\,B_z \tag{1} \label{1}\]

You know that $c = E_y/B_z$ and you can rewrite the above equation if the charge is the charge on an electron as

\[F_{Bx} = \frac{eE_y\,v_y}{c} = \frac{F_{Ey}\,v_y}{c} \tag{2} \label{2}\]

where $eE_y$ is the force on the electron due to the electric field. You know from the Newton's second law that the force is the rate of change of momentum. So, the magnetic force on the electron is the same thing as the momentum transfer. If you look at the expression $F_{Ey}\,v_y$, it is the work done per unit time called power or the rate of work done by the electric field.

It means the above equation suggests that the momentum absorbed is equal to the energy per unit time (power) divided by the speed of light. The momentum absorbed on the surface due to the magnetic field is equal to the rate of work done due to the electric field.

Let's consider that a plane electromagnetic radiation strikes a material surface of area $A$ perpendicular to the direction of propagation of the radiation. It time $dt$, the distance the electromagnetic radiation travels is $c\,dt$ and at the same time interval, the volume of $dV = Ac\,dt$ of electromagnetic radiation strikes the surface.

You can turn Equation \eqref{2} if you divide both sides by $dV$ into

\[\frac{dp}{dt\,A} = P = \frac{S}{c} \tag{3} \label{3}\]

where $S$ is the magnitude of the Poyinting vector. The rate of change of momentum is $dp/dt$ which is the rate at which the momentum is transferred to the absorbing material and this momentum change provides the radiation pressure. Here we initially (without explicitly saying) assumed that, the entire momentum is absorbed in the surface, but if the surface is a perfect reflector and for the normal incidence, the radiation pressure is twice that given by Equation $\eqref {3}$, that is $P = 2S/c$.

If you consider a total absorbance case, the average radiation pressure, $P_\text{av}$ is the average Poyinting vector divided by the speed of light , that is $P_\text{av} = S_\text{av} / c = I/c$ where $I$ is the intensity of the radiation.

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