# Can radiation pressure do work, if so does the magnetic field participate?

I understand that the magnetic field cannot do work because the Lorentz force is always perpendicular to the displacement. In my optics book(Hecht Ch3) these equations appear:

Electromagnetic pressure, P, is equal to the energy density of a light wave.

$$P = \frac{1}{2}(\epsilon_0E^2 +\frac{1}{\mu_0}B^2)$$

And the force exerted by a beam of radiation is

$$AP = \frac{\Delta p}{\Delta t}$$, where A is the area of an absorbing surface (I assume the answer to my question is the same for a reflecting surface).

Now, there must be something wrong with writing this:

$$W = F \Delta s = AP\Delta s$$, if some kinetic energy is imparted to a massive object by the light.

My confusion has to do with the fact that light seems to do work, but if we calculate it by the pressure we would have to attribute half of the work to the magnetic field. How do we attribute the work done in this case, given that(is this safe to assume about the surface?) there are no charges? (any 'force' due to light has nothing to do with $$q \vec{E}$$ or $$q \vec{v} \times \vec{B}$$). Maybe I should have had the same question about classical waves on a string and the energy they carry?

• The Lorentz force F=q(E+VxB) . I think you mean the VxB magnetic term does no work as this force is perpendicular to velocity V. Nov 9, 2020 at 1:29
• Not sure why you say no charge? Radiation pressure works by acting on movable charge in material, either conduction electrons or on induced dipole charges (which arise from microscopic charges at atomic and molecular levels). Nov 9, 2020 at 1:40

we would have to attribute half of the work to the magnetic field.

This is not so.

First, strictly speaking, due to definition of work, mechanical work is not attributed to a field, it is attributed to some force. This force may be due to a field or something else.

If we are talking about Lorentz force on charged particle, then only electric part of the Lorentz force can do work, the magnetic part cannot.

Just because the energy expression for EM energy has magnetic field in it does not imply magnetic part of the Lorentz force does work on some material body. What it means that when electric force does work on some material body, this work results in decrease of EM energy and this can manifest as change in electric field, or change in magnetic field, or both.

An example of the first process, when parallel plate capacitor discharges through a resistor, Coulomb electric forces of the capacitor work against friction forces in the resistor. This requires the charges from one plate to move out, go through the resistor and end up on the other plate. So this process results in decrease of charge on the capacitor plate and thus decrease of EM energy in the capacitor. This decrease manifests as decrease of electric field between the plates of the capacitor.

An example of the second process, when energized inductor discharges through a resistor, induced electric forces of the inductor work against the friction forces in the resistor. This requires the mobile charges carrying the current to slow down (otherwise the induced field would not appear). So this process results in decrease of electric current and thus decrease of energy in the inductor. This decrease manifests as decrease of magnetic field strength inside the inductor. But the work was done by electric forces, not magnetic forces.

In your example of light pushing on some wall, there are both electric and magnetic fields, and if the wall accepts some energy from them via Maxwell pressure, both electric and magnetic field strength near the wall will decrease. But it is only the electric forces that do work on charged particles. Electric work can decrease both electric and magnetic field that region of space.

Radiation pressure can certainly do work, I've seen a demonstrationin which a specially prepared pinwheel was propelled by a laser. It did so be absorbing (or possibly reflecting) photons, taking their momentum for itself.

You cannot really assume there's no charges, as every matter is composed of charged particles that do interact with electromagnetic waves. If you had some matter that truly has no charge (like dark matter), it would be completely transparent to light, there would be no absorption and no momentum transfer.

Analyzing how electromagnetic wave interracts with matter on a microscopic level is difficult but not impossible. It displaces the elemental charges in the matter (and both electric and magnetic fields take part in it) and the movement of these charges creates a new electromagnetic wave that interferes with the orignal wave. The overall effect depends on the structure of the material and it may be described as refraction, absorption or deflection, among other, more exotic effects. This will affect how much momentum is actually transfered to the material; note for example that deflection transfers twice the amount of momentum than absorption, since it involves emitting a wave with opposite momentum.

Anyway even if only electric field can generate a force that changes the energy of a particle, it doesn't mean that only electric field is depleted. The electromagnetic wave is a dynamical system, and it should be considered as a whole; as its electrical field genrates work, it's total energy decreases, and because of the law of electromagnetism it will mean that both electric and magnetic fields are depleted.

The absorption of photons is a process better described by Quantum Field Theory. When you use Δp to compute the final minus the initial linear momentum carried by the light beam, you are not computing any sort of slowing down of the beam. The Δp arises from the change in the number of wave packets.

We do not know exactly what happens during the destruction of the photon, but we can apply conservation laws to the beginning and end of the photon destruction. That way, the question of the force exerted by the light is reduced to compute the linear momentum carried by the light. That is basically the equation AP=Δp/Δt.

Hence, the real question is 'why is the linear momentum carried by the light equal to its energy?' (Throw some c=1 wherever needed.) The answer is simple: because it moves at the speed of light.

Again, one cannot use many of the classical concepts to the process of absorption (or reflection) of light. In particular, I do not know whether F=qv×B is valid during the destruction of the photon. Since that is information we do not need to apply the conservation laws, we should not care.

The E and B field in a classical elecromagnetic wav is not the same as the static electric and magnetic field, the fields cycle following Maxwell's equations.

The energy density of a light wave can also be given by:

$$η=ε_0Ε^2$$

So when considering interactions with matter it is the electric field that is used for calculations of energy transfer, and that is how radiation pressure is treated, because it can interact with the spill over electric fields of matter.

Of course, as the other answer states, one can use the photon framework , the quantum mechanical one, but usually ,as the classical frame is consistent with the quantum mechanical, one uses the classical unless individual photon interactions are important.