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The average energy density of a electromagnetic wave is given as

$U_{average}$ =$\frac{e_0E^2}{2}$

My textbook also claims that "electromagnetic waves incident on a surface exert a force on the surface" Doesn't the above statement mean that EM waves have "mass", being a wave by itself should also have some "velocity". Shouldn't the electromagnetic waves also have a kinetic energy? Why is the Kinetic energy of a electromagnetic wave not included in the expression of its average energy?

I have just started learning the theory of electromagnetic wave, Any help is appreciated :)

Shashaank
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  • Why should EM waves carrying momentum imply they have mass? – HTNW Sep 12 '22 at 14:12
  • Because the linear momentum is defined as mass times velocity? and force is defined as mass times acceleration – Shashaank Sep 12 '22 at 14:13
  • That is the Newtonian definition of mass, which is invalid for relativistic physics. The theory of EM requires relativity (relativity was discovered by inspecting EM). Mass is relativistically defined as the magnitude of 4-momentum, and relativity allows nonzero vectors to have zero magnitude. – HTNW Sep 12 '22 at 14:15
  • Ohh I ain't introduced to the relativistic physics, I am sorry – Shashaank Sep 12 '22 at 14:16
  • Is there any book where I can learn the Electromagnetism properly? – Shashaank Sep 12 '22 at 14:17
  • @HTNW Will you please elaborate it a bit – Shashaank Sep 12 '22 at 14:22
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    Your book seems fine. You can do EM without knowing relativity. You just have to treat the EM field as special (since it is always relativistic) and therefore different from normal objects (which you have studied nonrelativistically). I don't have references to books that describe EM relativistically from the start and I don't think that's necessary. – HTNW Sep 12 '22 at 14:26
  • Your textbook is probably fine. They are introducing concepts. In more advanced E&M the concept of the Maxwell Stress tensor, field momentum are introduced in more detail and can relate the transfer of momentum from the field to the object, but this is not often covered in a lot of detail in undergrad courses. – UVphoton Sep 12 '22 at 16:29
  • If you do want to learn relativistic physics, I can recommend Thomas Moore's A Traveler's Guide to Spacetime (now out of print), or his Six Ideas that Shaped Physics: Unit R which contains much of the same text but edited & revised. – Michael Seifert Sep 12 '22 at 17:44

2 Answers2

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When a particle applies force on a surface in case of an impact, it means that the particle has had momentum.

Particle can have momentum without having a mass, this is actually the case for photons (the quanta of light). That is the reason light can apply pressure on a surface without having a mass.

Since photons have momentum, they have kinetic energy, too. The kinetic energy of a photon is given by: $$E=pc$$ where $p$ is the momentum of the photon and $c$ is the speed of light.

But photon is a quantum mechanical notion while electromagnetic waves can be described without quantum mechanics. It can be proven that classical EM fields carry momentum and its density can be found from: $$\mathbf{P}=\epsilon_0 \mathbf{E}\times \mathbf{B} $$

Hossein
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  1. Mass (density) of an electromagnetic wave. This depends on what you mean by "mass". The standard meaning people put into the term is often the rest mass. In this case, it is zero. This can be immediately inferred to from the fact that as the energy flux (the energy density you have wrote times the speed of light) happens to be proportional to the Poynting vector (=momentum density). $E \propto |p|$ is equivalent to saying that the rest mass is zero. On the other hand, if you define your mass (density) as energy (density) divided by c$^2$, then ${\it this}$ mass will be non-zero.

  2. The energy density expression for EM wave does have a kinetic energy term. What you wrote is just a handy formula useful for calculations. The actual expression for the energy density reads: $$\mathscr{E}_{EM}=\frac{1}{2}\left(\epsilon_0{\vec{E}^2+\frac{\vec{B}^2}{\mu_0}} \right)$$ Now Maxwell's equations can be viewed as dynamical equations of motion, with the above expression being the Hamiltonian. In this formulation the electric field $\vec{E}$ will be the generalized coordinates, and the magnetic field $\vec{B}$ turns out to be its conjugated momentum. So if by kinetic energy you mean the part of of Hamiltonian that depends on generalized momenta, then the second (magnetic) part of the energy density expression is just it.

P.S. In a freely propagating EM wave in vacuum the magnetic part of the energy is exactly equal to the electric one. Hence for computational purposes it is often dropped, and the electric part is simply multiplied by a factor of 2. The $\frac{1}{2}$ factor in your expression is still there, because by $E$ people typically mean the peak electric field (which is exactly $\sqrt{2}$ times the r.m.s. average electric field).

John
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