Total Energy of an electromagnetic wave

Question

This question may sound stupid. But why when we calculate the total energy density (energy/volume), carried by an electromagnetic wave $u_T$, we add both $u_B$ + $u_E$.
Where $u_B = \frac{B^2}{2\mu_0}$ and $u_B = \frac{\epsilon_0 E^2}{2}$
From what I understand (Which I feel is wrong), the magnetic field cannot do work, and thus it transfers its energy to another field, induced Electric Field, in order to do work, this question can further explain my point here.
Then my question is if in electromagnetic waves the E field is induced by the changing B field and vice versa, wouldn't that mean that the energy we refer to as $u_B$ the same as $u_E$ and thus making the sum of them as summing the same thing twice?
Note: I'm not yet into quantum mechanics, I know everything changed there, but I have not yet studied it.

Answer 1

This comes right from Maxwell's equations as follows:

There is a vector identity that $$(\nabla \times \vec E) \cdot \vec B -(\nabla \times \vec B) \cdot \vec E = \nabla \cdot (\vec E \times \vec B)$$ This is a vector identity so it holds for all vector fields. Now, if we substitute in the microscopic forms of Faraday's law and Ampere's law (in natural units to avoid all of the constants) then we get $$-\vec B \cdot \frac{\partial}{\partial t} \vec B - \vec E \cdot \vec J - \vec E \cdot \frac{\partial}{\partial t}\vec E= \nabla \cdot (\vec E \times \vec B)$$ $$ 0 = \frac{1}{2} \frac{\partial}{\partial t} E^2 + \frac{1}{2} \frac{\partial}{\partial t} B^2+ \nabla \cdot (\vec E \times \vec B) + \vec E \cdot \vec J$$ $$0=\frac{\partial}{\partial t} u + \nabla \cdot \vec S + \vec E \cdot \vec J $$ where $u=\frac{1}{2}(E^2+B^2)$ is interpreted as the microscopic electromagnetic field energy density, and $\vec S = \vec E \times \vec B$ is interpreted as the microscopic electromagnetic field energy flux density. You cannot get rid of the $B^2$ term since it falls right out of Maxwell's equations.

It may help to think of a loop of superconducting wire, like a MRI magnet. There is a very large current with a very large magnetic field which has a very large energy density as described above. Because it is superconducting there is no voltage when the current is steady. However, in an emergency the field can be quenched. When this happens the large amount of energy in the field gets dissipated into the wire and then into the liquid helium and liquid nitrogen. This causes a rapid boil-off of the cryogens and is very loud and dramatic when it happens. Since boiling the cryogens that quickly requires a substantial amount of energy and since the only source of that energy was the magnetic field, then it is clear that the magnetic field energy is a real thing, as suggested by the theory, and cannot be neglected.

Answer 2

You could say the same thing of a mass oscillating on a spring. The total energy there is $$ E= \frac 12 m \dot x^2 +\frac 12 k x^2, $$ and the energy passes from one term to the other during each period. We still need both terms to get the energy.

Answer 3

A magnetic field can do work - for example, when it makes a compass needle point due north.

Answer 4

A couple of comments:

1. The B field of the EM wave does not induce the E field any more than the E-field induces the B field. Instead, a time-varying B field always comes with a spatially varying E field so that its spatial circulation is not zero, and also a time varying E field always comes with a spatially varying B field so that its spatial circulation is not zero.
2. all commercial/industrial "electric" motors/generators are based on the work done by the magnetic field in them; yeah I know the nomenclature is confusing. There are eletrostatic motors/generators in which the static electric field does the work but these are not really commercial gadgets https://en.wikipedia.org/wiki/Electrostatic_generator.

Answer 5

This is not a stupid question, instead is a very very clever question. The main reason is that there are problems in Maxwell's theory of radiation electromagnetic field. We know that Maxwell's previous theory was proved by experiments. For example, the magnetic quasi-static equation is the theory proved by experiments. $$\nabla\times\boldsymbol{H}=\boldsymbol{J}$$

$$\nabla\times\boldsymbol{E}=-\frac{\partial}{\partial t}\boldsymbol{B} $$ From the above we can obtained vector potential,

$$\boldsymbol{A}=\frac{\mu_{0}}{4\pi}\iiint_{V}\frac{\boldsymbol{J}}{r}dV $$

and similar scale potential $\phi$. We can obtained electric field and magnetic field as the following,$$\boldsymbol{B}=\nabla\times\boldsymbol{A}$$

$$\boldsymbol{E}=-\frac{\partial}{\partial t}\boldsymbol{A}-\nabla\phi $$ Maxwell's theory is to add displacement current to the magnetic quasi-static theory. However, after the displacement current is added, can the electric field and magnetic field defined in the same way be called electric field and magnetic field? The following is Maxwell's equation, $$ \nabla\times\boldsymbol{H}=\boldsymbol{J}+\frac{\partial}{\partial t}\boldsymbol{D}$$

$$\nabla\times\boldsymbol{E}=-\frac{\partial}{\partial t}\boldsymbol{B} $$

from that we can obtained the retarded potential,

$$\boldsymbol{A}^{(r)}=\frac{\mu_{0}}{4\pi}\iiint_{V}\frac{\boldsymbol{J(t-r/c)}}{r}dV$$

$$\boldsymbol{B}^{(r)}=\nabla\times\boldsymbol{A}^{(r)} $$

$$\boldsymbol{E}^{(r)}=-\frac{\partial}{\partial t}\boldsymbol{A}-\nabla\phi$$ I think that after the displacement current is added, the curl of the magnetic vector potential is not necessarily a magnetic field, nor is it necessarily an electric field. Therefore, when the displacement current is added, the Poynting theorem becomes worthless. As far as I know, Maxwell's contemporaries, such as Kirchhoff's paper in 1855 and Lorenz's paper in 1867, did not use the concept of electromagnetic field when studying radiated electromagnetic field. For example, Kirchhoff and Lorenz are in the current $$ \boldsymbol{J}=\sigma\boldsymbol{E}$$ They use the following directly $$\boldsymbol{J}=\sigma\boldsymbol{(-}\frac{\partial}{\partial t}\boldsymbol{A}-\nabla\phi) $$ if they need the following formula,

$$ \nabla\times\boldsymbol{B}=\mu_{0}\boldsymbol{J}$$

they directly writte

$$\nabla\times(\nabla\times\boldsymbol{A})=\mu_{0}\boldsymbol{J}$$ At first, I thought that Maxwell was really clever and used the concepts of electric field and magnetic field. This makes formulas and concepts easy. But then I thought about it. Kirchhoff, is Lorenz really that stupid? Actually, the were not. They clearly know that the magnetic vector potential and the scalar potential have changed, and this change will cause the curl of the magnetic vector potential to change. This change makes the expressions of electric and magnetic fields not necessarily applicable.

Hence the question the above $\boldsymbol{E}^{(r)},\boldsymbol{B}^{(r)}$ can still be called electric field and magnetic field? I do not think so. $\boldsymbol{B}^{(r)}$ can only be called curl of the magnetic vector potential，$\boldsymbol{E}^{(r)}$ is not electric field, but it can only be $-\frac{\partial}{\partial t}\boldsymbol{A}^{(r)}-\nabla\phi^{(r)}$.

In this case after introduced the displacement current, the Poynting theorem become meaningless.

In this case after introduced the displace ment current, the Poynting theorem become meaningless. And hence, the following $$\frac{1}{2\mu_{0}}|\nabla\times\boldsymbol{A}^{(r)}|^{2} $$ $$\frac{1}{2\epsilon_{0}}|\boldsymbol{(-}\frac{\partial}{\partial t}\boldsymbol{A}^{(r)}-\nabla\phi^{(r)})|^2 $$are also not the energy.