# How can I apply Conservation of Energy in electromagnetism?

How can I apply C.O.E to a system that applies magnetic & electric fields at the same time to do work, and convert energy from one form to another?

Let assume we have a conductor that moves within a magnetic field(electric motor) how can C.O.E be applied so that input $E$ = output $E$ ?

Or in general a wire that has current flowing, and the work done to do that. And the work done by the magnetic forces to move it. And in many other applications... Is there a general way to calculate it?

It seems a bit difficult.

• You might be interested to read about the Poynting theorem May 22 '14 at 22:03

Maxwell's equations are based on COE of electromagnetism. All electromagnetic phenomena are covered under maxwell's equation.

The following link covers the relationship between maxwell's equations and COE http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

In modern electrodynamics you have energy-momentum tensor which is conserved - especially it's time component is conserved, what is for conservation of energy.

Long story short, you have for both forms of fields - $\vec{E}$ and $\vec{B}$ - a density of total energy, $d\tau=\frac{1}{2}\Big[\varepsilon_0\vec{E}^2+\frac{1}{\mu_0}\vec{B}^2\Big]d^3x.$ I think (but I can be wrong) we can split this density into two parts - the electric and magnetic ones. Now all we need to do is integrate both over all space and voila - we have equation for dependence of electric and magnetic energies.

There are two kinds of energy conservations laws, the first is based on the Poynting theorem. The Poynting theorem is,

$$$$-\iint_{\Gamma}(\boldsymbol{E}\times\boldsymbol{H})\cdot\hat{n}d\Gamma=\iiint_{V}(\boldsymbol{J}\cdot\boldsymbol{E}+\boldsymbol{E}\cdot\frac{\partial}{\partial t}\boldsymbol{D}+\boldsymbol{H}\cdot\frac{\partial}{\partial t}\boldsymbol{B})dV\label{eq:10}$$$$

Assume there are $$N$$ electric current elements, $$\boldsymbol{J}_{i}$$ $$i=1,\cdots N$$. According to the superposition principle,

$$$$\boldsymbol{E}=\sum_{i=1}^{N}\boldsymbol{E}_{i}\label{eq:60}$$$$ $$$$\boldsymbol{H}=\sum_{i=1}^{N}\boldsymbol{H}_{i}\label{eq:70}$$$$ $$$$\boldsymbol{J}=\sum_{i=1}^{N}\boldsymbol{J}_{i}\label{eq:80}$$$$

There is energy conservation law of $$N$$ electric current sources,

$$$$-\sum_{i=1}^{N}\sum_{j=1}^{N}\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{j})\cdot\hat{n}d\Gamma=\sum_{i=1}^{N}\sum_{j=1}^{N}\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j}+\boldsymbol{E}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{D}_{j}+\boldsymbol{H}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{B}_{j})dV\label{eq:90}$$$$

The above formula is obtained by direct substitute the superposition principle to Poynting theorem. Hence, the above can be called as the Poynting theorem of $$N$$ current elements.

However there is another kind of energy conservation law, the mutual energy principle (you can google mutual energy principle''), which is,

$$$$-\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{j})\cdot\hat{n}d\Gamma=\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j}+\boldsymbol{E}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{D}_{j}+\boldsymbol{H}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{B}_{j})dV\label{eq:30}$$$$

The mutual energy principle is sufficient and necessary condition of the $$N$$ groups of the Maxwell equations: $$$$\left(\begin{array}{c} \nabla\times\boldsymbol{E}_{i}=-\frac{\partial}{\partial t}\boldsymbol{B}_{i}\\ \nabla\times\boldsymbol{H}_{i}=\boldsymbol{J}_{i}+\frac{\partial}{\partial t}\boldsymbol{D}_{i} \end{array}\ \ \ \ \ \ \ \ i=1,\cdots N\right)\label{eq:40}$$$$

Please notice that the above $$N$$ is at least 2. That means for the mutual energy principle the current elements at least 2. If there is only one current element the mutual energy principle can not be build.

From mutual energy principle we can obtained another energy conservation law:

$$$$\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\intop_{t=-\infty}^{\infty}dt\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j})dV\equiv\sum_{i=1}^{N}\sum_{j=1,j

The above energy conservation law is self-explanatory. Since $$\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j})dV$$ is the power of current element $$\boldsymbol{J}_{j}$$ give to $$\boldsymbol{J}_{i}$$. $$\iiint_{V}(\boldsymbol{J}_{j}\cdot\boldsymbol{E}_{i})dV$$ is the power of current element $$\boldsymbol{J}_{i}$$ give to $$\boldsymbol{J}_{j}$$. If $$\boldsymbol{J}_{i}$$ get some power, then $$\boldsymbol{J}_{j}$$ will loss the same amount. Hence, in general there is, $$$$\intop_{t=-\infty}^{\infty}dt\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{j})dV=-\intop_{t=-\infty}^{\infty}dt\iiint_{V}(\boldsymbol{J}_{j}\cdot\boldsymbol{E}_{i})dV\label{eq:20}$$$$

The two kinds of energy conservation looks very similar. But there is the major difference, the difference is at the summation,

$$$$\sum_{i=1}^{N}\sum_{j=1}^{N}\ \ \ \ \ \ vs\ \ \ \ \ \ \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}\label{eq:100}$$$$

Please think which of the above summation is correct? Actually both the energy conservation law are correct. If both of them are correct the difference of them, i.e.,

$$$$-\sum_{i=1}^{N}\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{i})\cdot\hat{n}d\Gamma=\sum_{i=1}^{N}\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{i}+\boldsymbol{E}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{D}_{i}+\boldsymbol{H}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{B}_{i})dV\label{eq:110}$$$$ should has all terms as zero, that is,

$$$$\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{i})\cdot\hat{n}d\Gamma=0\label{eq:120}$$$$ $$$$\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{i})dV=0\label{eq:130}$$$$ $$$$\iiint_{V}(\boldsymbol{E}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{D}_{i}+\boldsymbol{H}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{B}_{i})dV=0\label{eq:140}$$$$

However, the above means that, $$$$\boldsymbol{E}_{i}\equiv0\label{eq:150}$$$$ $$$$\boldsymbol{H}_{i}\equiv0\label{eq:160}$$$$

This is clear wrong. The electromagnetic field cannot be 0 every where. This means the electromagnetic field theory obtained a conflict. The theory is not self-consistent!

This problem is solved by the theory based on the mutual energy principle ans self-energy principle. The self-energy principle tells us that the electromagnetic wave corresponding to the self-energy are all collapse back, hence, they do not transfer the energy.

The wave collapse back is done through the time-reversal wave. Wave collapse back is different concept compare with the wave collapse in quantum mechanics. Wave collapse is collapse the target of the wave. Hence the wave is started from a source and collapse to an sink. However the wave collapse back is that the wave is started from the source and go back to the source through a time-reversal process.

The time-reversal process can be described by time-reversal transform which is,

$$$$\boldsymbol{E,H}\rightarrow\boldsymbol{e},\boldsymbol{h}\label{eq:161}$$$$

$$$$\frac{\partial\boldsymbol{E}}{\partial t},\frac{\partial\boldsymbol{H}}{\partial t}\rightarrow-\frac{\partial\boldsymbol{e}}{\partial t},-\frac{\partial\boldsymbol{h}}{\partial t}\label{eq:162}$$$$

$$$$\boldsymbol{J}=\rho\boldsymbol{v}=\rho\frac{d\boldsymbol{x}}{dt}\rightarrow-\rho\frac{d\boldsymbol{x}}{dt}=\boldsymbol{j}\label{eq:163}$$$$

$$$$\left(\begin{array}{c} \nabla\times\boldsymbol{E}=-\frac{\partial}{\partial t}\boldsymbol{B}\\ \nabla\times\boldsymbol{H}=\boldsymbol{J}+\frac{\partial}{\partial t}\boldsymbol{D} \end{array}\right)\rightarrow\left(\begin{array}{c} \nabla\times\boldsymbol{e}=\frac{\partial}{\partial t}\boldsymbol{b}\\ \nabla\times\boldsymbol{h}=-\boldsymbol{j}-\frac{\partial}{\partial t}\boldsymbol{d} \end{array}\right)\label{eq:164}$$$$

$$\boldsymbol{e},\boldsymbol{h},\boldsymbol{j}$$ satisfy the time-reversal Maxwell equations. The time-reversal wave can cancel the energy flow of the self energy flow, that means, there is,

$$$$\iint_{\Gamma}(\boldsymbol{E}_{i}\times\boldsymbol{H}_{i})\cdot\hat{n}d\Gamma+\iint_{\Gamma}(\boldsymbol{e}_{i}\times\boldsymbol{h}_{i})\cdot\hat{n}d\Gamma=0\label{eq:120-1}$$$$ $$$$\iiint_{V}(\boldsymbol{J}_{i}\cdot\boldsymbol{E}_{i})dV-\iiint_{V}(\boldsymbol{j}_{i}\cdot\boldsymbol{e}_{i})dV=0\label{eq:130-1}$$$$ $$$$\iiint_{V}(\boldsymbol{E}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{D}_{i}+\boldsymbol{H}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{B}_{i})dV-\iiint_{V}(\boldsymbol{e}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{d}_{i}+\boldsymbol{h}_{i}\cdot\frac{\partial}{\partial t}\boldsymbol{b}_{i})dV=0\label{eq:140-1}$$$$

The above formula guarantees that the self energy terms do not transfer energy. It is referred as the self-energy principle. But in this case, we do not need $$\boldsymbol{E}_{i}\equiv0$$, $$\boldsymbol{H}_{i}\equiv0$$. Hence, we have,

$$$$\boldsymbol{E}_{i}\neq0\label{eq:170}$$$$ , $$$$\boldsymbol{H}_{i}\neq0\label{eq:180}$$$$

When we added the time-reversal wave, $$\boldsymbol{e},\boldsymbol{h},\boldsymbol{j}$$, the conflict is solved. Hence tactically the electromagnetic field is not only the retarded wave, but also the advanced wave and the time-reversal waves. The time reversal wave has the one corresponding to the retarded wave and the time-reversal wave corresponding to the advanced wave. Hence the electromagnetic field has 4 different waves. The advanced wave violated the causalities, most engineer do not believe it. But there are many scientists support this idea, for example Wheeler and Feynman, John Cramer, Lawrence M. Stephenson. The two time-reversal waves are introduced by the author in the theory about the mutual energy and self-energy principle.

After introduce the self-energy principle above, the two kinds of energy conservations are all hold.