Conservation of energy in an electrical circuit can be expressed by Ampere's law $$\nabla \times \textbf{B} = \mu_o \textbf{J} + \epsilon_o \mu_o \frac {\partial \textbf{E}} {\partial t}$$ when both sides of the equation are dotted with $\textbf{E}$ and the input power is placed on the left side of the equation while the dissipated power (the output power) is placed on the right hand side of the equation

$$\textbf{E} \cdot (\nabla \times \textbf{B}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \mu_o \textbf{J} $$

$$ \textbf{E} \cdot \frac {1} {\mu_o} (\nabla \times \textbf{B}) - \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \textbf{J} $$

$$ \textbf{E} \cdot (\nabla \times \textbf{H}) - \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \textbf{J} $$

The units of all terms are $\left[ \frac {W} {m^3} \right]$, which is energy per unit time per cubic meter.

If we now need to get the Poynting theorem from the above equation then we have to add on both sides of the equation the term $+\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t}$ which from Faraday's law dotted with $\textbf{B}$ is equal to $-\textbf{B} \cdot(\nabla \times \textbf{E})$

$$\textbf{E} \cdot (\nabla \times \textbf{B}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t} = \textbf {E} \cdot \mu_o \textbf{J} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t}$$

$$ \textbf{E} \cdot (\nabla \times \textbf{H}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} -\textbf{B} \cdot(\nabla \times \textbf{E}) = \textbf{E} \cdot \mu_o \textbf{J} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t}$$

$$ \textbf{E} \cdot (\nabla \times \textbf{H}) -\textbf{B} \cdot(\nabla \times \textbf{E}) = \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t} + \textbf{E} \cdot \mu_o \textbf{J} $$

$$ -\nabla \cdot \textbf{S} = \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} + \frac {\textbf{B}} {\mu_o} \cdot \frac {\partial \textbf{B}} {\partial t} + \textbf{E} \cdot \textbf{J} $$

where $\textbf{S} = \frac {1} {\mu_o} \textbf{E} \times \textbf{B} = \textbf{E} \times \textbf{H}$ $\left[ \frac {W} {m^2} \right]$ is the Poynting's vector. Doesn't the above make the Poynting vector redundant in proving CoE, the latter appearing to follow directly from Ampere's law alone?

Notice, it is not true that

$$ \textbf{E} \cdot (\nabla \times \textbf{H}) - \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \textbf{J} $$

"just states some relation" between vector fields the way it is not true that

$$ \frac{\partial}{\partial t} \biggl[ \frac{1}{2} \biggl( \epsilon_0 \textbf{E}^2 + \frac{\textbf{B}^2}{\mu_0}\biggr) \biggr] $$

"just states some relation" between vector fields. These state $\textit{change in energy density of electromagnetic field}$. As shown above, obtaining of the Poynting vector is only a result of a mathematical manipulation of an equation which already shows balance of these power densities and therefore the Poynting vector and the theorem connected with it are redundant.

  • $\begingroup$ How does Ampere's law per se express conservation of energy? All I see in Ampere's law are two fields, no energy whatsoever. $\endgroup$
    – Siyuan Ren
    Feb 15, 2012 at 2:51

2 Answers 2


It is not an answer, but more a hint. From school I remember the simple problem which forces to accept the concept of a Poynting vector. If one considers two charged particles moving in perpendicular directions and write energy/momentum conservation for this system, the solution will contain Poynting vector explicitly. In some simple cases (like particles moving in ONE direction or non-interacting particles moving in static field) you may obviously skip this concept.


Karsus Ren is right. The Ampere's law just states some relation between three vector fields. And as you wrote, you had to use the Faraday's law to derive the conservation equation, so just the Ampere's law alone is not enough.

But my main point is, that It's not about proving the conservation of energy. What you just did, is only some mathematical manipulation of equations, which resulted in another equation. For some reasons, our favorite units are Watts and Joules, so we manipulated our equations in a way, that we get these units in them. We call the result "energy conservation equation", but It's nothing new or fundamental in any way. Whatever equations you have, as long as you have sufficient set of units in them, you can always do some manipulation and arrive to some "energy conservation equation".

So by deriving energy conservation equation, you just stated something you already knew, in a somewhat different way, that is useful to you in some situations. The only purpose of Poynting vector is, that It's handy substitution, because you can interpret it as a density of energy flow.

When you see the energy conservation equation, you see that It's reasonable to call the term $$ \epsilon_0 \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} + \frac {\textbf{B}} {\mu_0} \cdot \frac {\partial \textbf{B}} {\partial t} = \frac{\partial}{\partial t} \biggl[ \frac{1}{2} \biggl( \epsilon_0 \textbf{E}^2 + \frac{\textbf{B}^2}{\mu_0}\biggr) \biggr] \equiv \frac{\partial u }{\partial t} $$

the change in energy density of electromagnetic field. Again, this is just a way we decided to call it. Because it has units of energy, it has electric and magnetic fields in it and it appears in the conservation equation. By using this handy substitution, you get Poynting theorem:

$$ \frac{\partial u }{\partial t} + \nabla \cdot \textbf{S} = - \textbf{J} \cdot \textbf{E} $$

By integrating over arbitrary region in space and applying the divergence theorem, you get:

$$ \frac{\partial}{\partial t} \int_V u \,\,dV = - \oint_{\partial V} \textbf{S} \cdot \textbf{n} \,\, dS - \int_V \textbf{J} \cdot \textbf{E}\,\, dV $$

It says, that change in energy of electromagnetic field in a given area (left side), can happen either by flow of $\textbf{S}$ through the boundary of said area, or by interaction of electric field with electric current, that we happen to call Joule heat. Forgetting about the last term, you see, that the Poynting vector behaves like it was really a density of energy flow. In fluid mechanics, you have exactly the same equation only with different variables, It's the continuity (mass conservation) equation:

$$ \frac{dm}{dt} = \frac{\partial}{\partial t} \int_V \rho \,\,dV = - \oint_{\partial V} \textbf{v} \cdot \textbf{n} \,\, dS $$

And we base the flow-type terminology on this analogy. But you always have to be careful with the interpretation. Energy isn't something real or material. Same for its flow. Energy is a property of things. But thinking in terms of energy flows (and using Poynting vector) helps us understand and better visualize the issue in our brains and perhaps makes various practical computations prettier.

  • $\begingroup$ As I can't comment on the question above, I have to do it here: Dear ganzewoort, perhaps you should have commented here, rather than modyfying your question, thus disrupting the chronology of posts. As of the subject matter, you could have clearly read in my reply that I basically agree with you. Poyinting vector is rendundant in your sense. However, in the same sense the energy conservation law is also redundant, as it is fully derived from Maxwell equations, therefore states nothing new. $\endgroup$
    – airguru
    Feb 20, 2012 at 16:32

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