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I'm trying to figure out an example from a textbook (Demtröder -- Experimentalphysik 2, pg. 198) where the energy transport caused by a current is depicted:

Assume you have a wire (with some resistance $R$) and a current $I$ flowing through the wire. The wire will emit energy in form of heat (with $\dot W = I^2 \cdot R$). It states that since the electric field $\mathbf E$ is parallel to to the wire (i.e. the current), and the magnetic field $\mathbf B$ is tangential to the wire, the Poynting vector $\mathbf S$ must point radially (and orthogonally) into the wire.

The book goes on to argue, that therefore, the energy to replenish the "lost" heat-energy flows radially from the outside into the wire. My question is now, where does that energy come from? I would have thought that it comes from the source of the current (e.g. a battery) and travels through the wire to the point where the heat is emitted. However, this is explicitly stated to be wrong. Could you please shed some light on the issue?

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2 Answers 2

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The Poynting vector shows that the energy isn't transmitted through the wires; it's transmitted through the surrounding space, as in this picture:

enter image description here

To simplify the picture, most of the heat in this circuit is produced by the blue light bulb on the right side. The energy flows from the battery "through the air" to the light bulb. The energy flows are given by the white arrows in red ellipses. The electric field isn't just parallel to the wire; because the current (charges) must ultimately get back somewhere in space, the electric field goes "mostly" from a piece of wire to another piece of wire which is oriented oppositely. On the picture above, the electric field is given by the thin red arrows (from the top to the bottom). Similarly, the thin green lines are magnetic field lines (around the wires, as you probably expected).

It was taken from this page:

http://www.furryelephant.com/content/electricity/visualizing-electric-current/surface-charges-poynting-vector/

It's important to realize that many of the flows indicated by the Poynting vector are partially fictitious. In fact, you can have just a static electric field induced by an electric charge at point $X_{\rm EL}$ on top of a static magnetic field from an end point of a long bar magnet located at $X_{\rm MG}$. And the Poynting vector will tell you that energy is running in loops around the axis connecting $X_{\rm EL}$ with $X_{\rm MG}$, in the surrounding vacuum. This is not a problem. One may also define the Poynting vector (and the whole stress-energy tensor) differently so that the flows will be different.

However, it's equally important to realize that the energy is locally conserved with the Poynting vector being the flux. In particular, in the vacuum, the energy just "flows" through the vacuum and whenever the Poynting lines gets denser, they must become longer, and vice versa. There are no sinks or sources of energy in the vacuum. This conservation law

$$ \frac{\partial \rho_{\rm energy}}{\partial t} + {\rm div}\,\,\vec S =0 $$

can be proved by multiplying Maxwell's equations by fields and combining the equations appropriately. In the wire, there is an extra term from the heat creation in the conductor etc.

Again, in these situations, one shouldn't overreact. One shouldn't imagine energy as "some kind of liquid" that pushes all things. Energy is just an abstract quantity that is conserved. When it just runs in some loops in the vacuum, it is not a problem. Whether such an energy flow does some work on charges etc. depends on the detailed equations, the Lorentz force, Maxwell's equations, and so on: one shouldn't try to guess such influences from the flow of energy only.

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Thanks for this explanation. It really helped me. Just one more thing: If you can reduce energy to your abstract conserved quantity, where you can take out as much as you like, as long as you put in the same amount somewhere else, how does that play with the law that an effect (e.g. heating the wire) cannot travel faster than the speed of light (measured from its source -- e.g. the battery)? (I totally understand if this is asking a bit too much in an addendum.) –  bitmask Jan 28 '12 at 17:23
    
You're welcome, bitmask. Maxwell's equations may be explicitly proven to respect relativity, including the condition that signals never propagate faster than light. I didn't really say that you may subtract energy from any point and return it anywhere. On the contrary, I quoted the differential equation with ${\rm div}\,\,\vec S$ which is manifestly local: if energy disappears from some place, it is because it is flowing to a nearby point as shown by the $\vec S$ vector. The conservation is local. I just wanted to say that energy doesn't have to have a "material carrier". –  Luboš Motl Jan 28 '12 at 17:37
    
In general relativity, by the way, the energy conservation law becomes problematic and when we want to include energy carried by gravitational waves etc., it cannot really be written locally in any meaningful way. But in non-gravitational physics, we can do it. When I said that energy was an abstract quantity, I meant that even in the vacuum, $\vec E, \vec B$ may be nonzero and generic, and therefore $\vec S$ is probably nonzero, too. Energy is flowing even though there are no "particles" over there. Nothing wrong about it. My focus was the word "abstract", like "numerical". –  Luboš Motl Jan 28 '12 at 17:41
    
Oh, one more addition. In your comment (3 above me), you may have referred to my remark about the various ambiguous definitions of the Poynting vector. I meant that one may redefine the Poynting vector in such a way that the conservation law above will still hold (and the total energy will be unchanged) but the flows (circulation of energy) will be different. But this freedom to redefine $\vec S$ really means that the energy flux density vector, when loosely defined, is ambiguous and unphysical. That makes it obvious that such subtle changes of $\vec S$ can't be used to send any signals. –  Luboš Motl Jan 28 '12 at 17:46
    
Imagine that you have two definitions of $\vec S$ such that both of them obey the continuity equation with the divergence above but the values of $\vec S$ are different. These two configurations of $\vec S$ may describe the very same physical situation. This means that $\vec S$ can't be "directly measured"; the role of $\vec S$ is to have a vector to be used in "accounting" to prove that the energy is locally conserved. Concerning the ambiguity, see en.wikipedia.org/wiki/… –  Luboš Motl Jan 28 '12 at 17:50

You can zoom in and look at just the wire; if you do that E and J are parallel(given some resistivity, we can call it z-hat) and B is parallel to the surface of the wire(we can call that theta-hat), so by definition you do get a poynting vector into your wire(z cross theta = -r). This is a transfer of energy from the flowing charges to the heat in the wire. The "outside" giving the energy is seemingly bunk; maybe the book means it comes from whatever is causing your current flow, because that's really where it comes from.

You'll notice that if there is no resistivity then your E would be zero (E=J*resistivity), and that would mean your poynting vector would also be zero. That's a superconductor, there's no heating there because there's no resistivity. Resistance/resistivity is basically a way to account for electron collisions, but that's not seemingly important to your question.

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