Question

How can one prove that electromagnetism is conservative? (Can this be done directly from Maxwell's equations?)

Answer 1

I know there are instances in electromagnetism induction when we find the electric field by itself to be very obviously non-conservative, though I am talking about electromagnetism as a whole. Any interaction with the electric or magnetic fields, even with induction, do not result in any violations of the conservation of energy. How can we prove this?

Answer 2

Something to think about, I can't prove it. When considering a broadcast transmitter. Does a receiver 100 miles away experience a loss of received signal if additional receivers are placed and operated at 50 miles?

Answer 3

I don't see why not? If the additional receivers are placed at the midpoint..

Answer 4

(If they don't physically block the signals that would otherwise be sent, then no)

Answer 5

I was just wondering about that. Like lets say there was a field of unterminated antennas, and then they were all loaded with receivers, would this impact the level of signal further out.....(not considering the path blocking)

Answer 6

If they were not blocking, no. Electromagnetic waves are independent of other electromagnetic waves and carry energy with them. What happens to one electromagnetic wave somewhere else won't affect the strength of another taking a different course.

Answer 7

Not to mention that light is also an electromagnetic wave...an light source doesn't get darker just because you shine it on something black.

Answer 8

can you please elaborate the meaning of "electromagnetism is conservative"?
i think only fields are conservative or non conservative..how can electromagnetism be conservative??

Answer 9

Like, energy is never created or destroyed. You wouldn't initially figure that given that, for instance, during electromagnetic induction, you can't assign a voltage field to the electric field. Though, somehow, energy is still conserved (charges affected by the electric field will send out magnetic fields, which the interact as electric fields, I figure, and in the end everything cancels out to conserve energy). How can we prove that?

Answer 10

It's only conservative if you consider the energy lost via radiation. If you're talking exclusively about the energy of a particular charged object being influenced by electric and magnetic fields, energy would go unaccounted for because the accelerating charge would radiate electromagnetic waves.

Answer 11

I want to account for the radiation energy... I'm fine with the energy in any form, as long as it can be expressed quantitatively. I don't see any easy way to prove this though.

Answer 12

The impedance of free space is purported to be \[Z _{o}=377 Ohms \] Or \[120\pi\] however I do not understand the exact nature of this impedance. I do know a radar transmitting/receiving horn is designed to match this impedance from the characteristic impedance of the waveguide. I have never seen whether this impedance is reactive or resistive (which I am sure it is not) or a combination of both capacitive and inductive reactance (which I believe it might be) If it is wholly reactavie then no energy is destroyed but exchanged.

Answer 13

I am exceedingly tired at the moment, but if I can summon the strength, I'll type something up in a little while. In the meantime, look for an article on the Electromagnetic Stress-Energy Tensor if you want to beat me to the punch.

Answer 14

Silly me -- we don't need tensors for this (thankfully, I'm not in the mood).
The Electromagnetic energy density inside a volume element is
\[ u_{em} = \frac{\epsilon_0 E^2}{2} + \frac{B^2}{2 \mu_0} \]
Therefore,
\[ \frac{\partial}{\partial t}u_{em} = \epsilon_0 \vec{E}\cdot \frac{\partial \vec{E}}{\partial t} + \frac{1}{\mu_0}\vec{B}\cdot \frac{\partial \vec{B}}{\partial t} \]

Maxwell's Equations tell us that
\[\frac{\partial \vec{E}}{\partial t} = \frac{1}{\epsilon_0 \mu_0} \left(\nabla \times \vec{B}\right) - \frac{1}{\epsilon_0}\vec{J} \]
and that
\[ \frac{\partial \vec{B}}{\partial t} = -\left(\nabla \times \vec{E}\right) \]

Substituting that in and simplifying a bit, we find that

\[\frac{\partial}{\partial t} u_{em} = \frac{1}{\mu_0} \left[ \vec{E}\cdot \left(\nabla \times \vec{B}\right) - \vec{B}\cdot \left(\nabla \times \vec{E}\right) \right] - \vec{E}\cdot \vec{J} \]

Rewriting this with a handy vector identity, we find

\[\frac{\partial}{\partial t} u_{em} = -\nabla \cdot \frac{\vec{E}\times \vec{B}}{\mu_0} - \vec{E}\cdot \vec{J} \]

Observing the definition of the Poynting Vector

\[ \vec{S} = \frac{\vec{E}\times \vec{B}}{\mu_0} \]

We can write the following conservation law:

\[\frac{\partial}{\partial t} u_{em} + \nabla \cdot \vec{S} + \vec{E}\cdot \vec{J}=0 \]

This can be interpreted as follows: The rate of loss of electromagnetic energy inside a volume is equal to the flux of electromagnetic energy out of the volume (carried by EM waves) plus the rate at which work is done on charges within the volume.

Answer 15

Wow! I am very impressed! Thank you so much for the response! Very clear, and seems to be just what I was looking for!

Answer 16

No problem! Additionally, when we apply relativity to electromagnetism, we discover the magical fact that there are frames of reference in which we can interpret magnetic fields as the manifestations of relativistic electric fields, so it's all one big bundle.

Answer 17

I have heard about that! Quite fascinating indeed!