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Electric and magnetic field energy

  1. Feb 11, 2009 #1
    For electrostatics one can work out the energy of a charge configuration in terms of its associated field:

    [tex]
    U = \epsilon_0 \vec{E}^2/2
    [/tex]

    I've seen that the general energy of the field is given by:

    [tex]
    U = \epsilon_0 (\vec{E}^2 + c^2\vec{B}^2)/2
    [/tex]

    but am unsure how to show this. I suspect that a relativistic argument would be one way.

    Another way, which doesn't work (or I did it wrong), is a calculation of the Hamiltonian for the (non-proper time formulation) of the Lorentz force Lagrangian. I get an energy that only includes the q\phi part as in electrostatics. This actually makes some sense since doing a line integral against a perpendicular field won't contribute.

    I'd be interested to know two things:

    1) A high level non-math description of where the B^2 term comes from.

    2) Some hints for the math side of the question. What is a way (or some ways) that the E^2 + B^2 form of the field energy can be derived?
     
  2. jcsd
  3. Feb 11, 2009 #2
    Remember the relationship between energy and force. Force is the spatial gradient of potential energy.

    If you have the two fields E and B which do not vary over time, then

    [tex]F = \triangledown U = \triangledown \epsilon_0(E^2 + c^2 B^2)/2 = \epsilon_0(2 E + 2 c^2 B)/2 = \epsilon_0 (E^2 + c^2 B)[/tex]

    which is very similar to the form of the equation for the Lorentz force. The only part missing is the cross product of the charge's velocity, and I'm not sure exactly where that comes in. The equation you provided is missing any reference to the particle's velocity which (along with the charge) is necessary to go from magnetic potential to energy.
     
  4. Feb 11, 2009 #3
    Not for this velocity dependent potential. The closest you'll probably get to a potential in this case is the via the Lagrangian

    [tex]
    L = T - V = mv^2/2 -q\phi + q A \cdot v
    [/tex]

    application of the Euler Lagrange equations will then give you the Lorentz force. Paraphrasing the Euler Lagrange equations in english, I'd say a more accurate description would be "the spatial gradient of the Lagrangian (Generalized Force) equals the time derivative of the velocity gradient of the Lagrangian).
     
  5. Feb 11, 2009 #4

    clem

    User Avatar
    Science Advisor

    No. People have tried to do that but all have failed.
    1) Math is necessaryl at any level.
    2) It is derived in most EM textbooks, even low level ones, usually starting with
    [tex]dU/dt(matter)=\int{\bf j\cdot E}[/tex].
     
  6. Feb 11, 2009 #5

    jtbell

    User Avatar

    Staff: Mentor

    One common way to derive the electric field energy density is to calculate the work required to increase the charge on a parallel-plate capacitor from 0 to +Q and -Q on the two plates, and relate it to the final electric field inside the capacitor and the volume of the capacitor. This gives you the equation that you posted:

    [tex]u_E = \frac{1}{2} \epsilon_0 E^2[/tex]

    Similarly, you can derive the magnetic field energy density by calculating the work required to increase the current in a solenoid from 0 to I, and relate it to the final magnetic field inside the solenoid and the volume of the solenoid. This gives you

    [tex]u_B = \frac{1}{2} \frac{B^2}{\mu_0}[/tex]

    The total electromagnetic field energy density is the sum of these two. You can then eliminate either [itex]\epsilon_0[/itex] or [itex]\mu_0[/itex] using the relationship

    [tex]c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}[/tex]

    which can be derived in connection with electromagnetic waves.
     
  7. Feb 11, 2009 #6
    I didn't see it in my old purcell text, but did a derivation of something similar to what you have above, where I got (I believe)

    [tex]\partial U/\partial t + \nabla \cdot \bf P/c= -\bf j\cdot E}[/tex]

    where P was the Poynting vector. In that case I started with U as defined in my initial post, and derived the conservation equation as a consequence of applying Maxwell's equations.

    Now, by analogy, I see that the left hand side logically has the structure of a four vector energy momentum "divergence" so one could identify that with energy, but that's a whole lot different than identifying it as energy = force times distance as in the electrostatics case.
     
  8. Feb 11, 2009 #7
    Thanks. I'll ponder that and give it a try.

    It's not clear to me why this would have to be the direct sum of the two. For example, without some other guiding principle, why is it neccessary that there is no E,B cross terms?
     
  9. Feb 11, 2009 #8
    I think I understand it now.

    The - E . j term is a power density. For me I was able to obtain the energy/Poynting conservation relation by a purely mathematical manipulation of Maxwell's equations, but I didn't realize the significance of that last power density term. It can be related to work done by considering a line integral, moving a charge against the Lorentz force (and only the electric field ends up contributing to that integral)
     
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