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Mechanical energy in EM

  1. May 15, 2013 #1
    In griffith it is mentioned that the sum of the mechanical energy and electromagnetic field energy are conserved in a system, how is the mechanical energy defined in this case since a potential energy cannot be defined?
    Is it just the kinetic term by itself?
     
  2. jcsd
  3. May 15, 2013 #2

    Zag

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    HomogeneousCow, the energy is conserved in the sense that the kinetic energy of the charges plus the energy stored in the fields remain constant in time as long as you choose the boundary of your system to in infinity. Otherwise, if you choose your system to be finite, you would have a flow of energy across its boundary due to the emission of radiadion by the moving charges and, therefore, the energy of your system would not be conserved. This is basically what Poynting's Theorem states. (http://en.wikipedia.org/wiki/Poynting's_theorem)

    I'm sorry don't have much time to develop that more precisely right now, but as soon as I have time I will write a more complete answer for you.

    Best,
    Zag
     
  4. May 16, 2013 #3

    Jano L.

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    It may be done in more then one way. One can, as you say, forget about the potential energy. Then the only energy that is naturally called mechanical energy is the kinetic energy of he particles, given by Einstein's relation ##K = (\gamma-1)mc^2##. The other energy is the energy of the electromagnetic field.

    The other way is to adopt Coulomb potential energy into the mechanical energy. This potential energy is not exactly total electromagnetic energy, there is a correction, and one can say that this is the energy of the field.

    There may be other separations, one could perhaps include velocity dependent corrections into the mechanical energy, which is motivated by the Darwin Hamiltonian, but the remaining energy of the field will probably get more complicated.
     
  5. May 16, 2013 #4
    So the simplest form is just the kinetic energy plus the field energy?
     
  6. May 16, 2013 #5

    Jano L.

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    Actually, I made a mistake, one should rather take total energy of the particle ##\Epsilon = \gamma mc^2##, because this transforms nicely together with momentum via the Lorentz transformations.
     
  7. May 16, 2013 #6
    I don't understand. Why do you think that a potential energy cannot be defined?

    I wish I had the text you're speaking about. I have his first edition. What section are you referring to?

    The total energy is the sum of the mechanical energy + electromagnetic energy. The total mechanical energy is the sum of fhe kinetic energy and the potential energy. The potential energy of a charged particle in the EM field is defined as the quantity qU such that E = - grad U -@A/@t

    (I'm using "@" to represent the partial derivative symbol)
     
  8. May 16, 2013 #7
    Well see that definition of mechanical energy is conserved only in the electrostatic approximation.
    Since the electric field has a non-zero curl the line integral through it will depend on the entire path taken, thus a potential energy function cannot be defined
     
  9. May 16, 2013 #8
    Please recall my previous post where I wrote

    E = - grad U -@A/@t

    V = qU

    where

    U = Electric potential
    V = qU = potenial energy of charged particle in time varying EM field

    Since div B = 0 we can always find a vector function A such that B = curl A. Then Faraday's law in differential form can be written as

    curl (E + @A/@t) = 0

    Therefore since the curl is zero the vector field defined by E + @A/@t can be written as the gradient of a scalar function U. That's where

    E = - grad U -@A/@t


    comes from. That equation states that the curl of the electric field is the negative of the partial of Bwith respect to t. It can be shown that when this is the case, while the electric field is not the gradient of a scalar function, it is the sum of the gradient of a scalar function (the electric potential) minus the partial of the vector potential A with respect to t. Given that it can be shown that the energy of a particle moving through a time varying EM field equals the kinetic energy + the potential energy (= q*electric potential) and that's a function of time, i.e. the energy is defined but not conserved.

    If you have Classical Electrodynamics - Third Edition by John D. Jackson then you can look all this up in that text in section 6.2 pages 239-240. Griffiths should also explain all of this too.
     
  10. May 16, 2013 #9
    Anyways I am referring to page 348 on Griffiths 3rd edition Introduction to electromagnetism.

    Here he talks Poynting's theorem, I followed everything until the end where he replaces the rate at which work was done on the volume charge, dW/dt, with du/dt (integrated over the volume). Where u is the mechanical energy density of the matter field, could I get some more insight on this step.
     
    Last edited: May 16, 2013
  11. May 16, 2013 #10

    WannabeNewton

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    I assume you mean the step where he writes ##\frac{\mathrm{d} W}{\mathrm{d} t} = \frac{\mathrm{d} }{\mathrm{d} t}\int _\mathcal{V}u_{\text{mech}}d\tau##. What unsettles me about Griffiths' derivation is that you cannot take into account non-conservative forces using an energy density ##u_{\text{mech}}## since there is no way to construct a well defined state function for energy "associated" with a non-conservative force; the work done by the non-conservative forces is inherently a process function. The work-energy theorem from mechanics only explicitly brings the non-conservative forces into play when considering total work done since we can just speak of the work done by the non-conservative forces for the specific process (e.g. the path taken) but I don't see how Griffiths is taking this work into account since all he does is talk about the change in the mechanical energy (which I also find weird because in general the mechanical energy is defined as the sum total of the kinetic and potential energies and the work-energy theorem states that the work done by the conservative forces equals the change in the kinetic energy, or equivalently negative the change in the potential energy, and not the change in the mechanical energy).
     
  12. May 16, 2013 #11

    WannabeNewton

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    Personally, I like the following much better: ##\partial_{a}(T^{ab}_{\text{charges}} + T^{ab}_{\text{EM}}) = 0## as it is more encompassing and manifestly Lorentz covariant.
     
  13. May 16, 2013 #12
    This is starting to get jumbled up, could I get straight answers for each of the following questions

    1.In electrodynamics is the total energy of the system simply the kinetic energy of the particles plus the field energy?

    I think the entire affair is clearer if we just started with dW=q(E*J)dt, then we just expand q(E*J) the same in Griffiths treatment, we get the dW on one hand and the field energy and poynting flux terms on the RHS, multiplied by dt of course.
    Then we can integrate both sides with respect to time, and from there we are back at home, the LHS integrates out to a difference between kinetic energies, the RHS integrates out to a minus change in the field energy and the second RHS term is the minus energy lost through the boundaries.

    We then move the field difference over to the LHS and get a simple equation

    Total energy changed (Kinetic+field)=-total energy lost through the boundaries.

    The differential form makes me feel weird, the way he just turns dW/dt into du/dt

    I'm pretty sure I missed an integral sign somewhere up there so I hope you understand what I'm saying.
     
    Last edited: May 16, 2013
  14. May 16, 2013 #13

    WannabeNewton

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    Yes and that is a particular case of what I wrote in post #11. Also see the following passage from another textbook, it should clear everything up:

    field_energy_EM.png
     
  15. May 16, 2013 #14
    Ah I see, and the other indices just lead to the conservation of momentum. thanks !
    Which textbook is this extract from btw?
     
  16. May 16, 2013 #15

    WannabeNewton

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    "Gravitation: Foundations and Frontiers" - T.Padmanabhan (the passage is from the chapter on scalar and electromagnetic fields in special relativity, in particular page 92).
     
  17. May 16, 2013 #16
    Could you recommend a good book on the covariant formulation of EM? Preferably with lots of working and detail.
     
  18. May 16, 2013 #17

    WannabeNewton

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    The above book actually works out many things in gory detail (the calculations and such) and the chapter on EM in SR is quite detailed and has tons of interesting exercises (I haven't really attempted them yet).
     
  19. May 16, 2013 #18

    vanhees71

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    2016 Award

    One of the best books on this topic I know of is

    S. R. de Groot, L. G. Suttorp, Foundations of Electrodynamics, North-Holland Publishing Company, Amsterdam (1972)
     
  20. May 16, 2013 #19
    Are they the jackson kind?
     
  21. May 16, 2013 #20
    This is not really true if magnetic fields are involved. You would then need a velocity dependent potential energy.
     
    Last edited: May 16, 2013
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