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Work term in Poynting's theorem

  1. Aug 4, 2015 #1

    Dale

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    @cabraham

    I recently posted a short derivation of Poynting's theorem in https://www.physicsforums.com/threads/work-done-by-magnetic-field.825806/

    It reminded me of our conversations about Poynting's theorem in a motor. As we discussed earlier, the work term ##E \cdot J## includes all the work done on matter, including but not limited to the Ohmic losses. So I wanted to see if any further analysis of that term could be done to pull out the Ohmic losses.

    If a piece of material is moving at velocity ##v<<c## then we can make the following transformations:
    ##E' = E + v \times B##
    ##J' = J - \rho v##
    where the primes indicate values in the reference frame where the material is at rest.

    Substituting those in we have:
    ##E \cdot J = (E' - v \times B) \cdot (J' + \rho v)##
    ##=E' \cdot J' + E' \cdot \rho v - (v \times B) \cdot J' - (v \times B) \cdot \rho v##
    ##=E' \cdot J' + \rho v \cdot (E+ v \times B) - (J-\rho v) \cdot (v \times B)##
    ##=E' \cdot J' + \rho v \cdot E - J \cdot (v \times B)##

    So finally we end with
    ##E \cdot J=E' \cdot J' + v \cdot (\rho E + J \times B)##

    Since ##E'## and ##J'## are in the reference frame of the material, I would interpret the term ##E' \cdot J'## as being Ohmic losses. Then ##E \cdot J## contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.

    We had discussed earlier how ##E \cdot J## depends on ##B##, so I think that this calculation makes that dependency more explicit, and the final result is somewhat obvious in retrospect.

    I don't have a reference for this, so I don't know if someone else has already do this derivation, but I thought you might like to see it.
     
    Last edited: Aug 6, 2015
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  3. Aug 5, 2015 #2

    DrDu

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    I am not sure I understand what you are talking about. Maybe you should specify explicitly what material you consider (e.g. are you talking about macroscopic or microscopic electrodynamics). I thought of J being due to polarizable dipole whose COM is at rest. There may be a phase lag between J and E, so E.J may be non-Ohmic even in the rest frame.
     
  4. Aug 5, 2015 #3

    vanhees71

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    The derivation is not exact but only for the non-relativistic limit. Otherwise it's of course correct. Macroscopic electrodynamics is spoiled by almost all textbooks because it's not explicitly mentioned that the matter is treated non-relativistically, and thus the consituent equations are not relativistically covariantly formulated and thus cannot applied for relativistically moving media.

    The most simple relativistic phenomenological description is already by Minkowski. The electromagnetic field is represented by the antisymmetric four-tensor
    $$F^{\mu \nu}=\begin{pmatrix}
    0 & -\vec{E}^{T} \\
    \vec{E} & -\epsilon^{jkl} B^l
    \end{pmatrix}:=(\vec{E},\vec{B}),$$
    where ##(\vec{E},\vec{B})## is a notation in the 3+1-form wrt. to a fixed reference frame, which is particularly useful for electrodynamics although not manifestly covariant, but it's easy to translate the manifest covariant forms to this form, which then can be used to calculate things in the usual way known from the 3+1 formulation of non-relativistic E-dynamics (where I mean the usual non-relativistic treatment of matter; the electromagnetic field has no unique non-relativistic limit, but that's another story).

    In addition in phenomenological macroscopic electromagnetics one introduces the auxilliary fields ##\vec{D}## and ##\vec{H}## by splitting the charges and currents in ones belonging to the matter ("bound charge-current distributions") and external or free ones, and writes down the inhomogeneous Maxwell equations with these auxiliary fields with sources given by the free charge-current distribution only. Relativistically covariant these fields are also represented by an antisymmetric four-tensor,
    $$D^{\mu \nu}=(\vec{D},\vec{H}).$$
    One can also introduce electric an magnetic polarization fields via (in Heaviside-Lorentz, i.e., rationalized Gauss units)
    $$\vec{D}=\vec{E}+\vec{P}, \quad \vec{H}=\vec{B}-\vec{M},$$
    where one should note the nasty sign difference due to the classical errorneous interpretation of the magnetic field and flux density.

    The consituent equations for matter at rest read
    $$\vec{D} \stackrel{*}{=}\epsilon \vec{E}, \quad \vec{H} \stackrel{*}{=}\frac{1}{\mu} \vec{B}, \quad \vec{j} \stackrel{*}{=} \sigma \vec{E},$$
    where we assume an homogeneous medium. Usually one can then assume the material parameters ##\epsilon##, ##\mu##, and ##\sigma## to be only function on ##\omega## for harmonically time-dependent fields (neglecting not only anitropies of the medium (needed in crystal optics) but also spatial dispersion (needed in plasma physics)). The star above the equation sign indicates that this equation is valid only in the (local) restframe of the medium.

    Now it's easy to translate these consitituent equations into a Lorentz covariant form. To this end we just have to introduce the four-velocity field of the medium (as in relativistic hydrodynamics),
    $$(u^{\mu})=\gamma \begin{pmatrix} 1 \\ \vec{\beta} \end{pmatrix}, \quad \vec{\beta}=\frac{1}{c} \vec{v}(t,\vec{x}), \quad \gamma=\frac{1}{\sqrt{1-\vec{\beta}^2}}.$$
    In the (local) rest frame of the medium one has
    $$(u^{\mu}) \stackrel{*}=(1,0,0,0)^{T}.$$
    In covariant form we can write the consituent equations thus by projecting the four-tensors to ##u^{\mu}##. Note that then the material constants ##epsilon## etc. are always defined in the (local) restframe of the medium, and thus are scalars! It's easy to see that the consituent equations read
    $$D^{\mu \nu} u_{\nu} = \epsilon F^{\mu \nu} u_{\nu}, \quad (^\dagger D)^{\mu \nu} u_{\nu} =\frac{1}{\mu} (\dagger F)^{\mu \nu} u_{\nu}.$$
    The second set of equations contains the Hodge dual of the field tensors, defined by
    $$(^\dagger F)^{\mu \nu} = \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}$$
    with the Levi-Civita symbol defined such that ##\epsilon^{0123}=+1## and otherwise being totally antisymmetric unter permutation of its indices.

    In (3+1) notation the relativistically exact consistuent equations read
    $$\vec{D}+\vec{\beta} \times \vec{H}=\epsilon (\vec{E}+\vec{\beta} \times \vec{B}), \quad \vec{D}-\vec{\beta} \times \vec{H}=\frac{1}{\mu} (\vec{E}-\vec{\beta} \times \vec{H}).$$
    Note that these are the exact relativistic equations although it doesn't look so. The reason is that ##\vec{E}## etc. are not three-vectors which are part of a four-vector but components of the antisymmetric field tensors.

    Ohm's Law is tricky, because by definition in the rest frame of the medium, ##\vec{j}## means the conduction current. Strictly speaking from a microscopic point of view one would have to work with two media for, e.g., a metal as a conducting mediumL The one are the atoms with the bound electrons and the other are the "freely moving" conduction electrons, but that's not done in this macroscopic picture but one introduces a conduction current and a convection current, if the medium is charged, i.e., where
    $$c \rho_{0\text{free}}=u_{\mu} j_{\text{free}}^{\mu} \neq 0.$$
    ##\rho_{0\text{free}}## is the proper free charge density, as measured in the local restframe of the fluid. Then the consituent equation for the current reads
    $$j_{\text{free}}^{\mu}=j_{\text{cond}}^\mu+\rho_{0 \text{free}} u^{\mu}, \quad j_{\text{cond}}^{\mu}=\sigma F^{\mu \nu} u_{\nu}$$
    or in 3+1-form
    $$\vec{j}_{\text{cond}}^{\mu}=\sigma \gamma (\vec{E}+\vec{\beta} \times \vec{B}).$$
    Note that this conduction current also implies a contribution to the charge density in the frame where the medium is moving:
    $$j_{\text{cond}}^0=c \rho_{\text{cond}}=\vec{\beta} \cdot \vec{j}_{\text{cond}}.$$
    This you can also show by applying the (local) Lorentz boost from the local restframe, where ##j_{0,\text{cond}}^{\mu}=(0,\vec{j}_{0,\text{cond}})## to the observational frame, where the medium is moving.

    The only textbook I know, where this is elaborated in such a compact and clear form is unfortunately available in German only. It's a brillant new textbook, covering all theoretical physics for the undergrad studies (for a German, Austrian, or Swiss BSc):

    M. Bartelmann, B. Feuerbacher, T. Krüger, D. Lüst, A. Rebhan, A. Wipf, Theoretische Physik, Springer Verlag (2015)

    Of course you can also find this in other textbooks, but it's hard to collect in this compact form. I'd have to dig a bit to find an English reference.
     
    Last edited: Aug 5, 2015
  5. Aug 5, 2015 #4

    Dale

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    Good point. I was working with the microscopic Maxwells equations. So no polarization and no distinction between bound and free. I should rework it using the macroscopic equations and see if I get a similar result.
     
  6. Aug 5, 2015 #5

    Dale

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    Yes. That is true. I considered doing a relativistic derivation, but since the context from the previous discussion context was a motor I felt like it was overkill. I should have stated the assumption that ##v<<c##. I went back and put that in.
     
    Last edited: Aug 6, 2015
  7. Aug 5, 2015 #6

    vanhees71

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    There's also this tricky issue with the Abraham vs. the Minkowski Poynting vector ##\vec{E} \times \vec{H}## vs. ##\vec{D} \times \vec{B}##

    https://en.wikipedia.org/wiki/Abraham–Minkowski_controversy

    I'm not sure whether there's a clear resolution of this debate yet. Perhaps one must go back to a microscopic treatment, but that's quite complicated.
     
  8. Aug 5, 2015 #7

    DrDu

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    Strictly speaking? This has not been done in the last 100 years or so simply because since the advent of quantum mechanics it makes no sense to distinguish between free and bound electrons. The question about Ohm's law is rather about the phase of the current relative to the driving field.
     
  9. Aug 5, 2015 #8

    Dale

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    Yes. That debate was precisely the reason I started with the microscopic form. I couldn't determine how to judge for myself and there were a lot of convincing proponents on both sides.
     
  10. Aug 5, 2015 #9

    vanhees71

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    Sure, but it's an overkill to use many-body QED where simple effective models are sufficient ;-). E.g., it's an overkill to calculate the in-medium retarded photon propagator just to describe a classical electromagnetic wave propagating through a dielectric.

    You can of course derive the constituent equations from many-body QED. The usual ones follow in linear-response approximation, but that was not the point of my posting above, but to present a consistent set of effective equations, consistent with relativistic electrodynamics. You can save yourself (and your students!) a lot of trouble, if you treat electrodynamics as a relativistic theory, and that includes macroscopic electrodynamics. It's always easier to do the approximations of the non-relativstic treatment than to unlearn wrong concepts, when it comes to relativistic effects (like in the case of the unipolar generator, which is natural in the relativistic treatment but leads to endless discussions in the usual one).
     
  11. Aug 5, 2015 #10

    DrDu

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    I just wanted to say that in linear response the current of both the "bound" and "free" charges can be included in the same constituitive relation. The free currents being due to the real part of ##\sigma(\omega)## and the bound ones due to the imaginary part of the conductivity.
     
  12. Aug 5, 2015 #11

    vanhees71

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    Sure, this opens another can of worms ;-)). The distinction between free and bound charges is pretty arbitrary, and if you wish you can lump parts of the complex permittivity to the conductivity and vice versa. So you can, e.g., lump all charges into the charges of the medium (including the conduction electrons in a metal, to stay with my previous example). Then you have just a complex ##\epsilon## but ##\sigma=0##, and you can shift arbitrarily pieces from one to the other constitutive quantity. Macroscopic electrodynamics is not as simple as it looks at the first glance :-)).
     
  13. Aug 5, 2015 #12

    DrDu

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    You are right that the conventions are quite heterogeneous, but physics is clear. I prefer a description in terms of either ##\epsilon## or ##\sigma## (complex quantities in general) with the two being simply related by ##\sigma=i\omega \epsilon##which may only be inconvenient in electrostatics.
    More interesting is your neglect of spatial dispersion. Spatial dispersion is necessary to describe e.g. magnetic effects in terms of the E field alone.
    Anyhow one would expect that a Lorentz boost will transform a local relation between j and E into a non-local one, so this is not a covariant assumption.
     
  14. Aug 5, 2015 #13

    vanhees71

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    In the general case, one indeed has (already in linear response) non-local relations like (written in covariant notation)
    $$D_{\mu \nu}(x)=\int \mathrm{d}^4 x \epsilon^{\mu \nu \rho \sigma}(x,x') F_{\rho \sigma}(x')$$
    and anlogously for the dual vectors, if I remember right. I only can't remember, where I have seen this covariant treatment of the most general case of linear-response constitutive relations.
     
  15. Aug 5, 2015 #14

    Dale

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    How do you like this approach?

    http://arxiv.org/abs/physics/0005084
     
  16. Aug 6, 2015 #15

    vanhees71

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    On the one hand I like it, because it gives an interesting alternative perspective on electrodynamics, but on the other hand as an introductory lecture it's not so appealing, and as I stressed before, I'd prefer a relativistic treatment, particularly of the macroscopic electrodynamics, from the beginning, and the constitutive relations are not presented in a covariant (as I tried to summarize on hand of the most simple example of the Minkowski phenomenological model, which is closest to the usually used simplified constitutive relations in "non-relativistic" electrodynamics).
     
  17. Aug 6, 2015 #16

    DrDu

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    That's of course correct, but very redundant as on one hand you can express B in terms of E and on the other hand the definition of D is not unique.
    E.g. you can chose B=H and include all media effects in the relation between E and D (3-D). Probably it makes more sense to consider equivalently the relation between j and A which is unique up to the gauge freedom of A.
     
  18. Aug 6, 2015 #17

    Dale

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    The pedagogical problem with that is that most students encounter Maxwells equations first, before relativity.
     
  19. Aug 6, 2015 #18

    vanhees71

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    That's another "physics didactics sin". In my opinion, the right place for the introduction to special relativity, is in Theoretical Physics II (Analytical Mechanics), after the concept of symmetries in introduced on the example of Newtonian mechanics. The only problem with this approach is that you don't have the Maxwell equations to give the empirical arguments for the introdcuction of another space-time model. You just analyze the Principle of Inertia from the point of view of symmetries, leading to the Galilei-Newton and the Einstein-Minkowski space-time models as the only possible alternative. Then you can argue that the latter is empirically more successful in the context of electromagnetism without necessarily treat it in full glory first. You can, however, introduce the idea of the electromagnetic field as one example to describe a relativistically law of motion for point particles (of course neglecting the radiation-reaction problem).
     
  20. Aug 6, 2015 #19

    Dale

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    I was introduced to Maxwell's equations before then, in Intro Physics II. I think that is pretty common, so it is probably unrealistic to hope that you could introduce relativity before exposure to Maxwell's equations. Through life experiences, more people are personally familiar electromagnetic phenomena (like static electricity) than are familiar with relativistic phenomena (like time dilation).

    What you probably could do is to simply introduce the fully relativistic equations and mathematical structures without much discussion of reference frames and transformations. Then later when you introduce relativity and show the motivation for the mathematical structures it would be easy to see that Maxwell's equations are relativistic.
     
  21. Aug 6, 2015 #20

    Dale

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    If we do not negelct anisotropies etc. then does that change ##\epsilon## etc. into 3-tensors in the local rest frame? This neglecting anisotropies and so forth is something I would prefer to avoid in general derivations and wait until I have a specific scenario where I can determine if the assumption is justified or not.
     
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