Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confused about a derivation of Poynting's theorem

  1. Jul 17, 2012 #1
    I (and probably everyone) like to learn from books that describe the reasoning/philosophy behind the equations as much as possible, instead of "pluging this in and getting that" or even just dishing out "clearlys". There's a footnote to the word "clearly" in some online LaTex tutorial:

    Griffiths' "Introduction to Electrodynamics" is an excellent book, however, I feel I stumbled on a weak spot.
    Here is an excerpt, which is the source of confusion (start of a derivation of Poynting's theorem):

    Poynting.png

    _________________________________________________
    Not central to the point:

    The first equation should probably have one more "d":

    [itex]\mathrm{d}(\mathrm{d}W) = \mathrm{d}(\vec{F} \cdot \mathrm{d} \vec{l}) = \mathrm{d}q(\vec{E} + \vec{v} \times \vec{B}) \cdot \vec{v} \, \mathrm{d}t = \mathrm{d}q \, \vec{E} \cdot \vec{v} \, \mathrm{d}t[/itex] ,

    since we are talking about a differential volume (first d) and a differential time (second d). So that later [itex]\mathrm{d}q = \rho \, \mathrm{d}\tau[/itex], not [itex]q = \rho \, \mathrm{d}\tau[/itex], and everything works out with the integral. But that's not the point.
    _________________________________________________

    What bothers me: Equation (8.6) is fine, when there is an external electric field [itex]\vec{E}(\vec{r})[/itex] acting on a current density distribution [itex]\vec{J}(\vec{r})[/itex]. The right-hand side then properly describes the power of that external field (work of the surroundings per unit time) delivered to the charges. As far as I understand, the whole point of the field is to model the effects of the surroundings (other particles) on a particle. One should then not calculate the effects of a particle's field on itself.

    Example: There is a charge/current distibution somewhere. You would like to know how it affects another charge/current elsewhere. You use the Maxwell's equations to determine the [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] (caused by the first charge/current distribution) at the point of interest (elsewhere). You are then able to determine the effects of the fields on the second (another) charge/current using the Lorentz force law.

    But what is done in the excerpt is exactly that (calculating the effects of a particle's field on itself): In the first paragraph the fields [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] are said to be caused by a charge and current configuration. To get the power delivered by the fields (ofcourse (clearly :biggrin:) only E-field does work) to the current density [itex]\vec{J}[/itex] in a point of that configuration, I imagine we would need to first calculate the field in that point (caused by all the other points in the configuration --> integral) and then dot multiply it by [itex]\vec{J}[/itex] in that point. Then, to get the total power (over all points in the configuration), repeat (integrate) for all points in the configuration - the result would then be a double integral.

    Instead, it is implied in the last paragraph and equation, that the field [itex]\vec{E}[/itex] in the point of interest is caused only by [itex]\vec{J}[/itex] in that same point - since [itex]\vec{J}[/itex] is expressed from Ampere-Maxwell's law and substituted in.

    Perhaps the logic is, that every combination (of two points) appears twice in the double integral and the two appearances are exactly opposite, so the double integral evaluates to zero unless you also consider the effect of every point on itself? If so, that would work. But you would still be calculating the effect of particle's field on itself.

    Any insights?
     
  2. jcsd
  3. Jul 17, 2012 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member


    Are you saying that [itex]d\left( \mathbf{F} \cdot d \mathbf{l}\right) = d \mathbf{F} \cdot d \mathbf{l}[/itex] and [itex]d\left[ q \left( \mathbf{E}+ \mathbf{v} \times \mathbf{B} \right) \right] = dq \left( \mathbf{E}+\mathbf{v} \times \mathbf{B} \right)[/itex]? What happened to the product rule?:bugeye:

    The idea is that when he says "Here [itex]q=\rho d\tau[/itex]..." he is now looking at an infinitesimal volume of space so that the charge enclosed is just the charge density times the volume differential.

    Griffiths could perhaps make that clearer by using [itex]dq[/itex] in the preceding paragraph and equation, but it should really be understood by the time a student gets to chapter 8 of the text that [itex]\mathbf{F}=q( \mathbf{E}+\mathbf{v} \times \mathbf{B} )
    [/itex] applies only to point charges, or small enough volumes of charges that the fields, charge density and velocity are all constant over the volume.

    Certainly, I would not go around claiming that [itex]d^2W= dq \left( \mathbf{E}+\mathbf{v} \times \mathbf{B} \right) \cdot \mathbf{v} dt[/itex].

    See part (i) in section 2.4.4 (3rd edition).
     
    Last edited: Jul 17, 2012
  4. Jul 17, 2012 #3
    I should have used [itex]\partial[/itex]. So the first [itex]\partial[/itex] couples with infinitesimal volume [itex]\partial \tau[/itex], and the second with [itex]\partial t[/itex]. What I was getting at:

    [itex]\frac{\partial}{\partial \tau}(\frac{\partial W}{\partial t}) = \vec{E} \cdot \vec{J} \, ,[/itex]

    which you can't get to from the first equation in the excerpt if you use [itex]q = \rho \, \mathrm{d} \tau[/itex].

    I understand that, I just didn't like how there was a "d" on one side, but not on the other.
    I did not read the book sequentially. It was not in the literature for the course. I picked it up because I liked it, so I am reading different chapters as the need appears.

    I'll try to be more careful, so I don't cause any more allergic reactions. :biggrin:

    Thanks for the chapter reference. Will read it.
     
    Last edited: Jul 18, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Confused about a derivation of Poynting's theorem
  1. Poynting theorem (Replies: 2)

  2. Poynting theorem (Replies: 3)

Loading...