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Poynting's theorem and escaping particles

  1. Jun 23, 2014 #1
    I was thinking the other day and it looks like I need feedback if I am being stupid or crazy.

    My question is quite simple - does Poynting's theorem cover the situation of charged particles escaping through surface (just like radiation part does) of the volume over which we calculate?

    I honestly don't see it through some simple argument and I would like to avoid the necessity of boring calculations.

    For example:
    Two point charges
    [itex]T<0[/itex]: I have two standing particles with the same charge and a distance [itex]d[/itex] between them. No mechanical work being done, no change in fields inside chosen volume (with some big radius [itex]R[/itex]), no [itex]E\times B[/itex] escaping through surface.
    [itex]T>0[/itex]: I let them free and they start to move away from each other. Now the situation will be more interesting (and pain to calculate, at least for me). Particles accelerate and the bubble of changing fields will propagate from particles to the surface of our chosen volume. I really don't want to integrate fields of moving particles.
    [itex]T=+\infty[/itex]: Particles are in infinity and all the energy which was originally hidden in [itex]\approx \int_V E^2[/itex] have left our volume. Some of it changed into kinetic energy of those particles.

    And the rest must have crossed the surface as [itex]E\times B[/itex]. Well, maybe it did, it just doesn't "feel" right with my current state of intuition. Second example might be better.

    One point charge
    [itex]T<0[/itex]: Chunk of EM radiation goes in (it's not exactly classical electrodynamics but lets take photon). Photon is in [itex]T=0[/itex] completely absorbed by so far standing particle.
    [itex]T>0[/itex]: Particle moves with constant speed, so the shape of EM fields will be a bit easier.

    If I oversimplify this example and say that electron moves very very slow, then I could neglect [itex]B[/itex] field. And we can see where would such oversimplification lead.

    [itex]\approx j\cdot E[/itex] is no concern, after photon absorption there is no mechanical work to be done. All the original energy in [itex]\approx \int_V E^2[/itex] must leave through surface as [itex]E\times B[/itex]. And yet, is quite hard to do with negligible [itex]B[/itex].


    Two possibilities comes to my mind.
    1. Poynting's theorem covers a lots of situations, but it just isn't build for this. And we can avoid such situation all together if we choose infinite area over which we calculate (particles never reach surface).
    2. My intuition is wrong and all the energy from static fields really goes through surface as [itex]E\times B[/itex].
     
  2. jcsd
  3. Jun 23, 2014 #2

    WannabeNewton

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    You've written a lot, and I'm sorry if I can't reply to it all in detail, but even from the start it's quite clear that Poynting's theorem can handle that certainly well, unless of course I'm totally misunderstanding what you're describing. Are you looking at an integral form of the theorem? Look at the differential form instead and it will be entirely obvious to you. The differential form is given by ##\partial_t \mathcal{E} + \nabla \cdot S = -j_{f}\cdot E## which is just the continuity equation for the energy-momentum of the electromagnetic field. For charged particles leaving from the boundary of an infinitesimal volume, the energy flux carried by the particles away from the volume is entirely taken into account by the ## \nabla \cdot S## term which accounts for the flow into or out of the volume.
     
  4. Jun 23, 2014 #3
    Thanks for trying to answer, but sadly it didn't help me much. If we take finite volume, it doesn't really matters whether we take an integral form or differential one, so lets stick with your "infinitesimal" volume.

    Sure, if I just take the Poynting's theorem as a mantra, then all it's precisely as you said (the possibility number 2. in my original message) - it all has to leave through ## \nabla \cdot S## term, no other way.

    But do you actually see, how all the energy from originally static field of electron is "remelted" into ## \nabla \cdot S## when electron passes through the boundaries? If you can really visualize it in your head, then I have to bow before you.

    If I do a heresy for a moment and don't take Poynting's theorem as fundamental mantra, but instead look at it as:

    A. Guys, we need to make formula for energy-momentum conservation of EM fields.
    B. Sure, lets build one, what do we want from it?
    C. It has to be in harmony with Maxwell equations and follow some common sense.
    B. Ok, well then. What can happen to the energy of EM fields in some volume?
    A. It can be changed into mechanical work done on charged particles. ## \textbf{E} \cdot (\nabla \times \textbf{B}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \mu_o \textbf{J} ##
    C. That makes sense, what else?
    B. Energy and momentum can also escape as radiation. (Mixing first equation with ##\textbf{B} \cdot (\nabla \times \textbf{E}) + \textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t} = 0 ##)
    A. Nice, it looks like all of it.
    C. Yea, here I meanwhile wrote it into pretty formula.


    Guy D returns with new beers and looks on formula.

    D. Whats this? Sure, that makes sense and looks like it can be quite useful.
    A. Cheers.
    D. Wait a moment, I know we don't assign momentum to static E field without B (at least the net momentum is surely zero) and vice versa. But we won't argue it has energy. What about particles that goes through boundaries?
    B. Go away.


    Maybe it all really goes through ## \nabla \cdot S ## term. If so, it's truly nice, deep and not exactly intuitive result.

    Was the thing I am pointing towards taken into consideration in the derivation of Poynting's theorem?
    I am not aware of it, please direct me to proper materials if you know about it.
    If the theorem covers the situation I am describing, then again please direct me to proper materials.
    Or even better, show it here for everyone to see.
     
    Last edited: Jun 23, 2014
  5. Jun 23, 2014 #4

    Jano L.

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    The Poynting theorem
    $$
    \int_V \mathbf E\cdot \mathbf j\, dV = -\int_V\partial_t \left( \frac{1}{2}\epsilon_0E^2 + \frac{1}{2\mu_0}B^2 \right) \,dV - \oint_S d\mathbf S \cdot (\mathbf E\times \mathbf B/\mu_0),
    $$
    is valid as long as all the quantities are meaningful (integrals are convergent...), irrespective of whether matter crosses imaginary surface or not.

    However, your are right in your observation that the Poynting theorem has nothing to say about transfer of non-electromagnetic energy - the internal and kinetic energy of matter.

    When the term ##\mathbf j\cdot \mathbf E## is interpreted as work of the EM field on matter per unit volume and unit time, the theorem allows inference of expressions for energy of electromagnetic field and energy current density of that same field. But neither energy of matter nor density of its current can be inferred; in the theorem above, these are not specified.

    To describe explicitly the transfer of matter energy, one has to use equations of motion of matter (Maxwell's equations are not sufficient), for example Newtonian equations or Navier-Stokes equations, and infer energy and transfer of energy from these.

    The matter and field equations can then be used to derive energy transfer law similar to Poynting's where instead of the term ##\mathbf E\cdot \mathbf j## there are terms giving matter energy and flow of matter energy through the surface.

    Similar thing can be done for momentum of matter and field. The two equations then can be joined int he single tensor equation of energy-momentum transfer, where tensors of energy-momentum of matter and field are used.

    1) The Poynting theorem is valid in such situations, but of course it does not give full picture.
    2) Yes, transfer of energy of EM field is given by the Poynting vector. In case matter crosses surface, there is additional transfer of energy of matter, which is not given by the Poynting vector, but in principle can be derived from the equations of motion of matter. For example, if matter is incompressible fluid with density ##D##, pressure ##p## and velocity ##\mathbf v##, density of of matter energy is ##\epsilon_{matter}=\frac{1}{2}D v^2## and density current of matter energy is ##\mathbf S_{matter} = (p + \frac{1}{2}Dv^2)\mathbf{v}##.
     
    Last edited: Jun 23, 2014
  6. Jun 24, 2014 #5
    Ok, I have discussed this simultaneously on physics.stackexchange.com and it helped me to clear out the core of what I actually wanted to ask:

    If it is as I just described, then everything fits and there is no problem - P. theorem just doesn't deal with the energy in static fields.

    Indeed if it did, it would indicate, we could dig out Gauss's law out of Ampere's and Faraday's laws.

    Am I wrong?
     
  7. Jun 24, 2014 #6

    Jano L.

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    Poynting vector is dependent on the field. These are partially affected by velocity of charged matter, but they have also freedom on their own - the fields are not entirely determined by what matter does at the same time.

    No, Poynting expressions do not give total energy or total momentum corresponding to charged particle. They only give electromagnetic energy and momentum of field, including static field.

    There is another contribution to total energy and momentum due to kinetic energy and momentum ##m\mathbf v## of the particle.

    It deals with all EM energy including that of static field. It just does not deal with rest and kinetic energy of particles.
     
  8. Jun 24, 2014 #7
    Ok, not precise formulation on my part. No arguing about separate freedom of EM fields (independent waves, implications of retarded potentials and so on).

    I really should be more careful what I say. Could I say - P. vector caries the energy and momentum needed to "push" those originally static fields?

    EM wave goes in, part of it is spent in pushing the particle (mechanical momentum to particle, EM momentum to get moving originally static fields). Rest of the wave and some additional artifacts from acceleration goes out. Now the question is, what P. theorem sees when particle goes through the boundary of chosen area.

    Does it deal with energy of static fields completely? [itex]\approx \int_V E^2[/itex] leaves fully as the flux of Poynting vector when the particle leaves? Or does it register just the energy-momentum needed for "pushing" those static fields?

    Hope its clear what I am asking.
     
    Last edited: Jun 24, 2014
  9. Jun 24, 2014 #8
    And to prevent another possible misunderstanding.

    P. vector of independent radiation which goes away is one thing.

    P. vector of fields associated with moving point charge which has constant speed is another thing - the main contribution here I would say.
     
  10. Jun 24, 2014 #9

    Jano L.

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    I don't know what you mean. Fields are not usually thought of as being "pushed". Charged matter may be pushed by EM forces.

    The integral of the Poynting vector over the boundary surface gives electromagnetic energy that leaves the region through this surface per unit time. Another energy leaves due to particle leaving the region. This second contribution to energy drain is left unaccounted for by the Poynting theorem. The theorem is still valid, however.

    Yes.

    $$
    \int_V E^2dV
    $$

    does not "leave the region", it just changes value. When the particle leaves the region, some electromagnetic energy leaves that region too and this is accounted for exactly by the Poynting theorem.

    Again, it is not clear what do you mean.
     
  11. Jun 24, 2014 #10

    Jano L.

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    The Poynting theorem in the form I wrote it above is valid only if total fields are used to calculate all the quantities. You can define Poynting-like vector based on partial fields of one source, but then the Poynting theorem may not hold for it.
     
  12. Jun 24, 2014 #11
    Neutral particle (with some other properties) requires some energy-momentum to get moving with certain speed.

    Charged particle (with other properties unchanged) requires more energy-momentum to get moving with same speed as before.

    We could just increase the mass of the particle and be done with it. This difference in the mass is what I described as "pushing fields".
     
  13. Jun 24, 2014 #12
    Energy of fields (in the area) from particle (which have been in place A) changed its value to 0, because particle moved to far away B, where the energy of fields risen to the value of energy of fields from this moved particle. Sorry I described it as leaving...


    Cleaning everything from the table.

    That's why it seems to me P. theorem doesn't register ##E_1##.
     
    Last edited: Jun 24, 2014
  14. Jun 24, 2014 #13

    Jano L.

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    The Poynting theorem is not valid for point charged particles. Let us assume small ball of charge with finite charge density instead.

    Yes, but then also the rate of decrease of energy is close to zero.
     
  15. Jun 24, 2014 #14
    Now we are maybe getting somewhere, sorry I've wasted lots of your time on garbage before.

    Lets look on the result.

    Do I (with reduced speed) decrease the total flux that happens between ##(0,\infty)##, because I am reducing size of overall P. vector?

    The rate of decrease of energy slows down for sure too, but with enough time all the energy which was initially inside will leave.
     
  16. Jun 24, 2014 #15
    To answer my own question.

    When I look at final situation (energy moved completely from the chosen volume), I should be more careful how I integrate the flux of ##E\times B## (gets smaller) over time (gets longer).
     
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