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Poynting Vector Direction Confusion

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to understand why, in an example in Griffiths E&M (3rd ed, 8.1) says that the Poynting vector of a current carrying wire that is being heated via resistance (Joule heating), has a Poynting vector pointing radially inward. The E field is parallel to the wire, B field is circumferential with the right hand rule.


    2. Relevant equations



    3. The attempt at a solution
    I believe that the Poynting vector points in the direction of energy flow, and in the direction of the EM wave. But if the wire is radiating heat due to Joule heating, I just don't get why the Poynting vector is radially in, and not radially out. This should be so simple...
     
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  3. Aug 31, 2013 #2

    ehild

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    The wire radiates, because it is hot. But what did heat it up?

    ehild
     
  4. Aug 31, 2013 #3
    The moving charges (current) inside the wire with resistance.
     
  5. Sep 1, 2013 #4

    vanhees71

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    Think about what you just have calculated. It's true that the heat production is due to energy dissipation from the friction of the electrons in the conductor, but how have they been accelerated in the first place?
     
  6. Sep 1, 2013 #5

    rude man

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    P = E x H

    You have the directions of E and H inside the wire correct. Think of looking at a flat cross-section (hold the wire vertically and imagine having made the front half transparent). Let the current flow in direction j so E = E j, then H= +H k looking at the left hand side of the wire but H = -k looking at the right-hand side. You also know that the H field increases in magnitude as we move from the axis to the surface.

    So P = E x H gives a P field that points to the right from the left-hand side (+i direction) and to the left from the right-hand side (-i direction), diminishing in magnitude as the axis is approached from either side.

    Wish I were good at drawing but I'm not, hope you can visualize.

    EDIT: energy is supplied by the external e-m field and flows into the conductor to supply resistance losses in the conductor, so as the P field approaches the axis the energy loss approaches zero and so does the P field.

    If the wire were a perfect conductor there would be no radial P vector in the wire. No resistance loss, no radial P vector.
     
    Last edited: Sep 1, 2013
  7. Sep 1, 2013 #6

    BruceW

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    right, and what is 'forcing' those charges through the resistor?

    edit: I know, it is a slightly surprising answer. But think about a little section of wire. It is losing energy due to heat loss (don't worry about what kind, it could be convection, radiation, whatever). But also, we have a 'steady' system, where the drift speed of the electrons stays the same. The energy in the little section of wire stays the same, so there must be some kind of energy being supplied to that little section of wire, right?
     
  8. Sep 1, 2013 #7
    I agree that energy must be being supplied to the wire to accelerate the charges down its length. But I guess I expected the energy to be coning down the length of the wire with the E field that is accelerating the charges. I believe the direction of the Poynting vector (cross product), but I just don't understand why it is such.

    What if the wire was in a vacuum and the E field contained entirely within its boundaries? Is this a plausible scenario?
     
  9. Sep 2, 2013 #8

    vanhees71

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    This is a quite common misconception. The Poynting vector tells you, where the energy accelerating the electrons in the wire come from, namely from the electromagnetic field. The Poynting vector gives the energy flow of the fields.
     
  10. Sep 2, 2013 #9
    Vector math does this for you. Thing is if you look at the charge of the center of the wire it won't make much sense. That's because the magnetic field is circular around the wire so there is no direction of the magnetic field if you think from the center of the wire. Look at the surface instead and imagine the direction of the magnetic field (Right handed). Now take that point and have your thumb face the E-field and straighten your fingers in the direction of the tangent magnetic field. Your palm will be facing the poynting vector.

    7 years of this **** and I still look like an idiot with my hands to double check with the right hand method.
     
    Last edited: Sep 2, 2013
  11. Sep 2, 2013 #10

    rude man

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    I don't think vacuum has anything to do with it.

    The energy to heat the wire comes from the e-m field outside the wire. The E field outside the wire points mostly in a radial direction (away from the conductor to some other part of the circuit) but there is a small tangential component equal to the E field within the wire (tangential E continuous).

    No way can you not have an E field outside the wire. Think of a path outside the wire between the top and bottom of the wire. There is a potential difference between those two points or there'd be no current. But that p.d. implies an E field of course, by p.d. = ∫E*ds.

    The Poynting vector outside the wire points mostly along the direction of current but there is a samll radial component pointing inwards as previously described.
     
    Last edited: Sep 2, 2013
  12. Sep 2, 2013 #11

    BruceW

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    maybe you are thinking too hard about it. You believe the direction of the Poynting vector. So just take this as the reason for why the energy comes in through the sides of the wire, and not up along the length of the wire. If it is 'as you expected' (where the energy goes down the length of the wire with the E field) then the energy is moving in the same direction as the E field. But we know this can't be true due to the Poynting vector.

    well the EM field has no problem travelling through a vacuum. Maybe instead you mean something like "what if the space around the wire was filled with a substance that is impermeable to EM radiation?" That's an interesting question. Maybe in this case, the energy would be carried by an evanescent wave of some kind?
     
  13. Sep 2, 2013 #12

    vanhees71

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    In the static case there are no waves that have to travel!
     
  14. Sep 2, 2013 #13

    BruceW

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    really? then how does the energy get from the battery to the resistor (or wire with resistance) ? In the static case, there is still energy lost as Ohmic heating. So there must be an energy influx (to the resistor). And I think this energy influx comes from the Poynting vector.
     
  15. Sep 2, 2013 #14

    vanhees71

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    Of course there is energy flux. To understand this issue, it's worth to solve the full problem for a "simple" example, which can be solved completely analytically.

    Let's take the somewhat idealized example of an infinitely long cylindrical wire with a DC running along it. Then you have to solve the stationary Maxwell equations (in Heaviside-Lorentz units)
    [tex]\vec{\nabla} \cdot \vec{E}=0, \quad \vec{\nabla} \cdot \vec{H}=0, \quad \vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \times \vec{H}=\vec{j}.[/tex]
    We introduce polar coordinates [itex](\rho,\varphi,z)[/itex]
    [tex]\vec{j}=\frac{i}{A} \Theta(a-\rho) \vec{e}_z. [/tex]
    Here [itex]a[/itex] is the radius of the wire and [itex]i[/itex] the current, and [itex]A[/itex] the cross section of the wire.

    From the vanishing of the curl of [itex]\vec{E}[/itex] we know that
    [tex]\vec{E}=-\vec{\nabla} \Phi.[/tex]
    For symmetry reasons we make the ansatz [itex]\Phi=\Phi(\rho,z).[/itex] Inside the wire we have
    [tex]\vec{E}=\frac{i A}{\sigma} \vec{e}_z=\text{const} \quad \text{for} \quad \rho<a. \quad (1)[/tex]
    This leads us to the ansatz
    [tex]\Phi(\rho,z)=\phi(\rho) z.[/tex]
    Now we have from the vanishing of the charge density everywhere
    [tex]\Delta \Phi=0 \; \Rightarrow \; [\rho \phi'(\rho)]'=0,[/tex]
    where the prime indicates derivatives wrt. [itex]\rho[/itex]. The general solution of this ODE is
    [tex]\phi(\rho)=C \ln (\rho/a)+D,[/tex]
    where [itex]C[/itex] and [itex]D[/itex] are appropriate constants, which may differ inside and outside of the Wire. From (1) we see that
    [tex]C=0, \quad D=-\frac{i}{A \sigma} \quad \text{for} \quad \rho<a.[/tex]
    Then for [itex]\rho>a[/itex] we use that the potential and the tangential components of [itex]\vec{E}[/itex] must be continuous at [itex]\rho=a[/itex]. This leads to
    [tex]D=-\frac{i}{\sigma A}, \quad C=0.[/tex]
    Thus we have
    [tex]\vec{E}=\frac{i}{A \sigma} \vec{e}_z \quad \text{everywhere}.[/tex]
    To find the magnetic field we use the vector potential
    [tex]\vec{H}=\vec{\nabla} \times \vec{A}[/tex]
    and the Coulomb-gauge condition
    [tex]\vec{\nabla} \cdot \vec{A}=0.[/tex]
    This yields
    [tex]\vec{\nabla} \times \vec{H}=-\Delta \vec{A}=\vec{j}.[/tex]
    Ths symmetry of the problem leads us to the ansatz
    [tex]\vec{A}=A_z(\rho) \vec{e}_z.[/tex]
    The curl reads
    [tex]\vec{H}=-A_z'(\rho) \vec{e}_{\varphi}.[/tex]
    Taking the curl again gives
    [tex]\vec{\nabla} \times \vec{H}=\frac{1}{\rho} (\rho A_z')' \vec{e}_z.[/tex]
    The prime always denotes a derivative wrt. [itex]\rho[/itex]
    Then we have
    [tex](\rho A_z')'=\frac{i}{A} \rho \quad \text{for} \quad \rho<a.[/tex]
    Integration gives
    [tex]A_z'=\frac{i}{2A} \rho + \frac{C}{\rho},[/tex]
    and another integration
    [tex]A_z=-\frac{i}{4A} \rho^2 + C \ln(\rho/a)+D.[/tex]
    Since [itex]A_z[/itex] should have no singularities at [itex]\rho=0[/itex] we must have [itex]C=0[/itex], and we also can set [itex]D=0[/itex] for [itex]\rho<a[/itex]. This leads to
    [tex]\vec{H}=\frac{i}{2A} \rho \vec{e}_z \quad \text{for} \quad \rho<a.[/tex]
    For [itex]\rho>R[/itex] we find
    [tex]A_z=C \ln(\rho/a)+D.[/tex]
    The continuity of [itex]A_z[/itex] and [itex]H_{\varphi}[/itex] leads to (using [itex]A=\pi a^2[/itex])
    [tex]A_z=\frac{i}{4A} a^2 + \frac{i}{2 \pi} \ln(\rho/a), \quad \vec{H}=\frac{i}{2 \pi \rho} \vec{e}_{\varphi} \quad \text{for} \quad \rho \geq a.[/tex]
    The Pointing vector outside of the wire is
    [tex]\vec{S}=\vec{E} \times \vec{H}=\frac{i}{\sigma A} \frac{i}{2\pi \rho} \vec{e}_z \times \vec{e}_{\varphi}=-\frac{I^2}{2 \pi \sigma A \rho} \vec{e}_{\rho}.[/tex]
    The Poynting vector is always directed radially inwards the wire. The total power (energy per unit time) flowing into a piece of the wire with length [itex]L[/itex] gives
    [tex]P_L=2 \pi L a \frac{I^2}{2 \pi \sigma A a}.[/tex]
    To bring this is a more familiar form we note that
    [tex]R=\frac{L}{\sigma A} \; \Rightarrow \; [/tex]
    is the resistance of the piece of wire of length L under consideration, which leads to
    [tex]P_L=\frac{I^2}{2R},[/tex]
    as it should be, because that's precisely the power dissipated through the resistance of the wire!
     
  16. Sep 2, 2013 #15

    BruceW

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    That looks like a nice derivation. I'll read it more carefully tomorrow. I did notice, in the last equation I think the 1/2 is there by accident? So, the conclusion for this 'simple' example is that there is an EM wave which supplies energy to the wire, whose Poynting vector points (cylindrical) radially into the wire. I'm still not sure what you meant that there are no waves that travel. Did you mean this EM wave is not a travelling wave? I forget the definition of travelling wave.
     
  17. Sep 3, 2013 #16

    vanhees71

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    Sure! The final formula of course reads
    [tex]P_L=\frac{I^2}{R}.[/tex]
    For some reason, I'm not allowed to correct the previous posting.

    @Admins: Why can't one correct typos in ones own postings after some time? I'd find it better to be able to do so, because mistakes might misguide also people who read the postings later and then don't see a correction in a later posting.
     
  18. Sep 3, 2013 #17

    rude man

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    Did you not mean I^2 R?
     
  19. Sep 3, 2013 #18

    rude man

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    I take issue with this statement and part of the derivation. The purpose of the Poynting vector is to indicate direction of energy travel. If σ → ∞ there is in fact no ρ component at all, either inside the wire or outside. Outside the wire the Poynting vector is predominantly along the direction of current. Least that's what I thought I learned ...

    I did not follow your derivation step-by-step, however, so I can't point to any specific error.
     
  20. Sep 3, 2013 #19

    vanhees71

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    The whole derivation doesn't make sense for [itex]\sigma \rightarrow \infty[/itex]. You can, however take that limit for the Poynting vector, which gives indeed 0 as it should, because for zero resistance there is no energy dissipation and thus no power needed to sustain the current!

    As the calculation shows tha Poynting vector is NOT directed along the current. That's the whole point of the calculation! If you don't believe me, you find a very detailed discussion in Sommerfeld's lectures on theoretical physics vol. III :-). There he treats the somewhat more complete task to evaluate the same thing for a coax cable, but the fundamental thing is the same as in my more simplified example.
     
  21. Sep 3, 2013 #20

    rude man

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    I don't have access to the Sommerfeld paper (could it be that his S is not your S? What's the date of 1st publication of that paper? Maybe his S is not E x H? In any case, if S outside the wire were not in the direction of the wire we would not have cable TV!

    I in turn will cite H H Skilling, Fundamentals of Electric Waves, 2nd edition, pp. 133 ff.

    BTW you apparently never derived the E field for outside the wire, using the inside value instead thruout your derivation. I suppose that's what led you to conclude that S points radially outside the wire as well as inside.

    The predominant component of E outside the wire is due to surface charges placed on the wire when and as current is established. The E field points from those charges to other similar charges elsewhere in the circuit, and is typically >> E inside the wire. So the E field points predominantly in the ρ direction, with only a small z component to satisfy continuity across the metal-space boundary.
     
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