Poynting Vector Direction Confusion

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Homework Statement



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.


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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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  • #2
ehild
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The wire radiates, because it is hot. But what did heat it up?

ehild
 
  • #3
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The moving charges (current) inside the wire with resistance.
 
  • #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?
 
  • #5
rude man
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Homework Statement



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.


Homework Equations





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...

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.
 
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  • #6
BruceW
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The moving charges (current) inside the wire with resistance.
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?
 
  • #7
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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?
 
  • #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.
 
  • #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.
 
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  • #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.
 
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  • #11
BruceW
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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.
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.

teroenza said:
What if the wire was in a vacuum and the E field contained entirely within its boundaries? Is this a plausible scenario?
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?
 
  • #12
vanhees71
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In the static case there are no waves that have to travel!
 
  • #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.
 
  • #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!
 
  • #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.
 
  • #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.
 
  • #17
rude man
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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.

Did you not mean I^2 R?
 
  • #18
rude man
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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]

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.
 
  • #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.
 
  • #20
rude man
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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.

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.
 
  • #21
vanhees71
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Sommerfeld is a textbook. I have only the German version at hand, so perhaps it doesn't make to much sense to put the pages online. It is really a somewhat surprising result that the energy flow is perpendicular to the direction of the cable, but it's a well-known fact of basic Maxwell theory!

BTW: For the coaxial cable the Poynting vector in the free space between the inner and the outer conductor has a component in direction of the axis. I've put the calculation also online, but in a my German Physics FAQ. Perhaps you can understand it, because there are many formulae:

http://theory.gsi.de/~vanhees/faq/coax/coax.html

You find the simpler but somewhat less practical example of the single wire treated in here as the limit [itex]b \rightarrow \infty[/itex] of the coax-cable calculation.

For the full coax-cable calculation it is shown explicitly that the entire energy transport is not along the cable but by the field in the free space between the conductor! Maybe that's another surprising result for many of you (at least I was surprised, when I've seen this calculation at the first time :-)).

There is also a fully relativistic treatment, taking into account the self-consistent Hall effect. This was a discussion in a German newsgroup (at a time, where the usenet was still useful for physics discussions :-)).

Cable TV has nothing to do with this, because there you have a waveguide, where the wave is traveling along the axis of the cable.

The Poynting vector is always [itex]\vec{S}=\vec{E} \times \vec{B}[/itex] (and it's [itex]\vec{B}=\vec{H}[/itex] in vacuo (in the here used Heaviside-Lorentz units), of course also in Sommerfeld's book and in my calculation in this thread, and it's the energy current density of the electromagnetic field.

The TEM wave in a (e.g., coaxial-cable) wave guide is like a plane wave in free space, and there the Poynting vector is in direction of the wave vector. That's a different non-static situation!
 
  • #22
rude man
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Sommerfeld is a textbook. I have only the German version at hand, so perhaps it doesn't make to much sense to put the pages online.
It's all well and good to quote innumerable sources. We can all play that game. I have too. And, for the record, I do read German.

But, you haven't answered my question about your derivation of the E, hence the S, fields outside the conductor. You never addressed that issue.

Take an example: two long parallel wires, length L each, conjoined top and bottom by arbitrarily short wires. A very thin source of emf is placed in the circuit at the half-way point of one wire, forming a closed circuit with finite current i.

The E field inside the wire is of course j/σ. What do you think the E field between those two very close-together wires is? Answer: >> j/σ and directed mostly between the wires, i.e. almost perpendicular to the inside-the-wire E field.

Then S is therefore also not pointing predominantly radially but along the wires.
 
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  • #23
rude man
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@teroenza: I hope you realize that, despite the bickering, we all agree on the Poynting vector's direction inside the wire, and the reason for it.

It's outside the wire that disagreement continues ...
 
  • #24
rude man
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OK, for the benefit of our non-German-reading fans, I will excerpt a section of this paper and paraphrase:

Da weiter als Tangentialkomponente entlang der Leiterflächen stetig sein muß, muß im Außenraume, Gebiet 4, ebenfalls das konstante elektrische Feld herrschen.

'Tangentialkomponente' means 'tangential components', which, for the E field, as I have pointed out, have to be equal just inside & outside the conductor. That's all that that statement says. Trust me.

But the E field outside the conductor comprises much more than that tangential component. I hope I have shown with some persuasion in my previous post (#22) that the outside E field can be, and typically is, much larger than that tangential component.
 
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  • #25
vanhees71
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But I have calculated the electric field everywhere, inside and outside of the conductor in both my posting and in the FAQ article. I don't know what else to do to convince you.
 
  • #26
rude man
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But I have calculated the electric field everywhere, inside and outside of the conductor in both my posting and in the FAQ article. I don't know what else to do to convince you.

My take is that you merely recognized the need for continuity of the tangential component and concluded that that is all there is to the external E field. But I have tried to show that your wire must be part of a circuit, and that implies static surface charges, potential changes along the circuit, and hence an E field on top of the tangential component connecting those surface charges.

I think maybe the problem here is that you're thinking of an isolated wire and I'm thinking of the wire as part of a circuit. In any case, we're probaly just muddying the waters for our OP, and we all agree on the E and S vectors inside the wire, so I'm going to leave it at that.
 
  • #27
BruceW
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yeah, for sure, in a real circuit, the energy must travel through the space between the battery and the little bit of wire we are considering. So the Poynting vector is not 'radial everywhere'. I'm guessing this is what you mean.
 
  • #28
vanhees71
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Sigh. The coax cable is indeed more realistic as I stressed a lot of times. I think you can unerstand that on my FAQ page although the text is in German. The "formula density" should be high enough for that:

http://theory.gsi.de/~vanhees/faq/coax/node5.html

It's of course also an academic situation because of the infinitely long wire. As stressed in the Section I just link to you can see it as the solution for a finitely long wire, where the length is much larger than the outer radius of the coax, so that boundary effects can be neglected. This approximation should be find to see precisely, how the Poynting vector behaves in the various regions (in the interior of the inner conductor; in the free space between the conductor; in the outer condutor; and outside of the cable).

The boundary conditions, i.e., continuity of the tangential components of the electric field, on the cylinder mantle are crucial to find the solution. That's all described in the calculation of the electromagnetic field in the Sections before.
 
  • #29
rude man
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yeah, for sure, in a real circuit, the energy must travel through the space between the battery and the little bit of wire we are considering. So the Poynting vector is not 'radial everywhere'. I'm guessing this is what you mean.

More or less. I don't know about the 'little bit' of wire. But yeah, the smaller the circuit the bigger the E field perpendicular to the wire in relation to the internal E field. The external field will be >> the internal one and more so for smaller circuits since V = -∫E ds so for the same potential difference at some remote point along the wire V we get a bigger E for a smaller loop.

@vanhees, I see no significance in forming an analogy with coax wire. This is a single-wire circuit.
 
  • #30
vanhees71
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Which geometry do you have in mind? If you want a complete closed and finite circuit, maybe a toroidal shape is solvable. I have to check when I have som time over the weekend. The problem is that it is not easy to solve even simple looking examples analytically.

That's why the coax cable is used as the most simple example, and in fact it teaches you that the energy flow is not along the wire inside (how can it since the current and thus also the electric field inside the wire is tangential to the wire, and the energy flow (Poynting vector) is perpendicular). In the case of the coax cable the energy transfer along the wire is due to the electromagnetic field in the free space between the inner and outer conductor.
 

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