Talk:Poynting vector

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1 Definition
2 Non-zero Poynting vector in static situtation
3 Poynting vector around a wire
4 Standard textbooks
- 4.1 Free and non-free space
5 Alternate Expressions for Poynting Vector
6 Radiation in a DC circuit

[edit] Definition

Is the definition

S = E x H

or is it the complex conjugate of H

S = E x H*

Anyone know which it is? --H2g2bob 21:58, 23 May 2006 (UTC)

In nonquantum physics (which is what we are talking about here), the directional components of E and H are real numbers. So H=H* for any actual physical field. However, it is common in electrical engineering to apply a complex Fourier decomposition to the real fields and talk about complex valued sinusoidal waves. So the equations are modified to a complex form for handling such waves. JRSpriggs 03:45, 24 May 2006 (UTC)

The poynting vector is ExH which gives the instantaneous energy flow crossing unit area normal to the direction of flow per unit time. However, you may have seen ExH* in the expression for the time averaged poynting vector which is 1/2 Re(ExH*) and is the average energy of the electromagnetic wave crossing unit area (perpendicular to the vector) per unit time. —Preceding unsigned comment added by 62.31.207.26 (talk) 20:01, 6 May 2008 (UTC)

[edit] Non-zero Poynting vector in static situtation

Let's imagine a simple situation: We have two vertical parallel metallic plates, carrying charges +Q and -Q, generating a homogeneous electrical field E. From above on a thin cord we hang between these plates a bar magnet, which eventually gets to rest. So nothing moves, nothing happens. Because there are electric and magnetic fields present, we can assign to any point in space a Poynting vector, which says, that at (almost) each point there is flowing energy.

I have difficulty to imagine and believe this. Is there any experimental evidence about this energy flow? Has it been measured? Please don't come with explanations from quantum physics. The Poynting vector was introduced in classical times in the end of nineteenth century. —Preceding unsigned comment added by Pschultz (talk • contribs)

Just because there is no radiation does not mean that there should be no linear momentum in the electromagnetic field, even if it adds up to zero. It may be necessary to store angular momentum. JRSpriggs 06:40, 7 September 2006 (UTC)

The energy can be considered to be circulating in closed loops. The electric and magnetic field exist also outside the plates. The energy can be considered to be continually circulating between the outside of the plates and the space in between them. However the net transfer of energy between any two points is zero. A better explanation is given briefly in Richard Feynman's "Lectures in Physics" volume 2 page 27-8. Grottlu 16:56, 29 December 2006 (UTC)grottlu

Another useful reference is Examples of Momentum Distributions in the Electromagnetic field and in Matter by W. H. Furry, published in American Journal of Physics 37, June 1969. A couple of quotes from this paper:

"Some recent papers [1-3] in this Journal have emphasised the importance of accepting the Poynting vector as a description of the space distribution of energy flux and momentum density in the electromagnetic field. As these authors have remarked, the deprecatory attitude toward the Poynting vector traditionally taken in textbooks is not only unfortunately misleading, but actually entirely incorrect. The example of angular momentum in a static field given by Pugh and Pugh [2] is particularly conclusive evidence for this."

"The idea that the Poynting vector is not to be taken seriously as a detailed distribution of energy flow and momentum density is a relic from the time when the energy-balance theorem for the electromagnetic field was taken as an isolated result. The single theorem indeed remains valid if one changes the Poynting vector by adding to it any vector whose divergence is zero. Such a change is clearly forbidden, however, by the theorems for momentum and angular momentum. The example given by Pugh and Pugh [2] conclusively vindicates the Poynting vector for the notorious case of a charged magnet for which generations of textbook writers declared that there could obviously be no actual energy flow."

The references [1-3] are [1] R. H. Romer, Amer. J. Phys. 34, 772 (1966); 35, 445 (1967); [2] E. M. Pugh and G. H. Pugh, Amer. J. Phys. 35, 153 (1967); [3] M. G. Calkin, Amer. J. Phys. 34, 921 (1966). A description of the Pugh & Pugh example can also be found in Albert Shadowitz's book The Electomagnetic Field (pp 426-429 in the 1988 Dover edition) and is interesting in that it describes a case where the discrepancy in mechanical angular momentum is exactly matched by the angular momentum of the static electromagnetic field. The angular momentum of the em field depends on the Poynting vector, showing the profound importance of this vector, even in the static case.AQv (talk) 17:26, 14 January 2008 (UTC)

[edit] Poynting vector around a wire

I removed this statement, but it was reverted: For example, the Poynting vector near an ideally conducting wire is parallel to the wire axis - so electric energy is flowing in space outside of the wire. The Poynting vector becomes tilted toward wire for a resistive wire, indicating that energy flows from the e/m field into the wire, producing resistive Joule heating in the wire. In an ideally conducting straight wire, there is no potential difference along the wire since otherwise the current would be infinite. That is simply Ohm's law. Hence, the Poynting vector ExH must be zero. If the wire has an ohmic resistance, E is parallel to the wire and H is circular around the wire. In this case, S=ExH will point into the wire, rather than being tilted towards the wire. I checked Jackson, Classical electrodynamics, 2nd edition, and it appears that more is wrong in this article. In the discussion of the Poynting vector, it states:

$\frac{\partial u}{\partial t} + \mathbf{\nabla} \cdot \mathbf{S} = - \mathbf{J}\cdot\mathbf{E}$

(warning: Gaussian/CGS units; u is the electromagnetic energy density) and "Since only the divergence appears in the conservation law, the Poynting vector is arbitrary to the extent that the curl of any vector field can be added to it." Hence, it is not generally correct at all to interpret the Poynting vector as an energy flow per se. Furthermore, this was removed for "being untrue": For an electromagnetic wave propagating in vacuum, the irradiance can be written as

$I = \frac{1}{\mu_0 c} E^2 = \frac{\epsilon_0 c}{2} E^2,$

where E is the amplitude of the electric field, $ε 0$ is the permittivity of free space.. For this I can simply refer you to Jackson, equation 7.13. Han-Kwang 07:58, 8 July 2007 (UTC)

What do you mean CGS? That's the way Poynting's theorem looks in SI units. AQv (talk) 15:55, 16 January 2008 (UTC)

You say you're quoting from the 2nd edition of Jackson's Classical electrodynamics. It was publised in 1975, but do you know that there is a third edition, from 1998? I haven't got access to it myself right now, but have you checked if the formulation about the arbitrariness of the Poynting vector is the same in this later edition? AQv (talk) 17:53, 16 January 2008 (UTC)

Given the price and how rarely I look up things I never bothered buying the new edition. I'll see tomorrow if someone at work (a physics research lab) has the new edition. Han-Kwang (t) 18:33, 16 January 2008 (UTC)

[edit] Explanation of revert by JRSpriggs

(1) You dropped a factor of two in your irradiance equation, since $\mu_0 \epsilon_0 c^2 = 1 \!$ is an identity.

(2) You said "In an ideally conducting straight wire, there is no potential difference along the wire since otherwise the current would be infinite. That is simply Ohm's law. Hence, the Poynting vector ExH must be zero.". This is false and shows that you do not know what you are talking about. The E field would not be along the wire, it is perpendicular to the wire (pointing away from it or towards it). Since the B field wraps around the wire, the S which is their cross product points along the wire. How else could the electrical energy from the source be delivered to the load?

(3) Also your statement about S being indeterminate by a constant is false. S represents the flux of energy (or density of linear momentum) and is part of the stress-energy tensor which is the source of the gravitational field. Consequently, it must have a definite value at each event in space-time.

(4) There is no basis for introducing square-roots where you introduced them.

I will not waste my time looking for additional errors. These are enough to justify reverting you. JRSpriggs 02:28, 9 July 2007 (UTC)

[edit] Han-Kwang's rebuttal

I based most of my edits on Jackson's Classical Electrodynamics. If you doubt the validity of this standard reference, you should provide a reference.

(1) Agreed, thank you for pointing this out to me.

(2) The E field would not be along the wire, it is perpendicular to the wire (pointing away from it or towards it). -- The current in an ohmic wire follows the eq. J = sigma E, which means that the field in general is along the wire. However, I think I see now what you mean. Since in a typical application, there is another wire somewhere else at a different potential, the is a field perpendicular to the wire. However, this is not generally the case, so such a statement should be clarified. For example, there is no potential difference with a superconducting loop in a magnet. In any case, the current article statement Poynting vector becomes tilted toward wire for a resistive wire, indicating that energy flows from the e/m field into the wire, producing resistive Joule heating in the wire. is misleading, since the Poynting vector should be used in Poynting's theorem, in conjunction with the energy density u and the current dissipation J.E.

(3) Quoting from Jackson page 237: The vector S, representing energy flow is called the Poynting vector. It is given by S=c/4pi (ExH) [CGS units] and has the dimensions of (energy/area x time). Since only its divergence appears in the conservation law, the Poynting vector is arbitrary to the extent that the curl of any vector field can be added to it. Such an added term can, however, have no physical consequences. Hence it is customary to make the specific choice (eq 6.109).

(4) I am not sure what you mean. Do you mean this one?

$\langle S \rangle = \frac{1}{2 \mu_0 c} \sqrt{\frac{\epsilon_r}{\mu_r}} E_0^2$

It is based on the CGS equation 7.13 in Jackson: S = c/8pi * sqrt(epsilon/mu) * |E0^2| * n, where n is the unit vector in the direction of propagation. Did I make an error in the conversion into SI units?

Han-Kwang 06:50, 9 July 2007 (UTC)

Jackson's paper Surface charges on circuit wires and resistors play three roles, Am. J. Phys. 64(7), July 1996, pp 855-870, supports the idea of a perpendicular component of the electric field outside the wires of a circuit. It's caused by surfaces charges on the wire, not by a distant source. AQv (talk) 16:52, 4 January 2008 (UTC)

yes, but that is already discussed below. (whether you prefer to interpret the field as a result of surface charge or potential difference is a matter of taste, not something fundamentally different) Han-Kwang (t) 00:45, 7 January 2008 (UTC)

[edit] Wire discussion

I made the following diagram as an illustration to point (2).

Shown are a couple of ways to send current through a closed loop of wires, with the E, H, and S fields in blue, black, and red, respectively. I think it is clear that a statement along the lines of "the Poynting vector is parallel to the wire" is only correct in very specific cases. Han-Kwang 09:35, 10 July 2007 (UTC)

[edit] jrs to hw

(1) You are welcome.

(2) Do not forget that the context of my statement was that the wire was non-resistive. If it has resistance, then the Poynting vector will tilt towards the wire since it absorbs some of the power. It is true that I was assuming that the wire leads to a load which consumes most of the power. Your drawings are helpful in showing that the situation may be more complicated. If you want to add qualifications to the paragraph to make it correct, then go ahead. But I object to your statement that the vector was zero, which is also not true generally. Even if there is not a second wire, the current has to return through the chassis, Earth, water, air, or whatever. So an electric field will exist between the wire and that other conductor.

(3) Standard text or not, I think that Jackson is just wrong, if he says "Since only its divergence appears in the conservation law, the Poynting vector is arbitrary to the extent that the curl of any vector field can be added to it. Such an added term can, however, have no physical consequences.". The stress-energy tensor could not transform correctly under coordinate transformations as required by the theory of relativity, if the "added term" were present, nor could the Einstein field equation be satisfied. I have never seen any other text mention this alleged indeterminacy of the Poynting vector. And it would clearly be pointless to even look for a refutation of his statement, just as it would be pointless to look for a reference which said "The Poynting vector is not made of green cheese.".

(4) Yes, that is the equation I meant. I am not complaining about the units. I am concerned about how you justify the square-root. Please explain how you (or Jackson) derive that equation. JRSpriggs 05:37, 11 July 2007 (UTC)

[edit] HK's reply

I'm glad to see that we can have a reasonable discussion to improve the article.

(2) I guess it will be an diagram similar to the the middle diagram above. I'll be without internet for a while, so I will defer this to a couple of weeks from now, except of course if you feel like making better illustrative diagrams yourself.

(3) Special relativity is rather far away from my daily expertise. However, Jackson mentions the stress-energy tensor in section 12.16 and says that the "covariant generalization of the differential conservation law of [Poynting's theorem] is

$\partial_{\alpha} T^{\alpha\beta} = 0 \,$ ".

The tensor T contains lots of ExB terms, as well as an E^2+B^2 term (equivalent to u in Poynting's theorem). You probably know all this, but I'm trying to summarize what I understand from reading this. I'll assume that you're correct that replacing ExB by ExB+curl C (for any C) will result in an inconsistent stress tensor. However, I see no contradictions here, since Jackson writes the stress tensor components for example as (leaving out the 4pi factors)

$T^{0i} = (E\times B) + \nabla \cdot (A_i E)$

and not as

$T^{0i} = \mu S + \nabla \cdot (A_i E).$

Are there any authors who write the stress tensor as in the latter equation? In any case, there is no μ and/or H in the stress tensor either. I assume that special relativity (did you say gravitational field???) doesn't deal with macroscopic μ and ε parameters.

(4) Leaving out the cumbersome 4pi etc. prefactors and vectors,

$B_0 = \sqrt{\mu\epsilon} E_0$

$\langle S \rangle = E_0 H_0/2 = E_0 (1/\mu) B_0 / 2 = \sqrt{\epsilon/\mu} E_0^2/2$

Han-Kwang 06:53, 11 July 2007 (UTC)

[edit] jrs to hk

(3) I did not mention "special"; I was referring to general relativity. The correct expression for the mixed time-space components of the electromagnetic stress-energy tensor in an inertial frame of reference is

$T^{0i} = T^{i0} = S_{i} = \frac{1}{\mu_0}(E\times B)_{i} \!$

If you mean the A in $+ \nabla \cdot (A_i E)$ to refer to the vector potential, then this would violate gauge invariance as well as having no justification that is apparent to me. JRSpriggs 03:44, 14 July 2007 (UTC)

[edit] hk to jrs

OK, general relativity is definitely outside my expertise, so I can't comment much on that. But I still would like to point out that the definition of the Poynting vector on electromagnetic stress-energy tensor (i.e. ExB/mu0) is different from that on this page (ExH). In a linear medium, the difference is $μ r$ , with things like hysteresis it gets worse. It might be best if we explain in this article that there are different definitions of the Poynting vector, with slightly different properties, depending on which field within physics.

It could something along the lines of:

In classical electrodynamics, as described by Jackson, S=ExH; its definition is arbitary to the extent of that the curl of any field can be added to it. In general relativity, S=ExB/mu0. It appears in the e.m. stress-energy tensor and because of gauge invariance, the curl arbitrariness does not apply here.

Han-Kwang 14:01, 31 July 2007 (UTC)

[edit] Standard textbooks

Example of incorrect picture of the electromagnetic wave.

Standard textbooks sometimes does define Poynting vector as $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ , probably for sake of simplicity, or more probably for sake of tradition (just as they call H, not B, the magnetic field for sake of tradition). However, Poynting vector is $\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$ and magnetizing field is $\mathbf{H} = \frac{\mathbf{B}}{\mu_0 \mu_r}$ , which gives $\mathbf{S} = \mathbf{E} \times \mu_r \mathbf{H}$ if one wants to express the Poynting vector in terms of E and H.

One example to show that standard text books can indeed be colectively be wrong is the picture of electromagnetic wave in which magnetic and electric fields are in phase, which is incorrect, but it is unfortunately much more often used than correct picture.

If standard textbooks do some errors we don't have to blindly copy those errors. This is one of the many examples how obeying the tradition can be harmful. Thank you for understanding. --78.0.23.69 (talk) 16:52, 21 December 2007 (UTC)

E and B (or H) are in phase in a plane wave. What are you talking about? Han-Kwang (t) 00:40, 7 January 2008 (UTC)

I wonder if the current definition in the article of the Poynting vector as S=Exμ_rH isn't wrong, and that S=(1/μ₀)ExB should be valid only for free space, while S=ExH covers both free space and cases with μ_r $\neq$ 1. According to Encyclopaedia Britannica's online article about the Poynting vector: "The Poynting vector S is defined as to be equal to the cross product S=(1/μ)ExB, where μ is the permeability of the medium through which the radiation passes", and I've never seen S=Exμ_rH anywhere in literature. AQv (talk) 17:37, 7 January 2008 (UTC)AQv (talk) 15:02, 8 January 2008 (UTC)

I suspected exactly the same thing (it was changed recently). According to Jackson (standard reference!), it's ExH, but according to Griffiths "Introduction to electrodynamics, 2nd ed.", it's ExB/mu0. It's also like that according to hyperphysics, but there free space is explicitly mentioned. The mu0 in Griffiths' version comes from Ampere's law, which is curl B = mu_0 J without a mu_r appearing. (It's like that on Wikipedia, in Griffiths, and Jackson after CGS->SI conversion). Anyway, since I have 2 references for mu0 and 1 reference for mu0*mur, and I don't feel like redoing the derivations myself, I decided to leave it as it is now. Han-Kwang (t) 17:58, 8 January 2008 (UTC)

I can think of two reasons why S=ExB/μ₀ is often found in textbooks. One is that free space is often assumed even when not explicitly stated. The other is that for many diamagnetic and paramagnetic substances μ_r is close enough to 1, making the free space approximation of S as ExB/μ₀ acceptable in these cases. (Being quite inexperienced as an editor I'd rather wait, too, before making any changes, at least until some consensus has been reached.) AQv (talk) 18:37, 8 January 2008 (UTC)

I've removed the mu_r from the definition and added a remark that it is for free space, such that it is certainly correct, although maybe incomplete. When we've sorted this out, we can add a version for materials with mu_r > 1. Han-Kwang (t) 23:50, 8 January 2008 (UTC)

At hyperphysics free space isn't even implicitly mentioned. Only thing that has to with free space mentioned there is sentence Electromagnetic waves carry energy as they travel through empty space. which only says that em waves doesn't need non-free space to propagate. --83.131.29.161 (talk) 09:36, 12 January 2008 (UTC)

Notice how c is used several times on that page. It stands for the speed of light in vacuum, and the way it is used shows that free space conditions are in fact implied. If you follow their link to speed of light, you'll notice that they say the speed of light is 299,792,458 m/s, forgetting to mention that this is the speed of light in vacuum only (which it obviously is). AQv (talk) 18:27, 14 January 2008 (UTC)

[edit] Free and non-free space

Perhaps this Jackson guy was having free space in mind when he said "S=ExH". --83.131.77.67 (talk) 13:50, 3 February 2008 (UTC)

[edit] Alternate Expressions for Poynting Vector

There was no mention of the possibility of alternate expressions for poynting vector, e.g. a term which is a curl of some vector can be added to E x H keeping the Poynting Theorem valid. I think this bit is mentioned in Jordan Balmain. Also I have a paper by one Mr. Slepian in which he discusses several cases. Welt^anschaunng 19:09, 7 February 2008 (UTC)

Well, if you read a bit further up you'll see a long discussion about that topic. Apparently this invariance is not correct anymore in general relativity. There was some strong opposition against mentioning that invariance and I didn't pursue it anymore. Han-Kwang (t) 19:29, 7 February 2008 (UTC)

[edit] Radiation in a DC circuit

Recently this has been added to the article:

However, it is also known that power cannot be radiated without accelerated charges, i.e. time varying currents. Since we are considering DC (time invariant) currents here, radiation is not possible.

But the radiation can originate from atoms and molecules even if there are no large scale AC currents. In a DC circuit powered by an ordinary battery the energy primarily originates from the chemical electrode reactions. In a way one could think of the changes in electron states involved in these reactions as a kind of (sub)microscopic charge accelerations, not necessarily involving any macroscopic AC currents. Besides, any DC current will always create heat that radiates from a wire as electromagnetic heat radiation. To say that energy can't radiate within, or from, a DC circuit is simply wrong. In fact, materia kept at a temperature above absolute zero, it could be an ice cube at 5 K, will give of some thermal radiation in the form of electromagnetic waves, without there being any large scale AC currents involved. —Preceding unsigned comment added by AQv (talk • contribs) 17:53, 12 March 2008 (UTC)

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