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CURIOSITIES IN THEORETICAL PHYSICS – V Electromagnetic Angular Momentum T. PADMANABHAN IUCAA, Pune 411007 ([email protected]) You would have certainly learnt that electromagnetic field possesses energy and momentum. The usual expressions for energy per unit volume (U) and momentum per unit volume (P) are U= 1 1 ( E 2 + B 2 ); P = (E × B) 8π 4 πc (1) What is not stressed adequately in text books is that electromagnetic field − and pretty simple ones at that − also possesses angular momentum. Just as the electromagnetic field can exchange its energy and momentum with charged particles, it can also exchange its angular momentum with a system of charged particles often leading to rather surprising results. In this installment, we shall explore one such example. One simple configuration in which Physics Education • January − March 2007 exchange of angular momentum occurs is shown in Figure 1, discussed in volume II of Feynman Lectures in Physics.1 A plastic disk, located in the x – y plane, is free to rotate about the vertical z-axis. On the disk is embedded a thin metallic ring of radius a carrying a uniformly distributed charge Q. Along the zaxis, there is a thin long current carrying solenoid producing a magnetic field B contributing a total flux Φ. This initial configuration is completely static with a magnetic field B confined within the solenoid and an electric field E produced by the charge located on the ring. Let us suppose that the current source is disconnected leading the magnetic field to die down. The change in the magnetic flux will lead to an electric field which will act tangential to the ring of charge thereby giving it a torque. Once the magnetic 285 field has died down, this torque would have resulted in the disk spinning about the z-axis with a finite angular momentum. Feynman, as usual, makes a song and dance about this problem but it is pretty much obvious that the angular momentum in the initial field is what appears as the mechanical angular momentum of the rotating disk in the The angular momentum of the final rotating disk is easy to compute. The rate of change of angular momentum dL/dt due to the torque acting on the ring of charge is along the z-axis and so we only need to compute its magnitude. This is given by dL Q Q ∂Φ E. dl = − = aQE = 2π 2 πc ∂ t dt ∫ (2) Here E is the tangential electric field generated due to the changing magnetic field and the last equality follows from the Faraday’s law. Integrating this equation and noting that the initial angular momentum of the disc and the final magnetic flux are zero, we get 286 final stage. What is really important and interesting is to work this out and explicitly verify that the angular momentum is conserved (which Feynman doesn't do!). Let us see what is involved. (There is large literature on this problem not all of which is illuminating; one place to start the search is from ref.[2].) L= Q Φ initial 2πc (3) It is interesting that the final angular momentum depends only on the total flux and not on other configurational details. We now need to show that the initial static electromagnetic configuration had this amount of stored angular momentum. I will first do this in a rather unconventional manner and then indicate the connection with the more familiar approach. To do this, let us recall that the canonical momentum of a charge q located in a magnetic field is given by p – (q/c)A where A is the vector potential related to the magnetic field by B =−∇×A and p is the usual momentum. This suggests that one can Physics Education • January − March 2007 associate with charges located in a magnetic field, a momentum (q/c)A. For a distribution of charge, with a charge density ρ, the field momentum per unit volume will be (1/c)ρA. Hence, to a charge distribution located in a region of vector potential A, we can attribute an angular momentum LA = ∫ 1 d 3 xρ(x)[x × A (x)] c (4) In our problem, the charge distribution is confined to a ring of radius a and there is negligible magnetic field at the location of the charge. But the vector potential will exist outside the solenoid and the above expression can be non zero. To compute this, let us use a cylindrical coordinate system with (r,θ,z) as the coordinates. We will choose a gauge in which the vector potential has only the tangential component; that is, only Ap is non zero. Using ∫ A.dl = Φ (5) where Φ is the total magnetic flux, we get 2πrAp = Φ for a line integral of A around any circle. Hence Ap = Φ/(2πr). This can be written in a nice vectorial form as A= Φ ( z$ × r ) 2 πr 2 (6) where ẑ is the unit vector in the z-direction. When we substitute this expression in Eq.(4) and calculate the angular momentum, the integral gets contribution only from a circle of radius a. Using further the identity, r × ( z$ × r ) = z$ r 2 we get the result that LA = Q Φ initial z$ 2πc (7) which is exactly the final angular momentum which we computed in Eq.(3). Rather nice! This elementary derivation, as well as the expression for electromagnetic angular Physics Education • January − March 2007 momentum in Eq.(4) raises several intriguing issues. On the positive side, it makes vector potential a very tangible quantity, something which we learnt from relativity and quantum mechanics but could never be clearly demonstrated within the context of classical electromagnetism. In the process, it also gives a physical meaning to the field momentum (q/c)A which is somewhat mysterious in conventional approaches. On the flip side, one should note that A, by very definition, is gauge dependent and one would have preferred a definition of electromagnetic angular momentum which is properly gauge invariant. It is, of course, possible to write down another expression for the electromagnetic angular momentum which is more conventional. Using the definition of the field momentum density, one can define an angular momentum by LA = = ∫ 1 d 3xρ(x)[x × A (x)] LEM c ∫ 1 d 3x[x × (E × B ) 4 πc (8) which just replaces the momentum density ρA/c in Eq.(4) by E×B/4πc. It is trivial to verify that, as momentum densities, these two expressions are, in general, unequal. But what is relevant, as far as our computation goes, is the integral over the whole space of these two expressions. If these two expressions differ by terms which vanish when integrated over whole space, then we have an equivalent, gauge invariant, definition of field angular momentum. It turns out that this is indeed the case in any static configuration if we choose to describe the magnetic field in a gauge which satisfies ∇.A = 0. One can then show that 1 1 (E × B) α = (E × (∇ × B )) α 4π 4π 287 = ρA α + ∂ β + ∂ βV βα (9) where Vβα is a complicated second rank tensor built out of field variables, ∂α is a condensed notation for the partial derivative ∂/∂xα and we are using the convention that Greek indices are summed over 1,2,3. While one can provide a proof of Eq.(9) using vector identities (you should try it out!), it is a lot faster and neater to use four dimensional notation and special relativity to get this result. Let me briefly outline this for those who are familiar with the four dimensional notation. We begin with the expression for the momentum density of the electromagnetic field in terms of the stress tensor Tab of the electromagnetic field. The T00 component of this tensor is proportional to the energy density of the electromagnetic field while the T0α is proportional to the momentum density Pα. More precisely, 1 T0α = (E × B ) α = cP α 4π (10) On the other hand, the electromagnetic stress tensor can be written in terms of the four dimensional field tensor Fab (with the convention that Latin letters range over 0,1,2,3) in the form T0α = − 1(1 / 4 π) F αβ F0β . We will manipulate this expression using the facts that (i) the configuration is static and (ii) the vector potential satisfies the gauge condition ∇ ⋅ A = ∂ α A α = 0 to prove Eq.(9). Using the definition of the field tensor in terms of the four vector potential, Fij = ∂ i A j − ∂ i A j , we can write T0α = − =− 288 1 αβ 1 α β β α F F0β = − ( ∂ A − ∂ A ) F0β 4π 4π ∂ β F0β 1 α β 1 β ( ∂ A ) F0β + ∂ ( F0β A α ) − A α 4π 4π 4π =− 1 1 β ∂ ( F0β A α ) ( − ∂ α A β ∂ β Aα ) + 4π 4π − Aα ∂ β F0β 4π (11) To arrive at the second line we have used the chain rule for differentiation and to obtain the third line we have used ∂0Aβ=0 since the configuration is time independent. We next use the result ∂βF0β = −∇.E = −4πρ in the last term and another application of chain rule in the first term, using the gauge condition ∇.A = ∂αAα = 0. This gives 1 T0α = ρA α + ∂ β [ A0 ∂ α Aβ − A α ∂ β A0 ] (12) 4π We thus find that cP α = ρA α + ∂ βV βα ; V βα ≡ 1 [ A0 ∂ α Aβ − A α ∂ β A0 ] 4π (13) which proves the equivalence between the two expressions for electromagnetic momentum density, when used in integrals over all space, provided the second term vanishes sufficiently fast. For the case we are discussing, this is indeed true. Now that everything appears to be well settled, I will leave you with two questions to worry about. First, let us take one more look at the final configuration. The solenoid has no current and it produces no magnetic field. But we do have a ring of charge which is rotating around, which constitutes a current loop! Surely this current produces a magnetic field. We also have the electric field due to the charged ring itself. Together, won’t this configuration have some field angular momentum? Then our angular momentum balance seems to go for a toss! Second, let us note that nowhere in the problem we really worried about the radius l of the solenoid in the initial configuration. So the results should hold Physics Education • January − March 2007 even if we take the radius to be infinitesimally small with a correspondingly large magnetic field, keeping the flux πBl2 constant. But now the magnetic field is confined to the z-axis. (Mathematically, it is given by Bz(x,y,z) = Φδ(x)δ(y).) The electric field on the z-axis is along the z-axis as well − which means that E×B vanishes and there is no angular momentum according to Eq.(8) while Eq.(4), of course gives the same angular momentum as before! What is going on? Physics Education • January − March 2007 References 1. Feynman, R.P., Feynman Lectures in Physics, Volume II; section 17-4, McGraw-Hill, New York (1970). 2. J.M. Agqirregabiria, A.Hernandez, (1981) Eur.J.Phys., 2, 168. 289 290 Physics Education • January − March 2007