Today in Physics 217: magnetic vector potential Potentials, and the z magnetic vector potential dB2 A Arbitrariness of the r1 dB1 potential: the Coulomb r2 gauge ƒƒƒƒƒƒ x A and the Biot-Savart law dA into page What is A? Magnetic boundary conditions 8 November 2002 Physics 217, Fall 2002 1 Potentials Recall the Helmholtz theorem (lecture, 16 September): any vector function F can be expressed as FW=−∇∇U + × 11 where UDd()rr= ()′′τ 4π ∫ rr− ′ Scalar and vector V potential 11 Wd()rCr= ()′′τ 4π ∫ rr− ′ V and under the assumptions that, as r →∞, FF→⋅→×→0,rr22∇∇ 0, and F0 . 8 November 2002 Physics 217, Fall 2002 2 Potentials (continued) We have also discussed two immediate consequences of the Helmholtz theorem : Irrotational fields. If a vector function is such that F =−∇U, then all of the following are true: ∇×F =0. b ∫ Fl⋅ d is independent of path, given a and b. a v∫ Fl⋅=d 0 (thus "irrotational"). In electrostatics, the electric field E is irrotational, and is the gradient of the (scalar) potential V. 8 November 2002 Physics 217, Fall 2002 3 Potentials (continued) Solenoidal fields. If a vector function is such that FW=×∇ , then all of the following are true: ∇ ⋅=F 0. ∫ Fa⋅ d is independent of surface, given the boundary C. S v∫ Fa⋅=d 0. In magnetostatics, the magnetic field B is solenoidal ()∇ ⋅=B 0 , and is the curl of the magnetic vector potential: BA=×∇ . 8 November 2002 Physics 217, Fall 2002 4 Potentials (continued) Recall also that although F is a unique solution to the differential equations, the potentials U and W are not, just because the field depends upon the gradient and curl of these potentials, and: if a constant is added to U, the result has the same gradient, because the gradient of a constant is zero. Similarly, if a gradient is added to W, the result has the same curl, because the curl of a gradient is zero. Thus we can add an arbitrary constant to a scalar potential, and the gradient of an arbitrary scalar function to a vector potential, without changing the physics. To make special choices of these “offsets” can make certain problems easier to set up. A set of such choices is called a gauge. 8 November 2002 Physics 217, Fall 2002 5 Arbitrariness of A: the Coulomb gauge A is arbitrary in the sense that one can add a gradient to it without changing B, because the curl of any gradient is zero: ∇∇∇∇×()AA +——λλ =×+× =× A. As was the case for V, there is a conventional reference point, corresponding to our conventional choice ofVr()r →→∞0 as . The convention comes from a consideration of Ampère’s Law: ∇∇∇×=××=BAAA——() ⋅ −∇2 4π = J . c If — ⋅ A were zero, then Ampère’s law would yield an expression resembling the Poisson equation. In electrodynamics this choice is called the Coulomb gauge. 8 November 2002 Physics 217, Fall 2002 6 Arbitrariness of A: the Coulomb gauge (continued) 4π — ⋅=AAJ0, ⇒∇2 =− c which might be handy, and also makes magnetostatics look even more like electrostatics. But is there always a scalar function λ such that its gradient can be added to A0 to make a new vector potential with zero divergence? Yes. Here’s why. Let AA=+0 —λ , where — λ is chosen such that 22 ——⋅==⋅AA0 00 +∇⇒∇=−⋅λλ — A 8 November 2002 Physics 217, Fall 2002 7 Arbitrariness of A: the Coulomb gauge (continued) But this is just a Poisson equation, and we know it has a solution, so the appropriate function λ exists. We even know what it is, because it is governed by the same relation as the electrostatic scalar potential, for which the general solution is known by Coulomb’s law: 1 —′′⋅ Ar() λτ= 0 d ′ . 4π ∫ r (That turns out not to be why they call it the Coulomb gauge, though.) 8 November 2002 Physics 217, Fall 2002 8 A and the Biot-Savart law We showed on Wednesday, using the Biot-Savart law, that 1 Jr()′ Br()=×— dτ ′ . c ∫ r V Apparently, 1 Jr()′′ρ () r Ar()==dVττ′′cf. () r d. c ∫∫rr V This turns out already to be divergenceless: 1 Jr()′ ——⋅=⋅Ad()r τ ′ Apply Product Rule #5: c ∫ r V 8 November 2002 Physics 217, Fall 2002 9 A and the Biot-Savart law (continued) = 0 111 Jr()′′ doesn t ———⋅=Ar() ⋅+⋅ Jr()′′ Jr () dτ ′ c ∫ rr depend on r V 11 11 =−Jr()′′ ⋅— dτ ′ because ——=− ′ c ∫ r rr V = 0 11Jr()′ =−——′′′′ ⋅ − ⋅Jr()dτ magnetostatics! c ∫ rr V 1 Jr()′ =− ⋅da′ divergence theorem c v∫ r S = 0 by definition, J = 0 everywhere on S. 8 November 2002 Physics 217, Fall 2002 10 What is the magnetic vector potential? Unlike V, which we think of as work per unit charge, there’s no obvious mechanical interpretation of A. Momentum per unit charge (times c) comes closest. For instance, the canonical momentum of a charged particle in an electromagnetic field is q ppA=− , canonical c and as such it appears frequently in the equations of quantum mechanics. Also unlike V, it’s a vector. Since B is solenoidal it can’t have a scalar potential. Thus A is much less useful in magnetostatic calculations than V is in electrostatics. 8 November 2002 Physics 217, Fall 2002 11 Magnetic boundary conditions Consider an infinite plane covered by a current with constant surface density K. What is B? z dB2 r1 dB1 r2 L δ δ x ƒƒƒƒƒƒ x dA into page K Solve with Ampère’s law with the loop as shown, as the vertical components cancel and the horizontal ones add. 8 November 2002 Physics 217, Fall 2002 12 Magnetic boundary conditions (continued) 4π B ⋅=dIA v∫ c enclosed 4π BL++00 BL += KL c 22ππKK ⇒=Bxˆˆ()zz >0, − x() < 0. cc 4π ⇒−=×BB Kzˆ . above below c This problem is the paradigm for obtaining the boundary conditions for magnetostatic boundary-value problems: 8 November 2002 Physics 217, Fall 2002 13 Magnetic boundary conditions (continued) If δ→ 0, Ampère’s law becomes Babove 4π L ()BBLKL&&, above−= , below , c δ as the contributions from the δ perpendicular components of B vanish as the sides of the loop approach zero size. Bbelow Thus 4π BB&&−= K. , above , below c For the components perpendicular to the surface, use — ⋅=B 0, with a Gaussian pillbox that has very short sides: 8 November 2002 Physics 217, Fall 2002 14 Magnetic boundary conditions (continued) Bda⋅=0 v∫ A BABA⊥⊥, above−= , below δ δ Combine with result for parallel components: 4π BB−=× Knˆ . above below c (cf. EEabove−= below 4.)πσ nˆ Similarly, AAabove−= below 0, ∂∂4π ∂ AAabove−=−≡⋅ below K nˆ — . ∂∂nn cn ∂ 8 November 2002 Physics 217, Fall 2002 15.