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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 F 0 .

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 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 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 , 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 ? 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 111  Jr()′′ doesn t ———⋅=Ar()  ⋅+⋅ Jr()′′ Jr ()dτ ′ c ∫ rr depend on r V  11 11  =−Jr()′′ ⋅— dτ ′ because ——=− ′  c ∫ r rr  V = 0 11Jr()′ =−——′′′′ ⋅ − ⋅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