Introduction to Electrodynamics · Problem 5.31

Magnetic Fields in Matter — David J. Griffiths

Physics · magnetic-vector-potential

Problem

(a) Complete the proof of Theorem 2, Sect. 1.6.2. That is, show that any divergenceless vector field F\mathbf{F} can be written as the curl of a vector potential A\mathbf{A}. What you have to do is find AxA_x, AyA_y, and AzA_z such that (i) ∂Az/∂y−∂Ay/∂z=Fx\partial A_z/\partial y - \partial A_y/\partial z = F_x; (ii) ∂Ax/∂z−∂Az/∂x=Fy\partial A_x/\partial z - \partial A_z/\partial x = F_y; and (iii) ∂Ay/∂x−∂Ax/∂y=Fz\partial A_y/\partial x - \partial A_x/\partial y = F_z. Here's one way to do it: Pick Ax=0A_x = 0, and solve (ii) and (iii) for AyA_y and AzA_z. Note that the "constants of integration" are themselves functions of yy and zz—they're constant only with respect to xx. Now plug these expressions into (i), and use the fact that ∇⋅F=0\nabla \cdot \mathbf{F} = 0 to obtain Ay=∫0xFz(x′,y,z) dx′;Az=∫0yFx(0,y′,z) dy′−∫0xFy(x′,y,z) dx′.A_y = \int_0^x F_z(x', y, z)\, dx'; \quad A_z = \int_0^y F_x(0, y', z)\, dy' - \int_0^x F_y(x', y, z)\, dx'. (b) By direct differentiation, check that the A\mathbf{A} you obtained in part (a) satisfies ∇×A=F\nabla \times \mathbf{A} = \mathbf{F}. Is A\mathbf{A} divergenceless? [This was a very asymmetrical construction, and it would be surprising if it were—although we know that there exists a vector whose curl is F\mathbf{F} and whose divergence is zero.] (c) As an example, let F=yx^+zy^+xz^\mathbf{F} = y\hat{x} + z\hat{y} + x\hat{z}. Calculate A\mathbf{A}, and confirm that ∇×A=F\nabla \times \mathbf{A} = \mathbf{F}. (For further discussion, see Prob. 5.53.)

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