A point charge is not an ordinary charge-density function. Its field has a singularity, and its equation must be interpreted by integration against smooth probes. This chapter builds that interpretation, then derives Coulomb’s field and the magnetic field of a steady current with their actual hypotheses.
We use the derivatives and vector identities of chapter 2, the integrals and integral theorems of chapter 3, and the maximum principle of chapter 4. The electric and magnetic fields are vector-valued functions \(E,B\); charge density \(\rho\) is scalar and current density \(J\) is vector-valued. The positive constants \(\varepsilon_0,\mu_0\) are the vacuum permittivity and permeability in SI units. These are physical inputs, not mathematical theorems. A domain here means a nonempty open connected subset of Euclidean space. Compare Alphabeta Math’s fields, sources, and SI conventions.
5.1. Test functions and distributions
For continuous functions on bounded regions, our volume integral is the integral built in chapter 3. For a general measurable function, “integrable” and “locally integrable” below mean Lebesgue integrable and integrable on each compact subset, respectively. The definitions and full construction, including absolute integrability, are in the measure chapter and its L¹ section; readers unfamiliar with those definitions should read those sections and the intervening integral construction first. Our explicit singular kernels can alternatively be integrated by puncturing a ball and taking an absolutely convergent improper integral. Changing a value at a single point does not change the integral, because points have volume zero.
Definition 5.1. Test functions and distributions
A test function on an open set \(\Omega\subset\mathbb R^d\) is a real \(C^\infty\) function \(\varphi\) whose support, the closure of \(\{x:\varphi(x)\ne0\}\), is a compact subset of \(\Omega\). Their vector space is \(C_c^\infty(\Omega)\). For a tuple \(\alpha=(\alpha_1,\ldots,\alpha_d)\) of nonnegative integers, write \(|\alpha|=\sum\alpha_j\) and \(\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_d^{\alpha_d}\).
A distribution \(T\) is a linear rule assigning a real number \(T(\varphi)\) to each test function, with this continuity condition: for each compact \(K\subset\Omega\), some integer \(m\ge0\) and constant \(C\) satisfy
\[|T(\varphi)|\le C\max_{|\alpha|\le m}\sup_K|\partial^\alpha\varphi|\qquad(\operatorname{supp}\varphi\subset K).\]This describes the test-function topology concretely: on a common compact support, convergence means uniform convergence of every derivative. For fixed \(K\), the quantities \(p_m(\varphi)=\max_{|\alpha|\le m}\sup_K|\partial^\alpha\varphi|\) are seminorms: they obey \(p_m(a\varphi)=|a|p_m(\varphi)\) and \(p_m(\varphi+\psi)\le p_m(\varphi)+p_m(\psi)\). These follow from the same scalar rules for absolute values. Neighborhoods of zero in this topology contain a set defined by finitely many conditions \(p_{m_j}(\varphi)<\varepsilon_j\). The bound is equivalent to continuity on each fixed-support space. Indeed it implies continuity directly. Conversely, continuity at zero supplies such a neighborhood on which \(|T(\varphi)|<1\); take the largest \(m_j\) and the smallest positive \(\varepsilon_j\), and rescale \(\varphi\) to get the displayed inequality. If \(p_m(\varphi)=0\), arbitrary rescalings stay in that neighborhood, forcing \(T(\varphi)=0\).
A test function exists. Use \(e^z=\sum_{n=0}^\infty z^n/n!\), with its convergence, addition, and derivative laws proved in the complex-arithmetic chapter. Put \(b(s)=e^{-1/s}\) for \(s>0\), and \(b(s)=0\) otherwise. Repeated differentiation for \(s>0\) gives a polynomial in \(1/s\) times \(e^{-1/s}\). Every such expression tends to zero at zero: set \(z=1/s\) and use \(e^z\ge z^N/N!\) with \(N\) larger than the polynomial degree. Induction using the derivative definition at zero therefore proves \(b\in C^\infty\). The function \(b(1-|x|^2)\) is a nonzero test function supported in the closed unit ball. Translation and rescaling give one inside any open ball.
Definition 5.2. Functions, point masses, and derivatives
A locally integrable \(f\) defines its regular distribution \(T_f(\varphi)=\int f\varphi\,dV\). The Dirac distribution at \(a\in\Omega\) is \(\delta_a(\varphi)=\varphi(a)\). The distributional derivative is \((\partial_jT)(\varphi)=-T(\partial_j\varphi)\). Repetition defines higher derivatives, including \((\Delta T)(\varphi)=T(\Delta\varphi)\). A distributional equation means equality on every test function.
Alphabeta Math: regular distribution · Dirac delta · distributional derivative
Verification.
For support in \(K\), \(|T_f(\varphi)|\le(\int_K|f|)\sup|\varphi|\), and \(|\delta_a(\varphi)|\le\sup|\varphi|\). These are order-zero continuity bounds. A derivative has the bound of \(T\) with order increased by one. For \(f\in C^1\), one-variable integration by parts in each coordinate, followed by integration in the remaining coordinates, gives \(-\int f\partial_j\varphi=\int(\partial_jf)\varphi\); no boundary term survives the compact support. Thus the new derivative agrees with the ordinary derivative. Mixed distributional derivatives commute because mixed derivatives of smooth test functions commute. These arguments also explain the minus sign.
□Examples. In one dimension the step function \(H(x)=0\) for \(x<0\), \(H(x)=1\) for \(x>0\), has distributional derivative \(\delta_0\): \(-\int_0^\infty\varphi'(x)\,dx=\varphi(0)\). Its value at zero is irrelevant. For a smooth constant function the derivative is zero. Hence a derivative can be a distribution even where no ordinary derivative exists.
Why a point mass is not a locally integrable function
Suppose \(T_f=\delta_0\) for locally integrable \(f\). Normalize the bump above to obtain \(0\le\psi\le1\), \(\psi(0)=1\), and support in a fixed ball. Set \(\psi_\varepsilon(x)=\psi(x/\varepsilon)\). Then \(1=\delta_0(\psi_\varepsilon)=\int f\psi_\varepsilon\), whose absolute value is at most \(\int_{|x|\le\varepsilon}|f|\), tending to zero. To justify the last limit without an implicit interchange: on a fixed ball, choose \(M\) so that \(\int_{|f|>M}|f|<\eta\); this follows by monotone convergence of \(|f|\mathbf1_{|f|\le M}\) to \(|f|\). The remaining integral is at most \(M\operatorname{vol}(B_\varepsilon)\to0\). This is a contradiction.
□5.2. The point-source kernel
Theorem 5.3. A fundamental solution for −Δ in three dimensions
Define \(G(x)=1/(4\pi|x|)\) for \(x\ne0\), with any value at zero. It is locally integrable and satisfies \(-\Delta T_G=\delta_0\). Such a kernel is called a fundamental solution: its distributional image is a unit point source. Off zero, \(\nabla G=-x/(4\pi|x|^3)\) and \(\Delta G=0\).
Alphabeta Math: fundamental-solution convention · harmonicity off the pole · point-source identity
Worked example for Theorem 5.3
Counterexample: Decay and punctured equations do not exclude an extra dipole source
Proof.
The spherical-coordinate volume element is \(r^2\sin\theta\,dr\,d\theta\,d\phi\), derived in chapter 3 from the coordinate determinant. Thus
\[\int_{|x|<\varepsilon}G\,dV=\frac{\varepsilon^2}{2},\qquad\int_{|x|<\varepsilon}|\nabla G|\,dV=\varepsilon.\]Both singularities are locally integrable. Differentiation off zero gives the stated gradient, and \(\partial_j(x_j/r^3)=r^{-3}-3x_j^2r^{-5}\); their sum is zero. Let \(\varphi\) be a test function and choose \(R\) with its support strictly inside \(B_R\). Integrate on \(\varepsilon<r<R\). Apply the divergence theorem to \(G\nabla\varphi-\varphi\nabla G\), whose divergence is \(G\Delta\varphi-\varphi\Delta G\). On the outer sphere the integrand vanishes. The outward normal of the punctured domain on its inner sphere is \(-e_r\); there \(\partial_nG=1/(4\pi\varepsilon^2)\), and \(\partial_n\varphi=-\partial_r\varphi\). Consequently
\[-\int_{\varepsilon<r<R}G\Delta\varphi\,dV=\frac1{4\pi\varepsilon}\int_{r=\varepsilon}\partial_r\varphi\,dS+\frac1{4\pi\varepsilon^2}\int_{r=\varepsilon}\varphi\,dS.\]The first term is bounded in absolute value by \(\varepsilon\sup|\nabla\varphi|\). The second is the spherical average of \(\varphi\); its difference from \(\varphi(0)\) is at most \(\sup_{|x|\le\varepsilon}|\varphi(x)-\varphi(0)|\to0\). The omitted volume integral is bounded by \(\sup|\Delta\varphi|\,\varepsilon^2/2\). Passing to the limit proves \(-\int G\Delta\varphi=\varphi(0)\), exactly the distributional identity.
□The ordinary gradient above is also the distributional gradient: punctured integration by parts leaves an inner boundary term bounded by \((4\pi\varepsilon^2)(1/(4\pi\varepsilon))\sup|\varphi|=\varepsilon\sup|\varphi|\to0\). Volume integrals converge by the two bounds. The next derivative requires a point-source contribution; differentiating the off-pole formula and then forgetting the pole would incorrectly give \(-\Delta G=0\) everywhere.
5.3. Electrostatics and Coulomb’s field
Definition 5.4. Electrostatic equations
In a time-independent vacuum situation, the electric equations are \(\nabla\cdot E=\rho/\varepsilon_0\), \(\nabla\times E=0\). A scalar potential is a function \(\Phi\) with \(E=-\nabla\Phi\). It obeys \(-\Delta\Phi=\rho/\varepsilon_0\). For singular charges interpret these equations distributionally, component by component.
Alphabeta Math: electrostatic regime · electrostatic potential · vacuum Maxwell postulate
Proposition 5.5. The isolated point-charge solution
For charge \(q\) at the origin, \(\rho=q\delta_0\),
\[\Phi(x)=\frac{q}{4\pi\varepsilon_0|x|},\qquad E(x)=\frac{q}{4\pi\varepsilon_0}\frac{x}{|x|^3}\quad(x\ne0)\]solve the electrostatic equations distributionally. They are rotationally symmetric and decay at infinity. Among potentials whose difference from this \(\Phi\) extends to a classical harmonic function on all of \(\mathbb R^3\) and tends to zero at infinity, this solution is unique. The extension condition excludes an additional singular source at the charge.
Alphabeta Math: Coulomb solution
Worked example for Proposition 5.5
Counterexample: Decay and punctured equations do not exclude an extra dipole source
Proof.
The fundamental-solution identity proves the divergence equation, and commutation of distributional derivatives proves the curl of \(-\nabla\Phi\) is zero. The gradient computation gives the field formula. Rotations preserve radius and rotate the vector \(x\), so the symmetry holds. Uniqueness in the declared class follows from the maximum principle on increasingly large balls applied to the harmonic difference tending to zero.
□There is another derivation with explicit symmetry. Require \(E(Qx)=QE(x)\) for every rotation \(Q\) and \(x\ne0\). Rotations fixing \(x\) force \(E(x)\) to be parallel to \(x\); rotations taking one direction to another force \(E=f(r)e_r\). Away from the charge, the divergence equation gives \((r^2f)'=0\), either by coordinate differentiation or by the spherical flux formula. Thus \(f=C/r^2\). The distributional unit-source normalization above, or the flux through a sphere, gives \(4\pi C=q/\varepsilon_0\). This uses rotational symmetry as an additional isolation condition. Gauss’s law alone does not give the pointwise radial field.
A counterexample to using Gauss alone. Add any constant vector \(E_0\) to the Coulomb field. Its divergence and curl are zero, so both local electrostatic equations still hold. Its total flux through a sphere is zero, because opposite surface elements have opposite normals. Thus the same charge and same spherical flux coexist with a different field. The added field violates decay and rotational symmetry. Those conditions select an isolated-source solution.
5.4. Which fields have potentials?
Definition 5.6. Star-shaped domains and vector potentials
An open domain \(\Omega\) is star shaped about \(a\) if the segment \(a+t(x-a)\), \(0\le t\le1\), stays in \(\Omega\) for every \(x\in\Omega\). A vector potential for \(B\) is a vector field \(A\) with \(B=\nabla\times A\). Replacing \(A\) by \(A+\nabla\chi\) does not change \(B\), because the curl of a gradient is zero. Such freedom is called gauge freedom. The condition \(\nabla\cdot A=0\) is the Coulomb gauge.
Alphabeta Math: star-shaped domain · potential conventions · gauge transformation
Theorem 5.7. Explicit potentials on a star-shaped domain
Translate the center to zero. If \(F\in C^1(\Omega,\mathbb R^3)\) and \(\nabla\times F=0\), then
\[p(x)=\int_0^1 F(tx)\cdot x\,dt,\qquad \nabla p=F.\]If \(B\in C^1\) and \(\nabla\cdot B=0\), then
\[A(x)=\int_0^1t\,[B(tx)\times x]\,dt,\qquad\nabla\times A=B.\]These formulas prove existence globally on this domain, not on every domain with holes.
Proof.
For a fixed \(x\), its segment is compact inside the open domain, so nearby segments lie in a common compact neighborhood. The integrands and derivatives are uniformly continuous there; the one-variable difference quotient and fundamental theorem of calculus justify differentiation under the integral. Curl zero says \(\partial_jF_i=\partial_iF_j\). Consequently
\[\partial_jp=\int_0^1\left(F_j(tx)+t\sum_i x_i\partial_jF_i(tx)\right)dt=\int_0^1\frac d{dt}[tF_j(tx)]\,dt=F_j(x).\]For the second formula, direct expansion of the three curl components gives, with \(V(x)=B(tx)\),
\[\nabla\times(V\times x)=2V+(x\cdot\nabla)V-x(\nabla\cdot V).\]For example, the first component is \(\partial_2(V_1x_2-V_2x_1)-\partial_3(V_3x_1-V_1x_3)=2V_1+\sum_jx_j\partial_jV_1-x_1\sum_j\partial_jV_j\); the other two follow by relabeling indices. Since \(\nabla\cdot V=t(\nabla\cdot B)(tx)=0\), multiplying by \(t\) gives \(2tB(tx)+t^2(x\cdot\nabla B)(tx)=\frac d{dt}[t^2B(tx)]\). Integrate from zero to one. The lower endpoint is zero by continuity.
□Examples and topology failures. For constant \(B_0\), the formula gives \(A=(B_0\times x)/2\). For constant \(F_0\), \(p=F_0\cdot x\). On space with the \(z\)-axis removed, the field \(F=(-y,x,0)/(x^2+y^2)\) has curl zero but circulation \(2\pi\) around a unit circle. It cannot be a gradient, whose closed-path integral is zero by the fundamental theorem for line integrals. On \(\mathbb R^3\setminus\{0\}\), the divergence-free field \(B=x/|x|^3\) cannot have a global \(C^1\) vector potential: its flux through a sphere is \(4\pi\), while any global curl has zero sphere flux. For the latter statement, apply Stokes on the upper and lower hemispheres; their equator integrals cancel. Thus divergence zero is a local condition, and global potential existence also depends on the domain.
5.5. Smooth extended sources and convolution
Definition 5.8. Newtonian potentials
For a smooth compactly supported scalar source \(f\) on \(\mathbb R^3\), its Newtonian potential is
\[(G*f)(x)=\int_{\mathbb R^3}G(x-y)f(y)\,dV_y=\int_{\mathbb R^3}G(z)f(x-z)\,dV_z.\]The star denotes convolution, this translation integral. For a vector source integrate each component. Smooth means \(C^\infty\) here. Compact support means the source vanishes outside one bounded closed set.
Alphabeta Math: Newtonian potential · Poisson identity · decay
Proposition 5.9. Differentiation without a hypersingular integral
For \(f\in C_c^\infty(\mathbb R^3)\), \(U=G*f\) is smooth, \(\partial^\alpha U=G*(\partial^\alpha f)\), and \(-\Delta U=f\). Its first derivative can also be written
\[\nabla U(x)=\int\nabla G(x-y)f(y)\,dV_y.\]If the support lies in \(B_R\), then \(U(x)\to0\) and \(\nabla U(x)\to0\) as \(|x|\to\infty\). These first-derivative integrals converge absolutely and give continuous fields.
Proof: Proposition 5.9. Differentiation without a hypersingular integral
Proof.
On a compact set of \(x\)'s, the translated supports of every derivative of \(f(x-z)\) lie in one bounded set of \(z\)'s. They are bounded there, and \(G\) is integrable there. The integral of a difference quotient minus \(\partial_jf(x-z)\) is bounded by \(\int G\) times the uniform error of that difference quotient, which tends to zero by the fundamental theorem of calculus and uniform continuity of \(\partial_jf\). Iteration proves the derivative formula and continuity, using the same bound for ordinary differences. For each fixed \(x\), \(z\mapsto f(x-z)\) is a test function. The fundamental-solution identity therefore gives directly
\[-\Delta U(x)=-\int G(z)\Delta_z[f(x-z)]\,dV_z=f(x).\]To shift the first derivative to the kernel, integrate by parts over \(|z|>\varepsilon\), with an outer sphere beyond the translated support. The inner boundary term is bounded by \(\varepsilon\sup|f|\), and tends to zero. The omitted gradient-kernel integral has absolute value at most \(\varepsilon\sup|f|\); the omitted \(G\partial_jf\) integral is at most \(\varepsilon^2\sup|\partial_jf|/2\). Hence \(\int G(z)\partial_jf(x-z)=\int\partial_jG(z)f(x-z)\), with the correct sign because \(\partial_{z_j}f(x-z)=-\partial_jf(x-z)\).
For \(|x|>2R\), \(|x-y|\ge|x|/2\) on the source support, so
\[|U(x)|\le\frac{\int|f|}{2\pi|x|},\qquad|\nabla U(x)|\le\frac{\int|f|}{\pi|x|^2}.\]The derivative formula with derivatives on the smooth source already proves continuity. A direct singular-kernel check is useful: the contribution from \(|x-y|<\varepsilon\) to a gradient integral is at most \(\sup|f|\varepsilon\), uniformly in \(x\). Away from that ball the kernel is continuous and uniformly bounded on compact sets. Using a continuous cutoff equal to zero for \(r\le\varepsilon\) and one for \(r\ge2\varepsilon\), the cutoff integral is continuous and its uniform error is at most \(2\varepsilon\sup|f|\). A uniform limit of continuous functions is continuous. This proves the claim without differentiating the \(r^{-2}\) kernel again.
□The Hessian has a nonintegrable singularity. For example, \(\partial_3^2(1/r)=(3x_3^2-r^2)/r^5\). On the angular band \(|x_3|\le r/2\), its absolute value is at least \(1/(4r^3)\); its band on the unit sphere has area \(2\pi\), by integrating \(\sin\theta\) over \(\pi/3\le\theta\le2\pi/3\). Its absolute volume integral therefore bounds a positive multiple of \(\int_0^\varepsilon dr/r\), whose divergence follows from the dyadic interval estimate below. Explicitly, on each interval \([\varepsilon2^{-(k+1)},\varepsilon2^{-k}]\), the integrand is at least \(2^k/\varepsilon\) and the interval length is \(\varepsilon2^{-(k+1)}\), so its integral is at least \(1/2\). Adding arbitrarily many disjoint intervals forces divergence. An unqualified second differentiation under the original singular integral is therefore invalid. Moving derivatives to the smooth source, as above, avoids that error.
Extended charge. For \(\rho\in C_c^\infty\), set \(\Phi=(G*\rho)/\varepsilon_0\) and \(E=-\nabla\Phi\). Then \(\nabla\cdot E=\rho/\varepsilon_0\), \(\nabla\times E=0\), and both fields decay. If another classical potential with the same source decays at infinity, their difference is an entire harmonic function tending to zero, hence zero by the maximum principle. This proves uniqueness of the decaying potential for these extended sources. It does not assert a classical value of the point-charge field at its pole.
5.6. Magnetostatics and Biot–Savart
Theorem 5.10. The decaying field of a smooth compact steady current
Let \(J\in C_c^\infty(\mathbb R^3,\mathbb R^3)\) satisfy \(\nabla\cdot J=0\). Define
\[A=\mu_0G*J,\qquad B=\nabla\times A.\]Then \(\nabla\cdot A=0\), \(-\Delta A=\mu_0J\), and
\[\nabla\cdot B=0,\qquad\nabla\times B=\mu_0J,\qquad B(x)=\frac{\mu_0}{4\pi}\int J(y)\times\frac{x-y}{|x-y|^3}\,dV_y.\]The last expression is the Biot–Savart formula. The vector potential and magnetic field are smooth and decay at infinity. A steady current is time independent; its zero divergence expresses absence of accumulating charge. The displayed two magnetic equations are the magnetostatic vacuum Maxwell equations.
Alphabeta Math: steady-current regime · vector potential · Biot–Savart · kernel curl identity
Worked example for Theorem 5.10
Counterexample: A smooth compact current need not be a steady Maxwell source
Proof.
Proposition 5.9 applies to each component. In particular \(\nabla\cdot A=\mu_0G*(\nabla\cdot J)=0\), and \(-\Delta A=\mu_0J\). The identities \(\nabla\cdot(\nabla\times A)=0\) and \(\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-\Delta A\), proved in chapter 2, give the magnetic equations. For the integral, component differentiation gives \(\nabla\times[G(x-y)J(y)]=\nabla_xG(x-y)\times J(y)\). Since \(\nabla G=-(x-y)/(4\pi|x-y|^3)\), antisymmetry of the cross product changes this to the displayed integrand. The first-derivative kernel proof and its \(\varepsilon\)-bound apply componentwise, so the integral is justified even at points inside the current support. Decay follows from Proposition 5.9.
□Uniqueness holds among \(C^2\) magnetic fields tending uniformly to zero at infinity and satisfying these equations: the difference \(D\) has both divergence and curl zero. The curl-of-curl identity gives \(\Delta D=0\); each component is an entire harmonic function tending to zero, hence vanishes. Gauge freedom still allows different potentials for the same field unless the potential gauge and boundary conditions are also fixed.
Examples. A compactly supported smooth divergence-free current can be made by choosing a nonconstant test function \(\psi\) and setting \(J=\nabla\times(0,0,\psi)=(\partial_y\psi,-\partial_x\psi,0)\). Its divergence is zero by equality of mixed derivatives, so the theorem applies. A compact source that has nonzero divergence does not meet the hypothesis: the constructed \(A\) generally has nonzero divergence, and its curl satisfies \(\nabla\times B=\mu_0J+\nabla(\nabla\cdot A)\) rather than the magnetostatic equation.
5.7. A straight wire and the thin-wire idealization
An infinite wire is not compactly supported. We treat it separately, with its own symmetry and boundary assumptions. Use cylindrical radius \(r=\sqrt{x^2+y^2}\), unit vectors \(e_r=(x,y,0)/r\), \(e_\phi=(-y,x,0)/r\), and \(e_z=(0,0,1)\), for \(r>0\).
Proposition 5.11. A finite-radius infinite straight current
Let \(J=j(r)e_z\), where \(j:[0,\infty)\to\mathbb R\) is continuous and zero for \(r>a\). Its total current is \(I=2\pi\int_0^a j(s)s\,ds\). Seek a field invariant under translations in \(z\) and rotations about the axis, bounded near the axis, with no externally imposed axial field and decaying transversely as \(r\to\infty\). Then
\[B(r)=b(r)e_\phi,\qquad b(r)=\frac{\mu_0}{r}\int_0^rj(s)s\,ds,\qquad B(R)=\frac{\mu_0I}{2\pi R}e_\phi\quad(R\ge a).\]The last formula gives the boundary field at the conductor’s surface when \(R=a>0\).
Proof.
The symmetry permits \(B=b_r(r)e_r+b_\phi(r)e_\phi+b_z(r)e_z\). Coordinate differentiation of these unit vectors gives \(\nabla\cdot B=(rb_r)'/r\), \((\nabla\times B)_\phi=-b_z'\), and \((\nabla\times B)_z=(rb_\phi)'/r\); the radial curl is zero. The equations therefore force \(rb_r\) constant, \(b_z\) constant, and \((rb_\phi)'=\mu_0rj\). Boundedness at the axis eliminates the \(1/r\) radial term and the integration constant in \(rb_\phi\); transverse decay eliminates the constant axial field. Integration yields the formula. It defines a \(C^1\) field even at the axis: write \(B=h(r)(-y,x,0)\), where \(h(r)=\mu_0r^{-2}\int_0^rj(s)s\,ds\). Continuity of \(j\) implies \(h(r)\to\mu_0j(0)/2\) and \(rh'(r)=\mu_0j(r)-2h(r)\to0\). The first limit proves differentiability at the axis; the second makes the extra coordinate derivative terms tend to zero, so those derivatives are continuous there. Equivalently, Ampère’s integral law on a disk gives \(2\pi r b(r)=\mu_0\int_{\mathrm{disk}}J\cdot e_z\,dS=2\pi\mu_0\int_0^rj(s)s\,ds\). This step uses the symmetry and regularity just specified. For \(r\ge a\), the integral equals \(I/(2\pi)\).
□Uniform-current example. An ideal piecewise-constant profile \(j=I/(\pi a^2)\) inside the wire and zero outside gives \(b=\mu_0Ir/(2\pi a^2)\) inside and \(b=\mu_0I/(2\pi r)\) outside. It has a jump in current at \(a\), so the differential equations hold classically on each side and distributionally across the interface. The field is continuous at \(a\), and hence has no surface-delta contribution in its first derivative: integration by parts on the two sides produces opposite interface boundary terms with the same field value, so they cancel. For a continuous profile the preceding proposition applies directly.
Definition 5.12. An ideal line current
The infinitely thin straight wire is the vector distribution defined, for a vector-valued test function \(\Psi\), by
\[J_{\mathrm{line}}(\Psi)=I\int_{\mathbb R} e_z\cdot\Psi(0,0,s)\,ds.\]A vector distribution means a tuple of scalar distributions, paired with the corresponding tuple of test functions. On a compact support contained in a cylinder of length \(L\), its pairing is bounded by \(|I|L\sup|\Psi|\), proving the distribution continuity condition. This line current is not a volume-density function. Its divergence is zero: on a scalar test function \(\varphi\), \((\nabla\cdot J_{\mathrm{line}})(\varphi)=-I\int\partial_s\varphi(0,0,s)\,ds=0\). Compact support supplies zero endpoint values.
Alphabeta Math: weak sources · thin-current limit · ideal line-current field
The corresponding line-source Biot–Savart integral, evaluated away from the line at \(x=(r,0,z)\), is convergent and elementary:
\[B(x)=\frac{\mu_0I}{4\pi}\int_{-\infty}^{\infty}\frac{e_z\times(r,0,z-s)}{[r^2+(z-s)^2]^{3/2}}\,ds=\frac{\mu_0I}{2\pi r}e_\phi.\]The cross product is \(r e_\phi\); the scalar integral is \(2/r^2\), since an antiderivative in \(u=s-z\) is \(u/(r^2\sqrt{r^2+u^2})\). This is a source-measure integral, not the literal substitution of \(J\,dV\) by \(I\,d\ell\) for ordinary functions.
Verification across the line and the thin-radius limit
Put \(C=\mu_0I/(2\pi)\), so \(B=C e_\phi/r\). Its size \(|C|/r\) is locally integrable in three dimensions: on a cylinder of length \(L\) and radius \(\varepsilon\), its absolute integral is \(2\pi L|C|\varepsilon\). It has ordinary divergence and curl zero away from the axis. On a punctured cylinder, integrate against compactly supported test functions. Since its normal component on the inner cylinder is zero, integration by parts for divergence gives zero, even in the limit. For a vector test function \(\Psi\), the identity \(\nabla\cdot(B\times\Psi)=\Psi\cdot(\nabla\times B)-B\cdot(\nabla\times\Psi)\) gives
\[\int_{r>\varepsilon} B\cdot(\nabla\times\Psi)\,dV=-\int_{r=\varepsilon}(n\times B)\cdot\Psi\,dS=C\int\!\int_0^{2\pi}\Psi_z(\varepsilon\cos\phi,\varepsilon\sin\phi,z)\,d\phi\,dz,\]because the inner outward normal is \(-e_r\), so \(n\times B=-Ce_z/\varepsilon\), and \(dS=\varepsilon\,d\phi\,dz\). Other boundary terms vanish by the test support. The omitted volume integral tends to zero by the preceding \(O(\varepsilon)\) bound. Uniform continuity of \(\Psi\) on its compact support makes the right side tend to \(2\pi C\int\Psi_z(0,0,z)\,dz\). The left side is the distributional curl pairing (the two derivative minus signs and antisymmetry give this positive pairing). Thus \(\nabla\times B=\mu_0J_{\mathrm{line}}\), and \(\nabla\cdot B=0\) distributionally.
For the uniform-radius wires above, \(J_a=I e_z/(\pi a^2)\) on \(r<a\). Pairing with \(\Psi\) gives \(I\) times the disk average of \(\Psi_z\), integrated in \(z\); uniform continuity makes this tend to the line pairing as \(a\downarrow0\). The finite-radius magnetic field equals \(B\) outside \(r=a\). Inside, its absolute integral on a cylinder is bounded by \(L|\mu_0I|a/3\), while that of \(B\) there is \(L|\mu_0I|a\). Their difference therefore tends to zero in local absolute integral, and hence as distributions. This proves the ideal limit with its equations, rather than assigning an ordinary density to a line.
□The infinite wire has \(1/r\) transverse decay but is unchanged along its axis; it does not decay as \(|x|\to\infty\) in every direction. Its vector potential would involve a divergent \(\int ds/\sqrt{r^2+s^2}\) if we copied the compact-source Newtonian formula. The magnetic integral converges, and its distributional verification above supplies the correct separate argument.
Static sources produce Poisson equations. Time dependence adds coupled evolution equations and a finite propagation speed, the subject of chapter 6.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. A test function and a point derivative
Applies to: Definition 5.1, Definition 5.2.
Let \(\psi(x)=e^{-1/(1-|x|^2)}\) on \(|x|<1\), zero outside, in \(\mathbb R^3\). This is a test function: every derivative approaching the unit sphere is a polynomial in the coordinates and \((1-|x|^2)^{-1}\) times the exponential, which tends to zero faster than every such power. Thus its extension is smooth with compact support. The point mass \(\delta_0\) is linear and satisfies \(|\delta_0(\varphi)|\le\sup|\varphi|\), so it has distribution order zero on every compact test support. Its derivative satisfies \((\partial_1\delta_0)(\varphi)=-\partial_1\varphi(0)\), bounded by the first-derivative seminorm. On \(\varphi=x_1\psi\) this gives \(-\psi(0)=-e^{-1}\), a nonzero distributional derivative. The locally integrable function 1 defines \(T_1(\varphi)=\int\varphi\); its derivative is zero, since integrating a compactly supported partial derivative gives zero. These verify both regular and singular distribution rules.
Further proof details
To justify the smooth extension at the sphere, put \(s=1-|x|^2\). Each derivative from inside is a finite sum of polynomial coordinate factors times \(s^{-m}e^{-1/s}\), with finite \(m\). For any integer \(N>m\), the exponential series gives \(e^{1/s}\ge s^{-N}/N!\), hence \(s^{-m}e^{-1/s}\le N!s^{N-m}\to0\) as \(s\downarrow0\). All these derivatives therefore extend by zero continuously. Induction on derivative order, using the one-variable difference quotient across the boundary in a coordinate direction, shows the extended derivatives are the derivatives of the extension. Thus the bump is genuinely \(C^\infty\), and its support is the closed unit ball, a compact subset of \(\mathbb R^3\).
Worked example 2. A translated point charge with its weak source
Applies to: Theorem 5.3, Definition 5.4, Proposition 5.5.
Place charge \(q\ne0\) at \(a=(1,0,0)\), and use the translated fundamental solution \(G_a(x)=1/(4\pi|x-a|)\). Its local absolute integral on a radius-\(\eta\) ball is \(\int_0^\eta r\,dr=\eta^2/2\), finite. Off \(a\), differentiating gives \(\Phi=qG_a/\varepsilon_0\), \(E=q(x-a)/(4\pi\varepsilon_0|x-a|^3)\), zero curl, and zero divergence. The outward flux on any sphere about \(a\) is \(q/\varepsilon_0\). Integrating by parts outside a small sphere and letting its radius tend to zero gives \(-\int E\cdot\nabla\varphi=(q/\varepsilon_0)\varphi(a)\); the boundary average of \(\varphi\) tends to \(\varphi(a)\), and the omitted local integral tends to zero since \(|E|\) is locally integrable. Hence \(\operatorname{div}E=(q/\varepsilon_0)\delta_a\), checking the fundamental-solution and electrostatic signs.
At distance 2 the field magnitude is one quarter of that at distance 1, and \(\Phi,E\) decay at infinity. If another potential differs by a globally harmonic function tending to zero at infinity, apply the maximum principle to that difference and its negative on arbitrarily large balls; the boundary suprema tend to zero, proving uniqueness in exactly the asserted class.
Further proof details
The weak divergence calculation has a definite boundary sign. Write \(C=q/(4\pi\varepsilon_0)\). On the region \(|x-a|>\eta\), the inner outward normal is \(-e_r\), so compact support of \(\varphi\) and the exterior zero divergence give
\[-\int_{|x-a|>\eta}E\cdot\nabla\varphi\,dV=\frac C{\eta^2}\int_{|x-a|=\eta}\varphi\,dA.\]The right side is \(4\pi C\) times the spherical average, tending to \((q/\varepsilon_0)\varphi(a)\) by continuity. The omitted integral is bounded in absolute value by \(4\pi|C|\eta\sup|\nabla\varphi|\), so it vanishes. The distributional curl is zero too: \(E=-\nabla\Phi\), and \((\partial_i\partial_jT)(\varphi)=T(\partial_j\partial_i\varphi)\), making mixed distribution derivatives commute. These facts prove both weak electrostatic equations, including their singular parts.
Worked example 3. Radial formulas for constant and gradient fields
Applies to: Definition 5.6, Theorem 5.7.
On the star-shaped domain \(\mathbb R^3\), let \(F=(2x,2y,2z)\), which has zero curl. The radial formula gives \(p(x)=\int_0^1 2t|x|^2dt=|x|^2\), whose gradient is \(F\). For the divergence-free constant field \(B=e_z\), it gives \(A=\int_0^1t(e_z\times x)dt=(-y/2,x/2,0)\). Differentiation gives \(\operatorname{curl}A=e_z\) and \(\operatorname{div}A=0\), so this is a Coulomb-gauge vector potential. Taking \(\chi=xz\) changes it to \(A+\nabla\chi=A+(z,0,x)\); the added gradient has zero curl and divergence \(\Delta\chi=0\), preserving both \(B\) and this gauge. The explicit segment \(tx\), \(0\le t\le1\), lies in the domain, so the global radial hypotheses are satisfied.
Worked example 4. A compact smooth source with an exact potential
Applies to: Definition 5.8, Proposition 5.9, Theorem 5.10.
Use the bump \(\psi\) from example 1 and put \(f=-\Delta\psi\), a smooth compactly supported scalar source (it need not be nonnegative). Integration by parts in the distribution pairing gives \(G*f=G*(-\Delta\psi)=(-\Delta G)*\psi=\delta_0*\psi=\psi\). The derivatives act on the smooth compact test function, so this identity introduces no nonintegrable second derivative of \(G\). Thus \(U=G*f=\psi\), \(-\Delta U=f\), and \(\partial^\alpha U=G*\partial^\alpha f\). Both the potential and gradient even vanish outside a compact set, illustrating the Newtonian convolution and its differentiation and decay assertions.
For a nonzero steady current, set \(A_0=(\psi_y,-\psi_x,0)\) and \(J=-\Delta A_0/\mu_0\). These are smooth and compactly supported, and \(\operatorname{div}A_0=\psi_{xy}-\psi_{yx}=0\), so \(\operatorname{div}J=0\). The same convolution calculation gives \(\mu_0G*J=A_0\). Its magnetic field is \(B=(\psi_{xz},\psi_{yz},-\psi_{xx}-\psi_{yy})\). Direct derivatives give \(\operatorname{div}B=0\), \(\operatorname{curl}B=(-\partial_y\Delta\psi,\partial_x\Delta\psi,0)=\mu_0J\). This checks all the magnetostatic equations and the Coulomb gauge in Theorem 5.10. Taking the curl of \(G*J\) yields its Biot–Savart integral: \(\nabla_xG(x-y)\times J(y)=J(y)\times(x-y)/(4\pi|x-y|^3)\); the \(|x-y|^{-2}\) singularity is locally integrable in three dimensions.
Further proof details
Here the inverse convolution identity is an application of the proved distribution equation to an explicit test function, rather than a formal movement of a singular derivative. For each fixed \(x\), \(z\mapsto\psi(x-z)\) is smooth and compactly supported. Consequently
\[\int G(z)(-\Delta\psi)(x-z)\,dz=(-\Delta T_G)(\psi(x-\cdot))=\psi(x).\]For \(x\) in a compact neighborhood, all these test supports lie in one bounded region, and every derivative of \(\psi(x-z)\) is uniformly bounded there. The integrable majorant is a constant times the locally integrable \(G(z)\), so differentiating under the integral is justified by the difference-quotient bound from the mean value theorem followed by dominated convergence. This gives the stated derivative identities. The same argument applies to each component of \(A_0\) in place of \(\psi\), establishing the exact current potential. To transfer a first derivative from a smooth source to \(G\), integrate by parts outside \(|z|=\eta\): the inner boundary contribution is bounded by a constant times \(\eta^2/\eta=\eta\), and tends to zero; the remaining derivative kernel is bounded by a constant times \(|z|^{-2}\), locally integrable. This proves the first-derivative Biot–Savart calculation with a controlled singularity.
Worked example 5. A continuous finite-radius current profile
Applies to: Proposition 5.11.
Choose \(a>0,j_0>0\), and \(j(r)=j_0(1-r^2/a^2)\) for \(0\le r\le a\), zero for \(r>a\). This is continuous, including at \(a\), as the proposition requires. Its total current is \(I=2\pi j_0\int_0^a(s-s^3/a^2)ds=\pi j_0a^2/2\). The specified symmetric field is
\[b(r)=\begin{cases}\mu_0j_0(r/2-r^3/(4a^2)),&0<r\le a,\\\mu_0j_0a^2/(4r),&r\ge a.\end{cases}\]The values agree at \(a\), are bounded and tend to zero at the axis, and decay as \(r\to\infty\). The cylindrical curl is \((rb)'/r=\mu_0j(r)\), with zero divergence. Near the axis the Cartesian field is \(\mu_0j_0(1/2-r^2/(4a^2))(-y,x,0)\), so there is no singularity there. At the surface \(b(a)=\mu_0j_0a/4=\mu_0I/(2\pi a)\), explicitly checking the boundary field and all symmetry and regularity qualifications.
Worked example 6. Testing an ideal line current
Applies to: Definition 5.12.
Let \(I\ne0\) and take a vector test function \(\Psi(x,y,z)=\eta(x)\eta(y)\eta(z)e_z\), where \(\eta(s)=e^{-1/(1-s^2)}\) for \(|s|<1\), zero outside. The line-current pairing is \(I\eta(0)^2\int_{-1}^1\eta(s)ds\), nonzero and finite. On any compact test support of axial length \(L\), its absolute pairing is at most \(|I|L\sup|\Psi|\), proving continuity. Its divergence pairs with a scalar test function as \(-I\int\partial_z\varphi(0,0,z)dz=0\), by compact support.
The associated azimuthal field off the axis is \(B=\mu_0I e_\phi/(2\pi r)\). On a radius-\(R\) circle, circulation is \(2\pi R\,\mu_0I/(2\pi R)=\mu_0I\). Its curl is zero off the axis; the nonzero circulation is supplied by the singular line source, so applying a smooth curl theorem across the axis would omit its hypotheses.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Decay and punctured equations do not exclude an extra dipole source
Related statements: Theorem 5.3, Proposition 5.5.
Claim refuted. The Coulomb potential is determined just by decay, harmonicity away from the charge, and its total sphere flux.
Proof.
Let \(G=1/(4\pi|x|)\) and \(H=\partial_zG=-z/(4\pi|x|^3)\) off zero, interpreted as the locally integrable function with that formula. The local integral of \(|H|\) is finite because it is bounded by a constant times \(|x|^{-2}\). It is harmonic off zero: differentiating \(\Delta G=0\) gives \(\Delta H=0\). It decays like \(|x|^{-2}\), but is nonzero. Distributional derivatives commute directly on test functions, so \(-\Delta H=\partial_z\delta_0\); on \(\varphi=z\psi\), for a bump with \(\psi(0)\ne0\), this equals \(-\psi(0)\ne0\).
On a sphere, write \(H=-\cos\theta/(4\pi r^2)\). The added electric radial component from \(-\nabla H\) is \(-\cos\theta/(2\pi r^3)\); its sphere flux is zero because \(\int_0^\pi\cos\theta\sin\theta d\theta=0\). Thus adding a nonzero multiple of \(H\) to a Coulomb potential retains decay, the exterior equations, and total flux, while adding a dipole distribution at the origin. Its difference from Coulomb does not extend to a globally smooth harmonic function. This is exactly the extra-singularity exclusion in Proposition 5.5; a total-charge flux cannot detect all multipoles.
Counterexample 2. A smooth compact current need not be a steady Maxwell source
Related statements: Theorem 5.10.
Claim refuted. Every smooth compactly supported vector current is admissible in the magnetostatic construction without checking its divergence.
Proof.
Let \(\psi=e^{-1/(1-|x|^2)}\) inside the unit ball and zero outside, and set \(J=\nabla\psi\). The bump proof in the worked examples shows \(J\) is smooth and compactly supported. At zero, \(\partial_i\psi=0\) and \(\partial_i^2\psi=-2e^{-1}\), by differentiating \(-1/(1-|x|^2)\); hence \(\operatorname{div}J(0)=\Delta\psi(0)=-6/e\ne0\). If a \(C^2\) magnetostatic field satisfied \(\operatorname{curl}B=\mu_0J\) near zero, taking divergence would give \(0=\mu_0\operatorname{div}J\), a contradiction. The missing hypothesis of Theorem 5.10 is source conservation, not regularity or support. Such a current would require time-dependent charge and the full Maxwell system instead.
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