Absence of Magnetic Monopoles

Derivation

Absence of Magnetic Monopoles

D-051 Home PU-102 Threads fields · symmetry Depends on Biot–Savart Law from the Current Force Law, divergence-theorem

Statement

The magnetostatic field produced by any bounded, steady current distribution through the Biot–Savart law is divergence-free everywhere: \( \nabla \cdot \mathbf{B} = 0 \) identically. Equivalently, the net magnetic flux through every closed surface vanishes, so magnetic field lines never begin or end, and no isolated magnetic charge (monopole) can act as a source of \( \mathbf{B} \).

Why it matters

This is Maxwell's second equation, \( \nabla \cdot \mathbf{B} = 0 \). Here it is not postulated but derived: unlike Gauss's law \( \nabla\cdot\mathbf{E}=\rho/\varepsilon_0 \), the magnetic equation carries no source term, and this asymmetry is forced by the very form of the Biot–Savart field — in which the source is rotated by a cross product rather than radiated outward from a charge.

The vanishing divergence is exactly what guarantees a globally single-valued vector potential \( \mathbf{B} = \nabla \times \mathbf{A} \), the starting point of every gauge treatment of magnetism, of magnetic flux quantisation, and of the classification of magnetic field lines as closed loops.

Assumptions

The field obeys the Biot–Savart law with a steady current density \( \mathbf{J}(\mathbf{r}') \).If the current is not steady (\( \partial \rho / \partial t \neq 0 \)), Biot–Savart is no longer exact and one must use the full retarded fields; the conclusion \( \nabla\cdot\mathbf{B}=0 \) survives, but this particular proof does not.

The source current density is spatially bounded (localised) and integrable.If \( \mathbf{J} \) does not fall off fast enough, the vector-potential integral may diverge and the interchange of \( \nabla \) with \( \int d^3r' \) becomes unjustified.

The field point \( \mathbf{r} \) may lie inside the source, and the singular \( 1/|\mathbf{r}-\mathbf{r}'| \) kernel is handled as an improper integral.If the mild singularity at \( \mathbf{r}=\mathbf{r}' \) is treated carelessly, one may wrongly conclude that differentiation under the integral fails; a principal-value / distributional argument keeps the result exact even at interior points.

Fields are ordinary (twice-differentiable) functions away from the sources, so mixed partial derivatives commute.If \( \mathbf{A} \) were not \( C^2 \), the identity \( \nabla\cdot(\nabla\times\mathbf{A})=0 \) — which relies on \( \partial_i\partial_j = \partial_j\partial_i \) — could fail.

Derivation

\[ \mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \times \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^{3}} \, d^3 r' \]

Starting point: the Biot–Savart law for a volume current, taken as an established prior result. A

\[ \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^{3}} = -\nabla\!\left( \frac{1}{|\mathbf{r}-\mathbf{r}'|} \right) \]

The Coulomb-type kernel is minus the gradient (with respect to the field point \( \mathbf{r} \)) of the inverse distance. Direct differentiation of \( |\mathbf{r}-\mathbf{r}'|^{-1} \) gives this for all \( \mathbf{r}\neq\mathbf{r}' \). B

\[ \mathbf{B}(\mathbf{r}) = -\frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \times \nabla\!\left( \frac{1}{|\mathbf{r}-\mathbf{r}'|} \right) d^3 r' \]

Substitute the gradient form of the kernel into the integrand. Purely algebraic rewriting. A

\[ \nabla \times \left( \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \right) = \nabla\!\left( \frac{1}{|\mathbf{r}-\mathbf{r}'|} \right) \times \mathbf{J}(\mathbf{r}') = \mathbf{J}(\mathbf{r}') \times \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^{3}} \]

Vector identity \( \nabla\times(f\,\mathbf{J}) = (\nabla f)\times\mathbf{J} + f\,(\nabla\times\mathbf{J}) \). Since \( \mathbf{J}(\mathbf{r}') \) is constant with respect to the field-point operator \( \nabla \), the term \( f\,\nabla\times\mathbf{J} \) vanishes; the remaining term reproduces the Biot–Savart integrand (the two sign flips cancel). B

\[ \mathbf{B}(\mathbf{r}) = \nabla \times \underbrace{\left[ \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, d^3 r' \right]}_{\displaystyle \mathbf{A}(\mathbf{r})} = \nabla \times \mathbf{A} \]

The operator \( \nabla \) acts on \( \mathbf{r} \) while the integral runs over \( \mathbf{r}' \); with a localised, integrable source the two commute, so the curl passes outside the integral. Thus \( \mathbf{B} \) is the curl of a well-defined vector potential \( \mathbf{A} \). B

\[ \nabla \cdot \mathbf{B} = \nabla \cdot (\nabla \times \mathbf{A}) \]

Take the divergence of both sides of the result of Step 5. A

\[ \nabla \cdot (\nabla \times \mathbf{A}) = \varepsilon_{ijk}\,\partial_i \partial_j A_k = 0 \quad\Longrightarrow\quad \boxed{\;\nabla \cdot \mathbf{B} = 0\;} \]

The divergence of any curl vanishes identically: \( \partial_i\partial_j A_k \) is symmetric in \( i,j \) (mixed partials commute for a \( C^2 \) field) while \( \varepsilon_{ijk} \) is antisymmetric, so their contraction is zero term by term. No property of \( \mathbf{A} \) beyond smoothness is used. C

Result

\[ \nabla \cdot \mathbf{B} = 0 \qquad\Longleftrightarrow\qquad \oint_{S} \mathbf{B} \cdot d\mathbf{A} = 0 \ \text{ for every closed } S \]

Reading. The magnetic field of any steady current has zero divergence at every point. By the divergence theorem this is equivalent to the statement that the net magnetic flux through any closed surface is zero: whatever flux enters a closed surface must leave it. Field lines have no sources or sinks — they close on themselves. There is no magnetic charge for \( \mathbf{B} \) to diverge from, i.e. no magnetic monopole.

Units check. \( \mathbf{B} \) has units of tesla (T), so \( \nabla\cdot\mathbf{B} \) has units \( \mathrm{T\,m^{-1}} \). On the source side, \( \mu_0 \) is \( \mathrm{T\,m\,A^{-1}} \) and \( \mathbf{J} \) is \( \mathrm{A\,m^{-2}} \); the kernel \( (\mathbf{r}-\mathbf{r}')/|\mathbf{r}-\mathbf{r}'|^3 \) is \( \mathrm{m^{-2}} \) and \( d^3r' \) is \( \mathrm{m^3} \), giving \( \mathrm{T\,m\,A^{-1}}\cdot\mathrm{A\,m^{-2}}\cdot\mathrm{m^{-2}}\cdot\mathrm{m^{3}} = \mathrm{T} \) for \( \mathbf{B} \); one further \( \nabla \) (\( \mathrm{m^{-1}} \)) makes \( \nabla\cdot\mathbf{B} \) come out in \( \mathrm{T\,m^{-1}} \), consistently equal to the identically-zero right-hand side.

Limiting cases

Single current filament. Reducing \( \mathbf{J}\,d^3r' \to I\,d\boldsymbol{\ell}' \) recovers the wire form of Biot–Savart; the same \( \mathbf{B}=\nabla\times\mathbf{A} \) structure holds, so \( \nabla\cdot\mathbf{B}=0 \) still.
Far field / magnetic dipole. At \( r \gg \) source size the field becomes the dipole \( \mathbf{B}\propto (2\cos\theta\,\hat{\mathbf{r}}+\sin\theta\,\hat{\boldsymbol\theta})/r^3 \); explicit spherical divergence gives exactly zero (see Worked example 2).
No current, \( \mathbf{J}=0 \). Then \( \mathbf{B}=0 \) (or a source-free harmonic field), and \( \nabla\cdot\mathbf{B}=0 \) trivially.
Static limit of electrodynamics. The result is the \( \partial_t\to 0 \) case of the exact Maxwell equation \( \nabla\cdot\mathbf{B}=0 \), which in fact holds even when currents vary.

Breaks when

Hypothetical magnetic charge exists. If nature contained monopoles of density \( \rho_m \), the equation would generalise to \( \nabla\cdot\mathbf{B}=\mu_0\rho_m \neq 0 \). The Biot–Savart law would then be incomplete — it builds \( \mathbf{B} \) purely as a curl and structurally cannot produce a divergence — and this derivation would no longer describe the full field.
Non-steady currents with the naive static kernel. When \( \partial\rho/\partial t\neq 0 \), charge continuity forbids \( \nabla\cdot\mathbf{J}=0 \), the instantaneous Biot–Savart form is not the true field, and one must use retarded potentials. The conclusion \( \nabla\cdot\mathbf{B}=0 \) still holds, but the proof above (which assumed the static kernel) breaks.
Singular or distributional sources handled formally. Idealised surface currents or point dipoles introduce \( \delta \)-function contributions; if the differentiation under the integral is done without care at \( \mathbf{r}=\mathbf{r}' \), spurious non-zero divergences can appear as artefacts of the idealisation.

Failure modes

Confusing divergence with curl. Writing \( \nabla\cdot\mathbf{B}=\mu_0\mathbf{J} \) — but that is Ampère's law for the curl, \( \nabla\times\mathbf{B}=\mu_0\mathbf{J} \). The divergence of \( \mathbf{B} \) is always zero; the current sources its curl.
Treating \( \mathbf{J}(\mathbf{r}') \) as depending on the field point. Applying \( \nabla \) (which acts on \( \mathbf{r} \)) to \( \mathbf{J}(\mathbf{r}') \) and keeping a spurious \( \nabla\times\mathbf{J} \) term in Step 4.
Sign slip in the kernel. Forgetting the minus sign in \( (\mathbf{r}-\mathbf{r}')/|\mathbf{r}-\mathbf{r}'|^3 = -\nabla(1/|\mathbf{r}-\mathbf{r}'|) \), which propagates into the wrong sign of \( \mathbf{A} \) (though it does not affect the final zero).
Believing a radial field \( \mathbf{B}\propto\hat{\mathbf{r}}/r^2 \) is allowed. Such a “monopole” field has non-zero divergence and cannot arise from Biot–Savart (see Problem 5).
Assuming “closed field lines” is the definition. \( \nabla\cdot\mathbf{B}=0 \) is a local statement about sources, a consequence, not a definition; a divergence-free field can have field lines that never close (they may wind ergodically).

Discussion

The proof isolates the geometric origin of the monopole ban. In the Biot–Savart law the source \( \mathbf{J} \) enters through a cross product with the Coulomb kernel. Because that kernel is itself a gradient, the whole integrand is a curl, and \( \mathbf{B} \) inherits the structure \( \mathbf{B}=\nabla\times\mathbf{A} \). A curl can never have divergence, so no configuration of steady currents can build a field that spreads out of a point. Contrast the electric case: there the source \( \rho \) multiplies the same kernel directly (a gradient, not a curl), so \( \mathbf{E}=-\nabla V \) has \( \nabla\cdot\mathbf{E}=\rho/\varepsilon_0\neq 0 \). Cross product versus scalar product is precisely the difference between “no monopoles” and Gauss's law.

Physically, \( \nabla\cdot\mathbf{B}=0 \) is a conservation statement about flux. The equivalent integral form \( \oint_S \mathbf{B}\cdot d\mathbf{A}=0 \) says magnetic flux is solenoidal: any tube of field lines carries the same flux along its length, so lines cannot terminate. This is why a bar magnet cut in two yields two dipoles rather than an isolated north and south pole — the field lines threading the magnet must close externally regardless of where you cut.

The result also underwrites the entire vector-potential formalism. Helmholtz's theorem guarantees that a field with zero divergence can be written globally as the curl of some \( \mathbf{A} \); \( \nabla\cdot\mathbf{B}=0 \) is exactly the licence for this. All of magnetostatics, the Aharonov–Bohm effect, and gauge theory then rest on the existence of \( \mathbf{A} \).

The exactness of the identity \( \nabla\cdot(\nabla\times\mathbf{A})=0 \) is topological, not merely computational: it is the statement \( d\circ d = 0 \) for differential forms (here \( \mathbf{B}=d\mathbf{A} \) as a 2-form, so \( d\mathbf{B}=0 \)). Dirac showed that even if a monopole existed, quantum mechanics would tolerate it only through a singular string in \( \mathbf{A} \), and consistency (single-valued wavefunctions) would then force charge quantisation \( eg = 2\pi n\hbar \). So the classical “no monopole” result and the quantum quantisation of electric charge are two faces of the same geometry.

Common misconceptions. (i) “No monopoles” is often taken to be an experimental fact only; here it is a theorem given Biot–Savart. (ii) \( \nabla\cdot\mathbf{B}=0 \) does not mean \( \mathbf{B}=0 \) in current-free regions — it constrains sources, not magnitude. (iii) It does not forbid magnetic dipoles, which are the true elementary magnetic sources.

Worked examples

Example 1 — Divergence of the infinite straight-wire field.

\[ \mathbf{B}(s) = \frac{\mu_0 I}{2\pi s}\,\hat{\boldsymbol\varphi} \]

Field of an infinite straight wire (prior result); only the azimuthal component is non-zero and it depends on \( s \) alone. A

\[ \nabla\cdot\mathbf{B} = \frac{1}{s}\frac{\partial(s\,B_s)}{\partial s} + \frac{1}{s}\frac{\partial B_\varphi}{\partial\varphi} + \frac{\partial B_z}{\partial z} \]

Cylindrical divergence. Here \( B_s=B_z=0 \), so only the middle term can survive. A

\[ \nabla\cdot\mathbf{B} = \frac{1}{s}\frac{\partial}{\partial\varphi}\!\left(\frac{\mu_0 I}{2\pi s}\right) = 0 \]

\( B_\varphi \) is independent of \( \varphi \), so its \( \varphi \)-derivative is zero. B

\[ I = 10\ \mathrm{A},\quad s = 0.050\ \mathrm{m} \ \Rightarrow\ B_\varphi = \frac{(4\pi\times10^{-7})(10)}{2\pi(0.050)} = 4.0\times10^{-5}\ \mathrm{T} \]

Numeric magnitude to show the field is genuinely non-zero (40 µT) even though its divergence vanishes. A

\[ \nabla\cdot\mathbf{B} = 0,\qquad B_\varphi = 40\ \mu\mathrm{T} \ \text{at } s=5\ \mathrm{cm} \]

Reading. A strong, real field with zero divergence: the closed circular field lines neither start nor end. Flux into any closed cylinder around the wire exactly cancels the flux out.

Example 2 — Divergence of the magnetic dipole field.

\[ B_r = \frac{\mu_0 m}{4\pi}\,\frac{2\cos\theta}{r^{3}},\qquad B_\theta = \frac{\mu_0 m}{4\pi}\,\frac{\sin\theta}{r^{3}},\qquad B_\varphi = 0 \]

Standard far-field dipole components for moment \( m \) along the polar axis. A

\[ \nabla\cdot\mathbf{B} = \frac{1}{r^{2}}\frac{\partial(r^{2}B_r)}{\partial r} + \frac{1}{r\sin\theta}\frac{\partial(\sin\theta\,B_\theta)}{\partial\theta} \]

Spherical divergence; the \( \varphi \)-term drops since \( B_\varphi=0 \) and nothing depends on \( \varphi \). A

\[ \frac{1}{r^{2}}\frac{\partial}{\partial r}\!\left(\frac{\mu_0 m}{4\pi}\frac{2\cos\theta}{r}\right) = -\frac{\mu_0 m}{4\pi}\,\frac{2\cos\theta}{r^{4}} \]

Radial term: \( r^2 B_r \propto 1/r \), whose \( r \)-derivative gives \( -1/r^2 \). B

\[ \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\!\left(\frac{\mu_0 m}{4\pi}\frac{\sin^{2}\theta}{r^{3}}\right) = \frac{\mu_0 m}{4\pi}\,\frac{2\cos\theta}{r^{4}} \]

Polar term: \( \partial_\theta(\sin^2\theta)=2\sin\theta\cos\theta \), and dividing by \( r\sin\theta \) leaves \( 2\cos\theta/r^4 \). B

\[ \nabla\cdot\mathbf{B} = -\frac{\mu_0 m}{4\pi}\frac{2\cos\theta}{r^{4}} + \frac{\mu_0 m}{4\pi}\frac{2\cos\theta}{r^{4}} = 0 \]

The two terms are equal and opposite for all \( r>0 \) and all \( \theta \): exact cancellation. C

\[ \nabla\cdot\mathbf{B}_{\text{dipole}} = 0 \quad (r>0) \]

Reading. Even the field that most “looks like” it emanates from a pole — the dipole — has strictly zero divergence away from the origin. Using Earth's moment \( m\approx8.0\times10^{22}\,\mathrm{A\,m^2} \) at \( r=6.4\times10^{6}\,\mathrm{m} \), \( \theta=0 \): \( B_r = \dfrac{(10^{-7})(2)(8.0\times10^{22})}{(6.4\times10^{6})^3}\approx 6.1\times10^{-5}\,\mathrm{T} \) — a real 61 µT field, yet source-free.

Problems

(A) Show explicitly, by differentiating, that \( \dfrac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^{3}} = -\nabla\!\left(\dfrac{1}{|\mathbf{r}-\mathbf{r}'|}\right) \) for \( \mathbf{r}\neq\mathbf{r}' \), and hence that its curl vanishes there.
Solution
Let \( u=|\mathbf{r}-\mathbf{r}'| = \sqrt{(x-x')^2+(y-y')^2+(z-z')^2} \). Then \( \partial u/\partial x = (x-x')/u \). So \( \partial(1/u)/\partial x = -u^{-2}\,(x-x')/u = -(x-x')/u^3 \), and likewise for \( y,z \). Hence \( \nabla(1/u) = -(\mathbf{r}-\mathbf{r}')/u^3 \), i.e. \( (\mathbf{r}-\mathbf{r}')/u^3 = -\nabla(1/u) \). Being minus the gradient of a scalar, it is a gradient field, and \( \nabla\times\nabla(1/u)=\mathbf{0} \) identically for \( \mathbf{r}\neq\mathbf{r}' \). Thus \( \nabla\times\big[(\mathbf{r}-\mathbf{r}')/u^3\big]=\mathbf{0} \).
(B) Using index (Levi-Civita) notation, prove \( \nabla\cdot(\nabla\times\mathbf{A})=0 \) for any \( C^2 \) field \( \mathbf{A} \).
Solution
\( (\nabla\times\mathbf{A})_k = \varepsilon_{kij}\partial_i A_j \). Then \( \nabla\cdot(\nabla\times\mathbf{A}) = \partial_k(\varepsilon_{kij}\partial_i A_j) = \varepsilon_{kij}\,\partial_k\partial_i A_j \). Because \( \mathbf{A}\in C^2 \), \( \partial_k\partial_i A_j = \partial_i\partial_k A_j \) is symmetric under \( k\leftrightarrow i \), while \( \varepsilon_{kij} \) is antisymmetric under \( k\leftrightarrow i \). Relabel the dummy indices \( k\leftrightarrow i \): \( \varepsilon_{kij}\partial_k\partial_i A_j = \varepsilon_{ikj}\partial_i\partial_k A_j = -\varepsilon_{kij}\partial_k\partial_i A_j \). The quantity equals its own negative, hence it is zero.
(B) A field is claimed to be \( \mathbf{B} = \beta\, s\,\hat{\mathbf{s}} \) (radially outward in cylindrical coordinates), with \( \beta = 2.0\times10^{-3}\ \mathrm{T\,m^{-1}} \). Compute \( \nabla\cdot\mathbf{B} \) and decide whether this can be a magnetostatic field.
Solution
\( \nabla\cdot\mathbf{B} = \tfrac{1}{s}\partial_s(s\,B_s) = \tfrac{1}{s}\partial_s(s\cdot\beta s) = \tfrac{1}{s}\partial_s(\beta s^2) = \tfrac{1}{s}(2\beta s) = 2\beta \). Numerically \( 2\beta = 4.0\times10^{-3}\ \mathrm{T\,m^{-1}}\neq 0 \). Since \( \nabla\cdot\mathbf{B}\neq0 \), this violates \( \nabla\cdot\mathbf{B}=0 \) and cannot be produced by any steady current via Biot–Savart. (In the fictitious-monopole convention it would require a volume magnetic charge density \( \rho_m = (\nabla\cdot\mathbf{B})/\mu_0 = 4.0\times10^{-3}/(4\pi\times10^{-7}) \approx 3.2\times10^{3}\ \mathrm{A\,m^{-2}} \).)
(B) Verify the integral form directly for the straight-wire field \( \mathbf{B}=\dfrac{\mu_0 I}{2\pi s}\hat{\boldsymbol\varphi} \): show the net flux through a closed coaxial cylinder (radius \( a \), length \( L \), with caps) is zero.
Solution
Split the closed surface into the curved side and the two flat caps. Curved side: outward normal is \( \hat{\mathbf{s}} \), but \( \mathbf{B}=B_\varphi\hat{\boldsymbol\varphi} \) is purely azimuthal, so \( \mathbf{B}\cdot\hat{\mathbf{s}}=0 \) — zero flux. Caps (top and bottom): outward normals are \( \pm\hat{\mathbf{z}} \), and \( \mathbf{B}\cdot\hat{\mathbf{z}}=0 \) since there is no \( z \)-component — zero flux. Total \( \oint\mathbf{B}\cdot d\mathbf{A}=0 \), consistent with \( \nabla\cdot\mathbf{B}=0 \). This holds for any \( a,L \); e.g. \( a=3\ \mathrm{cm},\,L=10\ \mathrm{cm},\,I=5\ \mathrm{A} \) still gives exactly \( 0\ \mathrm{Wb} \).
(C) Consider a proposed monopole field \( \mathbf{B}=\dfrac{\mu_0 g}{4\pi}\dfrac{\hat{\mathbf{r}}}{r^{2}} \). Show that the total outward flux through any enclosing sphere equals \( \mu_0 g \), independent of radius, so \( \nabla\cdot\mathbf{B}=\mu_0 g\,\delta^{3}(\mathbf{r}) \); then state why Biot–Savart can never generate it.
Solution
Away from the origin, \( \nabla\cdot(\hat{\mathbf{r}}/r^2)=\tfrac{1}{r^2}\partial_r(r^2\cdot r^{-2})=0 \). But the flux through a sphere of radius \( R \) is \( \oint\mathbf{B}\cdot d\mathbf{A} = \tfrac{\mu_0 g}{4\pi}\tfrac{1}{R^2}\cdot 4\pi R^2 = \mu_0 g \), the same for every \( R \). A field divergence-free everywhere except a point yet with non-zero, radius-independent enclosed flux must carry a delta-function source: \( \nabla\cdot(\hat{\mathbf{r}}/r^2)=4\pi\delta^3(\mathbf{r}) \), hence \( \nabla\cdot\mathbf{B}=\mu_0 g\,\delta^3(\mathbf{r}) \) — a genuine magnetic charge \( g \) at the origin. Biot–Savart writes \( \mathbf{B}=\nabla\times\mathbf{A} \) with a globally single-valued \( \mathbf{A} \), and \( \nabla\cdot(\nabla\times\mathbf{A})=0 \) even distributionally for smooth \( \mathbf{A} \); therefore no steady current can produce this field — a monopole would demand a singular (Dirac-string) \( \mathbf{A} \) that Biot–Savart never yields.