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If $g$ is smooth and $\nabla g\ne 0$ on $S=\{g=0\}$, transverse local coordinates give $⟨\delta (g),\phi ⟩={\int }_S\phi /|\nabla g|dS$. Thus ${\delta }_S=|\nabla g|\delta (g)$ as distributions. In particular ${\delta }_{S^1}=2\delta (|x{|}^2-1)$. Omitting this Jacobian confuses the delta of a defining function with a geometric surface delta distribution.

Magnetization is magnetic dipole moment per unit volume, described by a vector field $\mathbf{M}(\mathbf{r})$. Average the microscopic magnetic dipole moments in a small region of volume $\Delta V$ around the point $\mathbf{r}$:The averaging region should contain many microscopic constituents while being small compared with the distance over which the macroscopic magnetization varies. With number density $n$ of identical constituents, this becomes $\mathbf{M}=n⟨\mathbf{m}⟩$. Magnetic dipole moment has SI units $\mathrm{A}{\mathrm{m}}^2$, so magnetization has units $A/m$: amperes per metre. The average is a vector sum, not a sum of magnitudes.^[ref]

Alignment can change magnetization without changing the magnitude of any individual magnetic dipole moment. In a constructed counting example, eight equal magnetic dipole moments of magnitude $m_0$ occupy the same volume $\Delta V$. Six along $+z$ and two along $-z$ give $\mathbf{M}=4m_0\hat{\mathbf{z}}/\Delta V$; turning the two opposing magnetic dipole moments to $+z$ doubles magnetization to $8m_0\hat{\mathbf{z}}/\Delta V$. Both arrangements have the same sum of magnitudes, $8m_0$. Changing the individual magnetic dipole moments provides another possible response; their alignment and their magnitudes are distinct ingredients. Microscopic spin contributes magnetic dipole moment without being a literal rotating charged ball.

Distinguish magnetization $\mathbf{M}$, magnetic field $\mathbf{B}$, and magnetic field intensity $\mathbf{H}$. In SI,Here ${\mu }_0$ is the vacuum permeability, and $\mathrm{T}$ denotes the tesla. Magnetization describes matter's magnetic dipole moment density; magnetic field enters the Lorentz force; magnetic field intensity separates the free electric current contribution in the macroscopic equations. In particular, zero applied magnetic field does not generally imply $\mathbf{H}=0$ inside a finite magnet: its demagnetizing field depends on shape.^[ref]

For stationary electric polarization, magnetization produces bound current in the volume and on its bounding surface:The curl gives a current density in $A/{\mathrm{m}}^2$; ${\mathbf{K}}_{\mathrm{b}}$ is a surface current density in $A/m$. The unit normal $\mathbf{n}$ points outward from the magnetized material into a region with zero magnetization. Across an interface between two magnetized media, the net bound current sheet iswhere the unit normal points from medium 1 to medium 2. Both faces contribute: medium 2 has the opposite outward normal vector. This fixes the cross product sign. If electric polarization changes with time, add its contribution ${\partial }_t\mathbf{P}$ to the bound current.^[ref]

A short proof outline explains why both contributions are needed. Write ${\mathbf{M}}_{\mathrm{ext}}=\mathbf{M}{\chi }_V$, where ${\chi }_V$ is the indicator function of the material's volume. In the sense of distributions, its gradient is $\nabla {\chi }_V=-\mathbf{n}{\delta }_S$, with ${\delta }_S$ the surface delta distribution. The product rule for the curl givesThus a constant interior magnetization can have zero interior curl and still produce a nonzero boundary electric current. Equivalently, neighboring ideal microscopic current loops cancel along shared edges while uncompensated edges remain on the outside. This bound current representation predicts the same macroscopic magnetic field as the magnetization distribution; it does not require a wire or free charge transport around the magnet. Its usefulness is that the familiar Biot-Savart law can now calculate the magnetic field from a specified magnetization.

A time-dependent average magnetization need not vary smoothly under a smooth drive. The Barkhausen effect resolves collective jumps through their induced electromotive force; Barkhausen pulse area explains how that derivative readout relates to the underlying change.

For a moving level surface, $n=\nabla S/|\nabla S|$, $v_n=-S_t/|\nabla S|$, and the surface delta distribution is ${\delta }_s=|\nabla S|\delta (S)$. Piecewise classical conservation laws acquire a distributional flux defect $[b]\cdot n-v_n[a]$. Vanishing of this defect is the Rankine-Hugoniot condition.

Let $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$, so the Lighthill stress tensor is $T_{ij}^{-}+D_{ij}H(S)$. Write $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. The distributional derivative of the Heaviside step function gives ${\partial }_{i}H(S)=S_i\delta (S)$. Applying the product rule a second time yields the surface sources of a discontinuous Lighthill stress tensor:Repeated indices are summed. The first line gives the two bulk acoustic quadrupole distributions. The second line consists of additional surface delta distributions and their derivatives, all supported on the shock wave $S=0$. For a symmetric Lighthill stress tensor, the first two coefficients of $\delta (S)$ combine to $2D_{ij,i}{S}_j$ after relabelling indices.

A discontinuity therefore adds both a surface-delta source and a surface-delta-derivative source. The level function need not be a signed distance: retaining its derivatives makes the expression invariant under a smooth reparametrization of the same surface. If the two bulk fields are only continuously differentiable, their second derivatives in the first line are read weakly; no classical second derivative is being assumed. The coefficients multiplying ${\delta }^{\prime }(S)$ are kept as extensions, not replaced prematurely by traces.

Put ${\rho }^{\prime }=\rho -{\rho }_0$, and take the retarded acoustic Green function for ${\partial }_t^2-c_0^2{\nabla }^2$:This selects the causal, outgoing acoustic density perturbation, with no additional incoming homogeneous wave equation solution. Convolving the acoustic dipole forcing with this Green function and moving its spatial derivative outside the integral givesThe surface delta distribution converts the spatial integral to the moving surface. At fixed surface labels $(p,q)$, setThe radial Mach number enters the moving-surface retarded Jacobian, because $dr/d\tau =-\hat{r}\cdot v$ and henceLet ${\tau }^{\ast }$ be a root of the retarded time equationThe delta change-of-variable rule now givesHere $h_p{h}_q$ is the surface-area factor from the orthogonal surface coordinates. For a subsonic surface, $|v|<c_0$, the retarded equation is monotone in $\tau$ and has one root when the motion is defined for the required past times. For more general motion, sum the displayed contribution over all simple retarded roots. A root with $1-M_r=0$ requires a separate limiting treatment; the simple-root formula does not apply there.

Choose the shock wave normal to point from $S<0$ to $S>0$:The latter identity follows by differentiating $S({\mathbf{x}}_s(t),t)=0$ along the moving surface. Write $[a]=a^{+}-a^{-}$ and $[\mathbf{b}]={\mathbf{b}}^{+}-{\mathbf{b}}^{-}$. The distributional derivative of the Heaviside step function is ${\partial }_{t}H(S)=S_t\delta (S)$ and $\nabla H(S)=\nabla S\delta (S)$. The regular terms cancel using the conservation equations on each side, leavingThus the explicitly requested vector isOnly its normal component matters; adding tangential velocity to the surface parametrization changes neither result. The invariant surface delta distribution is ${\delta }_s=|\nabla S|\delta (S)$, so the formula is the moving-interface conservation jump identity ${\partial }_{t}a+\nabla \cdot \mathbf{b}=([\mathbf{b}]\cdot \mathbf{n}-v_n[a]){\delta }_s$.

The test-function space is $\mathcal{D}({\mathbb{R}}^n)=C_c^{\infty }({\mathbb{R}}^n)$. Its topology is the strict inductive limit topology of the spaces ${\mathcal{D}}_K$ of smooth functions supported in a fixed compact set $K$, each with the seminorms $p_{K,j}(\phi )={\max }_{|\alpha |\le j}{\sup }_K|{\partial }^{\alpha }\phi |$. In particular, a sequence converges in $\mathcal{D}$ precisely when its supports lie in one compact set and all its derivatives converge uniformly. The indices $\alpha$ use multi-index notation.

The distribution space ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ consists of continuous complex-linear forms on $\mathcal{D}$. Equivalently, for every compact $K$ there are $C_K$ and an integer $q_K\ge 0$ such thatWe use distributional convergence: $u_j\to u$ means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{D}$. The pairings are bilinear, with no complex conjugation. A test function $f$ is identified with its regular distribution $⟨f,\phi ⟩=\int f\phi$.

For a distribution $u$ and a test function $\psi$, their smoothing convolution with a test function isIt is a smooth function, with ${\partial }_x^{\alpha }(u\ast \psi )(x)=⟨u_y,{\partial }^{\alpha }\psi (x-y)⟩$. Indeed, for $x$ in a compact neighborhood all translated tests have support in one compact set, and their difference quotients converge in the space of test functions. Notice that $u\ast \psi$ need not be compactly supported.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }(B_1)$ with $\int \rho =1$, and write ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$. If $\tilde{\rho }(x)=\rho (-x)$, thenThe right-hand test tends to $\phi$ in $\mathcal{D}$: its supports lie in $\mathrm{supp}\phi +{\overline{B}}_1$ for $\epsilon \le 1$, and every derivative converges uniformly by the approximate-identity argument. Thus the smooth regularizations converge to $u$ as distributions.

To obtain actual compactly supported approximants, choose a smooth cutoff function $\chi$ equal to one on $B_1$ and supported in $B_2$, and setFor each fixed test function $\phi$, the cutoff is identically one on its support once $j$ is large. Hence $⟨f_j,\phi ⟩=⟨u,{\tilde{\rho }}_{1/j}\ast \phi ⟩\to ⟨u,\phi ⟩$. This proves $\mathcal{D}$ is dense in ${\mathcal{D}}^{\prime }$, and in fact establishes the density of test functions in distributions and gives a convergent approximating sequence for each distribution. The expanding cutoff is essential when $u$ has noncompact support of a distribution.

For the radial limit, take a test function $\phi$ and introduce $t=r^2-1$ in polar coordinates. The Jacobian $rdr=dt/2$ givesWhen $\phi$ is supported away from the origin, $A$ vanishes near $t=-1$ and extends to a compactly supported smooth function on the whole line. The folded sine approximation to a Dirac delta is exposed by setting $B(s)=A(s)+A(-s)$: integration by parts givesby the Riemann-Lebesgue lemma. ThereforeThis surface delta distribution is arclength measure on the unit circle. By the level-set normalization of a surface delta, it is $2\delta (|x{|}^2-1)$: the factor two cancels the gradient magnitude $|\nabla (|x{|}^2-1)|=2$ on the circle. It is not twice arclength measure.

For a test function that can meet the origin, the endpoint cannot be discarded. Now $A(-1)=\pi \phi (0)$. Its right derivative is integrable, with $A^{\prime }(-1+s)=(\pi /4)\Delta \phi (0)+O(s)$ from differentiated Taylor expansion: the angular average cancels odd Taylor terms, so $A(t)=\pi \phi (0)+(\pi /4)(1+t)\Delta \phi (0)+O((1+t{)}^2)$ near $t=-1$. Applying integration by parts separately on the negative and positive intervals givesThe last two integrals tend to zero by the Riemann-Lebesgue lemma. Thus the radial quadratic oscillation defect in two dimensions gives the stronger oscillating point-mass defect formulaChoose $\phi$ supported in $B_{1/2}$ with $\phi (0)=1$. Its pairing is $-\pi \cos m+o(1)$, which does not converge. For completeness, if $\cos m$ had a limit $L$, the recurrence $\cos (m+1)+\cos (m-1)=2\cos 1\cos m$ would force $L=0$, whereas the even subsequence identity $\cos (2m)=2{\cos }^{2}m-1$ would then force $L=-1$. Consequently there is no limit in ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.

For the geometric surface delta distribution on the radius-$a$ sphere in ${\mathbb{R}}^3$, $⟨u_a,\varphi ⟩=a^2{\int }_{S^2}\varphi (a\omega )d\sigma (\omega )$. Angular integration gives the Fourier transform ${\hat{u}}_a(\xi )=4\pi a\sin (a|\xi |)/|\xi |$, with removable value $4\pi a^2$ at zero. The convolution of distributions with a compactly supported factor gives the regular distributionThis density has total mass $16{\pi }^2{a}^2{b}^2$. Values on endpoint spheres do not affect the distribution. When $a=b$, the inverse-distance singularity is locally integrable in three dimensions and is not an additional point mass. The annulus expresses the triangle inequality for the sum of two vectors of fixed lengths.

Write a piecewise smooth Lighthill stress tensor as $T_{ij}=T_{ij}^{-}+D_{ij}H(S)$, where $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$ and the regular level surface $S=0$ is a shock wave. The distributional derivative of the Heaviside step function givesHere repeated indices are summed, $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. Keep the coefficients as smooth extensions before multiplying distributions: prematurely replacing the coefficient of ${\delta }^{\prime }(S)$ by its surface trace can lose a $\delta (S)$ contribution. The extra surface delta distributions and their derivatives are supported entirely on the discontinuity.