Chapter 1 — Mathematical Preliminaries

Quote

The enormous usefulness of mathematics in the natural sciences is something bordering on the mysterious.

— Eugene Wigner (1960)

Motivation

This chapter collects the mathematical machinery used throughout electrodynamics: vector calculus in Cartesian and curvilinear coordinates, index notation, integral theorems, delta functions, Fourier analysis, and the Helmholtz decomposition. None of this is new — but having it condensed and cross-referenced saves constant trips back to math references.

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Coordinate Systems and Differential Operators

Cartesian Coordinates $(x,y,z)$

Volume element and operators:

$${\mathrm{d}}^{3}r=\mathrm{d}x\mathrm{d}y\mathrm{d}z$$ $$\nabla =\hat{\mathbf{x}}\frac{\partial }{\partial x}+\hat{\mathbf{y}}\frac{\partial }{\partial y}+\hat{\mathbf{z}}\frac{\partial }{\partial z}$$ $$\nabla \cdot \mathbf{F}=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$$ $${\nabla }^{2}f=\frac{{\partial }^{2}f}{\partial x^2}+\frac{{\partial }^{2}f}{\partial y^2}+\frac{{\partial }^{2}f}{\partial z^2}$$ $$[\nabla \times \mathbf{F}{]}_i={ϵ}_{ijk}{\partial }_j{F}_k$$

Spherical Coordinates $(r,\theta ,\varphi )$

$${\mathrm{d}}^{3}r=r^2\sin \theta \mathrm{d}r\mathrm{d}\theta \mathrm{d}\varphi$$ $$\nabla f=\hat{\mathbf{r}}\frac{\partial f}{\partial r}+\hat{\mathbf{\theta }}\frac{1}{r}\frac{\partial f}{\partial \theta }+\hat{\mathbf{\varphi }}\frac{1}{r\sin \theta }\frac{\partial f}{\partial \varphi }$$ $$\nabla \cdot \mathbf{F}=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{F}_r)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta F_{\theta })+\frac{1}{r\sin \theta }\frac{\partial F_{\varphi }}{\partial \varphi }$$ $${\nabla }^{2}f=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial f}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}f}{\partial {\varphi }^2}$$ $$\nabla \times \mathbf{F}=\frac{1}{r\sin \theta }\left[\frac{\partial }{\partial \theta }(\sin \theta F_{\varphi })-\frac{\partial F_{\theta }}{\partial \varphi }\right]\hat{\mathbf{r}}+\frac{1}{r}\left[\frac{1}{\sin \theta }\frac{\partial F_r}{\partial \varphi }-\frac{\partial }{\partial r}(r{F}_{\varphi })\right]\hat{\mathbf{\theta }}+\frac{1}{r}\left[\frac{\partial }{\partial r}(r{F}_{\theta })-\frac{\partial F_r}{\partial \theta }\right]\hat{\mathbf{\varphi }}$$

Cylindrical Coordinates $(\rho ,\varphi ,z)$

$${\mathrm{d}}^{3}r=\rho \mathrm{d}\rho \mathrm{d}\varphi \mathrm{d}z$$ $$\nabla f=\hat{\mathbf{\rho }}\frac{\partial f}{\partial \rho }+\hat{\mathbf{\varphi }}\frac{1}{\rho }\frac{\partial f}{\partial \varphi }+\hat{\mathbf{z}}\frac{\partial f}{\partial z}$$ $$\nabla \cdot \mathbf{F}=\frac{1}{\rho }\frac{\partial }{\partial \rho }(\rho F_{\rho })+\frac{1}{\rho }\frac{\partial F_{\varphi }}{\partial \varphi }+\frac{\partial F_z}{\partial z}$$ $${\nabla }^{2}f=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial f}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^{2}f}{\partial {\varphi }^2}+\frac{{\partial }^{2}f}{\partial z^2}$$ $$\nabla \times \mathbf{F}=\left(\frac{1}{\rho }\frac{\partial F_z}{\partial \varphi }-\frac{\partial F_{\varphi }}{\partial z}\right)\hat{\mathbf{\rho }}+\left(\frac{\partial F_{\rho }}{\partial z}-\frac{\partial F_z}{\partial \rho }\right)\hat{\mathbf{\varphi }}+\frac{1}{\rho }\left(\frac{\partial (\rho F_{\varphi })}{\partial \rho }-\frac{\partial F_{\rho }}{\partial \varphi }\right)\hat{\mathbf{z}}$$

Curl in General Orthogonal Coordinates

In an orthogonal coordinate system $(u_1,u_2,u_3)$ with scale factors $h_i=|\partial \mathbf{r}/\partial u_i|$, the curl is the determinant:

$$\nabla \times \mathbf{F}=\frac{1}{h_1{h}_2{h}_3}\left|\begin{array}{ccc}{h}_1{\hat{\mathbf{e}}}_1& h_2{\hat{\mathbf{e}}}_2& h_3{\hat{\mathbf{e}}}_3\\ \partial /\partial u_1& \partial /\partial u_2& \partial /\partial u_3\\ h_1{F}_1& h_2{F}_2& h_3{F}_3\end{array}\right|$$

The $h_i$ are the metric coefficients (not Christoffel symbols, though related). For spherical: $h_r=1$, $h_{\theta }=r$, $h_{\varphi }=r\sin \theta$. For cylindrical: $h_{\rho }=1$, $h_{\varphi }=\rho$, $h_z=1$.

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Einstein Summation Convention

Repeated indices are summed over:

$$A_i{B}_i:=\sum _{i=1}^3{A}_i{B}_i=\mathbf{A}\cdot \mathbf{B}$$

This makes vector identities into index algebra. The key objects are ${\delta }_{ij}$ and ${ϵ}_{ijk}$.

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Kronecker Delta and Levi-Civita Symbol

$${\delta }_{ij}=\{\begin{array}{ll}1& i=j\\ 0& i\ne j\end{array}{ϵ}_{ijk}=\{\begin{array}{ll}+1& ijk\text{ is an even permutation of }xyz\\ -1& ijk\text{ is an odd permutation of }xyz\\ 0& \text{any repeated index}\end{array}$$

Essential Identities

Identity	Formula
Dot product of basis vectors	${\hat{\mathbf{e}}}_i\cdot {\hat{\mathbf{e}}}_j={\delta }_{ij}$
Trace	${\delta }_{kk}=3$
Selector	$V_k{\delta }_{kj}=V_j$
Partial derivative of position	${\partial }_k{r}_j={\delta }_{jk}$
Cross product	$[\mathbf{V}\times \mathbf{F}{]}_i={ϵ}_{ijk}{V}_j{F}_k$
Curl	$[\nabla \times \mathbf{A}{]}_i={ϵ}_{ijk}{\partial }_j{A}_k$
Contraction to zero	${\delta }_{ij}{ϵ}_{ijk}=0$
Full contraction	${ϵ}_{ijk}{ϵ}_{ijk}=6$

Additional 3D Identities

Fourier representation:

$$\delta (\mathbf{r}-{\mathbf{r}}^{\prime })=\frac{1}{(2\pi {)}^3}\int {\mathrm{d}}^{3}k{e}^{i\mathbf{k}\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })}$$

Surface delta function — for a surface $S$ defined by $g({\mathbf{r}}_S)=0$:

$$\int {\mathrm{d}}^{3}rf(\mathbf{r})\delta [g(\mathbf{r})]={\int }_S\mathrm{d}S\frac{f({\mathbf{r}}_S)}{|\nabla g({\mathbf{r}}_S)|}$$

Second derivative identity:

$$\frac{\partial }{\partial r_k}\frac{\partial }{\partial r_m}\frac{1}{r}=\frac{3r_k{r}_m-r^2{\delta }_{km}}{r^5}-\frac{4\pi }{3}{\delta }_{km}\delta (\mathbf{r})$$

Why the Delta Term in the Second Derivative

The traceless tensor $3r_k{r}_m/r^5$ integrates to zero over any sphere centered at the origin — but ${\partial }_k{\partial }_m(1/r)$ must be consistent with ${\nabla }^2(1/r)=-4\pi \delta (\mathbf{r})$. Taking the trace ($k=m$, summing) of the left side must give $-4\pi \delta (\mathbf{r})$, which forces the $-\frac{4\pi }{3}{\delta }_{km}\delta (\mathbf{r})$ contact term. This becomes important in the theory of dipole fields.

TODO: Add link to dipole theory.

The Master Identity

The single most useful identity in all of vector calculus via indices:

$${ϵ}_{ijk}{ϵ}_{ist}={\delta }_{js}{\delta }_{kt}-{\delta }_{jt}{\delta }_{ks}$$

Almost every vector identity can be derived from this by choosing appropriate indices and contracting.

Proof of the $ϵ$-$ϵ$ Identity

Starting from ${ϵ}_{ijk}={\hat{\mathbf{e}}}_i\cdot ({\hat{\mathbf{e}}}_j\times {\hat{\mathbf{e}}}_k)$, the product ${ϵ}_{ijk}{ϵ}_{ist}$ sums over $i$. Since ${ϵ}_{ijk}$ is nonzero only for distinct $i,j,k$ and distinct $i,s,t$: the index $i$ is fixed once $j,k$ are chosen, leaving two possible matchings of $(j,k)$ to $(s,t)$. This gives exactly ${\delta }_{js}{\delta }_{kt}-{\delta }_{jt}{\delta }_{ks}$.

Generalization: No Repeated Indices

When there are no repeated indices to sum:

$${ϵ}_{\ell ki}{ϵ}_{mpq}=\left|\begin{array}{ccc}{\delta }_{\ell m}& {\delta }_{\ell p}& {\delta }_{\ell q}\\ {\delta }_{km}& {\delta }_{kp}& {\delta }_{kq}\\ {\delta }_{im}& {\delta }_{ip}& {\delta }_{iq}\end{array}\right|$$

Determinant via Levi-Civita

For a $3\times 3$ matrix $C$:

$$\det C={ϵ}_{ijk}{C}_{i1}{C}_{j2}{C}_{k3}={ϵ}_{ijk}{C}_{1i}{C}_{2j}{C}_{3k}$$

A related identity used in the tensor section:

$${ϵ}_{\ell mn}\det C={ϵ}_{ijk}{C}_{i\ell }{C}_{jm}{C}_{kn}$$

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Vector Identities

Cartesian Only

Index-based proofs of vector identities work only in Cartesian coordinates. The strategy: derive the identity in Cartesian indices, then pull the result back into a coordinate-invariant expression. The final identity then holds in any coordinate system.

This is also why it’s often useful to choose $\hat{\mathbf{z}}$ along a special direction (e.g., a dipole moment or external field), work out the algebra, and then re-express the answer in coordinate-free form.

Useful Identities

Product rules:

$$\nabla \cdot (f\mathbf{g})=f\nabla \cdot \mathbf{g}+\mathbf{g}\cdot \nabla f$$ $$\nabla \times (f\mathbf{g})=f\nabla \times \mathbf{g}-\mathbf{g}\times \nabla f$$

Double cross products:

$$\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$$ $$\nabla \times (\mathbf{A}\times \mathbf{B})=\mathbf{A}\nabla \cdot \mathbf{B}-(\mathbf{A}\cdot \nabla )\mathbf{B}+(\mathbf{B}\cdot \nabla )\mathbf{A}-\mathbf{B}\nabla \cdot \mathbf{A}$$

Deriving $\nabla \times (\mathbf{A}\times \mathbf{B})$ by Indices

$$[\nabla \times (\mathbf{A}\times \mathbf{B}){]}_i={ϵ}_{ijk}{\partial }_j(\mathbf{A}\times \mathbf{B}{)}_k={ϵ}_{ijk}{\partial }_j{ϵ}_{k\ell m}{A}_{\ell }{B}_m={ϵ}_{ijk}{ϵ}_{k\ell m}{\partial }_j(A_{\ell }{B}_m)$$

Using ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$:

$$=({\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }){\partial }_j(A_{\ell }{B}_m)={\partial }_j(A_i{B}_j)-{\partial }_j(A_j{B}_i)$$

Expanding by product rule and reassembling into vector form gives the result.

Mixed dot/cross products:

$$(\mathbf{A}\times \mathbf{B})\cdot (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot \mathbf{C})(\mathbf{B}\cdot \mathbf{D})-(\mathbf{A}\cdot \mathbf{D})(\mathbf{B}\cdot \mathbf{C})$$ $$(\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot \mathbf{C}\times \mathbf{D})\mathbf{B}-(\mathbf{B}\cdot \mathbf{C}\times \mathbf{D})\mathbf{A}$$

Decomposing a Vector Along an Arbitrary Direction

For any vector $\mathbf{b}$ and unit vector $\hat{\mathbf{n}}$:

$$\mathbf{b}=(\mathbf{b}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}+\hat{\mathbf{n}}\times (\mathbf{b}\times \hat{\mathbf{n}})$$

The first term is the component parallel to $\hat{\mathbf{n}}$; the second is the perpendicular part. This is used constantly in boundary condition problems.

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Functions of $\mathbf{r}-{\mathbf{r}}^{\prime }$

Let $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }=(x-x^{\prime })\hat{\mathbf{x}}+(y-y^{\prime })\hat{\mathbf{y}}+(z-z^{\prime })\hat{\mathbf{z}}$ with $R=|\mathbf{R}|$. Then:

$$\nabla f(R)=f^{\prime }(R)\hat{\mathbf{R}}\nabla \cdot \mathbf{g}(R)={\mathbf{g}}^{\prime }(R)\cdot \hat{\mathbf{R}}\nabla \times \mathbf{g}(R)=\hat{\mathbf{R}}\times {\mathbf{g}}^{\prime }(R)$$

Primed ↔ Unprimed Swap

Since ${\nabla }^{\prime }=\hat{\mathbf{x}}\partial /\partial x^{\prime }+\hat{\mathbf{y}}\partial /\partial y^{\prime }+\hat{\mathbf{z}}\partial /\partial z^{\prime }$ and $R$ depends on $\mathbf{r}-{\mathbf{r}}^{\prime }$:

$${\nabla }^{\prime }f(R)=-\nabla f(R)$$

This is used constantly when flipping between source coordinates (${\mathbf{r}}^{\prime }$) and field point coordinates ($\mathbf{r}$). Convention: primed coordinates refer to sources, unprimed to the point where we evaluate fields/potentials.

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The Convective Derivative

Let $\phi (\mathbf{r},t)$ be a scalar function of space and time. An observer at a fixed point records $\partial \phi /\partial t$. An observer moving along a trajectory $\mathbf{r}(t)$ with velocity $\mathbf{\upsilon }(t)=\dot{\mathbf{r}}(t)$ records:

$$\frac{\mathrm{d}\phi }{\mathrm{d}t}=\frac{\partial \phi }{\partial t}+(\mathbf{\upsilon }\cdot \nabla )\phi$$

For a vector function $\mathbf{g}(\mathbf{r},t)$:

$$\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}=\frac{\partial \mathbf{g}}{\partial t}+(\mathbf{\upsilon }\cdot \nabla )\mathbf{g}$$

What This Actually Means

Calling $\partial /\partial t$ a “partial” and $\mathrm{d}/\mathrm{d}t$ a “total” derivative is standard but misleading. What’s really going on:

• $\partial \phi /\partial t$ is the rate of change at a fixed point in space — the field evolves around you.
• $(\mathbf{\upsilon }\cdot \nabla )\phi$ is the change you pick up by moving through a spatially varying field — even if it’s static.
• $\mathrm{d}\phi /\mathrm{d}t$ is the rate of change along a worldline $\mathbf{r}(t)$.

The parameter $t$ plays a dual role: it parameterizes both the system’s evolution (through $\partial /\partial t$) and the observer’s trajectory (through $\mathbf{\upsilon }$). The convective derivative combines both.

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Multidimensional Taylor Expansion

$$f(\mathbf{r}+\mathbf{ϵ})=\exp (\mathbf{ϵ}\cdot \nabla )f(\mathbf{r})=\left[1+\mathbf{ϵ}\cdot \nabla +\frac{1}{2!}(\mathbf{ϵ}\cdot \nabla {)}^2+\cdots \right]f(\mathbf{r})$$

Differentials in the Exponent

The notation $\exp (\mathbf{ϵ}\cdot \nabla )$ is formal — it means “apply the differential operator $\mathbf{ϵ}\cdot \nabla$ repeatedly.” This is not exponentiation of a number; it’s the operator Taylor series. But it’s extremely useful as a bookkeeping device: translation in space is generated by the gradient operator, just as translation in time is generated by $\partial /\partial t$.

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The Jacobian

The determinant of the Jacobian matrix relates volume elements when changing variables:

$${\mathrm{d}}^{N}x=|J(\mathbf{x},\mathbf{y})|{\mathrm{d}}^{N}y=\left|\det \frac{\partial x_i}{\partial y_j}\right|d^{N}y$$

Spherical Coordinates

$x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$, $z=r\cos \theta$ gives $|J|=r^2\sin \theta$, recovering $d^{3}r=r^2\sin \theta drd\theta d\varphi$.

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Integral Theorems

The Divergence (Gauss’) Theorem

For a vector function $\mathbf{F}(\mathbf{r})$ defined in a volume $V$ enclosed by a surface $S$ with outward normal $\hat{\mathbf{n}}$:

$${\int }_V{\mathrm{d}}^{3}r\nabla \cdot \mathbf{F}={\oint }_S\mathrm{d}\mathbf{S}\cdot \mathbf{F}$$

Intuition

The divergence $\nabla \cdot \mathbf{F}$ at a point measures the net “source strength” — how much more flux leaves a tiny volume than enters it. Summing over all such volumes, interior faces cancel (flux leaving one cell enters the neighbor), and only the flux through the outer boundary $S$ survives.

Variants from $\mathbf{F}=\mathbf{c}\psi$ and $\mathbf{F}=\mathbf{A}\times \mathbf{c}$ (with $\mathbf{c}$ an arbitrary constant vector):

$${\int }_V{\mathrm{d}}^{3}r\nabla \psi ={\oint }_S\mathrm{d}\mathbf{S}\psi$$ $${\int }_V{\mathrm{d}}^{3}r\nabla \times \mathbf{A}={\oint }_S\mathrm{d}\mathbf{S}\times \mathbf{A}$$

Examples: Two Surface Integrals

Let $S$ bound a volume $V$.

(a) ${\oint }_{S}d\mathbf{S}=0$

Set $\psi =1$ in the gradient variant: ${\int }_V{d}^{3}r\nabla (1)={\oint }_{S}d\mathbf{S}$. Since $\nabla (1)=0$, the result follows.

(b) $\frac{1}{3}{\oint }_{S}d\mathbf{S}\cdot \mathbf{r}=V$

Apply the divergence theorem to $\mathbf{F}=\mathbf{r}$: ${\int }_V{d}^{3}r\nabla \cdot \mathbf{r}={\oint }_{S}d\mathbf{S}\cdot \mathbf{r}$. Since $\nabla \cdot \mathbf{r}=3$, we get $3V={\oint }_{S}d\mathbf{S}\cdot \mathbf{r}$.

Green’s Identities

Choose $\mathbf{F}=\phi \nabla \psi$ in the divergence theorem → Green’s first identity:

$${\int }_V{\mathrm{d}}^{3}r[\phi {\nabla }^2\psi +\nabla \phi \cdot \nabla \psi ]={\oint }_S\mathrm{d}\mathbf{S}\cdot \phi \nabla \psi$$

Exchange $\phi \leftrightarrow \psi$ and subtract → Green’s second identity:

$${\int }_V{\mathrm{d}}^{3}r[\phi {\nabla }^2\psi -\psi {\nabla }^2\phi ]={\oint }_S\mathrm{d}\mathbf{S}\cdot [\phi \nabla \psi -\psi \nabla \phi ]$$

Where These Are Used

Green’s identities are the backbone of potential theory. The first identity proves uniqueness of solutions to Laplace’s equation. The second identity is the basis for Green function methods. The vector analogues appear in radiation theory.

Vector analogues: Choose $\mathbf{F}=\mathbf{P}\times \nabla \times \mathbf{Q}$ and use $\nabla \cdot (\mathbf{A}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{A}-\mathbf{A}\cdot \nabla \times \mathbf{B}$:

$${\int }_V{\mathrm{d}}^{3}r[\nabla \times \mathbf{P}\cdot \nabla \times \mathbf{Q}-\mathbf{P}\cdot \nabla \times \nabla \times \mathbf{Q}]={\oint }_S\mathrm{d}\mathbf{S}\cdot (\mathbf{P}\times \nabla \times \mathbf{Q})$$

Exchange $\mathbf{P}\leftrightarrow \mathbf{Q}$ and subtract:

$${\int }_V{\mathrm{d}}^{3}r[\mathbf{Q}\cdot \nabla \times \nabla \times \mathbf{P}-\mathbf{P}\cdot \nabla \times \nabla \times \mathbf{Q}]={\oint }_S\mathrm{d}\mathbf{S}\cdot [\mathbf{P}\times \nabla \times \mathbf{Q}-\mathbf{Q}\times \nabla \times \mathbf{P}]$$

Stokes’ Theorem

For a vector function $\mathbf{F}(\mathbf{r})$ on an open surface $S$ bounded by a closed curve $C$:

$${\oint }_S\mathrm{d}\mathbf{S}\cdot \nabla \times \mathbf{F}={\oint }_C\mathrm{d}\mathbf{\ell }\cdot \mathbf{F}$$

$C$ is traversed in the direction given by the right-hand rule with the thumb along $d\mathbf{S}$.

Intuition

The curl $\nabla \times \mathbf{F}$ at a point measures the local circulation density. Summing over all tiny loops tiling $S$, adjacent edges cancel (shared edges are traversed in opposite directions), leaving only the boundary $C$.

Variants from $\mathbf{F}=\mathbf{c}\psi$ and $\mathbf{F}=\mathbf{A}\times \mathbf{c}$:

$${\oint }_S\mathrm{d}\mathbf{S}\times \nabla \psi ={\oint }_C\mathrm{d}\mathbf{\ell }\psi$$ $${\oint }_C\mathrm{d}\mathbf{\ell }\times \mathbf{A}={\int }_S\mathrm{d}{S}_k\nabla A_k-{\int }_S\mathrm{d}\mathbf{S}(\nabla \cdot \mathbf{A})$$

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The Delta Function

Not a Function

$\delta (x)$ is a distribution (generalized function) — it is defined only by what it does under an integral sign. It cannot be evaluated pointwise.

One Dimension

Defining property:

$${\int }_{-\infty }^{\infty }\mathrm{d}xf(x)\delta (x-x_0)=f(x_0)$$

Other properties:

$$\delta (ax)=\frac{1}{|a|}\delta (x),a\ne 0$$ $$\delta [g(x)]=\sum _m\frac{1}{|g^{\prime }(x_m)|}\delta (x-x_m),g(x_m)=0,g^{\prime }(x_m)\ne 0$$

Proof of $\delta (ax)=\delta (x)/|a|$

For $a>0$: substitute $u=ax$, $du=adx$:

$$\int dxf(x)\delta (ax)=\int \frac{du}{a}f(u/a)\delta (u)=\frac{f(0)}{a}$$

For $a<0$: the substitution reverses the limits, picking up an extra minus sign, giving $1/|a|$.

Integral Representation

$$\delta (x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }dk{e}^{ikx}$$

Proof via Regularization

Define $D(x)={\lim }_{m\to \infty }\frac{\sin mx}{\pi x}$. For any smooth $f(x)$:

$${\int }_{-\infty }^{\infty }dxf(x)\frac{\sin mx}{\pi x}=\underset{m\to \infty }{\lim }{\int }_{-\infty }^{\infty }\frac{dy}{m}f\left(\frac{y}{m}\right)\frac{\sin y}{\pi }\cdot \frac{m}{y}=f(0)\cdot \frac{1}{\pi }{\int }_{-\infty }^{\infty }dy\frac{\sin y}{y}$$

The assertion follows if ${\int }_{-\infty }^{\infty }dy\frac{\sin y}{y}=\pi$. Trick:

$${\int }_{-\infty }^{\infty }dy\frac{\sin y}{y}=2{\int }_0^{\infty }dy\frac{\sin y}{y}={\int }_0^{\infty }dy\sin y{\int }_0^{\infty }d\nu e^{-\nu y}=2{\int }_0^{\infty }\frac{d\nu }{1+{\nu }^2}=\pi$$

(Exchanging integration order, then using ${\int }_0^{\infty }dy{e}^{-\nu y}\sin y=1/(1+{\nu }^2)$.)

Connection to Plemelj's Formula

The regularized representation of the step function

$$\Theta (x)=\underset{ϵ\to 0}{\lim }\frac{i}{2\pi }{\int }_{-\infty }^{\infty }dk\frac{e^{-ikx}}{k+iϵ}$$

is intimately connected to the Sokhotski–Plemelj formula:

$$\underset{ϵ\to 0}{\lim }\frac{1}{x\pm iϵ}=\mathcal{P}\frac{1}{x}\mp i\pi \delta (x)$$

where $\mathcal{P}$ denotes the Cauchy principal value. The delta function picks out the pole contribution; the principal value gives the “regular” part.

Step Function and Sign Function

$$\Theta (x)=\{\begin{array}{ll}0& x<0\\ 1& x>0\end{array}\frac{\mathrm{d}\Theta }{\mathrm{d}x}=\delta (x)$$ $$\mathrm{sgn}(x)=\frac{\mathrm{d}}{\mathrm{d}x}|x|=\{\begin{array}{ll}-1& x<0\\ +1& x>0\end{array}=-1+2{\int }_{-\infty }^x\mathrm{d}y\delta (y)$$

Three Dimensions

$${\int }_V{\mathrm{d}}^{3}rf(\mathbf{r})\delta (\mathbf{r}-{\mathbf{r}}^{\prime })=\{\begin{array}{ll}f({\mathbf{r}}^{\prime })& {\mathbf{r}}^{\prime }\in V\\ 0& {\mathbf{r}}^{\prime }\notin V\end{array}$$

In curvilinear coordinates, the delta function must integrate to unity over all space so:

$$\delta (\mathbf{x}-{\mathbf{x}}_0)=\frac{1}{|J(\mathbf{x},\mathbf{y})|}\delta (\mathbf{y}-{\mathbf{y}}_0)$$

Half-Space Integral

${\int }_0^{\infty }dr\delta (r)=1$, not $1/2$. The delta function is defined by what it does to test functions, and the standard convention (consistent with ${\nabla }^2(1/r)=-4\pi \delta (\mathbf{r})$) requires the full weight at $r=0$.

An Important Identity For Electrostatics

$${\nabla }^2\frac{1}{r}=-\nabla \cdot \frac{\hat{\mathbf{r}}}{r^2}=-4\pi \delta (\mathbf{r})$$

Proof

For $r\ne 0$: direct computation gives ${\nabla }^2(1/r)=0$ (using ${\nabla }^{2}f=(r^2{f}^{\prime }{)}^{\prime }/r^2$ with $f=1/r$).

At $r=0$: integrate ${\nabla }^2(1/r)$ over a sphere of radius $R$ centered at the origin:

$${\int }_V{\mathrm{d}}^{3}r{\nabla }^2\frac{1}{r}={\oint }_S\mathrm{d}\mathbf{S}\cdot \nabla \frac{1}{r}={\oint }_{S}d\mathbf{S}\cdot \left(-\frac{\hat{\mathbf{r}}}{r^2}\right)=-\frac{1}{R^2}\cdot 4\pi R^2=-4\pi$$

A function that vanishes everywhere except at the origin and integrates to $-4\pi$ is $-4\pi \delta (\mathbf{r})$.

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Fourier Analysis

Fourier Series

Every periodic function $f(x+L)=f(x)$ has a Fourier series:

$$f(x)=\sum _{m=-\infty }^{\infty }{\hat{f}}_m{e}^{i2\pi mx/L},{\hat{f}}_m=\frac{1}{L}{\int }_0^L\mathrm{d}xf(x)e^{-i2\pi mx/L}$$

Fourier Transform

For non-periodic functions, the sum becomes an integral:

$$f(x)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\mathrm{d}k\hat{f}(k)e^{ikx}\hat{f}(k)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\mathrm{d}xf(x)e^{-ikx}$$

**Full spacetime (note sign flip convention in temporal direction):

$$f(\mathbf{r},t)=\frac{1}{(2\pi {)}^2}\int {\mathrm{d}}^{3}k{\int }_{-\infty }^{\infty }\mathrm{d}\omega \hat{f}(\mathbf{k}|\omega )e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$$ $$\hat{f}(\mathbf{k}|\omega )=\frac{1}{(2\pi {)}^2}\int {\mathrm{d}}^{3}r{\int }_{-\infty }^{\infty }\mathrm{d}tf(\mathbf{r},t)e^{-i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$$

For real $f$: $\hat{f}(k)={\hat{f}}^{\ast }(-k)$.

Discrete → Continuous

The Fourier series has discrete modes because periodicity imposes a cutoff: the longest wavelength is $L$, so $k_m=2\pi m/L$. For a non-periodic function, $L\to \infty$, the spacing $\Delta k=2\pi /L\to 0$, and the sum becomes an integral. The factor $1/L$ in ${\hat{f}}_m$ becomes $dk/(2\pi )$.

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Time-Averaging Theorem

Let $A(\mathbf{r},t)=\mathbf{a}(\mathbf{r})e^{-i\omega t}$ and $B(\mathbf{r},t)=\mathbf{b}(\mathbf{r})e^{-i\omega t}$ where $\mathbf{a},\mathbf{b}$ are complex-valued. Then:

$$⟨\mathrm{Re}[A]\mathrm{Re}[B]⟩=\frac{1}{2}\mathrm{Re}[\mathbf{a}{\mathbf{b}}^{\ast }]$$

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Tensors (Brief)

A tensor of rank $n$ is an object whose components transform under rotation $A_{ij}$ by acquiring one factor of $A$ per index:

Rank	Object	Transformation
0	Scalar $f$	$f^{\prime }(\mathbf{r})=f(\mathbf{r})$
1	Vector $V_i$	$V_i^{\prime }=A_{ij}{V}_j$
2	Dyadic $T_{ij}$	$T_{ij}^{\prime }=A_{ik}{A}_{jm}{T}_{km}$

A vector is characterized by preservation of its length under orthogonal transformations: $V_i^{\prime }{V}_i^{\prime }=A_{ij}{V}_j{A}_{ik}{V}_k=V_k{V}_k$ (using $A_{ij}{A}_{ik}={\delta }_{jk}$).

A dyadic is a rank-2 tensor composed of juxtaposed (not multiplied) vectors:

$$\mathsf{T}={\hat{\mathbf{e}}}_i{T}_{ij}{\hat{\mathbf{e}}}_j$$

The unit dyadic has $I_{ij}={\delta }_{ij}$, so $\mathsf{I}={\hat{\mathbf{e}}}_i{\hat{\mathbf{e}}}_i$, and $\mathbf{v}\cdot \mathsf{I}=\mathsf{I}\cdot \mathbf{v}=\mathbf{v}$.

Pseudovectors

The cross product $\mathbf{m}=\mathbf{a}\times \mathbf{b}$ transforms as:

$$m_i^{\prime }=(\det A)A_{iq}{m}_q$$

Under proper rotations ($\det A=+1$), this is identical to a vector. Under reflections or inversions ($\det A=-1$), an extra minus sign appears. Such objects are called pseudovectors (or axial vectors).

	Polar × Polar	Axial × Polar	Axial × Axial
Result	Axial	Polar	Axial

Example: $\mathbf{B}$ is a Pseudovector

$\mathbf{r}$ is polar → $\mathbf{v}=d\mathbf{r}/dt$ is polar → $\mathbf{j}=\rho \mathbf{v}$ is polar. The gradient $\nabla$ transforms like $\mathbf{r}$, so it’s polar. From $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$: the curl of $\mathbf{B}$ is polar, so $\mathbf{B}$ must be axial (polar × axial = polar).

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The Helmholtz Theorem

Statement

An arbitrary vector field $\mathbf{C}(\mathbf{r})$ that vanishes faster than $1/r$ as $r\to \infty$ can be uniquely decomposed as:

$$\mathbf{C}=\underset{\nabla \cdot ()=0}{\underbrace{\nabla \times \mathbf{F}}}-\underset{\nabla \times ()=0}{\underbrace{\nabla \Phi }}$$

with

$$\mathbf{F}(\mathbf{r})=\frac{1}{4\pi }\int d^3{r}^{\prime }\frac{{\nabla }^{\prime }\times \mathbf{C}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|},\Phi (\mathbf{r})=\frac{1}{4\pi }\int d^3{r}^{\prime }\frac{{\nabla }^{\prime }\cdot \mathbf{C}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

This holds for both static and time-dependent fields.

Physical Intuition

• $\nabla \cdot$ extracts the longitudinal (irrotational) part: $\nabla \cdot \mathbf{C}$ determines $\Phi$, and $-\nabla \Phi$ reconstructs the curl-free piece.
• $\nabla \times$ extracts the transverse (solenoidal) part: $\nabla \times \mathbf{C}$ determines $\mathbf{F}$, and $\nabla \times \mathbf{F}$ reconstructs the divergence-free piece.

In other words, sources and vortices determine the field.

For a scalar field, knowing $\nabla \phi$ determines $\phi$ (up to a constant).

Proof: Existence

Start from $\mathbf{C}(\mathbf{r})=\int d^3{r}^{\prime }\mathbf{C}({\mathbf{r}}^{\prime })\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. Use $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })=-\frac{1}{4\pi }{\nabla }^2\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$:

$$\mathbf{C}(\mathbf{r})=-\frac{1}{4\pi }\int d^3{r}^{\prime }\mathbf{C}({\mathbf{r}}^{\prime }){\nabla }^2\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Apply the double-curl identity ${\nabla }^2=\nabla (\nabla \cdot )-\nabla \times (\nabla \times )$ to split this into two terms and plop $\mathbf{C}({\vec{r}}^{\prime })$ since it’s constant with respect to $\mathbf{r}$.

Use $\nabla f(|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-{\nabla }^{\prime }f(|\mathbf{r}-{\mathbf{r}}^{\prime }|)$ to turn derivatives into those over ${\mathbf{r}}^{\prime }$:

$${\nabla }^{\prime }\cdot \left[\frac{\mathbf{C}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}\right]=\frac{{\nabla }^{\prime }\cdot \mathbf{C}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}-\mathbf{C}({\mathbf{r}}^{\prime })\cdot {\nabla }^{\prime }\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

(and similarly for curl). This generates four terms. The two “total derivative” terms convert to surface integrals at infinity via the divergence theorem:

$$\oint d\mathbf{S}\cdot \frac{\mathbf{C}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}\text{and}\oint d\mathbf{S}\times \frac{\mathbf{C}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Both vanish if $\mathbf{C}$ decays faster than $1/r$. The remaining two terms give $\nabla \times \mathbf{F}-\nabla \Phi$.

Proof: Uniqueness

Suppose $\nabla \cdot {\mathbf{C}}_1=\nabla \cdot {\mathbf{C}}_2$ and $\nabla \times {\mathbf{C}}_1=\nabla \times {\mathbf{C}}_2$. Let $\mathbf{W}={\mathbf{C}}_1-{\mathbf{C}}_2$. Then $\nabla \cdot \mathbf{W}=0$ and $\nabla \times \mathbf{W}=0$, so the double-curl identity gives ${\nabla }^2\mathbf{W}=0$.

Apply Green’s first identity with $\phi =\psi =W$ (any Cartesian component):

$${\int }_V{d}^{3}r|\nabla W{|}^2={\oint }_{S}d\mathbf{S}\cdot W\nabla W$$

The surface integral vanishes as $S\to \infty$ (since $\mathbf{C}$ decays). Therefore $\nabla W=0$ everywhere, and since $W\to 0$ at infinity, $W=0$, i.e., ${\mathbf{C}}_1={\mathbf{C}}_2$.

Helmholtz Variants

Variant I: Scalar Field in a Finite Volume

For a scalar field $\phi (\mathbf{r})$ in a volume $V$ bounded by $S$:

$$\phi (\mathbf{r})=-\nabla \cdot \frac{1}{4\pi }{\int }_V{\mathrm{d}}^3{r}^{\prime }\frac{{\nabla }^{\prime }\phi ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}+\nabla \cdot \frac{1}{4\pi }{\oint }_S\mathrm{d}{S}^{\prime }\frac{\phi ({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Proof trick: Pull one ${\nabla }^{\prime }$ outside the integrand in the Helmholtz starting expression, then integrate by parts — the volume piece gives ${\nabla }^{\prime }\phi$ and the surface piece gives $\phi {\nabla }^{\prime }(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)$ on $S$, which is exactly the second term.

Source: D.A. Woodside, Journal of Mathematical Physics 40, 4911 (1999).

Variant II: Divergence-Free and Curl-Free in a Finite Volume

If $\nabla \cdot \mathbf{Z}=0$ and $\nabla \times \mathbf{Z}=0$ everywhere in a simply connected volume $V$ bounded by $S$, then $\mathbf{Z}(\mathbf{r})$ can be found everywhere in $V$ from its values on $S$ alone:

$$\mathbf{Z}(\mathbf{r})=\frac{1}{4\pi }\nabla {\oint }_S\mathrm{d}{S}^{\prime }\frac{\hat{\mathbf{n}}({\mathbf{r}}^{\prime })\cdot \mathbf{Z}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}-\frac{1}{4\pi }\nabla \times {\oint }_S\mathrm{d}{S}^{\prime }\frac{\hat{\mathbf{n}}({\mathbf{r}}^{\prime })\times \mathbf{Z}({\mathbf{r}}^{\prime })}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}$$

Proof trick: Start from the finite-volume Helmholtz decomposition (which has both volume and surface integrals). The two volume integrals involve ${\nabla }^{\prime }\cdot \mathbf{Z}$ and ${\nabla }^{\prime }\times \mathbf{Z}$, which both vanish by assumption — only the surface terms survive.

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Lagrange Multipliers

To extremize $f(x,y)$ subject to the constraint $g(x,y)=\text{const.}$, extremize the unconstrained function:

$$F(x,y,\lambda )=f(x,y)-\lambda g(x,y)$$

How This Actually Works

Treat $\lambda$ itself as a variable. Then $\partial F/\partial x=0$ and $\partial F/\partial y=0$ give $\nabla f=\lambda \nabla g$, while $\partial F/\partial \lambda =0$ recovers the constraint $g=0$ (if the constraint is $g=c\ne 0$, just redefine $g\to g-c$). You’ve traded a constrained problem in 2 variables for an unconstrained problem in 3, where the extra variable’s equation of motion is the constraint.

Geometrically: At a constrained extremum, $\nabla f$ must be parallel to $\nabla g$ — otherwise there’s a component of $\nabla f$ along the constraint surface, meaning you could slide along $g=\text{const.}$ and still increase $f$. The multiplier $\lambda$ is exactly the proportionality constant.

When $\lambda$ Becomes a Function

In many physics problems, the constraint must hold at every point in space — e.g., requiring a charge distribution to produce a given potential everywhere on a surface. Then $\lambda$ is promoted to a function $\lambda (\mathbf{r})$, and the “unconstrained” functional becomes:

$$F[\rho ]=U[\rho ]-\int d^{3}r\lambda (\mathbf{r})g[\rho (\mathbf{r})]$$

The variation $\delta F/\delta \rho =0$ now gives a local equation at each point. The simplest example: finding the charge distribution on a conductor that minimizes electrostatic energy subject to $\phi =V$ on the surface — this is Thomson’s theorem (Ch. 7).

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Problems

Foundations

1.7 — A Representation of the Delta Function

Show that $D(x)={\lim }_{m\to \infty }\frac{\sin mx}{\pi x}$ is a representation of $\delta (x)$ by proving ${\int }_{-\infty }^{\infty }dxf(x)D(x)=f(0)$.

Solution Sketch

Change variables $y=mx$, so $f(x)\to f(y/m)$ and $\sin (mx)/(\pi x)\to m\sin (y)/(\pi y)$. As $m\to \infty$, $f(y/m)\to f(0)$, leaving $f(0)\cdot \frac{1}{\pi }{\int }_{-\infty }^{\infty }dy\frac{\sin y}{y}$.

Evaluating the Dirichlet integral Trick: Write $1/y={\int }_0^{\infty }d\nu e^{-\nu y}$ (for $y>0$), exchange integration order, compute the Laplace transform of $\sin y$:

$${\int }_{-\infty }^{\infty }\frac{\sin y}{y}dy=2{\int }_0^{\infty }\frac{d\nu }{1+{\nu }^2}=2\cdot \frac{\pi }{2}=\pi$$

1.12 — Unit Vector Practice

Express the derivatives of spherical unit vectors $\partial \hat{\mathbf{r}}/\partial \theta$, $\partial \hat{\mathbf{r}}/\partial \varphi$, $\partial \hat{\mathbf{\theta }}/\partial \theta$, $\partial \hat{\mathbf{\theta }}/\partial \varphi$, $\partial \hat{\mathbf{\varphi }}/\partial \theta$, $\partial \hat{\mathbf{\varphi }}/\partial \varphi$ in terms of $\hat{\mathbf{r}},\hat{\mathbf{\theta }},\hat{\mathbf{\varphi }}$.

Solution Sketch

Direct calculation using $\hat{\mathbf{r}}=\hat{\mathbf{x}}\sin \theta \cos \varphi +\hat{\mathbf{y}}\sin \theta \sin \varphi +\hat{\mathbf{z}}\cos \theta$ and similarly for $\hat{\mathbf{\theta }},\hat{\mathbf{\varphi }}$. Differentiate component by component and re-express in the spherical basis:

	$\partial /\partial \theta$	$\partial /\partial \varphi$
$\hat{\mathbf{r}}$	$\hat{\mathbf{\theta }}$	$\sin \theta \hat{\mathbf{\varphi }}$
$\hat{\mathbf{\theta }}$	$-\hat{\mathbf{r}}$	$\cos \theta \hat{\mathbf{\varphi }}$
$\hat{\mathbf{\varphi }}$	$0$	$-\sin \theta \hat{\mathbf{r}}-\cos \theta \hat{\mathbf{\theta }}$

Intuition for a few entries:

• $\partial \hat{\mathbf{r}}/\partial \theta =\hat{\mathbf{\theta }}$: increasing $\theta$ tilts $\hat{\mathbf{r}}$ toward the equator — that’s the $\hat{\mathbf{\theta }}$ direction.
• $\partial \hat{\mathbf{\varphi }}/\partial \theta =0$: $\hat{\mathbf{\varphi }}=-\hat{\mathbf{x}}\sin \varphi +\hat{\mathbf{y}}\cos \varphi$ has no $\theta$-dependence — it lives in the $xy$-plane regardless of polar angle.
• $\partial \hat{\mathbf{\varphi }}/\partial \varphi =-\sin \theta \hat{\mathbf{r}}-\cos \theta \hat{\mathbf{\theta }}$: rotating in $\varphi$ swings $\hat{\mathbf{\varphi }}$ inward, with components toward both the radial direction and the pole.

1.13 — Compute the Normal Vector

Given a surface $\Phi (x,y,z)=x^2/a^2+y^2/b^2+z^2/c^2=1$ (an ellipsoid), compute the outward unit normal $\hat{\mathbf{n}}$.

Solution Sketch

$$\hat{\mathbf{n}}=\frac{\nabla \Phi }{|\nabla \Phi |}=\frac{(x/a^2)\hat{\mathbf{x}}+(y/b^2)\hat{\mathbf{y}}+(z/c^2)\hat{\mathbf{z}}}{\sqrt{x^2/a^4+y^2/b^4+z^2/c^4}}$$

Check: When $a=b=c$, this reduces to $\hat{\mathbf{n}}=\frac{x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}}{\sqrt{x^2+y^2+z^2}}=\frac{\mathbf{r}}{r}=\hat{\mathbf{r}}$, as expected for a sphere.

Delta Function and Fourier

1.16 — Densities of States

Prove the delta function identity $\delta [g(x)]={\sum }_n\frac{1}{|g^{\prime }(x_n)|}\delta (x-x_n)$ and use it to compute densities of states.

Solution Sketch

Near each zero $x_n$ where $g(x_n)=0$, Taylor expand: $g(x)\approx g^{\prime }(x_n)(x-x_n)$. Then locally $\delta [g(x)]=\delta [g^{\prime }(x_n)(x-x_n)]=\frac{1}{|g^{\prime }(x_n)|}\delta (x-x_n)$, using $\delta (ax)=\delta (x)/|a|$. Summing over all zeros gives the identity. In physics, $g(x)$ is typically an energy-momentum relation $E-E(\mathbf{k})$, and the sum over zeros gives the density of states.

Vector Calculus Identities

1.3 — Divergence and Curl Identities

Prove: (a) $(\mathbf{A}\times \mathbf{B})\cdot (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot \mathbf{C})(\mathbf{B}\cdot \mathbf{D})-(\mathbf{A}\cdot \mathbf{D})(\mathbf{B}\cdot \mathbf{C})$ (b) $\nabla \cdot (\mathbf{f}\times \mathbf{g})=\mathbf{g}\cdot (\nabla \times \mathbf{f})-\mathbf{f}\cdot (\nabla \times \mathbf{g})$ (c) $(\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot \mathbf{C}\times \mathbf{D})\mathbf{B}-(\mathbf{B}\cdot \mathbf{C}\times \mathbf{D})\mathbf{A}$

Solution Sketch

All follow from the master identity ${ϵ}_{ijk}{ϵ}_{ist}={\delta }_{js}{\delta }_{kt}-{\delta }_{jt}{\delta }_{ks}$.

(a): $(\mathbf{A}\times \mathbf{B}{)}_i(\mathbf{C}\times \mathbf{D}{)}_i={ϵ}_{ijk}{A}_j{B}_k{ϵ}_{ist}{C}_s{D}_t=({\delta }_{js}{\delta }_{kt}-{\delta }_{jt}{\delta }_{ks})A_j{B}_k{C}_s{D}_t=A_s{C}_s{B}_t{D}_t-A_t{D}_t{B}_s{C}_s$.

(b): ${\partial }_i({ϵ}_{ijk}{f}_j{g}_k)={ϵ}_{ijk}({\partial }_i{f}_j)g_k+{ϵ}_{ijk}{f}_j({\partial }_i{g}_k)=g_k(\nabla \times \mathbf{f}{)}_k-f_j(\nabla \times \mathbf{g}{)}_j$.

(c): Let $\mathbf{E}=\mathbf{C}\times \mathbf{D}$. Then $(\mathbf{A}\times \mathbf{B})\times \mathbf{E}=\mathbf{B}(\mathbf{A}\cdot \mathbf{E})-\mathbf{A}(\mathbf{B}\cdot \mathbf{E})$ by the BAC-CAB rule.

1.4 — Levi-Civita Proofs

(a) Show ${ϵ}_{ijk}={\hat{\mathbf{e}}}_i\cdot ({\hat{\mathbf{e}}}_j\times {\hat{\mathbf{e}}}_k)$. (b) Prove $\mathbf{a}\times \mathbf{b}={ϵ}_{ijk}{\hat{\mathbf{e}}}_i{a}_j{b}_k$.

Solution Sketch

(a): For $ijk=xyz$: $\hat{\mathbf{x}}\cdot (\hat{\mathbf{y}}\times \hat{\mathbf{z}})=\hat{\mathbf{x}}\cdot \hat{\mathbf{x}}=1={ϵ}_{xyz}$. Any even permutation gives $+1$, odd gives $-1$, repeated index gives $0$ (since ${\hat{\mathbf{e}}}_j\times {\hat{\mathbf{e}}}_j=0$).

(b): $(\mathbf{a}\times \mathbf{b}{)}_i={ϵ}_{ijk}{a}_j{b}_k$ by definition of the cross product in index notation. Attach ${\hat{\mathbf{e}}}_i$ to get the vector.

1.11 — Integral Identities

Assume $\phi (\mathbf{r})$ and $|\mathbf{G}(\mathbf{r})|$ vanish faster than $1/r$ as $r\to \infty$. (a) If $\mathbf{F}=\nabla \phi$ and $\nabla \cdot \mathbf{G}=0$, show $\int d^{3}r\mathbf{F}\cdot \mathbf{G}=0$. (b) If $\mathbf{F}=\nabla \phi$ and $\nabla \times \mathbf{G}=0$, show $\int d^{3}r\mathbf{F}\times \mathbf{G}=0$. (c) Prove ${\int }_V{d}^{3}r\mathbf{P}=-{\int }_V{d}^{3}r\mathbf{r}(\nabla \cdot \mathbf{P})+{\oint }_{S}dS(\hat{\mathbf{n}}\cdot \mathbf{P})\mathbf{r}$.

Solution Sketch

(a): Integrate by parts: $\int d^{3}r\nabla \phi \cdot \mathbf{G}=\oint d\mathbf{S}\cdot \phi \mathbf{G}-\int d^{3}r\phi \nabla \cdot \mathbf{G}$. Surface term vanishes (decay); volume term vanishes ($\nabla \cdot \mathbf{G}=0$).

(b): Use $\nabla \times (\phi \mathbf{G})=\nabla \phi \times \mathbf{G}+\phi \nabla \times \mathbf{G}$. Since $\nabla \times \mathbf{G}=0$: $\int d^{3}r\nabla \phi \times \mathbf{G}=\int d^{3}r\nabla \times (\phi \mathbf{G})=\oint d\mathbf{S}\times \phi \mathbf{G}=0$.

(c): Consider ${\partial }_j(P_j{r}_i)$ and apply the divergence theorem. The product rule gives $P_i+r_i{\partial }_j{P}_j$, so $P_i={\partial }_j(P_j{r}_i)-r_i\nabla \cdot \mathbf{P}$. Integrate over $V$ and convert the first term to a surface integral.

Physical Significance of (c)

The integral of a polarization $\mathbf{P}$ over a volume can be rewritten in terms of $\nabla \cdot \mathbf{P}$ (bound charge) and the surface term (bound surface charge), weighted by position $\mathbf{r}$.