Maxwell's equations compare fields with their spatial and temporal rates of change. A field assigns a number or an arrow to each point. This chapter builds the algebra of arrows and then makes “rate of change in space” precise. We assume single-variable real analysis and calculus, Euclidean distance and basic topology, and ordinary real and complex arithmetic. No linear algebra or multivariable calculus is assumed.
Positions are triples \(x=(x_1,x_2,x_3)\in\mathbb R^3\). An open set \(U\) contains a small ball about each of its points; all derivatives below are taken on open sets. A scalar field is a function \(f:U\to\mathbb R\), and a vector field is a function \(F:U\to\mathbb R^3\). A time-dependent field is a function on an open subset of \(\mathbb R\times\mathbb R^3\); time derivatives hold position fixed.
For completeness, the functions \(\sin t\) and \(\cos t\) used in examples can be defined by \(e^{it}=\cos t+i\sin t\). Their series construction, derivatives, and identities are proved in the exponential and trigonometry prerequisite; only that lemma is needed here.
Compactness arguments used later require the elementary Euclidean compactness theorem. A proof from nested boxes and subsequences, including boundedness and extrema for continuous functions, is given in this prerequisite lemma. It applies equally to real coordinates; no quantum mechanics is required.
2.1. Arrows, lengths, and orientation
Definition 2.1. Vector operations
Add triples and multiply them by real numbers coordinate by coordinate. Write \(e_1=(1,0,0),e_2=(0,1,0),e_3=(0,0,1)\). Every triple is uniquely \(\sum_i x_i e_i\). A linear map \(A:\mathbb R^n\to\mathbb R^m\) obeys \(A(au+bv)=aAu+bAv\); its matrix entries are the coordinates of \(Ae_j\), so \((Au)_i=\sum_j A_{ij}u_j\). Matrix multiplication represents composition by substitution in this formula.
\[u\cdot v=\sum_{i=1}^3u_iv_i,\qquad |u|=\sqrt{u\cdot u}.\]The dot product is symmetric and linear in each argument. Vectors are orthogonal when their dot product is zero. A vector is a unit vector when its length is one. An orthonormal basis is three pairwise orthogonal unit vectors: if these are \(a_i\), then \(v=\sum_i(v\cdot a_i)a_i\). The proof immediately below establishes this expansion without assuming a dimension theorem.
Alphabeta: Euclidean inner product, Cauchy–Schwarz and norm.
Length inequalities. For \(v\ne0\), nonnegativity of \(|u-tv|^2\) at \(t=(u\cdot v)/|v|^2\) gives \(|u\cdot v|\le |u||v|\), the Cauchy–Schwarz inequality. It also holds for \(v=0\). Consequently \(|u+v|^2\le(|u|+|v|)^2\), proving the triangle inequality. Equality in Cauchy–Schwarz for nonzero vectors holds precisely when they are scalar multiples, because the displayed square then vanishes. For any orthonormal family \(a_1,\ldots,a_k\), expansion of a square gives \(0\le|v-\sum_j(v\cdot a_j)a_j|^2=|v|^2-\sum_j(v\cdot a_j)^2\). This is the finite Bessel inequality. Four orthonormal vectors in \(\mathbb R^3\) would be impossible: apply the inequality to each standard \(e_i\) and add, obtaining \(\sum_{j=1}^4|a_j|^2\le\sum_{i=1}^3|e_i|^2\), or \(4\le3\). For three orthonormal vectors, the residual \(v-\sum_j(v\cdot a_j)a_j\) is orthogonal to all three; if nonzero, normalizing it would provide an impossible fourth orthonormal vector. Hence it is zero, proving the basis formula.
Example. \((1,2,0)\cdot(2,-1,3)=0\). A force \(F\) and a displacement \(v\) have scalar product \(F\cdot v\), which measures the force component along the displacement times its length. A large force perpendicular to the displacement has zero dot product. The dot product is not coordinatewise multiplication: its result is one number.
Definition 2.2. Cross product and signed volume
Our standard basis has positive orientation, specified by \(e_1\times e_2=e_3\). Define the determinant of three columns by \(\det[u,v,w]=(u\times v)\cdot w\). It is the signed volume of the parallelepiped spanned by those columns; the unsigned volume is its absolute value. An orthonormal basis \((a,b,n)\) is positively oriented if \(n=a\times b\).
Alphabeta: Cross product, Determinants.
Proposition 2.3. Cross-product identities
The scalar triple product is unchanged by a cyclic permutation and changes sign under exchange of two entries. The cross product is bilinear but is not associative.
Alphabeta: Cross-product algebra, Cross-product length.
Proof and the geometric meaning of area
Antisymmetry follows by exchanging the factors in each component. Dotting with \(u\) or \(v\) gives six terms that cancel in pairs. Expanding the length square gives \(\sum_{i<j}(u_iv_j-u_jv_i)^2\). Expansion of \((\sum u_i^2)(\sum v_i^2)-(\sum u_iv_i)^2\) gives exactly that sum. For the triple identity, its first component is
\[u_2(v_1w_2-v_2w_1)-u_3(v_3w_1-v_1w_3) =v_1(u\cdot w)-w_1(u\cdot v).\]The second and third follow by cyclic index permutation. Expanding the six terms of \((u\times v)\cdot w\) proves the scalar triple assertions. To interpret area, project \(v\) onto \(u\ne0\): \(v_\perp=v-u(u\cdot v)/|u|^2\). The parallelogram has base \(|u|\) and height \(|v_\perp|\), whose product is \(|u\times v|\) by the proved identity. Dotting its normal with \(w\) multiplies this area by signed perpendicular height, proving the volume interpretation.
Example and counterexample. \(e_1\times e_2=e_3\), so \((e_1\times e_2)\times e_2=-e_1\), whereas \(e_1\times(e_2\times e_2)=0\). Parentheses matter. Parallel vectors have cross product zero even when both have large lengths.
2.2. Differentiating several variables
Definition 2.4. Partial and total derivatives
The partial derivative \(\partial_j f(p)\) is the ordinary derivative at zero of \(t\mapsto f(p+te_j)\). The total derivative of \(f:U\subset\mathbb R^n\to\mathbb R^m\) at \(p\) is a linear map \(Df(p)\) such that
\[f(p+h)=f(p)+Df(p)h+r(h),\qquad \forall\epsilon>0\ \exists\delta>0:\quad 0<|h|<\delta\Longrightarrow |r(h)|\le\epsilon |h|.\]This is also written \(r(h)=o(|h|)\). The estimate must hold for all directions at once. The derivative is unique: subtract two candidate expansions, take \(h=tv\), divide by \(|t|\), and let \(t\to0\). Evaluating along \(e_j\) shows its matrix is \((\partial_j f_i)_{ij}\), called the Jacobian matrix. The notation \(C^k\) means all coordinate partial derivatives through order \(k\) exist and are continuous; \(C^0\) means continuous, and \(C^\infty\) means \(C^k\) for every finite \(k\).
Alphabeta: Partial derivatives, Total derivative, Cᵏ maps.
Counterexample. Set \(f(x,y)=xy/(x^2+y^2)\) away from zero and \(f(0,0)=0\). Both partial derivatives at zero vanish because both axis restrictions vanish. Along \(x=y\ne0\), its value is \(1/2\), so it is not even continuous at zero, hence not totally differentiable. Existence of the coordinate slopes does not guarantee a common local linear approximation.
Theorem 2.5. Continuous partial derivatives suffice
If every first partial derivative is continuous in a neighborhood of \(p\), the function is totally differentiable at \(p\), with the Jacobian just described.
Alphabeta: Continuous partials imply differentiability.
Proof with a uniform remainder bound
First take a scalar function. Change coordinates one at a time from \(p\) to \(p+h\) inside a small box contained in \(U\). The one-dimensional fundamental theorem of calculus gives
\[f(p+h)-f(p)=\sum_{j=1}^n\int_0^{h_j}\partial_jf(p+h_1e_1+\cdots+h_{j-1}e_{j-1}+te_j)\,dt.\]Given \(\epsilon>0\), continuity supplies \(\delta>0\) for which each derivative in that ball differs from \(\partial_j f(p)\) by at most \(\epsilon/\sqrt n\). Every intermediate point has distance at most \(|h|\) from \(p\). Subtracting \(\sum_j\partial_j f(p)h_j\) bounds the remainder by \((\epsilon/\sqrt n)\sum_j|h_j|\le\epsilon|h|\). For \(m\) components use tolerance \(\epsilon/\sqrt{mn}\) for each partial; each component remainder is at most \(\epsilon|h|/\sqrt m\), so their combined Euclidean length is at most \(\epsilon|h|\).
Example. For \(f(x,y,z)=x^2y+\sin z\), \(Df(p)h=2p_1p_2h_1+p_1^2h_2+\cos(p_3)h_3\). Total differentiability does not require continuous derivatives nearby: \(g(x)=x^2\sin(1/x^2)\), extended by zero, has derivative zero at zero but an unbounded derivative nearby: for \(x\ne0\), \(g'(x)=2x\sin(1/x^2)-2\cos(1/x^2)/x\). Along \(x_n=(2\pi n)^{-1/2}\), this equals \(-2/x_n\), which tends to negative infinity. Thus \(C^1\) is sufficient, stronger than differentiability.
Theorem 2.6. Chain rule
If \(f\) is differentiable at \(p\) and \(g\) is differentiable at \(f(p)\), then
\[D(g\circ f)(p)=Dg(f(p))Df(p).\]In particular, for a differentiable path \(\gamma(t)\) and scalar field \(f\), \((f\circ\gamma)'(t)=\sum_j\partial_j f(\gamma(t))\gamma'_j(t)\). The directional derivative in direction \(v\) is \(Df(p)v\), the derivative of \(s\mapsto f(p+sv)\) at zero.
Alphabeta: Total chain rule, Directional derivatives, Derivative algebra.
Every matrix has a bound \(|Av|\le M|v|\): Cauchy–Schwarz on each row permits \(M=(\sum_{ij}A_{ij}^2)^{1/2}\). Write \(f(p+h)-f(p)=Ah+r(h)=k\). For small \(h\), \(|k|\le(M+1)|h|\), so \(k\to0\). The second expansion gives \(g(f(p)+k)-g(f(p))=Bk+s(k)\), with \(|s(k)|/|k|\to0\). Its remainder after \(BAh\) is \(Br(h)+s(k)\), bounded by \(M_B|r(h)|+o(|k|)=o(|h|)\). If \(k=0\), its remainder term is zero, so division by \(|k|\) is unnecessary. This proves the rule. Applying the scalar product rule to each component also proves \(\partial_j(fg)=f\partial_jg+g\partial_jf\) and the product rule for dot and cross products.
Example. If \(f(x,y)=x^2+y^2\) and \(\gamma(t)=(\cos t,\sin t)\), the chain rule gives \(2\cos t(-\sin t)+2\sin t\cos t=0\). Although both coordinates move, this scalar is constant on the circle.
Theorem 2.7. Equality of mixed partial derivatives
For a \(C^2\) function, \(\partial_i\partial_j f=\partial_j\partial_i f\). The order matters in the notation until this hypothesis is established.
Alphabeta: Clairaut–Schwarz theorem.
Proof by a rectangular increment
Fix all other coordinates and put
\[R(h,k)=f(p+he_i+ke_j)-f(p+he_i)-f(p+ke_j)+f(p).\]For positive small \(h,k\), two applications of the one-dimensional fundamental theorem express this either as \(\int_0^h\int_0^k\partial_j\partial_i f(p+se_i+te_j)\,dt\,ds\), or as \(\int_0^k\int_0^h\partial_i\partial_j f(p+se_i+te_j)\,ds\,dt\). No interchange theorem is needed: these are separately obtained expressions for the same increment. Each expression divided by \(hk\) differs from its derivative's value at \(p\) by at most that derivative's maximum deviation on the shrinking rectangle. Continuity makes both deviations tend to zero, proving equality.
Example. For \(f(x,y)=x^2y^3\), both \(\partial_x\partial_yf\) and \(\partial_y\partial_xf\) equal \(6xy^2\).
Counterexample without the hypothesis. Extend \(f(x,y)=xy(x^2-y^2)/(x^2+y^2)\) by zero at the origin. Along \(y=0\), \(\partial_y f(x,0)=x\); along \(x=0\), \(\partial_x f(0,y)=-y\). Therefore \(\partial_x\partial_y f(0,0)=1\) and \(\partial_y\partial_x f(0,0)=-1\). Both exist at that point, but the continuous-second-partial hypothesis fails.
2.3. The four differential operators
Definition 2.8. Gradient, divergence, curl, and Laplacian
The gradient and curl are vector fields; divergence and the scalar Laplacian are scalar fields. For a vector field, \(\Delta F\) means applying the scalar Laplacian to each component. The symbol \(\nabla\) is a convenient collection of derivatives, not an ordinary numerical vector: derivatives act on everything specified after them. A field's full derivative is its nine-entry Jacobian; divergence and curl retain only certain combinations.
Alphabeta: Jacobian and gradient, Divergence and curl, Laplacian.
Worked example for Definition 2.8
Counterexample: Divergence and curl alone do not determine a field
The directional derivative is \(\nabla f(p)\cdot v\). For unit \(v\), Cauchy–Schwarz bounds it by \(|\nabla f(p)|\), attained by \(v=\nabla f(p)/|\nabla f(p)|\) when the gradient is nonzero. This justifies “steepest increase.” A tangent direction to a level surface \(f=c\) is orthogonal to the gradient because the chain rule gives derivative zero along any path in that surface.
Examples. For \(f=|x|^2\), \(\nabla f=2x\) and \(\Delta f=6\). For \(F=(x,y,z)\), divergence is \(3\) and curl is zero. For \(F=(-y,x,0)\), divergence is zero and curl is \((0,0,2)\). For \(F=(x,-y,0)\), both divergence and curl vanish, although the derivative sends \((h_1,h_2,h_3)\) to \((h_1,-h_2,0)\). Replacing this field by \(10^9F\) makes its variation arbitrarily fast while curl stays zero. Curl measures an antisymmetric part of spatial change, not its total size.
Proposition 2.9. Differential identities
For \(C^2\) fields,
\[\nabla\times\nabla f=0,\qquad \nabla\cdot(\nabla\times F)=0,\qquad \nabla\times(\nabla\times F)=\nabla(\nabla\cdot F)-\Delta F.\]For \(C^1\) fields,
\[\nabla\cdot(fF)=\nabla f\cdot F+f\nabla\cdot F,\qquad \nabla\cdot(F\times G)=G\cdot(\nabla\times F)-F\cdot(\nabla\times G).\]Alphabeta: Curl of a gradient, Divergence of a curl, Cross-product differential identities.
Worked example for Proposition 2.9
Counterexample: Divergence and curl alone do not determine a field
The first identity is the three mixed-partial differences. The second is six second derivatives that cancel in pairs by Theorem 2.7. The first component of the third is
\[\partial_2(\partial_1F_2-\partial_2F_1)-\partial_3(\partial_3F_1-\partial_1F_3) =\partial_1(\partial_1F_1+\partial_2F_2+\partial_3F_3)-\Delta F_1.\]Cyclic index permutation proves the other two components. The scalar product rule proves the fourth formula by summing \(\partial_j(fF_j)\). Expanding \(\partial_1(F_2G_3-F_3G_2)\) and its cyclic companions and grouping terms differentiated on \(F\) versus on \(G\) proves the last formula. These derivations explain why the tempting ordinary-vector identity with \(\nabla\) happens to work in the curl-curl case.
Counterexample to a common picture. For \(F=(-y,x,1)\), \(\nabla\times F=(0,0,2)\), and \(F\cdot(\nabla\times F)=2\). Curl need not be perpendicular to the field. For \(F=(0,x^2,0)\), curl is \((0,0,2x)\) and curl of curl is \((0,-2,0)\), which agrees with \(\nabla(\nabla\cdot F)-\Delta F\).
2.4. What depends on coordinates?
Proposition 2.10. Rigid coordinate changes
A matrix \(Q\) is orthogonal when \(Q^{\mathsf T}Q=I\), where transpose exchanges rows and columns and \(I\) is the identity. Its columns are an orthonormal basis. In coordinates \(x=b+Qy\), write \(\widetilde f(y)=f(b+Qy)\) and \(\widetilde F(y)=Q^{\mathsf T}F(b+Qy)\). Then
\[\nabla_y\widetilde f=Q^{\mathsf T}\nabla_xf,\quad \nabla_y\cdot\widetilde F=\nabla_x\cdot F,\quad \Delta_y\widetilde f=\Delta_x f,\] \[\nabla_y\times\widetilde F=(\det Q)Q^{\mathsf T}(\nabla_x\times F).\]All fields on the right are evaluated at \(b+Qy\). A proper rotation has \(\det Q=1\); an orientation-reversing orthogonal map has \(\det Q=-1\), as does a reflection. Curl changes as a vector under proper rotations and acquires an extra orientation sign under reflection.
Alphabeta: Orthogonal operators.
Proof without assuming coordinate invariance
Orthogonality gives \(|Qv|=|v|\), \(Q^{-1}=Q^{\mathsf T}\), and \(\sum_jQ_{ij}Q_{kj}=\delta_{ik}\), where \(\delta_{ik}\) is one when \(i=k\) and zero otherwise. Chain rule gives \(\partial_{y_j}=\sum_iQ_{ij}\partial_{x_i}\), proving the gradient formula and, after a second application and the displayed sum, the Laplacian formula. For divergence expand \(\sum_j\partial_{y_j}\sum_kQ_{kj}F_k\) and use that same sum.
For completeness, the determinant's six-term formula is alternating and multilinear. Expanding each column of \(QA\) in the columns of \(Q\), terms with repeated columns vanish; the six remaining terms give \(\det(QA)=\det Q\det A\). The six-term formula also gives \(\det Q^{\mathsf T}=\det Q\). Applying multiplicativity to \(Q^{\mathsf T}Q=I\) gives \((\det Q)^2=1\). The triple-product definition now implies \((Qu)\times(Qv)=(\det Q)Q(u\times v)\): dot both sides with arbitrary \(Qw\) and use multiplicativity and preservation of dot products. Expanding the curl formula with \(\partial_{y_j}=\sum_iQ_{ij}\partial_{x_i}\) uses this same cross-product identity for the coordinate basis vectors and proves its transformation law.
Example. Rotating the axes by 90 degrees about \(z\) leaves the numerical divergence of \((x,y,z)\) equal to three. The components of its arrows change. For a reflection \(x_1=-y_1,x_2=y_2,x_3=y_3\), the swirl \((-x_2,x_1,0)\) becomes \((y_2,-y_1,0)\); its curl changes from \(2e_3\) to \(-2e_3\), exactly the extra sign.
The next chapter proves that divergence is local outward flux per volume and that each curl component is local circulation per area. Those limits supply geometric meaning to these coordinate formulas. They also give the correct formulas when the coordinate basis itself changes from point to point.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Dot products, cross products, and orientation
Applies to: Definition 2.1, Definition 2.2, Proposition 2.3.
Let \(u=(1,1,0),v=(0,1,1)\). Their dot product is 1, their lengths are both \(\sqrt2\), and \(u\times v=(1,-1,1)\). Its dot products with \(u,v\) vanish and its squared length is \(3=2\cdot2-1^2\). With \(w=e_3\), the signed volume \((u\times v)\cdot w=1\), while swapping \(u,v\) gives \(-1\). The cyclic products \((v\times w)\cdot u\) and \((w\times u)\cdot v\) also equal 1. Directly \(u\times(v\times w)=(1,-1,0)=v(u\cdot w)-w(u\cdot v)\). Nonassociativity is visible from \(e_1\times(e_1\times e_2)=-e_2\) but \((e_1\times e_1)\times e_2=0\). Bilinearity follows by distributing each coordinate product, e.g. \((2u+v)\times v=2(1,-1,1)\).
The vectors \((1,1,0)/\sqrt2,(-1,1,0)/\sqrt2,e_3\) are a positive orthonormal basis. Taking dot products with them reconstructs \((x,y,z)\) with coefficients \((x+y)/\sqrt2,(-x+y)/\sqrt2,z\). Thus the coordinate and basis assertions are checked as well as the cross-product identities.
Worked example 2. A derivative checked in every direction
Applies to: Definition 2.4, Theorem 2.5, Theorem 2.6, Theorem 2.7.
For \(f(x,y,z)=x^2y\), at \(p=(1,2,0)\) the Jacobian is \((4,1,0)\). Its remainder at increment \((h,k,\ell)\) is \(2h^2+2hk+h^2k\). If \(r=\sqrt{h^2+k^2+\ell^2}\le1\), its absolute value is at most \(5r^2\), so the error divided by \(r\) tends to zero. All partials are polynomials, hence continuous of every order; the sufficient hypothesis of Theorem 2.5 holds. Along \(\gamma(t)=(1+t,2-t,0)\), the derivative at zero is \((4,1,0)\cdot(1,-1,0)=3\); expanding \((1+t)^2(2-t)\) gives the same result. The mixed partials \(\partial_y\partial_x f=2x=\partial_x\partial_y f\) agree. This verifies total differentiability, regularity, the path chain rule, and the mixed-partial theorem with their hypotheses explicit.
Worked example 3. Checking all the differential identities
Applies to: Definition 2.8, Proposition 2.9.
Let \(f=x^2+y^2+z^2\), \(F=(xy,yz,zx)\), \(G=e_z\). Then \(\nabla f=(2x,2y,2z)\), \(\Delta f=6\), \(\operatorname{div}F=x+y+z\), and \(\operatorname{curl}F=(-y,-z,-x)\). A second curl is \((1,1,1)\), also \(\nabla\operatorname{div}F-\Delta F\), since \(\Delta F=0\). The divergence of its first curl is zero, and the curl of \(\nabla f\) is zero, by direct partial derivatives.
The divergence product rule gives \(\operatorname{div}(fF)=2x^2y+2y^2z+2z^2x+f(x+y+z)\), exactly the result of differentiating its three components. Also \(F\times G=(yz,-xy,0)\), whose divergence is \(-x\); this equals \(G\cdot\operatorname{curl}F-F\cdot\operatorname{curl}G=-x\). Every field here is smooth, so all stated derivative hypotheses hold.
Worked example 4. Curl under a reflection
Applies to: Proposition 2.10.
Take \(Q=\operatorname{diag}(-1,1,1)\), \(b=0\), so \(x_1=-y_1,x_2=y_2,x_3=y_3\) and \(\det Q=-1\). For \(F=(-x_2,x_1,0)\), \(\widetilde F=(y_2,-y_1,0)\). Its divergence is zero and its curl is \((0,0,-2)\), exactly \((\det Q)Q^T(0,0,2)\). Omitting the orientation sign would give the wrong answer. For \(f=x_1^2+2x_2^2+3x_3^2\), the reflected scalar has gradient \((2y_1,4y_2,6y_3)=Q^T\nabla_xf(Qy)\) and Laplacian 12 in either chart. This checks the gradient, divergence, Laplacian, and axial curl rules under a nontrivial rigid change.
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. Divergence and curl alone do not determine a field
Related statements: Definition 2.8, Proposition 2.9.
Claim refuted. Knowing both divergence and curl determines a vector field without boundary or decay information.
Proof.
On all of \(\mathbb R^3\), compare \(F=0\) with \(G=(y,x,0)\). Both are smooth, and \(\operatorname{div}G=\partial_xy+\partial_yx=0\), while \(\operatorname{curl}G=(0,0,\partial_xx-\partial_yy)=0\). Thus their divergence and curl agree everywhere. Yet \(G(1,0,0)=e_y\ne0\), and its Jacobian has nonzero entries \(\partial_yG_1=\partial_xG_2=1\). In fact \(G=\nabla(xy)\) and \(\Delta(xy)=0\). Adding the gradient of this harmonic scalar leaves both derivative summaries unchanged. A uniqueness claim therefore requires additional boundary or behavior-at-infinity hypotheses; the local definitions impose none.
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