Transverse fields, helicity and angular momentum

In the whole-space componentwise convention, the absolute-tail estimate in the Definition makes the integer-radius integrals of each real or imaginary component a real Cauchy sequence. Its convergence follows from [F8], and the same uniform tail estimate gives that limit for all real radii tending to infinity. This argument applies separately to each of the finitely many components and introduces no choice principle. Since $K$ is compact and excludes $0$, it has positive distance from the origin. Thus extending $a$ by zero across a neighborhood of $0$ gives a smooth compactly supported map on all of ${\mathbb{R}}^3$. For any compact time interval $J$, each function $f_t(k):=a(k)\exp (-ic\mathrm{\mid }k\mathrm{\mid }t)$ and each of its $t$-derivatives is smooth and supported in $K$; all its derivatives are uniformly bounded on $K\times J$. Every integral in the Definition is therefore an ordinary Riemann integral of a continuous compactly supported function. Differentiating in $x$ or $t$ under the integral is justified because the corresponding difference quotients converge uniformly on $K$ for each fixed $(x,t)$: the phase derivatives in $(x,t)$ are bounded because $k$ ranges over $K$ and $t$ lies in $J$, while the complex exponential has unit modulus. The one-variable mean-value estimate for the exponential is therefore uniform in $k$. The absolute-integral estimate in [F5] then bounds the difference between the integrals by the volume of a containing cube times the uniform error, which tends to zero. Iterating this argument for the resulting derivatives proves existence and continuity of every mixed space-time derivative.

For multi-indices $\alpha ,\beta$, differentiating in $x$ multiplies the integrand by $(ik{)}^{\beta }$. To control a factor $x_j$, integrate the $k_j$-derivative of the product of the compactly supported factor and $\exp (ik\cdot x)$ over a containing cube. The one-dimensional FTC in [F5] gives zero boundary terms because the factor vanishes near both endpoints; Riemann-Fubini in [F5] integrates that identity over the other two coordinates. Repeating this argument transfers every monomial $x^{\alpha }$ to derivatives in $k$, leaving an integral of ${\mathrm{\partial }}_k^{\alpha }((ik{)}^{\beta }{f}_t(k))$ times a unit-modulus exponential, up to the fixed factor of modulus one produced by integration by parts. Its absolute value is uniformly bounded in $x$ and locally uniformly in $t$. The same argument applies after any $t$-derivative, since it only adds the smooth factor $(-ic\mathrm{\mid }k\mathrm{\mid }{)}^q$ on $K$. By [F1], $A_{+}$, $A$, $E$, and $B$ are therefore Schwartz in $x$ at each $t$, with the claimed local bounds.

Expanding ${\prod }_{j=1}^3(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^2$ gives finitely many absolute monomials, each controlled by a Schwartz seminorm in [F1]. Hence each Schwartz component $h$ is bounded by $C_h{\prod }_j(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^{-2}$. Applying this bound to $E$ and $B$ gives a constant $C\ge 0$, depending on the fixed $t$, such that $0\le u(x,t)\le CG(x)$, where $G(x):={\prod }_{j=1}^3(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^{-4}$; nonnegativity follows from [F4]. For $R>0$, the ordinary one-dimensional Riemann integral and [F5] give ${\int }_{-R}^R(1+\mathrm{\mid }s\mathrm{\mid }{)}^{-4}ds=2{\int }_0^R(1+s{)}^{-4}ds=\frac{2}{3}(1-(1+R{)}^{-3})\le \frac{2}{3}\mathrm{.}$ Riemann-Fubini in [F5] therefore gives ${\int }_{Q_R}G{d}^{3}x={(\frac{2}{3}(1-(1+R{)}^{-3}))}^3\le (2\mathrm{/}3{)}^3$, and so $0\le W_R(t):={\int }_{Q_R}u(x,t)d^{3}x\le C(2\mathrm{/}3{)}^3$ for every $R>0$. If $R^{\mathrm{\prime }}>R$, cutting $Q_{R^{\mathrm{\prime }}}$ at the six coordinate planes $x_j=\pm R$ decomposes it into finitely many rectangles, including $Q_R$; finite-slice additivity and nonnegativity of the remaining cell integrals in [F5] give $W_R(t)\le W_{R^{\mathrm{\prime }}}(t)$. Thus $\{W_R(t):R>0\}$ is nonempty and bounded above. Let $W(t)$ be its supremum, which exists by [F8]. For every $\epsilon >0$, the least-upper-bound property gives some $R_0>0$ with $W(t)-\epsilon <W_{R_0}(t)$; for all $R\ge R_0$, monotonicity gives $W(t)-\epsilon <W_R(t)\le W(t)$. Hence ${\lim }_{R\to \mathrm{\infty }}{W}_R(t)=W(t)<\mathrm{\infty }$, which is the whole-space energy defined above.

Put $\gamma :=(2\pi {)}^{-3\mathrm{/}2}$. Direct differentiation gives $\mathrm{div}A=2\gamma \mathrm{Re}\int i(k\cdot a(k))\exp (i(k\cdot x-c\mathrm{\mid }k\mathrm{\mid }t))d^{3}k=0$. The identity $\phi =0$ and [F2] show that $(\phi ,A)$ represents exactly the displayed $(E,B)$ and is in Coulomb gauge.

Put $q(k,x,t):=\exp (i(k\cdot x-c\mathrm{\mid }k\mathrm{\mid }t))$ and $\gamma :=(2\pi {)}^{-3\mathrm{/}2}$. By [F3] and real-linearity of $\mathrm{Re}$, $E=2\gamma \mathrm{Re}\int ic\mathrm{\mid }k\mathrm{\mid }a(k)q{d}^{3}k$ and $B=2\gamma \mathrm{Re}\int i(k\times a(k))q{d}^{3}k$. Differentiating each integrand using [F3], [F7], and step 1.1 gives $\mathrm{div}E=2\gamma \mathrm{Re}\int -c\mathrm{\mid }k\mathrm{\mid }(k\cdot a(k))q{d}^{3}k=0$, $\mathrm{div}B=2\gamma \mathrm{Re}\int -k\cdot (k\times a(k))q{d}^{3}k=0$, $\mathrm{curl}E+{\mathrm{\partial }}_{t}B=2\gamma \mathrm{Re}\int [-c\mathrm{\mid }k\mathrm{\mid }k\times a(k)+c\mathrm{\mid }k\mathrm{\mid }k\times a(k)]q{d}^{3}k=0$, and $\mathrm{curl}B-c^{-2}{\mathrm{\partial }}_{t}E=2\gamma \mathrm{Re}\int [-k\times (k\times a(k))-\mathrm{\mid }k{\mathrm{\mid }}^{2}a(k)]q{d}^{3}k=0$. The last equality uses the cross-product identity in [F7] and $k\cdot a(k)=0$. These are the source-free equations in [A1], proving that the integrated fields satisfy the adopted smooth Maxwell branch.

The exponent $k\cdot x-c\mathrm{\mid }k\mathrm{\mid }t$ is dimensionless. Since $d^{3}k$ has units ${\mathrm{m}}^{-3}$, the assigned $[a]=\mathrm{T}{\mathrm{m}}^4$ gives $[A]=\mathrm{T}\mathrm{m}$; then $E=-{\mathrm{\partial }}_{t}A$ has units $\mathrm{T}\mathrm{m}{\mathrm{s}}^{-1}=\mathrm{V}{\mathrm{m}}^{-1}$, and $B=\mathrm{curl}A$ has units $\mathrm{T}$. These agree with [F6].

Since $n\cdot n=1$, the formula $P_{T}v=v-n(n\cdot v)$ gives $n\cdot P_{T}v=0$ and $P_{T}v=v$ whenever $v\in T_k$; hence $P_T$ is idempotent with range $T_k$. Also $n\times n=0$, so $Cn=0$. Write $N_{n}v:=n\times v$. Its real matrix is skew-symmetric, so $N_n^T=-N_n$; with the Hermitian product of [F3], $(i{N}_n{)}^{\ast }=(-i)N_n^T=i{N}_n$, hence $C(k)$ is Hermitian. The vector triple product in [F2] gives $C(k{)}^{2}v=-n\times (n\times v)=v-n(n\cdot v)=P_T(k)v$. Also $P_T(k{)}^{\ast }=P_T(k)$ because $n$ is real, $P_T(k{)}^2=P_T(k)$, its range is $T_k$, and $C(k)n=0$.

The cross product with $n$ is perpendicular to $n$, so $P_{T}C=C$; also $C(n(n\cdot v))=0$, so $C{P}_T=C$. Thus $P_{T}C=C{P}_T=C$. Since $C^2=P_T$, direct multiplication gives $P_{\sigma }^2=\frac{1}{4}(P_T+2\sigma C+C^2)=\frac{1}{2}(P_T+\sigma C)=P_{\sigma },P_{+1}{P}_{-1}=\frac{1}{4}(P_T-C^2)=0,P_{+1}+P_{-1}=P_T\mathrm{.}$ Both are Hermitian by step 1.1. To verify their ranks, let $e_1,e_2$ be the first two coordinate unit vectors. They are linearly independent, so at least one, say $e_j$, is not parallel to $n$. Set $u:=(e_j-(e_j\cdot n)n)\mathrm{/}\mathrm{\mid }{e}_j-(e_j\cdot n)n\mathrm{\mid }$; its denominator is positive, and $u$ is a real unit vector perpendicular to $n$. Put $v:=n\times u$. The norm identity in [F2] gives $\mathrm{\mid }v\mathrm{\mid }=1$, and the determinant identity gives $u\cdot v=0$; the triple-product identity gives $n\times v=-u$. Thus $u,v$ span the real transverse plane. For $w_{+}:=(u+iv)\mathrm{/}\sqrt{2}$ and $w_{-}:=(u-iv)\mathrm{/}\sqrt{2}$, expansion gives $⟨w_{+},w_{+}{⟩}_H=⟨w_{-},w_{-}{⟩}_H=\frac{1}{2}(\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }v{\mathrm{\mid }}^2)=1$ and $⟨w_{+},w_{-}{⟩}_H=\frac{1}{2}(\mathrm{\mid }u{\mathrm{\mid }}^2-\mathrm{\mid }v{\mathrm{\mid }}^2-2i(u\cdot v))=0$. The relations $n\times u=v$ and $n\times v=-u$ give $C{w}_{+}=w_{+}$ and $C{w}_{-}=-w_{-}$. Since some coordinate $n_j\mathrm{\ne }0$, the equation $n\cdot z=0$ uniquely determines $z_j$ from the other two complex coordinates, so $T_k$ has complex dimension two. Hence $w_{+},w_{-}$ span $T_k$, and the complementary orthogonal projectors $P_{+1},P_{-1}$ each have rank one there. The finite coordinate choice is only for this fixed $k$; no family of choices is made.

The coefficients of $P_T(k)$ and $C(k)$ are smooth functions of $k\mathrm{\ne }0$ because $n=k\mathrm{/}\mathrm{\mid }k\mathrm{\mid }$ is smooth there. Step 2.1 shows that $P_{\sigma }(k)$ has range in $T_k$, so multiplication by it sends a smooth transverse amplitude supported in a compact set away from zero to another such amplitude. Since $U_t$ multiplies by the scalar $e^{-ic\mathrm{\mid }k\mathrm{\mid }t}$, $P_{\sigma }(k)e^{-ic\mathrm{\mid }k\mathrm{\mid }t}a(k)=e^{-ic\mathrm{\mid }k\mathrm{\mid }t}{P}_{\sigma }(k)a(k)$ pointwise, proving $M_{\sigma }{U}_t=U_t{M}_{\sigma }$.

For $R$ as in [A1], the map $R_{\ast }a$ is smooth and its support is the rotated compact set $R(K)$, which still avoids zero. Moreover $k\cdot Ra(R^{T}k)=(R^{T}k)\cdot a(R^{T}k)=0$ by $R^{T}R=I$, so $R_{\ast }a$ belongs to the amplitude class. Preservation of orientation gives $R(n\times z)=(Rn)\times (Rz)$: for every real $w$, $(Rn\times Rz)\cdot Rw=\det [Rn,Rz,Rw]=\det (R)\det [n,z,w]=(n\times z)\cdot w=(R(n\times z))\cdot Rw$, and nondegeneracy of the Euclidean inner product identifies the two vectors. Orthogonality gives $\mathrm{\mid }Rk\mathrm{\mid }=\mathrm{\mid }k\mathrm{\mid }$. Therefore $P_T(Rk)=R{P}_T(k)R^T$ and $C(Rk)=RC(k)R^T$, hence $P_{\sigma }(Rk)=R{P}_{\sigma }(k)R^T$. Substituting this identity into the definitions of $M_{\sigma }$ and $R_{\ast }$ gives $M_{\sigma }{R}_{\ast }a(k)=P_{\sigma }(k)Ra(R^{T}k)=R{P}_{\sigma }(R^{T}k)a(R^{T}k)=R_{\ast }{M}_{\sigma }a(k)$. This proves the proper-rotation commutation claim; the determinant condition is what preserves, rather than reverses, the helicity label.

For every Schwartz component $f$ and every integer $N$, the finite expansion of ${\prod }_{j=1}^3(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^N$ is controlled term-by-term by the seminorms in [F1], so $\mathrm{\mid }f(x)\mathrm{\mid }\le C_N{\prod }_j(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^{-N}$; the same holds for its derivatives. Choose $N=4$. Each component of each vector integrand defining $J_{\mathrm{k}\mathrm{i}\mathrm{n}}$, $L_{\mathrm{c}\mathrm{a}\mathrm{n}}$, and $S_{\mathrm{c}\mathrm{a}\mathrm{n}}$ is a sum of products of two Schwartz components, with at most one coordinate factor. Therefore, for each such vector integrand $V$, ${\sum }_{j=1}^3\mathrm{\mid }{V}_j(x)\mathrm{\mid }\le C_{V}G(x)$ for some $C_V\ge 0$, where $G(x):={\prod }_{j=1}^3(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^{-7}$. By [F5], for $R>0$ the one-dimensional factor satisfies ${\int }_{-R}^R(1+\mathrm{\mid }s\mathrm{\mid }{)}^{-7}ds=\frac{1}{3}(1-(1+R{)}^{-6})\le \frac{1}{3}$ and tends to $1\mathrm{/}3$. Riemann-Fubini [F4] gives ${\int }_{Q_R}G{d}^{3}x={(\frac{1}{3}(1-(1+R{)}^{-6}))}^3\le 1\mathrm{/}27$. Thus the nonnegative cube integrals $A_R(V):={\int }_{Q_R}{\sum }_{j=1}^3\mathrm{\mid }{V}_j(x)\mathrm{\mid }{d}^{3}x$ are bounded above by $C_V\mathrm{/}27$ and are nondecreasing in $R$ by finite-slice additivity and monotonicity [F6]. The set $\{A_R(V):R>0\}$ is nonempty (take $R=1$) and bounded above, so its finite real supremum $A(V)$ exists by [F7]. For every $\epsilon >0$, choose $R_0$ with $A_{R_0}(V)>A(V)-\epsilon$; then $A(V)-\epsilon <A_R(V)\le A(V)$ for all $R\ge R_0$, proving that $A_R(V)$ has a finite limit. Hence each $V$ is absolutely integrable in the preceding Definition's sense, and its signed component integrals have finite cube-exhaustion limits by the proved absolute-tail implication in [F1]. This proves that all three functionals are well-defined finite vectors.

The integrand of $J_{\mathrm{k}\mathrm{i}\mathrm{n}}$ is $x$ (metres) crossed with the momentum density $g$ of [F2]. Since $g$ has units $\mathrm{k}\mathrm{g}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$, its integral after multiplication by $x$ has units $\mathrm{k}\mathrm{g}{\mathrm{m}}^2{\mathrm{s}}^{-1}=\mathrm{J}\mathrm{s}$. For each canonical term, ${\epsilon }_{0}EA$ has units ${\epsilon }_0{E}^2\mathrm{s}$ because $A$ has the units of $E$ times time by [F3]. From $[{\epsilon }_0]=\mathrm{F}{\mathrm{m}}^{-1}=\mathrm{C}{\mathrm{V}}^{-1}{\mathrm{m}}^{-1}$ and $1\mathrm{C}\mathrm{V}=1\mathrm{J}$, ${\epsilon }_0{E}^2$ has units $\mathrm{J}{\mathrm{m}}^{-3}$; integration over volume again gives $\mathrm{J}\mathrm{s}$. These are the units asserted in the Definition.

For $i\in \{1,2,3\}$, expand the cross product by [F3]: $(E\times \mathrm{curl}A{)}_i={\sum }_{j,k,m,n}{ϵ}_{ijk}{ϵ}_{kmn}{E}_j{\mathrm{\partial }}_m{A}_n$ $={\sum }_j{E}_j{\mathrm{\partial }}_i{A}_j-{\sum }_j{E}_j{\mathrm{\partial }}_j{A}_i$. Indeed, if either index pair repeats, both sides of ${\sum }_k{ϵ}_{ijk}{ϵ}_{kmn}={\delta }_{im}{\delta }_{jn}-{\delta }_{in}{\delta }_{jm}$ are zero. For distinct $i,j$, only the remaining coordinate $k$ contributes; the left side is $+1$ when $(m,n)=(i,j)$, $-1$ when $(m,n)=(j,i)$, and zero otherwise, exactly as on the right. Multiplying by ${\epsilon }_0{\sum }_{m,i}{ϵ}_{\mathrm{\ell }mi}{x}_m$ and summing over $i,m$ makes the first term exactly $(L_R{)}_{\mathrm{\ell }}$ and leaves $(J_R{)}_{\mathrm{\ell }}-(L_R{)}_{\mathrm{\ell }}=-{\epsilon }_0{\sum }_{m,i,j}{\int }_{Q_R}{ϵ}_{\mathrm{\ell }mi}{x}_m{E}_j{\mathrm{\partial }}_j{A}_i{d}^{3}x$.