Generalized Maxwell equations for exterior-algebra multivectors in (k, n) space-time dimensions

Abstract

This paper presents an exterior-algebra generalization of electromagnetic fields and source currents as multivectors of grades r and $r-1$, respectively, in a flat space-time with n space and k time dimensions. Formulas for the Maxwell equations and the Lorentz force for arbitrary values of r, n, and k are postulated in terms of interior and exterior derivatives, in a form that closely resembles their vector-calculus analogues. These formulas lead to solutions in terms of potentials of grade $r-1$, and to conservation laws in terms of a stress–energy–momentum tensor of rank 2 for any values of r, n, and k, for which a simple explicit formula is given. As an application, an expression for the flux of the stress–energy–momentum tensor across an $(n+k-1)$-dimensional slice of space-time is given in terms of the Fourier transform of the potentials. The abstraction of Maxwell equations with exterior calculus combines the simplicity and intuitiveness of vector calculus, as the formulas admit explicit expressions, with the power of tensors and differential forms, as the formulas can be given for any values of r, n, and k.

Appendix A: Proofs related to the stress–energy–momentum tensor

1.1 Appendix A.1: Distributive properties of the interior product

Proof of (22)

To prove this equation, we first expand the vectors and multivectors in the left-hand side of (22) in terms of their components to get

$$\begin{aligned} \left( {\mathbf {v}}\wedge {\mathbf {w}}\right) \cdot \left( {\mathbf {w}}' \wedge {\mathbf {v}}' \right)&= \left( \sum _{i,I:i\not \in I} v_i w_I \sigma (i,I) {\mathbf {e}}_{i+I} \right) \cdot \left( \sum _{j,J:j\not \in J} v_j' w_J' \sigma (J,j) {\mathbf {e}}_{j+J} \right) \end{aligned}$$

(A.1)

$$\begin{aligned}&= \sum _{\begin{array}{c} i,I:i\not \in I \\ j,J:j\not \in J \end{array}} v_i w_I v_j' w_J' \sigma (i,I) \sigma (J,j) \varDelta _{i+I, j+J}. \end{aligned}$$

(A.2)

At this point, we separate the cases $i=j$ and $i\ne j$, namely

$$\begin{aligned} \left( {\mathbf {v}}\wedge {\mathbf {w}}\right) \cdot \left( {\mathbf {w}}' \wedge {\mathbf {v}}' \right)&= \sum _{i,I} v_i v_i' w_I w_I' \sigma (i,I) \sigma (I,i) \varDelta _{ii} \varDelta _{ii} \nonumber \\&\quad + \sum _{i\ne j, I,J} v_i v_j' w_I w_J' \sigma (i,I) \sigma (J,j) \varDelta _{ii} \varDelta _{jj} \varDelta _{I{\setminus } j,J {\setminus } i}, \end{aligned}$$

(A.3)

and it is easy to note that the first term can be written as

$$\begin{aligned} \sum _{i,I} v_i v_i' w_I w_I' \sigma ^2(i,I) (-1)^r \varDelta _{ii} \varDelta _{ii} = (-1)^r \left( {\mathbf {v}}\cdot {\mathbf {v}}' \right) \left( {\mathbf {w}}\cdot {\mathbf {w}}' \right) . \end{aligned}$$

(A.4)

Regarding the second term, we first prove the equality

$$\begin{aligned} \sigma (i,I) \sigma (J,j)=\sigma (I{\setminus } j, j) \sigma (i, J{\setminus } i) , \end{aligned}$$

(A.5)

which we can be rewritten in the form

$$\begin{aligned} \sigma (I{\setminus } j, j) \sigma (i,I) = \sigma (i, J{\setminus } i) \sigma (J,j) . \end{aligned}$$

(A.6)

Fig. 1

Visual aid for the identity among permutations in Eq. (A.6)

In fact, with the visual help of Fig. 1 it is easy to see how the sign of the permutation on the left-hand side of (A.6) is the same as that on the right-hand side. Then, the second term of the expression (A.3) can be written

(A.7)

$\square$

Proof of (24)

To prove this identity, we write separately the explicit expressions of the left and the right side of (24). On the left side, we get

(A.8)

(A.9)

while on the right side we result is

(A.10)

(A.11)

Fig. 2

Visual aid for the identity among permutations in Eq. (A.12)

For the expressions (A.8) and (A.10) to be identical, it is sufficient that

$$\begin{aligned} \sigma (I{\setminus } b{\setminus } a, a) \sigma (b,I{\setminus } b) = \sigma (b, I{\setminus } a{\setminus } b) \sigma (I{\setminus } a,a) \end{aligned}$$

(A.12)

is satisfied. With the aid of Fig. 2, we notice that both sides represent the signature of two possible permutations reordering the list $(b,I{\setminus } a{\setminus } b, a)$ into I. This proves (24). $\square$

Proof of (27)

We prove this relation by first writing separately the three terms of the equation. On the left-hand side, we get

(A.13)

(A.14)

The central term is

(A.15)

(A.16)

Thus, it is easy to check that $\varDelta _{i+I,J}=\varDelta _ {ii}\varDelta _ {J\setminus i,I}$, so that the first two terms of (27) coincide. Regarding the third term on the left hand side, we obtain

(A.17)

(A.18)

which corresponds to the first two expressions since $\varDelta _{i, J\setminus I}\varDelta _ {II} = \varDelta _{i+I,J}=\varDelta _ {ii}\varDelta _ {J\setminus i,I}$. $\square$

Finally, it is easy to check that $\varDelta _{i+I,J}=\varDelta _ {ii}\varDelta _ {J{\setminus } i,I}$, so that (27) is proved. $\square$

1.2 Appendix A.2: Interior derivative of the tensor

For the sake of compactness, we define the bitensors ${\mathbf {T}}_\odot = \mathbf{F} \odot \mathbf{F}$ and to prove the following identities

(A.19)

(A.20)

Using equations (13), (14), (16) and (17), we write ${\mathbf {T}}_\odot$ and explicitly in terms of components. That is, we obtain

(A.21)

(A.22)

and

(A.23)

(A.24)

We start by computing the interior derivative given by

(A.25)

After some mathematical manipulations, Eq. (A.25) is expanded as

(A.26)

Since $\varDelta _{I{\setminus } i, J{\setminus } j}$ is nonzero only if $J{\setminus } j = I{\setminus } i$, we can use this condition in the relation $\sigma (j,J{\setminus } j) = \sigma (j, I{\setminus } i) = \sigma (I{\setminus } i, j) (-1)^{r-1}$ such that (A.26) becomes

(A.27)

Exchanging the indices i and j and the labels I and J, we get the following simplified expression

(A.28)

A similar reasoning can be followed for the operation given by

(A.29)

The former equation can be expanded as

(A.30)

and after a few manipulations using the properties of the signatures as done to obtain (A.27), from (A.30) we have

(A.31)

Simplifying terms, we obtain

(A.32)

We next derive explicit forms for the operations and . We start by noting that can be expanded in terms of component as

(A.33)

(A.34)

As a consequence, we have that

(A.35)

Noting that $A {\setminus } (B{\setminus } j)$ is a single element i such that $A{\setminus } i = B{\setminus } j$, we finally get

(A.36)

(A.37)

Since (A.37) corresponds exactly to (A.28), we proved (A.19).

Similarly, we next write the operation and write it out in terms of components, i.e.,

(A.38)

(A.39)

Again noting that $j+B{\setminus } A = i$ and that $A+i = B+j$, from (A.39) we obtain

(A.40)

proving the equivalence with (A.32) and therefore proving (A.20).

Combining (A.19) and (A.20) and defining the stress–energy–momentum tensor ${\mathbf {T}}$ of the Maxwell field $\mathbf{F}$ as , we find that the interior derivative of the tensor ${\mathbf {T}}$ satisfies the following formula

(A.41)

Therefore, from the Maxwell equations, i.e., $\varvec{\partial }\wedge \mathbf{F} = 0$ and , we recover the conservation law for energy–momentum relating the Lorentz force ${{\mathbf {f}}}$ (64) and the stress–energy–momentum tensor,

(A.42)

1.3 Appendix A.3: Explicit formulas for the tensor components for generic r

Starting with (69), we note that

(A.43)

(A.44)

for any pair of i and j, and therefore,

$$\begin{aligned} \mathbf{F} \odot \mathbf{F} \bigr |_{ij}&= \frac{1}{2}\sum _{I,J\in {\mathcal {I}}_{r}}\sigma (I{\setminus } i,i)\sigma (j,J{\setminus } j)F_I F_J {\mathbf {e}}_{I{\setminus } i}\cdot {\mathbf {e}}_{J{\setminus } j} \end{aligned}$$

(A.45)

$$\begin{aligned}&= \frac{1}{2}\sum _{L\in {\mathcal {I}}_{r-1}:i,j\not \in L}\sigma (L,i)\sigma (j,L)F_{i+L} F_{j+L} \varDelta _{LL}, \end{aligned}$$

(A.46)

where we have defined the set L such that $I{\setminus } i = J{\setminus } j = L$. The summation contains $\smash {\left( {\begin{array}{c}k+n-2\\ r-1\end{array}}\right) }$ nonzero terms. For $i = j$, and using that $\sigma (L,i)\sigma (i,L) = (-1)^{r-1}$, it can be evaluated as

$$\begin{aligned} \mathbf{F} \odot \mathbf{F} \bigr |_{ii}&= (-1)^{r-1}\frac{1}{2}\sum _{L\in {\mathcal {I}}_{r-1}:i\notin L}F_{i+L}^2\varDelta _{LL}, \end{aligned}$$

(A.47)

Moving on to (70), we note that

$$\begin{aligned} {\mathbf {e}}_{i}\wedge \mathbf{F}&= \sum _{I\in {\mathcal {I}}_{r}} F_I\sigma (i,I){\mathbf {e}}_{i+I} \end{aligned}$$

(A.48)

$$\begin{aligned} \mathbf{F} \wedge {\mathbf {e}}_{j}&= \sum _{J\in {\mathcal {I}}_{r}} F_J\sigma (J,j){\mathbf {e}}_{j+J}, \end{aligned}$$

(A.49)

and we study the cases such that the lists $i+I$ and $j+J$ coincide. First, if $i = j$, we have to sum over $I = J$ such that $i \notin I$, i.e.,

(A.50)

(A.51)

because $\sigma (i,I)\sigma (I,i) = (-1)^r$. Second, if $i\ne j$, we can find an $L \in {\mathcal {I}}_{r-1}$ such that $L = I{\setminus } j = J {\setminus } i$ and $i,j\ne L$. Then,

(A.52)

(A.53)

The product $\sigma (i,L+j)\sigma (L+i,j)$ is equal to $\sigma (L,i)\sigma (j,L)$. To prove it, we write the relation $\sigma (i,L+j)\sigma (L+i,j) = \sigma (L,i)\sigma (j,L)$ and we first multiply both sides by $\sigma (L+i,j)\sigma (j,L)$ to obtain $\sigma (i,L+j)\sigma (j,L) = \sigma (L+i,j)\sigma (L,i)$, namely the permutations sorting the lists (i, j, L) and (L, i, j), respectively. Secondly, we note that $\sigma (i,j,L) = (-1)^{2(r-1)}\sigma (L,i,j) = \sigma (L,i,j)$.

Combining (A.46) and (A.53) into , we have

$$\begin{aligned} T_{ii}&= \frac{(-1)^{r}}{2}\varDelta _{ii}\Biggl (\sum _{I\in {\mathcal {I}}_{r}: i \in I}F_I^2\varDelta _{II} - \sum _{I\in {\mathcal {I}}_{r}: i \notin I}F_{I}^2\varDelta _{II}\Biggr ) \end{aligned}$$

(A.54)

$$\begin{aligned} T_{ij}&= -\sum _{L\in {\mathcal {I}}_{r-1}}\sigma (L,i)\sigma (j,L)F_{i+L} F_{j+L} \varDelta _{LL}. \end{aligned}$$

(A.55)

1.4 Appendix A.4: Stokes theorem for the interior derivative of a bitensor

Considering a bitensor field

$$\begin{aligned} {\mathbf {T}}= \sum _i T_{ii}{\mathbf {u}}_{ii} + \sum _{i< j}T_{ij}{\mathbf {u}}_{ij} , \end{aligned}$$

(A.56)

then the Stokes theorem we wish to prove states that

(A.57)

The proof will follow the reasoning operating for the vector field [11, Sect. 3.5], so we start expanding the integrand on the right-hand side

(A.58)

(A.59)

The last term can be rewritten by changing the indices $i\longleftrightarrow j$ and using the property of symmetry $T_{ji}=T_{ij}$ as

$$\begin{aligned} \sum _{i>j}\sum _{I_{n+k-1} : j\not \in I} T_{ij} \,\mathrm {d}x_I \sigma (j,I) \, {\mathbf {e}}_i , \end{aligned}$$

(A.60)

and finally, the three terms can be compacted in

(A.61)

Then, taking the exterior derivative [14, Sect. 36.B], we get

(A.62)

(A.63)

On the other hand, regarding the right-hand side, we first evaluate

(A.64)

(A.65)

(A.66)

and the differential

$$\begin{aligned} \left( \,\mathrm {d}^{n+k} {\mathbf {x}}\right) ^{\scriptscriptstyle {\mathcal {H}}^{-1}}= \left( \sum _{L_{n+k}} \,\mathrm {d}x_L {\mathbf {e}}_L \right) ^{\scriptscriptstyle {\mathcal {H}}^{-1}}= \sum _{L_{n+k}} \,\mathrm {d}x_L. \end{aligned}$$

(A.67)

Then, we get

(A.68)

namely (A.63) and thereby proving the stated Stokes’ Theorem.

1.5 Appendix A.5: Flux of the stress–energy–momentum tensor

The flux of the field (65) across the boundary $\partial {\mathcal {V}}_\ell ^{k+n}$, denoted by $\varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}})$, is given by the integral in (85),

(A.69)

The r.h.s. of (A.69) is computed w.r.t. $x_{\ell ^c}$, being $\ell ^c$ the set of indices excluding $\ell$.

We next write the flux (A.69) in terms of the Fourier transform of ${\mathbf {F}}$, denoted as ${\hat{\mathbf{F }}} (\varvec{\xi })$ as in (51). Assuming that the Fourier transform of ${\mathbf {F}}$ is supported only in the set $\varvec{\xi }\cdot \varvec{\xi }=0$, as postulated in (57), we express ${\mathbf {F}}$ as

$$\begin{aligned} \mathbf{F} ({\mathbf {x}}) = \int _{-\infty }^{+\infty }\dots \int _{-\infty }^{+\infty } \,\mathrm {d}^{k+n}\varvec{\xi }\, \delta ( \varvec{\xi }\cdot \varvec{\xi }) \, e^{j2\pi \varvec{\xi }\cdot {\mathbf {x}}} \, {\hat{\mathbf{F }}} (\varvec{\xi }). \end{aligned}$$

(A.70)

Inserting (A.70) in (68) and using the linearity properties of $\odot$ and , we obtain that the stress–energy–momentum tensor ${\mathbf {T}}$ can be written as

(A.71)

Using (A.71) back in (A.69), we obtain that

(A.72)

Since the integration w.r.t. $x_{\ell ^c}$ only acts on the exponential term in (A.72), interchanging the integration order and using the definition of the delta function, we can find the inverse Fourier transform of the exponential as

$$\begin{aligned} \int _{-\infty }^{+\infty } \dots \int _{-\infty }^{+\infty } \,\mathrm {d}x_{\ell ^c} \, e^{j2\pi (\varvec{\xi }+\varvec{\xi }')\cdot {\mathbf {x}}} = \prod _{m\ne \ell } \delta (\xi _m + \xi _m') e^{j2\pi (\xi _{\ell } + \xi '_{\ell } )x_{\ell } \varDelta _{\ell \ell }}. \end{aligned}$$

(A.73)

Plugging the r.h.s. of (A.73) in (A.72) and defining $\varphi (\varvec{\xi },\varvec{\xi }')$ as

(A.74)

we write the flux as

$$\begin{aligned} \varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}}) = \int _{-\infty }^{+\infty } \dots \int _{-\infty }^{+\infty } \,\mathrm {d}^{k+n} \varvec{\xi }\,\mathrm {d}^{k+n} \varvec{\xi }'\, \delta (\varvec{\xi }\cdot \varvec{\xi }) \delta (\varvec{\xi }'\cdot \varvec{\xi }') \prod _{m\ne \ell } \delta (\xi _m + \xi _m') \varphi (\varvec{\xi }, \varvec{\xi }'). \end{aligned}$$

(A.75)

In order to solve the integration w.r.t. $\xi _\ell$, we rewrite the condition $\varvec{\xi }\cdot \varvec{\xi }=0$ as

$$\begin{aligned} \varDelta _{\ell \ell }\xi _\ell ^2 + \varvec{\xi }_{{\bar{\ell }}}\cdot \varvec{\xi }_{{\bar{\ell }}} = 0, \end{aligned}$$

(A.76)

where $\varvec{\xi }_{{\bar{\ell }}} = \varvec{\xi }-\xi _\ell {\mathbf {e}}_\ell$. We can solve this equation for $\xi _\ell$ as long as $-\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}}\cdot \varvec{\xi }_{{\bar{\ell }}} \ge 0$, in which case we define $\chi _\ell$ as the positive root of the equation

$$\begin{aligned} \chi _\ell ^2 = -\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}}\cdot \varvec{\xi }_{{\bar{\ell }}}, \end{aligned}$$

(A.77)

and we thus take for $\xi _\ell$ the two possible values $\xi _\ell = \pm \chi _\ell$. We similarly have the analogous versions of (A.76) and (A.77) for $\xi _\ell '$.

Using [16, p. 184] w.r.t. the integration variables $\xi _\ell$ and $\xi _\ell '$ and the limitation in the integration range, Eq. (A.75) is expressed as

$$\begin{aligned} \varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}}) = \int _{\begin{array}{c} \varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}} \cdot \varvec{\xi }_{{\bar{\ell }}}\le 0\\ \varDelta _{\ell \ell } \varvec{\xi }'_{{\bar{\ell }}} \cdot \varvec{\xi }'_{{\bar{\ell }}}\le 0 \end{array}} \,\mathrm {d}\xi _{\ell ^c} \,\mathrm {d}\xi '_{\ell ^c}\, \frac{1}{4\chi _\ell \chi '_\ell } \prod _{m\ne \ell } \delta (\xi _m + \xi _m') \sum _{\xi _\ell = \pm \chi _\ell , \xi '_\ell = \pm \chi _\ell '} \varphi (\varvec{\xi }, \varvec{\xi }'). \end{aligned}$$

(A.78)

Since the remaining Dirac delta function conditions imply that $\xi '_m=-\xi _m$ for every $m\ne \ell$, we also have that $\xi '_\ell =\pm \chi _\ell$. To further deal with the four terms in the summation in (A.78), we define the vectors

$$\begin{aligned} \varvec{\xi }_+&=(\xi _0, \ldots , \xi _{\ell -1}, \chi _\ell , \xi _{\ell +1}, \ldots , \xi _{k+n-1}), \end{aligned}$$

(A.79)

$$\begin{aligned} \varvec{\xi }_-&=(\xi _0, \ldots , \xi _{\ell -1},-\chi _\ell , \xi _{\ell +1}, \ldots , \xi _{k+n-1}), \end{aligned}$$

(A.80)

and the counterparts $\varvec{\xi }'_+=-\varvec{\xi }_-$ and $\varvec{\xi }'_-=-\varvec{\xi }_+$. Using these definitions to solve the integration w.r.t. $\xi '_{\ell ^c}$, we obtain that

$$\begin{aligned} \varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}}) = \int _{\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}} \cdot \varvec{\xi }_{{\bar{\ell }}}\le 0} \,\mathrm {d}\xi _{\ell ^c} \, \frac{1}{4\chi _\ell ^2} \bigl ( \varphi (\varvec{\xi }_+,\varvec{\xi }_+') + \varphi (\varvec{\xi }_+,\varvec{\xi }_-') + \varphi (\varvec{\xi }_-,\varvec{\xi }_+') + \varphi (\varvec{\xi }_-,\varvec{\xi }_-') \bigr ). \end{aligned}$$

(A.81)

It remains to study the four summands in (A.81) by exploiting the properties of exterior algebra. We start by writing $\varphi (\varvec{\xi }_+,\varvec{\xi }_+')$ from its definition in (A.74) and use the fact that ${\hat{\mathbf{F }}}(\varvec{\xi }'_+)={\hat{\mathbf{F }}}(-\varvec{\xi }_-)={\hat{\mathbf{F }}}^*(\varvec{\xi }_-)$ to obtain

(A.82)

where for the sake of clarity we defined the tensor ${\mathbf {T}}_1$ as

(A.83)

Using the definitions of $\odot$ and in (69) and (70), respectively, and the identity (22), we may write the ijth component of ${\mathbf {T}}_1$ as

(A.84)

(A.85)

It will prove convenient to study Eq. (A.85) in terms of the $(r-1)$-vector potential ${\mathbf {A}}$, which is related to ${\mathbf {F}}$ as in (48), or in the Fourier domain,

$$\begin{aligned} {\hat{\mathbf{F }}} (\varvec{\xi }) = 2\pi j \, \varvec{\xi }\wedge {\hat{{\mathbf {A}}}} (\varvec{\xi }), \end{aligned}$$

(A.86)

where ${\hat{{\mathbf {A}}}} (\varvec{\xi })$ denotes the Fourier transform of ${\mathbf {A}}$. Substituting (A.86) in the definitions of $\alpha _{1,ij}$ and $\beta _{1,ij}$ in (A.85) and using the identity (21), we obtain that

(A.87)

(A.88)

We start by expanding $\alpha _{1,ij}$. Using the identity (26), we get

(A.89)

Computing the products, rearranging terms, and using the relations (22), (24)–(25) and (27) in various places, we obtain

(A.90)

We next simplify the terms of the form and . To do so, we note that $\varvec{\xi }_+$ and $\varvec{\xi }_-$, respectively, given in (A.79) and (A.80) are related as

$$\begin{aligned} \varvec{\xi }_+ = \varvec{\xi }_- + 2\chi _\ell {\mathbf {e}}_\ell . \end{aligned}$$

(A.91)

Recalling that the gauge condition in the Fourier domain is given by

(A.92)

equations (A.91) and (A.92) imply that

(A.93)

(A.94)

Similar relations are obtained for the right interior product. Applying (A.93) and (A.94) into (A.90), we obtain for $\alpha _{1,ij}$ that

(A.95)

For $\beta _{1,ij}$, we first use (22) directly into (A.88) so that it is written as

(A.96)

(A.97)

In view of (A.91) and $\varvec{\xi }_+ \cdot \varvec{\xi }_+ = 0$, the $\varvec{\xi }_+\cdot \varvec{\xi }_-$ term in the previous equation equals

$$\begin{aligned} \varvec{\xi }_+\cdot \varvec{\xi }_- = \varvec{\xi }_+\cdot \varvec{\xi }_+ - 2\chi _\ell \,\varvec{\xi }_+\cdot {\mathbf {e}}_\ell = -2 \varDelta _{\ell \ell } \chi ^2_\ell . \end{aligned}$$

(A.98)

Therefore, using (A.93), (A.94) and (A.98) we finally obtain

(A.99)

Although we derived expressions of $\alpha _{1,ij}$, $\alpha _{1,ji}$ and $\beta _{1,ij}$ in (A.95) and (A.99), needed to obtain $T_{1,ij}$ in (A.85) for arbitrary ij, we are only interested in such terms containing the $\ell$th component, since $\varphi (\varvec{\xi }_+,\varvec{\xi }_+')$ in (A.82) involves computing the quantity

(A.100)

We start with the first case in which $i=\ell \ne j$. Using that , from (A.95) we get

(A.101)

Similarly, we also have

(A.102)

Furthermore, the fact that $\varDelta _{ij}=0$ for $i\ne j$ implies from (A.99) that

$$\begin{aligned} \beta _{1,\ell j} = 0. \end{aligned}$$

(A.103)

Combining (A.101), (A.102) and (A.103) in the initial expression of $T_{1,\ell j}$ in (A.85), using that $\xi _{\pm ,\ell } = \pm \chi _\ell$, and writing the right interior products as left interior products, we get

(A.104)

We note that the last two summands in the former equation trivially cancel out, whereas the remaining two also do so because $\xi _{-,j}=\xi _{+,j}$ for $j\ne \ell$. Hence,

$$\begin{aligned} T_{1,\ell j} = 0 , \qquad j\ne \ell . \end{aligned}$$

(A.105)

We continue with the second case $j=\ell \ne i$ and we have, as in the first case

(A.106)

(A.107)

and

$$\begin{aligned} \beta _{1,i\ell } = 0. \end{aligned}$$

(A.108)

We rearrange (A.106), (A.107) and (A.108) in the initial expression (A.85), using $\xi _{\pm ,\ell = \pm \chi _\ell }$ and that $\xi _{-,i}=\xi _{+,i}$ for $i \ne \ell$, we obtain It results to be zero since for $i \ne \ell$ we have $\xi _{-,i}=\xi _{+,i}$, namely

$$\begin{aligned} T_{1,i\ell } = 0 , \qquad i\ne \ell . \end{aligned}$$

(A.109)

Regarding the last case $i=j=\ell$, we evaluate $\alpha _{1,\ell \ell }$ from (A.95) writing all the right interior products as left interior products

(A.110)

and from (A.99)

(A.111)

We substitute $\alpha _{1,\ell \ell }=\beta _{1,\ell \ell }$ from (A.110) and $\gamma _{1,\ell \ell }$ from (A.110) into $T_{1,\ell \ell }$ (A.85) and considering that $\xi _{\pm _\ell } = \pm \chi _\ell$, we directly get

$$\begin{aligned} T_{1,\ell \ell } = 0 . \end{aligned}$$

(A.112)

In conclusion, from (A.100) we realize that

(A.113)

We continue studying the second summand in (A.81). We consider $\varphi (\varvec{\xi }_+,\varvec{\xi }_-')$ using its definition in (A.74) joint with the fact that ${\hat{\mathbf{F }}}(\varvec{\xi }'_-) = {\hat{\mathbf{F }}}(-\varvec{\xi }_+) ={\hat{\mathbf{F }}}^*(\varvec{\xi }_+)$ and we get

(A.114)

where the tensor ${\mathbf {T}}_2$ is defined as

(A.115)

As for ${\mathbf {T}}_1$, we use the definitions of $\odot$ and in (69) and (70) and the identity (22) and we spread out the ijth component of ${\mathbf {T}}_2$, which is written

(A.116)

and we substitute the Maxwell field in terms of the potential in the Fourier domain thanks to (A.86), so that we find

(A.117)

(A.118)

We start from (A.117) and we use the relation (26) after writing . Thus, we get

(A.119)

Again, carrying out all the products, and applying (22) and (26), we obtain

(A.120)

We note that (24) implies

(A.121)

and we can simplify $\alpha _{2,ij}$ thanks to the facts that $\varvec{\xi }_+ \cdot \varvec{\xi }_+ = 0$ and to the gauge condition from (A.92), obtaining

$$\begin{aligned} \alpha _{2,ij} = 4\pi ^2 (-1)^{r-1} \varDelta _{ii}\varDelta _{jj}\xi _{+,i}\xi _{+,j}{ \hat{{\mathbf {A}}}}(\varvec{\xi }_+)\cdot {\hat{{\mathbf {A}}}}^*(\varvec{\xi }_+). \end{aligned}$$

(A.122)

Hence, as a consequence,

$$\begin{aligned} \alpha _{2,ji} = 4\pi ^2 (-1)^{r-1} \varDelta _{ii}\varDelta _{jj}\xi _{+,i} \xi _{+,j}{\hat{{\mathbf {A}}}}(\varvec{\xi }_+)\cdot {\hat{{\mathbf {A}}}}^*(\varvec{\xi }_+). \end{aligned}$$

(A.123)

Regarding $\beta _{2,ij}$, we write $\varvec{\xi }_+ \wedge {\hat{{\mathbf {A}}}}^*(\varvec{\xi }_+) = (-1)^{r-1}{\hat{{\mathbf {A}}}}^*(\varvec{\xi }_+) \wedge \varvec{\xi }_+$ and we apply again (22) in (A.118) so that we immediately verify that it vanishes

(A.124)

As a consequence, the result for $T_{2,ij}$ in (A.116) is

$$\begin{aligned} T_{2,ij}= 8\pi ^2 (-1)^{r-1} \xi _{+,i}\xi _{+,j}{\hat{{\mathbf {A}}}}(\varvec{\xi }_+)\cdot {\hat{{\mathbf {A}}}}^*(\varvec{\xi }_+) \end{aligned}$$

(A.125)

and we finally evaluate the tensor ${\mathbf {T}}_2$ as

$$\begin{aligned} {\mathbf {T}}_2 = 8\pi ^2 \sum _{i\le j} (-1)^{r-1} \xi _{+,i}\xi _{+,j}\bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi }_+)\bigr \vert ^2 {\mathbf {u}}_{ij} . \end{aligned}$$

(A.126)

We move on to the third term of (A.81). The equality ${\hat{\mathbf{F }}}(\varvec{\xi }'_+)={\hat{\mathbf{F }}}^*(\varvec{\xi }_-)$ allows us to write $\varphi (\varvec{\xi }_-,\varvec{\xi }_+')$ as

(A.127)

where we defined the tensor ${\mathbf {T}}_3$ as follow

(A.128)

Comparing (A.128) and (A.115), we note that the difference is only in the presence of $\varvec{\xi }_-$ instead of $\varvec{\xi }_+$. Thus, the mathematical steps are identical, including that the relation $\varvec{\xi }_+ \cdot \varvec{\xi }_+ = 0$ has its counterpart $\varvec{\xi }_- \cdot \varvec{\xi }_- = 0$. The gauge conditions in (A.92) can also be written as

(A.129)

So, in analogy with the result obtained in (A.126), the final expression for ${\mathbf {T}}_3$ is

$$\begin{aligned} {\mathbf {T}}_3 = 8\pi ^2 \sum _{i\le j} (-1)^{r-1} \xi _{-,i}\xi _{-,j}\bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi }_-)\bigr \vert ^2 {\mathbf {u}}_{ij} . \end{aligned}$$

(A.130)

We conclude the evaluation of the initial integral in (A.81) computing $\varphi (\varvec{\xi }_-, \varvec{\xi }_-')$. From its definition in (A.74) and using ${\hat{\mathbf{F }}}(\varvec{\xi }'_-)={\hat{\mathbf{F }}}^*(\varvec{\xi }_+)$, we get

(A.131)

defined, as in the previous cases,

(A.132)

We can further expand (A.132) in components which would appear

(A.133)

As in the analysis of $T_{1,ij}$, we express $\alpha _{4,ij}$ and $\beta _{4,ij}$ in terms of the potential

(A.134)

(A.135)

We can now note that the differences between (A.132) and (A.83) are in $\varvec{\xi }_+$ exchanged with $\varvec{\xi }_-$, ${\hat{\mathbf{F }}}(\varvec{\xi }_+)$ with ${\hat{\mathbf{F }}}(\varvec{\xi }_-)$ and ${\hat{\mathbf{F }}}^*(\varvec{\xi }_-)$ with ${\hat{\mathbf{F }}}^*(\varvec{\xi }_+)$. So, the differences among (A.134) and (A.135) with respect to (A.87) and (A.88) are, in addition to the aforementioned, ${\hat{{\mathbf {A}}}}(\varvec{\xi }_+)$ interchanged with ${\hat{{\mathbf {A}}}}(\varvec{\xi }_-)$ and ${\hat{{\mathbf {A}}}}^*(\varvec{\xi }_-)$ with ${\hat{{\mathbf {A}}}}^*(\varvec{\xi }_+)$. We consider (A.91) and the gauge condition (A.129), and we replace the conditions (A.93) and (A.94) with

(A.136)

(A.137)

If we follow the same procedure applied to obtain ${\mathbf {T}}_1$ with the conditions (A.129), (A.136) and (A.137), we can rapidly state

(A.138)

Then, we write the integral for the flux in (A.81) substituting the definition (A.74). Removing the first and the last summands thanks to (A.113) and (A.138), it results

(A.139)

Using (A.126) and (A.130), we get

(A.140)

We consider that $\chi _\ell ^2 = \xi _{+,\ell }^2 = \xi _{-,\ell }^2$ and we expand the product thanks to (85) and then use that $\sum _i \xi _{\pm ,i}{\mathbf {e}}_i = \varvec{\xi }_\pm$, so that (A.140) is

$$\begin{aligned} \varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}}) = (-1)^{r} \pi ^2 \sigma (\ell , \ell ^c) \int _{\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}} \cdot \varvec{\xi }_{{\bar{\ell }}}\le 0} \,\mathrm {d}\xi _{\ell ^c} \biggl ( \frac{\varvec{\xi }_{+}}{\xi _{+,\ell }}\bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi }_+)\bigr \vert ^2 +\frac{\varvec{\xi }_{-}}{\xi _{-,\ell }}\bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi }_-)\bigr \vert ^2 \biggr ) .\quad \end{aligned}$$

(A.141)

As an aside, we may use [16, p. 184] and that $\chi _\ell = \xi _{+,\ell } = -\xi _{-,\ell }$ to undo the step leading to (A.78) to recover the Dirac delta function

$$\begin{aligned} \varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}}) = (-1)^{r} 2\pi ^2 \sigma (\ell , \ell ^c)\int _{-\infty }^{+\infty } \dots \int _{-\infty }^{+\infty } \,\mathrm {d}^{k+n} \varvec{\xi }{{\,\mathrm{sgn}\,}}(\xi _\ell )\varvec{\xi }\bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi })\bigr \vert ^2\delta (\varvec{\xi }\cdot \varvec{\xi }). \end{aligned}$$

(A.142)

Returning to (A.141), we split the integral into $I_{\ell ,+}+I_{\ell ,-}$, where

$$\begin{aligned} I_{\ell ,\pm } = \int _{\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}} \cdot \varvec{\xi }_{{\bar{\ell }}}\le 0} \,\mathrm {d}\xi _{\ell ^c} \frac{\varvec{\xi }_{\pm }}{\xi _{\pm ,\ell }}\bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi }_\pm )\bigr \vert ^2. \end{aligned}$$

(A.143)

Taking into account that $\mathbf{A }({\mathbf {x}})$ is real, we may express the squared modulus of ${\hat{{\mathbf {A}}}}(\varvec{\xi }_-)$ as $\bigl |{\hat{{\mathbf {A}}}}(\varvec{\xi }_-)\bigr |^2={\hat{{\mathbf {A}}}}(-\varvec{\xi }_-){\hat{{\mathbf {A}}}}^*(-\varvec{\xi }_-)$. Therefore, the integral $I_{\ell ,-}$ becomes

$$\begin{aligned} I_{\ell ,-} = \int _{\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}} \cdot \varvec{\xi }_{{\bar{\ell }}}\le 0} \,\mathrm {d}\xi _{\ell ^c} \frac{\varvec{\xi }_{-}}{\xi _{-,\ell }} {\hat{{\mathbf {A}}}}(-\varvec{\xi }_-){\hat{{\mathbf {A}}}}^*(-\varvec{\xi }_-) . \end{aligned}$$

(A.144)

Changing the integration variables according to $\varvec{\xi }_{{\bar{\ell }}} \rightarrow \varvec{\zeta }_{{\bar{\ell }}} =-\varvec{\xi }_{{\bar{\ell }}}$, together with the definition $\varvec{\zeta }_\pm = (\zeta _0, \ldots c,\zeta _{\ell -1},\pm \chi _\ell ,\zeta _{\ell +1}, \ldots c,\zeta _{n+k-1})$, yields

$$\begin{aligned} I_{\ell ,-} = \int _{\varDelta _{\ell \ell } \varvec{\zeta }_{{\bar{\ell }}} \cdot \varvec{\zeta }_{{\bar{\ell }}}\le 0} \,\mathrm {d}\zeta _{\ell ^c} \frac{\varvec{\zeta }_{+}}{\zeta _{+,\ell }} {\hat{{\mathbf {A}}}}(\varvec{\zeta }_+){\hat{{\mathbf {A}}}}^*(\varvec{\zeta }_+) . \end{aligned}$$

(A.145)

Since (A.145) is formally equivalent to $I_{\ell ,+}$, the flux can be rewritten as

$$\begin{aligned} \varPhi _{\partial {\mathcal {V}}_\ell ^{k+n}} ({\mathbf {T}}) = (-1)^{r} 2\pi ^2 \sigma (\ell , \ell ^c) \int _{\varDelta _{\ell \ell } \varvec{\xi }_{{\bar{\ell }}} \cdot \varvec{\xi }_{{\bar{\ell }}}\le 0} \,\mathrm {d}\xi _{\ell ^c} \frac{\varvec{\xi }_{+}}{\xi _{+,\ell }} \bigl \vert {\hat{{\mathbf {A}}}}(\varvec{\xi }_+)\bigr \vert ^2 . \end{aligned}$$

(A.146)