The Aharonov–Bohm effect in a closed flux line

Abstract

The Aharonov–Bohm (AB) effect was convincingly demonstrated using a micro-sized toroidal magnet but it is almost always explained using an infinitely-long solenoid or an infinitely-long flux line. The main reason for this is that the formal treatment of the AB effect considering a toroidal configuration turns out to be too cumbersome. But if the micro-sized toroidal magnet is modelled by a closed flux line of arbitrary shape and size then the formal treatment of the AB effect is exact, considerably simplified, and well-justified. Here we present such a treatment that covers in detail the electromagnetic, topological, and quantum-mechanical aspects of this effect. We demonstrate that the AB phase in a closed flux line is determined by a linking number and has the same form as the AB phase in an infinitely-long flux line which is determined by a winding number. We explicitly show that the two-slit interference shift associated with the AB effect in a closed flux line is the same as that associated with an infinitely-long flux line. We emphasise the topological nature of the AB phase in a closed flux line by demonstrating that this phase is invariant under deformations of the charge path, deformations of the closed flux line, simultaneous deformations of the charge path and the closed flux line, and the interchange between the charge path and the closed flux line. We also discuss the local and nonlocal interpretations of the AB effect in a closed flux line and introduce a non-singular gauge in which the vector potential vanishes in all space except on the surface surrounded by the closed flux line, implying that this vector potential is zero along the trajectory of the charged particle except on the crossing point where this trajectory intersects the surface bounded by the closed flux line, a result that questions the alleged physical significance of the vector potential and thereby the local interpretation of the AB effect.

Appendices

Appendix A. Derivation of Eq. (4)

Consider the potential of the closed flux line defined in Eq. (7)

$$\begin{aligned} {\varvec{A}}=\frac{\varPhi }{4\pi } \nabla \times \oint _{\mathcal {C}} \frac{d {\varvec{x}}'}{|{\varvec{x}} - {\varvec{x}}'|}. \end{aligned}$$

(A1)

The curl of this potential and the use of the identity $\nabla \times (\nabla \times {\varvec{F}}) = \nabla (\nabla \cdot {\varvec{F}})- \nabla ^2 {\varvec{F}}$ yield

$$\begin{aligned} \nabla \times {\varvec{A}}= & {} \frac{\varPhi }{4\pi }\nabla \times \bigg [\nabla \times \bigg ( \oint _{\mathcal {C}} \frac{d{\varvec{x}}'}{|{\varvec{x}}-{\varvec{x}}'|} \bigg )\bigg ]\nonumber \\= & {} \frac{\varPhi }{4\pi } \bigg [\nabla \oint _{\mathcal {C}} \nabla \cdot \bigg ( \frac{d{\varvec{x}}'}{|{\varvec{x}}-{\varvec{x}}'|} \bigg ) -\oint _{\mathcal {C}}\nabla ^2 \bigg (\frac{ d{\varvec{x}}'}{|{\varvec{x}}-{\varvec{x}}'|} \bigg )\bigg ]. \end{aligned}$$

(A2)

Inserting $\nabla \cdot (d {\varvec{x}}'/|{\varvec{x}}-{\varvec{x}}'|) = -\nabla '(1/|{\varvec{x}}\!-\!{\varvec{x}}'|) \cdot d{\varvec{x}}'$ and $\nabla ^2( 1 / |{\varvec{x}}-{\varvec{x}}'|)=-4 \pi \delta ({\varvec{x}} - {\varvec{x}}')$ in Eq. (A2), we obtain

$$\begin{aligned} \nabla \times {\varvec{A}}= \frac{\varPhi }{4\pi } \bigg [-\nabla \oint _{\mathcal {C}} \nabla ' \bigg ( \frac{1}{|{\varvec{x}}-{\varvec{x}}'|} \bigg )\cdot d{\varvec{x}}' +4\pi \oint _{\mathcal {C}}\delta ({\varvec{x}}- {\varvec{x}}')\,d{\varvec{x}}'\bigg ]. \end{aligned}$$

(A3)

The first term on the right-hand side vanishes because $\oint _{\mathcal {C}}\nabla '(1/|{\varvec{x}}- {\varvec{x}}'|)\cdot d {\varvec{x}}'=0$ on account of the gradient theorem and the fact that $1/|{\varvec{x}}- {\varvec{x}}'|$ is a single-valued function of ${\varvec{x}}'$. Thus, Eq. (A3) becomes

$$\begin{aligned} \nabla \times {\varvec{A}}= \varPhi \oint _{\mathcal {C}}\delta ({\varvec{x}}- {\varvec{x}}')\,d{\varvec{x}}'={\varvec{B}}, \end{aligned}$$

(A4)

which shows that the curl of Eq. (7) yields the magnetic field given in Eq. (4).

Appendix B. Derivation of Eq. (8)

Consider the potential of the closed flux line defined in Eq. (7)

$$\begin{aligned} {\varvec{A}}=\frac{\varPhi }{4\pi } \nabla \times \oint _{\mathcal {C}} \frac{d {\varvec{x}}'}{|{\varvec{x}} - {\varvec{x}}'|}. \end{aligned}$$

(B1)

Using the Stokes theorem in the closed line integral of this potential we obtain

$$\begin{aligned} \oint _{\mathcal {C}}\frac{d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|}= \int _{\mathcal {S}}d{\varvec{S}}'\times \nabla '\bigg ( \frac{1}{|{\varvec{x}}- {\varvec{x}}'|} \bigg ), \end{aligned}$$

(B2)

where $\mathcal {S}$ is the surface enclosed by the curve $\mathcal {C}.$ Making use of the relations $\nabla \times (d {\varvec{S}}'/|{\varvec{x}}- {\varvec{x}}'|)= -d {\varvec{S}}' \times \nabla (1/|{\varvec{x}}-{\varvec{x}}'|)$ and $\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)=-\nabla '(1/|{\varvec{x}}- {\varvec{x}}'|)$ in Eq. (B2), we obtain

$$\begin{aligned} \oint _{\mathcal {C}}\frac{d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|}= \nabla \times \int _{\mathcal {S}}\frac{d{\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}, \end{aligned}$$

(B3)

which allows us to write Eq. (B1) as

$$\begin{aligned} {\varvec{A}}= \frac{\varPhi }{4\pi }\bigg [ \nabla \times \bigg (\nabla \times \int _{\mathcal {S}}\frac{d{\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}\bigg )\bigg ]. \end{aligned}$$

(B4)

The use of the identity $\nabla ^2{\varvec{F}} = \nabla (\nabla \cdot {\varvec{F}})-\nabla \times (\nabla \times {\varvec{F}})$ in Eq. (B4) gives

$$\begin{aligned} {\varvec{A}}=\frac{\varPhi }{4\pi }\bigg [ \nabla \int _{\mathcal {S}}\nabla \cdot \bigg (\frac{d{\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}\bigg )- \int _{\mathcal {S}}\nabla ^2\bigg (\frac{d{\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}\bigg )\bigg ]. \end{aligned}$$

(B5)

Considering the results $\nabla \cdot (d {\varvec{S}}'/|{\varvec{x}}- {\varvec{x}}'|)=\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)\cdot d {\varvec{S}}'$ and $\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)=-\nabla '(1/|{\varvec{x}}- {\varvec{x}}'|)$ in Eq. (B5), it becomes

$$\begin{aligned} {\varvec{A}}= \frac{\varPhi }{4\pi }\bigg [ -\nabla \int _{\mathcal {S}}\nabla '\bigg (\frac{1}{|{\varvec{x}}- {\varvec{x}}'|}\bigg ) \cdot d{\varvec{S}}'- \int _{\mathcal {S}}\nabla ^2\bigg (\frac{d{\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}\bigg )\bigg ]. \end{aligned}$$

(B6)

Using $\nabla '(1/|{\varvec{x}}- {\varvec{x}}'|)=({\varvec{x}}- {\varvec{x}}')/|{\varvec{x}}- {\varvec{x}}'|^3$ and $\nabla ^2(1/|{\varvec{x}}- {\varvec{x}}'|)=-4\pi \delta ({\varvec{x}}- {\varvec{x}}')$ in Eq. (B6), we obtain

$$\begin{aligned} {\varvec{A}}=\frac{\varPhi }{4\pi }\bigg [\nabla \int _{\mathcal {S}}\frac{({\varvec{x}}'-{\varvec{x}})\cdot d{\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|^3} + 4\pi \int _{\mathcal {S}}\delta ({\varvec{x}}-{\varvec{x}}')\,d{\varvec{S}}'\bigg ]. \end{aligned}$$

(B7)

The first integral is identified with the single-valued solid angle $\varOmega _0$ defined in Eq. (9) while the second integral is identified with the surface vector Dirac delta ${\varvec{\delta }}_{\mathcal {S}}$ specified in Eq. (10). Thus, we get Eq. (8): ${\varvec{A}} = \varPhi \nabla \varOmega _0/(4\pi ) + \varPhi {\varvec{\delta }}_{\mathcal {S}}.$

Appendix C. Proof of Eq. (11)

The proof of Eq. (11) will be developed in two parts. In the first part we will explicitly demonstrate that the circulation of the gradient of $\varOmega _0$ along an arbitrary closed path C vanishes $\oint _{C}\nabla \varOmega _0\cdot d {\varvec{x}}=0$. In the second part we will transform this circulation using the Stokes theorem $\oint _{C}\nabla \varOmega _0\cdot d {\varvec{x}}=0=\int _{S}\nabla \times \nabla \varOmega _0 \cdot d {\varvec{S}}$ to show $\nabla \times \nabla \varOmega _0=0$. This last result will be used to demonstrate Eq. (11).

Let us obtain a suitable form of the gradient of $\varOmega _0$. The gradient of Eq. (9) gives

$$\begin{aligned} \nabla \varOmega _0 = \nabla \int _{\mathcal {S}}\frac{({\varvec{x}}' -{\varvec{x}})\cdot d {\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|^3}. \end{aligned}$$

(C1)

Using $\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)= -({\varvec{x}}- {\varvec{x}}')/|{\varvec{x}}- {\varvec{x}}'|^3$, Eq. (C1) becomes

$$\begin{aligned} \nabla \varOmega _0 = -\nabla \int _{\mathcal {S}}\nabla '\bigg (\frac{1}{|{\varvec{x}}- {\varvec{x}}'|}\bigg )\cdot d {\varvec{S}}'. \end{aligned}$$

(C2)

Considering the relations $\nabla '(1/|{\varvec{x}}- {\varvec{x}}'|)=-\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)$ and $\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)\cdot d {\varvec{S}}'=\nabla \cdot (d {\varvec{S}}'/|{\varvec{x}}- {\varvec{x}}'|)$ in Eq. (C2), we obtain

$$\begin{aligned} \nabla \varOmega _0 = \nabla \int _{\mathcal {S}}\nabla \cdot \bigg (\frac{d {\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}\bigg ). \end{aligned}$$

(C3)

When the identity $\nabla (\nabla \cdot {\varvec{F}}) = \nabla \times (\nabla \times {\varvec{F}}) + \nabla ^2{\varvec{F}}$ is used in Eq. (C3), it becomes

$$\begin{aligned} \nabla \varOmega _0 = \nabla \times \bigg [ \nabla \times \int _{\mathcal {S}}\frac{d {\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}\bigg ] + \nabla ^2 \int _{\mathcal {S}}\frac{d {\varvec{S}}'}{|{\varvec{x}}- {\varvec{x}}'|}. \end{aligned}$$

(C4)

Inserting $\nabla \times (d {\varvec{S}}'/|{\varvec{x}}- {\varvec{x}}'|)= -d {\varvec{S}}' \times \nabla (1/|{\varvec{x}}-{\varvec{x}}'|)$ together with $\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)=-\nabla '(1/|{\varvec{x}}- {\varvec{x}}'|)$ in the quantity within the brackets and using $\nabla ^2(1/|{\varvec{x}}- {\varvec{x}}'|)=-4\pi \delta ({\varvec{x}}- {\varvec{x}}')$ on the second term of Eq. (C4), we obtain

$$\begin{aligned} \nabla \varOmega _0 = \nabla \times \bigg [ \int _{\mathcal {S}}d {\varvec{S}}' \times \nabla '\bigg (\frac{1}{|{\varvec{x}}- {\varvec{x}}'|} \bigg )\bigg ] -4\pi \int _{\mathcal {S}}\delta ({\varvec{x}}- {\varvec{x}}')\,d{\varvec{S}}'. \end{aligned}$$

(C5)

The quantity within the brackets in Eq. (C5) can be transformed into a closed line integral via the Stokes theorem

$$\begin{aligned} \int _{\mathcal {S}}d {\varvec{S}}' \times \nabla '\bigg (\frac{1}{|{\varvec{x}}- {\varvec{x}}'|} \bigg )= \oint _{\mathcal {C}}\frac{d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|}, \end{aligned}$$

(C6)

where $\mathcal {C}$ is the boundary of $\mathcal {S}.$ When Eq. (C6) and the surface vector Dirac delta $\int _{\mathcal {S}}\delta ({\varvec{x}}- {\varvec{x}}')d{\varvec{S}}'={\varvec{\delta }}_{\mathcal {S}}$ given in Eq. (10) are used in Eq. (C5), it takes the form

$$\begin{aligned} \nabla \varOmega _0 = \nabla \times \oint _{\mathcal {C}}\frac{d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|} -4\pi {\varvec{\delta }}_{\mathcal {S}}. \end{aligned}$$

(C7)

Considering $\nabla \times (d {\varvec{x}}'/|{\varvec{x}} - {\varvec{x}}'|)= \nabla (1/|{\varvec{x}}- {\varvec{x}}'|)\times d {\varvec{x}}'$ and $\nabla (1/|{\varvec{x}}- {\varvec{x}}'|)=({\varvec{x}}'- {\varvec{x}})/|{\varvec{x}}- {\varvec{x}}'|^3$ we can write Eq. (C7) as

$$\begin{aligned} \nabla \varOmega _0 = \oint _{\mathcal {C}}\frac{({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|^3} -4\pi {\varvec{\delta }}_{\mathcal {S}}, \end{aligned}$$

(C8)

which is a suitable form of the gradient of $\varOmega _0$. Let us now take the circulation to Eq. (C8) along an arbitrary closed path C

$$\begin{aligned} \oint _{C}\nabla \varOmega _0\cdot d {\varvec{x}} = \oint _{C}\oint _{\mathcal {C}}\frac{[({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}']\cdot d {\varvec{x}}}{|{\varvec{x}}- {\varvec{x}}'|^3} -4\pi \oint _{C}{\varvec{\delta }}_{\mathcal {S}}\cdot d {\varvec{x}}. \end{aligned}$$

(C9)

Making use of the relation $[({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}']\cdot d {\varvec{x}}=({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')$ in the first term of the right-hand side of Eq. (C9), we obtain

$$\begin{aligned} \oint _{C}\nabla \varOmega _0\cdot d {\varvec{x}} =4\pi \bigg [\frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}\bigg ] -4\pi \bigg [\oint _{C}{\varvec{\delta }}_{\mathcal {S}}\cdot d {\varvec{x}}\bigg ]. \end{aligned}$$

(C10)

The first quantity within the brackets is the linking number l defined by Eq. (21). The second quantity within the brackets is another equivalent form of the linking number defined by Eq. (27). Since C corresponds to the same path in the two closed line integrals on the right-hand side of Eq. (C10) then it follows that

$$\begin{aligned} \oint _{C}\nabla \varOmega _0 \cdot d{\varvec{x}}= {\left\{ \begin{array}{ll} 4\pi l - 4\pi l =0&{} \hbox {if} \,C \,\hbox {encloses} \,\mathcal {C} \\ 0 &{}\text {otherwise} \end{array}\right. } \end{aligned}$$

(C11)

We observe that regardless of the path C we have the vanishing of the circulation

$$\begin{aligned} \oint _{C}\nabla \varOmega _0 \cdot d{\varvec{x}}=0, \end{aligned}$$

(C12)

which is the first step in the proof of Eq. (11). In the second step we transform the left-hand side of Eq. (C12) into a surface integral via the Stokes theorem

$$\begin{aligned} \oint _{C}\nabla \varOmega _0 \cdot d{\varvec{x}}= \int _{S} \nabla \times \nabla \varOmega _0 \cdot d{\varvec{S}}, \end{aligned}$$

(C13)

where S is the surface enclosed by C. Equations (C12) and (C13) imply

$$\begin{aligned} \oint _{C}\nabla \varOmega _0 \cdot d{\varvec{x}}= 0=\int _{S} \nabla \times \nabla \varOmega _0 \cdot d{\varvec{S}}. \end{aligned}$$

(C14)

Since this result holds for any path C then it follows that the second equality in Eq. (C14) is valid for any surface S implying the vanishing of the curl of the gradient of $\varOmega _0$ in all space

$$\begin{aligned} \nabla \times \nabla \varOmega _0=0. \end{aligned}$$

(C15)

To show Eq. (11) let us write Eq. (C14) in index notation

$$\begin{aligned} \oint _{C}\partial _{k}\varOmega _0\,dx^k= 0=\int _{S}\varepsilon _{kmn}\partial ^m\partial ^n\varOmega _0\,dS^k. \end{aligned}$$

(C16)

Consider now the antisymmetric tensor $dS_{ij}=\varepsilon _{ijk}dS^k$ representing an infinitesimal element of the surface S. In terms of $dS_{ij}$ we may write the differential surface vector in the following form $dS^k=(1/2)\varepsilon ^{kij}dS_{ij}$. Using this result together with the identity $\varepsilon _{kmn}\varepsilon ^{kij}=\delta ^{i}_{m}\delta ^{j}_{n}-\delta ^{j}_{m}\delta ^{i}_{n}$ we obtain $\varepsilon _{kmn}\partial ^m\partial ^n\varOmega _0\,dS^k=(1/2)(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega _0 dS_{ij}$, which is used in the second equality in Eq. (C16) to obtain the relation

$$\begin{aligned} 2\oint _{C}\partial _{k}\varOmega _0\,dx^k= 0=\int _{S}\,(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega _0 \,\,dS_{ij}. \end{aligned}$$

(C17)

Since the first equality is valid for any path C then the second equality is valid for any surface S and this implies Eq. (11): $(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega _0=0$ in all space.

Appendix D. Proof of Eq. (15)

Our approach to show Eq. (15) is as follows. We will show that the circulation of the gradient of the solid angle $\varOmega$ along an arbitrary closed path C is non-vanishing: $\oint _{C}\nabla \varOmega \cdot d {\varvec{x}}\ne 0$. Then we will transform this circulation via the Stokes theorem $\oint _{C}\nabla \varOmega \cdot d {\varvec{x}}=\int _{S}\nabla \times \nabla \varOmega \cdot d {\varvec{S}}$ to show $\nabla \times \nabla \varOmega = {\varvec{\delta }}_{\mathcal {C}}$. We will use this result to demonstrate Eq. (15).

Using Eq. (14), the circulation of the gradient of $\varOmega$ takes the form

$$\begin{aligned} \oint _{C}\nabla \varOmega \cdot d {\varvec{x}}= \oint _{C}\nabla \varOmega _0 \cdot d {\varvec{x}} + 4\pi \oint _{C}{\varvec{\delta }}_{\mathcal {S}}\cdot d {\varvec{x}}. \end{aligned}$$

(D1)

We can transform the first circulation on the right-hand side of Eq. (D1) using Eq. (C10). This gives

$$\begin{aligned} \oint _{C}\nabla \varOmega \cdot d {\varvec{x}}= & {} \oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3} -4\pi \oint _{C}{\varvec{\delta }}_{\mathcal {S}}\cdot d {\varvec{x}} + 4\pi \oint _{C}{\varvec{\delta }}_{\mathcal {S}}\cdot d {\varvec{x}} \nonumber \\= & {} 4\pi \bigg [\frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}\bigg ] .\qquad \qquad \qquad \quad \quad \,\,\,\,\, \end{aligned}$$

(D2)

The quantity within the brackets is the Gauss linking number defined in Eq. (21). Therefore

$$\begin{aligned} \oint _{C} \nabla \varOmega \cdot d{\varvec{x}}= {\left\{ \begin{array}{ll} 4\pi l&{} \hbox {if} \,C \,\hbox {encloses} \,\mathcal {C} \\ 0 &{}\text {otherwise} \end{array}\right. } \end{aligned}$$

(D3)

Using the Stokes theorem we can transform the left-hand side of Eq. (D3),

$$\begin{aligned} \oint _{C}\nabla \varOmega \cdot d {\varvec{x}}=\int _{S}\nabla \times \nabla \varOmega \cdot d {\varvec{S}}, \end{aligned}$$

(D4)

where C is the boundary of the surface S. When C does not enclose $\mathcal {C}$ then from Eq. (D3) we have

$$\begin{aligned} \oint _{C}\nabla \varOmega \cdot d {\varvec{x}}=0=\int _{S}\nabla \times \nabla \varOmega \cdot d {\varvec{S}}, \end{aligned}$$

(D5)

and thus $\nabla \times \nabla \varOmega =0$ locally holds for any surface S not pierced by $\mathcal {C}.$ However, this result does not hold in all space (i.e. globally) because if C encloses $\mathcal {C}$ then the left-hand side of Eq. (D5) is non-vanishing and from Eq. (D3) we obtain

$$\begin{aligned} \oint _{C}\nabla \varOmega \cdot d {\varvec{x}}=4\pi l=\int _{S}\nabla \times \nabla \varOmega \cdot d {\varvec{S}}, \end{aligned}$$

(D6)

and thus the relation $\nabla \times \nabla \varOmega \ne 0$ holds. To find the explicit form of this relation, we use Eq. (14): $\nabla \varOmega = \nabla \varOmega _0 + 4\pi {\varvec{\delta }}_{\mathcal {S}}$ and therefore $\nabla \times \nabla \varOmega = 4\pi {\varvec{\delta }}_{\mathcal {C}}$ where we have used Eq. (C15): $\nabla \times \nabla \varOmega _0=0$ and Eq. (12): $\nabla \times {\varvec{\delta }}_{\mathcal {S}}= {\varvec{\delta }}_{\mathcal {C}}$, where ${\varvec{\delta }}_{\mathcal {C}}=\oint _{\mathcal {C}}\delta ({\varvec{x}}- {\varvec{x}}')d {\varvec{x}} '$ is a line Dirac delta along the closed path $\mathcal {C}$ which forms the boundary of $\mathcal {S}$. Thus

$$\begin{aligned} \int _{S}\nabla \times \nabla \varOmega \cdot d {\varvec{S}}=4\pi \int _{S}{\varvec{\delta }}_{\mathcal {C}}\cdot d {\varvec{S}}, \end{aligned}$$

(D7)

which implies

$$\begin{aligned} \nabla \times \nabla \varOmega = 4\pi {\varvec{\delta }}_{\mathcal {C}}. \end{aligned}$$

(D8)

To prove Eq. (15), we write Eq. (D4) in index notation

$$\begin{aligned} \oint _{C}\partial _{k}\varOmega \,dx^k=\int _{S}\varepsilon _{kmn}\partial ^m\partial ^n\varOmega \,\,dS^k. \end{aligned}$$

(D9)

Consider now the antisymmetric tensor $dS_{ij}=\varepsilon _{ijk}dS^k$ representing an infinitesimal element of the surface S. Using this result and the identity $\varepsilon _{kmn}\varepsilon ^{kij}=\delta ^{i}_{m}\delta ^{j}_{n}-\delta ^{j}_{m}\delta ^{i}_{n}$, we obtain $\varepsilon _{kmn}\partial ^m\partial ^n\varOmega \,dS^k=(1/2)(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega dS_{ij}$, which is used in the second equality in Eq. (D9), obtaining

$$\begin{aligned} 2\oint _{C}\partial _{k}\varOmega \,dx^k=\int _{S}(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega \,\,dS_{ij}. \end{aligned}$$

(D10)

When the path C does not enclose the curve $\mathcal {C}$ then from Eq. (D3) we have

$$\begin{aligned} 2\oint _{C}\partial _{k}\varOmega \,dx^k=0=\int _{S}(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega \,\,dS_{ij}, \end{aligned}$$

(D11)

which implies $(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega =0$ for any surface S not pierced by $\mathcal {C}.$ In this case $\varOmega$ is locally single-valued. However, this is not the global case for if C encloses $\mathcal {C}$ then the left-hand side of Eq. (D10) is non-vanishing and from Eq. (D3) we obtain

$$\begin{aligned} 2\oint _{C}\partial _{k}\varOmega \, dx^k=8\pi l=\int _{S}(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega \,\,dS_{ij}. \end{aligned}$$

(D12)

which implies $(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega \ne 0$ when C encircles $\mathcal {C},$ or equivalently stated, when C crosses S. To find the explicit form of $(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega$ we use Eqs. (D12) and (D9) to obtain

$$\begin{aligned} \int _{S}(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega \,\,dS_{ij}=2\int _{S}\varepsilon _{kmn}\partial ^m\partial ^n\varOmega \,\,dS^k. \end{aligned}$$

(D13)

Equation (D8) in index notation reads $4\pi ({\varvec{\delta }}_{\mathcal {C}})_k=\varepsilon _{kmn}\partial ^m\partial ^n\varOmega$. This result and $dS^k=(1/2)\varepsilon ^{kij}dS_{ij}$ give the relation $\varepsilon _{kmn}\partial ^m\partial ^n\varOmega \,dS^k=2\pi \varepsilon ^{ijk}({\varvec{\delta }}_{\mathcal {C}})_kdS_{ij}$ so that Eq. (D13) reduces to

$$\begin{aligned} \int _{S}\,(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega \,\,dS_{ij}= 4\pi \int _{S} \varepsilon ^{ijk}({\varvec{\delta }}_{\mathcal {C}})_k \,dS_{ij}, \end{aligned}$$

(D14)

and this implies Eq. (15): $(\partial ^i\partial ^j - \partial ^j \partial ^i)\varOmega =4\pi \varepsilon ^{ijk}({\varvec{\delta }}_{\mathcal {C}})_k.$

Appendix E. Proofs of Eqs. (74), (77), and (80)

The proof of Eq. (74) is based on a proof given by Gelca [41]. Similar proofs for Eqs. (77) and (80) will be given. Our general strategy is as follows: we will apply topological transformations to the linking number l (i.e. deformations of the associated curves in l) and show that these transformations leave the linking number invariant.

Proof of Eq. (74). Consider the linking number of the curves ${{\mathbb {C}}}$ and $\mathcal {C}$

$$\begin{aligned} \frac{1}{4\pi }\oint _{{{\mathbb {C}}}}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l({{\mathbb {C}}},\mathcal {C}). \end{aligned}$$

(E1)

Let C be a closed path encircling the curve $\mathcal {C}$ and let us deform the path C into the path $C'$ via the transformation $C \rightarrow C'$ and let ${{\mathbb {C}}}=C\cup (-C')$ be the union of C and $(-C')$ which bounds the surface ${{\mathbb {S}}}$ traced by C while being deformed into $C'$. Accordingly, ${{\mathbb {C}}}=C\cup (-C')=\partial {{\mathbb {S}}}$ where $\partial {{\mathbb {S}}}$ is the boundary of ${{\mathbb {S}}}$. We also assume C and $C'$ encircle the same number of times $\mathcal {C}$. Using the properties $\oint _{{{\mathbb {C}}}\,=\,C\,\cup \,(-C')} = \oint _{C} + \oint _{-C'}$ and $\oint _{-C'} = -\oint _{C'}$ it follows that Eq. (E1) can be decomposed as

$$\begin{aligned} \frac{1}{4\pi }\oint _{{{\mathbb {C}}}}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=\frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3} - \frac{1}{4\pi }\oint _{C'}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}, \end{aligned}$$

(E2)

or equivalently,

$$\begin{aligned} l({{\mathbb {C}}},\mathcal {C})= l(C,\mathcal {C}) - l(C',\mathcal {C}), \end{aligned}$$

(E3)

where

$$\begin{aligned} \frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C,\mathcal {C}), \quad \frac{1}{4\pi }\oint _{C'}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C',\mathcal {C}), \end{aligned}$$

(E4)

are the linking numbers of C and $\mathcal {C}$, and $C'$ and $\mathcal {C}$, respectively. Therefore if $l({{\mathbb {C}}},\mathcal {C})=0$ then $l(C,\mathcal {C})=l(C',\mathcal {C})$ and this would proof Eq. (74). In Appendix C we demonstrated Eq. (C8) which can be re-arranged to obtain the relation

$$\begin{aligned} \oint _{\mathcal {C}}\frac{({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|^3}= \nabla \varOmega _0(\mathcal {S}) + 4\pi {\varvec{\delta }}_{\mathcal {S}}, \end{aligned}$$

(E5)

where $\varOmega _0(\mathcal {S}) = \int _{\mathcal {S}}\{({\varvec{x}}' - {\varvec{x}})\cdot d {\varvec{S}}'/|{\varvec{x}}- {\varvec{x}}'|^3 \}$ is the single-valued solid angle subtended by $\mathcal {C}$ and ${\varvec{\delta }}_{\mathcal {S}}=\int _{\mathcal {S}}\delta ({\varvec{x}}- {\varvec{x}}')d {\varvec{S}}'$ is the surface vector Dirac delta defined along the surface $\mathcal {S}$ bounded by $\mathcal {C}.$ Using Eq. (E5) in the left-hand side of Eq. (E1) and $[({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}']\cdot d {\varvec{x}}=({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')$ it follows

$$\begin{aligned} l({{\mathbb {C}}},\mathcal {C})= \frac{1}{4\pi }\oint _{{{\mathbb {C}}}}\nabla \varOmega _0(\mathcal {S})\cdot d{\varvec{x}} + \oint _{{{\mathbb {C}}}}{\varvec{\delta }}_{\mathcal {S}}\cdot d {\varvec{x}}. \end{aligned}$$

(E6)

The first line integral in the right-hand side vanishes because $\varOmega _0(\mathcal {C})$ is a single-valued function. Applying the Stokes theorem to the second line integral in the right-hand side, we obtain

$$\begin{aligned} l({{\mathbb {C}}},\mathcal {C})= \int _{{{\mathbb {S}}}}{\varvec{\delta }}_{\mathcal {C}}\cdot d {\varvec{S}}, \end{aligned}$$

(E7)

where ${{\mathbb {S}}}$ is the surface bounded by ${{\mathbb {C}}}$ and we have used Eq. (12): $\nabla \times {\varvec{\delta }}_{\mathcal {S}} = {\varvec{\delta }}_{\mathcal {C}}$ where ${\varvec{\delta }}_{\mathcal {C}}= \oint _{\mathcal {C}}\delta ({\varvec{x}}- {\varvec{x}}')d {\varvec{x}}'$ is a line vector Dirac delta defined along $\mathcal {C}.$ The surface ${{\mathbb {S}}}$ corresponds to the surface traced by the path C while being deformed into the path $C'$ and therefore the curve $\mathcal {C}$ never crosses the surface ${{\mathbb {S}}}$. Accordingly, the function ${\varvec{\delta }}_{\mathcal {C}}$ vanishes along the surface ${{\mathbb {S}}}$ and therefore $\int _{{{\mathbb {S}}}}{\varvec{\delta }}_{\mathcal {C}}\cdot d {\varvec{S}}=0$ which gives $l({{\mathbb {C}}},\mathcal {C})=0$. This result and Eq. (E3) imply $l(C,\mathcal {C})=l(C',\mathcal {C})$ and this proves Eq. (74).

Proof of Eq. (77). Consider the linking number of the curves C and ${{\mathcal {C}}}$

$$\begin{aligned} \frac{1}{4\pi }\oint _{C}\oint _{{{\mathcal {C}}}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C,{{\mathcal {C}}}). \end{aligned}$$

(E8)

Let C be a closed path encircling the curve $\mathcal {C}$. Let us deform the curve $\mathcal {C}$ into the curve $\mathcal {C}'$ via the transformation $\mathcal {C} \rightarrow \mathcal {C}'$ and let ${{\mathcal {C}}}=\mathcal {C}\cup (-\mathcal {C'})$ be the union of $\mathcal {C}$ and $(-\mathcal {C'})$ which bounds the surface ${{\mathcal {S}}}$ traced by $\mathcal {C}$ while being deformed into $\mathcal {C}'$. It follows that ${{\mathcal {C}}}=\mathcal {C}\cup (-\mathcal {C'})=\partial {{\mathcal {S}}}$ where $\partial {{\mathcal {S}}}$ is the boundary of ${{\mathcal {S}}}$. We assume C encircles the same number of times $\mathcal {C}$ and $\mathcal {C}'$. Using the properties $\oint _{{{\mathcal {C}}}\,=\,\mathcal {C}\,\cup \,(-\mathcal {C}')} = \oint _{\mathcal {C}} + \oint _{-\mathcal {C}'}$ it follows that Eq. (E8) can be decomposed as

$$\begin{aligned} \frac{1}{4\pi }\oint _{C}\oint _{{{\mathcal {C}}}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=\frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3} - \frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}'}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}, \end{aligned}$$

(E9)

or equivalently,

$$\begin{aligned} l(C,{{\mathcal {C}}})= l(C,{\mathcal {C}})- l(C,\mathcal {C}'), \end{aligned}$$

(E10)

where

$$\begin{aligned} \frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C,\mathcal {C}), \quad \frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}'}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C,\mathcal {C}'), \end{aligned}$$

(E11)

are the linking numbers of C and $\mathcal {C}$, and C and $\mathcal {C}'$, respectively. Therefore if $l(C, {{\mathcal {C}}})=0$ then $l(C,\mathcal {C})=l(C,\mathcal {C}')$ and this would prove Eq. (77). Following the same line of arguments that led to Eq. (C8), it follows that we can make the replacement $\mathcal {S}\rightarrow {\mathcal {S}}$ in Eq. (C8) and obtain

$$\begin{aligned} \oint _{{\mathcal {C}}}\frac{({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}'}{|{\varvec{x}}- {\varvec{x}}'|^3}= \nabla \varOmega _0({{\mathcal {S}}}) + 4\pi {\varvec{\delta }}_{{\mathcal {S}}}, \end{aligned}$$

(E12)

where $\varOmega _0({{\mathcal {S}}}) = \int _{{\mathcal {S}}}\{({\varvec{x}}' - {\varvec{x}})\cdot d {\varvec{S}}'/|{\varvec{x}}- {\varvec{x}}'|^3 \}$ is the single-valued solid angle function subtended by the curve ${\mathcal {C}}$ and ${\varvec{\delta }}_{{\mathcal {S}}}=\int _{{\mathcal {S}}}\delta ({\varvec{x}}- {\varvec{x}}')d {\varvec{S}}'$ is the surface vector Dirac delta defined along the surface ${\mathcal {S}}$ bounded by ${\mathcal {C}}.$ Using Eq. (E12) and the relation $[({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}']\cdot d {\varvec{x}}=({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')$ we obtain

$$\begin{aligned} l(C,{{\mathcal {C}}})= \frac{1}{4\pi }\oint _{C}\nabla \varOmega _0({{\mathcal {S}}})\cdot d {\varvec{x}} + \oint _{C}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}. \end{aligned}$$

(E13)

The first line integral on the right-hand side vanishes because $\varOmega _0({\mathcal {S}})$ is single-valued. On the other hand, the surface ${\mathcal {S}}$ corresponds to the surface traced by the curve $\mathcal {C}$ while being deformed into the curve $\mathcal {C}'$ and therefore the path C never crosses the surface ${\mathcal {S}}$. Consequently, the function ${\varvec{\delta }}_{{\mathcal {S}}}$ vanishes along the path C so that $\oint _{C}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}=0$ and this gives $l(C,{{\mathcal {C}}})=0.$ This result and the right-hand side of Eq. (E10) imply $l(C,{\mathcal {C}})=l(C,\mathcal {C}')$, result that proves Eq. (77).

Proof of Eq. (80). Consider the linking number of the curves ${\mathbb {C}}$ and ${\mathcal {C}}$

$$\begin{aligned} \frac{1}{4\pi }\oint _{{{\mathbb {C}}}}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l({{\mathbb {C}}},{\mathcal {C}}). \end{aligned}$$

(E14)

Let C be a closed path encircling the curve $\mathcal {C}$. Let us simultaneously deform the path C into the path $C'$ via the transformation $C \rightarrow C'$ and deform the curve $\mathcal {C}$ into the curve $\mathcal {C}'$ via the transformation $\mathcal {C}\rightarrow \mathcal {C}'.$ Let ${{\mathbb {C}}}=C\cup (-C')$ be the union of C and $(-C')$ which bounds the surface ${{\mathbb {S}}}$ traced by C while being deformed into $C'$ and let ${{\mathcal {C}}}=\mathcal {C}\cup (-\mathcal {C'})$ be the union of $\mathcal {C}$ and $(-\mathcal {C'})$ which bounds the surface ${{\mathcal {S}}}$ traced by $\mathcal {C}$ while being deformed into $\mathcal {C}'$. Accordingly, ${{\mathbb {C}}}=C\cup (-C')=\partial {{\mathbb {S}}}$ where $\partial {{\mathbb {S}}}$ is the boundary of ${{\mathbb {S}}}$ and ${{\mathcal {C}}}=\mathcal {C}\cup (-\mathcal {C'})=\partial {{\mathcal {S}}}$ where $\partial {{\mathcal {S}}}$ is the boundary of ${{\mathcal {S}}}$. We assume the path C encircles $\mathcal {C}$ and $\mathcal {C}'$ the same number of times the path $C'$ encircles $\mathcal {C}$ and $\mathcal {C}'$. Using $\oint _{{{\mathbb {C}}}\,=\,C\,\cup \,(-C')}\oint _{{\mathcal {C}}\,=\,\mathcal {C}\,\cup \,(-\mathcal {C}')}=(\oint _{C} - \oint _{C'})(\oint _{\mathcal {C}} - \oint _{\mathcal {C}'})=\oint _{C}\oint _{\mathcal {C}} - \oint _{C}\oint _{\mathcal {C}'} -\oint _{C'}\oint _{\mathcal {C}} + \oint _{C'}\oint _{\mathcal {C}'}$, $\oint _{{{\mathbb {C}}}\,=\,C\,\cup \,(-C')}=\oint _{C} + \oint _{-C'}$, $\oint _{-C'}=-\oint _{C'}$, $\oint _{{{\mathcal {C}}}\,=\,\mathcal {C}\,\cup \,(-\mathcal {C}')}=\oint _{\mathcal {C}} + \oint _{-\mathcal {C}}$, and $\oint _{-\mathcal {C}'}=-\oint _{\mathcal {C}'}$, we can decompose Eq. (E14) as

$$\begin{aligned} \frac{1}{4\pi }\oint _{{{\mathbb {C}}}}\oint _{{\mathcal {C}}}\frac{({\varvec{x}} \!-\! {\varvec{x}}')\!\cdot \! (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}= & {} \frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3} -\frac{1}{4\pi } \oint _{C}\oint _{\mathcal {C}'}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}\nonumber \\&-\frac{1}{4\pi }\!\oint _{C'}\oint _{\mathcal {C}}\frac{({\varvec{x}} \!-\! {\varvec{x}}')\!\cdot \! (d{\varvec{x}} \!\times \!d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3} \!+\! \frac{1}{4\pi }\!\oint _{C'}\oint _{\mathcal {C}'}\frac{({\varvec{x}} \!-\! {\varvec{x}}')\!\cdot \! (d{\varvec{x}} \!\times \! d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}, \qquad \end{aligned}$$

(E15)

or equivalently,

$$\begin{aligned} l({\mathbb {C}},{{\mathcal {C}}})= l(C,{\mathcal {C}})-l(C,{\mathcal {C}'})-l(C',{\mathcal {C}})+l(C',{\mathcal {C}'}), \end{aligned}$$

(E16)

where the corresponding linking numbers are defined by

$$\begin{aligned} \frac{1}{4\pi }\oint _{C}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C,\mathcal {C}),\quad \frac{1}{4\pi } \oint _{C}\oint _{\mathcal {C}'}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C,\mathcal {C}'), \end{aligned}$$

(E17)

$$\begin{aligned} \frac{1}{4\pi }\oint _{C'}\oint _{\mathcal {C}}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C',\mathcal {C}), \quad \frac{1}{4\pi }\oint _{C'}\oint _{\mathcal {C}'}\frac{({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')}{|{\varvec{x}}- {\varvec{x}}'|^3}=l(C',\mathcal {C}'). \end{aligned}$$

(E18)

Now, we have the result $l(C',\mathcal {C})=l(C,\mathcal {C})$ because of Eqs. (E3) and (E7) (which follows from the transformation $C\rightarrow C'$). Also, we have the result $=l(C,{\mathcal {C}})=l(C,{\mathcal {C}'})$ because of Eq. (E10) and (E13) (which follows from the transformation $\mathcal {C}\rightarrow \mathcal {C}'$). Using these results Eq. (E16) reduces to

$$\begin{aligned} l({\mathbb {C}},{{\mathcal {C}}})= l(C',{\mathcal {C}'})-l(C,{\mathcal {C}}). \end{aligned}$$

(E19)

Accordingly, if the left-hand side of Eq. (E19) vanishes then $l(C,{\mathcal {C}})=l(C',{\mathcal {C}'})$ and this would prove Eq. (80). Using Eq. (E12) together with $[({\varvec{x}}' - {\varvec{x}})\times d {\varvec{x}}']\cdot d {\varvec{x}}=({\varvec{x}} - {\varvec{x}}')\cdot (d{\varvec{x}} \times d{\varvec{x}}')$ we can write

$$\begin{aligned} l({\mathbb {C}},{{\mathcal {C}}})= \frac{1}{4\pi }\oint _{{{\mathbb {C}}}}\nabla \varOmega _0({{\mathcal {S}}})\cdot d {\varvec{x}} + \oint _{{{\mathbb {C}}}}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}. \end{aligned}$$

(E20)

The first line integral on the right-hand side vanishes because $\varOmega _0({\mathcal {S}})$ is single-valued. This result and the relations $\oint _{{{\mathbb {C}}}\,=\,C\,\cup \,(-C')}=\oint _{C} + \oint _{-C'}$ and $\oint _{-C'}=-\oint _{C'}$ yield $l({\mathbb {C}},{{\mathcal {C}}})=\oint _{C}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}-\oint _{C'}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}.$ The surface ${\mathcal {S}}$ corresponds to the surface traced by the curve $\mathcal {C}$ while being deformed into the curve $\mathcal {C}'.$ Accordingly, neither the path C nor the path $C'$ cross the surface ${\mathcal {S}}$ along which ${\varvec{\delta }}_{{\mathcal {S}}}$ is non-vanishing. Therefore $\oint _{C}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}=0$ and $\oint _{C'}{\varvec{\delta }}_{{\mathcal {S}}}\cdot d {\varvec{x}}=0$ which implies $l({\mathbb {C}}, {{\mathcal {C}}})=0.$ This result and Eq. (E20) give $l(C,{\mathcal {C}})=l(C',{\mathcal {C}'})$ and this proves Eq. (80).