Vector potential

In vector analysis, a vector potential is a vector field whose rotor is equal to a given vector field. It is similar to a scalar potential , which is defined as a scalar field whose gradient is equal to a given vector field.

Formally, if $\mathbf {v}$ ${\ displaystyle \ mathbf {v}}$ $\ mathbf {v}$ - vector field, vector potential is called a vector field $\mathbf {A}$ ${\ displaystyle \ mathbf {A}}$ $\ mathbf {A}$ such that

\mathbf {v} =\nabla \times \mathbf {A} .

{\ displaystyle \ mathbf {v} = \ nabla \ times \ mathbf {A}.}

{\ mathbf {v}} = \ nabla \ times {\ mathbf {A}}.

If a $\mathbf {A}$ ${\ displaystyle \ mathbf {A}}$ $\ mathbf {A}$ is the vector potential for the field $\mathbf {v}$ ${\ displaystyle \ mathbf {v}}$ $\ mathbf {v}$ then from identity

\nabla \cdot (\nabla \times \mathbf {A} )=0

{\ displaystyle \ nabla \ cdot (\ nabla \ times \ mathbf {A}) = 0}

\ nabla \ cdot (\ nabla \ times {\ mathbf {A}}) = 0

(rotor divergence is zero) follows

\nabla \cdot \mathbf {v} =\nabla \cdot (\nabla \times \mathbf {A} )=0,

{\ displaystyle \ nabla \ cdot \ mathbf {v} = \ nabla \ cdot (\ nabla \ times \ mathbf {A}) = 0,}

\ nabla \ cdot {\ mathbf {v}} = \ nabla \ cdot (\ nabla \ times {\ mathbf {A}}) = 0,

i.e $\mathbf {v}$ ${\ displaystyle \ mathbf {v}}$ $\ mathbf {v}$ should be a solenoidal vector field .

For any solenoidal vector field that satisfies certain conditions, there is a vector potential. In particular, its existence depends on the region in which the field is defined - in the case of a multiply connected region, the potential of the vortex field usually does not exist.

Theorem

Let be

\mathbf {v} :\mathbb {R} ^{3}\to \mathbb {R} ^{3}

{\ displaystyle \ mathbf {v}: \ mathbb {R} ^ {3} \ to \ mathbb {R} ^ {3}}

{\mathbf {v}}:{\mathbb R}^{3}\to {\mathbb R}^{3}

- twice continuously differentiable solenoidal vector field . Let's pretend that $\mathbf {v} \left(\mathbf {x} \right)$ ${\ displaystyle \ mathbf {v} \ left (\ mathbf {x} \ right)}$ $\mathbf {v} \left(\mathbf {x} \right)$ decreases fast enough when $\|\mathbf {x} \|\rightarrow \infty$ ${\ displaystyle \ | \ mathbf {x} \ | \ rightarrow \ infty}$ $\|\mathbf {x} \|\rightarrow \infty$ . Define

\mathbf {A} (\mathbf {x} )={\frac {1}{4\pi }}\nabla \times \int \limits _{\mathbb {R} ^{3}}{\frac {\mathbf {v} (\mathbf {y} )}{\left\|\mathbf {x} -\mathbf {y} \right\|}}\,d\mathbf {y} .

{\ displaystyle \ mathbf {A} (\ mathbf {x}) = {\ frac {1} {4 \ pi}} \ nabla \ times \ int \ limits _ {\ mathbb {R} ^ {3}} {\ frac {\ mathbf {v} (\ mathbf {y})} {\ left \ | \ mathbf {x} - \ mathbf {y} \ right \ |}} \, d \ mathbf {y}.}

{\mathbf {A}}({\mathbf {x}})={\frac {1}{4\pi }}\nabla \times \int \limits _{{{\mathbb R}^{3}}}{\frac {{\mathbf {v}}({\mathbf {y}})}{\left\|{\mathbf {x}}-{\mathbf {y}}\right\|}}\,d{\mathbf {y}}.

Then $\mathbf {A}$ ${\ displaystyle \ mathbf {A}}$ $\mathbf {A}$ is a vector potential for $\mathbf {v}$ ${\ displaystyle \ mathbf {v}}$ $\mathbf{v}$ , i.e

\nabla \times \mathbf {A} =\mathbf {v} .

{\ displaystyle \ nabla \ times \ mathbf {A} = \ mathbf {v}.}

\nabla \times {\mathbf {A}}={\mathbf {v}}.

A generalization of this theorem is the Helmholtz expansion , according to which any vector field can be represented as the sum of a solenoidal vector field and a vortex-free vector field .

The ambiguity of the choice of potential

The vector potential of a solenoidal vector field is determined ambiguously. If a $\mathbf {A}$ ${\ displaystyle \ mathbf {A}}$ is a vector potential for $\mathbf {v}$ ${\ displaystyle \ mathbf {v}}$ also it is

\mathbf {A} +\nabla m,

{\ displaystyle \ mathbf {A} + \ nabla m,}

Where $m$ ${\ displaystyle m}$ - any continuously differentiable scalar function. This is a consequence of the fact that the gradient rotor is zero.

In electrodynamics, this gives ambiguity in determining the potentials of the electromagnetic field and is solved by imposing an additional calibration condition on the potential.

Vector Potential in Physics

Maxwell Equations

One way to write Maxwell's equations is to formulate it in terms of vector and scalar potentials. Vector potential $\mathbf {A}$ ${\ displaystyle \ mathbf {A}}$ is introduced in such a way that

\mu _{0}\mathbf {H} =\mathbf {B} =\operatorname {rot} \mathbf {A}

{\ displaystyle \ mu _ {0} \ mathbf {H} = \ mathbf {B} = \ operatorname {rot} \ mathbf {A}}

(in SI system).

In this case, the equation $\operatorname {div} \mathbf {B} =0$ ${\ displaystyle \ operatorname {div} \ mathbf {B} = 0}$ satisfied automatically.

Expression substitution for $\mathbf {A}$ ${\ displaystyle \ mathbf {A}}$ at

\operatorname {rot} \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}

{\ displaystyle \ operatorname {rot} \ mathbf {E} = - {\ frac {\ partial \ mathbf {B}} {\ partial t}}}

leads to the equation

\operatorname {rot} \left(\mathbf {E} +{\frac {\partial \mathbf {A} }{\partial t}}\right)=0,

{\ displaystyle \ operatorname {rot} \ left (\ mathbf {E} + {\ frac {\ partial \ mathbf {A}} {\ partial t}} \ right) = 0,}

according to which, just as in electrostatics, a scalar potential is introduced. However now in $\mathbf {E}$ ${\ displaystyle \ mathbf {E}}$ both scalar and vector potential contribute:

\mathbf {E} =-\operatorname {grad} \;\varphi -{\frac {\partial \mathbf {A} }{\partial t}}.

{\ displaystyle \ mathbf {E} = - \ operatorname {grad} \; \ varphi - {\ frac {\ partial \ mathbf {A}} {\ partial t}}.}

From the equation $\operatorname {rot} \mathbf {H} =\mathbf {j} +{\frac {\partial \mathbf {D} }{\partial t}}$ ${\ displaystyle \ operatorname {rot} \ mathbf {H} = \ mathbf {j} + {\ frac {\ partial \ mathbf {D}} {\ partial t}}}$ should

\operatorname {rot} \;\operatorname {rot} \mathbf {A} =\mu _{0}\mathbf {j} +\epsilon _{0}\mu _{0}{\frac {\partial }{\partial t}}\left(-\operatorname {grad} \;\varphi -{\frac {\partial \mathbf {A} }{\partial t}}\right).

{\ displaystyle \ operatorname {rot} \; \ operatorname {rot} \ mathbf {A} = \ mu _ {0} \ mathbf {j} + \ epsilon _ {0} \ mu _ {0} {\ frac {\ partial} {\ partial t}} \ left (- \ operatorname {grad} \; \ varphi - {\ frac {\ partial \ mathbf {A}} {\ partial t}} \ right).}

Using equality $\operatorname {rot} \;\operatorname {rot} \mathbf {A} =\operatorname {grad} \;\operatorname {div} \mathbf {A} -\nabla ^{2}\mathbf {A}$ ${\ displaystyle \ operatorname {rot} \; \ operatorname {rot} \ mathbf {A} = \ operatorname {grad} \; \ operatorname {div} \ mathbf {A} - \ nabla ^ {2} \ mathbf {A} }$ , the equations for the vector and scalar potentials can be written as

\Delta \mathbf {A} -\operatorname {grad} \left(\operatorname {div} \mathbf {A} +{\frac {1}{c^{2}}}{\frac {\partial \varphi }{\partial t}}\right)-{\frac {1}{c^{2}}}{\frac {\partial ^{2}\mathbf {A} }{\partial t^{2}}}=-\mu _{0}\mathbf {j} ,

{\ displaystyle \ Delta \ mathbf {A} - \ operatorname {grad} \ left (\ operatorname {div} \ mathbf {A} + {\ frac {1} {c ^ {2}}} {\ frac {\ partial \ varphi} {\ partial t}} \ right) - {\ frac {1} {c ^ {2}}} {\ frac {\ partial ^ {2} \ mathbf {A}} {\ partial t ^ {2 }}} = - \ mu _ {0} \ mathbf {j},}

\Delta \varphi +{\frac {\partial }{\partial t}}\operatorname {div} \mathbf {A} =-{\frac {\rho }{\epsilon _{0}}}.

{\ displaystyle \ Delta \ varphi + {\ frac {\ partial} {\ partial t}} \ operatorname {div} \ mathbf {A} = - {\ frac {\ rho} {\ epsilon _ {0}}}. }

The physical meaning of the vector potential

In classical electrodynamics, the vector potential has often been interpreted as a quantity that does not have a direct physical meaning, formally introduced only for the convenience of calculations, although the vector potential comes in such a direct way in the structure of action for classical electrodynamics that this suggests its fundamental nature.

In quantum theory, this has a transparent physical meaning of the direct influence of the vector potential on the phase of the wave function of a particle moving in a magnetic field. Moreover, it was possible to carry out quantum experiments, which showed that the vector potential is accessible to measurement that is quite direct in a sense (at least, we are talking about the fact that the vector potential can affect the quantum particle in an observable manner even when the magnetic field strength in the regions accessible to the particle is everywhere equal to zero, that is, the magnetic field cannot affect the particle through tension, but only directly through the vector potential; see Aaronov-Bohm effect ).

Just as the scalar potential is associated with the concept of energy , the vector potential reveals a close connection with the concept of momentum . So, in the case of a quick shutdown of the magnetic field, the particle located in it receives an additional momentum qA.