# Helmholtz theorem. Scalar and vector potentials

## Electromagnetic theory, lecture — III.

### Helmholtz theorem. Vector and Scalar Potentials.

This lecture, the web version of the third lecture given in the electromagnetic theory paper of the physics honors degree class, was delivered on 9th January 2018.

All electromagnetic theory lectures of this series, will be found here.

Topics covered

A. Formalism of electrodynamics — fundamental theorem

B. Application of Helmholtz theorem — to electrostatics

C. Application of Helmholtz theorem — to magnetostatics

### Formalism of electrodynamics

#### Helmholtz theorem

In our previous lectures, lecture – I and lecture — II, we discussed the Maxwell’s equations in free space and in material media. We also discussed the boundary conditions which helps us solve for the $\vec{E}$ and $\vec{B}$ fields.

The $\vec{E}$ and $\vec{B}$ fields are vector fields and Maxwell equations involve differentiation of these vectors. So we have to deal with curl $\vec{\nabla}\times$ and divergence $\vec{\nabla}\cdot$, of these vectors — or vector fields, apart from any boundary conditions to be satisfied by these, as we saw in our last lecture linked.

The Helmholtz theorems provides some useful relations and properties which helps us determine the $\vec{E}$ and $\vec{B}$ fields uniquely. It is  important for us to understand the mathematical rigor and the physical vigor these equations provide in having an unified formal treatment that all of electrodynamics espouses.

One needs a basic proficiency in vector calculus to be able to grasp the inherent significance of these methods. To begin with one can go through the slides herevector calculus,  to have a reasonable hold on the vector calculus concepts.

There is a theorem which deals with the implications of curl of a vector.

#### Theorem 1

If curl of a vector field is zero — everywhere, then $\vec{F}$ can be written as the gradient $\vec{\nabla}$ of a scalar field, also known as the scalar potential function $V$.

$\boxed{\vec{\nabla}\times\vec{F}=0\hspace{15pt}\Leftrightarrow \hspace{15pt}\vec{F}=-\vec{\nabla}V}$

Here the minus sign is just a convention.

a. Such fields, — $\vec{F}$ here, are known as curl-less or “irrotational” fields.

b. The curl is zero everywhere. I.e. $\vec{\nabla}\times\vec{F}=0$

c. The quantity $\int\limits_a^b \vec{F}\cdot d\vec{l}$ is independent of path — of integration, as long as limits $a$ and $b$ are fixed.

d. The quantity $\oint \vec{F}\cdot d\vec{l}=0$ for closed paths or loops of integration.

e. The vector $\vec{F}$ can be written as the gradient of some scalar such that, $\vec{F} = - \vec{\nabla} V$.

f. If any one of the  above statement is valid, it ensures all the others to be valid. That implies $c \implies a,\,b,\,d$ and so on.

g. Scalar potential $V$ is not unique. Any function independent of position can be added to $V$ without changing $\vec{F}$

Now there is a theorem which deals with the implications of divergence of a vector.

#### Theorem 2

If divergence $\vec{\nabla}\cdot$, of a vector field $\vec{F}$ is zero — everywhere, then $\vec{F}$ can be expressed as the curl of a vector potential function $\vec{A}$.

$\boxed{\vec{\nabla}\cdot\vec{F}=0\hspace{15pt}\Leftrightarrow \hspace{15pt}\vec{F}=\vec{\nabla}\times \vec{A}}$

a. Such fields — $\vec{F}$ here, are known as divergence-less or “solenoidal” fields.

b. The divergence is zero everywhere. I.e. $\vec{\nabla}\cdot\vec{F}=0$.

c. The quantity $\int\vec{F}\cdot d\vec{a}$ — is independent of surface of integration, as long as the boundary of the surface — a line, is fixed.

d. The quantity $\oint \vec{F}\cdot d\vec{a}=0$ for closed surface of integration.

e. The vector $\vec{F}$ can be written as the curl of some vector such that, $\vec{F} = \vec{\nabla}\times \vec{A}$.

f. If any one of the  above statement is valid, it ensures all the others to be valid. That implies $b \implies a,\,c,\,d$ and so on.

g. Vector potential $\vec{A}$ is not unique as gradients of any scalar function is curl-less. Gradients of any scalar function $f$ can be added to the  vector $\vec{A}$ and this transformation would leave the vector $\vec{F}$ unaltered.

I.e. if $\vec{A}'=\vec{A} + \vec{\nabla}f$ both new potential $\vec{A}'$ and old potential $\vec{A}$ would give rise to the same $\vec{F}$.

There is another mathematical result which has implications for both of the above theorems and we will state it.

Any general vector irrespective of whether this vector is divergence-less or not, and whether this vector is curl-less or not, can always be written as the sum of two vectors,

a. one vector which is the curl of some vector and

b. another vector which is the divergence of some scalar function.

In symbols: $\boxed{\vec{F} = - \vec{\nabla} V + \vec{\nabla}\times \vec{A}}$.

### Application of the above theorems to 3 situations

We will study how the above theorems have a great deal of implication for almost all of the field of electromagnetic theory. These theorems are together known as Helmholtz theorems.

But we will study them under 3 situations — gradually moving up the ladder of generality, beginning with our familiar knowledge of the static electric and magnetic fields.

a. Electrostatics

b. Magnetostatics

and

c. Electrodynamics

#### Electrostatics

When electric charges are stationary the resulting electric fields lead to electrostatic behavior. Under these conditions, electric fields — $\vec{E}$, are always “irrotational“, i.e. curl-less, thus they satisfy conditions of — theorem 1, of Helmholtz theorem.

Because $\vec{\nabla}\times\vec{E}=0$, all the other conditions are satisfied and a potential function known as scalar potential ( $V$ ) can be defined.

$\boxed{\vec{\nabla}\times\vec{E}=0 \hspace{15pt}\Rightarrow \hspace{15pt}V(\vec{r}\,) = -\int\limits_{Ref}^{|\,\vec{r}\,|} \vec{E}\cdot d\vec{l}}$

Here Ref — for reference, on the lower limit of the integral sign stands for a particularly chosen location for the potential, from where all calculations are to be made. There are two standards for this idea to be implemented;

a. if charge distributions are finite, reference can be set to $|\vec{r}\,| = \infty$

b. if charge distributions are not finite, other means are to be explored to calculate $V(\vec{r}\,)$ E.g. one should not set $|\vec{r}\,| = \infty$ on the lower limit of integration, in place of Ref, for an infinitely long cylinder.

The above boxed equation which we can define because curl of the electrostatic field $\vec{E}$ is zero, is an integral form of the scalar potential, $V$. The differential form of definition of scalar potential is then;

$\boxed{\vec{E}=-\vec{\nabla} V}$

In the electrostatics conditions, the potential formulation serves a very useful role, as 1 scalar equation is to be solved in lieu of 3 vector equations — we do not have to bother about the 3 vector components, which makes the problem cumbersome.

The two Maxwell equations — one vector and one scalar, $\vec{\nabla}\cdot \vec{E} = \frac{\rho}{\epsilon_0}$ and $\vec{\nabla}\times\vec{E}=0$ which are 4 equations in total, give rise to the Poisson’s equation.

Poisson’s equation: $\boxed{\nabla^2 V = -\frac{\rho}{\epsilon_0}}$

When source term is absent i.e. $\rho = 0$, Poisson’s equation turns into Laplace equation.

Laplace equation$\boxed{\nabla^2 V = 0}$

Potential can now be calculated as:

$\boxed{V(\vec{r}\,)=\frac{1}{4\pi \epsilon_0} \int \frac{\rho(\vec{r}\,')}{r_s}d \tau '}$

where $r_s \equiv |\vec{r}_s|= |\vec{r}-\vec{r}\,'|$$\vec{r}_s$ is known as the separation vector. Now electric field can be calculated from $\vec{E}=-\vec{\nabla}V$ The expression is nothing but the Coulomb’s law.

#### Magnetostatics

Magnetostatics problems follow a similar logic. The magnetostatic nature is defined by stationary currents or steady-currents. That means charge densities that produce the field must not vary in time.

Mathematically: $\frac{\partial \rho}{\partial t} = 0$. When used in the equation of continuity: $\frac{\partial \rho}{\partial t} + \vec{\nabla}\cdot\vec{j}= 0$, this means: $\vec{\nabla}\cdot\vec{j}= 0$. Thus the condition for magnetostatics is given by: $\vec{\nabla}\cdot\vec{j}= 0$.

The second Maxwell equation is: $\vec{\nabla}\cdot\vec{B}= 0$, i.e. magnetic fields are divergence-less in all situations.

According to theorem 2 of Helmholtz theorem then, magnetic field can always be written as curl of a vector potential $\vec{A}$, i.e. $\vec{B} = \vec{\nabla} \times \vec{A}$.

$\vec{A}$ is known as vector potential or magnetic vector potential.

By Ampere’s law of Maxwell equations — i.e. $\vec{\nabla}\times\vec{B}= \mu_0 \vec{j}$, we have: $\vec{\nabla}\times\vec{B}= \vec{\nabla} \times (\vec{\nabla}\times \vec{A})= \vec{\nabla}(\vec{\nabla}\cdot\vec{A}) - \nabla ^2 \vec{A}=\mu_0 \vec{j}$.

We already discussed that the magnetic vector potential $\vec{A}$ is not unique. Because gradient of any scalar is curl-less.

By this freedom we can always find scalar $\lambda$, so that $\vec{\nabla}\cdot\vec{A}=0$. If original vector potential $\vec{A}$ is not divergence-less we can add $\vec{\nabla} \lambda$ to it, to make it divergence-less.

Thus, $\vec{A}\,'= \vec{A} + \vec{\nabla} \lambda$ and $\vec{\nabla}\cdot\vec{A}\,'=\vec{\nabla}\cdot\vec{A}+\nabla^2 \lambda$.

For $\vec{\nabla}\cdot\vec{A}\,'=0$ we have $-\vec{\nabla}\cdot\vec{A}=\nabla^2 \lambda$. This is nothing but Poisson’s equation, as we have seen.

Remember Poisson’s equation for electrostatics: $\nabla^2 V = - \frac{\rho}{\epsilon_0}$. So $\vec{\nabla}\cdot\vec{A}$ behaves like source: $\frac{\rho}{\epsilon_0}$.

As in the case of electrostatics case if $\vec{\nabla}\cdot\vec{A}\to 0 \hspace{10pt}at\,\infty$ we have a solution to the Poisson’s equation: $\nabla^2 \lambda = -\vec{\nabla}\cdot\vec{A}$.

This solution is given by:

$\boxed{\lambda = \frac{1}{4\pi} \int \frac{\vec{\nabla}\cdot \vec{A}}{r_s} d\tau'}$.

If the source term $\vec{\nabla}\cdot\vec{A}$ does not go to zero, ( at $\infty$ ) we have to use other means as in the case of electrostatics.

It is always possible to find $\lambda$, to make $\vec{A}$ divergence-less.

Thus Ampere’s law becomes: $\nabla^2\vec{A}=-\mu_0 \vec{j}$. This is also Poisson’s equation — actually 3 Poisson’s equations, because its a vector equation.

If $\vec{j}\to 0\,\hspace{10pt}at\, \infty$ the solution is:

$\boxed{\vec{A}(\vec{r}\,)=\frac{\mu_0}{4\pi} \int \frac{\vec{j}(\vec{r}\,')}{r_s}d\tau'}$

Thus for steady currents we have:

$\boxed{\vec{B} = \vec{\nabla}\times\vec{A}=\frac{\mu_0}{4\pi}I\int \frac{d\vec{l}\,' \times \hat{r}_s}{r_s^2}}$

We will discuss the 3rd situation of applicability of the Helmholtz theorems to the most general case of electrodynamics in the next lecture, — lecture — IV.

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