# Scalar and Vector Potential

I am a physics undergraduate student currently studying electromagnetics. I have previously studied electrostatics and magnetostatics yet the concept of scalar potential, $V$ and the vector potential, A have eluded me.

I understand Maxwell's equations and relevant formulas to calculate them in certain situations and how to go between these quantities and the E and B fields. But I do not understand them conceptually.

I would like to understand their meaning and purpose. If somebody has a good analogy to view these quantities and how they relate to the E and B fields this would be even better.

Please could somebody answer this question and if possible please avoid using too math heavy (some math is expected) approach, so that I can clearly read and understand the concepts presented.

• Analogy? You can represent $\bf E$ as the scalar gradient of $\phi$, the scalar potential. Want to do this for magnetic field? Well, its circulation is not always zero; but its divergence is. Then you can represent it as curl of another field $\bf A$, the vector potential. It's so simple as that and is crystal clear too. What is bothering you actually? – user36790 Apr 13 '16 at 6:58
• Related: physics.stackexchange.com/q/45796 and links therein. – user36790 Apr 13 '16 at 6:59
• Thank you for replying. As I said I am familiar with the equations that relate the quantities and I am familiar with their properties. What I would like to know is what does it mean to have a scalar and vector potential? – user105149 Apr 13 '16 at 7:00
• That link is very confusing – user105149 Apr 13 '16 at 7:02
• There is no "physical" significance to the potentials precisely because they are unphysical variables - they contain a certain redundancy that manifests in gauge symmetry, see this answer of mine and the link therein. – ACuriousMind Apr 13 '16 at 10:08

The vector potential has a divergence of zero; we can obtain some intuition by considering the geometry required by the divergence theorem: the volume integral of the divergence of the vector potential is zero for any volume, hence the total net flux through any surface is zero.

So given your specific conditions, you can imagine geometric boundaries and apply this flux rule. This, along with the other conditions, provides some insight into the situation.

The next step is to consider how the curl of the vector potential becomes the magnetic field.

First of all, we are fundamentally interested in E & B. But essentially, it comes down to the fact that only E and B are physically measurable and so $\phi$ and A are mainly only considered as mathematical constructs, but this isn't always true - they can be conceptualised.

I apologise that I cannot give you an immediate physical insight, but I can only first explain via the math and then show what it means.

By the helmholtz theorem, which is really a mathematical construct rather than a physical insight, shows that we can rewrite E & B as a combination of a vector potential & scalar potential.

The theorem reads that any vector field (which E & B are) can be written as:

$$\mathbf{ F} = - \boldsymbol{\nabla} \phi +\boldsymbol{\nabla} \times \mathbf{A}$$

So we can rewrite E and B as

$$\mathbf{ E} = - \boldsymbol{\nabla} \phi +\boldsymbol{\nabla} \times \mathbf{A}$$ $$\mathbf{ B} = - \boldsymbol{\nabla} \phi +\boldsymbol{\nabla} \times \mathbf{A}$$

But we know that in electrostatic situations the curl of the E field is zero. In this case the electric field is conservative and only determined by the gradient of the potential. So,

$$\mathbf{ E} = - \boldsymbol{\nabla} \phi$$

The fact that B =$\boldsymbol{\nabla} \times \mathbf{A}$ is a little bit more involved. We know that no magnetic monopoles exist so there can be no sinks or sources so

$$\boldsymbol{\nabla} \cdot \mathbf{B} = 0$$

There is also a second part of the Helmoltz theorem which gives that

$$\phi(\mathbf{r})=\frac{1}{4\pi}\int\int\int\frac{\boldsymbol{\nabla}\cdot\mathbf{B}}{|\mathbf{r-r'}|}\textrm{d}V'$$

and

$$\mathbf{A}(\mathbf{r})=\frac{1}{4\pi}\int\int\int\frac{\boldsymbol{\nabla}\times\mathbf{B}}{|\mathbf{r-r'}|}\textrm{d}V'$$

We can now see that $\phi(\mathbf{r})$ must be zero. Which means that

$$\mathbf{ B} = \boldsymbol{\nabla} \times \mathbf{A}$$

By working with $\phi$ and A it can greatly make our mathematical constructs simpler and make it easy for us to calculate the E and B fields. I can't convince you of this immediately, but hopefully you can see how the above expressions look simpler than the Biot Savart law, for instance.

The minus sign in our expression for E reflects the fact that positive charges move from high potential to low potential, so in the direction opposite the steepest increase in the potential function. This is purely conventional. Further we now that the E field is conservative. You can take any route from a to b and the value of the below integral is the same:

$$\phi(\mathbf{r})=-\int_O^r \mathbf{E}\cdot \textrm{d}\mathbf{l}$$

with O as the place where we say potential is taken to be 0. This might arise to some confusion as we don't define a route, but, because the E field is conservative, it won't matter. We can place the zero point whereever we want - we can even add a constant, because we are only interested in the derivative because the E field is the only thing physically measurable.

In most electrostatic situations it will be easier to calculate the electric potential than the E field directly. Note, if we do have moving charge, we'll have a B field from Maxwell-Amperes law. This means that the curl of the E field is no longer zero.

Do not confuse electric potential with potential energy. The electric potential has one particular value at each point in space, independent of whatever charge you might place there, and its gradient gives you the electric field (which is the force per unit charge, also independent of the actual amount of charge placed at that point). To determine the potential energy (derivation not given) U=q$\phi$

So, now to your physical interpretation for A. We know we can have a physical interpretation for the scalar potential, but is A anything special?

For most purposes it's fine to think of the vector potential as a convenient mathematical tool without any physical meaning, but it does have a physical interpretation. By inspection, we can see that A has units of momentum per charge, and we can think of A as the "momentum per unit charge" that's stored in the electromagnetic field. This is analogous to potential energy, but its a potential momentum.

Further, The vector potential is an important quantity in many areas of modern physics (superconductivity, Aharonov-Bohm effect, Josephson junctions, SQUIDS, etc.). For students of Lagrangian mechanics, the canonical momentum for a charged particle in an electromagnetic field is given by p = mv + qA. In the relativistic formulation of electrodynamics, A can be written as a 4-vector. As 4-vectors, both of these quantities transform between inertial reference frames according to the Lorentz transformations. Until you come across these situations, probably OK to think of it as nothing more than a mathematical convenience.

• No, you cannot think of the vector potential as the "momentum per unit charge stored" because the vector potential is not unique. I can add any gradient of a scalar function to it and the result is still a valid vector potential. – ACuriousMind Apr 13 '16 at 10:10
• I agree, like we add any function whose gradient is zero to the electric potential, we can add any function whose curl vanishes with no effect on B. We exploit this freedom to eliminate the divergence of A (divA=0). Therefor A does not admit a simple physical interpretation in terms of potential energy per unit charge. However, in some contexts it can be interpreted as momentum per unit charge. See M.D Semon and J.R Taylor Am .J Phys 64, 1361 (1996) where they conclude that qA can be viewed as potential momentum. – Tomi Apr 13 '16 at 11:10

The ${\bf E}$ and ${\bf B}$ fields are like a force field ${\bf F}({\bf x})$ - they physically push charged particles around. The potentials are like the potential energy function $V({\bf x})$ that you get by integrating the force field with respect to position. Just as you can calculate ${\bf F}({\bf x})$ by taking the (negative) gradient of $V({\bf x})$, you can calculate ${\bf E}$ and ${\bf B}$ by taking (slightly more complicated) first partial derivatives with respect to space and time. Just as shifting $V({\bf x})$ by an overall constant doesn't change anything physically important, because the constant drops out when you take the derivative, so do the E&M potentials contain (slightly more complicated) mathematical "pieces" that cancel out when you take the partial derivatives, so you can freely change those parts without affecting the physically important ${\bf E}$ and ${\bf B}$ fields.