Electrodynamics/Four-Vectors

Elecromagnetic Four-Potential

This page will introduce the Four-Potential, and the Four-Current notations, as well as the d'Alembertian, which is used when studying these topics under the theoretical framework of Special relativity. These constructs, while a little confusing for some people, are fundamental to the way in which modern physicists study electric and magnetic fields, and therefore are worth learning. These topics may be introduced a little early here, although this chapter will not be rigorous. Instead, we will provide some common results here, and explain them thoughtout the next few chapters.

Potentials, As They Stand

Electric Potential (V), and Magnetic Vector Potential (A) are given by:

$\mathbf {E} =-\nabla V$

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

Using Gauss's Law for Electricity:

$\nabla \cdot \mathbf {E} ={\frac {\rho }{\epsilon _{0}}}$

and then substituting E with the Gradient of V. We see that:

$\nabla \cdot \nabla V=-{\frac {\rho }{\epsilon _{0}}}$

or:

$\nabla ^{2}V=-{\frac {\rho }{\epsilon _{0}}}$

And from Coulomb's Law we see that:

$V={\frac {1}{4\pi \epsilon _{0}}}\int {\frac {\rho }{\tau }}{d\tau }$

Using a similar technique in Magnetostatics gives:

$\nabla ^{2}\mathbf {A} =-\mu _{0}\mathbf {J}$ and

$\mathbf {A} ={\frac {\mu _{0}}{4\pi }}\int {\frac {\mathbf {J} }{\tau }}d\tau$

Four Potentials and Four Currents

We define a vector called the Four Potential as:

$A^{\mu }=\left({\frac {V}{c}},\mathbf {A} \right)=(V,A_{x},A_{y},A_{z})$

And another called the Four Current, which instead of V has ρ, the charge density, and instead of A, has J the current density:

$J^{\mu }=(c\rho ,\mathbf {J} )=(\rho ,J_{x},J_{y},J_{z})$

Where in each case c is the speed of light.

Both the Four-Potential and the Four-Current are vectors with four scalar values. What each of these values represents will be made clear at a later point.

Bringing it Together

You may have noticed that the equations for A with J and V with ρ are pretty much the same, and that the Four Potential and Current contains only these. We can combine these into a single equation as such:

$\nabla ^{2}A^{\mu }=-\mu _{0}J^{\mu }$

and:

$A^{\mu }={\frac {\mu _{0}}{4\pi }}\int {\frac {J^{\mu }}{\tau }}d\tau$

Relativity

The rules of relativity state that a current now cannot produce a magnetic field at a distance instantaneously. The effects of the current may travel at the speed of light at the fastest, and the field may not change any faster then that. If you've never looked at Special relativity before, this may be a good time to do so.

The d'Alembertian

We obviously need some term which compensates for the fact that nothing travels faster than light. So far our Laplacian operator looks like this:

$\nabla ^{2}={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}$

But now we need to bring it into Minkowski Space (A way of describing Relativistic Space). We can notice that the Laplacian cannot be applied to a four vector, and that the Laplacian is not invariant under Lorentz Transformations. To correct this we use the d'Alembertian. We define the d'Alembertian as such:

$\Box =\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}$

The d'Alembertian reduces to the Laplacian (∇²) if we aren't worried about time dependence, or relativity. There are a number of different ways to denote the d'Alembertian, depending on what text you read. Here are a number of different methods that are used in common texts:

$\Box =\Box ^{2}=\nabla _{\mathbf {M} }=\partial _{i}\partial ^{i}=\partial ^{2}=\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}$

Note

This wikibook will use the $\Box$ notation for the d'Alembertian, for simplicity and ease of authoring.

We end up with our equation in terms of the d'Alembertian, as such:

$\Box A^{\mu }=-\mu _{0}J^{\mu }$

We also have to bring in a time term into the integral form of the equation. It becomes:

$A^{\mu }={\frac {\mu _{0}}{4\pi }}\int {\frac {J^{\mu }(r',t')}{|r-r'|}}d^{3}r'$