Are the Maxwell's equations enough to derive the law of Coulomb?

Question

Are the 8 Maxwell's equations enough to derive the formula for the electromagnetic field created by a stationary point charge, which is the same as the law of Coulomb $$ F~=~k_e \frac{q_1q_2}{r^2}~? $$ If I am not mistaken, due to the fact that Maxwell's equations are differential equations, their general solution must contain arbitrary constants. Aren't some boundary conditions and initial conditions needed to have a unique solution. How is it possible to say without these conditions, that a stationary point charge does not generate magnetic field, and the electric scalar potential is equal to

$$\Phi(\mathbf{r})=\frac{e}{r}.$$

If the conditions are needed, what kind of conditions are they for the situation described above (the field of stationary point charge)?

Answer 1

The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law.

More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law $$ \nabla\cdot\mathbf{E}=\rho/\epsilon_0\quad\Leftrightarrow\quad \oint_S\mathbf{E}\cdot\mathrm{d}\mathbf{a}=Q/\epsilon_0, $$ where $Q$ is the total charge enclosed by the (arbitrary) surface $S$.

To derive Coulomb's law, consider the electric field of a single point particle, with nothing else in the universe. Because of isotropy (which must be added as an additional postulate), the electric field at a sphere of radius $r$ centred on the charge must be radial and with the same magnitude throughout. That means the integral is trivial and the electric field must be $$\mathbf{E}=\frac{Q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}.$$

Coupled with Lorentz's force law at zero velocity for the test particle (since Coulomb's law only holds in electrostatics) this yields Coulomb's law.

It is not obvious that this highly symmetric situation can give the general electrostatic force for multiple particles. This follows from the superposition principle, which is very much at the heart of classical electrodynamics, and which can be obtained from the linearity of Maxwell's equations. This gives you the field for a single source; add the fields for all the individual sources and you'll get the field for the collection of sources.

Answer 2

The exact derivation goes as follows. You start from Gauss' Law, integrate on both sides over some volume V:

$$ \stackrel{\tiny div}{\vec{\nabla}}\cdot\vec{\mathbf{E}}=\frac{1}{\epsilon_0}\rho \,\,\,\,\,\,\,\,\,\,\,\Big/\iiint\limits_V\,d^3\vec{r} $$ Then switch to integration over a closed surface, and also note that total charge inside this volume is Q: $$ \iiint\limits_V\stackrel{\tiny div}{\vec{\nabla}}\cdot\vec{\mathbf{E}}\,\,d^3\vec{r}=\oint\vec{\mathbf{E}}\cdot d\vec\sigma=\iiint\limits_V\frac{1}{\epsilon_0}\rho\,\,d^3\vec{r}=\frac{Q}{\epsilon_0} $$ Now you need to note that the volume of integration is quite arbitrary and so is the surface, so we will use a sphere. You can describe the integral over a sphere using: $$ \frac{Q}{\epsilon_0}=\oint\vec{\mathbf{E}}\cdot d\vec\sigma=\int\limits_{\phi=0}^{\phi=2\pi}\int\limits_{\theta=0}^{\theta=\pi}\mathbf{E}\hat{\vec{n}}\cdot\hat{\vec{n}}\,R\,d\phi\,R\,d\theta=4\pi R^2\mathbf{E}\,\,\,\,\,\,\Big/\frac{1}{4\pi R^2} $$ And so you obtain: $$ \mathbf{E}=\frac{Q}{4\pi \epsilon_0 R^2} $$ It should be: $$\vec{\mathbf{E}}=\frac{Q}{4\pi\epsilon_0 R^2}\hat{\mathbf{r}}$$ But I lost the normal vector along the way (I hope that someone can correct this and edit this post).

Now you use the Lorentz Force law (where $\vec{\mathbf{B}}=\vec 0$): $$ \vec{\mathbf{F}}_{lor}=q \vec{\mathbf{E}}+q \vec{\mathbf{V}}\times\vec{\mathbf{B}}=\frac{q\,Q}{4\pi\epsilon_0 R^2}\hat{\mathbf{r}} $$ And so you obtain the Coulomb force law.

Answer 3

If you are asking about the Coloumb's law for the electric fields, yes, which you can see others' answers.

If you are asking about the Coloumb's law for the electric force,

$$\text{NO!}$$

Maxwell equations do NOT tell you about how the force acting on the charges $q$ or currents $\vec{J}$.

Simply speaking, to FULLY understand classical E&M (i.e. one can determine the physics from a initial value problem to determine all its consequences - physics is all about to determine/predict the future), you need BOTH:

(1) Maxwell's equations

(2) Lorentz force law (Newtonian mechanics, E&M equivalence of Newtonian gravitation force.)

Punch line I: (1) and (2) are absolute different things.

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Lagrangian and variational principle E.O.M. viewpoint

However, if you start from a Lagrangian viewpoint, writing down the action: $$ S=\int (-\frac{1}{2} |f|^2 + A \wedge *J)=\int d^3xdt (-\frac{1}{4} f_{\mu\nu} f^{\mu\nu} + A_\mu J^\mu) $$ with the 2-form field strength $f_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ gives you E and M fields, you can determine Maxwell's equation from the equations of motions(E.O.M.) by doing the variation principle on the 1=form gauge field $A$. The source is a 1-form current $J=(\rho,\vec{J})$.

Maxwell's equations: E.O.M. respect to varying 1-form gauge field $A$

You obtain Maxwell equations by varying $A$: $$ d*F=J \;\;\;\text{-Gauss law for electricity, Maxwell-Ampere's law} $$ and $$ dF=d^2A=0 \;\;\;\text{-Gauss's law for magnetism, Maxwell–Faraday equation} $$

How about the Lorentz force law? You can do variation respect to the spatial coordinate $x^\mu=(t,x)$, and you need to specify which massive particle with mass $m$ experiencing the force $F$, which is $F=m \ddot{\vec{x}}$ by Newtonian mechanics. To specify massive particle in the Lagrangian/action, you just need to add its kinetic energy $\frac{1}{2}m \dot{\vec{x}}^2$.

Lorentz force law: E.O.M. respect to varying spacetime coordinates $x^\mu=(t,x)$

$$ S=\int d^3xdt (-\frac{1}{4} f_{\mu\nu} f^{\mu\nu} + A_\mu \wedge J^\mu+\frac{1}{2}m \dot{\vec{x}}^2) \to \int d^3xdt (+ q\Phi - q \dot{\vec{x}} \cdot \vec{A}+\frac{1}{2}m \dot{\vec{x}}^2) $$

you will derive Lorentz force law $$ m \ddot{\vec{x}}=q\vec{E}+q \dot{\vec{x}} \times \vec{B} $$

Punch line II: The action and Variational principles are very powerful to unite (1) Maxwell's equations and (2) Lorentz force law, in the same framework.

Answer 4

Given the Gauss law AND the Lorentz force, yes it's possible to derive Coulomb's law as it already been answered. So I think the question is if it's possible to derive it given ONLY the four Maxwell equations (and not the Lorentz force). The answer is still yes since the Lorentz force is equivalent to the Faraday law and it can be derive from it. The relation between Faraday law and Lorentz force is not trivial in 3D since the Faraday paradox arises (check wiki). On the other hand when electromagnetism is expressed in the covariant formulation no such paradox exists.