# Foundations of Maxwell equations

Suppose we have two particles with charge $$q_1$$, $$q_2$$ with velocity $$\mathbf{v}_1$$,$$\mathbf{v}_2$$ in positions $$\mathbf{r}_1,\,\mathbf{r}_2$$. We know that the particle 2 generates an electric field and a magnetic field and in position $$\mathbf{r}_1$$ are given by: $$\mathbf{E}_2 = kq_2 \frac{\mathbf{r}_1-\mathbf{r}_2}{||\mathbf{r}_1-\mathbf{r}_2||^3},\,\mathbf{B}_2 = \frac{\mu_0q_2}{4\pi} \frac{\mathbf{v}_2 \times (\mathbf{r}_1 -\mathbf{r}_2)}{||\mathbf{r}_1-\mathbf{r}_2||^3}\tag{01}$$ So by Lorentz force the particle 2 exerts a force on particle 1 given by $$\mathbf{F}_{21} = q_1(\mathbf{E}_2+\mathbf{v}_1\times \mathbf{B}_2)\tag{02}$$

Substituting the previous formula, we can obtain a formula depending only on $$q_1$$, $$q_2$$, $$\mathbf{v}_1$$,$$\mathbf{v}_2$$ and $$\mathbf{r}_1$$, $$\mathbf{r}_2$$. Is it possible to derive Maxwell's equations only from that formula? If I'm not wrong we can derive Gauss law for electric and magnetic field, Faraday-Neumann-Lenz and Ampere law. Maybe Maxwell law of current displacement is not possible. I think it is interesting if one can derive all classic electromagnetic field by a single formula like gravitational force.

Further I noted that the formula doesn’t obey to action and reaction that is $$\mathbf{F}_{21}$$ is not equal to $$-\mathbf{F}_{12}$$, maybe we must consider relativity or quantum field theory?

• I think you will run into some issues using electrostatic and magnetostatic equations for a scenario that is not electrostatic or magnetostatic. Oct 31, 2019 at 18:57
• In general, the fields generated by moving charges are found from the Liénard-Wiechert potential. You can substitute these fields into the Lorentz force law to obtain an equation of motion for the charges in terms of direct particle interaction. But you cannot derive Maxwell's equations from this method, since the Liénard-Wiechert formulas were derived from Maxwell's equations. Oct 31, 2019 at 19:14
• @TobErnack Maybe it is possible. The fact that the Liénard-Wiechert formulas were derived from Maxwell's equations doesn't necessary imply that Maxwell's equations can't be derived from Liénard-Wiechert formulas.
– asv
Nov 1, 2019 at 13:17
• You are right, I shouldn't have claimed that. But if you look at Maxwell's equations, they look simpler in form than the Liénard-Wiechert formulas, so it seems easier to discover Maxwell's equations first. Nov 1, 2019 at 14:33
• Concerning the apparent violation of Newton’s Third Law, this is explained via the concept of the fields themselves having momentum. See this question for details. Aug 25, 2021 at 13:43

The answer is definitely NO. The main reason is that from the very beginning your equations (01) for the electromagnetic field produced by an electric charge $$\,q\,$$ moving with velocity $$\,\mathbf v\,$$ are totally wrong even in the simple case of uniform translation ($$\mathbf v=\texttt{constant}$$).

Referring to my answer here Electric field associated with moving charge we note that the electromagnetic field produced by an arbitrarily moving electric charge $$\,q\,$$ is as follows (for convenience I copy herein the relevant equations)

\begin{align} \mathbf{E}(\mathbf{x},t) & \boldsymbol{=} \frac{q}{4\pi\epsilon_0}\left[\frac{(1\boldsymbol{-}\beta^2)(\mathbf{n}\boldsymbol{-}\boldsymbol{\beta})}{(1\boldsymbol{-} \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R^2} \right]_{\mathrm{ret}}\!\!\!\!\!\boldsymbol{+} \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\boldsymbol{\times}\left[(\mathbf{n}\boldsymbol{-}\boldsymbol{\beta})\boldsymbol{\times} \boldsymbol{\dot{\beta}}\right]}{(1\boldsymbol{-} \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{01.1}\label{eq01.1}\\ \mathbf{B}(\mathbf{x},t) & = \left[\mathbf{n}\boldsymbol{\times}\mathbf{E}\right]_{\mathrm{ret}} \tag{01.2}\label{eq01.2} \end{align} where \begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{02.1}\label{eq02.1}\\ \boldsymbol{\dot{\beta}} & = \dfrac{\boldsymbol{\dot{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{02.2}\label{eq02.2}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R} \tag{02.3}\label{eq02.3} \end{align} In equations \eqref{eq01.1},\eqref{eq01.2} all scalar and vector variables refer to the $$^{\prime}$$ret$$^{\prime}$$arded position and time.

Referring to my answer here Magnetic field due to a single moving charge we note that the electromagnetic field produced by an uniformly moving electric charge $$\,q\,$$ is as follows (for convenience I copy herein the relevant equations)

\begin{align} \mathbf{E}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right) & \boldsymbol{=}\dfrac{q}{4\pi \epsilon_{\bf 0}}\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}},\quad \beta\boldsymbol{=}\dfrac{\upsilon}{c} \tag{03a}\\ \mathbf{B}_{_{\mathbf{LW}}}\left(\mathbf{x},t\right) & \boldsymbol{=}\dfrac{1}{c^{ \bf 2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right)\vphantom{\dfrac{a}{\dfrac{}{}b}}\boldsymbol{=}\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\right)}{\left(1\!\boldsymbol{-}\!\beta^{\bf 2}\sin^{\bf 2}\!\phi\right)^{\boldsymbol{3/2}}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{{r}}}{\:\:\Vert\mathbf{r}\Vert^{\bf 3}} \tag{03b} \end{align} Equations (03) are relativistic. They come from the Lienard-Wiechert potentials.

I don’t provide an actual derivation of Maxwell from them, but the final two paragraphs outline how it might be accomplished from the other form (Jefimenko equations).

But even more in the spirit of the enquiry than that is just having electrodynamics from electrostatics to begin with: “If I'm not wrong we can derive [various Maxwell].. I think it is interesting if one can derive all classic electromagnetic field by a single formula..”

While we can’t get down to one formula, we can get them from two (unless we start with the single quantum-mechanical tensor that underlies both, but this answer is classical). In fact all of EM can be handled without fields, just with Coulomb and Bio-savat. And just as interestingly as getting them from two, is getting them as two. Maxwell’s equations have the fields on both sides of the equations.

Electric charges attract/repel. We can call the would-be force per unit-charge from them, the “electric field”. Currents also attract/repel. We can call the would-be force from a current on another current if it was present, the “magnetic field”.

But one must account for the fact that the effects of charges and of currents travel at the speed of light. Jefimenko’s equations are exactly this. (They are presented as calculating E and B with neither appearing on the RHS, but then the definition of fields as hypothetical forces could be used instead.)

Let’s start by clarifying a few basics about what Maxwell says and what is causal. Then mention and link to Jefimenko and discuss how it might derive Maxwell.

The four Maxwell Equations (five relationships) are best understood as:

1.Electric charges cause electric fields that converge/diverge at the charge: $$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}$$

(In other words, charges attract/repel: $$F_{1,2}= k_e \frac{q_1q_2 }{r^2}$$)

2.Currents cause magnetic fields that curl around the current: $$\underbrace{\nabla {\times} \vec{B} = \mu_0 \vec{J}}~\text{ } ~(+ \mu_0\varepsilon_0 \frac{\partial \vec{E}}{\partial t})$$

(In other words, currents attract/repel: $$F_{1,2}= \mu_0 \frac{\vec{I_1}\cdot\vec{I_2}L}{2\pi r}$$)

3.Magnetic field lines are always closed, with no sources or drains of field: $$\nabla \cdot \vec{B} =0$$

(Quite straightforward until here.)

4.The electric field curls around changes in the magnetic field: $$\nabla \times \vec{E} = \frac{-\partial \vec{B}}{\partial t}$$

This is a consequence, but during Maxwell is considered an additional relationship.

5.The magnetic field curls around changes in the electric field:

$$\underbrace{\nabla {\times} \vec{B} = \mu_0\varepsilon_0 \frac{\partial \vec{E}}{\partial t} }~\text{ } ~(+ \mu_0 \vec{J})$$

This is a consequence, but during Maxwell is considered an additional relationship.

From Jefimenko's equations we know that 4., 5. are not causal - not from the curl to the derivative nor vice versa. The terms are due to current variations affecting each field individually. Fields are tools. If fields (Maxwell) are used, the terms in 2 and 5 must both be included, even if they come from one object.

Time-varying application of this, Jefimenko Equations:

https://en.m.wikipedia.org/wiki/Jefimenko%27s_equations

We’re getting E and B at point r, time t, by integrating over every other point 𝑟′≠r, r in ℝ3, (as if r was (0,0,0) is one way to say it). We do require knowledge of the lagged state at r’ when the field left it (at c to travel ||r’-r||). So the integral is just a volume integral. On wikipeda they have ∫[...] 𝑑3𝑟′. in reality we might have ∫ [∫ (∫...𝑑𝑥) 𝑑𝑦] 𝑑𝑧 depending on functional form.

How would one go about showing Maxwell’s equations from Jefimenko? Starting with the differential forms of Maxwell as above, the first step would seem to be taking the appropriate curls, divergences, and time partials of the Jefimenko expressions and putting them in. Only $$\vec{J}$$ at $$r$$ remains, but as a primitive it can be used directly. The mentioned ∫ [∫ (∫...𝑑𝑥) 𝑑𝑦] 𝑑𝑧 form would make getting an initial expression for the $$\nabla$$ forms simple. The time partials will give second order derivatives of current and density, and for that perhaps again differentiating the Ampère–Maxwell equation and Gauss’ Law, respectively, will be needed.

If both theories are from the same foundations, the result will be true just mathematically, independent of physics. Of course, showing that is what would be a derivation of Maxwell. But any equation that arises while trying is true about electromagnetism, regardless of the theories’ foundations.

• I don't see how this answers the question: can Maxwell's equation be derived from the "fourth" equation (that the OP has not written) in the post. Aug 25, 2021 at 11:34
• Youre right. I took the question to be more than can we derive maxwell: “If I'm not wrong we can derive Gauss law for electric and magnetic field, Farady-Neumann-Lenz and Ampere law. Maybe Maxwell law of current displacement is not possible. I think it is interesting if one can derive all classic electromagnetic field by a single formula” But edited my answer to say what it addresses. Aug 25, 2021 at 13:13
• @garyp Ok I addressed deriving maxwell also. Mostly at very bottom. And said why the other Aug 25, 2021 at 13:45

Is it possible to derive Maxwell's laws ... only from that formula ... I think it is interesting if we can derive all classic electromagnetic field by a single formula ...

Hehl & Obulov provide a derivation of Maxwells equations starting only from relativistic charge conservation over any spacetime, this being a Lorentzian manifold. This principle is expressed in a 'single formula'. The usually equations are found after imposing a time field. This is detailed in their book, Foundations of Electromagnetism.

In a short paper on the Arxiv, they detail a similar derivation but over space. Here, they require both electric charge and magnetic flux conservation. This can be deduced from the preceding, as relativstic charge conservation implies electric charge and magnetic flux conservation. But here, they consider only space as opposed to time and hence they require two independent principles.

Their equations do not require a metric, but in this case they require four fields. When a spatial metric is imposed, the fields are reduced to two and the usual Maxwell equations are recovered.

The latter derivation is sketched out as follows. It uses the technology of differential forms but I use the term 'cofield' as a synonym for differential form as it's more closely aligned with the terminology in physics, aka fields. Thus a k-cofield is a differential k-form. In particular, a scalar cofield is a $$0$$-cofield and a coscalar cofield is a $$0'$$-cofield where $$k':= m-k$$.

Now, let $$M$$ be a spatial manifold, by this we mean a Riemannian manifold describing space. We let $$m$$ be it's dimension.

Let $$\rho$$ be the charge codensity. We call this a codensity because we describe it by by a coscalar cofield rather than a scalar cofield. This is because we want to avoid the use of the metric; and the Hodge star, which involves the metric, is required to transform a charge density into a charge codensity and it is the codensity rather than the density we integrate over a body to find its total charge (in the presence of a metric, the density is enough).

Now choose any closed ball $$B$$ of space $$M$$, then the total charge contained by $$B$$ is:

$$Q:= \int_B \rho$$

Now, since $$\rho$$ is a top cofield, it's exterior differential vanishes and hence it is a closed cofield. By Poincare's lemma, it will have a local potential $$D$$ which we call the electrical excitation. Thus:

$$dD = \rho$$, locally

This is Gauss's electric law. Notice it is only true locally. Now, global charge conservation means that we have current $$j$$ which is a cofield of degree $$m-1$$ and it satisfies:

$$\partial_t \int_B *\rho + \int_{\partial B} j = 0$$

We can rewrite this using Stoke's theorem as:

$$\partial_t \int_B *\rho + \int_B dj = 0$$

Since this is true for any closed ball $$B$$, we obtain the local form of charge conservation:

$$\partial_t *\rho + dj = 0$$

Locally, this is:

$$\partial_t dD + dj = 0$$

And this is equal to:

$$d(\partial_t D + j)= 0$$

The term in the brackets is again a closed form, so we can again introduce locally a potential for this term and which we call the magnetic excitation $$H$$. So:

$$dH = \partial_t D + j$$

This gives the Ampere-Maxwell law.  Thus we have derived the inhomogeneous Maxwell's equations purely from electric charge conservation and without the use of any metric and all this over any smooth orientated manifold. However, to write it in traditional terms through curl and div, we would require a metric. First we introduce their cofield variants of div and curl:

$$cocurl = *d$$ and $$codiv = *d*$$

These are related to the traditional versions through the raising and lowering operators:

$$curl = \sharp \circ cocurl \circ \flat$$ and $$div = \sharp \circ codiv \circ \flat$$

And also the electric and magnetic cofields $$E$$ and $$B$$ are defined by what Hehl & Obukov call the constitutative equations:

$$D := \epsilon_0. *E$$  and $$\mu_0.H:= *B$$

Recall,  Gauss's electric law is $$dD = \rho$$, locally. So replacing the electric excitation $$D$$ with the electric field we obtain,  $$d*E=\rho/\epsilon_0$$ and then applying the Hodge star, we further get $$(*d*)E =*\rho/\epsilon_0$$. This is just $$codiv(E) = *\rho/\epsilon_0$$. Then introducing the raising and lowering operators, we get $$\sharp \circ codiv \circ \flat \circ \sharp(E) = \sharp(*\rho)/\epsilon_0$$. And this reduces to:

$$div (E^{\sharp}) = (*\rho)^{\sharp}/\epsilon_0$$

This is the traditional electric Gauss equation once we recognise as we introduced $$E$$ as an electric cofield, then $$E^{\sharp}$$ is the electric field. And as we introduced $$\rho$$ as the electric charge codensity which is a coscalar field, then $$*\rho$$ is the electric charge density which is a scalar field and as the raising operator, $$\sharp$$, does nothing on scalar cofields, ie they are equivalently scalar fields. Thus renaming $$E^{\sharp} \rightarrow E$$ and $$(*\rho)^{\sharp} \rightarrow \rho$$, we get the traditional equation:

$$div(E) = \rho/\epsilon_0$$, local electric Gauss law.

But unlike the traditional equation this is valid on any Riemannian manifold considered as the spatial manifold. Similarly, for the Ampere-Maxwell law.

Hehl & Obukhov derive the homogeneous equations from the conservation of magnetic flux. Again, this does not require a metric. Instead, the electric and magnetic cofields are introduced independently and the conservation of magnetic flux is used to derive Gauss's magnetic law and Faraday's law. To find the traditional forms again we apply the constitutive laws as above.