Faraday's law, Lorentz force law and $3^{\rm rd}$ Maxwell Equation

I see a very serious conceptual problem in the way these three equations are related to one another, as it is usually shown in electrodynamics textbooks.

Usually one starts with experimental evidence: Faraday's law of induction

The electromotive force around a closed path is equal to the negative of the time rate of change of the magnetic flux enclosed by the path.

Briefly $$\mathcal{E} = - {d \over dt} {\large \Phi}_{\partial \Sigma}({\bf B})$$

$$\mathcal E$$ is the EMF and $${\large \Phi}_{\partial \Sigma} ({\bf B})$$ is the magnetic flux through any surface $$\Sigma$$ that has $$\partial \Sigma$$ as its boundary.

The next step is to show the mathematical necessity underlying the experiment in question, in order to give it an intrinsic truth. Proofs like this one show there are two equivalent (thanks Einstein) ways you can change the flux to induce the same EMF. Either:

• you move the magnetic source (so you're using the $$\rm 3^{rd}$$ Maxwell equation: $$\nabla\! \times \!\bf E = - {\partial {\bf B}\over \partial \rm t }$$) or
• you move the boundary $$\partial \Sigma$$ (here Lorentz force law kicks in: $${\bf F} = q {\bf v} \! \times \! {\bf B}$$)

Every time formulas or theorems appear in a proof, one expects them to be valid regardless of the particular proof in which they're used.

Now, we know that Lorentz force law is just the electric force seen from another system of reference (essentially it's Coulomb's law and Special Relativity; the basic idea is nicely shown here, in a simplified scenario).

But what is the origin of $$\nabla \! \times \! {\bf E} = - {\partial {\bf B} \over {\partial t}}$$ ? They say it comes from Faraday's law of induction obviously: \begin {align}\mathcal E \overset {\rm def}{=} \oint _{\partial \Sigma} {\bf E} \cdot d{\bf l} = \int_ {\Sigma} (\nabla \! \times \!{\bf E} ) \cdot {\hat {\bf n}}\ d\sigma= - {d \over dt} \int_{\Sigma} {\bf B} \cdot \hat {\bf n }\ d\sigma = \int_{\Sigma} \left ( - {\partial {\bf B} \over \partial t}\right) \cdot \hat {\bf n }\ d\sigma \end{align}

Isn't this logically incoherent?

• "Now, we know that Lorentz force law is just the electric force seen from another system of reference" - Is that true? Nov 18 '20 at 22:36
• @AlfredCentauri as far as I know, every magnetic field is produced by a stationary current somewhere (apart from em waves). And locally every current circuit is just a linear charge density moving at a constant velocity, so you can always find a specific frame of reference in which that bit of charge is at rest. The full expression is ${\bf F} = q ({\bf E} + {\bf v} \! \times \! {\bf B})$. Nov 18 '20 at 22:52
• @ric.san When there are multiple charges moving around in different directions, you cannot find a frame of reference in which all of them are at rest. So, you can't always see a magnetic field as an electric field by going to a different frame of reference. For example, the magnetic field produced by two charged particles moving in opposite directions doesn't transform to a pure electric field in any reference frame. Nov 18 '20 at 23:55
• Yes the differential equation comes from Faraday's law (inferred from experience) for stationary loops (the path of integration does not move). Why is that incoherent? Nov 19 '20 at 0:55

You can't derive physics from mathematical principles. You have to take certain things as experimental truths and go from there. Having said that, there are different experimental starting points that you can choose and arrive at the same true statements, i.e., you can show that different axioms are equivalent. Physicists, traditionally, like to focus on what are true statements rather than establishing the equivalence of different sets of axioms. However, sometimes it is fun to do so and even required for the sake of conceptual clarity in some cases. So, let's examine this case.

We will show that Faraday's law and Maxwell's $$3^{\text{rd}}$$ law are equivalent to each other.

• You can derive Faraday's law assuming the followings to hold true:

$$1$$. Gauss's law for magnetism, i.e., $$\nabla\cdot\vec{B}=0$$ which purely mathematically equivalent to $$\int\vec{B}\cdot d\vec{S}=0$$.
$$2$$. Lorentz force law, i.e., $$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$$.
3. Maxwell's $$3^{\text{rd}}$$ law, i.e., $$\nabla\times \vec{E}=-\frac{\partial\vec{B}}{\partial t}$$.

This has been done here as you already notice in your question.

• You can derive Maxwell's $$3^{\text{rd}}$$ law assuming the followings to hold true:

$$1$$. Gauss's law for magnetism, i.e., $$\nabla\cdot\vec{B}=0$$ which purely mathematically equivalent to $$\int\vec{B}\cdot d\vec{S}=0$$.
$$2$$. Lorentz force law, i.e., $$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$$.
We start the proof the same way the previously linked proof starts: \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{a} &= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} + \iint_{\Sigma} \mathbf{v}(\nabla \cdot \mathbf{B}) \cdot \mathrm{d}\mathbf{a} - \int_{\partial \Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\ \text{(Gauss's law for magnetism)}&= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell\\ \text{(Faraday's law)}\implies-\mathcal{E}&= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell\\ \text{(Lorentz force law)}\implies-\int_{\partial\Sigma}(\mathbf{E}+\mathbf{v}\times\mathbf{B})\cdot \mathrm{d}\boldsymbol\ell&= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell\\ \implies-\int_{\partial\Sigma}\mathbf{E}\cdot \mathrm{d}\boldsymbol\ell&= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}\\ \text{(Stoke's theorem)}\implies-\iint_{\Sigma}(\boldsymbol{\nabla}\times\mathbf{E})\cdot \mathrm{d}\mathbf{a}&= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}\\ \implies-\boldsymbol{\nabla}\times\mathbf{E}&=\dot{\mathbf{B}} \end{align*}
So, we have shown that Maxwell's $$3^{\text{rd}}$$ law implies Faraday's law and vice-versa and both proofs require assuming Gauss's law for magnetism and Lorentz force law. In other words, Maxwell's $$3^{\text{rd}}$$ law and Faraday's law are equivalent so long as Gauss's law for magnetism and Lorentz force law hold true.