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

Question

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?

Answer 1

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., $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$.
  3. Faraday's law.
  
  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.