Maxwell equations in Feynman's lectures

Question

I read the FLP vol. 2 and in the chapter 1, section 1.4 Feynman states from the physical point of view to form the intuition

$$c^2(\text{circulation of B around the curve} \ C)=\frac{d}{dt}(\text{Flux of E through the surface}\ S)+\frac{\text{Flux of electric current through}\ S}{\varepsilon_0}$$

and then we get the law:

$$c^2\nabla\times B=\frac{\partial E}{\partial t}+\frac{j}{\varepsilon_0}$$

I am confused because I thought that $\frac{d}{dt}(\text{Flux of E through the surface}\ S)$ should be equal to $\frac{\partial}{\partial t}(\nabla\cdot E)$ since it was stated before that the flux through the surface is represented by the dot product. Also why it is not $\frac{\nabla\cdot j}{\varepsilon_0}$ in the second term? Could you clarify please my doubts?

Answer 1

The flux of the electric field through a surface $\mathcal{S}$ is defined as:

\begin{equation} \Phi_E=\int_{\mathcal{S}}\boldsymbol{E}\cdot d\boldsymbol{a} \end{equation}

Thus \begin{equation} c^2\oint_{\Gamma}\boldsymbol{B}\cdot d\boldsymbol{l}=\frac{d}{dt}\int_{\mathcal{S}}\boldsymbol{E}\cdot d\boldsymbol{a}+\frac{1}{\epsilon_0}\int_{\mathcal{S}}\boldsymbol{J}\cdot d\boldsymbol{a} \tag{1} \end{equation} Apply Stokes theorem to LHS and taking the derivative inside the integral in RHS \begin{equation} c^2\int_{\mathcal{S}}(\nabla\times\boldsymbol{B})\cdot d\boldsymbol{a}=\int_{\mathcal{S}}\frac{\partial\boldsymbol{E}}{\partial t}\cdot d\boldsymbol{a}+\frac{1}{\epsilon_0}\int_{\mathcal{S}}\boldsymbol{J}\cdot d\boldsymbol{a} \end{equation} Since this must be valid for any surface $\mathcal{S}$, the integrands in LHS ans RHS must be equal and thus we get:

\begin{equation} c^2\nabla\times\boldsymbol{B}=\frac{\partial\boldsymbol{E}}{\partial t}+\frac{\boldsymbol{J}}{\epsilon_0} \tag{2} \end{equation}

Equations $(1)$ and $(2)$ both represent Ampére law with the term added by Maxwell (if you are wondering why it is named so, it is discussed in chapter 18 of FLP II), the former is the integral form, while the latter is the differential form of the same law. The difference between the two is that $(1)$ gives information about a specific path, while $(2)$ gives local information.

In chapter 2 and 3 of FLP II, you'll be introduced to some of the main rules of differential and integral calculus of vector fields, if you're not acquainted with it yet.

Answer 2

You correctly write the equation in the integral form as

$$c^2(\text{circulation of B around the curve} \ C)=\frac{d}{dt}(\text{Flux of E through the surface}\ S)+\frac{\text{Flux of electric current through}\ S}{\varepsilon_0}$$

which written as formula is $$c^2\oint_{C=\partial S} \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \frac{\mathrm{d}}{\mathrm{d}t} \iint_S \mathbf{E}\cdot\mathrm{d}\mathbf{S} + \frac{1}{\varepsilon_0}\iint_S \mathbf{J}\cdot\mathrm{d}\mathbf{S}$$

Note that the flux of a vector field across a surface is just the integral of the dot product of the field with the normal surface element. There is no divergence operator!

Now the key to go to the usual differential form is to write all the terms in the previous equation as integrals over the same domain.

Things look pretty good on the right-hand side because both terms are integrals over a surface $S$. We need to fix only the term on the left-hand side of the equation: we want to convert the integral over the curve $C$ to an integral over the surface $S$ (note that the curve $C$ is the "border" of $S$).

Luckily there is a very useful theorem, called the Stokes' theorem, which tells you that (under some assumptions) you can substitute the integral of a vector field over a closed line, with the integral of the rotor of such vector field over a surface having that line as its boundary. In a formula

$$ \oint_{C=\partial S} \mathbf{A}\cdot \mathrm{d}\mathbf{l} = \iint_S \nabla \times \mathbf{A}\cdot\mathrm{d}\mathbf{S} $$

If we apply this theorem to the left-hand side of the equation we have

$$\oint_{C=\partial S} \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \iint_S \nabla \times \mathbf{B}\cdot\mathrm{d}\mathbf{S}$$

At this point, we are all set, because your initial integral equation becomes

$$c^2\iint_S \nabla \times \mathbf{B}\cdot\mathrm{d}\mathbf{S} = \frac{\mathrm{d}}{\mathrm{d}t} \iint_S \mathbf{E}\cdot\mathrm{d}\mathbf{S} + \frac{1}{\varepsilon_0}\iint_S \mathbf{J}\cdot\mathrm{d}\mathbf{S} $$

Allow me to place the time derivative inside the integral (disclaimer mathematicians hate physicists for doing this trick all the without checking that all the hypotheses are fulfilled) so to have

$$\iint_S c^2\nabla \times \mathbf{B}\cdot\mathrm{d}\mathbf{S} = \iint_S \frac{\mathrm{d}\mathbf{E}}{\mathrm{d}t}\cdot\mathrm{d}\mathbf{S} + \frac{1}{\varepsilon_0}\iint_S \mathbf{J}\cdot\mathrm{d}\mathbf{S} $$

or

$$\iint_S \underbrace{\left[c^2\nabla \times \mathbf{B} - \frac{\mathrm{d}\mathbf{E}}{\mathrm{d}t} - \frac{\mathbf{J}}{\varepsilon_0}\right]}_0\cdot\mathrm{d}\mathbf{S} = 0 $$

Now the integral equation is verified over any surface $S$ only if the integrand is always zero, hence

$$c^2\nabla \times \mathbf{B} - \frac{\mathrm{d}\mathbf{E}}{\mathrm{d}t} - \frac{\mathbf{J}}{\varepsilon_0}=0$$ which is indeed what you wanted to prove!

Answer 3

An intuitive way to understand this is:

Let us find the curl of $B$ , as $\displaystyle \vec{B} = \mu_{0} \vec{v} \times \epsilon_{0}\vec{E} \ $ then assuming constant velocity,

$$\displaystyle \nabla \times \vec{B} = \mu_{0} \epsilon_{0} \nabla \times \vec{v} \times \vec{E} = (\nabla\cdot \vec{E}) \vec{v} $$

Now in current free space, $\displaystyle (\nabla\cdot \vec{E}) \vec{v} = \partial_{x^{i}}E_{i} \ \partial_{t}\vec{x^{i}} = \partial_{t}\vec{E}. $

That is how we get the result (in current free space) $\displaystyle \nabla \times \vec{B} = \mu_{0} \epsilon_{0} \partial_{t}\vec{E} $ .

The second case is when current exist : $\displaystyle (\nabla\cdot \vec{E}) \vec{v} = \frac {\rho}{\epsilon_{0} }\vec{v} = \frac {1}{\epsilon_{0}}\vec{J}. $