Maxwell Ampere law derivation

Question

I have seen a proof on the internet regarding the derivation of the maxwell ampere law in this link: Deriving the Ampère-Maxwell law and even though I am pretty much satisfied in the way he derives the equation at one step he is making something that i dont like . He almost proves the equation and reaches a spot where the rate of change of the electric field is described by the integral that is shown in the picture (I understand that by the final equation). I would like to know how can it be proven straight forward that the rate of change of the electric field is equal to that integral to complete the proof without using the sketchy way he uses to reach the final equation

$$\nabla\times\textbf{B} = \mu_0\textbf{J}(\textbf{r}) - \frac{\mu_0}{4\pi}\iiint_V\left[\textbf{J}(\textbf{r}^\prime)\cdot\nabla\right]\frac{(\textbf{r}-\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|^3}d^3\textbf{r}^\prime$$

Answer 1

We can do the math to put the second term on the RHS of equation 8 into a different form, but it does not mean that we will actually derive the Ampère–Maxwell equation.

The starting point was the Biot–Savart law, which is only valid for steady currents. With only knowledge of how steady currents generate magnetic fields, and pure mathematical reasoning, you cannot derive knowledge of how non-steady currents generate magnetic fields, which is present in the Ampère–Maxwell equation. What you can do is hypothesize a form for the Ampère–Maxwell equation, under the constraint that it must be consistent with the continuity equation and must reduce to the Biot–Savart law under the special case where current is steady. But that hypothesis must be confirmed by experimental evidence.

As such, the only thing we should expect about the term on the right-hand side of equation 8 is that it's consistent with the Biot–Savart law, which means it must vanish when the current is steady.

Explicitly, we may use the following "integration by parts" identity:

$$ \iiint_V (A \cdot \nabla) f \, \mathrm{d}^3x = \iint_{\partial V} fA \cdot \mathrm{d}a - \iiint_V f(\nabla \cdot A)\,\mathrm{d}^3x $$

Now

$$\iiint_V [J(r') \cdot \nabla] \frac{r - r'}{\|r - r'\|^3} \, \mathrm{d}^3 r' = - \iiint_V [J(r') \cdot \nabla'] \frac{r - r'}{\|r - r'\|^3} \, \mathrm{d}^3 r'$$

and by applying the integration by parts identity separately to each component of this latter expression, we obtain

$$-\frac{\mu_0}{4\pi} \iiint_V [J(r') \cdot \nabla] \frac{r - r'}{\|r - r'\|^3} \, \mathrm{d}^3 r' = \frac{\mu_0}{4\pi}\iint_{\partial V} \frac{r - r'}{\|r - r'\|^3} (J(r') \cdot \mathrm{d}a') - \left. \frac{\mu_0}{4\pi} \iiint_V \frac{r-r'}{\|r-r'\|^3}(\nabla' \cdot J(r')) \, \mathrm{d}^3 r' \right.$$

The first term on the right-hand side can be made to vanish by taking $V$ to be a large enough volume so that it encloses all currents. The second term can be simplified using the continuity equation so finally we obtain

$$ \nabla \times B = \mu_0 J + \frac{\mu_0}{4\pi} \iiint_V \frac{r-r'}{\|r-r'\|^3} \frac{\partial \rho(r')}{\partial t} \, \mathrm{d}^3 r' $$

Under some circumstances, the second term on the right-hand side is, in fact, equal to $\mu_0\epsilon_0 \, \partial E/\partial t$ (as we can recognize the integral as proportional to the time derivative of the electric field given by Coulomb's law) but this will not necessarily be the case in general. However, again, it doesn't really matter what this term is equal to in the non-steady current case, because the above equation is one that only holds when the currents are steady.

You could then proceed to hypothesize that maybe the correct equation in the presence of non-steady currents is $\nabla \times B = \mu_0 J + \mu_0\epsilon_0 \, \partial E/\partial t$. This is not really any different from hypothesizing this equation based on other means such as the "standard" one given in the blog post. It is not a derivation, it is just a way to form a hypothesis, which will need to be validated by experiment.

Answer 2

$\let\vec\mathbf$ I think there is a clearer way to get the result. I would start from the third equation above his equation (8), that is, \begin{equation}\label{e1}\tag{1} \nabla\left(\nabla\cdot\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right) = \nabla\left(J(\vec{r}^\prime).\nabla\left(\frac{1}{|\vec{r} - \vec{r}^\prime|}\right)\right) = -\nabla\left(J(\vec{r}^\prime).\nabla^\prime\left(\frac{1}{|\vec{r} - \vec{r}^\prime|}\right)\right), \end{equation} where we used the fact that $\nabla^\prime = -\nabla$. Now use the identity \begin{equation}\label{e2}\tag{2} \nabla^\prime\cdot\left(\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right) = J(\vec{r}^\prime).\nabla^\prime\left(\frac{1}{|\vec{r} - \vec{r}^\prime|}\right) + \frac{1}{|\vec{r} - \vec{r}^\prime|}\nabla^\prime\cdot\vec{J}(\vec{r}^\prime). \end{equation} For steady currents, $\nabla^\prime\cdot\vec{J}(\vec{r}^\prime) = 0$ so that \begin{equation}\label{e3}\tag{3} \nabla^\prime\cdot\left(\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right) = J(\vec{r}^\prime).\nabla^\prime\left(\frac{1}{|\vec{r} - \vec{r}^\prime|}\right). \end{equation} Using this in \eqref{e1} we get \begin{equation}\label{e4}\tag{4} \nabla\left(\nabla\cdot\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right) = -\nabla\left(\nabla^\prime\cdot\left(\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right)\right). \end{equation} The volume integral is over the primed coordinate. So that \begin{equation}\label{e5}\tag{5} \int dV^\prime \nabla\left(\nabla\cdot\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right) = -\int dV^\prime\nabla\left(\nabla^\prime\cdot\left(\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right)\right) = -\nabla\int dV^\prime\nabla^\prime\cdot\left(\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right). \end{equation} We use the divergence theorem on the right most term and choose the surface of integration so large that the current density on it is zero so that \begin{equation}\label{e6}\tag{6} \int dV^\prime \nabla\left(\nabla\cdot\frac{J(\vec{r}^\prime)}{|\vec{r} - \vec{r}^\prime|}\right) = 0. \end{equation}