Let us assume the validity of Ampère's circuital law $$\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$$where $\mathbf{B}$ is the magnetic field, $\gamma$ a closed path linking the current of intensity $I_{\text{linked}}$.
Can the Biot-Savart law $$\mathbf{B}=\frac{\mu_0}{4\pi}\oint\frac{Id\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}=\frac{\mu_0}{4\pi}\int_a^b I\boldsymbol{\ell}'(t)\times\frac{\mathbf{x}-\boldsymbol{\ell}(t)}{\|\mathbf{x}-\boldsymbol{\ell}(t)\|^3}dt$$where $\boldsymbol{\ell}:[a,b]\to\mathbb{R}^3$ is a parametrisation of a closed (or infinite) wire carrying the current $I$, be inferred without using Dirac's $\delta$, by using the tools of multivariate calculus and elementary differential geometry only, at least if we assume the validity of the Gauss law for magnetism or other of the Maxwell equations? All the proofs I have found (such as this, where, as far as I understand, $$\nabla^2\left[\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d^3r'\right]=-\mu_0\mathbf{J}(\mathbf{r}) $$ is derived by using the $\delta$) use the expression
$$\mathbf{B}=\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}\times\hat{\mathbf{r}}}{r^2}d^3x$$ and Dirac's $\delta$, but I wonder whether, both assuming a linear current distribution as when we use the expression of the magnetic field as$$
\mathbf{B}=\frac{\mu_0}{4\pi}\oint\frac{Id\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}$$ and assuming a tridimensional spatial current distribution, it is possible to prove the Biot-Savart law from Ampère's without the use of the $\delta$. I heartily thank any answerer.
Best Answer
I really like the proof contained in the paper
Derivation of the Biot-Savart Law from Ampere's Law Using the Displacement Currentfrom Robert Buschauer (2013)It's simple and it fulfills the role of convincing the reader.
Basically the author works with one point charge $q$ situated in origin of
Z azis$(0,0,0)$. He supposes a particle moving inZ axisto positiveZvalues with velocity $v$. He creates a magnetic field line in a arbitrarious circle with $c$ radius, by symmetry, with center in $(0,0,a)$. The angle between any point in the circle and the center of circle starting from origin $(0,0,0)$ is $\alpha$.Starting point is a part of
4th Maxwell's Equationof electromagnetism, the Ampere-Maxwell Law that consider changing electric flow with time in a area produces magnetic field circulation. This law generates a magnetic force that can be verified using special relativity that in another reference frame it's just a plain electric force.$$\oint B\, dl = \mu_0\epsilon_0 \; d/dt(\int_A E.dA)$$
In the left side, the solution consists of integrating the $\oint B dl$ in this circle (butterfly net ring). As $B$ is constant by symmetry, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \oint B\, dl = 2\pi c B \qquad\qquad$
(1)In the right side $\;[\;\mu_0 \epsilon_0 d/dt (\int_A E\; dA \,)\;],\;$ as the surface (butterfly net) we choose a sphere of radius $r$, to ensure that all points have the same value of electric field:
$$ E = q / 4\pi\epsilon_0r^2$$
Let's first calculate the right-hand integral in the right side. We adopted here a slightly different standard in spherical coordinate. Just to remember,the element for integration into spherical coordinates is $\; r^2 \sin \phi \, dr \, d\phi \, dq $
Let $\theta$ (
XY axis) vary from $0$ to $2\pi$ and by consider the angle $\phi$ with the vertical (Z axis) from 0 to $\alpha$.$$\Phi_E = \int_A E\; dA = q/4\pi \epsilon_0 r^2 \int_A dA = q/(4\pi \epsilon_0 r^2) r^2 \int_{0,2\pi} d\theta \int_{0,\alpha} \sin \Phi\; d\Phi = $$ $$q/4\pi \epsilon_0 2\pi ( -\cos \alpha + 1) = q/2\epsilon_0 (1 - cos\alpha)$$
Thus
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Phi_E = \mu_0 q /2 (1 - cos\alpha)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \Phi_E / dt = - q/2\epsilon_0 d \cos \alpha/dt\qquad$
(2)Putting $\alpha$ as a function of $z$, we have, by the chain rule:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d \cos \alpha/dt = (d \cos\alpha/dz) \; (dz/dt)\qquad$
(3)However as $z$ is decreasing with the motion at velocity $v$, we have
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad dz / dt = -v \qquad $
(4)On the other hand: $$ \cos \alpha = z / r = z / \sqrt{c^2 + z^2}$$
Using this online tool for derivation: $d \cos \alpha/dz = c^2/r^3$ where $r = \sqrt{c^2 + z^2}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad 2\pi c B = q \mu_0 /2 v (c^2/r^3)\qquad$ By (1),(2),(3),(4)
$$B = \mu_0 q v c / 4\pi c r^3$$
but $\quad\sin \alpha = c / r\quad$ so we can add $\quad \sin \alpha\; r / c$:
$$B = \mu_0 q v \sin \alpha /4\pi r^2 $$
Vectorizing we have a cross product:
$$B = \mu_0 q \; v\uparrow \times r\uparrow /4\pi r^3$$
In some infinitesimal point we can consider a element of electric current as a point charge, so we can add other charge points by integration (any force is addictive!) for using in real applications. Thus we have in scalar notation:
$$dB = \mu_0 dq \; v \; r \sin \alpha /4\pi r^2$$
Considering $\quad dq = i\;dt\quad$ and $\quad v = ds/dt\quad $, we finally have reached to
Biot-Savart law:$$dB = \mu_0 i \; ds \; r \sin \alpha /4\pi r^2$$