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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.

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    $\begingroup$ see: Is Biot-Savart law obtained empirically or can it be derived? $\endgroup$ – AccidentalFourierTransform Jan 8 '16 at 22:49
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    $\begingroup$ but... I just cant find any $\delta$ in that post at all... where are they? $\endgroup$ – AccidentalFourierTransform Jan 9 '16 at 10:29
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    $\begingroup$ well, it's true that we physicists tend to take a shortcut to that equation by using $\delta$'s, but that's not really necessary (you could try to ask how to solve $\nabla^2 f=g$ without $\delta$'s in math.SE, which is a very interesting question! mathematicians won't dare to use $\delta$'s unless they know precisely what they're doing). Also, $\int \frac{1}{r}\mathrm d^3r$ is well defined Riemann integral, because $\mathrm d^3r\sim r^2\mathrm dr$, so the integrand is not really singular ($\int\frac{1}{r}\mathrm d^3r\propto \int r \mathrm dr$). $\endgroup$ – AccidentalFourierTransform Jan 9 '16 at 13:36
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    $\begingroup$ It's not completely clear why you're so 'allergic' to the delta function. Elementary proofs do have their own special value, but rest assured that all the standard proofs based on the delta function can be made perfectly rigorous via the use of distribution theory, though you do need to be very careful in specifying which objects are functions and which objects are distributions. If you're looking for areas to self-study on the rigorous side of mathematical physics, a good understanding of distribution theory is a good place to go. $\endgroup$ – Emilio Pisanty Jan 15 '16 at 18:20
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    $\begingroup$ An additional note - to prove the Biot-Savart law, you definitely also need to start with the magnetic Gauss law as a premise, or you'll never get it - simply because your Biot-Savart magnetic field has zero divergence, which is not ruled out by the current premises of the question. $\endgroup$ – Emilio Pisanty Jan 15 '16 at 18:27
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You can take an infinite cylinder, the field due to it will be due to Ampere's law, equal to for a point at a distance of 'a' from the centre for $a>r$ equal to $\mu_o/2\pi a$ ($J$x $\pi r^2$). Now this can be written as written. Now the thing is you have to make the the volume go to zero, and to let the current remain the same you have to make $r \rightarrow0$ and at the same time $\rho\rightarrow \infty$ to keep the current constant.

Basic thing is you've to start to broaden your mathematical horizon for density fields, the density fields don't just behave like ordinary functions, they behave like generalised functions in nature, and you've to add dirac deltas into your mathematical vocabulary to describe them.

You can always be taking limits, but it is too cumbersome.

Also, after all these are just mathematical models, no real charges are point charges and no real currents have zero widths, these models are easier to calculate and are essentially what you call green's function. They help in solving the differential equations.

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  • $\begingroup$ Thank you for your answer! Could you explain the mathematical steps leading from Ampère's law to the Biot-Savart law? I haven't understood how you derive the formula [how can an expression for the field $\mathbf{B}$ (I haven't understood how you derive $\mathbf{B}$'s direction and magnitude) be derived without knowing anything else on the field, if we only know from Ampère's law that $\oint_\gamma\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$ (or that $\nabla\times\mathbf{B}=\mu_0\mathbf{J}$)?] you've written and how the Biot-Savart law is then derived from it... $\endgroup$ – Self-teaching worker Jan 14 '16 at 8:47
  • $\begingroup$ Are you trying to ask, how we use the symmetry arguments? $\endgroup$ – Meme Jan 14 '16 at 11:56
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    $\begingroup$ If you are having trouble thinking about curl, think of it like, just like differential equations describe a family of curves in 1-D, same way curl and divergence help to identify a family of vector functions/fields. They are in some way analog of curves of vectors. $\endgroup$ – Meme Jan 14 '16 at 11:58
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    $\begingroup$ I could write a mathematical derivation too but I don't think seeing couple of algebraic and limit calculations will convice you. A dirac delta function is a new object. The name function is a misnomer, you can try calling it the dirac delta object. It helps to describe density of a point charge, and if you feel this is too hand wavy, it is not. But still if you wish, you can check out green's function for a differential equation. The only use of dirac delta functions in electrostatics are to find the green's functions. Even though we take their solutions literally $\endgroup$ – Meme Jan 14 '16 at 12:03
  • $\begingroup$ There are no essentially no point charges in nature, neither are there 2-D sheets, neither perfect conductors, not perfectly linear dielectrics or infinite sheets, or infintely long and infinitely thing wires or charges. These are models, we can calculate field for easily. Some of the easy to calculate models for which we are able to solve maxwell's equations by various tricks like gauss divergence theorems etc. $\endgroup$ – Meme Jan 14 '16 at 12:07
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I really like the proof contained in the paper Derivation of the Biot-Savart Law from Ampere's Law Using the Displacement Current from 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 in Z axis to positive Z values 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 Equation of 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:

$$dB = \mu_0 i \; ds \; r­ \sin \alpha /4\pi r^2$$

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