- Mono Mole

Why does a current flowing through a wire generate a magnetic field? The answer lies in special relativity.

Consider a current flowing through a straight length of wire. In the rest frame of the wire, the positive ions (nuclei) are stationary, while the electrons are moving. Suppose a negative test charge is located at a distance from the wire, and moves parallel to the wire with the same velocity as the electrons. In this frame, the wire is electrically neutral overall because the number of positive charges equals the number of negative charges. Consequently, there is no net static electric field acting on the test charge in this frame. However, we observe that the test charge moves in response to a force towards the wire.

In the frame of the moving test charge, the test charge and the electrons are stationary, while the positive ions move in the opposite direction. Due to length contraction, the distance between the moving positive ions appears shorter to the test charge, leading to an increased positive charge density in the test charge’s frame. As a result, the wire has an apparent higher positive charge density compared to the negative charge density of the stationary electrons. This creates a net electric field, which exerts a force on the test charge towards the wire. In the rest frame of the wire, this force is referred to as a magnetic force. Therefore, both frames yield the same total force on the test charge, but it is described as a magnetic force in the wire’s rest frame and an electric force in the test charge’s frame. Since we can mathematically describe the magnetic force in the wire’s rest frame by defining a magnetic field, we say that a current flowing in a wire generates a magnetic field.

In summary, the electric field and the magnetic field are simply different observations of the same underlying phenomenon. Just as a coin has two faces that are inseparable yet different, electric and magnetic fields are inseparable parts of the same phenomenon, with their roles depending on the relative motion of the observer.

Question

Doesn’t length contraction also apply to the moving electrons in the wire’s rest frame? If so, why is there no net electric field acting on the test charge in that frame?

Answer

The rest frame of the wire is the initial premise. In this frame, the wire is observed to be electrically neutral, with length contraction already accounted for in this electrical neutrality. The question then shifts to how the situation differs in another reference frame, based on the initial premise.

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September 23, 2024

Laguerre polynomials

Laguerre polynomials $v(x)$ are a sequence of polynomials that are solutions to the Laguerre differential equation:

$x\frac{d^2v}{dx^2}+(1-x)\frac{dv}{dx}+nv=0\;\;\;\;\;\;\;\;420$

where $n$ is a constant.

When $x=0$ , eq420 simplifies to $\frac{dv}{dx}+nv=0$ . The solution to this first-order differential equation is $v(x)=Ae^{-nx}$ , which can be expressed as the Taylor series $v(x)=A\sum_{n=0}^{\infty}\frac{(-nx)^n}{n!}$ . This implies that eq420 has a power series solution around $x=0$ . To determine the exact form of the power series solution to eq420, let $v(x)=\sum_{j=0}^{\infty}a_jx^j$ .

Substituting $v$ , $v'$ and $v''$ in eq420 yields

$x\sum_{j=0}^{\infty}j(j-1)a_jx^{j-2}+(1-x)\sum_{j=0}^{\infty}ja_jx^{j-1}+n\sum_{j=0}^{\infty}a_jx^{j}=0\;\;\;\;\;\;\;\;421$

$\sum_{j=0}^{\infty}j^2a_jx^{j-1}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

Setting $j=j+1$ in the first sum,

$\sum_{j=-1}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=\sum_{j=0}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

$\sum_{j=0}^{\infty}\biggr\[(j+1)^2a_{j+1}+(n-j)a_j\biggr\]x^{j}=0\;\;\;\;\;\;\;\;422$

Eq422 is only true if all coefficients of $x$ in is 0 (see this article for explanation). So, $(j+1)^2a_{j+1}+(n-j)a_j=0$ , or equivalently,

$a_{j+1}=\frac{j-n}{(j+1)^2}a_j\;\;\;\;\;\;\;\;423$

Eq423 is a recurrence relation. If we know the value of $a_0$ , we can use the relation to find $a_1,a_2,...$ .

$\boldsymbol{\mathit{j}}$	Recurrence relation
$0$	$a_1=-na_0$
$1$	$a_2=\frac{1-n}{(1+1)^2}a_1=\frac{n(n-1)}{1^22^2}a_0$
$2$	$a_3=\frac{2-n}{(2+1)^2}a_2=-\frac{n(n-1)(n-2)}{1^22^23^2}a_0$
$\vdots$	$\vdots$

Comparing the recurrence relations, we have

$a_k=(-1)^k\frac{n(n-1)(n-2)\cdots (n-k+1)}{1^22^2\cdots k^2}a_0$

$a_k=(-1)^k\frac{n!}{(k!)^2(n-k)!}\;\;\;\;\;\;\;\;424$

where $a_0=1$ by convention (so that $L_n(0)=1$ ).

Letting $k=j$ in eq424 and substituting it in $v(x)=\sum_{j=0}^{n}a_jx^j$ yields the Laguerre polynomials:

$L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j\;\;\;\;\;\;\;\;425$

where we have replaced $v(x)$ with $L_n(x)$ .

The first few Laguerre polynomials are:

$L_0(x)=1$

$L_1(x)=-x+1$

$L_2(x)=\frac{1}{2}(x^2-4x+2)$

$L_3(x)=\frac{1}{6}(-x^3+9x^2-18x+6)$

Next article: Rodrigues’ formula for the Laguerre polynomials

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