Energy renormalization in a simple system

Considérons un petit morceau dl du fil, en intégrant partout la longeur L du fil considéré, on a: V(d,L)=\frac{\lambda}{2\pi\epsilon {0}}(lnL-lnd) => U(d,L)=eV(d,L)=\frac{e\lambda}{2\pi\epsilon {0}}(lnL-lnd) Par définition du sujet, on en déduit l'énergie électrostatique de l'électron: E(d)=U(d,L)-I(L)=\frac{-e\lambda}{2\pi\epsilon_{0}}lnd Avec d=1m, alors ln(d)=0, donc E(d)=0

Given it has been stated that the answer is an integer and universal constants with many decimal places are involved, it is easy to say that the answer of zero or unity would be the cleanest method.

However, aside from that inference, we see that using the long wire method, we can integrate the individual contributions of otential energy from small sections of wire to find the overall potential energy.

Such a section can be considered a point charge with charge λdy

I will refrain from using ϵo and use k instead due to ease of typing from this point onwards

The potential energy would therefore be given by kλedy/sqrt{d^2+y^2}

Where y is the distance between the midpoint and said section of wire.

Integrating with respect to y from -L/2 to L/2 gives us

U(d,L)=2kλe ln(sqrt[d^2+L^2/4] L/2d)

we know that d<<L, and this results in a value of 2kλe*ln(L/d)

This can be then expanded to obtain 2kλe*(ln(L)-ln(d))

With the first term being I(L) and the second term being W(d)

Here is where things get sticky

If we decide to let E(d)=W(d), we get -9.97223E-10J as the answer

However, we realise that since we are renormalising, we can change the zero point of the potential into anything we want, as long as the other terms remain constant. (I'm not sure why this point was chosen to be zero besides convenience to be brutally honest, I personally feel that using 1 as the zero point is mathematically nicer in terms of manipulations, and would result in the above non integer answer)

Not exactly a good solution by my standards (my IPhO trainers would've killed me years ago, but the best I could muster)

Let $x$ be the direction along the wire. The electrical potential energy can be calculated as:

$W(d,L)=\frac{\lambda e}{4\pi \epsilon_o} \int^{+L/2}_{-L/2} \frac{dx}{\sqrt{x^2+d^2}} = \frac{\lambda e}{4\pi \epsilon_o} \ln{(\frac{\frac{L}{2}+\sqrt{\frac{L^2}{4}+d^2}}{-\frac{L}{2}+\sqrt{\frac{L^2}{4}+d^2}})} \approx \frac{\lambda e}{4\pi \epsilon_o} \ln{(1+\frac{L^2}{d^2})} \approx \frac{\lambda e}{4\pi \epsilon_o} \ln{(\frac{L^2}{d^2})}$

$\Rightarrow W(d,L) = \frac{\lambda e}{2 \pi \epsilon_o} \ln{(L)} - \frac{\lambda e}{2 \pi \epsilon_o} \ln{( d)}$

So $I(L) = \frac{\lambda e \ln{(L)}}{2 \pi \epsilon_o}$ and $E(d)=-\frac{\lambda e \ln{( d)}}{2 \pi \epsilon_o}$ ; we get the value of the redefine electrical potential energy:

$E(d)=-\frac{\lambda e \ln{( d)}}{2 \pi \epsilon_o}=0(J)$

As we can see, $E(d)$ decreases as $d$ decreases, so the electron and the wire attract each other like what we expect!!! Also $I(L)$ goes to infinity if $L$ goes to infinity - it's an "infinite subtraction".

By applying this cut-off, one can see that we break the symmetry of the problem (if we move the electron parallel to the wire, the potential at the electron's place is changing; we don't have this problem with an infinite length wire). In Theoretical Physics, breaking a symmetry of a theory is a huge disadvantage, so the method of using cut-off is not always a preference. A trick that Physicists usually use in a renormalization problem is dimensional regulation - they change the number of dimension of the problem by a little bit. For example the number of dimension in our problem is $d=3$ , to use the dimensional regulation method one has to change the number of dimensions to $d=3-\epsilon$ , with $\epsilon$ is very small and go to zero, and the number of dimensions is not an integer anymore! By doing that, the symmetry of the theory will be preserved. We will have some discussions about how to apply this method in our future problems!