Energy renormalization in a simple system

If you put an electron near an infinitely long wire of uniform charge per unit length λ = 1 C / m \lambda = 1~C/m , the electron will be attracted toward the wire since the signs of charge are opposite. If you try to find the electrical potential energy U ( d ) U(d) at the electron's position d from the wire, you will find an infinite value. Correspondingly it's hard to show that's the potential energy will decrease if the electron is closer with the wire, since we're comparing infinity to infinity.

We get this infinity because we by default chose the potential energy to be zero very far from the wire. We could, however, compensate for the infinite potential energy at d by redefining the potential energy far away to be positive infinity. Amazingly, by adding a positive infinite value to a negative infinite value in just the right way, we can get a finite value for the redefined potential! This is the idea behind renormalization in quantum field theory.

Let's do this. First, let us cut off the length of the wire to be some very large L (with the electron equidistant from the ends of the wire) and set the potential energy at infinity to be zero (the usual case). Calculate the potential energy in this situation. Separate this potential energy U ( d , L ) U(d,L) into a part I ( L ) I(L) that does not depend on d and a part W ( d ) W(d) that does have d dependence, U ( d , L ) = I ( L ) + W ( d ) U(d,L)=I(L)+W(d) . Since all quantities are finite this is a well defined problem and we'll get some finite value for W(d). We will now redefine the potential energy to be E ( d ) = U ( d , L ) I ( L ) E(d)=U(d,L)-I(L) .

We can take the limit of this finite problem to get our problem with a well defined E(d) by letting I ( L ) = λ e ln ( L ) 2 π ϵ o I(L)=\frac{\lambda e \ln{(L)}}{2 \pi \epsilon_o} in the limit L L\rightarrow \infty . Hence we have effectively subtracted off the divergent piece and "renormalized" our electric potential energy.

Find the value of E(d) in Joules using this procedure if the electron is 1 m from the wire.

Note that the motion of the electron can be perfectly well calculated with this potential energy as only differences in potential energy matter. Hence that constant that we pulled out is irrelevant to the physics (which is why this whole procedure works).

Details and assumptions

  • The electron's charge is e = 1.6 × 1 0 19 ( C ) e=-1.6 \times 10^{-19}(C) and the electric permittivity is ϵ o = 8.85 × 1 0 12 ( F / m ) \epsilon_o=8.85 \times 10^{-12}(F/m) .


The answer is 0.

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3 solutions

Quan Dinh
May 20, 2014

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

Ang Yu Jian
May 20, 2014

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)

David Mattingly Staff
May 13, 2014

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

W ( d , L ) = λ e 4 π ϵ o L / 2 + L / 2 d x x 2 + d 2 = λ e 4 π ϵ o ln ( L 2 + L 2 4 + d 2 L 2 + L 2 4 + d 2 ) λ e 4 π ϵ o ln ( 1 + L 2 d 2 ) λ e 4 π ϵ o ln ( L 2 d 2 ) 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})}

W ( d , L ) = λ e 2 π ϵ o ln ( L ) λ e 2 π ϵ o ln ( d ) \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 ) = λ e ln ( L ) 2 π ϵ o I(L) = \frac{\lambda e \ln{(L)}}{2 \pi \epsilon_o} and E ( d ) = λ e ln ( d ) 2 π ϵ o E(d)=-\frac{\lambda e \ln{( d)}}{2 \pi \epsilon_o} ; we get the value of the redefine electrical potential energy:

E ( d ) = λ e ln ( d ) 2 π ϵ o = 0 ( J ) E(d)=-\frac{\lambda e \ln{( d)}}{2 \pi \epsilon_o}=0(J)

As we can see, E ( d ) E(d) decreases as d d decreases, so the electron and the wire attract each other like what we expect!!! Also I ( L ) I(L) goes to infinity if L 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 d=3 , to use the dimensional regulation method one has to change the number of dimensions to d = 3 ϵ d=3-\epsilon , with ϵ \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!

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