# Inverse Square Law in $D$ dimensions (two cases)

I am reading A. Zee "Quantum Field Theory in a Nutshell" and I have solved the problem about inverse square law in $D$ dimensions. Unfortunately, I have been confused with some results. Let me desribe my derivations briefly and focuse on results. The energy of interaction has the following form: $$E(r)=-\int\frac{d^Dk}{(2\pi)^D}\frac{e^{i{\bf k}\cdot{\bf r}}}{k^2+m^2}\equiv -W(r),$$ where $W(r)$ can be calculated using Schwinger parametrization (see Wiki) with the parameter $A=k^2+m^2$. Then, I have obtained the result: $$W(r)=\frac{1}{(2\pi)^{D/2}}\left(\frac{m}{r}\right)^{D/2-1}K_{D/2-1}(mr),$$ where $K_{\nu}$ is the Bessel functions 2nd kind. The result demonstrates correct answer for massive force carrier ($m\neq 0$) in 3D but I don't understand how to obtain $W(r)$ in 2D case with massive carrier because $K_0(mr)\neq \ln(mr)$. Moreover, my calculation falls down in case of massless carrier ($m=0$), it is easy to see it. Can anybody explain how to evaluate correct answers for massless carrier from my calculation?

• $(m/r)^{D/2-1}=1+\frac\epsilon2\log(m/r)$, with $d=2+\epsilon$. Mar 6 '18 at 21:47
• like the dimensional regularization? Mar 6 '18 at 22:32
• OK, let me try this. I expand $K_{\nu}$ and $(m/r)$ into series. Thus, I obtain: $$K_{0}(mr)-K_{0}(mr)\log\left(\frac{m}{r}\right)\epsilon+...$$ And...? It is not easy for me to see $\log(r)$ law for massless carrier. Mar 7 '18 at 7:14
• You should see it in Classical field theory by Mark Burgess. He has given these solutions. Mar 8 '18 at 16:31
• Dear Zohaib, can You be more specifif? Classical Covariant Fields by M. Burgess? Mar 8 '18 at 17:26

Thank You, AccidentalForierTransform & Sean E. Lake!

(1) To obtain the correct answer for massless carrier one can use Schwinger parametrization and obtain the following expression: $$E(r)=-\frac{2^{D/2-1}}{r^{D-2}}\Gamma\left(\frac{D}{2}-1\right)\frac{1}{2(2\pi)^{D/2}}.$$ (2) Unfortunately, both cases (massive and massless carriers) have "bad behavior" for $D=2$. The gamma function has the pole at $z=0$. To deal with it, one can use dimensional regularization: replace $D\rightarrow D+2\epsilon$. Thus, the integration measure is to change: $$\frac{d^{D}k}{(2\pi)^D}\rightarrow \frac{d^{D+2\epsilon}k}{(2\pi)^{D+2\epsilon}},$$ but with this replacement, one should correct the dimensionality and regularization parameter $\mu$. Finally, the measure has the following form: $$\frac{d^{D+2\epsilon}k}{(2\pi)^{D+2\epsilon}}\mu^{-2\epsilon}.$$ This regularization provides the physically correct answer. The gamma function should be expanded into series: $$\Gamma(\epsilon)\approx\frac{1}{\epsilon}-\gamma.$$ And the fraction $(1/(\mu r))^{\epsilon}$ should be expanded too: $$\left({\mu r}\right)^{-\epsilon}\approx 1 - \ln (\mu r)\epsilon.$$

Considering all the above, the answer is $$E(r)=\frac{1}{2\pi}\ln(\mu r),$$ which has the correct dimensionality (in contrast to $-\ln r/(2\pi)$ which is "unphysical" due to the logarithm of length).

1. dimensional regularization does not change the singularity character of the gamma function for $D=2$ because the expansion contains the pole at 0.
I'm also reading Zee's book. When attempting this question, I took a shortcut and considered the massless case $$(m=0)$$ from the get go. Then, I noticed that $$E(r)$$ is reduced to the Green's function for the $$D$$-dimensional Laplace equation. It is well known, or by Gauss' Law, one can find that $$\nabla E(r) = \frac{1}{S_{D-1}} \propto 1/r^{D-1}$$, where $$S_{D-1}$$ is the surface area of a $$D$$-dimensional sphere. Thus, $$E(r)\propto 1/r^{D-2}$$. In the case of $$D=2$$, $$\nabla E(r) \propto 1/r \implies E(r) \propto \ln(r)$$.