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?

  • 3
    $\begingroup$ $(m/r)^{D/2-1}=1+\frac\epsilon2\log(m/r)$, with $d=2+\epsilon$. $\endgroup$ Mar 6, 2018 at 21:47
  • $\begingroup$ like the dimensional regularization? $\endgroup$ Mar 6, 2018 at 22:32
  • $\begingroup$ 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. $\endgroup$ Mar 7, 2018 at 7:14
  • $\begingroup$ You should see it in Classical field theory by Mark Burgess. He has given these solutions. $\endgroup$ Mar 8, 2018 at 16:31
  • $\begingroup$ Dear Zohaib, can You be more specifif? Classical Covariant Fields by M. Burgess? $\endgroup$ Mar 8, 2018 at 17:26

2 Answers 2


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.
  2. the Schwinger parametrization is very convinient way to calculate propagator-type because it allows to avoid the charade with hyperpsherical coordinates
  3. of course, these tricks are easy for good physicists but I have not found any explanations and solutions for this problem
  • $\begingroup$ See also: limiting forms of Bessel functions in standard texts. e.g. dlmf.nist.gov/10.30 $\endgroup$ Mar 11, 2018 at 19:54

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 $D=2$, $\nabla E(r) \propto 1/r \implies E(r) \propto \ln(r)$.

  • $\begingroup$ If you would like to see calculations for D-dimensional case, notice me $\endgroup$ Aug 20, 2019 at 20:06
  • 1
    $\begingroup$ Additionly, your answer isn't correct completely. Indeed, you calculate log function of argument with length dimensionality. The argument of log functions should be dimensionless $\endgroup$ Aug 20, 2019 at 20:07
  • $\begingroup$ Ah you're right! thanks for pointing that out! $\endgroup$
    – bygolly
    Aug 21, 2019 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.