# Characteristic distance of integral by dimensional analysis?

We have the integral $$E = −\,{1 \over \left(\,2\pi\,\right)^{3}}\int_{\mathbb{R}^{3}}\, \mathrm{d}^{3}k\, {\mathrm{e}^{\mathrm{i}\vec{k}\cdot\left(\,\vec{x}_{1} - \vec{x}_{2}\,\right) } \over \vec{k}^{\,2} + m^{2}}$$

Could someone explain the following passage found in A. Zee, QFT in a Nutshell, 2nd edition page 28:

We identify $E$ as the potential energy between two static sources. Even without doing the integral, we see by dimensional analysis that the characteristic distance beyond which the integral goes to zero is given by the inverse of the characteristic value of $k$, which is $m$. Thus, we expect the attraction between the two sources to decrease rapidly to zero over the distance $1/m$. The range of the attractive force generated by the field $ϕ$ is determined inversely by the mass $m$ of the particle described by the field.

I have not heard the phrases characteristic distance/value before and how do we deduce its values through dimensional analysis?

• What are your thoughts on this so far? – sagittarius_a Aug 23 '16 at 20:29
• It has something to do with how fast it goes to zero or something, but google doesnt find anything on characteristic values so this is unorthodox terminology. – user63128 Aug 24 '16 at 9:40

Whenever you have an integral and you want to extract qualitative scaling information from it, the first step is to figure out a change of variables that makes all the constants in the integrand dimensionless. Note that $k$ and $m$ have the same dimensions (in units where $\hbar = c = 1$), so we should define a dimensionless dummy integration variable $\vec{\kappa} := \vec{k}/m$ so that $\vec{k} = m \vec{\kappa}$. Under this $u$-substitution, the integral becomes

$$E(\vec{\Delta x}) = -\frac{m^3}{(2 \pi)^3} \int d^3 \vec{\kappa} \frac{e^{i m \vec{\kappa} \cdot \vec{\Delta x}}}{m^2 (\vec{\kappa}^2 + 1)} = -\frac{m}{(2 \pi)^3} \int d^3 \vec{\kappa} \frac{e^{i \vec{\kappa} \cdot (m \vec{\Delta x})}}{\vec{\kappa}^2 + 1}.$$

Even without doing the integral, we can see that it only contains the product $m \vec{\Delta x}$, so it will evaluate to some (dimensionless) function $f(m \vec{\Delta x})$. The function must fall off to 0 at large $m \vec{\Delta x}$ in order for the Fourier transform to converge. Since it's dimensionless, we must introduce some characteristic length scale $\xi$ over which it decreases significantly, so that the function takes argument $f(\vec{\Delta x}/\xi)$. (For example, if $f$ were a decaying exponential, $\xi$ might be the length after which it falls off by a factor of $e$ from its initial value.) Comparing the two expressions above, we see that this characteristic length scale $\xi = 1/m$. In order to be more precise about what it means for $f$ to "decrease significantly" over the length $\xi$, you'd need to actually evaluate the integral.

What he means by the characteristic distance is the relevant distance scale obtained by using values of other dimensional parameters in the integral.

Since $k$ is a variable and $m$ is a constant having the dimension of $k$, therefore a mass scale is introduced into the problem, i.e $m$. The final integral therefore depends on $m$, which is the mass scale of the problem, and since [L] = [M]$^{-1}$, $1/m$ is the characteristic length scale.

Now the question is why is the integral cut off at around $k = m$, The reason is that the factor in the exponential becomes large when $k > m$, and as a result the rapidly increasing oscillatory behaviour of $k$ causes cancellation further beyond $k$, and so the remaining integral goes to zero very fast.

PS. If you do contour integration, you will get

$$E = \dfrac{1}{4\pi r} e^{-mr},$$ where $r=|x_1 - x_2|.$ As you can see the characteristic length scale is actually $1/m$

• Could you explain how you read of the characteristic length scale 1/m of the final integral? – user63128 Aug 24 '16 at 12:49
• @kalle it is similar to how you define the time constant of a LC circuit.The distance at which $E$ is scale invariant for all possible $m$ is given by $1/m$ and therefore $r = 1/m$ is a characteristic length scale of the system in question. – Bruce Lee Aug 24 '16 at 12:57
• Sorry I dont get it, what precisely do you mean by the distance at which E is scale invariant? – user63128 Aug 30 '16 at 15:24