July 19th, 2024.
Symmetry appears everything in physics. It is sometimes used to argue the construction of an object (say, constructing a spin 1/2 Lagrangian (density) through Lorentz invariance/symmetry), but other times we use it to break things down such as integrals.
If we have an integral over all $d $-dimensions and the integrand depends only on the radial distance, then all of the angular integration follows almost trivially.
In this post we will find a formula to do this quickly. The first to note is that the angular integration gives none other than the surface area of a $d $-dimensional unit sphere. We will show that this area is equal to
\[\text{Area} =\frac{2\pi ^{d/2}}{\Gamma \left( \frac{1}{2}d \right) } \tag{i}\]Where $d $, again, is the number of dimensions, and $\Gamma $ is the Euler gamma function:
\[\Gamma (z)=\int_{0}^{\infty } t^{z-1}e^{-t}dt \]This blog post serves a similar purpose the last blog (number 10). The advantage of using Eq. (i) to break down a $d $-dimensional integral is to get a 1-dimensional (radial) integral, which can be computed with less complications.
We will find the desired expression by performing the integral $\int d^{d}xe^{-x^2 } $ both in Cartesian and spherical coordinates.
We start by recalling that
\[\int_{-\infty }^{\infty } e^{-x^2 }dx = \sqrt{\pi } \]So the Cartesian integral is:
\begin{align*} \int_{-\infty }^{\infty } d^d x e^{-x^2 } &= \int_{-\infty }^{\infty } d^{d}xe^{-x_2 ^2 -\cdots -x_d^2 } \\ &= \sqrt{\pi }^{d} \tag{ii} \end{align*}For the spherical integral, we recall that
\[d^{d}x\rightarrow d\Omega dr r^{d-1} \]Where the $dr $ integral goes from $0 $ to $\infty $. Further, $x^2 \rightarrow r^2 $. We see that the integral in spherical coordinates is:
\begin{align*} \int d\Omega dr r^{d-1}e^{-r^2 } &= \int d\Omega \int_{0}^{\infty } dr\;r^{d-1}e^{-r^2 } \\ &= \Omega \int_{0}^{\infty } dr\;r^{d-1}e^{-r^2 }\\ \text{let $r^2 =u $, so $du=2rdr $} \;\;\;\;\; &= \Omega \int_{0}^{\infty } du\frac{1}{2}r^{d-2}e^{-u}\\ &= \frac{\Omega }{2} \int_{0}^{\infty } du (r^2 )^{\frac{d-2}{2}}e^{-u}\\ &= \frac{\Omega }{2} \int_{0}^{\infty } du u^{\frac{d-2}{2}}e^{-u}\\ &= \frac{\Omega }{2} \int_{0}^{\infty } du u^{\frac{d}{2}-1}e^{-u}\\ &= \frac{\Omega }{2}\Gamma (d/2) \tag{iii} \end{align*}Where $\Omega $ denotes the area of the unit $d $-dimensional sphere.
We now equate equations (ii) and (iii) to find:
\[\text{Area}\equiv \Omega =\frac{2\pi ^{d/2}}{\Gamma \left( \frac{1}{2}d \right) } \]As stated in the introduction.
So far we only have the area of the unit $d $-dimensional sphere, but what do we do with this? Consider an integral of the form:
\[\int d^{d}x\;f(x^2 ) \]Where $f $ only depends on the radial part. Then we turn the $d $ integrals $\int dx_i $ into a single integral over the variable $k^2 $:
\[d^{d}x = \frac{\pi ^{d/2}}{\Gamma (d/2)}(x^2 )^{\frac{d}{2}-1}d(x^2 ) \]Where the integral over $x^2 $ goes from $0 $ to $\infty $. It is often useful to make the trivial substitution $x^2 \rightarrow v $ or some other variable to avoid confusion with the 2 in the exponent acting just as a label rather than as an actual number.