# Averaging of dipolar coupling in solution due to tumbling

The angular part of dipolar splitting depends on $\theta$ as:

$$3\cos^2\theta - 1$$

In the liquid state or in solution molecular tumbling occurs which causes the NMR signal to become an average of all $\theta$ which is expressed mathematically as:

$$\int_0^\pi(3\cos^2\theta-1)\sin\theta\,\mathrm{d}\theta$$

where $\sin\theta$ is described, in my the book that I'm reading, as a weighting factor to take into account the probability of finding the system with an angle $\theta$

It is this weighting factor that confuses me. Surely all angles are equally likely? It looks to me like the $\theta$ part of the triple integral one would do to integrate over a sphere using polar coordinates (notice the limits on the integral). However, even if this is correct, I am still finding it difficult to picture this volume element $\sin\theta\,\mathrm{d}\theta$.

Please help me to rationalise why the weighting factor is there and also to visualise this $\sin\theta\,\mathrm{d}\theta$ volume element.

The $\sin(\theta)\,\mathrm{d}\theta$ comes from the surface element of a sphere of unit radius where $\theta$ is the polar angle measured from the north pole, as it were, to the equator. The surface element is $\mathrm{d}s = \sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$ where $\phi$ is the azimuthal angle measured is if longitude, i.e around the equator. The integral is $$\frac{1}{4\pi}\iint f(\theta)\sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$$ where $f(\theta)$ is the function you want to average. 