# Proton spin precession and alignment to magnetic field in NMR

Precession due to magnetic field occurs when the magnetic moment of a small coil is proportional to its angular momentum ($$\vec{m}=g\vec{L}$$). This is due to Euler momentum equation, which for such a system reads: $$\frac{d\vec{L}}{dt}=\vec{M}=\vec{m}\times \vec{B}=g\vec{L}\times\vec{B}$$and has the shape of a precession equation. In NMR theory we are told that a single proton can both precess around a magnetic field and align to it. This happens because of its spin, so in principle even in absence of orbital angular momentum. First of all: to justify the precession, do we need to generalize Euler equation and write that $$d\vec{J}/dt=\vec{M}$$ (where $$\vec{J}=\vec{L}+\hbar\vec{S})$$? Does this always hold? Secondly, what causes alignment? Is alignment just a statistical prediction? Does it always hold?

You will need to use the full angular momentum of the system considered, here $$\vec{J}=\vec{L}+\vec{S}$$ (note usually $$\vec{S}$$ is considered the spin, i.e. including the $$\hbar$$). This certainly holds for simple systems like this one, in more complex systems the specific type of spin-orbit coupling may need to be considered.
Alignment of protons in a magnetic field is caused by the interaction of the field with the proton's magnetic moment. The spin, and hence the moment, being parallel, or anti-parallel to the magnetic field vector result in different energy levels, therefore the proton's spin will show preference for the orientation with lower energy. At temperature $$T=0$$ all spins will settle in this orientation. At higher temperatures, thermal energy will cause a lot of the spins (up to 50%, or balanced, at very high temperatures) to switch to the non-preferred orientation; in this sense we're talking about a statistical phenomenon. In fact, at room temperature in easily achievable magnetic fields the relative imbalance for protons is only $$10^{-10}$$ .. $$10^{-8}$$, explaining why NMR signals are so small at room temperature.