Why thermal tumbling of molecules does not prevent the establishment of magnetization in NMR?

Question

Given the rapid thermal tumbling of water molecules at room temperature, theoretically speaking, why magnetic thermalization of proton nuclear spins is at all possible? Further, how can nuclear spins maintain alignment with the external field?

Proton in water at $1.5 \text{ T}$ magnetic field has the longitudinal/thermal relaxation rate, $T_1$, of about $4 \text{ s}$. As a rough estimate, we can take the microwave frequency of microwave ovens, $2.5\text{ GHz}$, as the inverse rotation timescale of water molecules. There is a nine orders of magnitude difference in timescales!

If we transfer to a coordinate system attached to a proton, the external magnetic field would appear randomly changing its orientation while maintaining the same magnitude. Given that the energy scale associate with the coupling of spin to the magnetic field is at best a few hundred megahertz (the gyromagnetic ratio of proton is 42.577 $\text{ MHz}/\text{T}$), the nuclear spin cannot possibly adiabatically remain aligned with the magnetic field.

Clearly, magnetization is established and maintained in NMR experiments and MRI procedures. So, what is the resolution of this seeming paradox?

Answer 1

why magnetic thermalization of proton nuclear spins is at all possible? ... There is a nine orders of magnitude difference in timescales!

The difference in time scales is not nearly as important as the difference in energy scales. The energy difference between the parallel and anti-parallel spin state is $h \gamma B$ where $h$ is Planck's constant, $\gamma$ is the gyromagnetic ratio, and $B$ is the field strength. The thermal energy is $kT$ where $k$ is Boltzmann's constant and $T$ is the temperature.

For body temperature and usual clinical field strengths that works out to about 10 parts per million. This means that out of a million spins the thermal polarization is only 10 spins.

If we transfer to a coordinate system attached to a proton, the external magnetic field would appear randomly changing its orientation while maintaining the same magnitude.

You really cannot learn much from that. That coordinate system is very non-inertial, so your typical reasoning about the behavior of physics in that frame will not be valid. It is not even "nicely non-inertial" like a rotating coordinate system, which can be helpful.

However, one thing that may help is to realize that magnetization is very transient on a quantum level. There are lots of interactions with other spins whereby the magnetization can be transferred from one to another. So just because an individual spin is jiggled doesn't mean that the magnetization disappears, it can instead be transferred. So while you may have only 10 spins per million polarized, which 10 spins those are changes rapidly.

Answer 2

The energy difference between (anti)aligned states is:

$$ \Delta E = \mu H=\hbar\gamma H= (7x10^-{10}\,{\rm eV/MHz})(43\,{\rm MHz/T})(1.5\,{\rm T})=4\times 10^{-8}\,{\rm eV}$$

and body temperature is around 1/40 electron volts. The relative probability of a state being occupied is given by the Boltzmann factors:

$$ p =e^{-\frac{\Delta E}{kT}} \approx 1-\frac{\Delta E}{kT}\approx 1-10^{-6}$$

So there is a very small thermal polarization. MRI's are on living samples, so that rules out cryogenics. There are MRIs with inhaled polarized $^3\vec{\rm He}$ that are not thermal distributions. The gas is optically pumped.

The reason it works is because $N_A$ is huge. The signal is suppressed by 1 million, but $10^{23}$ is a big number.

Answer 3

I'm thinking there is probably no reason to assume that the magnetic orientation of the nucleus should be effected by the “tumbling” of the molecule in which it resides. Is there a strong magnetic coupling between the nucleus and the molecule?