My question pertains to NMR (nuclear magnetic resonance).

$T_1$, the spin lattice relaxation time of protons in water is about 2.5 seconds. If you add some $CuSO_4$ (copper sulfate) to the water the spin lattice relaxation decreases dramatically. I.e. in a 1mM $CuSO_4$ solution, $T_1$ drops to about 0.46 seconds. Why is this? I've read that paramagnetic ions decrease spin lattice relaxation times, but how exactly does this work?


1 Answer 1


For starters, there are no simple answers in NMR relaxation, but nonetheless, here goes an attempt at one.

First point-- when you specify the relaxation time of anything, it's useful to give the polarizing B field strength, so that people can note the Larmor frequency. Why is this important? Well, here goes.

Relaxation happens because spins flip, and a spin flips because it feels oscillatory magnetic fields (polarized perpendicular to the main field of the NMR magnet) at its Larmor (i.e. NMR resonance) frequency. In normal water relaxation, imagine you are sitting on one proton of H2O. If you imagine yourself stationary, what do you see, magnetically speaking? The answer is: another proton (your partner in the water molecule) whipping about you in a rapid but irregular motion, due to rotational diffusion. Well, that little proton has a spin, so it is actually a tiny magnet whirling about you.

If you calculate the time dependent magnetic field due this motion, you will get stochastically varying fields in all three directions -- x, y, and z-- of your local coordinate system. Well, the average value of such a fluctuating field usually goes to zero over time, but the power spectrum of the field (calculated purely formally as the fourier transform of its time-correlation function) does not. To the extent that there is power 'broadcast' at the Larmor frequency, the proton on which you (the observer) are sitting will be caused to flip. These flips are the cause of relaxation. The amount of power at the Larmor frequency of interest depends upon the correlation time of the rotational diffusion. Roughly speaking, the shorter the correlation time, the more widespread the power, and the less of it any given frequency. Typical rotational correlation times for small molecules are on the order of a few picoseconds. For moderate size proteins (say 30,000 daltons), tens of nanoseconds. The exact form of the rotational correlation time depends specifically on the details of how the rotational diffusion equation is solved (in spherical coordinates.)

OK, so what about a paramagnetic ion? When a water molecule enters the first coordination sphere of such an ion, what it sees, magnetically speaking, changes radically. Instead of weak magnet diffusing rotationally, it now sees a very strong magnet (the paramagnetic ion).

The 'broadcast' power is correspondingly increased, and correspondingly, the rate of spin flips and therefore of relaxation. The residence time of water in the coordination sphere is not infinite, so, so to speak, the relaxation wealth gets spread around, as different water molecules come and go. (Sorry I don't have anything on the rate of such processes.)

Caveats-- and there are plenty. Relaxation is a quantum mechanical process, and what I have said does no justice to it. The simplest thing that makes any sense (in my view) is a paper by I. Solomon in Phys. Rev., back in 1955, on relaxation in HF (hydrogen fluoride). This will give a little taste of the quantum theory. Bloembergen is the father of relaxation, but I never found him easy to read or understand. Redfield (he was Bloembergen's post-doc) gave the theory the definitive form it finds today-- you can try Slichter's book for a summary of Redfield theory, but, if you are serious, you are going to have fight through the original literature.

Also, my use of the term 'broadcast power' is particularly treacherous, since we are dealing with dipolar near fields, which, strictly speaking, do not radiate, even though they exchange energy with neighboring spins.

Hope this helps.

Good luck.

  • $\begingroup$ Wow, thank you for your very detailed answer and for the references you provided! $\endgroup$
    – Dennis W
    Oct 26, 2013 at 18:01

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