I'm trying to understand the basic principles of Nuclear Magnetic Resonance reading this link but I have some doubts:

1) I have ever known that when protons aren't in a magnetic field, their spins are random oriented in the space. When the protons are in a magnetic field, their spin feel the effect of $ \tau= \mu \times B$ and are inclined to align themselves with the magnetic field. But they precede around B, they aren't oriented like B! Isn't it?

enter image description here

So I can't understand why in the link that I have reported, at the section "Spin Physics", paragraph "Energy Levels", there is written

When the proton is placed in an external magnetic field, the spin vector of the particle aligns itself with the external field,just like a magnet would. There is a low energy configuration or state where the poles are aligned N-S-N-S and a high energy state N-N-S-S.

(please, see the animation on the link)

It seems to me that the two interpretations are in contrast..

And, if I follow the interpretation that I have ever heard, I don't understand this step: (paragraph "$T_1$ process"))

At equilibrium, the net magnetization vector lies along the direction of the applied magnetic field Bo and is called the equilibrium magnetization Mo. In this configuration, the Z component of magnetization MZ equals Mo. MZ is referred to as the longitudinal magnetization."?

enter image description here

The only thing that came to my mind is: it is due to the fact that we are considering a lot of atoms and so the x,y components of the net magnetization vector are -on the average- equal to zero.. But I'm not sure that it could be right...

2) In the paragraph "Spin Relaxation" I dont't understand if the motions that influence $T_1$ are rotational motions at the Larmor frequency or any motion that causes a time varying field at the Larmor frequency..

And I'd like to understand if the loss of phase of the transverse magnetization is due to the action of the many molecules that rotates at a frequency less than and equal to the Larmor frequency..

Many thanks!

  • 1
    $\begingroup$ Regarding 1) the spins are quantised so they can't point parallel to $B$. They're instead at an angle of 54$^\circ$ (spin-1/2) or 45$^\circ$ (spin-1) to the field. Hence the precession. And your assumption regarding the x,y components averaging to zero is correct. I don't understand your question 2) though. $\endgroup$
    – lemon
    Oct 1, 2014 at 22:57

2 Answers 2


I don't consider myself an expert in MRI, but let me try (since nobody else has stepped up in the last hour...)

You are right with your first assertion: the spin precesses about the B vector (this is why you get resonance in the first place). However, on average there is a net component of the magnetic moment aligned with the B field. This is what gives rise to the change in energy - aligned or misaligned you will have either a drop or increase in energy.

It is this "average" magnetization $M_0$ that is equal to the component of the magnetic field along Z (the traditional direction of the magnetic field in an MRI). Yes - the X and Y component are on average zero, especially if you average over time (because of the precession of the individual nuclei).

As for your second point - T1 is called the spin-lattice relaxation as it refers to the way that the spin of a proton reverts to the "mean for the system"; in other words, when left alone the system will return to a certain number of up and down spins (net magnetization) which is different than the magnetization it had after the RF stimulus. So it is really "any motion" of the surrounding material at the right frequency - anything that can cause the spin transition. As your notes state, the higher the density of motions at the Larmor frequency, the shorter T1 (the more chances that the spin of an individual proton will be flipped).

Finally - transverse magnetization happens when the protons are all in phase, which you force by the RF pulse. As the protons experience slightly different local fields, they will end up precessing at slightly different phase which results in the loss of transverse magnetization. This is the T2 mechanism.

As was pointed out by CuriousOne, it is important to remember that the nuclear spins are subject to thermal equilibrium - there is only a small energy difference between the up and down states, so there will only be a small net magnetization (Boltzmann at work). There is something called hyperpolarization - a mechanism whereby nuclei (for example C13) are cooled to very low temperatures (below liquid helium, I believe) after which they can be hyper polarized - with the consequence that for a short time they exhibit a very strong MRI signal (up to $10^5\times$ stronger than at room temperature). Add to this the fact that C13 has a resonance at a different frequency than protons or any other nuclei in the human body, and you can briefly visualize organic molecules made with this technique with exquisite sensitivity (signal to noise ratio). I believe this is now used to image physiological processes such as cardiac or tumor metabolism (choline, acetate etc) in vivo without using ionizing radiation - an exciting new frontier in MRI. See for example this press release


This is perhaps a comment to the @Floris answer. (I can move it there.)
First about T1, It not only describes the decay of the Z-magnetization. But if you suddenly turn on a B field it also describes how long it takes the spins to become polarized in that direction. (Spins are not immediately polarized.)

Concerning the decay of x-y magnetization. (I think) This is only possible in the pulsed experiments. In that case there is an initial phase of the magnetization, because all the spins get flipped into the x-y plane at the same time. Then as sunrise said you can have spins in different magnetic environments have different frequencies. The biggest cause of the differences is typically spatial inhomogeneities in the applied field. (It's hard to make exactly the same B field everywhere.)

And then finally if you do have a perfect B field, decay in x-y is also limited by the T1 processes. (T2<=T1)

  • $\begingroup$ Thanks for you answer! But I'm not sure to have correctly understood what you said about T1. I have understood that after a time T1, the Z-component of the net magnetization, is equal to 63% of its value at the equilibrium. Does it indicate also the time that the spins take to align with the external field? $\endgroup$
    – sunrise
    Oct 2, 2014 at 13:50
  • 1
    $\begingroup$ Yes that's correct. If you turn on a B field, T1 describes the time it takes for the spins to align. If you leave the field on for a long time (long compared to T1) and then tun it off, T1 describes the time it takes the magnetization to decay. A story from the early days of NMR. One of the first people to look for NMR (don't know the name.) tried to see a signal in water. They didn't know T1 (of course they were first.) And assumed it would be fairly fast, they couldn't find a signal. Problem was that T1 is water can be several seconds! $\endgroup$ Oct 2, 2014 at 14:18
  • $\begingroup$ Hi George, I have a question (I ask you because you work with NMR apparatus.. I hope not to disturb you!). I haven't understood why "As transverse magnetization rotates about the Z axis, it will induce a current in a coil of wire located around the X axis".. could you help me? Many thanks again! $\endgroup$
    – sunrise
    Oct 9, 2014 at 20:51
  • $\begingroup$ (I can't understand why the magnetization induces current) $\endgroup$
    – sunrise
    Oct 9, 2014 at 21:34
  • 1
    $\begingroup$ @sunrise, expanding on Floris.. The magnetization is a little magnet. imagine it lying in the x-y plane and spinning about z-axis. A coil around the x-axis will see a changing magnetic field as first the magnet points one way (then perpendcitual to coil and not field) and then the other way. A changing B feild in a coil makes a current. What's important is that all the little magnetic moments in the atoms are in phase. (point in same direction at same time.) If they get out of phase there will be no induced current. $\endgroup$ Oct 10, 2014 at 12:37

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