Spin alignment in Magnetic Resonance Imaging

[Anonymouse]

I'm reading about the theory behind MRI, specifically from Farr's Physics for Medical Imaging, 2nd Edition, by Allisy-Roberts and Williams.

According to this book, the spins of the free protons in a sample line up, either parallel or antiparallel, with the externally applied magnetic field. In the case of a 1T field, there will be an excess of approximately 3 in every 1000000 protons aligned in the lower energy state parallel with the field, giving rise to a net magnetism M_z (where z represents the axis of the external field).

For MRI, a signal is generated by a set of RF coils that inject an RF pulse in a direction perpendicular to z. The book explains that this has the effect that some or all of the spin-up protons pick up energy, turn spin down, affecting M_z, which may be reduced, disappear or even reverse.

In particular, the book explains that for a so-called "180° pulse", an RF pulse of a certain total energy will give to each and every dipole exactly the energy required to tip them through 180°. This temporarily reverses the net magnetic vector M_z.

My question is, as far as I understand, if an RF pulse at the resonant frequency is directed towards the sample, then the chance of flipping a proton from the low to high energy state also applies to flipping a proton from the high to low energy state. Therefore, the best that could be achieved is a 50% equilibrium between the high and low energy states, giving M_z a minimum value of 0 (i.e. it cannot be negative).

How is it that "each and every dipole" will be flipped through 180°, thus reversing M_z? I think the dipoles will be flipped at random, and that M_z will be reduced, but cannot be reversed.

Am I misunderstanding something?

 

[Charles Link]

The 180 degree pulse is a special phenomenon, and differs from the resonant absorption. In the resonant absorption that occurs at the resonant frequency, the spins have higher population in the lower energy state from the static field $E=-\mu \cdot$ B , and they can be driven to nearly equal populations with an absorption of r-f energy occurring at the resonant frequency. $\\$ In the 180 degree r-f pulse (at the resonant frequency), in the frame rotating with the r-f field,(*), the r-f magnetic field becomes a DC magnetic field, and the strong DC field in the z-direction is effectively zero in the rotating frame. $\\$ (The equation $[\frac{dM}{dt}]_o=\frac{dM}{dt}+\omega \times M$, which holds for the derivative of any vector for how the time derivative in the laboratory frame is related to that in the rotating frame, along with $[\frac{dM}{dt}]_o=\gamma M \times B$ causes the effective DC field in the z-direction to be zero at resonance in the rotating frame where $\omega_o=-\gamma B$). $\\$ Thereby, the magnetization vector $M_z$ in the rotating frame "rotates about what appears to be a static magnetic field" (from the r-f magnetic field) in the x-y plane. If the duration of the r-f pulse is selected properly, it can be be turned off just as the magnetization $M_z$ points in the minus z direction. In the laboratory frame, this rotation of the $M_z$ vector will be a precession that eventually works its way to the minus z direction during the 180 degree r-f pulse. $\\$ For another reference on this, C.P. Slichter's text Principles of Magnetic Resonance gives a good treatment of it in Chapter 2.4. $\\$ (*) The r-f field is dived into two components: One rotating clockwise, and the other counterclockwise. In the frame rotating with one of the components, the other component rotates at twice the resonant frequency in the other direction and is assumed to have negligible effect.

 

[Anonymouse]

Thanks! That's exactly what I needed to know.

 

[Charles Link]

Just one additional input: After the 180 degree r-f pulse finishes, the magnetization will be $M=-M_o \hat{z}$. The static field in the +z direction will eventually cause the magnetization to reach a value of $M=+M_o \hat{z}$ once again. I'm not sure exactly what the time scale of this change is, but I think it occurs in something like 10 seconds to a minute. $\\$ Editing: A google showed a typical time to return to zero magnetization ($T_1$) is a couple hundred milliseconds. I believe these are for room temperature samples. If liquid nitrogen is used on the sample, this $T_1$ is likely to be much longer. $\\$ Additional editing: In the first method of working with the resonance, if you use a relatively weak r-f signal, you can get steady resonant absorption for an extended period, but if you use a somewhat strong r-f signal, you can easily make the spin populations nearly equal very quickly, and then you no longer get the resonant absorption. To get resonant absorption, you need an excess of spins in the lower energy state, i.e. aligned with the strong DC magnetic field. $\\$ Editing some more: An additional input: Normally with simple r-f absorption of the first kind mentioned above, you can drive the magnetism to zero, but you don't overpopulate the higher energy (spin opposite the static field) state. This 180 degree pulse is special in that it gives you an excess of spins in the opposite state. In section 2.8, C.P. Slichter treats the steady state case (much longer input duration of the r-f signal than a 180 degree pulse) of r-f absorption for a moderately weak r-f input signal. In the steady state case, resonant absorption occurs, but the value of $M_z$ is very nearly equal to $M_o$, i.e. the value it had before the r-f signal is applied. Meanwhile, in this steady state case, a rotating component of magnetization occurs in the x-y plane (rotating at the resonant frequency, or at the r-f frequency if the r-f signal is just off resonance), and a steady r-f input power (basically resonant absorption) is necessary to sustain it.

 

[Charles Link]

@Anonymouse Please see the edited additions to post #4 above.