Why does the magnetisation orient itself in the -y direction

[shedrick94]

In magnetic resonance, if we apply a 90 degree pulse in the x direction when we have a magnetisation orientated in the z direction. Why do we get the magnetisation then orientated in the -y direction immediately after the pulse?

I don't understand why it would not be in the +Y direction

 

[Charles Link]

Upon applying a 90 degree pulse, the entire magnetization, (from the nuclear spins), that has been rotated away from the z-direction, is precessing around in the X-Y plane if I remember the result correctly. (The large static field in the z-direction causes this precession. The large static field in the z-direction is effectively canceled only in the rotating frame. $M \times B_z$ will supply the necessary torque to cause the precession.) During the application of the r-f signal at resonance in the x-direction, the magnetization vector is precessing around in the laboratory frame, while in the rotating frame, it appears to simply make a 90 degree rotation from the z-direction to the X-Y plane. The direction the magnetization rotates in the rotating frame can be readily computed. Hopefully your textbook correctly computed it.

 

[Charles Link]

Besides what I mentioned in post #2, there are a couple of additional items that might be of interest. One important item in the derivations is the relationship between the time derivative of a vector $\vec{r}$ in the laboratory frame versus the rotating frame: $[d \vec{r}/dt]_o=d \vec{r} /dt +\omega \times \vec{r}$. One other item is that the angular momentum $J$ from magnetization $M$ is $M=\gamma J$, so that $dJ/dt=M \times B$ becomes $dM/dt=\gamma M \times B$. It should be noted in the last equation, that the solution for $M$ is that it rotates at frequency $\omega=-\gamma B$ about the vector $B$. Notice from the first equation above that $dM/dt=0$ in the rotating frame ($M$ is constant), but $[dM/dt]_o=\omega \times M$ in the laboratory frame.