Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why does the magnetisation orient itself in the -y direction

  1. May 7, 2016 #1
    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 dont understand why it would not be in the +Y direction
     
  2. jcsd
  3. May 7, 2016 #2

    Charles Link

    User Avatar
    Homework Helper

    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 cancelled 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.
     
    Last edited: May 7, 2016
  4. May 8, 2016 #3

    Charles Link

    User Avatar
    Homework Helper

    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.
     
    Last edited: May 8, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted