I have a confusion regarding MRI signal formation. So, as I understand it, we need to solve the Bloch equations for the excitation and the relaxation stages:

So, in the excitation stage, let us assume that we have an RF field excitation which is applied along with the slice selection gradient. The bloch equations are given by:

$$ \frac{dM}{dt} = \gamma M \times B - R $$

where $R$ is the vector accounting for the relaxation effects. Now, during excitation, the B1 field is on and as I understand it, the B vector is given as:

\begin{align} B_x &= B_1 \cos(\omega t)\\ B_y &= B_1 \sin(\omega t)\\ B_z &= z G_z \end{align}

The $B_z$ field is basically the gradient times the position offset from the origin and the $B_x$ and $B_y$ fields are the real and imaginary parts of the B1 waveform.

Now, when the B1 field is turned off, I think $B_x$ and $B_y$ components should be 0. However, let us assume that we have a readout gradient on when we are sampling the MR signal.

What will contribute to the $B_z$ field now? Assuming that the RF excitation is with a 90 degree pulse and after the excitation the magnetization vector lies completely along the $Y$ axes.


2 Answers 2


In the rotating reference frame, $B_z$ from the static magnetic field $B_0$ is transformed away, hence $B_z = 0$ if no gradient field is applied. All gradients change the $B_z$ value, as you described it with the slice selection gradient. This gradient causes $B_z$ to increase when you move along $z$. Note that it is irrelevant if you do this from the center of the magnet or somewhere close to the bore - you will have the same $z$-dependence at each position $(x,y)$.

For the readout-gradient (assuming it is along left-right direction that we'll call $x$), the same is true: $B_z = G_Rx$. If you go from left to right, $B_z$ changes. In this case, it is independent of the position along the bore, so moving along $z$ does not have an influence on the value, if you stay at the same $x$ coordinate.

A gradient field always acts along the $z$-component of the magnetic field. This is always a bit tricky, since gradients can be "applied along each direction".

  • 1
    $\begingroup$ it took me forever to figure this out. I still do not understand why the gradient always acts on the z-component of the field....So, this has to do with the transformations due to the rotating frame of reference? $\endgroup$
    – Luca
    Jul 4, 2016 at 15:38
  • 1
    $\begingroup$ No, they are designed that way. Have a look at mri-q.com/uploads/3/4/5/7/34572113/2583013_orig.gif?305 (I just found this image online). The field always points along $z$, but the change (i.e. mathematical gradient so to say) can be along any of the three directions. $\endgroup$
    – M529
    Jul 4, 2016 at 17:16
  • $\begingroup$ Wow...literally a case of a picture worth a thousand works! $\endgroup$
    – Luca
    Jul 4, 2016 at 21:24

It has been a while since I worked with these equations, but I think you should think about that is happening to the direction of the proton magnetization. $B$ is a vector that depends on where the magnetic dipole of the proton points.

Before $B_1$ is turned on, the proton will be aligned with the steady-state magnetic field. It will only have a $B_z$ component.

When $B_1$ is on, if $\omega$ is at the resonant frequency, the proton will process in the $xy$ plane. At the moment $B_1$ turns off, the proton is still in the $xy$ plane, and it continues to precess as it decays back to the $z$ direction. The rate at which it decays is determined by how the proton is bound - whether it is in a sugar, free water, or bone marrow, etc. While the proton is decaying is in a transient regime. After a while, the proton will again itself with the $z$ direction. Then it will be in the steady state regime.

When the proton magnetization is in the $xy$ plane, $B_z$ is equal to zero. After the decay, $B_z$ will return to its initial value.

This decay will be picked up by the MRI, location will be determined based on gradient information, and an intensity will be assigned to an appropriate voxel in an image.

Hope this can help.


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