4
$\begingroup$

In MRI we measure $T_1$ and $T_2$ relaxation times. $T_1$ relaxation time is defined as the time it takes for the longitudinal component of magnetization to recover $1-1/e$ of its original value, and $T_2$ relaxation time is the time it takes for the transverse component of magnetization to recover $1/e$ of its original value. Is there any significance to those two numbers adding to 1? ($1-1/e + 1/e = 1$)

As far as I understand, $T_2$ relaxation time is influenced by different atoms falling out of phase with each other. Was the reason for $1-1/e$ and $1/e$ that if the atoms would stay in phase, then $T_1$ would be equal to $T_2$, so that the difference between $T_1$ and $T_2$ relaxation time measures the degree of interaction? In that case, since $T_1$ and $T_2$ measure orthogonal components of a vector, shouldn't it have been $1/e$ and $\sqrt{1-1/e^2}$, so that the length of the vector becomes $1$ instead of the sum of the components becoming $1$?

$\endgroup$

1 Answer 1

3
$\begingroup$

$\newcommand{\xyp}{x-y \textrm{ plane}}$Notice that the magnitude of the magnetization does not need to be conserved, because the magnetization is a sum of all the nuclear magnetic moments, and these moments can cancel each other. For example, if $T_1$ is much much larger than $T_2$, then the magnetization will be near $0$ when $T_1 \gg t \gg T_2$. I will explain everything in more detail below.

Let me talk about what would happen in an ideal system. In an ideal system, if all the nuclear spins were oriented in the $\hat{x}$ direction and you applied a DC magnetic field in the $\hat{z}$ direction, all the spins would precess in the $\xyp$, so the direction of each spin would point in the $\xyp$ and at any point in time, all the spins would be pointing in the same direction. If you came back a year later, you would still find the spins precessing in the $\xyp$.

Now in real life the this doesn't happen. There are two things that go wrong: 1) The spins reorient along the external magnetic field. This is after all the lowest energy state. So in thermal equilibrium we expect that all the spins should point along the external magnetic field, and of course we expect any system to reach thermal equilibrium after a sufficiently long time.

There is another thing that can go wrong 2) Not all the spins precess at the same rate. If this were the only thing that went wrong, then after a long time the spins would all be in the $\xyp$, but they would be pointing in random directions so that there is no net magnetization in the $\xyp$. This thing that goes wrong is an example of "decoherence".

Now the two things that go wrong are independent, and they can happen at different rates. $T_1$ is the rate that the first thing that happens, and $T_2$ is the rate that the second thing happens. But what exactly do I mean by the rate that it happens? I need to give a little more detail about how they happen.

For the first one, the way the spins reorient is that the difference between the $z$ component of the sample's magnetic field and its equilibrium value decays exponentially. This $z$ component is initially zero. So if the equilibrium $z$ component of the sample's magnetic field is $B_{ze}$, and the external magnetic field is turned on at time $t_0$, then the $z$ time dependence of the $z$ component of the sample's magnetic field is $$B_z(t) = B_{ze}(1-e^{-(t-t_0)/T_{\ 1}}\ \ ).,$$ so $T_1$ is the time constant for the exponential decay. This formula serves as the definition for $T_1$, and it explains what I meant when I said $T_1$ is how fast the process happens. Notice that at $t=t_0 + T_1$, the $z$ component of the magnetic field is $1-e$ of what the equilibrium value is. Thus this criterion could be used for a simple definition if a detailed explanation of what is going on is not wanted.

Now let's talk about $T_2$. The component of the magnetic field in the $\xyp$ has some initial value $B_{xy0}$, but after time it decays to $0$. The exact time dependence of this decays is exponential, so $$B_{xy}(t) = B_{xy0}\ e^{-(t-t_0)/T_{\ 2}}.$$ So $T_2$ is the time constant for this exponential decay. This formula serves as the definition for $T_2$. Notice here that if $t= t_0 + T_2$, the component of the magnetic field in the $\xyp$ is $1/e$ of what it originally was. Thus $T_2$ could be defined by this simple criterion as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.