T1 relaxation

It is known by many experimental results that if a bunch of magnetic moments are placed in external magnetic field, a certain part of these moments will align to the direction of the field. If we are talking about quantum mechanical objects, for example spin-half particles, this means that in the equilibrium state of the system a bit more than half of the spins will be parallel to the external field, and a bit less than half of them will be antiparallel to it. The ratio of these two depends on their energy level in the magnetic field, which is of course determined by the projection of the angular momentum to the field:

$\label{energy_magnetic_field} E_m = - \gamma \hbar B_0 m = \pm \frac{1}{2} \gamma \hbar B_0$ (1)

And the equilibrium ratio of the parallel ( $N_+$ ) and antiparallel ( $N_-$ ) spins is the well-known Boltzmann factor with the energy level difference:

$\label{equilibrium_ratio_spins} \frac{N_-}{N_+} = \mathrm{e}^{- \tfrac{ \Delta E }{k_B T}} = \mathrm{e}^{ - \tfrac{ \gamma \hbar B_0} {k_B T} }$ (2)

If we alternate the ratio for example by applying an RF excitation the system will somehow return to the equilibrium state described in (2) after the excitation effect ceases. In a simple model where we assume the spins to be completely independent we can describe the relaxation method by two transition probabilities per unit time: the probability of a parallel spin becoming antiparallel $\big ( N_+ \to N_- \big )$ denoted by $W \hspace{-3pt} \downarrow$ , and the probability of the opposite, i. e. an antiparallel spin becoming parallel $\big ( N_- \to N_+ \big)$ denoted by $W \hspace{-3pt} \uparrow$ . Note that these two are generally not equal because of the environment of the spins end the existence of the external magnetic field. With these probabilities, the time derivatives of the spin population can be written as:

$\frac{ \mathrm{d} N_+} { \mathrm{d} t } = - N_+ W \hspace{-3pt} \downarrow + N_- W \hspace{-3pt} \uparrow$ (3)

$\frac{ \mathrm{d} N_-} { \mathrm{d} t } = N_+ W \hspace{-3pt} \downarrow - N_- W \hspace{-3pt} \uparrow$ (4)

The measureable quantity is the difference between these populations, so we will use the followings:

(5)

(6)

Using (3) and (4) the time derivative of the population difference will be:

$\label{spin_difference_derivative} \frac{ \mathrm{d} n} { \mathrm{d} t } = N \big (W \hspace{-3pt} \uparrow - W \hspace{-3pt} \downarrow \big ) - n \big (W \hspace{-3pt} \uparrow + W \hspace{-3pt} \downarrow \big )$ (7)

Whereof the equilibrium value of the population difference is

$\label{spin_difference_equilibrium} n_0 = \frac {N \big (W \hspace{-3pt} \uparrow - W \hspace{-3pt} \downarrow \big ) } {\big (W \hspace{-3pt} \uparrow + W \hspace{-3pt} \downarrow \big )}$ (8)

From this we define a time-dimension quantity as

$\label{T1_definiton} T_1 \equiv \frac{1} {\big (W \hspace{-3pt} \uparrow + W \hspace{-3pt} \downarrow \big )}$ (9)

Using this $T_1$ the time derivative of the population difference is as follows:

$\label{spin_difference_derivative2} \frac{ \mathrm{d} n} { \mathrm{d} t } = \frac{ n_0 -n} {T_1}$ (10)

With the solution of

$\label{spin_relaxation_T1} n(t) = n(0) \mathrm{e}^{- \tfrac{t}{T_1}} + n_0 \left ( 1 - \mathrm{e}^{- \tfrac{t}{T_1} } \right )$ (11)

As can be seen, the time dependence of the spin population difference - with the latter proportional to the longitudinal component of the magnetization pointing to the direction of the external field - shows exponential decay to the equilibrium state with a time constant $T_1$ . This effect is called longitudinal or $T_1$ relaxation.

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T1 relaxation

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Medical Imaging

Site Language: English