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© The scientific sentence. 2010

The NMR
1.Larmor frequency: The precession of the proton
In the absence of am external magnetic field, the magnetic
moment of spin are oriented randomly in space. When an
external magnetic field is applied, the magnetic moment of spin μ
will tend to align with the field, as shown in the folowing figure:
The manetic moment of spin is extended from the definition of
the angular magmetic momemt. Applyied for the electron, we have:
μ_{e/orbit} = eL/2m_{e} (1)
Where L is the angular momentum for the electron
Then:
μ_{e/spin} = geS/2m_{e} (2)
Where S is the spin angular momentum for the electron
g is called the Lande factor
g = 2.00232 for the electron
For the proton, let's write:
μ_{p/spin} = egS/2m_{p} (3)
g = 5.58569
The torque for the angular momentum is that
the force is for the linear momentum. We define the
vector torque τ applyied for the spin magnetic moment μ in a
magnetic field B_{0} as:
τ = μ x B_{0} = μ B_{0} sinθ (4)
Where &theta is the angle between μ and B_{0}.
Since dφ is small, We can write :
dS = S sinθ d&phi (5)
The variation of S in time can be written as:
τ = dS/dt = S sinθ dφ/dt
Using the equatios: (4) and (3), we get: (6)
τ = μ B_{0} sinθ = egS/2m_{p}B_{0} sinθ
Using (6) we get:
dφ/dt = eg/2m_{p}B_{0} (7)
Let's write this expression as follow:
ω_{0} = eg/2m_{p}B_{0}
= γB_{0} (8)
Where γ = eg/2m_{p}, (9)
called Gyromagnetic ratio.
Or ƒ_{0} = ω_{0}/2π = eg/4 π m_{p}B_{0}
Which is called the Larmor frequency.
The two possibles values of S are :
m_{s} = +1/2 or m_{s} =1/2
The related energy for each state is :
E_{1/2} =  μB_{0} (opposite the direction of the field : high energy).
E_{+1/2} = + μB_{0} (same as the direction of the field : low energy).
The difference is:
Δ = 2 μB_{0} = ħω
(according to the Planck relationship. ħ = h/2π , h is the Planck constant.)
Then:
ω = 2 μB_{0}/ħ (10)
We have the two following values for b = 1 Tesla (10,000 Gauss):
ω = 2.67 x 10^{8} s^{1}
ƒ = 42.58 MH_{z}
2. NMR: The principle
NMR stands for Nuclear Magnetic Resonace. That is
the effect of the spins is efficient only in the case of the
external frequency Ω is in the order of ω_{0} (Larmor):
Ω ≈ ω_{0}.
In this figure:
The sample to study is set inside the coil receiving the RF.
. B_{0} is the first magnetic field applyied.
. B_{1} is the second magnetic field generated
by FG: the frequency generator (Radiofrequency input)
. This field oscillates over the x axis.
. RFO is the radiofrequency output. It is the signal to
analyse by the imaging technique using Fourier transform
to deal with frequencies instead of time variable, that is
the MRI ( Magnetic Resonance Imaging).
3. The Bloch Equations:
The magnetization M_{0} of the substance is the average of
all the spin magnetic moments of each atom μ_{i}. We write:
M_{0}= Σμ_{i} which is pointed along of the z axis.
The second field B_{1}(t) exerts a second torque
on the moment M_{0} leading it to the y axis, as shown below:
The Bloch equations explain the evolution of the magnetization moment
during its precession. Since the magnetic field B_{1}(t) is
time dependant, so is the magnetization moment: M(t).
In the frame laboratory the magnetization vector M rotates
about y abd about z axix.
In the rotating frame the magnetization vector M rotates
about y axis.
Let's consider that the field B points toward certain
direction.
We have for the proton:
τ = μ x B = dS/dt (11)
From (9) and (3), we have:
γ S = μ
Then the relationship ( 11) becomes:
dμ/dt = μ x γB
The summation of μ over a unit volume gives:
dM/dt = M x γB (12)
That is M precesses about B with a frequency equal to ω = γB
This is the first equation of Bloch.
The important phenomenun related to the evolution of the
Magnetization vector is its relaxation following the
precession induced by the excitation.
Two relaxations occur: the longitudinal relaxation and the transverse relaxation.
3.1. The longitudinal relaxation
The longitudinal component of the magnetization M behaves
as follow:
dM_{z}/dt =  (M_{z}  M_{0})/T_{1} (13)
The solution of this equation is :
M_{z} = M_{0} + (M_{z}(0) M_{0})e^{t/T1}
Since M_{z}(0) = 0 ( at 90^{o}), we have then:
M_{z} = M_{0} ( 1 e^{t/T1}) (14)
T_{1} is called the spinlatice time constant. It
characterizes the return to equilibrium along the zdirection.
3.2. The transverse relaxation
It behaves according to :
dM_{xy}/dt =  M_{xy}/T_{2} (15)
Since M_{xy}(0) = M_{0}, the solution of this equation is:
M_{xy} = M_{0} e^{t/T1} (16)
T_{2} is called the spinspin time constant and characterizes
the decay of the transverse nuclear magnetization.
Combining the equations (12), (13) and (15), the general
Bloch equation becomes:
The first term of the above equation can be written as:
dM/dt = (M_{y}γB_{0},  M_{x}γB_{0}, 0)
The solution of the Bloch equation is:
M(t) = T R_{y}(ω_{0}t) M^{0} + (0, 0, M_{0} ( 1 e^{t/T1}) )
Where:
R_{y}(ω_{0}) is the rotation matrix about z.
M^{0} = (0,0, M_{0})
T = I.(e^{t/T2}, e^{t/T2}, e^{t/T1})
where I is the 3x3 unit matrix.
In the rotating frame, we can write:
M_{rot}(t) = (0, M_{rot}_{y}, M_{rot}_{z}) = (0, M_{0}sin ωt, M_{rot}cos ωt)
The first term from the Bloch equation can be written:
dM_{rot}/dt = M_{rot} x γ B_{eff}
Where B_{eff} = B_{rot}  ω/γ
The Bloch equation is the same in the two references, but
in the rotating frame B_{eff} influences M_{rot}.
