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Biomolecular NMR spectroscopy

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									Biomolecular NMR spectroscopy Exercise 1 1. Write equations of motion for Cartesian components of net magnetization M =  (where  = I) when the total angular momentum J = I evolves as
dJ  MB dt

in a magnetic field B. When multiplying both sides with  and breaking the cross product into the components gives dM x  M y  B z  M z B y dt dM y  M z B x  M x  B z . dt dM z  M x B y  M y  B x dt 2. Make a transformation to a frame of reference rotating with the angular velocity  about z-axis in the special case when Bz = Bo, Bx = By = 0. Use nomenclature o = -Bo and define the offset  =  – o. From the previous we set Bz = Bo, Bx = By = 0 and o = -Bo and to make the transformation to the rotating frame we replace Bo by Bo +  = -o +  =  to give
dM x  M y dt dM y  M x . dt dM z 0 dt

3. Extend the set of equations by including recovery to equilibrium (relaxation) with longitudinal and transverse rate constants R1 and R2 in the special case when Bz = Bo, Bx = By = 0. The recovery towards equilibrium is a first-order rate dM x   R2 M x dt dM y   R2 M y . dt dM z  R1 M o  M z  dt That we will simply add with the equations in 2. Note the recovery destroys the transverse components and drives the longitudinal towards Mo along the magnetic field. Solve your set of equations for the initial condition Mx(0) = Mo, My(0) = Mz(0) = 0. Plot Mx(t), My(t) and Mz(t).

The initial conditions tell us that the magnetization is along x-axis. Integration of the differential equations is easiest to do when writing M+ = Mx + iMy. gives M  (t )  M o exp(i  R2 )t . M z (t )  M o 1  exp( R1t )

4. Extend the set of equations in 2, i.e. omitting recovery, but including a transverse radiofrequency perturbation with the amplitude B1 given at rf. Write the set of equations in a matrix notation dM(t)/dt = RM(t) where R contains elements that depend on the offset  and the components of the effective field Br in the rotating frame. We take the previous equation in 1 to include an rf-field along x-axis that will give a rotation Bxr in the rotating frame. Then the set of three equations can be written in the matrix from which is particularly convenient 0   M x (t )   M x (t )   0   dM (t ) d      M (t ) 0 Bxr   M y (t ) .  y     dt dt r 0   M z (t )   M z (t )   0  Bx      5. Solve the equation of motion in 4 for M(p) when -B1p = 3/2 along x-axis in the special case when M(0) = Mz and  = 0. We simply rewrite the matrix with the given initial conditions. 0 0  0   M x ( t )  0 dM (t ) d    0  M (t ) 0 Bxr   0   y     dt dt r 0   M z (t )  M z (t )  0  Bx      d M (t )  0, dt x d r M y (t )  Bx M z (t ), dt d r M z (t )  Bx M y (t ) dt The equations are satisfied by a circular motion about the x-axis e.g. Mz(t) = Mocos(t) and Mosin(t). A pulse with 270 rotation will give Mz = 0 and My = Mo. 6. Continue to follow the faith of M(t) from M(p) under Bz = Bo including relaxation. The equations must show that the magnetization is initially along My with the length Mo. It will precess around z-axis while R2 will diminish the amplitude in the xy-plane and R1 will create longitudinal component (see. 4).


								
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