Help with math/physics of MRI!!! (docxter, please respond)

[devildoc2]

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Prove that d/dt M = M x y (B - BoZ), where both M and B are the magnetization and magnetic field in the rotating frame of reference, given that d/dtM = M x yB

M and B are vectors.

Is this some kind of geometric proof or what? I dont even know where to start.

 

[Docxter]

Prove that d/dt M = M x y (B - BoZ), where both M and B are the magnetization and magnetic field in the rotating frame of reference, given that d/dtM = M x yB

M and B are vectors.

Is this some kind of geometric proof or what? I dont even know where to start.

Are you sure you typed your question correctly? (I undersatnd that it's hard to type equations in plain text) I'm asking since what you have typed only applies if BoZ=0. In any case I think your question refers to the derivation of the simplified Bloch equation (in rotating frame) from the original governing Bloch equation. It's essentially done by applying a rotation transformation to the the euqation. I can't type it step-by-step because it's too long, but if you need more help, take a look at Chapter 6 in Nishimura's Principles of Magnetic Resionance Imaging.

 

[bigdan]

Good god above. I've been interested in Rads, but if I have to be able to handle physics like that, I'd better look at my other interests.

dc

 

[ears]

If you want to do rads, you can probably get through that stuff. Rads is definitely not limited to physicists...

 

[Whisker Barrel Cortex]

You don't need to know as much detail as that unless you are going into MR research.

 

[Docxter]

Good god above. I've been interested in Rads, but if I have to be able to handle physics like that, I'd better look at my other interests.

dc

You don't need this much physics to become a radiologist, even a good one. This type of physics is for people who want to do high level research in imaging technology and transfer them into the the clinical realm, i.e. the "bench to bedside" research analogy. MR physics to radiology is like what molecular genetics is to medicine/pathology.

 

[devildoc2]

OK docxter you are right I typed it wrong.

I finally figured it out. Took me awhile though. The key is to make conversions between lab frame and rotating frame of reference, make the substitutions and play with the original d/dt M = yM x B equation.

Here's the proof for all those interested:

Key:

* = dot product
x = cross product
M = lab frame (stationary) magnetization vector
Mrot = rotational frame magnetization vector
B = lab frame magnetic field vector
Brot = rotational frame magnetic field vector
y = gyromagnetic constant (42.5 MHz/T for hydrogen proton)
Bo = scalar for magnetic field strength
W = rotational frequency of proton in magnetic field
Wrot = rotational frequency vector in rotating frame
n = unit rotation vector about w

Given:

d/dt M = M x yB

Prove:

d/dt Mrot = Mrot x y(Brot - BoZ)


Proof:

d/dt M = M x yB

M = Mrot * n (Rotational frame dot product with unit vector rotation gives you the original lab frame of reference).

d/dt (Mrot * n) = (Mrot * n) x yB

Mrot * d/dt n + n * d/dt Mrot = (Mrot * n) x yB

d/dt n = - n x W (negative rotation about i, j, k axes assuming clockwise motion of rotation vector about W vector)

Mrot * (-n x W) + n * d/dt Mrot = (Mrot * n) x yB

n * d/dt Mrot = -Mrot * (-n x W) + (Mrot * n) x yB

n * d/dt Mrot = Mrot * (n x W) + (Mrot * n) x yB

n * d/dt Mrot = [(Mrot * n) x W] + [(Mrot * n) x yB]

n * d/dt Mrot = (Mrot * n) x (W + yB)

W = Wrot * n

n * d/dt Mrot = (Mrot * n) x [(Wrot * n) + yB]

B = Brot * n

n * d/dt Mrot = (Mrot * n) x [(Wrot * n) + y(Brot * n)]

n * d/dt Mrot = (Mrot * n) x [(Wrot + yBrot) * n]

n * d/dt Mrot = [Mrot x (Wrot + yBrot)] * n

d/dt Mrot = Mrot x (Wrot + yBrot)

Wrot = -WoZ, (assume frame is rotating about z/k axis at frequency Wo)

d/dt Mrot = Mrot x (-WoZ + yBrot)

Wo = yBo where Bo = scalar of static magnetic field strength

d/dt Mrot = Mrot x (-yBoZ + yBrot)

d/dt Mrot = Mrot x y(Brot - BoZ)

Q.E.D.

 

[bigdan]

Wow.

dc

 

[devildoc2]

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