Changing of frame II

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Few things have to say in advance.

  1. A Vector is NOT its coordinate
  2. A vector can only be coordinated when there is a frame.
  3. A frame is a set of “reference” vectors, which span the whole space. Those reference vectors are called basis of a frame.
  4. a transformation is on a vector or its coordinate. And it can be represented by a matrix.
  5. A Matrix should act on a coordinate or basis, but not a vector.


\hat{\alpha} = \begin {pmatrix} \hat{\alpha_1} \\ . \\ \hat{\alpha_n} \end{pmatrix} is the column vector of  basis reference vector.

\vec{u_{\alpha}} is the coordinate column vector in \alpha basis.

\vec{U} is the vector in space

\vec{V} is the transformed vector in space.

G and H are the matrix of transform.

G \cdot H \cdot G^{-1} has the same meaning of H , only the matrix representation of the transform is different due to different basis.

the Euler’s rotation can be illustrated by series of the diagram. each rotation of frame can be made by each G . but when doing real calculation, after we apply the matrix G  on the coordinate, the basis changed. when we using the fact that  a matrix can be regard as a frame transform or vector transform. we have follow:

This diagram can extend to any series of frame rotation. and the V_s \rightarrow X_s \rightarrow V_2 \rightarrow V_s triangle just demonstrate how 2 steps frame transform can be reduced to the vector transform in same frame.

i finally feel that i understand Euler angle and changing of frame fully. :D

HERE is a note on vector transform and frame transform.

detail treatment on Larmor Precession and Rabi Resonance

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a treatment on Larmor Precession and Rabi resonance

the pdf is a work on this topic. it goes through Larmor Precession and give example on spin-½ and spin-1 system.

then it introduce Density matrix and gives some example.

The Rabi resonance was treated by rotating frame method and using density matrix on discussion.

the last topic is on the relaxation.

the purpose of study it extensively, is the understanding on NMR.

the NMR signal is the transverse component of the magnetization.

Changing of frame

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sometimes, i will confuse on changing frame, especially between rotating frame and lab frame.

the notation plays an important role. for a rotating vector.

V_S' = R_n ( \omega t ) V_S

where  V_S is the static frame vector. R_n ( \theta ) is the rotation operator, with right -hand rotation is positive \theta around direction n . for clearance, when we think on a rotating, \omega is always positive.

R_n ( \theta ) \Rightarrow \begin {bmatrix} \theta > 0 & \rightarrow & + \\ \theta <0 & \rightarrow & - \end {bmatrix}

however, if we regard the vector does not rotate but is the frame rotating, which is equivalent as rotating backward ( left-hand or negative ). the expression is the same but with different notation.

V_R = R_n ( \omega_R t ) V_S

V_R is the rotating frame vector, \omega_R is the rate of rotating of the rotating frame respect to the static frame. for the rotating frame is rotating forward ( positive)

V_R = R_n ( - \omega_R t ) V_S

if both the vector is rotating an the rotating frame is rotating, in same direction, with different rate.

V_R' = R_n ( - \omega_R t ) V_S' = R_n ( - \omega_R t ) R_n( \omega t ) V_S

and combine the rotation operator.

V_R' = R_n ( (\omega - \omega_R ) t ) V_S

which make perfect sense.

the principle can apply to any frame transformation.

__________________  Examples _____________________

a rotation on positive \theta direction for the vector.

V_S = ( cos (\alpha ) , sin(\alpha ) )

V_S' = ( cos( \theta + \alpha ) , sin(\theta + \alpha) ) = V_R

(A_x)_S = (1,0)_S

(A_x)_R = ( cos ( \theta ) , sin ( \theta ) )_R

( A_x)_R = ( 1, 0 )_R = ( cos ( \theta ) , sin ( - \theta ) )_S

if we add the frame reference on as a subscript, things will be much clear. since the same vector can have different coordinate expression. so, we better mark the coordinate with reference system.

Special Relativity II

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We are going to talk about coordinate transform from “center of momentum frame” to “Laboratory frame”.

At the center of momentum [C.M.] frame, the total momentum is zero.

Refer here on Google Docs. for the Mathematica 7 code and calculation steps.

anyone want to have the .nb file, feel free to ask.

i am just discuss on the result.

  1. In the C.M. frame, energy of each particle reserved, there is no exchange in energy and momentum. after collision, they just change the moving direction.
  2. the energy is C.M frame is always smaller then Lab frame, or other fame. which is also from the face that, at the CM frame, total momentum is zero and the corresponding energy is the Proper Energy.
  3. In Lab frame, the scatter angle is always smaller than 90 degree for incident particle’s mass > target particle.
  4. In Lab frame, The larger the scatter angle, the smaller the momentum and larger the momentum transfer.