Ignore:
Timestamp:
Nov 3, 2017 3:55:22 PM (7 years ago)
Author:
Paul Kienzle <pkienzle@…>
Branches:
master, magnetic_scatt, release-4.2.2, ticket-1009, ticket-1094-headless, ticket-1242-2d-resolution, ticket-1243, ticket-1249, ticket885, unittest-saveload
Children:
6ccc55f
Parents:
08b9e331
Message:

translate equation images back into latex for sas_calculator_help

Location:
src/sas/sasgui/perspectives/calculator/media
Files:
11 deleted
1 edited

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  • src/sas/sasgui/perspectives/calculator/media/sas_calculator_help.rst

    r5ed76f8 r2f7ea43  
    2626intensity from the particle is 
    2727 
    28 .. image:: gen_i.png 
     28.. math:: 
     29 
     30    I(\vec Q) = \frac{1}{V}\left| 
     31        \sum_j^N v_j \beta_j \exp(i\vec Q \cdot \vec r_j)\right|^2 
    2932 
    3033Equation 1. 
     
    4649atomic structure (such as taken from a PDB file) to get the right normalization. 
    4750 
    48 *NOTE! $\beta_j$ displayed in the GUI may be incorrect but this will not 
     51*NOTE!* $\beta_j$ *displayed in the GUI may be incorrect but this will not 
    4952affect the scattering computation if the correction of the total volume V is made.* 
    5053 
     
    5659^^^^^^^^^^^^^^^^^^^ 
    5760 
    58 For magnetic scattering, only the magnetization component, $M_\perp$, 
    59 perpendicular to the scattering vector $Q$ contributes to the magnetic 
     61For magnetic scattering, only the magnetization component, $\mathbf{M}_\perp$, 
     62perpendicular to the scattering vector $\vec Q$ contributes to the magnetic 
    6063scattering length. 
    6164 
     
    6467The magnetic scattering length density is then 
    6568 
    66 .. image:: dm_eq.png 
     69.. math:: 
     70 
     71    \beta_M = \frac{\gamma r_0}{2 \mu_B}\sigma \cdot \mathbf{M}_\perp 
     72        = D_M\sigma \cdot \mathbf{M}_\perp 
    6773 
    6874where the gyromagnetic ratio is $\gamma = -1.913$, $\mu_B$ is the Bohr 
     
    8187.. image:: gen_mag_pic.png 
    8288 
    83 Now let us assume that the angles of the *Q* vector and the spin-axis (x') 
    84 to the x-axis are $\phi$ and $\theta_\text{up}$ respectively (see above). Then, 
     89Now let us assume that the angles of the $Q$ vector and the spin-axis ($x'$) 
     90to the $x$-axis are $\phi$ and $\theta_\text{up}$ respectively (see above). Then, 
    8591depending upon the polarization (spin) state of neutrons, the scattering 
    8692length densities, including the nuclear scattering length density ($\beta_N$) 
     
    8995*  for non-spin-flips 
    9096 
    91    .. image:: sld1.png 
     97.. math:: 
     98    \beta_{\pm\pm} = \beta_N \mp D_M M_{\perp x'} 
    9299 
    93100*  for spin-flips 
    94101 
    95    .. image:: sld2.png 
     102.. math:: 
     103    \beta_{\pm\mp} = - D_M(M_{\perp y'} \pm i M_{\perp z'}) 
    96104 
    97105where 
    98106 
    99 .. image:: mxp.png 
     107.. math:: 
    100108 
    101 .. image:: myp.png 
     109    M_{\perp x'} &= M_{0q_x}\cos\theta_\text{up} + M_{0q_y}\sin\theta_\text{up} \\ 
     110    M_{\perp y'} &= M_{0q_y}\cos\theta_\text{up} + M_{0q_x}\sin\theta_\text{up} \\ 
     111    M_{\perp z'} &= M_{0z} \\ 
     112    M_{0q_x} &= (M_{0x}\cos\phi - M_{0y}\sin\phi)\cos\phi \\ 
     113    M_{0q_y} &= (M_{0y}\sin\phi - M_{0y}\cos\phi)\sin\phi 
    102114 
    103 .. image:: mzp.png 
    104  
    105 .. image:: mqx.png 
    106  
    107 .. image:: mqy.png 
    108  
    109 Here the $M0_x$, $M0_y$ and $M0_z$ are the $x$, $y$ and $z$ 
    110 components of the magnetisation vector in the laboratory $xyz$ frame. 
     115Here the $M_{0x}$, $M_{0y}$ and $M_{0z}$ are 
     116the $x$, $y$ and $z$ components of the magnetisation vector in the 
     117laboratory $x$-$y$-$z$ frame. 
    111118 
    112119.. ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ 
     
    148155uses the Debye equation below providing a 1D output 
    149156 
    150 .. image:: gen_debye_eq.png 
     157.. math:: 
     158 
     159    I(|\vec Q|) = \frac{1}{V}\sum_j^N v_j\beta_j \sum_k^N v_k \beta_k 
     160        \frac{\sin(|\vec Q||\vec r_j - \vec r_k|)}{|\vec Q||\vec r_j - \vec r_k|} 
    151161 
    152162where $v_j \beta_j \equiv b_j$ is the scattering 
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