- Timestamp:
- Nov 28, 2017 9:21:33 AM (7 years ago)
- 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:
- 45bfe3f1
- Parents:
- 0fc5a03 (diff), 1b67f3e (diff)
Note: this is a merge changeset, the changes displayed below correspond to the merge itself.
Use the (diff) links above to see all the changes relative to each parent. - git-author:
- Steve K <smk78@…> (11/28/17 09:21:33)
- git-committer:
- GitHub <noreply@…> (11/28/17 09:21:33)
- Location:
- src/sas/sasgui/perspectives/calculator/media
- Files:
-
- 11 deleted
- 1 edited
Legend:
- Unmodified
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-
src/sas/sasgui/perspectives/calculator/media/sas_calculator_help.rst
r5ed76f8 r1b67f3e 26 26 intensity from the particle is 27 27 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 29 32 30 33 Equation 1. … … 46 49 atomic structure (such as taken from a PDB file) to get the right normalization. 47 50 48 *NOTE! $\beta_j$displayed in the GUI may be incorrect but this will not51 *NOTE!* $\beta_j$ *displayed in the GUI may be incorrect but this will not 49 52 affect the scattering computation if the correction of the total volume V is made.* 50 53 … … 56 59 ^^^^^^^^^^^^^^^^^^^ 57 60 58 For magnetic scattering, only the magnetization component, $ M_\perp$,59 perpendicular to the scattering vector $ Q$ contributes to the magnetic61 For magnetic scattering, only the magnetization component, $\mathbf{M}_\perp$, 62 perpendicular to the scattering vector $\vec Q$ contributes to the magnetic 60 63 scattering length. 61 64 … … 64 67 The magnetic scattering length density is then 65 68 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 67 73 68 74 where the gyromagnetic ratio is $\gamma = -1.913$, $\mu_B$ is the Bohr … … 81 87 .. image:: gen_mag_pic.png 82 88 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,89 Now let us assume that the angles of the $\vec Q$ vector and the spin-axis ($x'$) 90 to the $x$-axis are $\phi$ and $\theta_\text{up}$ respectively (see above). Then, 85 91 depending upon the polarization (spin) state of neutrons, the scattering 86 92 length densities, including the nuclear scattering length density ($\beta_N$) … … 89 95 * for non-spin-flips 90 96 91 .. image:: sld1.png 97 .. math:: 98 \beta_{\pm\pm} = \beta_N \mp D_M M_{\perp x'} 92 99 93 100 * for spin-flips 94 101 95 .. image:: sld2.png 102 .. math:: 103 \beta_{\pm\mp} = - D_M(M_{\perp y'} \pm i M_{\perp z'}) 96 104 97 105 where 98 106 99 .. image:: mxp.png107 .. math:: 100 108 101 .. 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 102 114 103 .. 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. 115 Here the $M_{0x}$, $M_{0y}$ and $M_{0z}$ are 116 the $x$, $y$ and $z$ components of the magnetisation vector in the 117 laboratory $x$-$y$-$z$ frame. 111 118 112 119 .. ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ … … 148 155 uses the Debye equation below providing a 1D output 149 156 150 .. 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|} 151 161 152 162 where $v_j \beta_j \equiv b_j$ is the scattering
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