Difference between revisions of "Talk:Unit cell"

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(Vectors (Wrong?))
(Vectors)
 
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According to:
 
According to:
 
* K. N. Trueblood, H.-B. Bürgi, H. Burzlaff, J. D. Dunitz, C. M. Gramaccioli, H. H. Schulz, U. Shmueli and S. C. Abrahams [https://scripts.iucr.org/cgi-bin/paper?S0108767396005697 Atomic Dispacement Parameter Nomenclature. Report of a Subcommittee on Atomic Displacement Parameter Nomenclature] ''Acta Cryst'' '''1996''', A52, 770-781. [https://doi.org/10.1107/S0108767396005697 doi: 10.1107/S0108767396005697]
 
* K. N. Trueblood, H.-B. Bürgi, H. Burzlaff, J. D. Dunitz, C. M. Gramaccioli, H. H. Schulz, U. Shmueli and S. C. Abrahams [https://scripts.iucr.org/cgi-bin/paper?S0108767396005697 Atomic Dispacement Parameter Nomenclature. Report of a Subcommittee on Atomic Displacement Parameter Nomenclature] ''Acta Cryst'' '''1996''', A52, 770-781. [https://doi.org/10.1107/S0108767396005697 doi: 10.1107/S0108767396005697]
 +
The vectors are written as:
 
:<math>\begin{array}{l}
 
:<math>\begin{array}{l}
 
\mathbf{a} = \begin{bmatrix}
 
\mathbf{a} = \begin{bmatrix}
Line 92: Line 93:
 
= \begin{bmatrix}
 
= \begin{bmatrix}
 
c \cos{\beta} \\
 
c \cos{\beta} \\
\frac{c}{\sin \gamma} \left( \cos \alpha - \cos \beta \cos \gamma \right) \\
+
c \frac{\cos\alpha -\cos\beta \cos\gamma }{\sin\gamma} \\
\frac{c}{\sin \gamma} \sqrt{ 1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma } \\
+
\frac{c}{\sin\gamma} \sqrt{1 + 2 \cos\alpha\cos\beta\cos\gamma - \cos^2 \alpha - \cos^2\beta -\cos^2\gamma}\\
 
\end{bmatrix}
 
\end{bmatrix}
 
\end{array}
 
\end{array}
 +
</math>
 +
This is, again, an equivalent expression. The equivalence can be show by:
 +
:<math>\begin{alignat}{2}
 +
c_z &= 1/c^{*} \\
 +
    &= \frac{c}{\sin\gamma} \sqrt{1 + 2 \cos\alpha\cos\beta\cos\gamma - \cos^2 \alpha - \cos^2\beta -\cos^2\gamma}\\
 +
    &= c \sqrt{\frac{1 + 2 \cos\alpha\cos\beta\cos\gamma - \cos^2 \alpha - \cos^2\beta -\cos^2\gamma}{\sin^2\gamma} }\\
 +
    &= c \sqrt{\frac{1}{\sin^2\gamma} + \frac{2 \cos\alpha\cos\beta\cos\gamma}{\sin^2\gamma} - \frac{\cos^2 \alpha}{\sin^2\gamma} - \frac{\cos^2\beta}{\sin^2\gamma} - \frac{\cos^2\gamma}{\sin^2\gamma} }\\
 +
    &= c \sqrt{\frac{1-\cos^2\gamma}{\sin^2\gamma} + \frac{2 \cos\alpha\cos\beta\cos\gamma}{\sin^2\gamma} - \frac{\cos^2 \alpha}{\sin^2\gamma} - \frac{\cos^2\beta}{\sin^2\gamma}  }\\
 +
    &= c \sqrt{\frac{\sin^2\gamma}{\sin^2\gamma} + \frac{2 \cos\alpha\cos\beta\cos\gamma}{\sin^2\gamma} - \frac{\cos^2 \alpha}{\sin^2\gamma} - \frac{\cos^2\beta(\sin^2\gamma+\cos^2\gamma)}{\sin^2\gamma}  }\\
 +
    &= c \sqrt{1 + \frac{2 \cos\alpha\cos\beta\cos\gamma}{\sin^2\gamma} - \frac{\cos^2 \alpha}{\sin^2\gamma} - \frac{\cos^2\beta\sin^2\gamma}{\sin^2\gamma} -\frac{\cos^2\beta\cos^2\gamma}{\sin^2\gamma}  }\\
 +
    &= c \sqrt{1 - \cos^2\beta \frac{\sin^2\gamma}{\sin^2\gamma} - \frac{\cos^2 \alpha}{\sin^2\gamma} + \frac{2 \cos\alpha\cos\beta\cos\gamma}{\sin^2\gamma} -\frac{\cos^2\beta\cos^2\gamma}{\sin^2\gamma}  }\\
 +
    &= c \sqrt{1 - \cos^2\beta  - \frac{1}{\sin^2\gamma}\left( \cos^2 \alpha - 2 \cos\alpha\cos\beta\cos\gamma + \cos^2\beta\cos^2\gamma \right)}\\
 +
    &= c \sqrt{1 - \cos^2\beta  - \frac{1}{\sin^2\gamma}\left( \cos\alpha - \cos\beta\cos\gamma\right)\left( \cos\alpha - \cos\beta\cos\gamma\right)}\\
 +
    &= c \sqrt{1 - \cos^2\beta  - \left(\frac{ \cos\alpha - \cos\beta\cos\gamma}{\sin\gamma}\right)^2 }\\
 +
 +
\end{alignat}
 
</math>
 
</math>
  

Latest revision as of 09:26, 14 November 2022

Extra math expressions

A given real-space cubic lattice will have dimensions:

Such that the position of any particular cell within the infinite lattice is:

Where h, k, and l are indices. The corresponding inverse-space lattice would be:

In the case where :

Vectors

There are many equivalent ways to define/construct the Cartesian basis for the unit cell in real-space. The unit cell vectors can be written as:

According to this, the vectors can be written as:

which is mathematically equivalent.

According to:

The vectors are written as:

This is, again, an equivalent expression. The equivalence can be show by:

TBD: Reciprocal vector components

Calculate q_hkl generally

    def q_hkl(self, h, k, l):
        """Determines the position in reciprocal space for the given reflection."""
        
        # The 'unitcell' coordinate system assumes:
        #  a-axis lies along x-axis
        #  b-axis is in x-y plane
        #  c-axis is vertical (or at a tilt, depending on beta)
        
        # Convert from (unitcell) Cartesian to (unitcell) fractional coordinates
        reduced_volume = sqrt( 1 - (cos(self.alpha))**2 - (cos(self.beta))**2 - (cos(self.gamma))**2 + 2*cos(self.alpha)*cos(self.beta)*cos(self.gamma) )
        #volume = reduced_volume*self.lattice_spacing_a*self.lattice_spacing_b*self.lattice_spacing_c
        a = ( self.lattice_spacing_a , \
                0.0 , \
                0.0  )
        b = ( self.lattice_spacing_b*cos(self.gamma) , \
                self.lattice_spacing_b*sin(self.gamma) , \
                0.0 )
        c = ( self.lattice_spacing_c*cos(self.beta) , \
                self.lattice_spacing_c*( cos(self.alpha) - cos(self.beta)*cos(self.gamma) )/( sin(self.gamma) ) , \
                self.lattice_spacing_c*reduced_volume/( sin(self.gamma) ) )
        
        # Compute (unitcell) reciprocal-space lattice vectors
        volume = np.dot( a, np.cross(b,c) )
        u = np.cross( b, c ) / volume # Along qx
        v = np.cross( c, a ) / volume # Along qy
        w = np.cross( a, b ) / volume # Along qz
        
        qhkl_vector = 2*pi*( h*u + k*v + l*w )
        qhkl = sqrt( qhkl_vector[0]**2 + qhkl_vector[1]**2 + qhkl_vector[2]**2 )
        
        return (qhkl, qhkl_vector)

                
        
    def q_hkl_length(self, h, k, l):
        
        qhkl, qhkl_vector = self.q_hkl(h,k,l)
        #qhkl = sqrt( qhkl_vector[0]**2 + qhkl_vector[1]**2 + qhkl_vector[2]**2 )
        
        return qhkl