GID instrumentation notes

Pseudo z-axis geometry

In this geometry, rather than tilting the whole goniometer (or bend the beam down), only the sample is tilted:

n c = 90o c  



The question is now, what is the exit angle  as a function of the detector arm angle ?

We introduce the rotational matrices for rotations about the x-axis and the z-axis:

骣1 0 0 骣cos(j )- sin( j ) 0 琪 琪 (j )= 0 cos( j ) - sin( j ) (j )= sin( j ) cos( j ) 0 R x 琪 R z 琪 琪 琪 桫0 sin(j ) cos( j ) 桫 0 0 1

The surface normal can now be written as:

0  0      n  () 0   sin() Rx         1  cos() 

The scattered wave vector is given by the detector position ():

 ( ) ( ) k f Rz Rx k i

0   2  k 1 k  with k i   and .    0 Evaluation yields

骣cos(d )- sin( d ) 0 骣 0 骣 - sin( d )cos( g ) 琪 琪 琪 k f =sin(d ) cos( d ) 0 cos( g ) = cos( d )cos( g ) k 琪 琪 琪 琪 琪 琪 桫0 0 1 桫 sin(g ) 桫 sin( g )

The exit angle can now be obtained using the scalar product 鬃, :

c sin(b )= cos( b ) = n , k f yielding the final formula

sin( )  cos()sin( )  sin()cos( )cos( )

Special cases: a) =0: sin() = cos() sin() - sin() cos() = sin(-) i.e.  = - b) =90o: sin() = cos() sin() i.e.  = for small  c) small ,,, arbitrary :  = - cos()

Formula c) will apply for most cases to the G2 setup with the horizontal diffractometer, where  is typically smaller than 10o. A small correction also applies to the in-plane scattering angle  as compared to the diffractometer angle . In order to calculate , we need to find the projections of the wavevectors ks onto the surface plane. This is easy for the incident beam:

骣0 骣 0 琪 琪 k/ k = (a ) 1 = cos( a ) s, i R x 琪 琪 琪 琪 桫0 桫 sin(a )

For the exit beam, we have to calculate the projection of kf onto the surface plane using the Gram-Schmidt orthogonalization scheme:

ks, f= k f - n, k f k f

Now  can be determined from the scalar product of the in-plane projections:

k, k cos(y ) = s, f s , i ks, f k s , i

I cheated for the last two steps and did the somewhat lengthy calculation with MathCAD’s symbolic math feature. The exact result is

cos(y )= cos( a )� cos( d ) sin( a ) tan( g )

Now we want to analyze special cases again, to check the formula:

a) =0: cos( cos(i.e., as it should be. b) small: same as a)

We confirmed that the -correction is minute for the typical values of  below 1º. In this case the  correction is not needed.