Sometimes to subtract the sample container from the sample+container data is not right, you may have a hint that something wrong happens if, for example, after container subtraction negative intensities are obtained. That happens because neutrons are absorbed by the sample and do not scatter any more at the backside of the container. We must then calculate the coefficients Assc and Arel of the following equation:

Sample(2theta,w)=1/A_{ssc}(S_{sample+container}(2theta,w)-A_{rel}*S_{container}(2theta,w)) (confer to Bee)

The calculations of these coefficients can be done by FRIDA1, and in general depend on 2theta and w. Before doing that some hints:

- To be sure that you do everything right, first you should calculate the transmission coefficient of your sample holder. To do that, we will calculate the sample holder as if it were the sample, i.e., setting the transmission coefficients of the sample holder to 1, putting the thickness of your sample holder as the thickness of the sample, and finally putting the information of the material of the sample holder (number density of scatterers, scattering cross section and absorption cross section).
**BE CAREFUL**The number density of scatterers and the scattering cross section MUST BE IN THE SAME “SCATTERER UNITS”, i.e., per molecule or per atom. If you have a molecule the simplest is to calculate the number density of molecules, and then give the scattering cross section of the whole molecule (SUM Xscatt total).- Another problem can arise when having in to account the coherent and incoherent scattering cross section. When the program is asking, it says “(coh+incoh)” but that is not really true. If your wavelength is big, you are looking (normally) at the incoherent part, therefore you should give only the incoherent part. But that is not true for bigger wavelengths, then you have incoherent plus coherent scattering. To do it correctly you should calculate which portion of your scattering is coherent and which portion is incoherent… Yes! not easy at all, but who said that neutrons experiments were easy??????
**Be careful again**When you calculate the self absortion coefficients, you are getting the points of the (2theta,w) grid from the file you are starting the _sac correction from. Let's say that we do it using the sample file… then when you multiply the A_{rel}*S_{container}you will probably get an error, because it is quite unlikely that sample and container were at the same position when measuring – therefore the points are most probably at different places. Therefore before continuing (or even before doing _sac) you should group the channels in the same grid. To do that use the mgr option.

- If your container has a transmission equal to 1, you can incorrectly think that you do not need to perform the SAC, that is (unfortunately) not true. In this case is A
_{ssc}=A_{rel}, and therefore:

S_{sample}(2theta,w)=1/A_{ssc}*S_{sample+container}(2theta,w)-S_{container}(2theta,w))

and because 0<A_{scc}<1, we will substract “less container”… In the following, We suppose that you have already calculated the transmission of your sample holder.

_sac | to call the Self Absortion Correction function. You can call it form any file with the variables (2theta,w). w has to be in meV – if it isn't (for example when evaluating backscattering data), use tu to convert the unit. |

0 | We will suppose that the whole sample was iluminated |

3 | we will calculate it form a hollow cylinder |

1.2 | outer radius in cm of the sample, which of course is equal to the inner radius of the outer container wall |

0.01 | thickness of the sample layer |

1 | Transmission of the outer container.It should be not away from 1. If you do not know it see how to calculate the transmission of your sample holder |

0.05 | Thickness of the inner container (if transmission is NOT 1!!!!), otherwise he will not ask because is transparent |

1 | transmission of the outer container |

0.06 | Thickness of the outer container, as before it will not be asked if the transmission is 1 |

0 | Fraction of back-scattered beam transversing |

7.984 | Number density of scatterers in units of 10h21 cm-1. If you have molecules decide to give values per molecule or per atom!!! read text. The values can be found on the website of the NIST http://www.ncnr.nist.gov/resources/n-lengths/ . You can also perform this calculation and the next two on their website http://www.ncnr.nist.gov/resources/sldcalc.html or with their program “dave” downloadable at www.ncnr.nist.gov/dave/. As an example we calculate it for glycerol partially deuterated: C_{3O2}H_{5}D_{3}, with density 1.261g/cc |

464.369 | Scattering cross section (coh+incoh) see text!!!! |

1.677 | absorption coefficient at 2200 m/sec, that is, for thermal neutrons lambda=1.798 |

y | everything is gonna be all right, hopefully |

80 | Angular meshes… the container+sample will be gridded in radial coordinates. This number and the following determines the grid in angle and radius. the bigger it is the more exact is the solution, and the longer it needs to calculate. It seems that a nice value for bopth is 80 |

80 | read the upper cell |

back to file menu | The program has calculated the coefficients A_{scc} and A_{rel}. If transmission=1 then only the A_{scc} will be calculated |

You should get a transmission of the sample of 0.89349… Now we have to make the calculation:

S_{sample}(2theta,w)=1/A_{ssc}(S_{sample+container}(2theta,w)-A_{rel}*S_{container}(2theta,w)). We will assume that 1 you have in file Nf your sample, in file Nas your coefficient, and in Nec your emptycan. Since we have used trnasmission=1, we will not use A_{rel}.

Ns | We will act on the sample file |

oy / y2 | we will act con y axis, and divide by an y of a second file |

Nas | We will divide by the Assc coefficient |

* | We have now in file No1 the divission, and we should now substract the empty can |

No1 | We will act in file No1 (Ssample/Assc) |

oy - y2 | We will substract the y of another file |

Nec | The empty can is in file Nec |

Nresult | That is the result… Please do compare it with a simple substraction to get a feeling of what we have done |

To calculate the transmission of the sample container… Please do not use without reading the introduction, the values are for our case, and are absolutely not correct in general

_sac | to call the Self Absortion Correction function. You can call it form any file. The program will simply make the calculus in the (2theta,w) space of your file) |

0 | We will suppose that the whole sample was iluminated |

3 | we will calculate it form a hollow cylinder |

1.2 | outer radius in cm of your sample holder |

0.1 | thickness of the sample layer, which in this case is the thickness of your sample holder |

1 | |

1 | transmission of the outer container set to one for the same reason |

0 | Fraction of back-scattered beam transversing |

60.26 | Number density of scatterers in units of 10 h 21 cm-3 , in our case was Aluminium… in fact that changes with T… that should not be so important… |

1.734 | Scattering cross section of Aluminium in barns |

0.231 | Absortion cross section of Aluminum for wavelength=1.798 in barns, Thermal neutrons… |

Then read what FRIDA says, transmission = 0.999857659876, and that is the transmission of your sample holder.

We will now try to group the channels in the same grid for two files…

mgr | to strat grouping the files |

1 | we will do it in a linear grid, and we will need to give the values of the 2theta,w grid |

0 | from this energy |

0.02 | in steps of 0.02 meV |

200 | to 200 meV. An error message will appear if too many points are chosen |

3 | You must decide wether to or to *** |

You may get an error message if the number of points is too big. If you want to study the inelastic part of the spectra is better to perform the grid in log scale…

… or