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Dream workflow: units are weird after Lorentz correction #465
According to the docstring, the corrections is computed as $d^4 \sin\theta$ (https://github.com/scipp/essdiffraction/blob/main/src/ess/powder/correction.py#L358).
This is indeed what is written in the gsas book:
This is weird because the other formulas listed in that document seem to be dimensionless?
But d^4 is obviously not dimensionless...
@celinedurniak Are you aware of a possible definition of $d$ that is dimensionless? I tried to look through the source code of GSAS but it's quite dense and I couldn't find a definition for d.
That paper shows the factor we use but does not seem to derive or explain it.
Regardless, the unit will cancel after vanadium normalisation.
But in general, I am confused about this correction. If we have identical binning in the sample and vanadium data, doesn't the correction cancel exactly? So do we only use it because we have different bins? How does this apply to using events in the sample and a histogram in vanadium?
My understanding here is quite rough, but this is how I understand it:
The Lorentz factor corrects for the time a Bragg peak is present in the detector (the time it satisfies the Bragg condition), different peaks will satisfy the Bragg condition for different amounts of time, and will therefore see different amounts of intensity during the measurement.
But the scattering from the vanadium is incoherent, so it doesn't need to satisfy the Bragg condition to scatter to the detector, so there's no need to compensate for that effect.
We can probably make the Lorentz factor dimensionless (normalizing by the maximum or something) because I think we're already missing constant factors in the expression anyway.
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jl-wynen commentedon Feb 3, 2026
@celinedurniak Are you aware of a possible definition of$d$ that is dimensionless? I tried to look through the source code of GSAS but it's quite dense and I couldn't find a definition for d.
celinedurniak commentedon Feb 3, 2026
There is also this publi https://journals.iucr.org/paper?S2053273322010427
jl-wynen commentedon Feb 4, 2026
That paper shows the factor we use but does not seem to derive or explain it.
Regardless, the unit will cancel after vanadium normalisation.
But in general, I am confused about this correction. If we have identical binning in the sample and vanadium data, doesn't the correction cancel exactly? So do we only use it because we have different bins? How does this apply to using events in the sample and a histogram in vanadium?
jokasimr commentedon Feb 4, 2026
I don't think the lorentz correction factor should be applied to the vanadium measurement,
@celinedurniak can you verify that?
jl-wynen commentedon May 22, 2026
Why would that be the case?
@celinedurniak ?
jokasimr commentedon May 22, 2026
My understanding here is quite rough, but this is how I understand it:
The Lorentz factor corrects for the time a Bragg peak is present in the detector (the time it satisfies the Bragg condition), different peaks will satisfy the Bragg condition for different amounts of time, and will therefore see different amounts of intensity during the measurement.
But the scattering from the vanadium is incoherent, so it doesn't need to satisfy the Bragg condition to scatter to the detector, so there's no need to compensate for that effect.
We can probably make the Lorentz factor dimensionless (normalizing by the maximum or something) because I think we're already missing constant factors in the expression anyway.