Anonymous at Wed, 1 May 2024 17:24:39 UTC No. 16155098

I have experimental data. When I use partial flux with Spectra taking into account the slit (as is done in real life) the calculation is very close to experimental data.

See pic related.

How can I take the slit into account for my calculation with the equation above?
Also, if I use a Monte Carlo ray tracing software, should anything be done other than the placement of a slit to get the partial flux?

Last remark that show the slit is the reason for the change: when I use partial flux with Spectra and put a huge slit (as if there is none) then we go back to the flux of the above equation.

Anonymous at Wed, 1 May 2024 20:03:58 UTC No. 16155215

bump

Anonymous at Wed, 1 May 2024 20:07:49 UTC No. 16155221

you sure do seem to like talking about yourself on social media

Anonymous at Wed, 1 May 2024 20:13:20 UTC No. 16155226

reposting the OP because I just realised the latex is broken:

Thanks to the help of an anon (who referred me to the Jackson textbook) and others a few weeks ago, I have managed to express the flux of a bending magnet like so:

[math]\frac{d_N}{\frac{d\omega}{\omega}} = \frac{I}{\hbar \omega_c^2} \omega \frac{9\sqrt{3}}{8 \pi} \int_{\frac{\omega}{\omega_c}}^{\infty} K_\frac{5}{3}(x)dx[/math]

this is expressed in the units of [math]photons/second/0.1\%bw[/math]

With a little python script, I can graph this as a function of the energy. To verify that the equation and code is correct I compare it with spectra (synchrotron sources software).
I end up with the same flux they do. All is good. I have normalised the fluxes.

See pic related.

When I select with Spectra the option partial flux which gives the user the option of adding a slit of a certain size at a certain distance from the source, I end up with a different flux (called partial photon flux).

First, I do not understand this.
I have both the angular divergence and the size of my beam at point 0(horizontal and vertical).
[math]beamsizeatD = 2*D*\tan \frac{\sigma'}{2}[/math] where D is the distance to the source and [math]\sigma'[/math] is the angular divergence (horizontal or vertical), I see that the beam should be smaller than the slit, therefore the slit shouldn't have an impact yet it does.

And even when I take into account the size of the beam [math]\sigma[/math] at point 0:
[math]\sigma + 2*D*\tan \frac{\sigma'}{2}[/math]
it's still smaller than the slit.

Anonymous at Wed, 1 May 2024 20:16:31 UTC No. 16155232

>>16155097
SPECTRA initializes your beam with a Gaussian profile, so there’s about 1/3 of your current at [math]r>sigma [/math] that could be getting intercepted.
Also the angular divergence of the electron beam is going to be less than that of the emitted photons.

Anonymous at Wed, 1 May 2024 20:19:53 UTC No. 16155239

>>16155232
thank you for your help

The beam does have a gaussian profile irl also I think

>Also the angular divergence of the electron beam is going to be less than that of the emitted photons.

Could you explain in more detail please?

Anonymous at Wed, 1 May 2024 21:34:29 UTC No. 16155355

>>16155232
is emittance of the beam related to your second sentence?

Anonymous at Wed, 1 May 2024 21:57:14 UTC No. 16155409

>>16155239
https://cds.cern.ch/record/375972/files/p1.pdf 3.5, also in jackson before you integrate out the solid angle from [math] \frac{dI}{d\omega d\Omega} [/math]
You end up with some angular spread of the photon distribution with width of approximately [math]1/\gamma [/math] radians, which is generally much larger than the beam angle. There's also larger spread for lower photon frequency/energy so you'll have a higher relative intensity in the tail of the distribution compared to one without a slit.

Anonymous at Wed, 1 May 2024 23:38:12 UTC No. 16155519

So finally we know, that it's just measurement, and it's wave and it been wave all the time?

??