🧵 Converting flux density (ph/s/mrad^2/0.1%bw) to flux (ph/s/0.1%bw)
Anonymous at Sat, 30 Mar 2024 12:46:02 UTC No. 16104688
Given the equation we have for the spectral flux density of a bending magnet (see pic related, taken from here: https://dx.doi.org/10.1119/1.361403
Can an anon help me out?
Anonymous at Sat, 30 Mar 2024 12:52:59 UTC No. 16104693
wtf is this schizophrenic notation?
Anonymous at Sat, 30 Mar 2024 14:40:38 UTC No. 16104814
any accelerator physics bros out there to help?
Anonymous at Sat, 30 Mar 2024 14:48:05 UTC No. 16104821
>>16104688
That's only the spectral density at [math]\theta=0[/math] you need to integral the full equation which is much more complicated. You end up with some integral over [math]K_{5/3}[/math]
Alternately, in the relativistic case where radiation is strongly peaked in the forward direction, you can just approximate the integrated flux as the expression in your pic multiplied by some small cutoff solid angle [math]2\pi \theta_c[/math]
I forget the exact expression for it, but im pretty sure it's in chapter 14 of Jackson
Anonymous at Sat, 30 Mar 2024 16:47:12 UTC No. 16104981
>>16104821
Thank you for your help.
So I need to integrate over psi and theta the equation (1) here: https://xdb.lbl.gov/Section2/Sec_2-
(Kim, referenced in the paper from where the pic in op is taken).
How does he get from (1) to (4) when he integrates over psi? A K_5/3 appears but I don't see how.
I checked out Jackson.
Here are two equations from chap 14.
Equation (14.79) is similar to Kim (1) equation. And Jackson does the integration over all angles and as you say we end up with K_5/3 in equation (14.91).
Still I'm having some difficulty in going from 14.91 to a flux in photon/sec/0.1%bw.
Anonymous at Sat, 30 Mar 2024 19:11:10 UTC No. 16105193
>>16104981
The derivation is pretty awful and involves obscure Airy/Bessel function identities. Wiedemann volume 2 and the original paper by Schwinger have it https://users.wfu.edu/natalie/s17ph
14.91 gives you the intensity for one electron, so you should be able to figure out the total flux from the beam current and energy. I'm pretty sure the answer is in the form [math]C E_{el} I_{el} \frac{\omega}{\omega_c} \int_{\omega/\omega_c}^\infty K_{5/3}(x) dx [/math] where C just contains physical constants and unit conversion factors, but I'm too lazy to work out what it actually is.
Anonymous at Sat, 30 Mar 2024 19:45:08 UTC No. 16105243
>>16105193
Thanks again. I'm having trouble figuring it out though.
In the same chapter, if we look a bit further on, Jackson talks about the fact that the equation (14.91) or (14.79) can be talked about in the language of photons. See pic related. (he divides by h_bar*w(omega))
Am I understanding here that the I in pic is I_el (ring current)?
I can make a (DW/W) appear so that I have the 0.1%Bandwith I want.
The N is number of photons.
And for the seconds?
I might (probably) be misunderstanding things here.
Thanks anon.
Anonymous at Sat, 30 Mar 2024 23:54:29 UTC No. 16105602
bump
Anonymous at Mon, 1 Apr 2024 03:48:38 UTC No. 16107283
not
Anonymous at Mon, 1 Apr 2024 03:50:51 UTC No. 16107286
>homework thread
Anonymous at Mon, 1 Apr 2024 15:47:16 UTC No. 16107869
>>16107286
Not a hw question.
I've managed with some help (using pic related) and the fact that dI = \h_bar * \omega * dN to go from equation (14.91) to (14.93).
I'm still trying to figure out how I can express 14.93 in units of Ph/sec/0.1%bw.
Anonymous at Mon, 1 Apr 2024 15:49:37 UTC No. 16107874
I cannot. The syncrotron niggers have ruined their maths. Try An Introduction to Synchrotron Radiation, he loved to use this niggermath