Line Profile Functions (Spectral Line Broadening)

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Aaron Parsons

Aaron Parsons

Күн бұрын

Пікірлер: 24
@stauffap
@stauffap Жыл бұрын
I've never come across a nicer explanation of this. Thank you! It's very understandable.
@xiaoqilu1353
@xiaoqilu1353 4 жыл бұрын
2:35 shouldn't the function N(t) be proportional to exp(-t/tau)? 3:19 sorry I'm still confused... I know what Fourier transform is, but I don't get the reasoning for why Fourier transform is the answer to natural line broadening here. Thanks!
@jacobvandijk6525
@jacobvandijk6525 2 жыл бұрын
No, because here tau = A10 . t (4:05) and A10 is in Hz. So tau is dimensionless, as it should be.
@xiaoqilu1353
@xiaoqilu1353 2 жыл бұрын
@@jacobvandijk6525 I guess you are right, but this seems to conflict with the previous claim at 2:35 that "tau as the half-life of a decay process".
@jacobvandijk6525
@jacobvandijk6525 2 жыл бұрын
@@xiaoqilu1353 Yes, that's an unfortunate remark. In his notation tau is a dimensionless decay constant: en.wikipedia.org/wiki/Half-life#Formulas_for_half-life_in_exponential_decay
@xiaoqilu1353
@xiaoqilu1353 2 жыл бұрын
@@jacobvandijk6525 Agree. Regarding the Fourier transform, I found another video that provides more details behind this argument: kzbin.info/www/bejne/fGm5maF4iZWpoZI.
@jacobvandijk6525
@jacobvandijk6525 Жыл бұрын
@ 2:29 Make that - 1/tau. Then we have exp(- (1/tau) . t) and a decaying exponential.
@jacobvandijk6525
@jacobvandijk6525 Ай бұрын
He is too arrogant to admit his error.
@faheemrajuu
@faheemrajuu 7 ай бұрын
Thank you for sharing your knowledge. Much appreciated.
@The_fusion_physics_guy
@The_fusion_physics_guy 6 ай бұрын
great video, really helped me sanity check something for my research!
@freakyfrequency2530
@freakyfrequency2530 10 ай бұрын
Best explanation!
@elizabethetheridge8192
@elizabethetheridge8192 9 жыл бұрын
Just a quick question, you said that anything that has a finite length in time has a width in frequency space, this makes me think that the width arises from the fact that it doesn't happen infinitely quickly, and hence that if it did happen instantaneously it would have no width. Why then does the width increase as the decay time decreases? Thanks
@pablofernandezesteberena7456
@pablofernandezesteberena7456 8 жыл бұрын
It's actually the other way around. When you have a wave package of a certain width, the narrower it is the wider the frequency interval you need. That's what's behind Heisenberg's Principle. If something happened infinitely fast (what's called a Dirac's Delta in time) it would need an infinite interval of frequencies.
@johngreen506
@johngreen506 5 жыл бұрын
Great explanation!!!
@hala7526
@hala7526 4 жыл бұрын
Never had such a beautiful explanation. Thanks for educating us.
@sujoysen8244
@sujoysen8244 3 жыл бұрын
Very nice explanation. Can you suggest a reference book?
@Higgsinophysics
@Higgsinophysics 6 жыл бұрын
Extremely well explained. Thank you
@KhalidBakri
@KhalidBakri 10 жыл бұрын
Great explanation. Well done sir
@krishvtrai
@krishvtrai 8 жыл бұрын
Thanks for the explanation.Good one👍
@live4Cha
@live4Cha 9 жыл бұрын
that has given me lots of thinking with no results. in the doppler term if you take the c inside the root (see the link bellow) you get the energy term (2kT) divided by mc2 which is the Einstein E-M equivalence. What its saying is that the square of shift of freq. times the rest energy is equal to square of initial freq. weighted times the field ()thermal energy. what does this mean physically? is the factor two right? two only right if we assme +/-v kzbin.info/www/bejne/ravLn3lsa7-io6cm38s
@ElliLovett
@ElliLovett 5 жыл бұрын
thanks for the help :)
@andersonribeiro5481
@andersonribeiro5481 5 жыл бұрын
Good job sir, Thank you very much indeed
@shotoyama9871
@shotoyama9871 5 жыл бұрын
Good video! I grasped what affects a line plofie. Thanks.
@lgbpinho
@lgbpinho 9 жыл бұрын
Thanks! I needed a quick intro do the Voigt profile! Stable distributions ftw .o/
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