Line Profile Functions (Spectral Line Broadening)

[@stauffap]

I've never come across a nicer explanation of this. Thank you! It's very understandable.

[@hala7526]

Never had such a beautiful explanation. Thanks for educating us.

[@faheemrajuu]

Thank you for sharing your knowledge. Much appreciated.

[@The_fusion_physics_guy]

great video, really helped me sanity check something for my research!

[@xiaoqilu1353]

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]

No, because here tau = A10 . t (4:05) and A10 is in Hz. So tau is dimensionless, as it should be.

[@xiaoqilu1353]

@@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]

@@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]

@@jacobvandijk6525 Agree. Regarding the Fourier transform, I found another video that provides more details behind this argument: kzbin.info/www/bejne/fGm5maF4iZWpoZI.

[@jacobvandijk6525]

@ 2:29 Make that - 1/tau. Then we have exp(- (1/tau) . t) and a decaying exponential.

[@jacobvandijk6525]

He is too arrogant to admit his error.

[@freakyfrequency2530]

Best explanation!

[@johngreen506]

Great explanation!!!

[@Higgsinophysics]

Extremely well explained. Thank you

[@krishvtrai]

Thanks for the explanation.Good one👍

[@KhalidBakri]

Great explanation. Well done sir

[@sujoysen8244]

Very nice explanation. Can you suggest a reference book?

[@elizabethetheridge8192]

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]

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.

[@shotoyama9871]

Good video! I grasped what affects a line plofie. Thanks.

[@andersonribeiro5481]

Good job sir, Thank you very much indeed

[@ElliLovett]

thanks for the help :)

[@lgbpinho]

Thanks! I needed a quick intro do the Voigt profile! Stable distributions ftw .o/

[@live4Cha]

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