/sci/ - Science & Math » Thread #13281757

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Anonymous Tue 15 Jun 20:43:13 2021 No.13281757 View Reply Original Report

Quoted By: >>13281772

Any quantum physics chads here that can explain this to me?
If spectral radiance units depend on the wavelength, then why is the wavelength on the x axis here? if the planck law equation is calculated using the wavelength, why does it produce a value that still depends on the wavelength?

Anonymous

Anonymous Tue 15 Jun 2021 20:45:16 No.13281766 Report

Quoted By:

isn't that saying that it would be possible for some reason to calculate the spectral radiands with a wavelength of 500nm, then you multiply that by a wavelength of 1000nm to cancel out the nm unit, and that gives you... what?

Anonymous

Anonymous Tue 15 Jun 2021 20:47:52 No.13281772 Report

Quoted By: >>13281817 >>13282216

>>13281757
It's a function.
It gives you the intensity as a function of wavelength for a given black body temperature.

That's just how it works.

Anonymous

Anonymous Tue 15 Jun 2021 21:10:39 No.13281817 Report

Quoted By: >>13281852

>>13281772
Well you sort of just proved what i'm talking about. The intensity is a function of wavelength and temperature, but the result doesn't have temperature units in it anywhere, but it does have wavelength units. there has to be a subtle reason behind this that I'm missing.

Anonymous

Anonymous Tue 15 Jun 2021 21:27:53 No.13281852 Report

Quoted By: >>13281992

>>13281817
>but the result doesn't have temperature units in it anywhere
look closer anon

Anonymous

Anonymous Tue 15 Jun 2021 22:08:28 No.13281992 Report

Quoted By: >>13282485

>>13281852
nah there is certainly something else going on as the y axis is still in units of wavelength, which specifically means if you multiply wavelength you get a meaningful result. for example a value on the y axis is giving you an energy ratio per nm metric, for each nanometer, this is how many kW per sr per m you have

there is some meaning to this that is more than just saying this is how it works

Anonymous

Anonymous Tue 15 Jun 2021 22:11:40 No.13282000 Report

Quoted By:

for each wavelength, you get an different amount of energy per unit wavelength. there has to be a deeper meaning to this.

Anonymous

Anonymous Tue 15 Jun 2021 23:32:54 No.13282216 Report

Quoted By: >>13282264

>>13281772
This is the answer.

Its a function. Its f(x) where x is the wavelength. That's it. You're reading too much into it.

This graph shows that the observed spectral radiance of light, over several wavelengths. Things that absorb or emit light typically (not lasers) absorb and emit in a Gaussian distribution, which is what you see here (actually it looks like two Gaussian distributions that are overlaid).

Its like a UV-Vis experiment where intensity is measured across multiple wavelengths of light.

Anonymous

Anonymous Tue 15 Jun 2021 23:46:11 No.13282264 Report

Quoted By: >>13282293 >>13282309 >>13282364 >>13282485

>>13282216
It's not the answer and you don't know what you're talking about. It's a gamma distribution, not gaussian. It's a function of two variables f(v, T) not just a function of x. You're really fucking dumb and you're opinion is not welcome. I give up on you retards, I'm better off just looking for online sources since nobody here knows anything.

Anonymous

Anonymous Tue 15 Jun 2021 23:55:58 No.13282293 Report

Quoted By:

>>13282264
>different curves show for different temperatures
are you the retarded one?

Anonymous

Anonymous Wed 16 Jun 2021 00:00:17 No.13282309 Report

Quoted By: >>13282469

>>13282264
A gamma distribution can be a Gaussian distribution. There are two variables. That's why the curves at different temperatures are different shapes.

Anonymous

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Anonymous Wed 16 Jun 2021 00:12:04 No.13282364 Report

Quoted By:

>>13282264
you just outed yourself as a pseud kek

Anonymous

Anonymous Wed 16 Jun 2021 00:58:25 No.13282457 Report

Quoted By: >>13282485

I didn't out myself I asked for help and found out the hard way you are all fucking idiots.

Anonymous

Anonymous Wed 16 Jun 2021 01:02:40 No.13282469 Report

Quoted By: >>13282499

>>13282309
>A gamma distribution can be a Gaussian distribution.
No it isn't you fucking insufferable liar, fucking cunt.

Anonymous

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Anonymous Wed 16 Jun 2021 01:07:28 No.13282485 Report

Quoted By: >>13282499 >>13282545

>>13281992
>which specifically means if you multiply wavelength you get a meaningful result
??? no it doesnt, youre not even close to interpreting this thing correctly

Planck's law gives this distribution of radiant flux for a given temperature, right? Meaning, the total power flux at a given temperature is the area under the associated curve (basically, integrating the curve from 0 to infinite wavelength). The radiant flux peaks at a given wavelength (this is called Wein's law). The units of the y axis are thus power/length^3 and the x axis is units of length. Alternatively, you could graphically depict Planck's law where the y axis is in units of power/(length^2*frequency) and the x axis is frequency. The associated flux at a temperature is then also the area under the curve.
The stephan-boltzmann law is directly obtained from planck's law with a little calculus.
>>13282264
>>13282457
fucking retard gets filtered by babby's first radiation

Anonymous

Anonymous Wed 16 Jun 2021 01:08:16 No.13282487 Report

Quoted By: >>13282499

what the fuck is going on here
>if spectral radiance units depend on the wavelength, then why is the wavelength on the x axis?
because you're seeing how the spectral radiance changes as a function of wavelength. it's like plotting the velocity (which depends on the units of time, meters per second) versus time on the x-axis.
you're just mistaking units for variables

Anonymous

Anonymous Wed 16 Jun 2021 01:11:48 No.13282499 Report

Quoted By:

>>13282485
>>13282487
Jesus christ OP is such a faggot it takes four anons telling him they're just units.

>>13282469
Wow you're a right cunt aren't you?

https://en.wikipedia.org/wiki/Normal-gamma_distribution
That's awkward.

Anonymous

Anonymous Wed 16 Jun 2021 01:25:14 No.13282545 Report

Quoted By: >>13282560 >>13282564

>>13282485
>Planck's law gives this distribution of radiant flux for a given temperature, right? Meaning, the total power flux at a given temperature is the area under the associated curve (basically, integrating the curve from 0 to infinite wavelength). The radiant flux peaks at a given wavelength (this is called Wein's law). The units of the y axis are thus power/length^3 and the x axis is units of length. Alternatively, you could graphically depict Planck's law where the y axis is in units of power/(length^2*frequency) and the x axis is frequency. The associated flux at a temperature is then also the area under the curve.
>The stephan-boltzmann law is directly obtained from planck's law with a little calculus.

Thanks, it does sound like you at least know what you are talking about more than the other trolls here. But you still didn't address the question I asked, which was what the additional spectral unit signifies, regardless if you plot it as frequency or wavelength? It clearly shows that the intensity of every frequency (or wavelength) is different, which is what I was trying to figure out. But no help here, thanks though for you at least this board might have a shot.

Anonymous

Anonymous Wed 16 Jun 2021 01:29:26 No.13282560 Report

Quoted By:

>>13282545
>But you still didn't address the question I asked, which was what the additional spectral unit signifies
dont know what the fuck this means
>It clearly shows that the intensity of every frequency (or wavelength) is different
yes, obviously. a blackbody at 4000 K gives off more red light than green and a blackbody at 1000 K gives off almost no visible light.

Nothing is wrong with any of the responses you got, btw, you are genuinely just slow.

Anonymous

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Anonymous Wed 16 Jun 2021 01:31:17 No.13282564 Report

Quoted By: >>13282589

>>13282545
>But you still didn't address the question I asked, which was what the additional spectral unit signifies
It doesn't have a tidy physical meaning. The meaning is literally "flux of radiative power per unit wavelength (or frequency) at a given temperature."
>It clearly shows that the intensity of every frequency (or wavelength) is different
yes, obviously. a blackbody at 4000 K gives off more red light than green and a blackbody at 1000 K gives off almost no visible light.

Nothing is wrong with any of the responses you got, btw, you are genuinely just slow

Anonymous

Anonymous Wed 16 Jun 2021 01:39:15 No.13282589 Report

Quoted By: >>13282592

>>13282564
>It doesn't have a tidy physical meaning.
Yes it does I just told you what it was. Per unit wavelength is a physical property. It explicitly relates power output in watts with the wavelength if area and solid angle are constant. And on top of that, the value changes with temperature as well.

>btw, you are genuinely just slow
projecting?

Anonymous

Anonymous Wed 16 Jun 2021 01:40:31 No.13282592 Report

Quoted By:

>>13282589
Great, so you understand Planck's law now, good job brainlet.

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