Thick as a planck and making a spectra of myself

[redbaldyhead]

for hv>>kT how does exp(hv/kT) compare to 1?
I understand hv >>KT leads to an exponential fall in brightness but why did Planck introduce 1 in his equation.
and only for values hv<<kT can this exponential be expanded!
Thanks for any help!

 

[Feldoh]

I have no clue how to help you, however that might be one of the catchiest titles I've ever seen XD

 

[CompuChip]

I'm not sure about the question, but you can always expand the exponential in
$$\exp(\hbar\nu / k T) = 1 + \hbar\nu / kT + \mathcal O((\hbar\nu / k T)^2)$$
where the last term indicates the error you would make if you would just terminate after the first two terms.
Now what happens if $\hbar\nu \gg k T$ and if $\hbar\nu \ll k T$?

BTW I agree, the title is not very descriptive but definitely caught my eye.

 

[redbaldyhead]

Thanks for looking; I have been looking at Rayleigh Jeans Formula and I need to understand what happens for hv>>kT

 

[klile82]

Catastrophe?

So, Rayleigh-Jeans says that the energy density is $$U(\nu) = \frac{8 \pi k T \nu^4}{(c^4)}$$ and Planck says $$U(\nu)d\nu = \frac{4 \hbar \nu^3}{c^3 (\exp(\hbar\nu / k T)-1)}$$.

When $$\hbar \nu >> kT, exp(\hbar \nu / kT) >> 1$$, so you can treat it as
$$U(\nu)d\nu = \frac{4 \hbar \nu^3}{c^3 \exp(\hbar\nu / k T)}$$

Now you can do a limit as $$\nu$$ goes to infinity for the Rayleigh-Jeans and the Planck equations and compare what happens at really high frequencies.

 

[redbaldyhead]

Thanks 82
I actually thought that the result would be of little consequence;
I'll look closely at the differentiation!
Obviously now that I see it, with h=constant and K=constant, any overall increase over 1 must mean the value of v increases against T which I assume doesn't change either because I should be looking at Dv against T. (D is supposed to be delta).
Is the 1 there because of the nature of exp(1)?
Will any of the differentiation has to employ the exponential rules or product rule?
Thanks Red

 

[olgranpappy]

for hv>>kT how does exp(hv/kT) compare to 1?

it's much larger than 1

I understand hv >>KT leads to an exponential fall in brightness but why did Planck introduce 1 in his equation.

to account for the fact that photons are bosons.

and only for values hv<<kT can this exponential be expanded!

in a taylor expansion about zero... true. But for hv>>kT the Planck formula itself can be expanded about zero in the variable $x=e^{-hv/kT}$ which is small and the first term of the expansion is given in klile82's post.

 

[olgranpappy]

Thanks 82
I actually thought that the result would be of little consequence;
I'll look closely at the differentiation!

what differentiation? do you mean to say "derivation"?

Obviously now that I see it, with h=constant and K=constant, any overall increase over 1 must mean the value of v increases against T which I assume doesn't change either because I should be looking at Dv against T. (D is supposed to be delta).
Is the 1 there because of the nature of exp(1)?
Will any of the differentiation has to employ the exponential rules or product rule?
Thanks Red

 

[redbaldyhead]

Thanks to all you good guys!

Is the Planck function mentioned above, valid for high frequency range where v>>kt/h?
What would a graph look like; v against kt/h?...an exponential rising upwards and rapidly from 0. How would the range on the y-axis appear 10^-1 to 10^-10 for example or the otherway about?

Red

 

[olgranpappy]

Thanks to all you good guys!

Is the Planck function mentioned above, valid for high frequency range where v>>kt/h?
What would a graph look like; v against kt/h?...an exponential rising upwards and rapidly from 0.

what the heck? have you read any of the previous posts? what was Planck's whole point?

 

[klile82]

Why the 1?

Is the 1 there because of the nature of exp(1)?

The 1 isn't a hack, it's there because of an identity used in the derivation. Basically, as you go through the derivation you wind up with a term that looks like the sum from n=0 to infinity of $$exp(-nh\nu / kT)$$. There's an identity that says that the sum from n=0 to inifity of x^n is \frac{1}{1-x} [/tex] as long as x< 1. That's where the 1 in the denominator comes from.

It's really helpful, if you're comparing the Rayleigh-Jeans equation to the Planck equation, to graph them as a function of v on the same graph. You should be able to see the difference right away. Just choose a value for T and use the standard values for c, k and h. You can do this easily in something like Mathematica. You can do this fairly easily in excel, too.