# Thick as a planck and making a spectra of myself

1. Mar 6, 2008

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!

2. Mar 6, 2008

### Feldoh

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

3. Mar 6, 2008

### 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.

4. Mar 6, 2008

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

5. Mar 6, 2008

### 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.

6. Mar 9, 2008

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

7. Mar 9, 2008

### olgranpappy

it's much larger than 1
to account for the fact that photons are bosons.
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.

8. Mar 9, 2008

### olgranpappy

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

9. Mar 9, 2008

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

10. Mar 9, 2008

### olgranpappy

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

11. Mar 9, 2008

### klile82

Why the 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.