Getting Planck's Law in terms of frequency from wavelength

[Kavorka]

Show that Planck's law expressed in terms of the frequency f is:
u(f) = (8πf²/c³)(hf/(e^hf/kT - 1))

from the equation:

u(λ) = (8πhcλ^-5)/(e^hc/λkT - 1)

When I do this algebraically by simply plugging in λ = c/f, I get:

u(f) = (8πhc^-4)/(f^-5(e^hf/kT - 1)

which clearly doesn't involve the correct powers of c and f.

This is the same thing I get when going back and putting n(λ) = 8πλ^^-4 in terms of the frequency and multiplying it by the average energy E bar = hf/(e^hf/kT - 1) to get u(f).

Going through all of these equations is confusing and I am having trouble putting it all together, but I figure that the problem is I need to go back and differentiate somewhere, since n(λ) is found from n(λ)dλ which involves the range between λ and λ+dλ, but it hasn't been converted to the range between f and f+df. I don't know how I would go about doing any of this...please help!

 

[DrClaude]

Planck's law is a distribution function, so you can simply start by considering $u(\lambda) d\lambda$ (radiance per unit wavelength times unit wavelength), which you can easily convert to $u(f) df$, for which you get $u(f)$.

 

[Kavorka]

You would have to put dλ in terms of df then right? If λ = c/f, then dλ = -c/f² df?

So I already found u(f) correctly (I think), so multiplying my equation for u(f) by df would get:

u(f)df = (8πhc^-4)/(f^-5(e^hf/kT - 1) (-c/f² df)

Dropping the df to get u(f) and simplifying:

u(f) = (-8πhc^-3)/(f^-3(e^hf/kT - 1) = (-8πhf³/c³)(e^hf/kT - 1)

Alright that is exactly the answer I need, except that my answer is negative because of the negative derivative. I'm not sure how to resolve that.

 

[DrClaude]

Alright that is exactly the answer I need, except that my answer is negative because of the negative derivative. I'm not sure how to resolve that.

If you integrate from wavelength $\lambda_1$ to $\lambda_2$, with $\lambda_2 > \lambda_1$, when happens when you convert that to frequency?

 

[Kavorka]

You'll integrate from f₁ to f₂ with f₁>f₂?

 

[DrClaude]

Yes, and considering that normally you want to integrate from low frequency to high frequency, what do you need to change?

 

[Kavorka]

Ah, I understand. So could I indicate this by writing that:

λ₂>λ₁ and f₁>f₂ thus when ∫u(λ)dλ → ∫u(f)df

∫f2f1 u(f)df = -∫f1f2 u(f)df

Thus u(f) = (-8πhf³/c³)(e^hf/kT - 1) taken from f₁ to f₂ =
u(f) = (8πhf³/c³)(e^hf/kT - 1) taken from f₂ to f₁

Or maybe a simpler way?

 

[DrClaude]

I don't think there is a simpler way to demonstrate it.

 

[Kavorka]

Thanks for your help!