# Getting Planck's Law in terms of frequency from wavelength

1. Mar 21, 2015

### Kavorka

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

from the equation:

u(λ) = (8πhcλ-5)/(ehc/λkT - 1)

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

u(f) = (8πhc-4)/(f-5(ehf/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/(ehf/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!

2. Mar 21, 2015

### Staff: Mentor

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 $u(f)$.

3. Mar 21, 2015

### 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(ehf/kT - 1) (-c/f2 df)

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

u(f) = (-8πhc-3)/(f-3(ehf/kT - 1) = (-8πhf3/c3)(ehf/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.

4. Mar 21, 2015

### Staff: Mentor

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

5. Mar 21, 2015

### Kavorka

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

6. Mar 21, 2015

### Staff: Mentor

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

7. Mar 21, 2015

### Kavorka

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

λ21 and f1>f2 thus when ∫u(λ)dλ → ∫u(f)df

f2f1 u(f)df = -∫f1f2 u(f)df

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

Or maybe a simpler way?

8. Mar 22, 2015

### Staff: Mentor

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

9. Mar 22, 2015