(Statistics) Blackbody spectrum in terms of wavelength?

[Ryaners]

This is a question about transforming a probability distribution, using the blackbody spectrum as an example.

Homework Statement
An opaque, non-reflective body in thermal equilibrium emits blackbody radiation. The spectrum of this radiation is governed by B(f) = af³ / (e^bf−1) , where a and b are constants and f is the frequency of the emitted light. Work out the corresponding distribution of wavelengths B(λ) using f = c/λ .

The attempt at a solution
I tried substituting f = c/λ into the given equation - which seemed like a good place to start - and came out with the following:
ac³ / λ³(e^bc/λ - 1).
Then when I looked up the actual spectrum in terms of wavelength (on HyperPhysics, here) it gives something else.
This is the HyperPhysics formula:
8πhc / λ⁵ ⋅ 1 / (e^hc/λkT - 1)
Comparing the constants in the frequency spectrum on HPhys with a & b in the formula in the question:
a = 8πh / c³
b = h / kT
Which then turns my answer of ac³ / λ³(e^bc/λ - 1) into:
8πh / c³ ⋅ c³ / λ³(e^hc/λkT - 1)
= 8πh / λ³ ⋅ 1 / (e^hc/λkT - 1)

It looks like I'm out by a factor of c / λ² compared to the 'real' wavelength spectrum - have I approached this the wrong way, or has the question just simplified the given formula in some way?

 

[SammyS]

This is a question about transforming a probability distribution, using the blackbody spectrum as an example.

Homework Statement
An opaque, non-reflective body in thermal equilibrium emits blackbody radiation. The spectrum of this radiation is governed by B(f) = af³ / (e^bf−1) , where a and b are constants and f is the frequency of the emitted light. Work out the corresponding distribution of wavelengths B(λ) using f = c/λ .

The attempt at a solution
I tried substituting f = c/λ into the given equation - which seemed like a good place to start - and came out with the following:
ac³ / λ³(e^bc/λ - 1).
Then when I looked up the actual spectrum in terms of wavelength (on HyperPhysics, here) it gives something else.
This is the HyperPhysics formula:
8πhc / λ⁵ ⋅ 1 / (e^hc/λkT - 1)
Comparing the constants in the frequency spectrum on HPhys with a & b in the formula in the question:
a = 8πh / c³
b = h / kT
Which then turns my answer of ac³ / λ³(e^bc/λ - 1) into:
8πh / c³ ⋅ c³ / λ³(e^hc/λkT - 1)
= 8πh / λ³ ⋅ 1 / (e^hc/λkT - 1)

It looks like I'm out by a factor of c / λ² compared to the 'real' wavelength spectrum - have I approached this the wrong way, or has the question just simplified the given formula in some way?

As a hint:

If f = c/λ, then what is the derivative, df/dλ ?

 

[Ryaners]

As a hint:

If f = c/λ, then what is the derivative, df/dλ ?

Ok, that's definitely got me thinking now :) I did some digging and found this formula for performing transformations of variables:
f(x) = g(y)⋅dy/dx
where I take it f(x) is the original form of the function, and g(y) is the function in the desired 'new' terms.

So in this case that would be B(f) = B(λ)⋅df/dλ... right? ***
⇒ B(λ) = B(f) / (df/dλ)

And as f = c/λ = c⋅λ^-1 ⇒ df/dλ = -c/λ² (is that correct?)

⇒ B(λ) = ac³ / λ³(e^bc/λ - 1) ⋅ -(λ²/c)
= - ac² / λ(e^bc/λ - 1)

I've definitely gone wrong there, haven't I? I can see that multiplying B(f) [= B(c/λ)] by df/dλ would get me to the formula I found on HPhys, but with a minus sign...?!

*** Edit *** Hang on, I can see I got dλ/df the wrong way around in B(f) = B(λ)⋅dλ/df above, which gives me:
B(λ) = ac³ / λ³(e^bc/λ - 1) ⋅ -(c / λ²)
= - ac⁴ / λ⁵(e^bc/λ - 1)
Now the only thing bugging me is that minus sign - what's going on there?

Thanks so much for your comment, it's really pointed me in the right direction!

 

[SammyS]

...

*** Edit *** Hang on, I can see I got dλ/df the wrong way around in B(f) = B(λ)⋅dλ/df above - now the only confusing thing is the minus sign...

Suppose λ₁ = c/f₁, and λ₂ = c/f₂ .

So if f₂ > f₁ , then λ₂ < λ₁ , and you would need to switch the order of the limits of integration, which changes the sign back to positive.

 

[Ryaners]

Suppose λ₁ = c/f₁, and λ₂ = c/f₂ .

So if f₂ > f₁ , then λ₂ < λ₁ , and you would need to switch the order of the limits of integration, which changes the sign back to positive.

Ahhhh of course, I forgot about that! Thank you so much, it's such a relief to 'get' that now. :)