Deriving Wien's Displacement Law from Planck's Law

[gnome]

Basically this problem is to derive Wien's displacement law from Planck's law.
Specifically:
a) Show that there is a general relationship between temperature and λ_max stating that Tλ_max = constant
and
b) Obtain a numerical value for this constant
[Hint: Start with Planck's radiation law and note that the slope of u(λ,T) is zero when λ = λ_max ]

So I obediently try to differentiate u(λ, T) with respect to λ:

I'll let C₁ = 8Πhc and C₂ = hc/(k_BT)

$$u(\lambda, T) = {C_1}\lambda^{-5} \left( e^\frac{C_2}{\lambda} - 1 \right)^{-1}$$

$$\frac{\partial u}{\partial \lambda} = <br /> C_1 \left((\lambda^{-5})(-1)({e^{\frac{C_2}{\lambda}} - 1)^{-2}({e^\frac{C_2}{\lambda}})(-\frac{C_2}{\lambda^2}) + (e^{\frac{C_2}{\lambda}} - 1)^{-1}(-5)\lambda^{-6}\right) = 0$$

I divide through by C₁ & get

$$\frac{C_2e^\frac{C_2}{\lambda}}{\lambda^7({e^\frac{C_2}{\lambda}} -1)^2} - \frac{5}{\lambda^6({e^\frac{C_2}{\lambda}} -1)} = 0$$

$$\frac{C_2{e^\frac{C_2}{\lambda}} - 5\lambda({e^\frac{C_2}{\lambda}} - 1)}{\lambda^7({e^\frac{C_2}{\lambda}} -1)^2} = 0$$

$$C_2{e^\frac{C_2}{\lambda}} - 5\lambda{e^\frac{C_2}{\lambda}} + 5\lambda = 0$$

$$let x = \frac{C_2}{\lambda} = \frac{hc}{{k_B}T\lambda}$$

$$x{\lambda}e^x - 5{\lambda}e^x + 5{\lambda} = 0$$

$$x = 5 - \frac{5}{e^x} = 5(1 - e^{-x})$$

Now I substitute back for x = hc/(kT&lambda;) and get:

$$\frac{hc}{{k_B}T{\lambda}} = 5(1 - {e^\frac{-hc}{{k_B}T{\lambda}}})$$

Oy!

This, amazingly, is exactly the expression I'm supposed to end up with (for part a), according to the answer in the book.

But how does this show that
$$T\lambda = constant$$

and how do I solve for this constant?

 

[jamesrc]

Well, this is kind of cheesy but:

You've got x = 5(1-e^-x)

So plug that into a graphing calculator

(y = x-5(1-e^-x), find x where y = 0)

Then you can write &lambda;T = hc/(kx), plugging in that value for x to find the value of the constant.

I think there is a better way to find that answer, but I'm not a mathematician and I don't remember it. Do you think that's good enough?

 

[cookiemonster]

The equation is transcendental. You have no choice but to solve it numerically.

As for how it's constant, the entire right hand of the equation is a constant and all but T and $\lambda$ are constants on the left.

cookiemonster

 

[gnome]

Yes, thanks, I did finally realize that everything but
T and lambda are constant so T*lambda must be constant.

But when I solve with the calculator
x = 5(1-e^(-x))
it gives me x=0
which, come to think of it, seems perfectly correct.

Except, it makes no sense, because how can it be that
$$\frac{hc}{{k_B}T{\lambda}} = 0$$
?

 

[cookiemonster]

It can't! That's why you have to discard that solution. It's impossible to get $\frac{hc}{{k_B}T{\lambda}} = 0$ because hc clearly is not zero. There is, however, another solution.

Try plotting y = x and y = 5(1-e^(-x)) on the same screen and find where the two intersect.

cookiemonster

 

[jamesrc]

Originally posted by cookiemonster
The equation is transcendental. You have no choice but to solve it numerically.

As for how it's constant, the entire right hand of the equation is a constant and all but T and $\lambda$ are constants on the left.

cookiemonster

Oh, right. I guess I'm not as cheesy as I thought. What a shame, I really like cheese.

 

[gnome]

That works -- after a bit of trial and error I get x = 4.965 as a pretty good approximation. That will let me solve for T&lambda;

Is that the "usual" technique for solving equations of this form?

How would you solve it if you didn't have a graphing device?

 

[cookiemonster]

I'd probably use bisection.

I'd guess that it lies between 4 and 5, test if the function y = 5(1-e^(-x)) - x crosses the x-axis between these points. y(4) is positive and y(5) is negative. Try y(4.5). If it's positive (it is), then we have a new lower bound, since the function crosses between 4.5 and 5. So try 4.75. You get the idea.

cookiemonster

 

[gnome]

Thanks, but I like your graph trick better.

Hope I remember it the next time a problem like this comes up.

PS:
James, if you graph
(y = x-5(1-e^(-x)), find x where y = 0)
on your calculator, you get (0,0).
(At least, that's all I get on mine.)

 

[Philcorp]

To solve this equation you could use an itterative method. You first assume the exponential term is small so that
$\frac{hc}{\lambda_mk_BT}=5$
Then you put that in as the argument of your exponent to get the next itteration of
$\frac{hc}{\lambda_mk_BT}=5(1-e^{-5})$
and continue this to get the root to desired accuracy.

 

[gnome]

Super!

It seems to take just a few iterations to get a good result, with very little thinking involved. What could be better?

Thanks.

 

[jamesrc]

Originally posted by gnome
Thanks, but I like your graph trick better.

Hope I remember it the next time a problem like this comes up.

PS:
James, if you graph
(y = x-5(1-e^(-x)), find x where y = 0)
on your calculator, you get (0,0).
(At least, that's all I get on mine.)

Really? I've got zero crossings at the origin and at x = 4.9651142317
(please forgive the obscene # of decimal places; I've got a root finder on my calculator (it probably uses Newton-Raphson or some other iterative method to find the root)). The function looks kind of like a v tilted to the right.

Anyway, none of that's important now since you solved it already. I hope I didn't add confusion.

 

[gnome]

Not at all, James. Your suggestions are always appreciated. How were you to know that I have a crappy calculator?