# Planck's Law

1. Mar 11, 2004

### 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 &lambda;max stating that T&lambda;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(&lambda;,T) is zero when &lambda; = &lambda;max ]

So I obediently try to differentiate u(&lambda;, T) with respect to &lambda;:

I'll let C1 = 8&Pi;hc and C2 = hc/(kBT)

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

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

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?

2. Mar 11, 2004

### 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?

3. Mar 11, 2004

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.

4. Mar 11, 2004

### 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
$$\frac{hc}{{k_B}T{\lambda}} = 0$$
?

5. Mar 11, 2004

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.

6. Mar 11, 2004

### jamesrc

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

7. Mar 11, 2004

### 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?

8. Mar 11, 2004

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.

9. Mar 11, 2004

### 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.)

Last edited: Mar 11, 2004
10. Mar 11, 2004

### 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.

11. Mar 11, 2004

### 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.

12. Mar 12, 2004

### jamesrc

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.

13. Mar 12, 2004

### gnome

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