According to Zettili, we can derive Wien's displacement law from Planck's energy density

$$ \tilde{u}(\lambda, T) = \frac{8\pi hc}{\lambda^{5}}\frac{1}{e^{hc/kT\lambda}-1} $$ where $\lambda$ is the wavelength and $T$ is the temperature. Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets

$$ \frac{\alpha}{\lambda} = 5(1-e^{-\alpha / \lambda}) $$

where $\alpha = hc/(kT)$. To solve this, Zettili goes on to say that we can write

$$ \frac{\alpha}{\lambda} = 5-\epsilon $$ so that the previously mentioned equation becomes $$ 5-\epsilon = 5-5e^{-5+\epsilon} $$ Apparently, $\epsilon \approx 5e^{-5}$ is a good answer. My first question is why do we use $5-\epsilon$ to rewrite the equation? What method is this? I've never seen that before and other resources just say that they got the values by solving numerically. My second question is why $\epsilon \approx 5e^{-5}$ when there is an extra factor of $e^{\epsilon}$ in the equation? Why can we ignore that factor?


1 Answer 1


Well this is sort of a Taylor expansion. Let us assume we do not take 5 from the beginning, what is the method behind? We want to solve something like: $$x=5(1-\mathrm e ^{-x})$$ Ok then, let us assume the solution is close to some value $x_0$ that we might guess, but only differs by a small $\epsilon\ll1$, i.e. $x=x_0-\epsilon$. We then have: $$x_0-\epsilon=5(1-\mathrm e ^{-x_0+\epsilon})$$ We then obviously have $$\epsilon=x_0-5+5\mathrm e ^{-x_0}\mathrm e^ \epsilon$$ with $\mathrm{e}^\epsilon\approx 1$. We need $\epsilon$ to be small, because that was our assumption in the first place.The value of $e ^{-x_0}$ is probably a small number, but what about $x_0-5$? Well this is small if $x_0=5$, as in this case this is zero. The result then must provide self consistently that $\epsilon$ is small. Otherwise the ansatz was wrong, but this is OK here. So it works.

Note that this requires a large factor for the parenthesis. 5 is OK, it would have been less precise for, e.g., 3 and would completely fail for 1.

To continue the idea let's see what we can do in case of 3. The True numerical solution for $x=3(1-\mathrm e^{-x})$ is $x=2.82144...$ but the simple idea from above gives $\epsilon=0.149...$, i.e. $x=3-0.149...=2.8506...$ so we are notably off. Our mistake is the assumption that $\mathrm e^\epsilon =1$ but with $0.149$ this is $1.16...$, i.e. $16\%$ off. So lets make a Taylor expansion of this. $\mathrm e ^\epsilon\approx 1+\epsilon$. Plugging this into the above for a factor of 3 we get

$$x_0-\epsilon=3-3\mathrm e ^{-x_0}(1+\epsilon)$$ as before (but now with 3) setting $x_0=3$ and solving for $\epsilon$ we get $$\epsilon=\frac{3 \mathrm e^ {-3}}{1-3 \mathrm e^{-3}}= 0.175...$$ and $3-\epsilon=2.8244$, reducing the error from $1.2\%$ to $0.4\%$

And just because it is sort of fun, let me even improve. Let's approximate $\mathrm{e}^{\epsilon}\approx 1+\epsilon+\epsilon^2/2$. This results in a quadratic equation for $\epsilon$ which gives us:

$$\epsilon= \frac 1 a -1-\sqrt{(\frac 1 a -1)^2-2}; \quad a=3\mathrm{e}^{-3}$$

The above result is actually an approximation of this as (just to practice Taylor series a little bit) $$\begin{eqnarray} \epsilon&= &(\frac 1 a -1)-\sqrt{(\frac 1 a -1)^2-2}\\ &= &(\frac 1 a -1)-(\frac 1 a -1)\sqrt{1-\frac{2}{(\frac 1 a -1)^2}}\\ &= &(\frac 1 a -1)\left(1-\sqrt{1-\frac{2}{(\frac 1 a -1)^2}}\right)\\ &\approx &(\frac 1 a -1)\left(1-\left(1-\frac{1}{(\frac 1 a -1)^2}\right)\right)\\ &=&\frac{1}{(\frac 1 a -1)}=\frac{a}{1-a} \end{eqnarray}$$ In any case, plugging in the value of $3$, the quadratic approximation gives $3-\epsilon=2.82162$, which is $0.006\%$

If you would have done this with the factor $5$ you'd get for the exact solution $ x= 4.96511423...$ and for the approximation $x\approx 4.96511448...$ providing a relative error of $5\cdot 10^{-8}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.