Help with the chain rule (physics example)

Question

I post this topic here, because I need help with the math.

I having problem with the chain rule in the derivation of the effective mass. I don't understand one step in the last equation below. Thanks in advance!

The derivation: Given: $$E=\frac{\hbar k^2}{2m} \text{,} \qquad F=\hbar\frac{dk}{dt}$$ $$ k=\frac{2\pi}{\lambda}\text{,}\qquad v_g=\frac{1}{\hbar}\frac{dE}{dk}$$ The constant $\hbar$ is $\frac{h}{2\pi}$.

Take derivative of $v_g$ with respect to $t$ so $$ \frac{dv_g}{dt}=\frac{1}{\hbar}\frac{d}{dt}\frac{dE}{dk}=\frac{1}{\hbar}\underbrace{ \frac{{d^2E}}{dk^2}\frac{dk}{dt} }_\text{Why and how?} =\frac{1}{\hbar^2}\frac{d^2E}{dk^2}F$$

Answer 1

Recall that the chain rule states that if $y = g(x)$, then $$\left.\frac{d}{dx}(f\circ g)\right|_{x_0} = \left.\frac{df}{dy}\right|_{g(x_0)}\cdot \left.\frac{dg}{dx}\right|_{x_0}$$

---

In your setup, $k$ is implicitly a function of time $t$. So let's make it a little more explicit. Let $\color{blue}{k} = \color{red}{k}(t)$ (that is we're "overloading" the symbol $k$ so that it can represent a $\color{blue}{\text{variable}}$ or a $\color{red}{\text{function}}$ -- this is so I don't have to make up a new variable name).

Then we have $$\frac{d}{dt}\left.\left(\frac{dE}{d\color{blue}{k}}\circ \color{red}{k}\right)\right|_{t} = \frac{d}{d\color{blue}{k}}\left.\left(\frac{dE}{d\color{blue}{k}}\right)\right|_{\color{red}{k}(t)}\cdot\left.\frac{d\color{red}{k}}{dt}\right|_{t} = \left.\frac{d^2E}{d\color{blue}{k}^2}\right|_{\color{red}{k}(t)}\cdot\left.\frac{d\color{red}{k}}{dt}\right|_{t}$$

by the chain rule.

Answer 2

This is what I thought;

$$ \frac{dv_g}{dt}=\frac{d}{dt}(\frac{dE}{dk}) \cdot \frac{1}{\hbar} $$

As $k$ is a function of $t$ (by the statements initially stated),then the chain rule may be applied;

$$ \frac{dv_g}{dt}=\frac{d}{dk}(\frac{dE}{dk}) \cdot \frac{dk}{dt} \cdot \frac{1}{\hbar} $$

$$\frac{dE}{dk} =f(k),\\ \therefore \frac{d}{dk}(\frac{dE}{dk}) = \frac{d^2E}{dk^2}$$

Hence the overall problem may be reduced to;

$$ \frac{dv_g}{dt}= \frac{d^2E}{dk^2} \cdot \frac{dk}{dt} \cdot \frac{1}{\hbar} $$

As $\frac{dk}{dt}=\frac{F}{\hbar}$, the overall result becomes;

$$ \frac{dv_g}{dt}= \frac{d^2E}{dk^2} \cdot \frac{F}{\hbar^2} \equiv \frac{1}{\hbar^2} \cdot \frac{d^2E}{dk^2} \cdot F $$

I think this is correct.

Hope this helps.