# Derivation of the group velocity

I know that the group velocity of a light pulse is defined as

$$\begin{split}v_g&=v_p\left(1+\frac{\lambda}{n}\frac{dn}{d\lambda}\right)\\ &=\frac{c}{n}\left(1+\frac{\lambda}{n}\frac{dn}{d\lambda}\right).\end{split} \tag{1}$$

On the other hand, it is also defined as

$$\begin{split}v_g &= \frac{c}{\left(n-\lambda\frac{dn}{d\lambda}\right)}\\ &=\frac{c}{n}\left(1-\frac{\lambda}{n}\frac{dn}{d\lambda}\right)^{-1}.\end{split}\tag{2}$$

$$0=\frac{\lambda^2}{n^2}\left(\frac{dn}{d\lambda}\right)^2.\tag{3}$$

After neither $\lambda$ nor $n$ is $0$, does that mean that $\left(\frac{dn}{d\lambda}\right)^2=0$? And how can I justify that?
Source: http://en.wikipedia.org/wiki/Group_velocity#Other_expressions
Alternative explanation: Wiki is crap?

• I don't see where in that article it gives $\frac{c}{n}\bigl(1 - \frac{\lambda}{n}\frac{\mathrm{d}n}{\mathrm{d}\lambda}\bigr)^{-1}$. Can you be more specific? – David Z Apr 13 '16 at 7:11
• Well, if $\lambda/n\ dn/d\lambda$ is small the two are equivalent to first order in this small parameter. – Nick P Apr 13 '16 at 7:35
• Possible duplicate of Derviation of group velocity – honeste_vivere Apr 13 '16 at 11:09
• @honeste_vivere: not so. Despite its title, that other Q is about deriving the definition of the group velocity. – L. Levrel Apr 13 '16 at 19:41
• Comment to the post (v5): OP is not quoting Wikipedia accurately. Eq. (2) should be $v_g=v_p \left(1-\frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}$, where $\lambda_0$ is the vacuum wavelength. – Qmechanic Apr 13 '16 at 22:09

The first equation correctly states that $$v_g = \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right).$$ But if you look at the wikipedia page that you linked to, you'll see that the second equation should read $$v_g = \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1},$$ where $\lambda_0$ is the wavelength in vacuum, and $$n = \frac{\lambda_0}{\lambda},$$ where $\lambda$ is the wavelength in the medium. The second equation can be derived from the first as follows: \begin{align} v_g &= \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right)\\ &= \frac{c}{n}\left[1 + \frac{\lambda}{n}\left(\frac{d\lambda}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \frac{\lambda_0}{n^2}\left(\frac{d(\lambda_0/n)}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \frac{\lambda_0}{n^2}\left(-\frac{\lambda_0}{n^2} + \frac{1}{n}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \left(-1 + \frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[\frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\left(-1 + \frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[\left(-\frac{\lambda_0}{n}\left(\frac{d\lambda_0}{dn}\right)^{-1} + 1\right)^{-1}\right]\\ &= \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}. \end{align}

Group velocity, for any kind of wave, is defined as $$\boxed{v_g=\frac{\mathrm dω}{\mathrm dk}}.$$

Phase velocity is defined as $v_p=\dfrac ωk$, and refraction index as $n=\dfrac c{v_p}$. So $ω=c\,\dfrac kn$. Hence, $$\mathrm dω=c\left(\frac{\mathrm dk}n-\frac{k\,\mathrm dn}{n^2}\right)=\frac{c\,\mathrm dk}{n}\left(1-\frac kn\frac{\mathrm dn}{\mathrm dk}\right).$$

Given that $k=\dfrac{2π}λ$, $\dfrac{\mathrm dk}k=-\dfrac{\mathrm dλ}λ$ (logarithmic differentiation), so that $$v_g=\frac{\mathrm dω}{\mathrm dk}=\frac cn\left(1+\frac λn\frac{\mathrm dn}{\mathrm dλ}\right).$$

Edit: this demonstration shows the first equation you cite is correct, the second is not.

• My point is that I do not have k in the first equation, but $\lambda$ instead – arc_lupus Apr 13 '16 at 8:47
• Please read the whole answer. Group velocity is defined as I stated, period. k=2π/λ so you can reformulate this with λ if you like, vg=-λ²/2π*dω/dλ. – L. Levrel Apr 13 '16 at 8:51
• That means that Wikipedia is apparently incorrect... – arc_lupus Apr 13 '16 at 9:12
• Edit 2: Found a solution for both possibilities, your answer might need an edit. – arc_lupus Apr 13 '16 at 9:38
• The group velocity is a vector, defined as:$$\mathbf{V}_{g} = \partial_{\mathbf{k}} \ \omega\left(\mathbf{k}\right) = \frac{c}{\omega} \left[ \hat{\mathbf{k}} \frac{\partial \omega}{\partial n} + \hat{\boldsymbol{\theta}} \frac{1}{n} \frac{\partial \omega}{\partial \theta} \right]$$ – honeste_vivere Apr 13 '16 at 11:24

Ok, I found an answer on my own:
For the first definition the answer from L. Levrel is sufficient. For the second definition one has to follow the following set of equations: $$\begin{split} \frac{1}{v_g}&=\frac{dk}{d\omega}\\ &=c^{-1}\left(n+\omega\frac{dn}{d\omega}\right)\\ &=c^{-1}\left(n-\lambda\cdot\frac{dn}{d\lambda}\right)\\ \Rightarrow v_g&=\frac{c}{n\left(1-\frac{\lambda}{n}\frac{dn}{d\lambda}\right)} \end{split}$$ which is what I wanted to get.
How to get from $\omega$ to $\lambda$: $$\begin{split} \omega&=\frac{2\pi c}{\lambda}\\ \Rightarrow\frac{dn}{d\omega}&=\frac{dn}{d\lambda}\frac{d\lambda}{d\omega}\\ &=\frac{dn}{d\lambda}\frac{d\lambda}{d\frac{2\pi c}{\lambda}}\\ &=\frac{1}{2\pi c}\frac{dn}{d\lambda}\frac{d\lambda}{d\frac{1}{\lambda}}\\ &=-\frac{\lambda^2}{2\pi c}\frac{dn}{d\lambda}\\ &=-\frac{\lambda}{\omega}\frac{dn}{d\lambda}\end{split}$$

• I'm curious to read how you get line 3 from line 2, that is, how you change variable $ω$ into $λ$. Remember that $ω=2πc/nλ$. Your answer might need an edit (or deletion) :-) – L. Levrel Apr 13 '16 at 19:37
• @L.Levrel: I added an explanation, I hope that helps. In general I had the impression that $\omega$ is not influenced by the refractive index, but the phase velocity is (by $v_p = \frac{\omega}{k_0\cdot n}$) – arc_lupus Apr 13 '16 at 20:38
• Suggestion to the answer (v2): Replace all appearances of $\lambda$ with the vacuum wavelength $\lambda_0$. – Qmechanic Apr 13 '16 at 22:18
• @L.Levrel: Do you have a source for $\omega=\frac{2\pi c}{n\lambda}$? I could not find anything for that... – arc_lupus Apr 14 '16 at 6:44
• By definition, $v_p=ω/k$. Since $v_p=c/n$ by definition of $n$, you find $ω=kc/n$. Of course $λ$ depends on $n$. See also Pulsar's answer. – L. Levrel Apr 14 '16 at 19:40