0
$\begingroup$

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

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

That leads to the equation

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

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?

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$
    – David Z
    Apr 13, 2016 at 7:11
  • 1
    $\begingroup$ Well, if $\lambda/n\ dn/d\lambda$ is small the two are equivalent to first order in this small parameter. $\endgroup$
    – Nick P
    Apr 13, 2016 at 7:35
  • 3
    $\begingroup$ Possible duplicate of Derviation of group velocity $\endgroup$ Apr 13, 2016 at 11:09
  • 2
    $\begingroup$ @honeste_vivere: not so. Despite its title, that other Q is about deriving the definition of the group velocity. $\endgroup$
    – L. Levrel
    Apr 13, 2016 at 19:41
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Apr 13, 2016 at 22:09

3 Answers 3

4
$\begingroup$

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: $$ \begin{align} v_g &= \frac{c}{n}\left(1 + \frac{\lambda}{n}\frac{dn}{d\lambda}\right)\\ &= \frac{c}{n}\left[1 + \frac{\lambda}{n}\left(\frac{d\lambda}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \frac{\lambda_0}{n^2}\left(\frac{d(\lambda_0/n)}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \frac{\lambda_0}{n^2}\left(-\frac{\lambda_0}{n^2} + \frac{1}{n}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[1 + \left(-1 + \frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[\frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\left(-1 + \frac{n}{\lambda_0}\frac{d\lambda_0}{dn}\right)^{-1}\right]\\ &= \frac{c}{n}\left[\left(-\frac{\lambda_0}{n}\left(\frac{d\lambda_0}{dn}\right)^{-1} + 1\right)^{-1}\right]\\ &= \frac{c}{n}\left(1 - \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}. \end{align} $$

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ My point is that I do not have k in the first equation, but $\lambda$ instead $\endgroup$
    – arc_lupus
    Apr 13, 2016 at 8:47
  • $\begingroup$ 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. $\endgroup$
    – L. Levrel
    Apr 13, 2016 at 8:51
  • $\begingroup$ That means that Wikipedia is apparently incorrect... $\endgroup$
    – arc_lupus
    Apr 13, 2016 at 9:12
  • $\begingroup$ Edit 2: Found a solution for both possibilities, your answer might need an edit. $\endgroup$
    – arc_lupus
    Apr 13, 2016 at 9:38
  • $\begingroup$ The group velocity is a vector, defined as:$$\mathbf{V}_{g} = \partial_{\mathbf{k}} \ \omega\left(\mathbf{k}\right) = \frac{c}{\omega} \left[ \hat{\mathbf{k}} \frac{\partial \omega}{\partial n} + \hat{\boldsymbol{\theta}} \frac{1}{n} \frac{\partial \omega}{\partial \theta} \right]$$ $\endgroup$ Apr 13, 2016 at 11:24
0
$\begingroup$

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 $\lambda$: $$\begin{split} \omega&=\frac{2\pi c}{\lambda}\\ \Rightarrow\frac{dn}{d\omega}&=\frac{dn}{d\lambda}\frac{d\lambda}{d\omega}\\ &=\frac{dn}{d\lambda}\frac{d\lambda}{d\frac{2\pi c}{\lambda}}\\ &=\frac{1}{2\pi c}\frac{dn}{d\lambda}\frac{d\lambda}{d\frac{1}{\lambda}}\\ &=-\frac{\lambda^2}{2\pi c}\frac{dn}{d\lambda}\\ &=-\frac{\lambda}{\omega}\frac{dn}{d\lambda}\end{split}$$

$\endgroup$
5
  • 1
    $\begingroup$ 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) :-) $\endgroup$
    – L. Levrel
    Apr 13, 2016 at 19:37
  • $\begingroup$ @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 $v_p = \frac{\omega}{k_0\cdot n}$) $\endgroup$
    – arc_lupus
    Apr 13, 2016 at 20:38
  • $\begingroup$ Suggestion to the answer (v2): Replace all appearances of $\lambda$ with the vacuum wavelength $\lambda_0$. $\endgroup$
    – Qmechanic
    Apr 13, 2016 at 22:18
  • $\begingroup$ @L.Levrel: Do you have a source for $\omega=\frac{2\pi c}{n\lambda}$? I could not find anything for that... $\endgroup$
    – arc_lupus
    Apr 14, 2016 at 6:44
  • $\begingroup$ 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. $\endgroup$
    – L. Levrel
    Apr 14, 2016 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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