- #1
jeebs
- 325
- 4
hi,
I am trying to derive the group velocity equation (actually its inverse):
[tex] \frac{1}{v_g} = \frac{1}{c}(n(\lambda) - \lambda\frac{dn}{d\lambda}) [/tex]
where n is the refractive index and depends on the light's wavelength.
I have started by saying that if [tex]v_g = \frac{\partial\omega}{\partial k} [/tex] then I can do [tex] \frac{dk}{d\omega} = \frac{dk}{d\lambda}\frac{d\lambda}{d\omega} [/tex].
I then say that [tex] k=\frac{2\pi}{\lambda} [/tex] and [tex] \omega=\frac{2\pi c}{n\lambda} [/tex], also that [tex] k=\frac{n\omega}{c} [/tex] .
When I've done the differentiation I have been using chain rules and product rules.
After messing around with this stuff all bloody weekend and calculating every possible derivative a dozen times I am certain that (among other things):
[tex] \frac{d\lambda}{d\omega} = 2\pi c(\frac{-1}{n^2\omega^2} - \frac{1}{n^2\omega}\frac{dn}{d\omega}) [/tex]
[tex] \frac{dk}{d\omega} = \frac{4\pi^2c}{\lambda^2n^2\omega^2}(n + \omega\frac{dn}{d\omega}) [/tex]
[tex] \frac{dn}{d\omega} = c(\frac{1}{\omega}\frac{dk}{d\omega} - \frac{k}{\omega^2}) [/tex]
[tex] \frac{dk}{d\lambda} = \frac{-2\pi}{k^2}[/tex]
there are more things I have calculated but I always seem to be going in circles, working out things that end up being in terms of just another "dx/dy".
after going through about 40 trees'worth of paper the closest I have got with this is to find that [tex] \frac{1}{v_g} = \frac{1}{c}(n - (\frac{4\pi^2 c}{n^2\omega^2\lambda}\frac{d\omega}{d\lambda} + \frac{4\pi^2 c}{n^3\omega\lambda}\frac{dn}{d\lambda})\frac{dn}{d\lambda})[/tex]
and to finish I'd need to show that the bit in the brackets is equal to [tex]\lambda[/tex] but I just can't. the fact that this is such a simple looking equation and that I have managed to spend over 12 hours in total tearing my hair out over this tells me I am probably overcomplicating things. Can anyone save my sanity here? I know this will be some simple trick that I'll feel like an idiot for not seeing.
thanks.
I am trying to derive the group velocity equation (actually its inverse):
[tex] \frac{1}{v_g} = \frac{1}{c}(n(\lambda) - \lambda\frac{dn}{d\lambda}) [/tex]
where n is the refractive index and depends on the light's wavelength.
I have started by saying that if [tex]v_g = \frac{\partial\omega}{\partial k} [/tex] then I can do [tex] \frac{dk}{d\omega} = \frac{dk}{d\lambda}\frac{d\lambda}{d\omega} [/tex].
I then say that [tex] k=\frac{2\pi}{\lambda} [/tex] and [tex] \omega=\frac{2\pi c}{n\lambda} [/tex], also that [tex] k=\frac{n\omega}{c} [/tex] .
When I've done the differentiation I have been using chain rules and product rules.
After messing around with this stuff all bloody weekend and calculating every possible derivative a dozen times I am certain that (among other things):
[tex] \frac{d\lambda}{d\omega} = 2\pi c(\frac{-1}{n^2\omega^2} - \frac{1}{n^2\omega}\frac{dn}{d\omega}) [/tex]
[tex] \frac{dk}{d\omega} = \frac{4\pi^2c}{\lambda^2n^2\omega^2}(n + \omega\frac{dn}{d\omega}) [/tex]
[tex] \frac{dn}{d\omega} = c(\frac{1}{\omega}\frac{dk}{d\omega} - \frac{k}{\omega^2}) [/tex]
[tex] \frac{dk}{d\lambda} = \frac{-2\pi}{k^2}[/tex]
there are more things I have calculated but I always seem to be going in circles, working out things that end up being in terms of just another "dx/dy".
after going through about 40 trees'worth of paper the closest I have got with this is to find that [tex] \frac{1}{v_g} = \frac{1}{c}(n - (\frac{4\pi^2 c}{n^2\omega^2\lambda}\frac{d\omega}{d\lambda} + \frac{4\pi^2 c}{n^3\omega\lambda}\frac{dn}{d\lambda})\frac{dn}{d\lambda})[/tex]
and to finish I'd need to show that the bit in the brackets is equal to [tex]\lambda[/tex] but I just can't. the fact that this is such a simple looking equation and that I have managed to spend over 12 hours in total tearing my hair out over this tells me I am probably overcomplicating things. Can anyone save my sanity here? I know this will be some simple trick that I'll feel like an idiot for not seeing.
thanks.
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