Trying to derive a group velocity equation?

[jeebs]

hi,
I am trying to derive the group velocity equation (actually its inverse):

$$\frac{1}{v_g} = \frac{1}{c}(n(\lambda) - \lambda\frac{dn}{d\lambda})$$
where n is the refractive index and depends on the light's wavelength.

I have started by saying that if $$v_g = \frac{\partial\omega}{\partial k}$$ then I can do $$\frac{dk}{d\omega} = \frac{dk}{d\lambda}\frac{d\lambda}{d\omega}$$.

I then say that $$k=\frac{2\pi}{\lambda}$$ and $$\omega=\frac{2\pi c}{n\lambda}$$, also that $$k=\frac{n\omega}{c}$$ .

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):

$$\frac{d\lambda}{d\omega} = 2\pi c(\frac{-1}{n^2\omega^2} - \frac{1}{n^2\omega}\frac{dn}{d\omega})$$

$$\frac{dk}{d\omega} = \frac{4\pi^2c}{\lambda^2n^2\omega^2}(n + \omega\frac{dn}{d\omega})$$

$$\frac{dn}{d\omega} = c(\frac{1}{\omega}\frac{dk}{d\omega} - \frac{k}{\omega^2})$$

$$\frac{dk}{d\lambda} = \frac{-2\pi}{k^2}$$

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 $$\frac{1}{v_g} = \frac{1}{c}(n - (\frac{4\pi^2 c}{n^2\omega^2\lambda}\frac{d\omega}{d\lambda} + \frac{4\pi^2 c}{n^3\omega\lambda}\frac{dn}{d\lambda})\frac{dn}{d\lambda})$$

and to finish I'd need to show that the bit in the brackets is equal to $$\lambda$$ 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.

 

[kreil]

Double check how you are taking those derivatives. For instance,

$$\frac{\partial}{\partial x} \left[ \alpha f(x) \beta g(x) \right] = \alpha \beta \left ( \frac{\partial f}{\partial x}g(x)+\frac{\partial g}{\partial x}f(x) \right )$$

where alpha and beta are constants

 

[jeebs]

i'm pretty sure that's the way I was doing them.

 

[kreil]

You say that

$$k=\frac{n\omega}{c} \implies \frac{dk}{d\omega}=\frac{1}{c} \left ( n + \omega \frac{dn}{d\omega} \right)$$

yet also that

$$\frac{dk}{d\omega} = \frac{4\pi^2c}{\lambda^2n^2\omega^2}(n + \omega\frac{dn}{d\omega})$$

 

[kreil]

Notice that

$$\frac{1}{v_g} = \frac{dk}{d\omega} = \frac{d}{d\omega} \frac{n\omega}{c} = \frac{1}{c} \left ( n + \omega \frac{dn}{d\omega} \right)$$

$$\omega \frac{dn}{d\omega} = \omega \frac{dn}{d\lambda} \frac{d \lambda}{d\omega}$$

now calculate this last term and you should be done.