How Does Lorentz's Equation Transform into Cauchy's Law?

[DannyJ108]

Homework Statement

Prove that when frequency is small, the Lorentz model for a refractive index in dielectrics becomes Cauchy's empirical law

Relevant Equations

$n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0 - \omega^2}$
$n(\lambda)=A+\frac B {\lambda^2}$

Hello fellow physicists,

I need to prove that when $\omega << \omega_0$, Lorentz equation for refractive indexes:
$n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0 - \omega^2}$
turns into Cauchy's empirical law:
$n(\lambda)=A+\frac B {\lambda^2}$
I also need to express A and B as a function of the Lorentz model parameters: $\omega_0$ and $\omega_p$, where:
$\omega_0$: resonant frequency
$\omega_p$: plasma frequency

Any idea on how I could approach and do this?

Thank you in advance!

 

[TSny]

Do you know how to approximate $\large \frac 1 {1-x}$ when $x \ll 1$?

 

[DannyJ108]

Do you know how to approximate $\large \frac 1 {1-x}$ when $x \ll 1$?

Yes, I do, but I don't know how I would get that expression with the equation I have. If $\omega << \omega_0$ , the equation would result in:
$n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0}$

 

[TSny]

If $\omega << \omega_0$ , the equation would result in:
$n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0}$

Whoa. That's going too far.

If $\omega \ll \omega_0$, how does $\frac {\omega^2} {\omega_0^2}$ compare to 1?

If you let $x = \frac {\omega^2} {\omega_0^2}$, can you rearrange the denominator in $n^2(\omega)$ to get a factor of $1-x$?

 

[berkeman]

Homework Statement:: Prove that when frequency is small, the Lorentz model for a refractive index in dielectrics becomes Cauchy's empirical law
Relevant Equations:: $n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0 - \omega^2}$
$n(\lambda)=A+\frac B {\lambda^2}$

Hello fellow physicists,

I need to prove that when $\omega << \omega_0$, Lorentz equation for refractive indexes:
$n^2(\omega) = 1 + \frac {\omega^2_p} {\omega^2_0 - \omega^2}$
turns into Cauchy's empirical law:
$n(\lambda)=A+\frac B {\lambda^2}$
I also need to express A and B as a function of the Lorentz model parameters: $\omega_0$ and $\omega_p$, where:
$\omega_0$: resonant frequency
$\omega_p$: plasma frequency

Any idea on how I could approach and do this?

Thank you in advance!

@DannyJ108 -- You are required to show more work at PF. Please use the hints provided by @TSny to motivate you to post your best efforts on this problem. Thank you.

 

[DannyJ108]

Whoa. That's going too far.

If $\omega \ll \omega_0$, how does $\frac {\omega^2} {\omega_0^2}$ compare to 1?

If you let $x = \frac {\omega^2} {\omega_0^2}$, can you rearrange the denominator in $n^2(\omega)$ to get a factor of $1-x$?

I have rearranged the equation to get:
$n^2(\omega)=1+\frac {\frac {\omega_p^2} {\omega_0^2}} {1-\frac {\omega^2} {\omega_0^2}}$

Since $\omega << \omega_0$, $\frac {\omega^2} {\omega_0^2}$ will be smaller than 1, so I can expand that fraction as a Taylor series to get:

$n(\lambda)=\sqrt{1+\frac {\omega_p^2} {\omega_0^2} + \frac {\omega_p^2} {\omega_0^4}4\pi^2 c^2 \frac {1} {\lambda^2}} (eq.1)$

Where I have used: $\omega=\frac {2\pi c} \lambda$

However, in Cauchy's law, $n$ is not squared:

$n(\lambda)=A+\frac B {\lambda^2}$

So I don't know how I could group the terms in $eq. 1$ to establish the $A$ and $B$ coefficients, I do however get the $\lambda^2$ in the denominator, so that's a good thing.

How could I proceed from here?

 

[TSny]

How would you approximate $\sqrt{a + b}$ assuming $b \ll a$?

 

[DannyJ108]

How would you approximate $\sqrt{a + b}$ assuming $b \ll a$?

But I'm not approximating $\sqrt{a + b}$ I'm approximating:
$\sqrt{1 + a + b}$

 

[TSny]

But I'm not approximating $\sqrt{a + b}$ I'm approximating:
$\sqrt{1 + a + b}$

Let $a' = 1+a$. Then you have $\sqrt{a' + b}$.