Frequency to wavelength question (thermal Doppler broadening)

[dartingeyes]

Homework Statement

I don't understand how to go from one equation to the other

Relevant Equations

provided in picture

I want to understand how the two equations in the picture are the same. I am confused because \nu=c/\lambda. This would mean that the term on the right side of the equation in the square root, call it x, would then become x^-1. The resulting equation being \Delta\lambda = \lambda x^-1. But somehow this is not the case?

 

[Orodruin]

What you are saying is only true for the relationship between frequency and wavelength themselves - not their small variation.

From $\omega \propto 1/\lambda$ follows that $\Delta\omega \propto \Delta\lambda /\lambda^2 \propto \Delta\lambda (\omega/\lambda)$.

 

[dartingeyes]

What you are saying is only true for the relationship between frequency and wavelength themselves - not their small variation.

From $\omega \propto 1/\lambda$ follows that $\Delta\omega \propto \Delta\lambda /\lambda^2 \propto \Delta\lambda (\omega/\lambda)$.

Thank you. This tracks for me.

 

[Orodruin]

A somewhat cleaner derivation:

The product $\omega\lambda$ is constant so $0=\Delta(\omega\lambda) \simeq \omega \Delta\lambda + \lambda\Delta\omega$. This directly leads to $\Delta\lambda/\lambda \simeq - \Delta\omega/\omega$.