Question on Feynman lectures volume 1 (wavelength and frequency)

[demonelite123]

in chapter 32 section 4 on electromagnetic waves, near the end he writes since $\lambda = \frac{2\pi c}{\omega}$ then $\Delta\lambda = \frac{2 \pi c \Delta \omega}{{\omega}^2}$. i understand the first equation he writes but then the second one i am having trouble convincing myself how he came up with that.

also earlier in chapter 30 on refraction, he also says the frequency is $\nu = \frac{c}{\lambda}$ so $\Delta \nu = \frac{c \Delta \lambda}{{\lambda}^2}$. similarly i don't understand how he comes up with the second equation. he doesn't seem to explain it and it's not obvious to me how he got that. can someone help clarify this? thanks!

 

[Ken G]

It sounds like you have not had calculus yet, when you do this will be much more clear. But in the mean time, you can still get it like this. If you have a function like f(x)=1/x (think of f as wavelength and x as frequency), and you want to know what changes in f will be, let x become x + dx, where dx is a small change in x, and call df the change in f that we are interested in. Then df = 1/(x+dx) - 1/x, correct? Now rewrite 1/(x+dx) as 1/x times 1/(1+dx/x), where dx/x << 1. Now here comes the key approximation, that works really well the smaller is dx/x:
I claim that 1/(1+dx/x) = 1-dx/x
to good approximation, and then note that if my claim is true, then df = dx/x², as in the above formulae.

To check my claim, cross multiply, and note that the claim requires 1 = (1+dx/x)*(1-dx/x) = 1 - (dx/x)². Can you see that (dx/x)² is indeed a very small error in this equation, and gets really small as dx/x gets smaller? That's the whole idea-- the expressions you quote are not exact, but they are very good approximations that get better and better as the change in frequency you are considering gets smaller and smaller.

 

[codelieb]

...then note that if my claim is true, then df = dx/x², as in the above formulae.

I don't mean to nitpick, but just want to mention that if f(x) = 1/x, then (even without calculus) one can see that f(x) decreases when x increases, and f(x) increases when x decreases, which means the that the proportion relating a change in f to a change in x has to be negative. In fact (using calculus) the correct relation is df = - dx/x². So Eq. (32.13), at the end of FLP Vol. I Chapter 32 section 3, which demonelite123 is asking about, and in which the minus sign is lacking, requires something more than calculus to explain it -- namely that, in this case, we are only interested in the width of the spectral lines so we can ignore the sign.

Mike Gottlieb
Editor, The Feynman Lectures on Physics

 

[demonelite123]

thank you both for your replies!

 

[sardar]

I can explain the derivation of the equations provided by Feynman in the context of electromagnetic waves. Let's start with the first equation, \lambda = \frac{2\pi c}{\omega}. This equation relates the wavelength (\lambda) of an electromagnetic wave to its frequency (\omega) and the speed of light (c). This equation is a fundamental relationship in physics and is derived from the wave equation for electromagnetic waves.

Now, to understand the second equation, \Delta\lambda = \frac{2 \pi c \Delta \omega}{{\omega}^2}, we need to consider the concept of wavelength and frequency as being related to each other. We can think of wavelength as the distance between two consecutive peaks or troughs of a wave, while frequency is the number of complete cycles of the wave that pass through a point in one second. Therefore, if the frequency of a wave changes by a small amount, \Delta \omega, the wavelength will also change by a small amount, \Delta \lambda.

Using the first equation, we can write \Delta\lambda = \frac{2\pi c}{\omega + \Delta \omega} - \frac{2\pi c}{\omega} = \frac{2 \pi c \Delta \omega}{{\omega}^2 + \omega \Delta \omega}. Now, assuming that \Delta \omega is very small compared to \omega, we can neglect the term \omega \Delta \omega and simplify the equation to \Delta\lambda = \frac{2 \pi c \Delta \omega}{{\omega}^2}. This is how Feynman derived the second equation.

Similarly, in the context of refraction in chapter 30, the frequency of an electromagnetic wave changes as it passes through a medium with a different refractive index. The equation \nu = \frac{c}{\lambda} relates the frequency (\nu) to the wavelength (\lambda) and the speed of light (c). Using a similar approach as above, we can derive the equation \Delta \nu = \frac{c \Delta \lambda}{{\lambda}^2}.

In summary, Feynman's equations are derived from the fundamental relationship between wavelength, frequency, and the speed of light in the context of electromagnetic waves. They may seem complex, but they are based on well-established principles in physics. I hope this explanation