# Question on feynman lectures volume 1 (wavelength and frequency)

1. Jan 10, 2012

### 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!

2. Jan 10, 2012

### 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.

3. Jan 11, 2012

### codelieb

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

Last edited: Jan 11, 2012
4. Jan 12, 2012

### demonelite123

thank you both for your replies!

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