What is the derivation of the Doppler effect for light using Taylor expansion?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 1K views
Skeptic.
Messages
5
Reaction score
0

Homework Statement



This isn't strictly a homework problem, but I didn't know where else to post this. I can't get the same derivation as my lecturer for the Doppler effect of light - which is shown in the attached file. If you cannot open this, I re-wrote it further down.

Homework Equations

The Attempt at a Solution


For the part in the red box, I thought one would do this via Taylor expansion, thus I expected the ##\frac{u}{c}## to be squared, i.e fr = fs##(1 \pm \frac{1}{2}\frac{u}{c}^2)(1 \pm \frac{1}{2}\frac{u}{c}^2)##. I can't see why this wouldn't be the case. Could someone please tell me why I'm wrong?

(In case you cannot open the file, my lecture notes say fr = fs##(1 \pm \frac{u}{c})^\frac{1}{2} (1 \pm \frac{u}{c})^\frac{-1}{2}##=##(1 \pm \frac{1}{2}\frac{u}{c})(1 \pm \frac{1}{2}\frac{u}{c})##)
 
Attachments
Physics news on Phys.org
McClaurin series of (1+x)1/2 = 1 + (1/2) x + ... Note that x is not squared in the second term of the right side.