The solutions given are in quite a brief format, where huge swathes of steps have been missed out. Often, it's just a few lines of algebra which is fairly straightforward, but today I don't seem to be able to "see" these, and have to work through them, blind.

One particular bit, which I've tried for the best part of 15 minutes, and have now given up on temporarily, is convincing myself that

\frac{1-v/c}{\sqrt{1-v^2/c^2}} = \sqrt{\frac{1-v/c}{1+v/c}}

This is a fairly fundamental step in calculating Relativistic Doppler shifts, since:

\nu^\prime = \nu\sqrt{\frac{1-v/c}{1+v/c}}

If anyone can tell me how those two things are equal, I'd love to know! Of course, maybe tomorrow it'll be clear. Today is not a good day to revise, but I have to do it anyway.

(1-v/c)/sqrt(1-v^2/c^2)=(1-v/c)/sqrt((1-v/c)(1+v/c))=(1-v/c)^(1-1/2)/sqrt(1+v/c)=sqrt((1-v/c)/(1+v/c))

ReplyDeleteWhere the identity 1-x^2=(1-x)(1+x) has been used ;)