Derivation of Doppler Shift from Frequency to Wavelenght

[IBY]

Homework Statement

Derive the doppler shift equation from the equation in terms of frequency to one in terms of wavelenght. Clues: frequency=c/lamda, use Taylor's expansion, velocity of source is much smaller than velocity of wave.

c-velocity of wave
v-velocity of source
f-frequency
lambda-wavelenght

Homework Equations

Doppler shift in terms of frequency
f'=f(1+\frac{v}{c}cos \theta)
Relationship between wavelenght and frequency
\lambda=\frac{c}{f}
Taylor's expansion
\sum_{n=0}^2 \frac{f'^n(f(1+\frac{v}{c}cos \theta)}{n!}x^n

The Attempt at a Solution

I used the wavelenght-frequency relationship equation in order to substitue for f in the doppler frequency equation to get:

\frac{c}{\lambda'}=\frac{c}{\lambda}(1+\frac{v}{c}cos \theta)

\frac{c}{\lambda'}=\frac{c}{\lambda}+\frac{c}{\lambda}\frac{v}{c}cos \theta

\frac{c}{\lambda'}=\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta

\lambda'=\frac{c}{\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta}

\lambda'=\lambda+\frac{\lambda c}{v cos\theta}

I used Taylor's expansion to the second degree, and I got:

\lambda'=\frac{\lambda c}{2v}x^2+{\lambda+\frac{\lambda c}{v}

I small angle approximate, so x^2=sin^2 (theta), use trig id to get sin^2 (theta)=1-cos^2 (theta)

\lambda'=\frac{\lambda c}{2v}-\frac{\lambda c}{2v}cos^2 \theta+\lambda+\frac{\lambda c}{v}

\lambda'=\frac{3\lambda c}{2v}-\frac{\lambda c}{2v}cos^2 \theta+\lambda

And now, I am dead in the water.

 

[gabbagabbahey]

\lambda'=\frac{c}{\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta}

\lambda'=\lambda+\frac{\lambda c}{v cos\theta}

You might want to double check this step

 

[IBY]

@gabbagabbahey
I don't see it. If one divides by fraction, aren't you supposed to multiply with the bottom part reciprocated?

 

[gabbagabbahey]

\frac{A}{A+B}\neq1+\frac{A}{B}

You can't divide up the denominator in this manner

 

[IBY]

If so, then I use Taylor's expansion on this?

\lambda'=\frac{\lambda c}{c+vcos \theta}

 

[gabbagabbahey]

I would divide both the denominator and numerator by c first:

\lambda'=\frac{\lambda}{1+\frac{v}{c}\cos\theta}=\lambda\left(1+\frac{v}{c}\cos\theta\right)^{-1}

The reason you want to do this is because you know v\ll c, so \frac{v}{c}\cos\theta will be a small number and you are left trying to Taylor expand (1+\text{small number})^{-1} which you can do easily.

 

[IBY]

Okay, one last fact checking (hopefully) before I go away for good.

So there is:

\lambda(1+\frac{v}{c}cos \theta)^{-1}

I derive it:

\lambda \frac{v}{c}sin \theta(1+\frac{v}{c}cos \theta)^{-2}x

I do it again, use the multiplication rule:

(\lambda (\frac{v}{c})^{2}sin^2 \theta(1+\frac{v}{c}cos \theta)^{-3}+\frac{\lambda \frac{v}{c}cos \theta(1+\frac{v}{c}cos \theta)^{-2}}{2})x^2

Small number, so sin of theta is zero and cos of theta is 1, x=sin so including all of the equation above, I am left with:

(\frac{\lambda \frac{v}{c}(1+\frac{v}{c})^{-2}}{2})sin^2 \theta+\lambda(1+\frac{v}{c})^{-1}

Trig id sin^2 theta is 1-cos^2 theta

(\frac{\lambda \frac{v}{c}(1+\frac{v}{c})^{-2}}{2})(1-cos^2 \theta)+\lambda(1+\frac{v}{c})^{-1}

For some reason, I feel like I went into a dead end.

 

[gabbagabbahey]

Okay, one last fact checking (hopefully) before I go away for good. :)

So there is:

\lambda(1+\frac{v}{c}cos \theta)^{-1}

I derive it:

\lambda \frac{v}{c}sin \theta(1+\frac{v}{c}cos \theta)^{-2}x

Huh?! With respect to what variable are you taking the derivative and why? Also, what is x in the above equation?

 

[IBY]

Hhhmm
Evidently, I made the mistaken assumption that the "theta" was the variable. Probably, I am making this more complicated than it is supposed to be.

 

[gabbagabbahey]

Well, if you have some function f(x) and x is small, then f(x)\approx f(0)+f'(0)x right?

You know that the quantity \frac{v}{c}\cos\theta is small, so why not call that quantity x?

When you do that, you have \lambda'=\lambda(1+x)^{-1} right? So define the function f(x)=(1+x)^{-1} and expand about small 'x'...make sense?

The reason you don't want to use \theta as your variable is because you don't know that \theta is small, so you would need to keep all terms in the Taylor expansion, which doesn't help you at all.

 

[IBY]

Okay, I think I got it. After I expanded it, I got:

\lambda'=\lambda(1+x)^{-1}-\lambda(1+x)^{-2}+\frac{2\lambda(1+x)^{-3}}{2}

x is small, so it is around 0

\lambda'=\lambda-\lambda x+\lambda x^2

x is v cos (theta)/lambda

\lambda'=\lambda-\lambda\frac{v}{c}cos \theta+\lambda(\frac{v}{c}cos \theta)^2

Well, it turns out I only need the first order expansion in order to get a parallel equation with wavelenght:

\lambda'=\lambda(1-\frac{v}{c}cos \theta)

So, I think I got it! Thanks for the help.

 

[gabbagabbahey]

Looks good to me!