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 wavelength. 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-wavelength

Homework Equations

Doppler shift in terms of frequency
$$f'=f(1+\frac{v}{c}cos \theta)$$
Relationship between wavelength 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 wavelength-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 wavelength:

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

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

 

[gabbagabbahey]

Looks good to me!