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
[tex]f'=f(1+\frac{v}{c}cos \theta)[/tex]
Relationship between wavelenght and frequency
[tex]\lambda=\frac{c}{f}[/tex]
Taylor's expansion
[tex]\sum_{n=0}^2 \frac{f'^n(f(1+\frac{v}{c}cos \theta)}{n!}x^n[/tex]

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:

[tex]\frac{c}{\lambda'}=\frac{c}{\lambda}(1+\frac{v}{c}cos \theta)[/tex]

[tex]\frac{c}{\lambda'}=\frac{c}{\lambda}+\frac{c}{\lambda}\frac{v}{c}cos \theta[/tex]

[tex]\frac{c}{\lambda'}=\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta[/tex]

[tex]\lambda'=\frac{c}{\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta}[/tex]

[tex]\lambda'=\lambda+\frac{\lambda c}{v cos\theta}[/tex]

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

[tex]\lambda'=\frac{\lambda c}{2v}x^2+{\lambda+\frac{\lambda c}{v}[/tex]

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

[tex]\lambda'=\frac{\lambda c}{2v}-\frac{\lambda c}{2v}cos^2 \theta+\lambda+\frac{\lambda c}{v}[/tex]

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

And now, I am dead in the water.

 

[gabbagabbahey]

[tex]\lambda'=\frac{c}{\frac{c}{\lambda}+\frac{v}{\lambda}cos \theta}[/tex]

[tex]\lambda'=\lambda+\frac{\lambda c}{v cos\theta}[/tex]

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]

[tex]\frac{A}{A+B}\neq1+\frac{A}{B}[/tex]

You can't divide up the denominator in this manner

 

[IBY]

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

[tex]\lambda'=\frac{\lambda c}{c+vcos \theta}[/tex]

 

[gabbagabbahey]

I would divide both the denominator and numerator by [itex]c[/itex] first:

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

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

 

[IBY]

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

So there is:

[tex]\lambda(1+\frac{v}{c}cos \theta)^{-1}[/tex]

I derive it:

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

I do it again, use the multiplication rule:

[tex](\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[/tex]

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:

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

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

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

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:

[tex]\lambda(1+\frac{v}{c}cos \theta)^{-1}[/tex]

I derive it:

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

Huh?! With respect to what variable are you taking the derivative and why? Also, what is [itex]x[/itex] 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 [itex]f(x)[/itex] and [itex]x[/itex] is small, then [itex]f(x)\approx f(0)+f'(0)x[/itex] right?

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

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

The reason you don't want to use [itex]\theta[/itex] as your variable is because you don't know that [itex]\theta[/itex] 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:

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

x is small, so it is around 0

[tex]\lambda'=\lambda-\lambda x+\lambda x^2[/tex]

x is v cos (theta)/lambda

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

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

[tex]\lambda'=\lambda(1-\frac{v}{c}cos \theta)[/tex]

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

 

[gabbagabbahey]

Looks good to me!