How does the Relativistic Doppler Effect Affect Observed Wavelengths?

[wahaj]

Homework Statement

A light source moves away from an observer at a speed v_s that is small compared to c. show that the fractional shift in the observed wavelength can be approximated by
\frac{\Delta \lambda}{\lambda} \approx \frac{v_2 }{c}

Homework Equations

f' = \frac{\sqrt{1+ \frac{v}{c} } } {\sqrt{1- \frac{v}{c} } } f
v = f \lambda

The Attempt at a Solution

First I know that v = v_s. if I put this into the above formula I get
f' = \frac{\sqrt{1+ \frac{v_s}{c} } } {\sqrt{1- \frac{v_s}{c} } } f
Since v_s is small compared to c the terms inside the square root signs will be approximately equal to 1 so I can say that
f' \approx f
f = \frac{v}{\lambda}
Since we are working with light, when it leave the light source it travels at v = c. So
f = \frac{c}{\lambda}
so far I have
f' = \frac{c}{\lambda}

this is where I get lost. I have no idea what to do next because I can't seem to quite figure out what f' would be. f' is the frequency of the light observed by the observer. In the observer's frame of reference this would simply be f given above. the equation I derived before implies that I am working in the source's frame of reference so I will need to adjust for f'. But I have no idea how to do that. I know that the velocity of light is fixed so the only way the observed frequency changes is because lambda changes by an amount say Δλ. but this still gets me no where. Could someone give me a hint?

 

[TSny]

You've gone too far with your approximation of the square roots. Keep the next higher (first-order) approximation. See binomial approximation

 

[wahaj]

so you are saying that I should do this using taylor series approximation? Sorry wikipedia doesn't work very well for when it comes to math. And what do you mean by gone too far?

 

[cepheid]

Yeah, it's a Taylor series approximation, and "gone too far" means that in your attempt above, you made this approximation only up to zeroth order, throwing away all of the higher order terms. As TSny says, you should make it to first order.

 

[wahaj]

I'm totally lost. Which equation am I supposed to take the approximation for. Speaking of which I never took an approximation in my attempt at the solution

 

[cepheid]

I'm totally lost. Which equation am I supposed to take the approximation for. Speaking of which I never took an approximation in my attempt at the solution

Yeah you did. You said that (1 + v/c) ≈ 1

You need to take a Taylor series expansion of the square root expressions, and then just keep the first few terms, under the assumption that v/c is small (but not so small that it can neglected entirely, like you did).

 

[wahaj]

yeah but I made that assumption without using calculus? I took the linear approximation it is as follows
\frac{\sqrt{1+ \frac{v_s}{c} } } {\sqrt{1 - \frac{v_s}{c} } } + \frac {3}{(v-c)^2 \sqrt { \frac {-(v+3)}{v-3} } } (v - v_s)

this might be wrong because I never fully understood what each letter in the approximation represents.
I don't see how this will help me though. My main concern was to adjust f' using relativity.

 

[TSny]

Binomial approximation to first order: $(1+x)^a \approx 1+ax$ for $x<<1$.

So, $\sqrt{1+x} = (1+x)^{1/2} \approx 1+\frac{1}{2}x$

and, $\frac{1}{\sqrt{1+x}} = (1+x)^{-1/2} \approx 1-\frac{1}{2}x$

 

[wahaj]

How's this for a proof?

\frac{ \sqrt{1 + \frac{v}{c} } } { \sqrt {1- \frac{v}{c} } } \approx \frac{1 + \frac{v}{2c} }{1 + \frac{v}{2c} } \approx 1
so
f&#039; = \frac{\sqrt{1+ \frac{v}{c} } } { \sqrt { 1 - \frac{v}{c} } } f

\lambda = \frac {c}{f}
\lambda &#039; = \frac{c + v_s}{f}
Δλ = λ&#039; - λ = \frac{c + v_s}{f} - \frac {c}{f}
Δλ = \frac{v_s}{f}
f = \frac{c}{λ}
Δλ = \frac{λ v_s } {c}
\frac {Δλ}{λ} \approx \frac{v_s}{c}

don't worry about the equal sign as opposed to the approximation sign in the proof, I'll fix that later

 

[TSny]

How's this for a proof?

\frac{ \sqrt{1 + \frac{v}{c} } } { \sqrt {1- \frac{v}{c} } } \approx \frac{1 + \frac{v}{2c} }{1 + \frac{v}{2c} } \approx 1

You have a mistake in the denominator. Note that if $\sqrt{1 + \frac{v}{c}} \approx 1+\frac{v}{2c}$ then $\sqrt{1 - \frac{v}{c} } \approx 1-\frac{v}{2c}$.

so
f&#039; = \frac{\sqrt{1+ \frac{v}{c} } } { \sqrt { 1 - \frac{v}{c} } } f

\lambda = \frac {c}{f}
\lambda &#039; = \frac{c + v_s}{f}

What is the justification for the last expression for $\lambda '$?
Suggestion: Use $f = c/\lambda$ and $f ' = c/\lambda '$ along with the equation $f' = \frac{\sqrt{1+ \frac{v}{c} } } { \sqrt { 1 - \frac{v}{c} } } f$ to find an expression for $\lambda ' /\lambda$. Then use the binomial approximation for the square roots. If you have a square root in a denominator, I would suggest you approximate it as $\frac{1}{\sqrt{1+v/c}} = (1+v/c)^{-1/2} \approx 1-v/2c$.

[Edit: So, $\frac{\sqrt{1-v/c}}{\sqrt{1+v/c}} =(1-v/c)^{1/2}(1+v/c)^{-1/2} \approx (1-v/2c)(1-v/2c)$ . Multiply out the right hand side and note that you can neglect $v^2/4c^2$ because it is a very small "second order" term]

 

[vela]

How's this for a proof?

\frac{ \sqrt{1 + \frac{v}{c} } } { \sqrt {1- \frac{v}{c} } } \approx \frac{1 + \frac{v}{2c} }{1 + \frac{v}{2c} } \approx 1

I'll just point out something that might already be obvious to you or will be in hindsight. In both this attempt and your original attempt, you managed to eliminate all of the dependence on v/c in your approximation and therefore ended up with f' = f. You don't want to do that because you want v/c to appear in your final result.

One suggestion to make the algebra simpler: Start off by showing that the Doppler relation in terms of wavelength is
$$\lambda' = \frac{\sqrt{1-v/c}}{\sqrt{1+v/c}} \lambda$$

 

[wahaj]

I actually did that and ended up with a reasonable answer. thanks for the help