"use geometry"Redshift derivation?by NanaToru Tags: astronomy, cosmology, doppler effect, geometry 

#1
Jan1113, 01:56 AM

P: 15

1. Use geometry to derive z=v/c where c is the speed of light and z = [v(obs)v(em)]/v(em).
2. Relevant equations None given... Though I am assuming that I am constant. 3. The attempt at a solution I believe it has to do with the Doppler effect which gives Δf = (Δv/c)f. I'm not entirely sure what it means by geometry though; am I looking at the geometry of the changes of light, cause that seems super bizarre to me. I feel so close to getting it but I feel like I'm missing something. 



#2
Jan1113, 06:40 AM

P: 937

What are v(obs) and v(em)? What about v? You cannot derive a formula without knowing what the different symbols mean




#3
Jan1113, 12:22 PM

P: 15

vobserved and vemitted, I'm assuming. It doesn't clarify much more than that




#4
Jan1113, 12:51 PM

Emeritus
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PF Gold
P: 5,198

"use geometry"Redshift derivation?
First of all, you have to be careful with your symbols. Physicists often use the Greek letter "nu" (##\nu##) for frequency, rather than f. This is not the same thing as ##v##, the velocity of the source of emission. Secondly, your equation for redshift, in terms of frequency, should be ##z = (\nu_{em}  \nu_{obs})/\nu_{obs}##, as opposed to what you have there.
As far as solving the problem, using "geometry" just involves thinking about the source of emission receding from the observer, and considering the observed vs. emitted frequency that results. To do this, consider two successive wavefronts of the EM wave. After all, the frequency is just how often wavefronts pass by. Have the first wavefront be emitted at t = 0, when the source is a distance d from the observer. Then the second wavefront is emitted at a time t = Δt_{em} later, this being the period between wavefronts, (which is the reciprocal of the emitted frequency). Note that at this later time, the source is now farther from the observer, so the second wavefront has a longer distance to travel than the first one did. At what time t_{1} does the first wavefront arrive at the obsever? At what time t_{2} does the second one arrive? What is the difference Δt_{obs} between them? How does this differ from Δt_{em}? So what is the frequency shift? 



#5
Jan1213, 07:17 PM

P: 15

Thanks; I'll try to be more careful... I'm guessing I should be able to find it on the symbols haha.
Anyway, here's a picture I drew (sorry, I'm on a tablet so it's kind of messy) So I'm guessing here that I can see that λ(obs)= λ(em) + x, and use that to figure out that z=x/cΔt = [itex]\nu[/itex] /c. Is there significance in the fact that the it's an isosceles triangle? 



#6
Jan1213, 07:55 PM

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PF Gold
P: 5,198

I don't think you need a spacetime diagram.
Did you try the method I suggested using wavefront arrival times? What is 'x' in your previous post ? 



#7
Jan1213, 08:56 PM

P: 15

Here's my attempt at what you described, but it's got me all mixed up again.
I think t_{2} = t_{1}+ some t_{Δx} Meaning that Δt_{obs}=t_{2}t_{1} = t_{1} + t_{Δx}  t_{1} = t_{Δx} Δt_{em}=t_{em} So if I replace z=[[itex]\nu[/itex]_{em}[itex]\nu[/itex]_{obs})/[itex]\nu[/itex]_{obs}, we wind up with z=[itex]\nu[/itex]_{em}/[itex]\nu[/itex]_{[itex]\Delta[/itex]x} and so [itex]\nu[/itex]=[itex]\lambda[/itex]/v? Is this the correct line of thinking so far? 



#8
Jan1313, 12:41 AM

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