Register to reply

Use geometry -Redshift derivation?

Share this thread:
NanaToru
#1
Jan11-13, 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.
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
clamtrox
#2
Jan11-13, 06:40 AM
P: 939
What are v(obs) and v(em)? What about v? You cannot derive a formula without knowing what the different symbols mean
NanaToru
#3
Jan11-13, 12:22 PM
P: 15
v-observed and v-emitted, I'm assuming. It doesn't clarify much more than that

cepheid
#4
Jan11-13, 12:51 PM
Emeritus
Sci Advisor
PF Gold
cepheid's Avatar
P: 5,196
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 = Δtem 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 t1 does the first wavefront arrive at the obsever? At what time t2 does the second one arrive? What is the difference Δtobs between them? How does this differ from Δtem? So what is the frequency shift?
NanaToru
#5
Jan12-13, 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?
Attached Thumbnails
image.jpg  
cepheid
#6
Jan12-13, 07:55 PM
Emeritus
Sci Advisor
PF Gold
cepheid's Avatar
P: 5,196
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 ?
NanaToru
#7
Jan12-13, 08:56 PM
P: 15
Here's my attempt at what you described, but it's got me all mixed up again.

I think t2 = t1+ some tΔx
Meaning that Δtobs=t2-t1 = t1 + tΔx - t1 = tΔx

Δtem=tem

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?
cepheid
#8
Jan13-13, 12:41 AM
Emeritus
Sci Advisor
PF Gold
cepheid's Avatar
P: 5,196
Quote Quote by NanaToru View Post
Here's my attempt at what you described, but it's got me all mixed up again.

I think t2 = t1+ some tΔx
Meaning that Δtobs=t2-t1 = t1 + tΔx - t1 = tΔx
You're thinking along the right lines, you're just not quite getting the details. Say t = 0 is when wavefront 1 is emitted. Also, t2 is when wavefront 2 arrives. t1 is when wavefront 1 arrives. I'm assuming that Δt_x is the travel time required for wavefront 2. Therefore t2 is not t1 + Δt_x, because that would mean that wavefront 2 wasn't emitted until after wavefront 1 arrived. Instead, t2 = Δt_em + Δt_x, since Δt_em is the time interval between the emission of the two wavefronts.


Register to reply

Related Discussions
In binary can we have a value with deci centi mili or more lower valued prefix? Computers 14
Could Redshift not result from the Doppler effect? General Physics 27
Confused about this blueshift/redshift paradox Special & General Relativity 4
Redshift first year astronomy Astronomy & Astrophysics 2