# Relativistic Doppler shift Problem

1. Jan 2, 2008

### graves248

A distant galaxy is moving away from us at a speed of 1.85x10^7m/s. Starting from the formula for the doppler shifted frequency, derive a formula for the change in wavelength and hence calculate the fractional redshift (λ'-λ₀)/λ₀ in the light from the galaxy.

I'm pretty confused at the moment as i've seen the formula i need written in two different ways:

f₀ =f $$\sqrt{}[(1+v/c)/(1-v/c)]$$

and

f₀ =f $$\sqrt{}[(c+v)/(c-v)]$$

Are these two different forms of the same equation, as I can't seen to derive one from the other?

I suppose then the relationship I need to use is v = f λ, but im confused about which values of velocity, wavelength etc this equation relies on in this instance.

Sorry if i've missed something obvious as I don't think this question should be too hard, but I just can't think where to start.
Thanks for any help

2. Jan 2, 2008

### Rainbow Child

It's the same formula. Multiply numerator and denominator inside the sqrt with $c$

3. Jan 2, 2008

### will.c

you're correct with the relationship between frequency and wavelength- now write your doppler shift formula in terms of the wavelength, and substitute the expression you get (solve for lambda in terms of lambda-naught) into the fractional redshift formula. it is actually a pretty clean result. Plug the given velocity in at the end, and voila! :)

4. Jan 3, 2008

### graves248

Cheers for the help.. this is what i did:

λ₀ = λ $$\sqrt{}[(c-v)/(c+v)]$$

and because

z = (λ₀ - λ)$$/$$λ

z = $$\sqrt{}[(c-v)/(c+v)]$$ - 1
But when i sub in my value for v, this gives me a negative redshift. Wouldn't this mean that the object was in fact moving towards me?

Have I done something wrong here? think i may have confused some lambdas for lambda noughts

5. Jan 3, 2008

### Rainbow Child

$f_i$ the initial frequency of the light emitted, thus $\lambda_i=\frac{c}{f_i}$

$f_o$ the frequency of the light we observe, thus $\lambda_o=\frac{c}{f_o}$

Then
$$f_o=\sqrt{\frac{c-v}{c+v}}\,f_i\Rightarrow \lambda_o=\sqrt{\frac{c+v}{c-v}}\,\lambda_i$$
thus the redshift

$$z=-1+\frac{\lambda_o}{\lambda_i}\Rightarrow z=-1+\sqrt{\frac{c+v}{c-v}}>0$$

6. Jan 9, 2008

### stevebd1

I'm currently looking at how the distance to stars is calculated using redshift and I've realised there are 4 types of redshift that need to be taken into account-

Doppler effect (which applies to < z =0.1)

$$r (distance) = \frac{v}{H} = \frac{zc}{H}$$

if z = 0.03

$$r (distance) = \frac{v}{H} = \frac{zc}{H} = \frac{0.03c}{71} = \frac{8993.775}{71} = 126.673 Mpc$$

Relativistic Doppler effect (which applies to > z =0.1)

equation as above but special relativity applies due to the motion of source close to the speed of light. z needs to be adjusted using the following equation-

$$z = \sqrt{\frac{1 + v/c}{1 - v/c}} -1$$

which provides the following equation-

$$v/c = \frac{(z + 1)^{2} - 1}{(z + 1)^{2} + 1}$$

therefore if z = 0.1

$$v/c = \frac{(0.1 + 1)^{2} - 1}{(0.1 + 1)^{2} + 1} = 0.095c$$

$$r (distance) = \frac{v}{H} = \frac{zc}{H} = \frac{0.095c}{71} = \frac{28480.2875}{71} = 401.13 Mpc$$

Expansion of Space (cosmological redshift)

The result of the stretching of space-time (as postulated by general relativity) and not by radial motion which occurs with the doppler effect. This may make objects appear redder than they should and would give a false sense of speed/ distance.

$$1 + z = \frac{^{a}now}{^{a}then}$$

Gravitational redshift

Only applies to black holes, neutron stars and possibly white dwarfs

Looking at the equations above, I'm under the impression that the equation for the relativistic doppler effect takes care of the cosmological redshift or is there another level of mathematics after applying the relativistic doppler equation in order to account for the cosmological redshift?

Steve

Last edited: Jan 9, 2008
7. Aug 27, 2010

### Nubi

Thank you for this posting but may i please ask for the algebra steps from:

Z=(sqr(1+v/c)/(1-v/c))-1 to,

v/c=((z+1)^2-1)/((z+1)^2+1)

8. Aug 27, 2010

### presbyope

Add 1 to both sides .. etc.

9. Aug 31, 2010

### Nubi

Yes "add one" and then I square which makes the equation:

(Z+1)^2=(1+v/c)/(1-v/c)

but how do I then rewrite it?

10. Sep 1, 2010

### presbyope

Multiply it out, group the v/c terms together and solve. You're almost there.

11. Sep 9, 2010

### Nubi

Pleace, "explicit math" would help..
If you know it:-D