Consider a galaxy with a red shift of 1. Use a standard candle measurement such as Saul Perlmutter's and you will determine the distance of the galaxy from us is around 10 billion light years. Using Hubble's Law and a Hubble constant of 71 km/second per megaparsec, we can determine that the galaxy is receding from us at a little over 0.8 times the speed of light. If we then calculate the red shift from the relativistic Doppler effect for this velocity, it is somewhat over 2. Why is there this difference? I could perhaps understand it if the red shift due to the relativistic Doppler effect was less than the measured red shift, because there could be other contributions, but how can it be greater?
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1$\begingroup$ What galaxy? Is this a real galaxy or a hypothetical one? $\endgroup$– WillOCommented Aug 15 at 21:56
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1$\begingroup$ Any galaxy that has a red shift of 1. $\endgroup$– John HobsonCommented Aug 15 at 23:51
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1$\begingroup$ See my answer to a related question here physics.stackexchange.com/a/819673/388464 Everything you need to know is in there. $\endgroup$– KDPCommented Aug 16 at 3:51
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4 Answers
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Measure the red shift of a galaxy at 1. Use a standard candle measurement and you will determine the distance of the galaxy from us at around 10 billion light years.
I get the predicted luminosity distance is equal to $\log(1+z)T_{now} =\log (2)\times 13.8 = $9.56 billion light years, so not much difference there. This equation comes from Ned Wright's cosmology website.
Using Hubble's Law we can determine that the galaxy is receding from us at a little over 0.8 times the speed of light.
Using an average figure for the current value of the Hubble constant of around 71 (km/s)/Mpc and a conversion factor of 1 Mpc $\approx$ 3,260,000 light years, the velocity of a galaxy 9.56 billion light years away works out as about 205,276 km/s. This equates to about 0.694c. A little under your figure.
Using an equation from my other answer that the recession velocity of the Hubble flow in terms of redshift (z) is equal to $v_r = \log(z+1) = \log(2) \approx 0.69c $. Again, not too far out.
If we then calculate the red shift from the relativistic Doppler effect for this velocity, it is somewhat over 2. Why is this?
Using the value of v=0.694c the calculated redshift from the relativistic Doppler effect is: $$z = \frac{1-v-\sqrt{1-v^2}}{v-1} = \frac{0.31-\sqrt{1-0.694^2}}{-0.31} \approx 1.3225$$
This is a fair amount under your estimate of z>2.
Personally, I think you are doing this the wrong way around. What we measure is the redshift and then we calculate the velocity. For a redshift of z=1, the recession velocity according to the relativistic Doppler shift equation is: $$v_r = \frac{z(z+2)}{z(z+2)+2}=\frac{3}{5} = 0.6c$$
It is well known that recession velocities using the expanding universe model and Hubble flow can appear to exceed the speed of light, so it does not make sense to do relativistic calculations using the expanding universe velocities. When done properly, it is hard to tell the relativistic model apart from the expanding universe model. The relativistic model predicts lower recession velocities for given redshifts, but predicts the exactly the same Doppler time dilation effect and very similar luminosity distances because the relativistic beaming effect greatly diminishes the apparent luminosity of receding objects which makes the objects appear to be further away than they really are. This is why galaxies that appear to be further away than the age of the universe would allow if recession velocities were constrained to be lower than the speed of light, can still be explained by the relativistic model.
I could perhaps understand it if the red shift due to the relativistic Doppler effect was less than the measured red shift
You have not really explained why you expect the relativistic prediction of redshift to be less than the redshift we observe, so I am not sure how to help you with that.
I used the relativistic formula ((1+v/c)/(1-v/c))^0.5 - 1 to calculate the relativistic red shift. Is this wrong?
If you solve (using the Wolfram Alpha Solver) $z = \sqrt{\frac{(1+v/c)}{(1-v/c)}} - 1$ for v/c you get:
$$\frac vc = \frac{(2 z + z^2)}{(2 + 2 z + z^2)}= \frac{z (2 + z)}{2 + z(2 + z)}$$
which is the equation I used. We should have got the same result.
This paper by Peacock gives a more refined equation for the redshift calculation that includes a term that allows for gravitational potential:
$$z = \left(1+\frac{\Delta \phi}{c^2}\right) \sqrt{\frac{(1+v/c)}{(1-v/c)}} - 1$$
where $\Delta \phi = −4\pi Gρr^2/6 $ is the difference in gravitational potential and $p$ is the average density of the enclosed volume. Using units such that c=1, this gives
$$ v = \frac{(z+1)^2 - (1+\Delta \phi)^2}{(z+1)^2 + (1+\Delta \phi)^2}$$
Wikpedia gives the critical density equation as $p = \frac{3H_0^2}{8\pi G}$ which yields a figure of $p= 9.47×10^{-27}$ when using a value of 71 for the Hubble constant. This gives a value of $\Delta \phi = −4\pi Gρr^2/6 \approx -1.21 × 10^{-16}$ for the difference in potential energy when assuming a distance of r =9.56 light years. Plugging this value into the velocity equation gives:
$$ v = \frac{4 - (1-1.21E-16)^2}{4 + (1-1.21E-16)^2} = 0.60000000000000007c$$ for the recession velocity using the relativistic calculations that take gravitational potential into account. It can be seen that at this distance the gravitational potential makes very little difference.
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$\begingroup$ I used the relativistic formula ((1+v/c)/(1-v/c))^0.5 - 1 to calculate the relativistic red shift. Is this wrong? $\endgroup$ Commented Aug 16 at 8:01
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$\begingroup$ I wouldn't expect the relativistic red shift to be greater than the red shift we observe. $\endgroup$ Commented Aug 16 at 8:08
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$\begingroup$ The luminosity distance for something at $z=1$ is more than 20 billion light years. The comoving distance, which is what goes into Hubble's law is about 11 billion light years. $\endgroup$– ProfRobCommented Aug 16 at 8:23
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$\begingroup$ Sorry, I was referring to the luminosity distance. $\endgroup$ Commented Aug 17 at 20:40
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I think I know the answer to this question but it will be very controversial.
If there is any large scale curvature to the universe, velocities of recession will not give rise to any Doppler effect.
This is because the velocity of expansion will be perpendicular to all the possible directions in which a photon can travel. Velocity, of sources or receivers, which is perpendicular to the direction of the travelling signal, does not give rise to any Doppler effect.
We can see this if we consider sound signals travelling around the surface of a balloon. If the balloon is expanding, at any point on the surface of the balloon, the velocity of the expansion will be perpendicular to any of the directions in which the sound can travel. There will be no Doppler effect for sound signals travelling around a balloon, due to expansion of the balloon.
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When evaluating Dopper shifts in a curved spacetime, you need to be careful because relative velocities are not uniquely defined in curved spacetimes. Relative velocities are angles in spacetime. If you draw two arrows at different positions on a curved surface, there is no unique way to define the angle between them.
In particular, the cosmological recession rate $Hd$ is not the correct relative velocity. Worse, it's not even a relative velocity in the first place! (It doesn't correspond to an angle in spacetime.) But even on conceptual grounds, $Hd$ says something about the motion between the observer today and the light source also today. How could the light carry information about what the source is doing long after the light was emitted?
The correct approach, if you want to use the Doppler formula, is to parallel transport the 4-velocity of the source along the path that the light takes in spacetime. This will give you the same result as the standard approach in cosmology of saying that the redshift is the ratio of scale factors between receipt and emission times.
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Hubble's law relates the scale factor to the rate of change of scale factor. Both of those quantities are defined at the same cosmic epoch and neither are directly measurable (except approximately for relatively low redshifts).
For example, if you want to use the value of the Hubble parameter now, this relates the proper distance (the distance measured by a string of observers between you and an object with a set of rulers) to the rate of change of that distance now.
The redshift is indirectly related to these quantities via a cosmological model. One can readily understand that, because when the light was emitted from the distant galaxy it was at a smaller distance than now and the rate of change of that distance also differs from that in the present universe. In particular, we need to know how the Hubble parameter changed in the past and this is related to the densities of matter, radiation and dark energy.
The bottom line is that the analysis you have presented is invalid. The relationship between cosmological redshift and the rate of change of proper distance is not given by the Special Relativistic Doppler effect formula.
