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I was trying to understand why we cannot explain the observed redshift of distant galaxies using special relativity and I came upon this article by Davis and Lineweaver.

Unfortunately when I arrive at section 4.2, where the authors explain why we cannot use special relativity to explain the observed redshift, I get stuck. In particular I don't understand this sentence:

"We calculate D(z) special relativistically by assuming the velocity in $v = HD$ is related to redshift via Eq. 2, so...".

What bothers me is the assumption that velocity is related to distance linearly. I was thinking that in a special relativistic model the basic assumptions were:

1)Relativistic Doppler shift formula $$ 1+z=\sqrt{\frac{1+v/c}{1-v/c}} $$ 2)Observed Hubble law $$ z=\frac{H}{c} d $$

Combining this two i get the following relation between velocity and distance $$ \sqrt{\frac{1+v/c}{1-v/c}}-1=\frac{H}{c} d $$ and not the one proposed in the article.

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  • $\begingroup$ You need to clarify your question to indicate what it is about this paper you don't understand. It seems pretty clear to me that Davis & Lineweaver are showing that a particular model doesn't work. Neither does $z \propto d$, as the briefest of glances at the type Ia supernovae data shows. $\endgroup$
    – ProfRob
    Commented Oct 11, 2015 at 7:08

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The Hubble parameter is defined to be $\dot{a}(t)/a(t)$, where $a$ is the scale factor of the universe. If you wished to have a model where redshifts were not due to expansion, but actually just due to things moving away from us (and this is what Davis & Lineweaver are doing in the section of paper you refer to), then you could assume that $H = v/d$ is an equivalent statement.

Then assuming that the redshift is only due to a velocity, then special relativity tells us that the redshift $z$ is given by $$ (1 + z)^2 = \frac{1 + v/c}{1 - v/c}$$ which can be rearranged to give eqn 2 in the reference you quote. $$ v = c \frac{(1+z)^2 -1}{(1+z)^2 +1}$$ Inserting $v=Hd$ gives $$ d = \frac{c}{H} \frac{(1+z)^2 -1}{(1+z)^2 +1}$$

The equation relating redshift and distance under the general relativistic universal expansion model is quite different to the relationship between redshift and distance in special relativity. The difference becomes apparent at high redshift, as explained in section 4.2 of the Davis & Lineweaver paper. Observations of course show that the relationship between distance and redshift is not the one derived above, which therefore favours the universal expansion interpretation of redshift.

You can of course always hypothesise some ad hoc relationship between $H$ and $d$ (or equivalently $H$ and $t$) to make a model to match the data. I think Davis & Lineweaver's aim was merely to show that the flattening of the $z$ vs $d$ relation cannot just be due to the non-linearity of the $z$ vs $v$ relationship in special relativity.

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  • $\begingroup$ I'm not so sure that we can mesaure the relationship $v=H d$, in fact we can mesaure redshifts not velocities. $\endgroup$
    – Cervantes
    Commented Oct 10, 2015 at 21:11
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    $\begingroup$ @Cervantes !! The whole point of the discussion in the article was to examine the hypothesis that the redshift is due to a velocity and that the velocity and redshift are related by the first equation above. In fact they are not related by that formula (see eqn.1 in the Lineweaver reference), which only approximately works at low redshift. $\endgroup$
    – ProfRob
    Commented Oct 10, 2015 at 21:20
  • $\begingroup$ If I can try to put @Cervantes's objection in my own words, perhaps the extrapolation of $v = Hd$ from low to high (0.8 in fig. 5 of the paper) redshifts isn't what we should be comparing to. Instead, perhaps we should have used $z = Hd/c$ (which agrees with the data just as well at very low redshift) and extrapolated this to higher redshift in an SR way, then let the velocities we derive from redshift obey whatever monotonic-but-not-simple law they have to to make this work. $\endgroup$
    – user10851
    Commented Oct 10, 2015 at 21:30
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    $\begingroup$ @Cervantes Yes. Redshift due to velocity and cosmological redshift are not distinguishable. But see here for a discussion of problems for anything but an expansion model. physics.stackexchange.com/a/186417 $\endgroup$
    – ProfRob
    Commented Oct 11, 2015 at 16:02
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    $\begingroup$ @Cervantes Another thing I just realised. If one adopted a $v=Hd$ model in flat space-time, the Hubble sphere would be at ~14 billion light years (for H=70 km/s per Mpc). Since we have not measured any source with such a distance, then that particular objection to the model is not really valid. $\endgroup$
    – ProfRob
    Commented Oct 12, 2015 at 11:24
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While $z=\frac{H}{c} d$ is more or less the relation that Hubble originally found, it doesn't hold out to arbitrarily large distances in FLRW cosmology. $v=HD$ does hold out to arbitrarily large distances, provided $v$ and $D$ are interpreted correctly as FLRW recessional velocity and FLRW spatial distance.

The real problem here is that the $v$ in $v=HD$ and the $v$ in $1+z=\sqrt{\frac{1+v/c}{1-v/c}}$ are not the same physical quantity. They shouldn't have used the same letter for them in a single paper, and they certainly shouldn't have punned on it by substituting one quantity into a formula that expects the other.

The $v$ in the special-relativistic formula is defined in terms of global inertial/Minkowski coordinates. In most FLRW cosmologies, you can't define global Minkowski coordinates because spacetime isn't flat, so you can't apply the special-relativistic formula because the $v$ in it is just meaningless.

However, in the zero-density ($\Omega=0$) limit of FLRW cosmology, spacetime is flat, you can define Minkowski coordinates on it, and the SR formula does work, out to arbitrary distances.

There are two zero-density FLRW cosmologies. One is boring: the scale factor $a$ is constant and $v_{\small\text{FLRW}} = v_{\small\text{SR}} = 0$ and both formulas give $z=0$. The other is much more interesting; it's called the Milne model and it describes a linearly expanding universe. In this case it turns out that $$v_{\small\text{SR}} / c = \tanh (v_{\small\text{FLRW}} / c)$$ which means that $v_{\small\text{FLRW}}$ is in SR terms the rapidity. If you plug this into the SR redshift formula you get, after a bit of manipulation, $$1+z = \exp(v_{\small\text{FLRW}} / c).$$

Meanwhile, in FLRW coordinates we have $a(t) = \dot at$ (for some constant $\dot a$) and, for objects moving with the Hubble flow, $$v_{\small\text{FLRW}} = a'(t)x = \dot ax$$ $$1+z = a(t_r)/a(t_e) = t_r/t_e$$ $$x = \int_{t_e}^{t_r} \frac{c\,\mathrm dt}{a(t)} = \frac{c}{\dot a} \int_{t_e}^{t_r} \frac{\mathrm dt}{t}$$ (using e and r subscripts for emission and reception respectively), and the integral of $1/t$ is $\ln t$, so $1+z = \exp(v_{\small\text{FLRW}} / c)$ as before.

The results agree because there is only one kind of redshift in general relativity, and the special-relativistic and cosmological formulas are special cases of it. Since they're different special cases, usually at most one of them is applicable to any given problem. But in the overlap of their zones of applicability, they're different coordinate descriptions of the same phenomenon, so they must agree.

I like Davis and Lineweaver's paper a lot, but they didn't catch every misconception about cosmology, not even all of their own misconceptions, and when they talk about intrinsic expansion of space they're just spreading another misconception. In reality there's no difference in GR between the relative motion of galactic superclusters and any other relative motion.

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  • $\begingroup$ Howdy. It seems to me, that we're on the same side of the fence. physics.meta.stackexchange.com/questions/14478 $\endgroup$
    – Marcin
    Commented Aug 8, 2023 at 6:28
  • $\begingroup$ @Marcin In the disputed question it seems like you're plugging the redshift of the CMBR into the SR redshift formula. The SR formula isn't valid in that context. There is just one kind of redshift and the SR formula is a special case of it, but it's the special case where $Ω=0$, while in real cosmology (ΛCDM), $Ω=1$. Length contraction and time dilation also don't really make sense when spacetime isn't flat. $\endgroup$
    – benrg
    Commented Aug 8, 2023 at 7:37
  • $\begingroup$ Can you explain how does $\Omega=1$ correspond to this plot in real cosmology? en.wikipedia.org/wiki/Expansion_of_the_universe#/media/… I totally reject your rejection of time dilation / length contraction in curved spacetime, because in the Planck scale spacetime is flat. That's the scale, on which I "measure" these properties in GR. $\endgroup$
    – Marcin
    Commented Aug 8, 2023 at 7:53
  • $\begingroup$ @Marcin That chart isn't well labeled. The red "acceleration" curve is the ΛCDM one and it also has $Ω=1$. In any case what matters is that $Ω$ is far from zero, not its exact value. The problem with t.d./l.c. is that (even in SR) you can get any value for them by picking a different local inertial frame. To get the "correct" value, you'd have to connect one of those local inertial frames to the motion of an object billions of (light) years away across curved spacetime, and that's the part that doesn't make sense. $\endgroup$
    – benrg
    Commented Aug 8, 2023 at 19:45
  • $\begingroup$ What values would you assign to the other curves, to correct the plot? across curved spacetime - that bothers me a lot and i'd be thankful if you explain - isn't spacetime flat on large scales due to roughly uniform density of matter/energy, even if we consider the density of dark energy? Does this curvature somehow reveals itself during the expansion? I think of the expansion as of an infinite series of snapshots - static, comoving frames, and in each and every one of them i regard the spacetime as flat. $\endgroup$
    – Marcin
    Commented Aug 8, 2023 at 20:13
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This is just the approximation that $\beta \equiv v/c \ll 1$.

Because, $\frac{1}{1-x} \approx 1 + x$

$$\left[ \frac{1+\beta}{1-\beta} \right]^{1/2} \approx \left[ (1 + \beta)^2 \right]^{1/2} = 1 + \beta$$

Thus, $\frac{v}{c} \approx \frac{H}{c}d$, and $$v \approx H\cdot d$$

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  • $\begingroup$ But in the $\beta\ll 1$ limit the SR and GR approach shouldn't be equivalent? $\endgroup$
    – Cervantes
    Commented Oct 10, 2015 at 21:15
  • $\begingroup$ @Cervantes why not? If you look at Fig.2 --- you'll see that the lines converge --- both the SR and GR predictions. The main reason (I think) for the difference between SR and GR predictions is due to acceleration being in the GR model. Acceleration is only important on larger scales --- where the difference becomes more apparent. $\endgroup$ Commented Oct 10, 2015 at 23:19
  • $\begingroup$ What i mean is that if we can make the assumption $\beta\ll 1$ both SR and GR should give the same results but we are interested in regimes where SR and GR gives different results so we can understand which interpretation is the correct one. $\endgroup$
    – Cervantes
    Commented Oct 10, 2015 at 23:52
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    $\begingroup$ @Cervantes First, it sounds like the question in your original post would be much better phrased as, "Why doesn't SR explain cosmological redshift". Second, the approximation that $\beta \ll 1$ only amounts to a 10% deviation at $\beta \approx 0.5$ --- which is far from enough to explain the discrepancy. I think you do bring up a good point, however, that a more rigorous demonstration would not have made this assumption. That would be a pretty easy calculation for you to do... $\endgroup$ Commented Oct 11, 2015 at 1:03
  • $\begingroup$ I don't think it would be so easy. What I should is: 1)Find a mathematical expression for the z-d relation observed even at hight redshift (cheating a little I could use the one given by GR but even in this case I still have some freedom in the choose of the cosmological model) 2)Find a v-d relation that explain this data (probably a very unlikely one) 3)Use this relation to repeat the calculation of Davis & Lineweaver $\endgroup$
    – Cervantes
    Commented Oct 11, 2015 at 9:16

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