I am trying to determine how an object's redshift (specifically, redshift due just to the expansion of the universe) changes in time. Starting with a definition of the Hubble parameter,

$$H \equiv \frac{\dot a}{a}$$

with $a$ being the scale factor, we can write

$$\dot a = Ha~.$$

We can calculate $\dot z$ in terms of $\dot a$. Since $a=(1+z)^{-1}$,

$$\dot a = -(1+z)^{-2}\dot z~.$$

Plugging $a$ and $\dot a$ into the first or second equation I wrote here we can find

$$\dot z = - H(1+z)~.$$

This negative sign is a bit surprising to me. I would have expected that $\dot z$ would have been positive, i.e., that an object's redshift increases with time. I would have expected this from the fact alone that the universe is expanding, but perhaps I am wrong in this thought. If so, please tell me how. However, the expansion of the universe is currently accelerating, and so I would expect from this as well that $\dot z$ would be positive, since at later times things will appear to be moving away from us at a faster rate than they are now. Is there some sort of cosmological constant dependence I did not take into account in my derivation above?

My question in summary: why is there a negative sign in the equation for $\dot z$? Did I derive the expression incorrectly? Or am I wrong in thinking it should be positive?


4 Answers 4


The redshift of a source actually changes in a more complicated way: when the source entered our cosmological horizon (i.e. at the moment its light reached Earth for the first time), its redshift was $\infty$, because it was located at the edge of our observable universe. Over time, this redshift then decreases to a minimum value, but eventually the expansion of the universe causes it to increase again. In the far future, all sources will be redshifted back to $\infty$ (in the Standard $\Lambda\text{CDM}$ Model).

Let's derive the correct formula. For more details, I refer to this post: https://physics.stackexchange.com/a/63780/24142

The Hubble parameter in the $\Lambda\text{CDM}$ Model is $$ H(a) = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0},a^{-2} + \Omega_{\Lambda,0}}\;, $$ with $\Omega_{K,0} = 1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0}$.

The observed redshift $z_\text{ob}=z(t_\text{ob})$ of a source at a time $t_\text{ob}$ is given by $$ 1 + z_\text{ob} = \frac{a_\text{ob}}{a_\text{em}}, $$ with $a_\text{ob} = a(t_\text{ob})$ the scale factor at the time of observation, and $a_\text{em} = a(t_\text{em})$ the scale factor at the time $t_\text{em}$, when the source emitted the light that was observed at $t_\text{ob}$. From this, we can write $a_\text{em}$ as a function of $z_\text{ob}$ and $a_\text{ob}$: $$ a_\text{em} = \frac{a_\text{ob}}{1 + z_\text{ob}}.\tag{1} $$ When the source moves with the Hubble flow, its co-moving distance remains constant: $$ D_\text{c}(z(t_\text{ob}),t_\text{ob}) = c\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{a^2H(a)} = \text{const}. $$ Therefore, if we treat $t_\text{ob}$ as a variable, the total derivative with respect to $t_\text{ob}$ is zero: $$ \dot{D}_\text{c} = \frac{\text{d} D_\text{c}}{\text{d} t_\text{ob}} = 0, $$ which means that, with Leibniz' integral rule, $$ \frac{\dot{a}_\text{ob}}{a_\text{ob}^2H(a_\text{ob})} = \frac{\dot{a}_\text{em}}{a_\text{em}^2H(a_\text{em})}. $$ or, with $H(a_\text{ob})= \dot{a}_\text{ob}/a_\text{ob}$, $$ \dot{a}_\text{em} = \frac{a_\text{em}^2}{a_\text{ob}}H(a_\text{em}).\tag{2} $$ We also have from eq. (1): $$ \dot{a}_\text{em} = \frac{\dot{a}_\text{ob}}{1 + z_\text{ob}} - \frac{a_\text{ob}\,\dot{z}_\text{ob}}{(1 + z_\text{ob})^2}. $$ Inserting this into eq. (2), we find $$ \dot{z}_\text{ob} = (1 + z_\text{ob})\frac{\dot{a}_\text{ob}}{a_\text{ob}} - \frac{a_\text{em}^2}{a^2_\text{ob}}(1 + z_\text{ob})^2H(a_\text{em}), $$ which simplifies to $$ \dot{z}_\text{ob} = (1 + z_\text{ob})H(a_\text{ob}) - H(a_\text{em}). $$ In particular, if we take the present day as the time of observation, we have $$ \dot{z} = (1+z)H_0 - H\!\left(\!\frac{1}{1+z}\!\right). $$ Since $H(a)$ decreases as a function of $a$, if follows that $\dot{z}_\text{ob} < 0$ if $z_\text{ob}$ is very large (and $a_\text{ob}$ is sufficiently small), and $\dot{z}_\text{ob} > 0$ if $z_\text{ob}$ is small or $a_\text{ob}$ is large.

This also means that there's a redshift at any time at which $\dot{z}_\text{ob} = 0$. Using the same values of the cosmological parameters as in my reference post, I find that this 'transition redshift' is currently $z=1.92$. In other words, the redshift of a galaxy with present-day redshift $z<1.92$ is increasing, while the redshift of a galaxy with $z>1.92$ is currently decreasing.

Also take a look at the diagram in my reference post: the dashed lines represent contours of constant $z_\text{ob}$ at a given time of observation; galaxies move vertically (dotted lines). You'll see the same thing: when a galaxy crosses the particle horizon, its redshift is $\infty$, after which it decreases, but in the (far) future it will increase again.

See also Eq. (11) in the paper Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Davis & Lineweaver.

  • $\begingroup$ Note that, in the question, $a_\mathrm{ob}\equiv 1$. $\endgroup$
    – Danu
    Oct 9, 2015 at 7:02

Note: In this post I'm analyzing the situation as the observer slides along the null path connecting to the source, i.e. the situation of receiving the same wavefront at different points along its light cone. If you want the evolution of one particular galaxy's redshift as we watch it over time, where the emission we catch at different times from the same location necessarily was emitted at different times along the galaxy's worldline, see Pulsar's answer.

It's all a matter of what time is being implied by those dots.

$z$ is the redshift we observe today of light emitted some time in the past. You're probably thinking that if the observer starts at the emission, then $z = 0$, and clearly $z$ increases as we move the observer forward in time.

However, $z$ as a function of $t$ is generally taken to mean the redshift observed at the fixed time $t_0$ today, as a function of the variable time $t$ when the light was emitted. As $t \to t_0$, the observed redshift should approach $0$ from the positive side, so $\dot{z}$ should be negative.

If you want to move the observer rather than the emitter, recall that the observed redshift obeys $$ 1 + z(t_1,t_2) = \frac{a_2(t_2)}{a_1(t_1)} $$ for light emitted at time $t_1$ with scale factor $a_1$ and observed at time $t_2$ with scale factor $a_2$. Then $$ \frac{\partial z}{\partial t_2} \bigg\vert_{t_1} = \dot{a}_2 \frac{\partial z}{\partial a_2} \bigg\vert_{a_1} = \frac{\dot{a}_2}{a_1}, $$ which is indeed positive in an expanding universe.

  • $\begingroup$ Sorry Chris, that's not entirely correct. If you only take the derivative of $a_2$, then you end up comparing the redshifts of different galaxies. If you want to know how the redshift of a single galaxy changes, you need to take the change in $a_1$ into account as well. See my post, and see also Eq. (11) in the Davis & Lineweaver paper, or page 13 in Weinberg's Cosmology. $\endgroup$
    – Pulsar
    Oct 8, 2015 at 19:35
  • $\begingroup$ I think we're just using different interpretations of the scenario, both of which are valid in their own way. I'm conceptually sliding the emitter and/or observer along the null path connecting them, whereas you're moving them along the Hubble flow. In other words, yes, my analysis doesn't track particular galaxies, but that's not the only way to interpret the question. Do you agree with my observation? $\endgroup$
    – user10851
    Oct 8, 2015 at 19:44
  • $\begingroup$ Well, the OP asks how an object's redshift changes with time, so I think he means a particular galaxy. I guess we have to ask him. Cosmology can be quite confusing :-) $\endgroup$
    – Pulsar
    Oct 8, 2015 at 19:50

Your derivation is correct, but you are getting confused in interpreting the result. To be clear, the fact that the expansion of the universe is accelerating is irrelevant to this problem because accelerating expansion has to do with $\ddot{a}$.

The fact that $\dot{z}<0$ might appear paradoxical because, as you say, an object's redshift increases with time. Let's break down this statement. Objects which are further away appear more redshifted due to the expansion of the universe. Why is this true? Well, because we are observing light that was emitted at a time when the scale factor of the universe was less than it is today, and the further away the light comes from, the further back in time it was emitted and the smaller the scale factor at that time.

Redshift is relative to the current scale factor of the universe. The full equation should read: \begin{equation} 1+z = \frac{a_0}{a}, \end{equation} where $a_0$ is the scale factor today, which we assume to be 1, and $a$ is the scale factor of some distant object whose redshift we want to know. It doesn't make much sense to talk about the time evolution of $a$ in this case, since the object whose light we are measuring existed at some point in the past when the scale factor of the universe was different than it is today. If we want to increase $a$, for a fixed expansion history, we have no choice but to move the object to a later time, when $a$ is larger, thereby moving the object to a lower redshift. This is the reason why you get a negative sign.

As you point out, in the future distant objects will be at higher redshifts than they are today. That is because the scale factor of the present $a_0$ will increase in the numerator of the above equation, forcing the redshift of objects at a fixed scale factor $a$ to increase as well.


As I understand it: As the object approach the horizon event, the delay to have news from it get higher and higher. When its apparent velocity is c (at horizon) you will never receive its light (BTW the redshift will be total: frequency=0 ). So there is an horizontal asymptot to 0 in the z(t) curve.

  • $\begingroup$ The question is asking about redshift due to the Hubble expansion of the Universe. $\endgroup$
    – Kyle Oman
    Oct 8, 2015 at 16:45
  • $\begingroup$ Yes. Concerning a given object that goes further and further due to expansion (or am I wrong ?). The event horizon I'm mentioning is the limit of visible Universe. $\endgroup$ Oct 8, 2015 at 16:50

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