cosmology - Does an object's redshift actually decrease with time?

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?

Answer

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.