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For positive matter and vacuum fractions with ${\Omega }_m+{\Omega }_{\Lambda }=1$, redshift drift vanishes when $s^2={\Omega }_m{s}^3+{\Omega }_{\Lambda }$, $s=1+z$. Factoring out the trivial observer root $s=1$ leaves ${\Omega }_m{s}^2-{\Omega }_{\Lambda }s-{\Omega }_{\Lambda }=0$, giving the displayed positive root of a polynomial. It lies at a positive redshift exactly when ${\Omega }_m<2/3$. It is independent of the Hubble constant at fixed density fractions and is not the redshift of the acceleration-to-deceleration transition.

At the present epoch, define $s=1+z$ and $F(z)=\dot{z}/H_0$. For flat matter plus a cosmological constant, neglecting radiation,In an Einstein-de Sitter universe, $E=s^{3/2}$, henceThe redshift drift is zero at $z=0$, has initial slope $F^{\prime }(0)=-1/2$, and decreases monotonically because $F^{\prime }=1-3\sqrt{s}/2<0$. Its second derivative is $-3/(4\sqrt{s})<0$, so the sketch bends downward, with $F\sim -z^{3/2}$ at large positive redshift.

A pure positive cosmological constant in a flat universe gives constant $H=H_0$, the expanding de Sitter model. ThusThe redshift drift is a straight line through the origin with positive slope. Here the source and observer expansion rates are equal; the factor $1+z$ in the observer contribution makes every positive-redshift comoving source drift toward still larger redshift.

For the specified flat matter-vacuum model,Its small-redshift behaviour is $F(z)=0.55z+O(z^2)$, since $q_0={\Omega }_{m,0}/2-{\Omega }_{\Lambda ,0}=-0.55$. At high redshift, matter dominates and $F\sim -\sqrt{0.3}{s}^{3/2}$, so the initially positive redshift drift must turn negative. To locate the crossing, square $s=\sqrt{0.3s^3+0.7}$; both sides are positive, so squaring introduces no positive-$s$ spurious root. Factorization givesBesides $z=0$, the physical nonzero redshift-drift root in flat matter-Lambda cosmology isThe redshift drift is positive for $0<z<z_{\ast }$ and negative for $z>z_{\ast }$. Its maximum satisfies $1=0.45s^2/\sqrt{0.3s^3+0.7}$ and lies near $z=0.949$, where $F\simeq 0.240$. This zero is distinct from the acceleration transition at $z_{\mathrm{acc}}=(2{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}{)}^{1/3}-1\simeq 0.671$: redshift drift compares expansion at two epochs, while acceleration is local to one epoch.

Use the observed wavelength in $d\lambda /\lambda =dv/c$. Since ${\lambda }_{\mathrm{obs}}=(1+z){\lambda }_{\mathrm{rest}}$,For an Einstein-de Sitter universe at $z=3$, $H(z)=H_0(1+z{)}^{3/2}=8H_0$, and the redshift drift is $\dot{z}=4H_0-8H_0=-4H_0$. With the rounded numerical values supplied in the paper,Therefore the spectroscopic velocity drift accumulated over ten years isThe negative sign means a shift toward shorter observed wavelength. Omitting the factor $1+z$ would overestimate the apparent velocity shift by four.

For fixed cosmological density parameters, the redshift drift has the form $\dot{z}=H_{0}F(z)$. Multiplying the Hubble constant changes the vertical scale of all three sketches but leaves their zeros unchanged. Thus the drift-zero redshift is independent of $H_0$ at fixed density fractions. Every model has the trivial $z=0$ zero; only the specified mixed model has a further positive zero, $z_{\ast }\simeq 2.089$. If instead physical matter and vacuum densities are held fixed while $H_0$ changes, their density fractions change too, so that is a different family of models.

Follow successive photons from the same comoving source. Their comoving radial distance is fixed, soIt follows that $d{t}_e/d{t}_0=a(t_e)/a(t_0)=1/(1+z)$. Taking the logarithmic derivative of $1+z=a(t_0)/a(t_e)$, with both endpoints allowed to vary, givesTherefore the observer-time redshift drift isHolding the emission time fixed would omit the second term and would not compare successive observations of the same source.

For successive wavecrests from the same comoving source, $1+z=a(t_0)/a(t_s)$. ThereforeThe same radial null path implies equality of the elapsed conformal time,Expanding both scale factors to first order gives the redshift drift