# Redshift - How can Z be greater than 1?

I'm having trouble understanding the equation for redshift: z = Δλ/λ ≈ Δf/f ≈ v/c

If z = v/c and c = speed of light, how can z>1 (as nothing can exceed the speed of light)?

• In what context do you expect z > 1? – The Photon May 22 '18 at 20:22
• Also notice that some of the "equalities" use $\approx$ instead of $=$. In particular $\Delta \lambda/\lambda \ne \Delta f/f$ unless $\Delta\lambda \ll \lambda$. – The Photon May 22 '18 at 20:23
• I read that 'the dwarf galaxy MACS0647-JD has a red shift of about z=11' – Umber May 22 '18 at 20:24

The formula $z \simeq v/c$ is only approximately true when $v \ll c$. Redshifts greater than 1 are possible if the redshift is caused by relativistic motion or by cosmological expansion.

The cosmological redshift is not a Doppler shift and should not be interpreted as such except perhaps at very small redshifts. It is caused by the expansion of space between the time when the light is emitted and when it is received by an observer. This expansion could be interpreted as a recession speed at small redshifts, but as you have surmised, that interpretation runs into trouble when redshifts become greater than 1. It is the expansion of space that allows things to apparently recede at greater than the speed of light. Your statement that "nothing can exceed the speed of light" is more nuanced in General Relativity and has received many questions and answers in these pages.

A redshift larger than 1 is also possible when relativistic motion is applied to a Doppler shift. The correct formula is $$z = \sqrt{\frac{c+v}{c-v}} -1,$$ which can become arbitrarily large as $v \rightarrow c$.

If $v \ll c$, then the above expression can be approximated by $$z = (1 + v/2c +...)(1 + v/2c -...) -1 \simeq v/c$$

$\frac{\Delta\lambda}{\lambda}\approx \frac{\Delta f}{f}$ is only true when $\Delta\lambda \ll \lambda$.

So it isn't true when $z\ge 1$.

• Can you give the correct formula if you think it isn't true? Redshifts greater than 1 are of course possible for relativistically moving objects or in cosmology. – Rob Jeffries May 23 '18 at 6:20
• @RobJeffries, $\Delta f = c/\lambda_2 - c/\lambda_1$. It doesn't work out to any special simple form involving just $f$ and $\Delta f$ as far as I can see (but I don't expect it to, due the nonlinear relationship) – The Photon May 23 '18 at 15:33
• Indeed, but I just don't see how this answers the question asked about how $z$ can exceed 1. Still, the OP has accepted it. – Rob Jeffries May 23 '18 at 16:31
• @RobJeffries, say $\lambda_1$ (the wavelength in the source's frame) is 1 cm and $\lambda_2$ (wavelength in the observer's frame) is 10 cm. (This applies to doppler shifts, I don't know enough about cosmic expansion to comment in that context) – The Photon May 23 '18 at 16:37
• Obviously if $\Delta \lambda > \lambda$ then $z>1$. – Rob Jeffries May 23 '18 at 16:40

$\Delta \lambda$ can be very large, much larger than $\lambda$, because the wavelength can be quite stretched out: a 21cm line earlier in the universe can be many meters now.