I'm slightly confused by an idea of redshift:

If we assume the distance between two points is given by:

$$d = R Δx$$

we can assume that

$$λ_0 = R_0 Δx$$

And when we receive the emission it will be:

$$λ = R Δx$$

So using the idea of redshift, we can deduce...

$$\frac{λ}{λ_0} = \frac{R}{R_0} $$

Now, here's where I'm confused. Assuming we detect the wavelength now from an emitted wavelength (some time ago). We are only getting a ratio between the scale factor then and now, in theory, wouldn't the wavelength be stretched by $R_0$ as well (at whatever time that is) when it was first emitted. So if we decide to use the spectral lines that we find on Earth, since they are the actual values, wouldn't it not match if we compared it to the wavelength initially emitted since it would have been stretched.

Another question is, Can we also measure the distance now? (without finding the distance at the time of emission, finding the speed (assume constant) till it reaches us and multiplying it by the time that has elapsed since then, and add the two and giving the distance now)


2 Answers 2


When calculating redshifts, we usually look for signature features in astronomical spectra, usually emission or absorption lines.

For example, the universe contains lots of hydrogen. From quantum mechanics, we know that hydrogen has many different energy states which are fixed. This means it can only emit photons with a particular set of wavelengths (these energy states are like a unique fingerprint for each element). So we know that hydrogen in the distant universe will emit photons with exactly the same wavelengths as we can measure in laboratories on Earth.

Here is a nice cartoon of the redshifting of spectral lines: redshifts You see that the pattern of lines stays the same, they are just shifted to redder (longer) wavelengths.

When light travels through the universe, the wavelengths of the photons are stretched as the universe expands, so the wavelength we measure on Earth $\lambda_{obs}$ will be larger than the original emitted wavelength $\lambda_{em}$ (and we generally know what $\lambda_{em}$ is because it will form part of this 'fingerprint'):

$$\frac{\lambda_{obs}}{\lambda_{em}} = 1 + z = \frac{R_0}{R} $$

The scale factor today $R_0 = 1$, so we can find the redshift $z$ and the scale factor of the universe when the light was emitted $R$.

  • $\begingroup$ Hi cnosam, thanks for your explanation. From what you said, we simply look the emitted and observed wavelengths simply as a comparison? right? $\endgroup$
    – user51515
    Commented Apr 28, 2016 at 5:57
  • $\begingroup$ Yes, we compare the wavelengths to find the scale factor of the universe when the light was emitted. $\endgroup$
    – cnosam
    Commented Apr 29, 2016 at 21:17

While cnosam's answer is completely correct, I don't know if it really solves your confusion. The key point is that, when a photon is emitted, it knows nothing about the current size of the Universe. It is emitted at a very specific wavelength given by quantum mechanics, not by cosmology. Traveling through expanding space subsequently increases its wavelength. It is the expansion that cause the redshift, and since prior to emission the photon did not exist, it hasn't experienced any redshift yet.

Thus, a photon emitted at rest wavelength $\lambda_0$ when the scale factor $a$ (what you call $R$, but I encourage you to start calling $a$ :) ) was $0.1$ and another photon emitted with $\lambda_0$ at $a=0.5$ are indistinguishible, and if they're observed when $a=0.2$ and $a=1$ (i.e. today), respectively, they will both have wavelength $2\lambda_0$.

If you lived in a crazy universe that were static for a billion years, then expanded by a factor of 2 in the fraction of a second, and then were static again, the photon would have its rest wavelength $\lambda_0$ for a billion years, then be stretched to $2\lambda_0$, and then stay at $2\lambda_0$.

  • $\begingroup$ Ok, thanks for the clarity. I'm just clarifying that it's simply a matter of ratios. The question is simply a random thought I had. $\endgroup$
    – user51515
    Commented Apr 28, 2016 at 6:35

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