I have been told that the peculiar velocity of an object in space is its velocity relative to the comoving frame. The formula is

$v_{pec}^i = \dot{a}(t)x^i$

where $x^i$ are the comoving coordinates and $a(t)$ is the scale factor that determines the expansion of the universe.

I am having trouble seeing exactly why this formula has this particular physical significance. To put it another way, suppose one is asked in a test to "find the velocity of an object relative to its comoving frame", how would one intuitively arrive at the above formula?

Finally, exactly why is the concept of a comoving frame useful to cosmologists? It is unlike any moving frame I have encountered before.

Any help is much appreciated.


1 Answer 1


I think you mistyped something in your formula. The one you have written down for the peculiar velocity does not have a dimension of velocity. Maybe you just forgot the dot. Let me explain how this is derived:

Usually one starts by introducing the comoving coordinates $x^i$ (coordinates that do by definition not change due to the expansion of the universe or any other behaviour of the scale factor $a$). These are related to the actual physical coordinates $x_{\mathrm{phy}}^i$ by $$x_{\mathrm{phy}}^i=a(t)x^i$$ where you have to note that you have to choose some normalisation of $a(t)$. For example choosing $a(t_0)=1$ means that the comoving coordinates are the physical coordinates NOW. Now you can calculate the derivative w.r.t to time and find two different velocities, since you have to apply the product rule: $$\frac{\partial}{\partial t}x_{\mathrm{phy}}^i=\dot{a}x^i+a\dot{x}^i$$ By multiplying $1=\frac{a}{a}$ into the first term and using the definition of physical coordinates and $H\equiv\frac{\dot{a}}{a}$, we arrive at $$\dot{x}^i_{\mathrm{phy}}=\frac{a}{a}\dot{a}x^i+a\dot{x}^i=Hx_{\mathrm{phy}}^i+a\dot{x}^i.$$ The first term is the velocity that is caused by the expansion, the so-called Hubble flow, while the second term is the peculiar velocity, i.e. $v_{\mathrm{pec}}=a\dot{x}^i$, which is caused by an actual change of the comoving coordinate (for example because of gravitational attraction due to a huge deviation of local energy density compared to averaged energy density, for example in the case of Andromeda and Milky-Way attracting each other).

It is useful to introduce comoving coordiates, because you investigate properties of components of the universe without always accounting for the change of $a$. For example if you want to understand structure formation, you can model this formation in comoving coordinates, and for example use a Fourier transformation to understand the behaviour of different comoving spatial modes, and only think about the ACTUAL size (at different times) of the structures in the end, when inserting $a$.

  • $\begingroup$ Thank you for your comments. I have fixed my mistake in the question. One point I would like to clear up is what, in simple terms, is meant by "energy density" and "huge derivation of local energy density compared to averaged energy density"? $\endgroup$
    – wrb98
    Apr 13, 2021 at 21:55
  • $\begingroup$ Maybe this was misleading in my answer, but you 'fixed' your answer the wrong way. For having the peculiar velocity, the dot has to be on the $x^i$. $\endgroup$
    – Koschi
    Apr 14, 2021 at 5:52
  • $\begingroup$ The metric, containing the scale factor $a$ describing the evolution of the whole universe, relies on the idea that on large scales, the universe is smooth and everywhere and in every direction the same. Therefore you 'take' all the content, which could be e.g. mostly non-relativistic matter, or mostly radiation, and average over it, which gives a certain energy amount per unit volume, which is called energy density. This energy density determines the behaviour of $a$. The actual deviations from this average were small in the early universe, but on small scales are huge now, as in galaxies. $\endgroup$
    – Koschi
    Apr 14, 2021 at 5:58
  • $\begingroup$ I just saw I originally wrote "derivation" instead of "deviation". This may have caused misunderstanding, sorry. $\endgroup$
    – Koschi
    Apr 14, 2021 at 6:00

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