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I'm currently reading the book Cosmology by Daniel Baumann, and in Chapter 2, I encountered a claim that I was unable to prove. To provide some context to my question, let's start with the expression it provides for the physical velocity of a particle:

$$\vec{v}_{phys}=\dfrac{d\vec{r}_{phys}}{dt}=\dfrac{d}{dt}(a\vec{r})=\dot{a}\vec{r}+a\dot{\vec{r}}=\dfrac{\dot{a}}{a}a\vec{r}+a\dot{\vec{r}}=H\vec{r}_{phys}+\vec{v}_{pec}$$

where:

  • $a=a(t)$ is the scale factor, which measures the expansion of the universe.
  • $H=\dfrac{\dot{a}}{a}$ is the Hubble parameter.
  • $H\vec{r}_{phys}$ is the Hubble flow.
  • $\vec{v}_{pec}=a\dot{\vec{r}}$ is the peculiar velocity of the particle.

Later on, it asks the reader to prove that the physical three-momentum, defined as $p^2=g_{ij}P^iP^j$, verifies:

  • $p=\dfrac{mv}{\sqrt{1-(v/c)^2}}$
  • $p$ is proportional to $a^{-1}$

This is easy to do, using the information that the geodesic equation provides, but then the book states that:

Since $p\propto a^{-1}$, free-falling particles converge onto the Hubble flow.

Why? I understand this would mean that, for free-falling particles, $\vec{v}_{pec}$ tends to zero as time passes, and I also suppose that the so-called by the book "physical peculiar velocity" $v$ that appears in the expression of $p$ is equal to $\dot{\vec{r}}$. But I don't see how to conclude from this that $\vec{v}_{pec}$ tends to zero.

Edited to provide more details:

If we invert algebraically that expression of the three-momentum $p$ in terms of the velocity $v$, we obtain the following expression for $v$ in terms of $p$:

$$v=\dfrac{p}{\sqrt{m+(p/c)^2}}$$

If we consider most cosmological objects and massive particles to be non-relativistic, we can neglect the denominator and say that, since $p\propto a^{-1}$:

$$v\simeq\dfrac{p}{\sqrt{m}}\propto p\propto a^{-1}$$

But then, if we consider that $v^2=g_{ij}\dot{x}^i\dot{x}^{j}$ denotes the same velocity $v$ than $\dot{\vec{r}}$ in the expression of the physical velocity, this means that:

$v_{pec}=a\cdot v\propto a\cdot a^{-1}=1$

So, $v_{pec}$ would not decrease with the expansion of the universe, which is what I interpreted from the phrase "free-falling particles converge onto the Hubble flow". Where is my mistake here?

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3 Answers 3

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Here's an intuitive picture. An object's peculiar velocity is its velocity with respect to nearby comoving observers (moving with the Hubble flow). Something with a large peculiar velocity will depart from its initial neighborhood, though, moving into places where the comoving observers have different velocities! This is why the peculiar velocity evolves even in the apparent absence of forces.

In particular, our fast-moving particle will overtake slower-moving comoving observers to end up in the neighborhood of observers moving closer to its own velocity. This is why the peculiar velocity decays.

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  • $\begingroup$ Why does it need to be a nearby observer? I think I understand it but I am not sure: comoving observers move along with the Hubble flow, that is, they follow the expansion of the universe. If they are too far away from the object whose velocity they are measuring, the velocity that the object and the observer have due to the expansion of the universe will differ signficantly, causing important relativistic effects that distort the measure of the velocity of the object. Right? But mathematically, what I obtain is that the peculiar velocity does not decay with a. This is what confuses me. $\endgroup$ Jan 12 at 11:03
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    $\begingroup$ It's not even a relativistic effect, peculiar velocities would decay in Newtonian physics if you also picked an expanding coordinate system. Sorry, haven't looked at the math closely enough at this point to see what's wrong. $\endgroup$
    – Sten
    Jan 12 at 20:12
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The velocity $v_{\text{pec}} = a \, \dot{r}$ is the velocity as measured by the local stationary observer located at $r$, close to the moving particle. This velocity is restricted by causality: $v_{\text{pec}} \le 1$ (in units of $c \equiv 1$). The velocity $v_{\text{rec}} = \dot{a} \, r$ isn't bounded since $r$ can be as large as you want (depending on the spatial geometry considered). So when $r$ is large, you get $v_{\text{rec}} \gg v_{\text{pec}}$, which defines the "Hubble flow" at large distances. In this case, the peculiar motion isn't much relevant and you may set $v_{\text{pec}} \approx 0$.

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You can write the total velocity of any "close" galactic object as

$$v = \frac{\partial}{\partial t}(a\chi) = \dot{a}\chi + a\dot{\chi} := v_{rec} + v_{pec}$$

Here $v_{rec}$: recession velocity, $v_{pec}$: peculiar velocity, $\chi$: comoving coordinate

If you look at "close" galaxy, you'll see the combination of the two velocities and usually it's not easy to separate one from the other.

In cluster scales or below, we can say that $v_{pec} \sim v_{rec}$ (Such as Andromeda galaxy getting closer to our galaxy). These peculiar velocities are caused by the gravitational effects. Now, if you go to much larger scales, then the gravitational effects becomes less important and we can say that $v_{pec} \ll v_{rec}$. Which is why, and also as a definition, we set $\dot{\chi}=0$.

free-falling particles converge onto the Hubble flow.

It means that gravitational attraction between the large scale objects (such as super-clusters) becomes less and less important as they loose momentum and they start to drift along with the Hubble flow as the universe expands.

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  • $\begingroup$ I don't quite get it. What is $v_{rec}$ in your argument? I don't see how to take distance into consideration as you say, or how do $p$ and $v_{pec}$ relate to each other. $\endgroup$ Jan 8 at 17:21
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    $\begingroup$ @WildFeather I updates my answer $\endgroup$
    – seVenVo1d
    Jan 9 at 11:09
  • $\begingroup$ Thank you, I understand what you mean, but I still don't get where that information comes from, for example I don't see how to obtain that $v_{pec}\ll v_{rec}$ at very large scales. I updated my question in an attempt to explain better where my problem appears in the calculations. $\endgroup$ Jan 10 at 19:07

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