Undergrad Understanding the property of the Comoving Coordinate

[Arman777]

In Weinberg's Cosmology, the comoving coordinate described as "A particle at rest in these coordinates will, therefore, stay at rest, so these are co-moving coordinates"

Now when we write the proper distance

$s = a(t)\chi$ where $\chi$ is the comoving coordinate.

Taking the time derivative

$$v =\frac{ds}{dt} = \dot{a(t)}\chi + a(t)\dot{\chi}$$

Here $V_H = \dot{a(t)}\chi$ which is due to Hubble flow and $v_p= a(t)\dot{\chi}$ is called the peculiar velocity. So according to the above description (the italic sentence), a particle will not stay at rest in these coordinates. In this sense, the comoving coordinates are not perfect (?)

To avoid this I guess we are assuming large comoving distances so that $V_H \gg v_p$, so that we can ignore the peculiar velocity?

 

[PAllen]

If a particle is at rest in the comoving coordinates, $\chi$ is constant by definition. Thus its derivative is zero, and there is no peculiar velocity.

 

[Arman777]

But observationally there's always peculiar velocity? I mean yes in the definition of the comoving coordinate we set $\dot{\chi} = 0$ but in the definition of the peculiar velocity we don't ($\dot{\chi} \ne 0$) . Is not this a contradictory. And I reccomended this solution.
"To avoid this I guess we are assuming large comoving distances so that $V_H \gg v_p$, so that we can ignore the peculiar velocity? "

More likely we are assuming $\chi \gg \dot{\chi}$ ?

 

[Orodruin]

But observationally there's always peculiar velocity? I mean yes in the definition of the comoving coordinate we set $\dot{\chi} = 0$ but in the definition of the peculiar velocity we don't ($\dot{\chi} \ne 0$) . Is not this a contradictory.

Observationally of what? A comoving observer by definition has zero peculiar velocity. If you observe something that has non-zero peculiar velocity it is just not a comoving object.

More likely we are assuming $\chi \gg \dot{\chi}$ ?

This makes no sense, $\chi$ and $\dot\chi$ have different dimensions so you cannot compare them.

 

[Arman777]

Observationally of what?

Peculiar velocity. There's a article about it https://arxiv.org/abs/1405.0105

This makes no sense, χχ\chi and ˙χχ˙\dot\chi have different dimensions so you cannot compare them.

Yes I should have said $V_H \gg v_p$ I guess

If you observe something that has non-zero peculiar velocity it is just not a comoving object.

Yes that's kind of my point. If every object has peculiar velocity then every object is not a comoving object. But we are making an approximation so that they become comoving, right ?

 

[Orodruin]

Peculiar velocity. There's a article about it https://arxiv.org/abs/1405.0105

I think you are misunderstanding: A comoving observer by definition has zero peculiar velocity. That does not mean that any particular observer must be a comoving observer or that galaxies need to be comoving. There is nothing here that can be observationally contradicted by seeing a galaxy that is not comoving.

If every object has peculiar velocity then every object is not a comoving object. But we are making an approximation so that they become comoving, right ?

No. It may be a good approximation for some objects and a bad approximation for some. In general there is going to be some distribution and that distribution will have some spread to it making more or less objects be close to comoving or not. For example, it is pretty clear that we are not comoving due to the dipole anisotropy in the CMB.

 

[Arman777]

Thanks, I understand it