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I have encountered two definitions of recession velocity that seem to refer to different concepts, but I am wondering whether the two are actually the same.

Definition 1

If we denote the scale factor, that measures the expansion of the universe, by $a=a(t)$, then we can express the physical velocity of an object in terms of comoving coordinates as follows, since $\vec{r}_{phys}=a\vec{r}$:

$$\dot{\vec{r}}_{phys}=\dot{a}\vec{r}+a\dot{\vec{r}}=\vec{v}_{rec}+\vec{v}_{pec}$$

where the first term is called the recession velocity and the second one is the peculiar velocity.

Definition 2

When doing the calculations that lead to the Hubble-Lemaître law, the recession velocity of a light source is considered to be:

$$v=cz$$

where $z$ is the redshift.

Are these two concepts different, or is there a way to obtain one expression from the other?

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1 Answer 1

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Let us take the FRW metric as

$$ds^2 = -c^2dt^2+R(t)^2[d\chi^2+S^2_k(\chi)d\Omega^2]$$

We can see that the radial distance $(d\Omega = 0)$ along a constant time-slice $(dt=0)$ is $$ds = Rd\chi$$ which by integrating we obtain $D = R\chi$.

If you differentiate it, by assuming $\dot{\chi}=0$, we obtain $v=HD$.

where $H=\frac{\dot{R}}{R}$

Now, consider a path of a photon $(ds=0)$ along a radial distance $(d\Omega=0)$. In this case FRW metric becomes $$cdt=R(t)d\chi$$ This equation shows that the he velocity of light is purely a peculiar velocity $c=Rd\chi/dt$.

To calculate the comoving coordinate between two points in the path of the photon, you need to leave $d\chi$ alone and integrate. In result you obtain $$\chi(\bar{z}) = \frac{c}{R_0H_0}\int_0^{\bar{z}}\frac{dz}{E(z)}$$

Now, in the above, we have found out that $v(z) = H_0D(z) \equiv H_0R_0\chi(z)$ where where $v(z)$ and $D(z)$ are the present day velocity and proper distance of a comoving galaxy at redshift $z$.

From here we obtain $$v(\bar{z}) = c\int_0^{\bar{z}}\frac{dz}{E(z)}$$

In very small $z$ values the integral becomes $v \approx cz$ but for large $z$ values it is given by the integral itself.

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  • $\begingroup$ What are you denoting by $E(z)$? I understand that $H(z)=R_0 H_0 E(z)$ but I am not sure. I suspect that because I have previously proven that the following equality holds: $$\chi(z)=c\int^{t_0}_{t_1}\dfrac{dt}{a(t)}=\int^z_0\dfrac{dz}{H(z)}$$ $\endgroup$ Jan 23, 2023 at 15:12
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    $\begingroup$ @WildFeather $H(z) = H_0E(z)$ $\endgroup$
    – seVenVo1d
    Jan 23, 2023 at 18:21
  • $\begingroup$ Thank you! May I ask how you arrive to this formula? $$\chi(\bar{z})=\dfrac{c}{R_0H_0}\int^{\bar{z}}_0\dfrac{dz}{E(z)}$$ The expression I deduced was the one in my first comment, so I don't understand where $R_0$ comes from in the denominator before the integral. Is it there because the upper limit of the integral is $\bar{z}$ instead of $z$? What is $\bar{z}$ in this case? $\endgroup$ Jan 24, 2023 at 10:36
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    $\begingroup$ @WildFeather I can try to explain but its kind of long. The $\bar{z}$ is just a dummy index to not confuse with the $z$. It's all about mathematics and has no cosmological importance. $\endgroup$
    – seVenVo1d
    Jan 24, 2023 at 15:52
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    $\begingroup$ @WildFeather In some books/articles people use $R(t)$ and some $a(t)$. It's about making the scale factor dimensionless or not...Please look at J. A. Peacock - Cosmological Physics, starting from chapter 3. You'll understand the idea more clearly.. $\endgroup$
    – seVenVo1d
    Jan 24, 2023 at 15:55

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