# Why is the relation between luminosity distance $d_L$ and comoving distance $\chi$ $d_L=\chi/a$?

The textbook (Scott Dodelson, Modern Cosmology, Section 2.2 Distance, Page 35-36) states the following:

Another way of inferring distances in astronomy is to measure the flux from an object of known luminosity. Recall that (forgetting about expansion for the moment) the observed flux $$F$$ a distance d from a source of known luminosity $$L$$ is $$$$F=\frac{L}{4\pi d^2} \tag{2.47}$$$$ since the total luminosity through a spherical shell with area $$4\pi d^2$$ is constant. How does this result generalize to an expanding universe? Again it is simplest to work on the comoving grid, this time with the source centered at the origin. The flux we observe is $$$$F=\frac{L(\chi)}{4\pi \chi^2 (a)} \tag{2.48}$$$$ where $$L(\chi)$$ is the luminosity through a (comoving) spherical shell with radius $$\chi(a)$$. To further simplify, let's assume that the photons are all emitted with the same energy. Then $$L(\chi)$$ is this energy multiplied by the number of photons passing through a (comoving) spherical shell per unit time. In a fixed time interval, photons travel farther on the comoving grid at early times than at late times since the associated physical distance at early times is smaller. Therefore, the number of photons crossing a shell in the fixed time interval will be smaller today than at emission, smaller by a factor of $$a$$. Similarly, the energy of the photons will be smaller today than at emission, because of expansion. Therefore, the energy per unit time passing through a comoving shell a distance $$\chi(a)$$ (i.e., our distance) from the source will be a factor of $$a^2$$ smaller than the luminosity at the source. The flux we observe therefore will be $$$$F=\frac{La^2}{4\pi \chi^2 (a)} \tag{2.49}$$$$ where $$L$$ is the luminosity at the source. We can keep Eq. (2.47) in an expanding universe as long as we define the luminosity distance $$$$d_L\equiv\chi/a \tag{2.50}$$$$

The questions that bother me are:

1. According to Dodelson's statements, there should be $$L(\chi)\propto \frac{1}{a^2}$$. Why does $$L(\chi)$$ equal to $$La^2$$ in Eq. $$(2.49)$$?
2. To my understanding, the physical distance is the comoving distance multiplied by the scale factor, i.e. $$d=a\cdot \chi$$. But, Eq. $$(2.50)$$ violates it obviously. Why?

Therefore, the number of photons crossing a shell in the fixed time interval will be smaller today than at emission, smaller by a factor of a.

The intensity of light can be written as $$L=\frac {nh\nu} {\Delta t}=\frac {nhc} {\lambda\Delta t}$$

The above sentence implies that

$$\frac{\lambda_{observed}}{\lambda_{emission}}=\frac{1}{a}\,\,(1)$$

Similarly, the energy of the photons will be smaller today than at emission, because of expansion.

And this sentence implies that

$$\frac{\Delta t_{observed}}{\Delta t_{emission}}=\frac{1}{a}\,\,(2)$$

So let us write the luminosity at $$t=$$emission

$$L_{emission}=\frac {nhc} {\lambda_e\Delta t_e}$$

And by using (1) and (2) we can write

$$L_{emission}=\frac {nhc} {a^2\lambda_o\Delta t_o}$$

Since $$L_{observed}=\frac {nhc} {\lambda_o\Delta t_o}$$

We can write

$$L_ea^2=L_o\,\,(3)$$

Let us define $$F_o$$

$$F_o=\frac {L_o}{4\pi X^2}$$

By using (3) we can write

$$F_o=\frac {L_ea^2}{4\pi X^2}\,\,(4)$$

so we can define a new distance or luminosity distance by setting

$$\frac{a^2}{X^2}=\frac {1}{d_L^2}$$

or $$d_L=X/a \,\,(5)$$

We can keep Eq. (2.47) in an expanding universe as long as we define the luminosity distance

to satisfy this equation we need a such form for (5). Its just definitons.

If you look at the Eqn (4). You can rewrite it as,

$$F_o=\frac {L_ea^2}{4\pi a^2_0X^2}$$ for $$d_p=a_0X$$ but $$a_0=1$$ so it just becomes

$$F_o=\frac {L_ea^2}{4\pi X^2}$$

• The $\Delta t_O$ is the same as the $\Delta t_E$. The Hubble Expansion doesn't change the time interval. Your equation (2) is not true. Apr 11, 2020 at 17:13