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If considering two galaxies $\mathcal{A}$ and $\mathcal{B}$ being separated by $d$ ($d$ being the comoving distance) and light is emmited from $\mathcal{A}$ at time $t_1$ and that light is received to galaxy $\mathcal{B}$ at time $t_2$ at which the distance between $\mathcal{A}$ and $\mathcal{B}$ grows to $a(t_2)\, d$ where $a(t)$ is the scale factor. Hence $c \,(t_2-t_1)$ (the distance travelled by the light in between $t_2$ and $t_1$) is equal to $a(t_2) \, d$ hence the comoving distance $d$ is $c \, (t_2-t_1)/a(t_2)$. But it's actually integral of $c \, dt/a(t)$ from $t_1$ to $t_2$. Will be someone please help me in sort out this discrepancy ??

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    $\begingroup$ Welcome to Physics! Please do not post formulae as plain text, but use MathJax instead. $\endgroup$ – ACuriousMind Jan 1 at 18:51
  • $\begingroup$ Don't confuse light traveling distance, proper distance and comoving distance. They're three distinct notions of distance in cosmology. $\endgroup$ – Cham Jan 1 at 19:08
  • $\begingroup$ The difference between proper and comoving distance I know ,but how is light travelling distance distinct from these two?? Please suggest $\endgroup$ – Apashanka Das Jan 1 at 19:35
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You can think of it like this: the distance traveled by a photon emitted at $\mathcal{B}$ is an integrated result of the full expansion history until $\mathcal{A}$ receives it. That is, as the photon travels, the scale factor changes and the distance the photon must travel is constantly affected.

But at small enough times your solution is correct. So if $t_1 = t_2 - {\rm d}t$ for a really tiny ${\rm d}t$, then the distance traveled by the photon according to your analysis is

$$ {\rm d}x = \frac{c(t_2 - t_1)}{a(t_2)} = \frac{c(t_2 - t_2 + {\rm d}t)}{a(t_2)} = \frac{c{\rm d}t}{a(t_2)} $$

Now you just need to add all these small changes up

and the result is

$$ d = \int_{t_1}^{t_2} \frac{c{\rm d}t}{a(t)} $$

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  • $\begingroup$ But sir what will be the integration limit on the LHS?? $\endgroup$ – Apashanka Das Jan 1 at 19:41
  • $\begingroup$ @ApashankaDas I realize you used another symbol, it should be $\int_0^{d}{\rm d}x = d$ $\endgroup$ – caverac Jan 1 at 19:43
  • $\begingroup$ But limit of LHS will be from x at t_1 to x at t_2,how is x defined here?! $\endgroup$ – Apashanka Das Jan 1 at 19:59
  • $\begingroup$ @ApashankaDas I will be from $x = 0$ at $t = t_1$ to $x = d$ at $t = t_2$ $\endgroup$ – caverac Jan 1 at 20:00
  • $\begingroup$ Ok that's fine but what is actually x defined here?? $\endgroup$ – Apashanka Das Jan 1 at 20:02
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I think they are equal and there's no discrepancy. But the problem is, you should consider $a(t)=C$, where $C$ is just a constant, to say that the distance traveled by light is $c(t_2-t_1)$. That's the crucial point.

Lets try this idea. At one hand you find $d=c(t_2-t_1)/a(t)$ notice that I write a(t).

For FLRW metric we can write $$ds^2=-c^2dt^2+a^2(t)[dr^2+S^2_k(r)d^2\Omega]$$ As you find it yourself for light $ds=0$ and $d\Omega=0$ so we are left with

$$c^2dt^2=a^2(t)dr^2$$

and

$$dr=cdt/a(t)$$ taking integral we have $$\int_0^d dr =\int _{t_1}^{t_2}cdt/a(t)$$Now, if we assume that a(t)=Constant then the equation just becomes, $$d =(t_2-t_1)c/a(t)$$

Which thats what we wanted to find.

We can take a(t) constant for small distances or for short period of times. If it wasn't constant then you cannot say $d=c(t_2-t_1)/a(t)$. But you can only say that proper distance is $$d=\int _{t_1}^{t_2}cdt/a(t)$$

Note: Since we assumed $a(t)$ is constant you can set it to the value of $a(t_{emission})$,if you want to.

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