# Flux in luminosity distance

I have gone through luminosity distance once again and found out that if $$L_s$$ is the energy emmited per unit time from the source ,then at any distance d the flux is $$F=\frac{L_s}{4πd^2}$$,for expanding universe d is the luminosity distance. And $$d^2=\frac{L_s}{4πF}....…....(1)$$ Now during expansion the flux recieved by the observer $$F=\frac{L_o}{4π(a_o\chi)^2} =\frac{L_s}{(1+z)^2(a_o\chi)^2}$$ $$\chi$$ being the comoving distance between source and observer for which $$d=(1+z)a_o\chi$$ Now my question is why this expression of flux is replaced in (1) ,are the two flux same ??and why are they??

• If someone please respond it would be beneficial to me .thank you – Apashanka Das Jan 31 at 16:05
• I want to help you out, but I did not understand your question – Reign Jan 31 at 16:11
• In d expression there is a flux F .my question is why is this flux taken equal to the received flux by an observer where the comoving distance between source and observer is $\chi$ and $L_s$ is the emmited energy per unit time by the source. – Apashanka Das Jan 31 at 16:12
• @Reign ,am I clear what I mean to say – Apashanka Das Jan 31 at 16:24
• F is always the observed flux ? And they are the same equation in the basis ? – Reign Jan 31 at 16:45

$$F(L_s,d)=\frac {L_s} {4\pi d^2}$$
So lets think an example of a distant galaxy and earth. This equation gives us the measured flux on earth and $$d$$ represents the distance between us. Now we can write this distance in terms of flux
$$d(F,L_s)=\sqrt{\frac {L_s} {4\pi F}}$$
Here again the flux is measured from earth and $$d$$ represents the distance between us and the galaxy. So the two flux are the same cause they are both measured on earth