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??

  • $\begingroup$ If someone please respond it would be beneficial to me .thank you $\endgroup$ Jan 31, 2019 at 16:05
  • $\begingroup$ I want to help you out, but I did not understand your question $\endgroup$
    – seVenVo1d
    Jan 31, 2019 at 16:11
  • $\begingroup$ 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. $\endgroup$ Jan 31, 2019 at 16:12
  • $\begingroup$ @Reign ,am I clear what I mean to say $\endgroup$ Jan 31, 2019 at 16:24
  • $\begingroup$ F is always the observed flux ? And they are the same equation in the basis ? $\endgroup$
    – seVenVo1d
    Jan 31, 2019 at 16:45

1 Answer 1


Flux is a function of distance and luminosity

$$F(L_s,d)=\frac {L_s} {4\pi d^2}$$

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


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