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Hubble’s law states that the recessional velocity is proportional to the distance. The further the observed distance, the further back in time one is observing. I want to know (generally) how this time dependence of observed light sources at different distances is accounted for in the law? Thx

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From the non relativistic Doppler effect $$\lambda_o=\lambda_e(1+\frac{v}{c})$$$$\frac{\Delta \lambda}{\lambda_e}=\frac{v}{c}$$$$z=\frac{v}{c}$$ A plot of redshift($cz$) vs. the physical distance $r(t)(a(t)\chi)$ is giving a linear relationship(best fit curve) implying $$cz \sim r(t)$$$$cz=v(t)=H(t)r(t)$$

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  • $\begingroup$ Thanks for the explanation but it’s not really getting to the heart of my question or I just don't fully understand the solution. Here is some clarification. With respect to the law in would state that a galaxy twice as far away would be receding twice as fast, but what isn’t mentioned is that this more distant galaxy was moving twice as fast but also twice the time in the past. The observed red shifted photons from the two galaxies have traveled two different distances and originated from two different times in the past. $\endgroup$ – Tachyon_Storm Mar 20 at 16:19
  • $\begingroup$ Yes it is, two photons from two galaxies are emmitted say at $\chi _1$ and $\chi_2$ where $\chi_1\gt \chi_2$ (comoving distance) and if are received today $t_0$ then the physical distance travelled by then would be $a(t)\chi_1$ and $a(t)\chi_2$ where the emmision time would be $t_{e1}$ and $t_{e2}$ where $t_{e1} \lt t_{e2}$ $\endgroup$ – Apashanka Das Mar 21 at 10:56

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