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7. Consider a galaxy whose light we see today at time $t_0$ that was emitted at time $t_e$. Show that the present proper distance to the galaxy (along a curve of constant $t_0$) is $d = a(t_0) \int_{t_e}^{t_0} dt/a(t)$.

          7. Consider a galaxy whose light we see today at time $t_0$ that was emitted at time $t_e$.
Show that the present proper distance to the galaxy (along a curve of constant $t_0$) is
$d = a(t_0) \int_{t_e}^{t_0} dt/a(t)$.
        
7. Consider a galaxy whose light we see today at time t0 that was emitted at time te.
Show that the present proper distance to the galaxy (along a curve of constant t0) is
d = a(t0) ∫te^t0 dt/a(t).

Added by Timothy L.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider a galaxy whose light we see today at time t0 that was emitted at time t. Show that the present proper distance to the galaxy (along a curve of constant t0) is √(3n/p).
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Transcript

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00:01 Hi, in this question they're saying that a galaxy is observed to be receding from us at 3 .9 % of the speed of light.
00:09 So we have v given as 3 .9 % of c.
00:14 So it will be 0 .039 times c which is speed of light.
00:19 Further, they are saying to calculate the approximate distance to the galaxy in light now according to hubble's law, the speed v at a galaxy is observed to move away from the earth is given as v equals to h0 into d where the value of h0 is equals to 70 km per second per mpc is a hubble's constant.
00:50 So here the value of v will be 0 .039 into 10 3 into 10 raised to power 5 km per second which is equals to 1 .17 into 10 raised to power 4 km per second...
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