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I wonder if someone can help me with the following problem.

I send a light beam to a distant galaxy which then bounces back to me.

I measure the travel time of the lightbeam using say a light clock of fixed size which I assume measures conformal time $\tau=\int dt/a(t)$.

Is the proper distance of the galaxy at the moment that I receive the lightbeam given simply by: $$L = \frac{c\ \tau}{2}?$$

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  • $\begingroup$ Are you sticking to special relativity, or do you want to consider general relativity? $\endgroup$ – John Rennie Jul 4 '14 at 9:45
  • $\begingroup$ I just want to use the flat FRW metric $ds^2 = -c^2dt^2 + a^2(t)(dx^2+dy^2+dz^2)$. $\endgroup$ – John Eastmond Jul 4 '14 at 9:54
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In Minkowski spacetime the one way light travel time to a galaxy at proper distance $\chi$ is just:

$$ t = \frac{\chi}{c} $$

so:

$$ \chi = ct $$

As you say. However in an FRW universe the travel time is given by a different equation so the proper distance is not simply $ct$. Let's assume all motion is in the $x$ direction, so the metric simplifies to:

$$ c^2ds^2 = -c^2dt^2 + a^2(t)dx^2 \tag{1} $$

We'll take our position to be $(0, 0)$ and the galaxy to be at $(0, \chi)$, and we'll adopt the usual convention that $a = 1$ at the current time. To get the proper distance we integrate $ds$, and since $dt = 0$ and $a = 1$ the proper distance is just:

$$ \Delta s = \int_0^\chi dx = \chi $$

Now let's calculate the time it takes the light beam to get from the galaxy back to us (i.e. one half of the journey). Light travels on a null geodesic so $ds = 0$ and putting this into the metric (1) and rearranging we get:

$$ \frac{dx}{dt} = \frac{c}{a(t)} $$

If the universe is static $a(t) = 1$ for all $t$, and we get $x = ct$ so you would be correct that the proper distance is equal to half the total travel time times $c$. But then with $a = 1$ we just have Minkowski spacetime so that's hardly surprising. To calculate the trajectory of the light we need to assume a form for $a(t)$ so let's make the approximation:

$$ a(t) = 1 + Ht $$

where $H$ is the current value of the Hubble constant. Then we get:

$$ \frac{dx}{dt} = \frac{c}{1 + Ht} $$

and this integrates to give us:

$$ \chi = \frac{c}{H} \log(H\tau + 1) $$

or rearranging this to get the travel time:

$$ \tau = \frac{exp(\frac{\chi H}{c}) -1}{H} \tag{2} $$

So the proper distance $\chi$ is not simply the travel time times $c$.

Just to reassure ourselves that we get the correct result in the limit of $H \rightarrow 0$, i.e. Minkowski spacetime, note that for small $H$:

$$ exp(\frac{\chi H}{c}) \approx 1 + \frac{\chi H}{c} $$

Put this back into equation (2) and we get:

$$ \tau \approx \frac{\chi}{c} $$

which is where we came in.

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  • $\begingroup$ did anyone ever measured the "the one way light travel time to" ... ? $\endgroup$ – Helder Velez Aug 27 '16 at 21:25

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