# A Roller Coaster Ride through Relativity/Appendix C

## Time on a distant star

How would you synchronise two clocks which are at rest but in different places?

One way to do it would be this:

• measure the distance d between the clocks
• set clock A to zero (but do not start it)
• set clock B to read time d/c (but again do not start it)
• start clock A and simultaneously send a pulse of light towards clock B
• when the pulse of light reaches B, start clock B
Synchronising two clocks

The two clocks will now be synchronised, at least to an observer at rest with respect to the two clocks. But what if I were to watch you synchronising a pair of clocks like this while you sailed by me (in the direction B to A) at a speed v? I would disagree about your calculation in two ways. Firstly the distance between the clocks would be contracted by a factor of γ. Secondly, while the light beam was travelling from A to B at what is to me a constant speed c, B would be racing towards the light beam at a speed v. To me, the light beam would be gaining on the clock at a relative speed of c + v so the time taken for the beam to get from A to B would be:

${\displaystyle T={{\sqrt {1-v^{2}/c^{2}}}~d \over c+v}}$

This can be rearranged and simplified as follows:

${\displaystyle T={l \over c}{{\sqrt {c^{2}-v^{2}}} \over c+v}}$
${\displaystyle T={l \over c}{\sqrt {(c+v)(c-v) \over (c+v)(c+v)}}}$
${\displaystyle T={l \over c}{\sqrt {c-v \over c+v}}}$

What this means is that while you set clock B to read l/c, I think you should have set it to the smaller time above. To put this another way, it seems to me that all your supposedly synchronised clocks ahead of you are running behind the true time and all those behind you are running ahead of the true time.

Now the difference in the two times is equal to:

${\displaystyle \Delta T={l \over c}-{l \over c}{\sqrt {c-v \over c+v}}}$

We can simplify this formula considerably if we assume that v is a lot smaller than c. First we rewrite the formula as follows:

${\displaystyle \Delta T={l \over c}(1-(1-v/c)^{1/2}(1+v/c)^{1/2})}$

hence using the Binomial Theorem we get:

${\displaystyle \Delta T={lv \over c^{2}}}$

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