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i have a question on which i've spent a fair amount of time thinking, but couldnt figure it out :

We observe (in rest frame) 2 events happens at the same time, one (Event1) happened on Earth and one (Event2) in the middle of milky way, which is 30000 light years apart. How fast should we move so that the Event2 happens 1 hour earlier than the Event 1 (and the way around)? And in which direction ?

Sorry for my English here, i tried, thanks for any help:D

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The answer follows straight from the Lorentz transformations.

Let the $x$-axis in the Earth's rest frame be along the line connecting events 1 and 2 as you describe, pointing from Event 1 to Event 2, and say there is a spaceship moving along this direction at relative velocity $v$. Let the coordinates of the 2 events be $(x_1, ct_0)$, $(x_2, ct_0)$ as seen in the Earth frame, and respectively $(x'_1, ct'_1)$, $(x'_2, ct'_2)$ as seen in the spaceship frame. We want to find $v$ so that $$ ct'_1 - ct'_2 = 1\;\text{lhr} $$ given that $$ x_2 - x_1 = 30000\;\text{lyr} $$ The Lorentz transformations tell us that $$ ct'_1 = \gamma(ct_0 - \beta x_1)\\ ct'_2 = \gamma(ct_0 - \beta x_2) $$ with $\beta = v/c$ and $\gamma = \sqrt{1 - \beta^2}$ as usual, so $$ ct'_1 - ct'_2 = - \beta\gamma (x_1 - x_2) $$ From this we get after some algebra $$ \gamma = \sqrt{1+\left(\frac{ct'_1 - ct'_2}{x_1 - x_2}\right)^2} $$ and $$ \beta = - \frac{1}{\gamma}\frac{ct'_1 - ct'_2}{x_1 - x_2} $$ Numerically, we have $$ - \frac{ct'_1 - ct'_2}{x_1 - x_2} = - \frac{1\;\text{lhr}}{1\;\text{lyr}} = - \frac{1}{24\;\times\;365} \approx -1.14\;\times\;10^{-4}\\ \gamma \approx 1 + 7.0\;\times\; 10^{-9} \approx 1\\ \beta \approx - \frac{ct'_1 - ct'_2}{x_1 - x_2} \approx - 1.14\;\times\;10^{-4}\\ v \approx - (1.14\;\times\;10^{-4}) \times (3\times 10^5 \text{km/s}) \approx -34\;\text{km/s} $$

So the spaceship should go at about $34$km/s in the negative direction of the $x$-axis, from Event 2 towards Event 1. Funny thing, the Earth's orbital velocity around the Sun is about $30$km/s.

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  • $\begingroup$ many thanks, i dont have time to review now, will reply to you later :D $\endgroup$ Dec 4 '15 at 12:07

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