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We've just started with relativity and I got a question regarding an exercise we got.

A spaceship passes by earth on its way to planet X, at the moment it passes by Anna is born on the spaceship. Can the spaceship travel fast enough so that it reaches planet X before Anna's first birthday(spaceships clock), and if so, how fast would it have to travel?

The answer says yes and it would have to travel very close to $c$.

What I don't get is how it can travel at less than $c$ and still travel a distance of 100 lightyears within one year in its own reference time.

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In the earth's frame of reference nothing can go 100 ly in less than 100 years. No surprise there.

In Anna's perspective nothing can go 100 ly in less than 100 years.

So what's the problem? How can Anna go 100ly in 1 year? The answer is that 100 Anna years is considerably different compared to 100 Earth years due to the fact that Anna and the earth have a high relative velocity, in this case the ratio specified by you is 100:1 due to time dilation. (the required velocity is 0.99995c)

So a year on earth is incidental to Anna only having aged half a week, and also that we on earth can definitively expect that Anna already travelled 1 ly of proper distance. After a century on earth, we would be dead but Anna would have reached planet X on her first birthday.

Apart from time dilation, you can also understand this problem using length contraction: that because Anna and the earth have a 100:1 Lorentz factor, 100 ly of proper distance perceived by Earth is only 1 ly perceived by Anna.

Neither of these perspectives are meaningless in any way, because it talks about the possibility for travel to nearby stars within one human lifespan. ALthough the technology to do it are possibly more than a human lifespans away. But who knows...

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Edit: I wrote this answer when I was tired and thought anna was born on the earth not the spaceship, and I don't plan on going back and changing it.

This is kind of a bad question the way you phrased it. Are you sure you're not just interpreting the question wrong?

First, remember that in the spaceship's frame the distance to planet X is length contracted, so it can get there (by its own clock) as fast as it wants. Just to get that out of the way.

You're right in that if we were talking about the Earth's frame of reference and the spaceship's clock, the answer would be no - however rapidly the spaceship travels, it cannot travel faster than c and so when Anna is one year old in the Earth's frame, the spaceship will always only be a bit less than one lightyear away.

However, if we're talking about the spaceship's frame of reference and the spaceship's clock ("spaceship's clock" alone is a bit ambiguous here. We also need to specify the frame of reference), the question changes. The distance between planet X and the spaceship undergoes length contraction so from its frame it gets there faster and faster as its speed increases. Likewise, time goes by more and more slowly on Earth as the spaceship's speed increases (or in spaceship frame of reference terms: as the spaceship stays still and the Earth moves faster and faster away). So those two effects mean that it can get there and as little time as you please has elapsed on Earth. From the spaceship's frame, it can get there well before Anna turns 1.

Is there a contradiction here? No, because the two locations are spacelike and causally separated and saying "x is happening at the same time as y" is relatively meaningless since they're so far away. It's technically correct but physically not too meaningful to say that "The spaceship arrived at the same time as Anna's first birthday in the spaceship's frame of reference", because as you noticed, in Anna's frame of reference the spaceship still has a long way to go (~99 lightyears) when she turns 1.

Try writing out the Lorentz transformation explicitly if it still baffles you.

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  • $\begingroup$ Thank you for the very well written answer. Though, I still wonder how the distance can undergo length contraction if the distance to planet X is 100 ly in the spaceships's frame of reference. I guess perhaps it boils down to a poorly phrased exercise. $\endgroup$ – user88747 Jan 27 '14 at 10:43

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