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Question about length contraction and motion


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17 hours ago, Dagl1 said:

if they aren't at rest (both moving along line A-B at the same speed so that the distance between them does not change),

Maybe this was already explained better but I didn't see it. An inertial object is only moving relative to something else (ie. an observer). If they're inertial and their distance isn't changing, then they're relatively at rest. If you then say that they're moving at the same speed but still at rest relative to each other, then that's relative to another frame, which you haven't mentioned.

In non-relativistic physics, you might assume you're talking about some universal frame, but probably no one else here would do that; if you only mention frames A and B and then a speed, I think everyone would assume you mean A's speed relative to B and vice versa. You would have to specify a third frame ("Earth frame" for example) for people to get what you mean.


Also, an observer is a frame of reference in SR. https://en.wikipedia.org/wiki/Observer_(special_relativity) This is because you observe ie. measure the same distances and times no matter where you are in a given inertial frame, so you don't have to distinguish different viewpoints in the same frame as different observers.


17 hours ago, Dagl1 said:

If I move towards a star at 80% the speed of light, and I measure the distance to be 6LY, then if I moved very slowly towards it it would have been 10LY? And measured from the star, I too am 6 LY away when moving at at 0.8c, right?

The first part sounds right........ if the star is moving towards you at 80% of the speed of light and is 6 LY away, then you accelerate instantly so it is at rest, it should now be 10 LY away (not "would have". It still is 6 LY as measured in your first frame, and is 10 LY in your second frame).

The second sentence is kinda wrong and this is where it gets fun! You could ignore this until you get the rest of the replies in the thread. Because of relativity of simultaneity, you and the star don't measure the distance between you as 6 LY at the same time. For example if the star is 6 LY away, approaching at 0.8c, it will take 6LY/.8c = 7.5 years for it to arrive. But as observed by the star, your clock is ticking at a rate of 0.6x its own. So while you measure 7.5 years to reach the star (from the event where you measure the star being 6 LY away), the star measures 12.5 years in its own frame, during which you travel 12.5 y * 0.8c = 10 LY. The star measures you as 10 LY away when (according to you!) you measure the star being 6 LY away.

You also measure the star's clock ticking at .6x your own. You and the star are symmetric: each of you measure the other as 6 LY away at the moment (in your own frame) that the other measures being 10 LY away.

Or another way to see it is: at the moment (according to you) that the star passes the 6 LY mark on your rulers (which are at rest relative to you), the star's rulers are length-contracted by a factor of 0.6, and you are passing the 10 LY mark on its ruler. At the event where you say "we're 6 LY away", the star measures you at the mark that's 10 LY away.

... But then, you might also see, if you're at the 10 LY mark on the star's ruler and it's your ruler that's length-contracted according to the star, then when you're at the star's 10 LY mark, the star is at a 16.666 (repeating of course) LY mark on your ruler! According to you, the event of you passing the star's 10 LY mark is simultaneous with the event of the star passing your ruler's 0.6 LY mark. According to the star, the event of you passing its 10 LY mark, is simultaneous with it passing your 16.666 LY mark. This is no problem because the relative simultaneity of distant events is different for different frames of reference.

There are a lot of ways to describe this, I edited it to try to simplify, others probably have clearer and simpler ways to say it.

Edited by md65536
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