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SpeedFreek
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I am pondering (or researching) and will bet back to you on this one vanDivX.. but I'm thinking that if length contraction is positive in your direction of travel, then length contraction would be negative in the opposite direction (i.e. it would make distances look larger). There doesn't need to be a specific term for that, the contraction would be either positive or negative. (This would be the same as <i>acceleration</i>, which in physics is referred to as acceleration whether you are speeding up or slowing down. What we might call deceleration is just negative acceleration, or acceleration in the opposite direction.) It would mirror what light does, as it is stretched (redshifted) when moving away from you, and is compressed (blueshifted) when moving towards you.<br /><br />We have to remember that when looking forwards from a relativistic ship, light-travel time or aberration actually make the distance in front look <i>larger</i>, and it is only when you subtract these factors from your on-board calculations that you could work out how much length has contracted in your direction of travel.<br /><br />This next bit is definitely correct (I have checked!) - if you travel 5 light years at 86.6% of light speed, the relativistic change factor is 2. A journey of 5 ly at 86.6% of c would look to an observer to take 5.77 years, just as you would expect. But on the ship, time runs half as fast and length contraction in your direction of travel causes the distance to your destination to look halved too. So you would travel 5 light years in 2.88 years, and your destination would (when you start, moving at 0.866c) look only 2.5 light years away, so there is no paradox in your observations from that ship.<br /><br />I am currently finding out what the view "out the back" would be, and will get back to you. <img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>