# Does traveling at relativistic speeds change distances?

**URL:** <https://boards.straightdope.com/t/does-traveling-at-relativistic-speeds-change-distances/616808>\
**Category:** Factual Questions\
**Created:** [March 26, 2012, 2:09am UTC](https://boards.straightdope.com/t/does-traveling-at-relativistic-speeds-change-distances/616808 "2012-03-26T02:09:52Z")\
**Posts on this page:** 4\
**Page:** 2

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**Author:** ![Bemused](https://avatars.discourse-cdn.com/v4/letter/b/46a35a/32.png) [@Bemused](https://boards.straightdope.com/u/Bemused)\
**Post date:** [March 27, 2012, 4:47pm UTC](https://boards.straightdope.com/t/does-traveling-at-relativistic-speeds-change-distances/616808/21 "2012-03-27T16:47:32Z")

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> [@naita](#):
>
> Try this. You’re interested in travelling 1 ly from Earth, as measured in Earth’s reference frame, but timed in your travelling reference frame. Your time _t_ at 0.707 c will be 0.707 _t[sub]0[/sub]_ (Earth time).
> 
> Your mixed reference frame speed will then be distance in Earth’s frame / time in yours: (0.707 c \* _t[sub]0[/sub]_)/(0.707 _t[sub]0[/sub]_) = c

So you are saying you solved it by inspection?

_shrug_ OK.

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**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\
**Post date:** [March 27, 2012, 5:35pm UTC](https://boards.straightdope.com/t/does-traveling-at-relativistic-speeds-change-distances/616808/22 "2012-03-27T17:35:21Z")

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> [@](#):
>
> And, since this is a 4th order equation, there are 4 roots. Only one of them will turn out to be the right one, and it should be obvious. This equation might have a doubled root. Not sure.

Yes, and those four roots are 1, 1, -1, and -1. The equation factors to (v^2-1)\*(v^2-1) = 0. This is not a difficult problem.

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**Author:** ![Bemused](https://avatars.discourse-cdn.com/v4/letter/b/46a35a/32.png) [@Bemused](https://boards.straightdope.com/u/Bemused)\
**Post date:** [March 27, 2012, 8:27pm UTC](https://boards.straightdope.com/t/does-traveling-at-relativistic-speeds-change-distances/616808/23 "2012-03-27T20:27:06Z")

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> [@Chronos](#):
>
> Yes, and those four roots are 1, 1, -1, and -1. The equation factors to (v^2-1)\*(v^2-1) = 0. This is not a difficult problem.

Yeah, that’s right. c = 1 light-year/year. I should have seen that.

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**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\
**Post date:** [March 27, 2012, 9:12pm UTC](https://boards.straightdope.com/t/does-traveling-at-relativistic-speeds-change-distances/616808/24 "2012-03-27T21:12:59Z")

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> [@Bemused](#):
>
> So you are saying you solved it by inspection?
> 
> _shrug_ OK.

No. I’m not saying that, I’m simply demonstrating that my solution was correct. The solution to which you responded “No, that’s not right.”

What I actually did was solve the equation a \* c \* t / (t \* sqrt(1 - ^(a\*c)^2/c^2)) = c

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