# Perspective

It's all in how you look at it

## Relativity

A science post ...

I seem to get questions along the lines of "How long would it take to reach a speed of half the speed of light accelerating at 1 g?" or "How long would it take to reach a certain star system accelerating at 1 g?" fairly frequently, so maybe I should post about special relativity.

First of all, it probably isn't covered in your typical science class - it wasn't covered in mine - but most scientific calculators do have the functions you'd need to answer these questions. If you've had third semester calculus you can actually prove the formulas I'll be using, if not then you'll just have to take them on faith.

It turns out that both of those questions involve hyperbolic functions.

If you accelerate at a constant rate, your velocity after t seconds is given by the formula v = c tanh(at/c) where a is your acceleration and c is the speed of light. For the first question above, v = 0.5c and a = g, so

0.5 = tanh(gt/c)
tanh-1(0.5) = gt/c
c tanh-1(0.5)/g = t

Many calculators these days include a number of physical constants, in case yours doesn't g = 9.81 m/s2 and c = 299792458 m/s (though for some reason my calculator actually calls it c0 ... (yes, we use the metric system for science here in the US - though if you prefer I could give you values in ft/sec as well)

So ... the answer is 16792466 seconds. Divide by 86400 to convert to days, you get 194.357 days (somewhat over 6 months).

Note that t is your time, not time as measured by someone on earth ... due to time dilation the time as measured by someone on earth would be longer - how much longer depending on how fast you were actually going. The formula for that is

T = c sinh(at/c)/a

(I'll let you work it out, then post my answer later.)

The formula for distance is the third hyperbolic function, hyperbolic cosine (aka cosh) as follows

s = c2 (cosh(at/c) - 1)/a

If you're going to try to answer a question like "How long would it take to reach α-Centauri accelerating at 1 g", keep in mind - you probably want to stop when you get there. Unless you're planning on colliding with the nearest planet that would mean turning around half way there - so you actually want to figure out how long it would take you to go half the distance, then take the same amount of time slowing down.

It being as late as it is now, I'll leave it at that. Okay ... I'm sure you could have used Wikipedia as easily as I did, but if you want to know they list the distance to α-Centauri as

4.365 ± 0.007 ly (1.338 ± 0.002 pc)

Mind you, the formula is in meters and seconds (as long as a and c are in meters and seconds) so you'll still have to convert that.

Felixclaudeb Wednesday, July 20, 2011 4:37:55 AM

That's very useful! Somehow, I failed to find this information by searching on my own. And as a science-fiction fan, it kind of matters to me. Thank you.

Stevesgunhouse Wednesday, July 20, 2011 4:48:07 AM

As I said, I haven't seen it in textbooks, but it is easy enough to prove. (Though come to think of it, back when I was in school most calculators didn't have hyperbolic functions. You could write it out with exponentials and logs of course, but these days there's little reason to.)

Mind you, a craft that can accelerate at 1 g for a year really is science fiction at this point ... but maybe someday.

Felixclaudeb Wednesday, July 20, 2011 6:07:16 PM

Even accelerating at 1g for 3-4 days is sci-fi right now, and that would get you to 1% of lightspeed -- a respectable velocity.

Stevesgunhouse Wednesday, July 20, 2011 6:17:04 PM

Respectable, but not "relativistic".

Casséus Jean Prosper Juniortouchatout Monday, March 26, 2012 12:09:50 PM

Hello! I recently created an
account under the pseudonym
Sworgman. After invitations to
have friends I found that my
account has been frozen. My
account, however, does not
contain lewd pictures. What to
do?

Stevesgunhouse Monday, March 26, 2012 5:28:02 PM

Someone reported it in the Feedback forum (in French, hence I couldn't read all of what was said), when I went to check it out it had already been banned so I can't say exactly why it was.

If you think it shouldn't have been banned, post in the Feedback forum asking for it to be reinstated.

Unregistered user Wednesday, September 18, 2013 9:34:36 AM

Anonymous writes: A spaceship that accelerates and decelerates at 1g could reach many stars in 2-3 years, and another 2-3 years to get back to earth. But when you get back, thousands of years may have passed on earth. In other words, when you time travel, there's no going home.

Felixclaudeb Wednesday, September 18, 2013 3:21:39 PM

Actually, my dear anonymous commenter, if you traveled at 99% of lightspeed, time would only be compressed about 8 times. Quite a bit, but you wouldn't get back home one thousand years later unless you traveled 125 years in your own subjective time -- and then you'd be long dead yourself. Not exactly a practical way to time-travel, at least not very far.

Stevesgunhouse Wednesday, September 18, 2013 3:55:51 PM

It isn't that bad, really. The formulas (and values for the physical constants) are listed above ... and remember that you do want to turn around half way there.

If it took 194.36 days to reach 1/2 c, then after 388.7 days you'll reach 0.8 c. Time dilation at that speed is only 5/3 (your home experiences 5 seconds for every 3 of yours). And that was after just over a year, so two years total turning around in the middle will only get you less than 3 years at your home. If you accelerate for 777 days (just over 2 years) (and decelerate another 2 years) you get to ... (8/5)/(41/25) = 40/41 c, at which point gamma (time dilation) is SQRT(1681/81) = 41/9 = 4.5555... as you see, you're not getting to "thousands of years on your home planet" quite yet. Now, a 20 year trip would be a different matter, but 5 years out and 5 years back won't be bad.

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