# Time as a function of distance for constant a

1. May 8, 2007

### DaveC426913

I want to make a little calculator that will
- take a distance (such as, say, 20.7 light years) and,
- with a given acceleration (say, 1g) followed by deceleration,
spit out the time elapsed.

I don't know how to fold Lorentz' velocity function into the formula for t as a function of d and a.

2. May 8, 2007

3. May 8, 2007

### Meir Achuz

$$d=[\sqrt{1+a^2t^2}-1]/a].$$
Solve for t.

4. May 8, 2007

### JesseM

Do you want the time elapsed by clocks on the ship, or in the frame where the ship was at rest at the beginning and end of the journey?

5. May 9, 2007

### DaveC426913

Yes!

(Both)

And nertz to me for neglecting to say so up front...

6. May 9, 2007

### DaveC426913

As Jesse wisely asks, is that ship time? or Earth time?

7. May 9, 2007

### JesseM

Just look at the page robphy pointed you to, all the terms are defined, and the equations you need are there. They say:
So you want to find the equations on that page which give you t or T if you already know d and a. Those would be:

t = sqrt[(d/c)^2 + 2d/a]

and

T = (c/a) ch-1 [ad/c^2 + 1]

These equations are for continuous uniform acceleration, so if you want to accelerate for half the trip and then decelerate for the second half, that's equivalent to double the time it would take to accelerate uniformly for half the distance. So the equations would become:

t = 2 * sqrt[([d/2]/c)^2 + 2d/a]

and

T = 2 * (c/a) ch-1 [a(d/2)/c^2 + 1]

ch-1 is the inverse hyperbolic cosine function, acosh. Here is an online calculator which understands the function acosh(x), but if you want to program your own calculator, this page says that $$acosh(x) = log[x + \sqrt{x^2 - 1}]$$.

Last edited: May 10, 2007
8. May 9, 2007

### DaveC426913

OK, so I kind suck at advanced math, making me the world's worst physics geek.

So I made a connect-the-dots graph from the data they provided. See attached. Four whole data poiints... But it gets me a number within 10%.

So, a 20.7ly journey is actually two symmetrical 10.35ly journeys, and a 10.35ly journey at 1g shows 3.2 years elapsed shiptime.

Thus, the whole journey lasts 6.4 years shiptime, while by Earth-time the journey lasts just a little more than 21 years.

Cool.

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9. May 9, 2007

### DaveC426913

On an loosely-related note, that page clears up some things I was never quite sure about in relativistiic travel.

It's a staple of sci stories that have long space journeys where they return to an Earth that's a zillion years older than when they left.

I'd never quite gotten the idea of whether something weird is happening to Earth-time, or ship-time or both. As if, by virtue of their trip, Earth-time had gone into overdrive and they had "missed" all those years.

But I'm beginning to see that even the ship's occupants can be sure that Earth's clock is "the right one" (forgive me Einstein).

A spaceship is scheduled for Star X, which everyone agrees is 83ly away. The spaceship occupants, once they get up to speed, see that their measurement of the distance to Star X is not fixed - it is not a journey of 83 light years, it is a journey of merely 5 years. Yet when they get to Star X, they see that Sol is once again 83ly away.

So, even *they* agree that the journey should - and did - take 83 years from a PoV stationary to both Sol and Star X.

So I can lay to rest my confusion. A journey of 100ly takes ~100 years - by Sol time AND by Star X time. The ship occupants acknowledge that it was they who slowed to a crawl while 83 years elapsed normally.

Or looking at it the other way, when the Niven's Rammer did his loop around the galactic core and came back "two million years in Earth's future", he really did make a trip (and one that he could plot and measure on a galactic map) whose duration should have been - and was - 2 million years.

Last edited: May 9, 2007
10. May 10, 2007

### Meir Achuz

In the equation I gave, t is the Earth time.

11. May 10, 2007

### MeJennifer

It's all in the proper quantities

Good thing!

The proper distance, which is the actual distance traveled by a traveler, is always smaller than the distance as measured by a reflecting light beam. So in other words, if the distance between two planets is X then any traveler must record a distance smaller than X.
Gives an interesting spin on Zeno's paradox doesn't it?

Proper acceleration changes proper distance, which is the reason that travelers traveling between two events can record a different proper time.

Last edited: May 10, 2007
12. May 10, 2007

### JesseM

I got 6.1 years and 22.6 years...one thing to remember is that, as noted on the relativistic rocket page, if you're using units of years for time and light-years for distance, then 1G acceleration is 1.03 ly/y^2.

Last edited: May 10, 2007
13. May 10, 2007

### DaveC426913

Wow thanks! That first one works perfectly.

The second one I'm trying to massage. This is what I get:

n = a*(d/2)/c^2 + 1
m = Math.log(n+Math.sqrt(n*n-1))
Te = 2 * (c/a) * m

Unfortunately, that spits out a number in the millions.

Last edited: May 10, 2007
14. May 10, 2007

### JesseM

Hmm, try having your program print out n and m as well as Te to give you a better idea of where it's going wrong. If I use your example of d = 20.7, then with a=1.03 and c = 1 I get:

n = 11.6605
m = 3.1475 (BTW, make sure you're using the natural log function 'ln' rather than log base 10)
Te = 6.11

15. May 10, 2007

### DaveC426913

Math.log(x) returns the natural log (base E) of x

What units is c in? I set it to 300000.

a = 1.03
d = 20
c = 300000;
n = a*(d/2)/c^2 + 1
ch1 = Math.log(n+Math.sqrt(n*n-1))
Te = 2 * (c/a) * ch1
alert(" Te:" + Te + " n:"+n +" ch1:"+ ch1 );

yields
Te:1026843.0140033511 n:3 1.7627417

Last edited: May 10, 2007
16. May 11, 2007

### JesseM

You can use whatever units you want, you just have to be consistent. If you want 1G acceleration, the a = 1.03 figure assumes you're using units of years for time and light-years for distance, in which case c = 1. If you want to use c = 300000, that means you're using units of kilometers for distance and seconds for time, in which case 1G acceleration would mean a = 0.0098 kilometers/second^2.
This would mean a distance of 20 kilometers and an acceleration of 1.03 km/s^2 = 105G, which is not what you wanted. But even with these numbers, something seems to be going wrong with your program's math:
If n = a*(d/2)/c^2 + 1, then you can just plug this into your calculator to see a*(d/2)/c^2 = 1.03*(10)/(300000)^2 = 1.144 * 10^-10, so n should be very close to 1. Have you checked the programming language's rules for entering equations? Maybe you need to have [d/2] instead of (d/2) of something minor like that. Anyway, I'd suggest entering different simple numbers (like a=1, d=1, c=1) to try to figure out what equation it's actually calculating, or break the equation up into more intermediate variables (like x = d/2, y = a*x, z = y/c^2) which it can print out so you can get a better idea of just where it's going wrong by double-checking the results with a calculator.

Last edited: May 11, 2007
17. May 12, 2007

### DaveC426913

I wish to verify something:

This:
T = 2 * (c/a) ch-1 [a(d/2)/c^2 + 1]

is actually
T = 2 * (c/a) * ch-1[a(d/2)/c^2 + 1]

Right?

18. May 12, 2007

### DaveC426913

Nope.I can't reconcile these two statements:

T = 2 * (c/a) h(a(d/2)/c^2 + 1)
where h(x) = log(x+x^2-1)

19. May 13, 2007

### JesseM

That's right.
h(x) is supposed to be log(x + sqrt(x^2 - 1)). So just calculate the value of a(d/2)/c^2 + 1, and let that be x in the second equation. For example, if d = 20.7, a=1.03 and c = 1, then a(d/2)/c^2 + 1 = 1.03*10.35 + 1 = 10.6605 + 1 = 11.6605, so according to this calculator acosh(11.6605) = 3.14751047236, and log(11.6605 + sqrt(11.6605^2 - 1)) = log(11.6605 + 11.6175) = log(23.278) = 3.1475.

Last edited: May 13, 2007