# Reaching Alpha Centauri in human lifetime

1. Jun 30, 2010

### JamesCallahan

Hello,

I'm new on this forum and I hope I will blend in nicely. I happen to have a couple of questions on the Alpha Centauri star in the constellation of Centaurus. I know the distance between this star and our milk way is huge, but suppose one could travel at the speed of light X 25, would it be feasible to reach the star or one of its planets within a human lifetime.

James Callahan, University of Georgia, Department of Psychology

2. Jun 30, 2010

### phyzguy

(1) Welcome!
(2) You say "the distance between this star and our Milky Way...". I think you are confused. The Milky Way galaxy is our galaxy, and is a collection of perhaps 100 billion stars. Both our sun, Alpha Centauri, and a whole bunch of other stars are members of the Milky Way galaxy.
(3) The distance to Alpha Centauri is about 4.2 light years, a huge distance in Earth terms, but very close in astronomical terms. This means that light takes 4.2 years to travel here from Alpha Centauri. So if you had a spaceship capable of going 1/2 the speed of light (physically possible, but far beyond our capabilities today), you could get there in about 8 years.
(4) As you get closer to the speed of light, your subjective time would slow down, so you could experience a time much less than 4.2 years if you got close enough to the speed of light. Look up "relativistic time dilation".
(5) It's impossible to go 25X the speed of light, or even to attain the speed of light.

3. Jun 30, 2010

### Staff: Mentor

There's no need for speeds faster than light (which is an unphysical assumption). The distance to Alpha Centauri is only about 4.5 light years. If he could just travel at about 0.41 times the speed of light (with respect to earth), a space traveler could get there in 10 years (according to his clocks--a bit longer according to earth clocks).

Last edited: Jun 30, 2010
4. Jun 30, 2010

### litup

Of course we know you can't go faster than light, at least that is the paradigm as we know it now. But, assuming some miracle happened, and we figured out some way to go 25X faster than light, you would make it to AC in just a bit over 2 months.

Or looking at it the other way 'round, you could go 25 light years in one year, 50 light years in 2 years, 500 light years in 20 years, so at that velocity, you could theoretically have a trip of 500 light years there and back within one human lifespan, 40 years in this case. That would be 1000 light years total trip distance.

There are about 15,000 stars within 100 light years of earth, so you could visit any one of them there and back within 8 years max assuming you were actually going 25X c and there was no time dilation. So if you had 15,000 spaceships you could visit every one and the final report on them would be ready within 10 years (assuming experts would need to massage the data brought back).

We could theoretically get so close to c (speed of light) that the people on board would THINK they are going faster than the speed of light. Get close enough to c and it could be a time dilation of 25 to 1. One year of ship time would equal 25 years of earth time.

The closer you get to c, the higher that time dilation, so closer yet to c could put time dilation at 100 to 1, 1 year of ship time, you would be going what you think is 100 times c and if you went out 100 light years and back (to a star 50 light years away) you would age only 1 year but you would find yourself 100 years in Earth's future, all your relatives would be dead.

So at a time dilation factor or 25:1, going to Alpha Centauri and back even though it might take only a couple of months on board the craft, say you spent 5 years exploring AC, you both would age 5 years in real time but the 4 months coming and going from Earth to AC and back would put yourself in your future.

That 2 month one way journey would still take 4.3 years according to earth clocks so the total time you think you took (and biologically and every other way you would be correct, you aged less) would be 5 years and 4 months.

If you had an identical twin left behind on Earth, he would be 8.6 years older. The 5 years you spent on AC would have been at the same relative clock rates so you both aged 5 years but the time you spent so close to c it made a time dilation of 25:1, you would have aged only two months so your time spent according to your clocks would have been 5 years and 4 months.

So suppose you were both exactly 20 years old when you left on your common birthday, you would be 25 years and 4 months old when you get back. Your brother would be 33 years and 7 months old. That would be a noticeable difference in your ages.

It would be real too, suppose you were both taking courses from the same software online style, where AI would grade your work and so forth. Suppose you had to read the equivalent of 1 book per week for your course work.

Since you had only 5 years and 4 months to do your work, you would only have been able to have read a couple hundred books.

Your brother would have been way ahead of you at home, having had time to have read over 700 books and he would have his Phd while you were still only one fourth the way through. Assuming you had to read 700 books for the whole course, say some kind of world history or some such.

Last edited: Jun 30, 2010
5. Jun 30, 2010

### qraal

Hi James

Alpha Centauri is composed of three separate stars, A, B & Proxima. The Milky Way, our Galaxy, contains our Sun, the Alpha Centauri system and over 100 billion other stars in a vast system of stars that's over 100,000 light years wide. Alpha Centauri's stars are a mere 4.2-4.4 light years away (depending on which component we're talking about) and can be reached within a human lifetime without the need to go faster than light.

Perhaps you meant 1/25 of the speed of light? In that case the trip takes 105-110 years, which is a bit longer than an average human lifespan of 70-80 years.

A point your other responders neglect to mention is the need to accelerate to high speed in order to make relativistic effects usefully noticeable. For example at an acceleration of just 1 gee (equivalent to Earth's surface gravity) the trip to Alpha Centauri A or B takes 6.035 years from the point of view of people staying on Earth. On board ship the journey takes just 3.58 years. Increasing the acceleration decreases the trip-time perceived by the crew quite significantly. If the crew can be protected against a full 25 gee, then the trip takes ~0.368 years (19 weeks.) From the viewpoint of the people at home it's ~4.437 years.

An accelerometer measures, obviously enough, the acceleration, moment to moment. An integrating accelerometer produces a 'speed' (called the rapidity in relativistic terminology) that is the acceleration multiplied by the time spent accelerating, adding up all those small moments over the whole trip. According to the ship-board integrating accelerometer the top speed measured is 1.85 times lightspeed for the 1 gee trip to Alpha Centauri and 4.6 times lightspeed for the 25 gee trip. To accelerate to x25 lightspeed, according to the integrating accelerometer of the ship, the time required is 24.2 years ship-time at 1 gee and the ship travels ~16 billion light years doing so. At 25 gees that trip takes 0.97 years and the ship travels 1.4 billion light years. From a stationary observer the ship never exceeds the speed of light.

That's relativity for you!

6. Jul 1, 2010

### litup

Hi, I wonder if you have the formulae for those numbers? It seems to me 1 g would get you close to c in about a year, so how are you deriving those numbers? Thanks.

7. Jul 1, 2010

### qraal

Hi
You have to convert between rapidity and speed. So if the rapidity, u, is 1 c, which is reached after 30,570,323 seconds, ship time T, at 1 gee, then the speed, v, is tanh(u) - in units of c (lightspeed, c = 1.)
Tanh(u) is sinh(u)/cosh(u), where sinh(u) is used to get the stationary observer time, while cosh(u) is the Lorentz gamma-factor, what old "Blake's 7" fans would call TDF (time distortion factors.)

I'll go into more detail if you're interested still and undaunted by hyperbolic trig functions.

Qraal

8. Jul 1, 2010

### qraal

Another point, often unappreciated, is the fact that news of a ship's arrival arrives back at the starting point always at a speed less than, but arbitrarily close to, half lightspeed. Thus stay at home observers receive news from explorers at a minimum elapsed time of twice the target's distance (when that's measured in light years.) Thus a probe launched off at 0.5c has an effective data return speed of 1/3 c, but a much lower energy cost than a probe expensively fired off at 0.9c for example. How one determines the most economical speed depends on how the waiting time for data is costed too.

Last edited: Jul 1, 2010
9. Jul 1, 2010

### litup

I would love to get all the math involved. Can you be more rigorous in your equations?That is to say, just write them out without the words? like v=Tanh(u)- what? Oh, the - is not a minus sign? So in that case V=tanh (u). Is that the same as the fitzgerald contraction?
What is the difference between speed (you mean velocity?) and 'rapidity'? If you are going to accelerate to say 0.8c there is not much in the way of time dilation, I don't think, right?

10. Jul 2, 2010

### qraal

Hi litup

Gamma is computed by

1/sqrt(1-(v/c)^2)

Which is a significant 5/3 at 0.8c some 2/3 higher than the Newtonian case.

There are several good websites on these issues. I'm trying to put together the equations in LaTeX format for easier reading, since ASCII and ExCel renderings aren't very readable.

Last edited: Jul 2, 2010
11. Jul 2, 2010

### qraal

And...

cosh(u) gives the gamma factor, where u = g.T for constant acceleration, g being acceleration measured by a shipboard accelerometer and T the acceleration time measured by the ship's clocks.

12. Jul 2, 2010

### litup

Are the trig functions you mentioned in degrees, grads or radians? My guess is radians. I am having a bit of trouble getting into radian mode with my simple casio, so I guess I will drag out my rusty but trusty HP48G:) You don't really need it for this simple stuff but its fun to dig it up every now and then.

13. Jul 2, 2010

### qraal

You guessed it - radians. Degrees are relative values not 'absolute' like radians.

But here's a secret... Hyperbolic trig functions are actually exponential functions. Look them up on Wikipedia for their full definitions.

14. Jul 2, 2010

### qraal

Ok, here goes some TeX...

$$\gamma=\frac{1}{\sqrt{1-\beta^2}}$$

where $$\beta=\frac{v}{c}$$

Last edited: Jul 2, 2010
15. Jul 2, 2010

### qraal

Thus...

$$\gamma=cosh(\frac{g.\tau}{c})$$

...where g & $$\tau$$ are in $$m.s^{-2}$$ & s respectively.

Funny getting used to how Firefox, my browser, handles the LaTeX files. I have to reload fully each time I hit the Preview button.

16. Jul 2, 2010

### qraal

$$v=c.tanh(\frac{g.\tau}{c})$$

$$x=\frac{c^2}{g}(cosh(\frac{g.\tau}{c})-1)$$

$$t=\frac{c}{g}sinh(\frac{g.\tau}{c})$$

17. Jul 2, 2010

### qraal

Those are the one way equations and need slight modification to be used for trips in which the vehicle accelerates then deccelerates to its destination. You can probably guess how to modify them.

18. Jul 2, 2010

### qraal

$$x=\frac{c^2}{g}(\sqrt{1+(\frac{g.t}{c})^2}-1)$$

$$v=\frac{g.t}{\sqrt{1+(\frac{g.t}{c})^2}}$$

$$a=\frac{dv}{dt}=\frac{g}{\sqrt[3/2]{1+(\frac{g.t}{c})^2}}$$

Last edited: Jul 2, 2010
19. Jul 2, 2010

### qraal

those equations produce the distance travelled, the speed of the vehicle and the acceleration seen by a 'stationary observer'. Acceleration in this case is the instantaneous rate of change of speed as seen by the observer in the same frame as the rest frame of the ship's home or destination. Of course everything is really in relative motion and across intergalactic distances the complication of general relativity arises, but 'stationary observer' is usually understood to mean the observer to whom the ship appears to be accelerating, but its point of origin does not i.e. in the same frame as the origin.

20. Jul 2, 2010

### qraal

And, in case you didn't figure it out...

$$cosh(\frac{g.\tau}{c}) = \sqrt{1-(\frac{g.t}{c})^2} = \gamma = \frac{1}{\sqrt{1-\beta^2}}$$