Article on High Speed Travel

1. Aug 11, 2012

schaefera

http://scienceblogs.com/startswithabang/2011/06/01/striving-for-the-speed-of-ligh/

In this article, the author claims that it would take a little over 1.25 years to reach 90% the speed of light if we accelerate at 1g.

However, if I let F=dp/dt, where p=γmv, and then integrate F with respect to t (assuming F=mg), I get that the velocity, v, will be reached at the time:

t=v/sqrt(g^2-g^2*t^2/c^2)... and plugging v=.9c into this, I get the result that it should take about 2 years and a day to reach the speed of .9c.

Whose calculation is wrong? If it is mine, where did I go wrong, and how can I correct it in my expression for time as a function of velocity?

2. Aug 12, 2012

ghwellsjr

According to The Relativistic Rocket article, it looks like you're both right. He is calculating the Proper Time, T, experienced by the traveler and you are calculating the Earth time in the non-accelerating frame.

3. Aug 12, 2012

schaefera

Thanks for the link-- it's great!

When I try to convert between measured time, t, and proper time, T, though I seem to run into a problem. I know that T=∫dT=∫(1/γ)dt. And since v=(gt)/sqrt(1+(gt/c)^2), I can plug this in and get a slightly nasty integral. After a lot of substitution, I end up wanting to integrate sec(u) du between 0 and arctan(gf/c) where f is the final time. I get something nasty with a natural log, but it comes out numerically to give me about 1.4 years which I'll say is where they rounded to get a year and a quarter. Thanks!

Last edited: Aug 12, 2012
4. Aug 12, 2012

schaefera

Using the proper time that the author of that article calculates, would that time be the actual time experience by a traveler, or merely the time an earth-bound observer would measure when using the proper equations to convert his time into the moving ship's frame?

For example, I calculate that the earth would measure 5.93 years for a trip to the nearest star while a traveler would measure 3.56 years. But does this exhibit the usual reciprocity of special relativistic calculations (eg traveler measures 5.93 and calculates 3.56 for the earth), or does the acceleration in his case negate the symmetry? Does a traveler really only age 3.56 years despite feeling like time runs normally for him?

5. Aug 12, 2012

Staff: Mentor

yes, because the amount of time that runs normally for him is 3.56 years. He sees 3.56 years tick by, by looking at his wristwatch, counting his heartbeats, watching paint dry... As long he's only looking at the local phenomena aboard his spaceship, where only 3.56 years passes. That's proper time, the time that he experiences on his path through spacetime.

Someone else taking a different path through spacetime will measure a different proper time on their path, even if the starting and ending points are the same. They're measuring proper time along the path that they took; it's a different path so in general it has a different proper time.

An analogy (and I must stress that this is an analogy - don't take it too seriously!): Suppose you and I are are both driving in our own cars between points A and B. Before we leave, we compare the odometer readings. We compare again at some time after we've arrived at the destination, and we see that my car has covered 600 miles and your car has covered only 500 miles. We expect that my car has experienced a bit more wear and is a bit closer to its next oil change, but we don't think anything strange has happened - we just conclude that you took a shorter route. We also still have no idea how far apart the two cities "really" are, although we do know that they are not more than 500 miles apart.

We just aren't used to thinking about time in the same way... we have this mental model that there's some giant clock up in the sky that we're all sharing. There isn't.

Last edited: Aug 12, 2012