If you get in the canonical 1G windowless spaceship and accelerate for 5 years as determined by your trusty wristwatch, then decelerate for 5 years, then decide its time to see where you are would the Newton way of calculating your distance from earth work?

Distance as predicted by Newton $= \frac12 at^2 = 1.22\times 10^{17} ~\mathrm{m}$ during acceleration and another $1.22\times 10^{17} ~\mathrm{m}$ during deceleration.

So when you get out of the rocket, and look back at earth, its 25.8 light years away, according to Newton. Max velocity = a*t = 5.15 times the speed of light.

Relativistic calculation:

T = 5 years --> shows

82.7ly + 82.7ly which is 165 ly - way farther than Newton's calculation.

I guess relativity is really different from Newtonian physics, and you can make measurements without another observer to conclude that something non-newtonian happened while on the ship.

I guess I answered my own question here. But if I'm wrong, please chime in.

I would think that most people (say those that went to first year physics) would assume that the relativistic calculation would show a shorter distance (c is a limiting speed). Some more investigation results in the thought that the time dilation and the gamma factor somehow 'cancel each other out'. But it turns out that relativity makes you go farther on the same tank of gas than Newton would say you need.

Leaving this as a question in the extremely likely circumstance that I made an error.


A key result from SR that must be kept in mind when making these kinds of comparisons is that, when speaking of elapsed time, there is more than one elapsed time to consider.

In a Newtonian universe, there is universal time so the elapsed time according to the astronaut equals the elapsed time according to mission control.

In a SR universe, the elapsed times are different due to time dilation. If the spacecraft accelerates* for 5 years according to the astronaut, the acceleration lasts much longer than 5 years according to mission control. This accounts for the longer distance travelled compared to the Newtonian calculation.

Conversely, if the spacecraft accelerates for 5 years according to mission control, the distance travelled will agree with the Newtonian calculation. However, the astronaut will record much less elapsed time to travel this distance.

*And, by the way, in a SR universe, we must also be careful to specify which acceleration we are talking about: the coordinate acceleration - the acceleration of the spacecraft according to mission control, or the proper acceleration - the acceleration as measured by the spacecraft's accelerometers.

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  • $\begingroup$ I was kind of pointing out how you don't have to enlist other observers at all. Most SR examples use two observers, the aged one one left behind vs the traveller. But you actually get farther with SR for the same proper time and fuel usage as you would with Newtonian mechanics. Put another way - the limiting speed 'gamma' factor is more than compensated for by the time dilation. $\endgroup$ – Tom Andersen Nov 3 '13 at 18:46

I would think that most people (say those that went to first year physics) would assume that the relativistic calculation would show a shorter distance (c is a limiting speed)

If you watched the ship accelerate then decelerate for five years in some hypothetical non-relativistic universe then it would indeed travel farther than it does in our universe. However in your question the five years is the time experienced by the occupants of the rocket, and this time is dilated relative to the people watching from Earth. So as viewed from Earth the trip would take longer than five years, which is why the rocket travels farther.

John Baez's article gives the equation for calculating the elapsed time for the unaccelerated observers:

$$ t = \frac{c}{a} \sinh \frac{aT}{c} $$

Taking $a$ as 9.81 m/sec$^2$ and the time measured on the rocket as five years this gives for one half of the trip:

$$ t \approx 2.66 \times 10^9 \text{seconds} \approx 84.4 \text{years} $$

Since the rocket spends most of flight at a velocity of nearly $c$ it shouldn't surprise you to find that the distance travelled is just a bit less than 84.4 light years.

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