Relativity and my Intergalactic Space Race

[StrangeChicke]

Sorry if this has been asked before, but I tried a search and couldn't find anything... so here goes my question.

Suppose I'm going to hold the 1st annual Intergalactic Space Race, but only 2 people show up to compete. One guy is in a SuperPlamasmonic X3000 Racer, capable of a accelerating very rapidly to 20% of c. The other guy, is actually Merlin the Wizard, and he is riding a magic dragon capable of going as slow or as fast as he wants it, up to 99.999% of c.

Let's say my race has us going out a ways and back for a total trip of 40ly. Now, here is where I am confused. If the x3000 is going .2c, it should take him ~200years in both his time-reference and Earth's. Merlin, has several options. He could choose to go a measly .1c, in which case it will take him ~400years, again time dilation would have little effect. He could also go .7c, in which case he'd easily beat the X3000, going 60 years his time, but 80 years Earth time. Or, he could go .99c, in which case he'd think he still won, taking 40years...but on Earth it would seem like he took 280years? In that case, the X3000 would have been back 80 years prior?

Anyway, I'm an economist, and my other economist friends and I were sitting around trying to act smart when we came up with this problem. We don't do real math, so I'm convinced I'm wrong...but on the other hand, physics is strange stuff to me. What I'd really like to know if my summery is correct, and if this is a real paradox or not.

-Thanks in Advance (P.S. I hit enter before I finished, sorry if you read this while it was incomplete)

 

[JesseM]

Sorry if this has been asked before, but I tried a search and couldn't find anything... so here goes my question.

Suppose I'm going to hold the 1st annual Intergalactic Space Race, but only 2 people show up to compete. One guy is in a SuperPlamasmonic X3000 Racer, capable of a accelerating very rapidly to 20% of c. The other guy, is actually Merlin the Wizard, and he is riding a magic dragon capable of going as slow or as fast as he wants it, up to 99.999% of c.

Let's say my race has us going out a ways and back for a total trip of 40ly. Now, here is where I am confused. If the x3000 is going .2c, it should take him ~200years in both his time-reference and Earth's. Merlin, has several options. He could choose to go a measly .1c, in which case it will take him ~400years, again time dilation would have little effect. He could also go .7c, in which case he'd easily beat the X3000, going 60 years his time, but 80 years Earth time. Or, he could go .99c, in which case he'd think he still won, taking 40years...but on Earth it would seem like he took 280years? In that case, the X3000 would have been back 80 years prior?

From the perspective of Earth's reference frame, time dilation basically just means his clocks slow down as he moves, but it doesn't affect his rate of movement in this frame. So if he goes 40 ly at 0.7c, in the Earth's frame it takes 40/0.7 = 57.143 years, and if he goes 40 ly at 0.99c, it takes 40/0.99 = 40.404 years in the Earth's frame. But as seen on Earth, his clock elapses less time because it's running more slowly...if it takes 57.143 years at 0.7c in the Earth's frame but his clock is slowed down by a factor of sqrt(1 - 0.7^2) = 0.71414, then his clock only advanced forward by 57.143*0.71414 = 40.808 years. Likewise, if it takes 40.404 years at 0.99c in the Earth's frame but his clock is slowed down by a factor of sqrt(1 - 0.99^2) = 0.14107, then his clock only advanced forward by 40.404 * 0.14107 = 5.6998 years.

You could also look at things from his perspective, where his own clock is running normally but the distance between the Earth and the distant landmark is shrunk by sqrt(1 - v^2/c^2)...in this case, if he sees the Earth and the landmark moving past him at 0.7c, then the distance between them is shrunk to 20*0.71414 = 14.2828 light years, so with the destination starting 14.28 ly away it takes 14.2828/0.7 = 20.404 years to get to the destination, and another 20.404 years to get back, for a total of 40.808 years. Likewise, if he sees the Earth and landmark moving at 0.99c, the distance between them is shrunk to 20*0.14107 = 2.8214 l.y., so the time to get from one to the other is 2.8214/0.99 = 2.8499 years, meaning the time to get there and back is 5.6998 years.

 

[StrangeChicke]

I think I understand, and even made a little excel graph to show it. But basically, it seems I was wrongly assuming the 'frames of reference' are not related/symmetrical. Thus, if the magic dragon did accelerate to near the speed of light, distance for him would shrink to maintain time? At .9c he'd run the 40ly race, but due to length contraction 40ly in his frame is roughly 17ly in our frame back here on Earth (his meters are 'smaller' than our meters in order for seconds to still be seconds). If I'm correct, that would mean at .9c he'd run the race in roughly ~45 years, but for him it would seem like it took ~20years (which is the time dilation effect). Or, using the twin paradox, we'd have aged 45 years but he'd have aged 20 years.

Anyway, that seems to be my understanding of special relativity in this regard.
Thanks for your help, you really saved my entire outlook from crashing down on my head.

 

[pervect]

I think your calculations are messed up, but you haven't given any details, really, of your calculations.

Note that if you are having a race, you want both contestants to run the same course. The logical way to do this is to set up the course in some particular frame of reference, we can call it the main frame. These would be, for instance, two buoys, which would be stationary relative to each other and maintain a constant distance apart. Call this distance, which is measured in the main frame, L.

Then the total trip time as measured in the main frame will be L/v, where v is the velocity.

THe total trip time as measured by the spaceship traversing the course will be shorter and equal to

$$\frac{L {\sqrt{1-(v/c)^2}}} {v}$$

So the faster contestant will ALWAYS finish the race first. (I'm not sure how you could think things would be otherwise, the only thing I can imagine is that you are making the faster contestant run a longer course, without realizing it).

 

[JesseM]

I think I understand, and even made a little excel graph to show it. But basically, it seems I was wrongly assuming the 'frames of reference' are not related/symmetrical. Thus, if the magic dragon did accelerate to near the speed of light, distance for him would shrink to maintain time?

You could put it that way...but note that in every frame, the length of an object moving in that frame will shrink by a factor proportional to the object's speed in that frame (the factor is $$\sqrt{1 - v^2/c^2}$$), and a clock moving in that frame will slow down by the same factor.

At .9c he'd run the 40ly race, but due to length contraction 40ly in his frame is roughly 17ly in our frame back here on Earth (his meters are 'smaller' than our meters in order for seconds to still be seconds).

It may be just the way you're saying it, but it sounds like you have it backwards--"in our frame back here on Earth" the distance from the Earth to the turnaround point is 20ly (since the distance was defined to be 20ly in our frame), but in his own frame the distance is smaller, because in his frame the Earth and turnaround point are moving at 0.9c, so considered as a single "object" (and you could imagine taking a long rod and placing one end on Earth and the other at the turnaround point), this length will be shrunk to 20*sqrt(1 - 0.9^2) = 8.7178 ly.

If I'm correct, that would mean at .9c he'd run the race in roughly ~45 years

In our frame it would take 20/0.9 = 22.222 years for him to go from Earth to the turnaround, so the whole trip would be twice this, or 44.444 years.

but for him it would seem like it took ~20years (which is the time dilation effect). Or, using the twin paradox, we'd have aged 45 years but he'd have aged 20 years.

That's about right, because of time dilation he would only experience 44.444*sqrt(1 - 0.9^2) = 19.373 years for the trip.