markfiend

I've seen a few discussions on the plausibility (or otherwise) of Relativity Theory, and I must admit to having a few problems. (Though the problems are surely with my understanding not with the theory!)

I have the equation for time dilation:

T = To /sqr(1-(v^2 / c^2))

(apologies that I can't work out how to set it out properly)

Now say I have observer A in a rocket traveling at 0.9c, travelling for 5 years.
She will travel 4.5 light years.

Observer B remains on observer A's home planet, and from his perspective, observer A's journey takes 11.47 years (if I've done my arithmetic right: see below) and so is at a speed of 0.39c

Is this right? Observer A observes herself traveling at 0.9c, while observer B observes her travelling at 0.39c And which value of the speed do you "plug into" the equation?

Arithmetic bits:
T = To /sqr(1-( (0.9c)^2/c^2))
T = To /sqr(1- 0.81)
T = To /sqr(0.19)
T = To /0.4359
T = 5/0.4359
T = 11.47 (4 significant figures)

Demosthenes

I'm no expert in relativity and I've hardly done any work with it, but your results don't sound right to me.

If observer A is travelling at 0.9c and travels 4.5 light years, but it can't have taken her five years from her perspective to get there. Because the time dilation due to 0.9c velocity would shrink her apparent journey to much less than five years, probably just a month or less.

An observer on earth by contrast would indeed see observer A taking 5 years to cross 4.5 light years.

I think you're confusing To and T. Five years should be T instead, try working out To and see what you get.

Jesse

You've got the time dilation equation backwards. If the rocket is travelling at 0.9c relative to the earth, then it should take 5 years to go 4.5 light-years according to earth-clocks and earth-rulers. The time dilation equation tells you that this journey will take less time from the perspective of someone on board the rocket. The time dilation equation is t' = t0/gamma, where t0 is the time in the rest frame and t' is the time in a frame moving relative to that one, and gamma = (1-(v/c)^2)^-1/2...so, using earth's frame as the rest frame, you'd solve for t', the time in the rocket's frame, using t' = 5 years/gamma, with v/c = 0.9...this gives t' = 5 years/2.29, or 2.18 years from the rocket's perspective. And remember that from the rocket's perspective the length from earth to the star will contract as well--instead of 4.5 light years, you can use the similar Lorentz contraction equation to show that an observer on the rocket will see the distance as only 4.5 l.y./2.29, or 1.96 l.y. Thus, the observer on the rocket will see the earth as moving away 1.96 l.y. in 2.18 years, so the earth is also moving at 0.9c relative to the rocket.

markfiend

Demosthenes: I think you're confusing To and T. Five years should be T instead, try working out To and see what you get.

Yeah, you're right. [Homer]Doh![/Homer]

Jesse: Thus, the observer on the rocket will see the earth as moving away 1.96 l.y. in 2.18 years, so the earth is also moving at 0.9c relative to the rocket.

I've plugged the numbers back in (the right places) and get 2.18 years too. So Einstein wasn't wrong

Thanks for the help. I did say it was probably my problem!

Shadowy Man

I think GPS satellites prove that Einstein did a pretty good job.

markfiend

I've looked at my OP again; Observer A travels for 5 years at 0.9c and ends up 4.5 ly away (in observer A's frame of reference)

Observer B sees A traveling for 11.47 years at 0.9c ending up 10.32ly away (in observer B's frame of reference)

But the distance varies with frame of reference (not speed); so no contradiction.

Is this right? The speed is independent of frame of reference?

contracycle

Weird shit you can do with relativity:

If we can build a wormhole ala Michio Kaku, then we can put one gate on earth and another on a ship.

Now we send the ship off, say, on a 10,000 ly hourney to another place.

The cool part is the subjective journey time for the gate on the ship is dilated; so (for arguments sake: these numbers are not accurate) this means that you can use the gate BEFORE it would arrive in the earth time-frame. Weird huh? Wormholes allow moivement through 4 domensions, not just three

Lobstrosity

If B sees A moving at velocity v, A sees B moving at velocity -v. By simple symmetry you know that each sees himself as stationary while perceiving the other to be moving at the same speed v.