Well, it can’t be measured for you. If your reference frame is a spaceship flying rapidly away from Earth, then you’ll measure Earth as being time dilated and length-contracted.
It just occurred to me (and I can describe it only klutzily): in the Twin Paradox, everybody talks about the older twin. Shouldn’t there be a shorter launch tower, or something?
ETA: or shorter Twin. (Not as a result of chronological growth.)
People talk about the older twin after the two twins are brought back together. They’re in the same frame of reference at that point, so there isn’t a shorter twin.
The “paradox” of the Twin Paradox comes from the fact that as the two twins separate, they will each see each other time dilated and length contracted, and yet it’s always the spacebound twin that has aged less than the earthbound twin once they reunite. The reason that the two paths aren’t symmetric is that the spacebound twin has to change frames of reference (accelerate) in order to return to Earth.
Suppose the universe is a Poincaré dodecahedral space, as was proposed some time back. In this universe, can’t I accelerate to a high velocity and then follow a geodesic back to my starting point? When I pass my twin who is older? I guess the answer must be that the topology of the space has general relativity like effects so that the traveling twin ends up younger? But is this correct?
Take a sheet of notebook paper and make a tube. If you make it straight, each line will meet itself in the overlapping part, and you’ll have a bunch of circles. But you can also make the paper a little skewed, so the lines on the left side of the paper meet the line above on the right side of the paper, and you end up with a line the spirals around the tube.
In the same way, if space is closed, then there’s a frame of reference where the spatial axis at a constant time in that frame forms a closed loop. For other frames of reference, moving WRT that frame, the spatial axis won’t meet as you go around the universe. So in a spatially closed universe, there is a special frame of reference that you don’t have in an infinite open universe. A twin in that frame where the spatial axis forms the closed loop will age more than a twin in a frame moving with respect to it.
That’s what I’ve always read. However, if I am in a spaceship, and I notice that my acceleration is decreasing, and it is because I am approaching the speed of light (as you said above) and therefore because of time dilation and length contraction, why can’t I use that acceleration measurement to determine the time dilation and length contraction?
Because you won’t notice that your acceleration is decreasing.
If that is the case, can you expand upon what you meant by the following:
If I won’t notice my acceration is decreasing, how will I find that the magnitude of my accleration will decrease (as quoted above)?
Are you saying that when protons are accelerated that their mass does not increase? Can mass be measured without changing momentum? When you smash a proton into something else it is interacting via electromagnetism, and when you bend it’s path it is interacting by gravitation, but in both cases is not the momentum of the proton changed?
As you fire your engines, the rest of the world seems to accelerate. You’ll notice that as the rest of the world goes faster, you have to fire the engines harder or longer to make it go a given amount faster. Say it takes you 1 second of max thrust to go 1 mph faster when you’re at rest with the rest of the world. When the rest of the world is whizzing by at 1 MPH short of C, it’ll take an infinite thrust time to make it go by 1 mph faster.
But if you jettisoned some stuff and then fired your engines for 1 second, you’d be moving away from your jetsam at 1mph (or a bit faster, since you’ve shed the mass of the jetsam).
Now, that jetsam was going at C-1 and now you’re going 1 mph faster, so addition no longer works? Well, no, because the units change. Jetsam moves at c-1 mph with respect to the world. You move at 1 mph with respect to the jetsam. With relativity, you can’t simply add different speeds with respect to different things. You have to apply the Lorentz transformation.
If you did have an infinite amount of time to spend, and infinite fuel to burn, when infinity is over, the rest of the world would be passing at C. 
Thanks for the reply. However, I’m speaking of myself and my rocket, not the rest of the world. If I was just in empty space, I can fire my rockets, measure my acceleration, use the amount of seconds I was accelerating at that rate, and calculate my velocity.
From what I read above, my acceleration would decrease as I approached the speed of light. If I noticed this decrease, I could use it to calculate my time dilation and length contraction.
But, again as state above, I cannot calculate my time dilation nor length contraction. So to what would I think is causing my decreasing acceleration?
And in this case also, this coyness, maiden, were no crime.
And again, you wouldn’t notice your decrease in acceleration. It’ll decrease with respect to some non-accelerating reference frame, but that doesn’t matter to you, because you’re not in that reference frame to see it.
So, in my theoretical spaceship, I am constantly measuring my acceleration, which will not show as decreasing to me. So after a certain amount of time, i will use this acceleration and time of acceleration to calculate my velocity. Somehow that calculation will be wrong, even though I won’t know that’s it’s wrong?
You’ll know it’s wrong because you’ll know that’s the wrong way to compute these velocities(*). If you say bye to a friend stationary with you on the rocket launch pad, and then you immediately fire your thrusters and keep your acceleration constant at 1g as measured within the ship, and you do that continuously for 1 year as measured within the ship, you are free to multiply those two numbers together and get something with units of velocity: (9.8 m/s/s) * (1 yr) = 3.09x10[sup]8[/sup] m/s (which coincidentally is nearly exactly the speed of light, to within 3%). But the velocity that gets you has nothing to do with anything, it turns out (and you can know that, even on your ship). In particular, the number you get doesn’t provide the speed that your friend back at base would see you having, and it doesn’t provide the speed you see your friend as having from the perspective of your ship.
() If all of the speeds involved stay small compared to the speed of light, then it’s approximately the right way to compute these velocities.*
I’m sure that’s true, but I’m not concerned about the speed my friend back at base sees me having, nor the speed I see my friend as having. I am asking about the speed I calculate myself as having, using my acceleration and my time of acceleration. When you say that the velocity calculation gets me nothing to do with anything, what does that mean?
And he also said, perhaps even more to the point here,
If any who deciphers best,
What we know not—ourselves—can know,
Let him teach me that nothing. This
As yet my ease and comfort is,
Though I speed not, I cannot miss.
Rereading that verse section, which is a difficult one, it really is appropriate as a whole to the relativity self/other reference thing. Speed is just the topper…
Actually, this gets back to what I was saying about proper velocity back in post 22. If you take your acceleration, multiplied by the time you’ve spent accelerating, what you’ll actually get is your proper velocity, which is allowed to get as big as you’d like.
So if somebody did that, and reported that they exceeded the speed of light, the response would be “no you didn’t”?