If I were to get within seconds of the heat death of the universe, and then accelerate to very close to c instantly, as I understand it, I should experience time considerably slower than a "stationary" (at least, in my reference frame) object observing me. What that leads me to conclude is that for me, it would be much longer until the heat death of the universe. Given that I accelerate to 0.99c and that for a stationary object, it would be 5 seconds until the heat death of the universe:
$$
t' = \frac{\Delta t}{\sqrt{1 – \frac{v^{2}}{c^{2}}}}
$$
$$
t' = \frac{5}{\sqrt{1 – 0.9801}} = 35.44
$$
I should (given my amateur-ish understanding of the equation) record 35.44 seconds before the death of the universe.
However, I can see two problems with this thought experiment (really, they're most likely problems with my thinking or my understanding):
- The heat death of the universe has to be absolute, ie. it makes no logical sense for all the universe's energy to be completely dissipated in one reference frame 45 seconds before another reference frame
- To get to 0.99c (even if I accelerated instantly, or there about) I would have to do work, which would dissipate energy. This seems to me to mean that I would experience the heat death even faster than the stationary observer, given that a stationary observer isn't dissipating energy in any amount close to how I would if I were going at 0.99c in my spaceship.
EDIT: My example of 5 seconds before heat death is an extreme example (I imagine it wouldn't be feasible, even if we could accelerate to 0.99c) but the concept should work at any time before heat death (again, this could be entirely wrong, please correct me)
Best Answer
(1) Heat death is no switch that flips, it's a gradual process. You can't time it sharply.
(2) There is no absolute time in special relativity, so it does make sense that it takes different times for different observers to reach the same event, even if they started out synchronized. This happens, for example, with the twin paradox.