Relativistic deceleration and energy It's common knowledge that due to relativistic effects, accelerating from 0.8c to 0.9c takes a lot more energy than accelerating from 0.1c to 0.2c.
However, what's the case with deceleration? Does it take more or less energy when you are closer to the speed of light?
Due to the conservation of energy it should take the same amount of energy to accelerate from 0 to 0.9c in perfect vacuum than to decelerate from 0.9c to 0. However, is the distribution among the various velocities the same for deceleration and acceleration?
 A: The energy of a relativistic body is given by:
$$ E^2 = p^2c^2 + m^2c^4 $$
where $m$ is the rest mass and $p$ is the relativistic momentum, which is given by:
$$ p = \frac{mv}{1 - v^2/c^2} $$
Using this you can easily calculate the energy as a function of velocity. As you have already worked out, it doesn't make any difference whether you are accelerating or decelerating - the energy difference between any two velocities is always the same.
A: vsz, we are talking about relativity here, aren't we? Therefore, acceleration and deceleration are relative as well. Not the very fact, obviously, because you can always measure your acceleration with an onboard accelerometer without referring to the outside world. But in order to measure your initial and final velocity you do need a frame of reference (different than your spaceship), and the appointment of such a frame is always arbitrary. 
This means that your zero speed is also relative, and so while you are decelerating toward that zero in one frame of reference, you are at the same time accelerating toward infinity in another frame. The other way round is also true, obviously ;-) 
The conclusion regarding energy is obvious, isn't it?
