So I was considering the following problem within the context of Special Relativity:
Given an object O, with initial velocity v, undergoing constant acceleration at a rate of a, I want to express the velocity as a function of time.
So from newtonian mechanics:
velocity (S) = initial-velocity (v) + acceleration*time (a*t)
However this makes no sense in context of special relativity since it suggests that given a particular acceleration and enough time it is possible to exceed the speed of light.
What I realized I needed was a mapping from newtonian velocity to its special relativistic equivalent. Which I derived as follows:
Kinetic Erel = m0∗c2/(1−v2rel/c2)1/2−m0∗c2
Kinetic Enewt = 1/2m0v2newt
Where vrel = relativistic velocity, vnewt = newtonian velocity, m0 = rest mass, c = speed of light.
Setting both equal to each other and dividing by m0 I find that: c2(1−v2relc2)12−c2=12v2newt
Adding c2 to both sides and raising to the power -1 I find: (1−vrel2c2)12c2=1c2+1/2v2newt
multiplying both sides by c^2, squaring both sides, subtracting 1, multiplying by -1, and taking the square root I now have: vrel=c⋅(1−c2c2+1/2v2newt)12
So given a velocity from a newtonian problem ex: 5 m/s I can convert to its energy equivalent in special relativity via this formula. Note that as newtonian velocity goes to infintiy relativistic velocity approaches c and at 0 both quantities are 0.
Given this framework I know for fact from earlier that newtonian velocity is given as:
v(initial) + at or using our previously defined units: v + at.
Therefore the relativistic velocity can be expressed as: vrel=c⋅(1−c2c2+1/2(v+at)2)12
Is this correct?
Answer
In special relativity, proper acceleration is defined as a=dudt,
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