homework and exercises - Velocity of an object undergoing homogenous acceleration

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 = $m_0*c^2/(1 - v_{\text{rel}}^2/c^2)^{1/2} - m_0*c^2$

Kinetic Enewt = $1/2 m_0 v_{\text{newt}}^2$

Where $v_\text{rel}$ = relativistic velocity, $v_{\text{newt}}$ = newtonian velocity, $m_0$ = rest mass, $c$ = speed of light.

Setting both equal to each other and dividing by $m_0$ I find that: $$\frac{c^2}{\left(1 - \frac{v_{\text{rel}}^2}{c^2}\right)^{\frac{1}{2}}} - c^2 = \frac{1}{2} v_{\text{newt}}^2$$

Adding $c^2$ to both sides and raising to the power -1 I find: $$\frac{\left(1 - \frac{vrel^2}{c^2}\right)^{\frac{1}{2}}}{c^2} = \frac{1}{c^2 + 1/2 v_{\text{newt}}^2}$$

multiplying both sides by c^2, squaring both sides, subtracting 1, multiplying by -1, and taking the square root I now have: $$v_{\text{rel}} = c\cdot \left(1 - \frac{c^2}{c^2 + 1/2 v_{\text{newt}}^2}\right)^{\frac{1}{2}}$$

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: $$v_{\text{rel}} = c \cdot \left(1 - \frac{c^2}{c^2 + 1/2(v + at)^2}\right)^{\frac{1}{2}}$$

Is this correct?

Answer

In special relativity, proper acceleration is defined as $$ a = \frac{du}{dt}, $$ where $$ u = \frac{dx}{d\tau} = v\frac{dt}{d\tau} $$ is the proper velocity, and $$ d\tau = dt\sqrt{1-v^2/c^2} $$ is the proper time. So $$ \frac{d}{dt}\left(\frac{v}{\sqrt{1-v^2/c^2}}\right) = a. $$ If we integrate this over a time interval $[0,t]$, we get, if $a$ is constant, $$ \frac{v}{\sqrt{1-v^2/c^2}} - \frac{v_0}{\sqrt{1-v_0^2/c^2}} = at, $$ with $v_0$ the initial velocity. If we define the constant $$ w_0 = \frac{v_0}{\sqrt{1-v_0^2/c^2}}, $$ then $$ v^2 = (1 - v^2/c^2)(at+w_0)^2, $$ so that we finally get $$ v(t) = \frac{at+w_0}{\sqrt{1+(at+w_0)^2/c^2}}. $$