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?

  • 1
    $\begingroup$ Hi, I've taken the liberty to typeset your math equations using LaTeX. $\endgroup$ – Lagerbaer May 28 '13 at 16:55

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 $$ v(t) = \frac{at+w_0}{\sqrt{1+(at+w_0)^2/c^2}}. $$

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  • $\begingroup$ Why is the proper acceleration defined as $\frac{du}{dt}$ and not $\frac{du}{d\tau}$? $\endgroup$ – Ufomammut Sep 16 '19 at 1:46
  • $\begingroup$ @Ufomammut See this post $\endgroup$ – Pulsar Sep 17 '19 at 20:50

In the context of Special Relativity, you do need to be careful about constraints such as "assume constant acceleration" without further qualification because, just as there is a need to distinguish between proper time and coordinate time, one must distinguish between proper acceleration (acceleration measured by an accelerometer) and coordinate acceleration, $\ddot x$.

Proper acceleration, like proper time, is frame invariant while coordinate acceleration is not.

It is perfectly acceptable to specify constant proper acceleration but, as you noted, specifying constant coordinate acceleration is inconsistent with Relativistic mechanics.

The constant proper acceleration solution is well known:

A simple problem is to solve the motion of a body which accelerates constantly. What does this mean? We don't mean that its acceleration as measured by an inertial observer is constant. We mean that it is moving so that the acceleration measured in an inertial frame travelling at the same instantaneous velocity as the object is the same at any moment. If it was a rocket and you were on board you would experience a constant G force. This problem can be solved in a number of ways. One is to use four-vector acceleration along its worldline which must have constant magnitude. Alternatively, the object is passing constantly from one inertial frame to another in such a way that its change of speed in a fixed time interval seen as a Lorentz boost is always the same. From our understanding of adding velocities we can see that the rapidity r of the object must be increasing at a constant rate a with respect to the proper time of the object T. The rapidity is related to velocity v by the equation

  v = c tanh(r/c)

From this we derive the equation

  v = c tanh(aT/c)
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