Modern Physics/Acceleration in Special Relativity

Classically we would talk about a particle at x(t) with acceleration dx²/d²t. In relativity we must treat time as just another coordinate, and use derivatives with respect to proper time, which means our notion of acceleration will change. We can assume the particle is moving slower than light.

Since τ increases monotonically with t (no time travel allowed) we can easily parameterise the path of the particle with τ rather than t. Derivatives with respect to these two variables then differ only by a factor of γ

$\frac{d}{d\tau }=\gamma \frac{d}{dt}$

which gives us the connection between the classical and relativstic formulae.

Lets suppose we have a particle moving on the path (x(τ),ct(τ)), where τ is the particle's proper time.

Its velocity four vector is

$\underset{\_}{u}=\frac{d}{d\tau }(x,ct)=\gamma (\dot{𝐱},c)$

In the particles rest frame this is (0,c), which has constant magnitude -c², but this scalar must be the same in all frames, thus

$\underset{\_}{u}\cdot \underset{\_}{u}=-c^2$

That is, the magnitude of the velocity is always constant.

Differentiating this, we can immediately say

$\underset{\_}{u}\cdot \underset{\_}{a}=\frac{d}{d\tau }\underset{\_}{u}\cdot \underset{\_}{u}=0$

so, velocity and acceleration are always perpendicular.

These two results are much simpler than in classical physics.

Differentiating the velocity gives us, after some algebra ,

$\underset{\_}{a}={\gamma }^4\ddot{x}(1,\frac{v}{c})$

when the motion is along the x-axis. Since the spatial and temporal components must be 3-vectors and scalars respectively we immediately see that for motion in arbitary directions,

$\underset{\_}{a}={\gamma }^4(\ddot{𝐱},\frac{\ddot{𝐱}\cdot 𝐯}{c})$

The magnitude of this is

$|\underset{\_}{a}|={\gamma }^3\ddot{x}$

which is the classical magnitude, corrected by a factor dependant on γ. When the velocity is much less than c this factor is approximately 1, so in that limit the magnitudes are the same, as they should be.

Knowing this, we can work out the equation of motion for a particle with constant acceleration a along the x-axis. For simplicity, we'll assume the initial velocity is 0.

Since the accleration is constant

$\frac{\dot{v}}{{(1-\frac{v^2}{c^2})}^{\frac{3}{2}}}=a$

Integrating once we get

$\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}=at$

or, on rearranging

$v=\frac{at}{\sqrt{1+a^2{t}^2/c^2}}$

If at is much less than c this gives a velocity of approximately at, identical to the classical result, but as t tends to infinity the velocity tends to c.

The position is

$x=d+\frac{c^2}{a}\sqrt{1+\frac{a^2{t}^2}{c^2}}$

where d is some constant. We can choose our coordinate system to make d zero.

Calling the interval between the origin and the particle, I, we have

$\begin{array}{ccccc}{I}^2& =& x^2& -& c^2{t}^2\\ & =& \frac{c^4}{a^2}(1+\frac{a^2{t}^2}{c^2})& -& c^2{t}^2\\ & =& \frac{c^4}{a^2}& & \end{array}$

Thus the interval between the particle and the origin is constant. Notice, this is the equation for an hyperbola, so we know the particle's trajectory.