Required math: calculus, hyperbolic functions
Required physics: relativity basics
A common misconception is that special relativity cannot handle accelerations. In fact, acceleration follows on naturally from the treatment of velocity that we’ve already seen.
The four-velocity is defined as the vector with unit magnitude along a particle’s world line, so in the particle’s rest frame it is
The magnitude is therefore
Now suppose that a particle does accelerate. In that case, we can have an inertial frame at any event in the particle’s life by defining the momentarily comoving reference frame or MCRF for short. This is a reference frame that, at a given event, has the same velocity as the particle. If the particle is accelerating, then the MCRF will change from one event to the next, but at each point it is always an inertial frame. If the particle makes an infinitesimal displacement then in its MCRF, the separation between the two events that define either end of this displacement is just an infinitestimal change in the proper time, . Therefore, the magnitude of the displacement is
If we take mathematical liberties with infinitesimals, we can now consider a vector which is the displacement vector divided by the increment in the proper time. In particular we can divide the magnitude equation by and get
which is the same as the definition of the four-velocity. Despite the dodgy mathematics involved in its definition, we arrive at an expression of the four-velocity as a derivative of a displacement with respect to proper time:
To introduce acceleration, we start with 2 and take its derivative with respect to :
where the last line arises because which is a constant. We can define the four-acceleration as the derivative of the four-velocity with respect to proper time:
Note that the four-acceleration is always orthogonal to the four-velocity. This does not mean that the direction of acceleration is always perpendicular to the velocity (although it can be, as in the case of circular motion). Remember that orthogonality with four-vectors is a more general concept that it is with three-vectors.
What this relation does mean, though, is that in the MCRF, we can write a specific form for . In the MCRF, we know that , so the orthogonality requires that , and we get
To see the relation between the four-acceleration and the more familiar Galilean acceleration, we can consider the case where a particle has a constant four-acceleration along the axis, with magnitude . In the MCRF, the four-vector . After an increment of proper time, the velocity of the particle will have changed to in the direction, so
The relation between proper time and the time in another inertial frame that is attached momentarily to the particle after the increment is given by the time dilation factor so that
In this case, to first order in , and , so again to first order we can write
which means that the constant acceleration is
and so is in the limit equivalent to the Galilean acceleration.
Now suppose we have a particle moving along the axis at a momentary speed . The four-velocity is found by applying the Lorentz transformation to the four-velocity in the MCRF, which is . We get upon transforming to the frame in which the particle is seen to have speed :
The four-acceleration is the derivative of this with respect to proper time , so we get
Since the acceleration in the MCRF is , in the frame moving at speed , we apply the Lorentz transformation as with the four-velocity, and get
Comparing this with the derivation above, we use the second component of the four-acceleration in each case to get
Assuming is a constant, we can integrate this equation to get (assuming )
Thus under constant acceleration, as , . Unlike the Galilean case where the velocity would increase without limit under constant acceleration, in relativity, the speed tends to the speed of light and never surpasses it, as must be the case.
The distance travelled after a given time can be found by another integration of the relation, since :
As an example, we can find how long it takes to reach a speed of . We can invert the relation above to give in terms of :
If the acceleration is (about 1g), we must first convert this to relativistic units. Since 1 sec = m, we get
To convert this back to ordinary units, we get
To find the elapsed proper time, we can start with
Integrating this, we get
In relativistic units, we find
For a trip to the centre of the galaxy, with we use 29 to work out the time, which comes out to essentially the same value since most of the journey the speed is so close to it makes little difference. We can then plug in the numbers and find that
In the inertial frame of the galaxy, the travel time is found by inverting the relation between and above:
This is time dilation in the extreme!