Relativistic Constant Acceleration

In special relativity, the co-ordinates used by two observers moving relative to one another are related to one another by the Lorentz transformation, which mixes up spatial and time co-ordinates. This complicates the addition law for velocities and so changes the nature of constant acceleration. While the Lorentz transformation can be expressed in terms of the relative velocity between the two observers, it is more neatly described as a hyperbolic rotation.

Consider an object experiencing constant acceleration a. Chose a frame of reference in which it is, at some moment, at rest; thereafter, its velocity will always be parallel to a; chose x-axis also parallel to a. With time parameter t in our chosen frame, let b(t) be the hyperbolic angle describing its movement, so that c.tanh(b(t)) is its apparent speed (in our chosen frame) as a function of t.

That it experiences constant acceleration says that, at each moment, in its instantaneous rest frame, its acceleration is our constant a. Over any short enough time interval q, its velocity at the end of the interval, in the frame of reference with respect to which it was at rest at the interval's start, is a.q plus terms of smaller order than q – i.e. if we vary q, and f(q) is the difference from a.q, then f(q)/q is tiny for all small enough q. Of course, q is measured in the instantaneous rest frame, so our chosen frame of reference sees this as a time interval q.cosh(b).

For small enough q, tanh(a.q/c) = a.q/c (since tanh'(0) = 1) so we can infer that b(t) changes by a.q/c while t changes by q.cosh(b), give or take terms of smaller order than q, yielding db/dt = a/c/cosh(b), whence d(sinh(b))/dt = cosh(b).db/dt = a/c which we can integrate to get sinh(b) = a.t/c plus a constant which we can make zero by suitable choice of when to take t = 0.

Notice that, no matter how long it continues accelerating, the body's speed will still be c.tanh(b) for some b; and tanh is bounded between −1 and +1, so the body's speed will always be smaller than the speed of light. Consequently, any displacement between positions along its path will be time-like.

Full solution

Shift the origin of our (t,x,y,z) co-ordinate system (without changing which objects it deems to be at rest) so that our object is at rest at t = x = y = z = 0. We have dx/dt = c.tanh(b) with sinh(b) = a.t/c, yielding

= c.tanh(b).dt
= c.c.sinh(b).db/a
= c.c.d(cosh(b))/a

whence a.x/c/c − cosh(b) is constant; we have x = 0 at t = 0, when sinh(b) = 0, so b = 0, yielding cosh(b) = 1, so we can infer 1 +a.x/c/c = cosh(b) = √(1+a.a.t.t/c/c). For small a.t/c, this is well approximated by 1 +a.x/c/c = 1 +a.a.t.t/c/c/2, which gives x = a.t.t/2, the classical solution.

The proper time of the accelerating body, s, has

= dt/cosh(b)
= c.db/a

giving us, with s measured from when b = 0, a.s/c = b; our hyperbolic angle simply grows at rate a/c, just like velocity in the classical model, when measured in the accelerating frame of reference. For small b, sinh(2.b) = 2.b, so this yields s = t, as we would expect. My python package's study/space/ provides an implementation of these computations.

Transformation law for accelerations

In a frame of reference which sees the constantly accelerating body as being, at some moment, at rest, its velocity is c.v = c.tanh(b), so its apparent acceleration is c.db/dt/square(cosh(b)) = a / power(3, cosh(b)) = a.power(3/2, 1−v.v). So a body with velocity c.v, accelerating at rate a parallel to v in its own frame, will have apparent acceleration a.(1−v.v).√(1−v.v). Another consequence of special relativity is that a body's mass increases when it is moving, by the factor 1/√(1−v.v), so the force acting on the body, with rest-mass m, is just m.a.(1−v.v), in its initial rest-frame (and something else must, as seen by that frame, be experiencing an opposite force equal to this, to keep the body accelerating), where m.a is the force in its instantaneous rest frame.

Now consider a body, experiencing acceleration a in its own rest frame, as seen by a non-accelerating observer, without assuming that the body's velocity is parallel to a. Pick an intermediate frame of reference in which the body's velocity and acceleration are parallel to one another and the observer's velocity is perpendicular to these; in this frame, use co-ordinates T for time, X in the direction of the body's acceleration and Y in the direction opposite to the observer's motion. Let the observer then use x = X as a co-ordinate varying perpendicular to the intermediate frame's relative motion and y, t as the transformed equivalents of Y, T. Let the body's velocity in the observer's frame then have components c.u in the y direction and c.v in the x direction. (By choice of direction of Y, u is non-negative; by choice of direction of X, a is positive; the sign of v is unspecified; none of which matters.)

Our intermediate frame sees the body's velocity vary with time, T, as c.tanh(b) with sinh(b) = a.T/c for some choice of moment to call T=0. The body's position in this frame is then given by a.X/c/c = cosh(b)−1, with the X-axis parallel to v, for some choice of where to mark X=0. In this frame, our body has no motion in the Y direction, so make Y = 0 at the body. Our observer's frame then has x = X and, for suitable choice of origin for y and t: y = Y.cosh(q) +c.T.sinh(q), t = T.cosh(q) +sinh(q).Y/c with tanh(q) = u. Our body has Y = 0, so

and the body's trajectory is

This last is, at least for small enough t, approximately a.t.t.(1 −u.u)/2 which, for small u, is approximately the a.t.t/2 that Newton or Galileo might have predicted. Differentiating these we get a velocity with y component c.u (as stipulated at the outset) and x component

= sinh(b).c.c.db/dt/a
= c.sinh(b)/cosh(b)/cosh(q)
= a.t.(1−u.u)/√(1+a.a.t.t.(1−u.u)/c/c)

which we're given to be c.v at our chosen moment, so c.v = a.t.(1−u.u)/cosh(b(t)). Squaring this and rearranging, we get

= c.c.v.v.cosh(b).cosh(b)
= v.v.(c.c +a.a.t.t.(1−u.u))


= a.a.t.t.(1−u.u).(1−u.u −v.v)


= c.v/a/√((1−u.u−v.v).(1−u.u))

at that moment. Differentiating our velocity again, we get an acceleration purely in the x direction of magnitude

= a.(1−u.u).d(t/√(1 +a.a.t.t.(1−u.u)/c/c))/dt

of form d(t/√(1 +k.t.t))/dt = 1/√(1 +k.t.t) −(1/2).t.(2.k.t)/power(3/2, 1 +k.t.t) = (1 +k.t.t −k.t.t)/power(3/2, 1 +k.t.t)

= a.(1−u.u)/power(3/2, 1 +a.a.t.t.(1−u.u)/c/c)

which, at the given moment when the component of velocity in this direction is c.v,

= a.(1−u.u)/power(3/2, 1 +v.v/(1−u.u−v.v))
= a.(1−u.u)/power(3/2, (1−u.u)/(1−u.u−v.v))
= a.(1−u.u−v.v).√((1−u.u−v.v)/(1−u.u)) = a.(1−u.u−v.v).√(1 −v.v/(1−u.u))
= (1−u.u−v.v).c.v/t/(1−u.u)
= (1 −v.v/(1−u.u)).c.v/t

agreeing (in its antepenultimate form, just before I brought c.v/t back into it) with our earlier result when u = 0. In the intermediate frame, our body was at rest at T = 0 = Y, when t = 0 = y; at that moment, the body has b = 0 so x = X = 0, too. Its average (observed) acceleration since then is c.v/t and its observed acceleration is just 1 −v.v/(1 −u.u) times this. If the body's rest-mass is m, its apparent mass is m/√(1 −u.u −v.v); multiplying this by acceleration, we get the force acting on it, m.a in its rest frame, but m.a.(1−v.v−u.u)/√(1−u.u) in the observing frame.

Notice that u.u+v.v is just the squared speed/c of the body at the moment under discussion, while u is just the component of the body's velocity/c perpendicular to a. Elsewhere, I'll more usually use v.c for the speed, i.e. v.v will replace u.u+v.v, while retaining u.c as the velocity component perpendicular to a.

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