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

dx
= 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).

The proper time of the accelerating body, s, has

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

giving us, with s measured from when b = 0, a.s/c = b. For small b, sinh(2.b) = 2.b, so this yields s = t, as we would expect. My python package's study/space/zoom.py 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).

Now consider a body experiencing acceleration a as seen in a frame with respect to which the body has, instantaneously, a velocity not parallel to a; break this velocity into components c.v parallel to a and c.u perpendicular to it. An intermediate frame moving with velocity c.u then 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. In our original frame, use co-ordinates: x varying parallel to v, y parallel to u and time co-ordinate t. Chose origin for t so that the body has y = 0 at t = 0; and let the intermediate frame's corresponding Y co-ordinate be measured from where the body is at some moment; since its velocity in this frame is in the x-direction, it will always have Y = 0. Since the relative velocity of the frames, c.u, is perpendicular to the direction of variation of x, we know x = X. The transformation between [y,t] and [Y,T] will be 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 t = T.cosh(q) and our body's trajectory is

Differentiating these we get a velocity with y component c.u (as stipulated at the out-set) and x component

d(cosh(b).c.c/a)/dt
= sinh(b).c.c.db/dT/(dt/dT)/a
= sinh(b).c/cosh(b)/cosh(q)
= a.T/cosh(q)/√(1+a.a.T.T/c/c)
= 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

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

whence

v.v.c.c
= a.a.t.t.(1−u.u).(1−u.u −v.v)

yielding

a.a.t.t.(1−u.u)/c/c
= v.v/(1−u.u−v.v)

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

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

agreeing with our earlier result when u = 0. 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.

Another consequence of special relativity is that a body's mass increases when it is moving, by the factor 1/√(1−v.v−u.u). Multiplying mass 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.

Elsewhere, I'll use 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.

Valid CSSValid HTML 4.01 Written by Eddy.