The Lorentz Transformation

It used to be supposed that, if two observers were in a state of steady motion relative to one another, applying a translation to what one sees at any given moment would convert it to what the other sees, with the translation growing linearly with time. Thus the space-time co-ordinate system each uses is obtained from the other's by a linear transformation called a shear, across the time axis.

This naïve assumed relationship became untennable when physics revealed that the speed of light, c, is the same for all observers. The transformation of co-ordinates which replaces it lies at the heart of special relativity.

The new transformation

Consider two observers, one of whom sees the other moving with velocity V.c; chose some point at some given moment of time; have each measure time from that moment and use the given point as origin of spatial co-ordinates – one's origin will be the position the other describes as V.c times time. Let each observer measure one spatial co-ordinate parallel to V and their others perpendicular to it.

Entangled co-ordinates

Let the variables used by the two frames for time and V-wards displacement be [t,x] and [T,X] for the two observers. Start by considering only events with the other spatial co-ordinates zero and suppose an arbitrary linear relation between [t,x] and [T,X]:

The speed of light is the same for both, so the sets of points described by c.t = x must also satisfy c.T = X; thus e+f.c = c.(a+b.c). Likewise, c.t = −x must correspond to c.T = −X, so e−f.c = −c.(a−b.c). Halving the sum and difference of these equations we get e = c.b.c and f.c = c.a. For V to have its claimed meaning, the set of positions at which X is zero must, with v as the magnitude of V, satisfy x = v.c.t, yielding 0 = X = (b.c +a.v).c.t for all t, whence b.c +a.v = 0, so b = −a.v/c, e = −a.v.c and we have T = a.(t −v.x/c), X = a.(x −v.c.t). Since the transformation must be symmetric between the two observers, aside from changing the sign of v, we must equally have t = a.(T +v.X/c) and x = a.(X +v.c.T). Substituting the former into the latter we find t = a.a.(t −v.x/c +v.x/c −v.v.t) = a.a.(1−v.v).t and x = a.a.(x −v.c.t +v.c.t −v.v.x) = a.a.(1−v.v).x and infer that a = 1/√(1−v.v). We can thus state the transformation of co-ordinates as

This is known as the Lorentz transformation, after the late Victorian scientist who pointed out that it would suffice to explain the non-result of the Michelson-Morley experiment – an attempt to detect our motion relative to the æther, or frame of reference in which Maxwell's equations take their simple form.

If we leave out the factors of √(1−v.v) and the v.x/c term in the first of these, we get the relationship previously presumed. Since c.v is the speed of relative movement, factor 1−v.v is almost indistinguishable from 1 for speeds tiny compared to c. Satellites in geostationary orbit have speeds of around 3 km/s or c/100000; the Earth's orbital speed about the Sun is about a factor of ten bigger and the Sun's motion relative to the cosmic microwave background is a further factor of about 10 bigger; our local group of galaxies, moving about twice as fast relative to that background, still only has v ≅ 1/480, yielding 1−v.v = 0.9999956, with square root 0.9999978, whose inverse is 1.0000022. Even such huge speeds (over 600 km/s) only produce a tiny perturbation: detecting the effects of more modest ones is a good deal harder.

To observe the v.x/c term's effect we would need accurately synchronized clocks far enough apart that, when viewed from a moving frame, the v.x/c term is not hidden by imprecision in the clocks. The Earth's diameter is about 1/24 light seconds and we can reasonably put sensitive equipment that far apart (we've managed a few cases of instruments further apart than that, but it's not yet easy or cheap); observed from low Earth orbit, with v around 1/40000, we would need to be able to discern microsecond discrepancies. Even this would have been difficult for the Victorians, and they couldn't put things into Earth orbit. Using the fastest-moving platforms that Victorian scientists could have hoped to use – trains, with c.v of order 100 km per hour, or around 30 m/s, so v is about one part in ten million – it would be necessary to discern discrepancies of order one part in a few hundred million. Thus the Lorentz transform's deviations from the prior theory were smaller than the sensitivity of existing measurements, allowing all the data, that had previously been understood as supporting the old theory, to stand equally in support of the new.

One consequence of the transformation is that the lengths of objects depend on velocities: if an object, of which some linear dimension is L when measured in its frame of rest, is observed moving at speed v.c parallel to the given dimension, then measurement of that linear dimension will yield L.√(1−v.v). [Proof: the two ends have positions X = 0 and X = L in its own frame of reference, for all T; these give values for x−v.t.c of 0 and L.√(1−v.v); for any given t, the two values of x thus differ by L.√(1−v.v).] This is known as the Lorentz-Fitzgerald contraction.

Equally, moving clocks must run slow by the same factor √(1−v.v). [Proof: using the clock's position to define X = 0, successive ticks happen with T = n.k for integer n and some fixed tick-interval k; the resulting values of t = (T +v.X/c)/√(1−v.v) are n.k/√(1−v.v) with tick-interval k/√(1−v.v), which exceeds k; so the clock appears to run slow, by the given factor.] This is known as time dilation. It implies that an unstable particle, decaying with some known half-life T in its rest frame, will decay with a slowed half-life T/√(1−v.v) when moving at speed v.c.

It is a remarkable tribute to the progress of measuring technology in the twentieth century that, at its start, none of the above discrepancies had the remotest chance of being measured (although the not totally unrelated precession of the orbit of Mercury had been measured, and was unexplained): while, by its end, all had been measured – and found to be in accord with the predictions of relativistic theory.

Sideways co-ordinates

Given that the effects of relative movement mess up our intuitions about independence between time and the spatial co-ordinate parallel to the relative motion, we have no immediate guarantee that our co-ordinates perpendicular to the motion aren't also entangled in the relationship between these two, nor that they are themselves unperturbed. So let us not even assume we know that distances from the x-and-X axis are unchanged, or that the two systems agree on which displacements are perpendicular to this axis (i.e. to V). However, our system is entirely symmetric under rotations about this axis, so it should suffice to consider one plane through the axis; we can probe the question by considering light trajectories in some such plane and the above tells us how the two frames are related to one another on the axis.

In the [t,x] frame, at t = 0, chose a plane in which the x-axis lies and, within that plane, use distance from the axis to define a co-ordinate y, positive on one side of the axis and negative on the other. For any value L for this y co-ordinate, consider the family of trajectories for flashes of light that pass through [c.t,x,y] = [0,0,L] either on their way to some point on the x-axis (at some positive t) or having been emitted from some point on the x-axis (at some negative t); these are parameterised by the angle, a, they make with the direction of V in the [x,y] plane and each has [x,y] = [0,L] +c.t.[Cos(a),Sin(a)], meeting the x-axis when 0 = y = L +c.t.Sin(a), so at [c.t,x] = −[1,Cos(a)].L/Sin(a) hence at [c.T,X] = −[1 +v.Cos(a), Cos(a) +v].L/Sin(a)/√(1−v.v). Since v is smaller than 1, we have some angle b with Cos(b) = v and Sin(b) = √(1−v.v) so that we can re-write this as

The sign of T at this start-point is the opposite of the sign of L/Sin(a), just as was the case for t. In the second frame, consider the event that is the meeting-point of all our flash trajectories, corresponding to [c.t,x,y] = [0,0,L]; we don't even know (although we may reasonably expect) that this has [c.T,X] = [0,0]. However, we can at least construct a plane of constant T through this event and the X axis. Define a co-ordinate Y in this plane that measures distance from the X-axis, positively on one side and negatively on the other, making this choice in such a way as to give the meeting-point's Y co-ordinate the same sign as L. (We are only interested in the case L ≠ 0, since we already dealt with the L = 0 case above; and this puts the meeting-point off the x-axis, which is the X axis; hence the meeting-point does also have a non-zero Y co-ordinate.) Let the meeting-point have [c.T,X,Y] = [K,J,H]; the family of light trajectories through it and the X-axis is then given by [X,Y] = [J,H] +(c.T−K).[Cos(e),Sin(e)], parameterised by angle e that this frame deems the light's velocity to make with V. Each such meets the X-axis when 0 = Y = H +(c.T−K).Sin(e), giving c.T = K −H/Sin(e) and X = J +(c.T−K).Cos(e) = J −H.Cos(e)/Sin(e); and the set of such events must be equal to the set of [c.T,X] events given above, parameterised by a.

Applying X = J +(c.T−K).Cos(e) to the earlier −[c.T,X] values, and scaling by Sin(a).Sin(b)/L, we obtain

−(Cos(a) +Cos(b))
= X.Sin(a).Sin(b)/L
= (J −K.Cos(e)).Sin(a).Sin(b)/L −(1 +Cos(b).Cos(a)).Cos(e)
= Sin(a).Sin(b).J/L −(1 +Cos(b).Cos(a) +Sin(a).Sin(b).K/L).Cos(e), whence
Sin(a).Sin(b).J/L +Cos(a) +Cos(b)
= (1 +Cos(b).Cos(a) +Sin(a).Sin(b).K/L).Cos(e), so
Cos(e)
= (Sin(a).Sin(b).J/L +Cos(a) +Cos(b))/(Sin(a).Sin(b).K/L +Cos(a).Cos(b) +1)

in which we can vary a freely yet must obtain a value between −1 and 1; this constrains J/L and K/L. Next, obtain:

H
= (K−c.T).Sin(e)
= (K +(1 +Cos(a).Cos(b)).L/Sin(a)/Sin(b)).√((Sin(a).Sin(b).K/L +Cos(a).Cos(b) +1)2 −(Sin(a).Sin(b).J/L +Cos(a) +Cos(b))2)/(Sin(a).Sin(b).K/L +Cos(a).Cos(b) +1)
= L.√((Sin(a).Sin(b).K/L +Cos(a).Cos(b) +1)2 −(Sin(a).Sin(b).J/L +Cos(a) +Cos(b))2)/Sin(a)/Sin(b)
= √((Sin(a).Sin(b).K +L.Cos(a).Cos(b) +L)2 −(Sin(a).Sin(b).J +L.Cos(a) +L.Cos(b))2)/Sin(a)/Sin(b)
= √((Sin(a).Sin(b).K +L.Cos(a).Cos(b) +L −Sin(a).Sin(b).J −L.Cos(a) −L.Cos(b)).(Sin(a).Sin(b).K +L.Cos(a).Cos(b) +L +Sin(a).Sin(b).J +L.Cos(a) +L.Cos(b)))/Sin(a)/Sin(b)
= √((Sin(a).Sin(b).(K −J) +L −L.Cos(a) −L.Cos(b) +L.Cos(a).Cos(b)).(Sin(a).Sin(b).(K +J) +L +L.Cos(a) +L.Cos(b) +L.Cos(a).Cos(b)))/Sin(a)/Sin(b)
= √((Sin(a).Sin(b).(K −J) +L.(1 −Cos(a)).(1 −Cos(b))).(Sin(a).Sin(b).(K +J) +L.(1 +Cos(a)).(1 +Cos(b))))/Sin(a)/Sin(b)
= √( Sin(a).Sin(a).Sin(b).Sin(b).(K −J).(K +J) +L.(K −J).Sin(a).(1 +Cos(a)).Sin(b).(1 +Cos(b)) +L.(K +J).Sin(a).(1 −Cos(a)).Sin(b).(1 −Cos(b)) +L.L.(1 +Cos(a)).(1 −Cos(a)).(1 +Cos(b)).(1 −Cos(b)) )/Sin(a)/Sin(b)

in the last term of which we can exploit (1 +Cos(u)).(1 −Cos(u)) = 1 −Cos(u).Cos(u) = Sin(u).Sin(u) for u in {a, b}:

= √( Sin(a).Sin(a).Sin(b).Sin(b).(K −J).(K +J) +L.(K −J).Sin(a).(1 +Cos(a)).Sin(b).(1 +Cos(b)) +L.(K +J).Sin(a).(1 −Cos(a)).Sin(b).(1 −Cos(b)) +L.L.Sin(a).Sin(a).Sin(b).Sin(b) )/Sin(a)/Sin(b)
= √(K.K −J.J +L.L +L.(K −J).(1 +Cos(a)).(1 +Cos(b))/Sin(a)/Sin(b) +L.(K +J).(1 −Cos(a)).(1 −Cos(b))/Sin(a)/Sin(b))

Now, this holds true for all e, hence for all a, so its value must be independent of a. As Cos(a) approaches 1, (1 −Cos(a))/Sin(a) is well approximated by −Cos'(a)/Sin'(a) = Sin(a)/Cos(a), which reaches 0 when Cos(a) is 1; likewise, near Cos(a) = −1, (1 +Cos(a))/Sin(a) ≈ Cos'(a)/Sin'(a) = −Sin(a)/Cos(a) also tends to zero. In each of these cases, one of the last two terms in the √(…) vanishes as Sin(a) approaches 0, but the other must grow without bound unless (given that v is smaller than 1, so Cos(b) isn't ±1) the relevant one of K −J and K +J is zero. For H to be independent of a, it cannot grow without bound at either end of a's range, so we must in fact have K +J = 0 = K −J, whence K = 0 = J. We then obtain H = L,

Cos(e)
= (Cos(a) +Cos(b))/(Cos(a).Cos(b) +1)
Sin(e)
= √(Cos(a).Cos(a).Cos(b).Cos(b) +2.Cos(a).Cos(b) +1 −Cos(a).Cos(a) −2.Cos(a).Cos(b) −Cos(b).Cos(b))/(Cos(a).Cos(b) +1)
= √(1 −Cos(a).Cos(a) −Cos(b).Cos(b) +Cos(a).Cos(a).Cos(b).Cos(b))/(Cos(a).Cos(b) +1)
= √((1 −Cos(a).Cos(a)).(1 −Cos(b).Cos(b)))/(Cos(a).Cos(b) +1)
= Sin(a).Sin(b)/(1 +Cos(a).Cos(b))

(Technically, the sign of Sin(e) is ambiguous – but a cursory inspection of the geometry of the situation requires it to have the same sign as Sin(a), which is what this formula gives, taking the positive sign for √(1−v.v) in Sin(b)'s specification.)

Now, H = L with J = 0 = K, independent of our choice of initial L, tells us that the y-axis and the Y-axis coincide and y = Y on this axis. It then follows (by simple linear geometry in the two frames of reference, assuming lines deemed straight by either are by the other also) that y and Y coincide everywhere and that, likewise, all other spatial directions perpendicular to V are the same for both frames of reference.

The Doppler shift

If we observe a clock from a position with fixed value of x, and its ticks are evidenced by light flashes, the interval between moments at which we see these flashes will change as the clock passes us. This is known as the Doppler shift. Chose [X,T] origin so that the clock is at X = 0 and ticks at moments T = n.k for integer n and some tick-interval k. Chose [x,t] origin so that our observer is at x = 0 and the clock passes the observer at t = 0. [In principle, the clock could pass the observer at a moment when it does not tick; this will make no practical difference, but for now presume that we adjust the clock by whatever fraction of a tick it takes to ensure it does tick at the moment of passing the observer.]

For t < 0, which is equally T < 0, the light flashes we see are from light travelling forwards from an [X,T] = [0,n.k] with n < 0; forward light from such events follows paths [x,t] = u.[c,1] + [v.c,1].n.k/√(1−v.v) for u varying and thus reaches x = 0 when u = −n.k.v/√(1−v.v) yielding t = n.k.(1−v)/√(1−v.v) = n.k.√((1−v)/(1+v)). For t > 0, the light flashes we see are from light traveling backwards from [X,T] = [0,n.k] with n > 0 along trajectories [x,t] = u.[−c,1] +[v.c,1].n.k/√(1−v.v) which hit x = 0 when u = n.v.k/√(1−v.v) yielding t = n.k.(1+v)/√(1−v.v) = n.k.√((1+v)/(1−v)). This is exactly the factor we would get for the t < 0 case if v were negated, so we can just say that the time interval between observations of ticks is k.√((1−v)/(1+v)) when a clock with tick-interval k is observed approaching at speed v, with recession expressed as approaching at negative speed.

Thus, while the clock (once we infer the time interval between its ticks, by correcting for the time taken by light pulses in reaching us) runs slow by the factor √(1−v.v), we observe ticks from it at a rate which, as it approaches, is fast by the factor √((1+v)/(1−v)) but, once it has passed, is slow by the inverse of this factor. As for clocks, so for any other periodic process in time: if some constituent of a distant star's photosphere perturbs its spectrum (from its natural black-body form) at some given frequency f (corresponding to 1/k) then the spectrum of the star, as we observe it, will be likewise perturbed at a frequency f.√((1+v)/(1−v)) if the star is moving towards us at speed v.c. For v > 0, a star actually moving towards us, this frequency is > f; the increase in frequency is called a blue-shift because blue light is at the high-frequency end of the spectrum of visible light. For v < 0, meaning the star is really moving away from us at (positive) speed −v.c, the frequency is < f; the decrease is called a red-shift because red is at the low-frequency end of the visible spectrum.

Classifying displacements

Notice that, in the [X,T] or [x,t] plane, X = 0 iff x = v.c.t and T = 0 iff c.t = x.v. Thus, for any position whose spatial co-ordinates, collectively, have smaller magnitude than c times its time co-ordinate, there is some velocity v.c = x/t in whose frame the given position is at the origin; likewise, for any position whose spatial magnitude is greater than c times time co-ordinate, there is a choice of velocity v.c = c.c.t/x in whose frame the given position has zero time-coordinate. In the first case, there is only one such frame of reference; but in the second case, relative to a frame which sees the position have zero time co-ordinate, any velocity perpendicular to the spatial position vector yields another frame which also sees the position have zero time co-ordinate. Thus, insisting that the lines x = ±c.t coincide with their analogues in any other frame of reference (i.e. that light goes at the same speed in all frames) leads to the conclusion that:

We may, thus, divide displacements in space-time into three classes:

light-like
the ratio of spatial to temporal components has magnitude c, regardless of frame of reference;
space-like
there are frames of reference which deem events separated by such a displacement simultaneous;
time-like
there is a frame of reference which deems events separated by such a displacement to happen in the same place – and we can sub-divide this type into forward and backward variants according as the given frame of reference deems the start of the displacement to happen before or after its end.

Hyperbolic Rotation

Notice that

c.c.T.T −X.X
= ( (c.t −v.x).(c.t −v.x) −(x −v.c.t).(x−v.c.t) )/(1−v.v)
= (c.c.t.t −2.v.c.t.x +v.v.x.x −x.x +2.x.v.c.t −v.v.c.c.t.t) / (1−v.v)
= c.c.t.t −x.x

so this combination of our co-ordinates is an invariant, just as rotating a (cartesian orthonormal) spatial co-ordinate system preserves the sum of squares of co-ordinates, i.e. the squared radius. In the spatial case, the change of co-ordinates is a rotation and preservation of the radial parameter is a characteristic property: so we may ask what analogue of rotation we can find for our invariant under the Lorentz transformation. For a rotation, we would have seen [Y,Z] = [y.Cos(A) −z.Sin(A), y.Sin(A) +z.Cos(A)] and described the rotation by the angle A. The sum of squares of Cos and Sin is 1 for all angles. Their hyperbolic analogues cosh and sinh (which can be obtained from looking at Cos and Sin of imaginary angles) are defined by

and the difference of their squares is always one: for any x,

cosh(x).cosh(x) −sinh(x).sinh(x)
= (exp(2.x) +2 +exp(−2.x) −exp(2.x) +2 −exp(−2.x)) / 4
= 1

making cosh and sinh natural substitutes for Cos and Sin. Indeed, a transformation of form

has Y.Y −Z.Z = y.y −z.z, as is easily verified. Using Y = c.T, y = c.t, Z = −X, z = −x puts the Lorentz transformation in this form with cosh(b) = 1/√(1−v.v) and sinh(b) = v/√(1−v.v). The ratio (: sinh(b)/cosh(b) ←b :) = tanh, analogous to the usual Tan of angles, gives us our speed as c.tanh(b), since v = sinh(b)/cosh(b) = tanh(b). Thus the Lorentz transformation is a hyperbolic rotation rather than the cross-time shear tacitly assumed by Galileo and Newton.

In these terms, the Doppler shift factor √((1+v)/(1−v)) for a source approaching the observer at speed v.c with v = tanh(b) becomes

√((1+tanh(b))/(1−tanh(b)))
= √((cosh(b) +sinh(b))/(cosh(b) −sinh(b)))
= √(exp(b) / exp(−b))
= exp(b)

so if we divide the observed frequency of a spectral line by the frequency it was at when produced, and take the log of the result, we get the hyperbolic angle whose tanh, when multiplied by c, gives the speed, towards us, of the source of the observed radiation. Equally, this can be computed as the difference in log between the two frequencies (each divided by some common unit). Furthermore, exp(b) = cosh(b) +sinh(b) = (1+v)/√(1−v.v) is just our Doppler shift ratio, so we can compute b directly from v as: b = log((1+v)/(1−v))/2; as long as v is smaller than 1, this is well-defined (and has the same sign as v).

Adding velocities

Now suppose we have – in addition to our first frame with co-ordinates [x, y, z, c.t] and second with [X, Y, Z, c.T] as before – a third moving at speed u.c in the direction of increasing X, as observed by the second frame. We can take as given that it has the same [y, z] = [Y, Z] side-ways co-ordinates as the previous two. Let its X-ward co-ordinate and time co-ordinate be W and S, respectively; so

and we can substitute our earlier transoformation into this; to obtain:

c.S
= (c.T −u.X)/√(1−u.u)
= (c.t −v.x −u.(x −v.c.t))/√(1−v.v)/√(1−u.u)
= ((1 +u.v).c.t −(u+v).x)/√(1−v.v)/√(1−u.u)
W
= (X −u.c.T)/√(1−u.u)
= (x −v.c.t −u.(c.t −v.x))/√(1−v.v)/√(1−u.u)
= ((1 +u.v).x −(u+v).c.t)/√(1−v.v)/√(1−u.u)

in which we're clearly going to have to shift out the factors of (1 +u.v), leaving (u+v)/(1 +u.v) in the rôle of relative speed. Indeed, this does yield

1 −(u+v).(u+v)/(1 +u.v)/(1 +u.v)
= ((1 +u.v).(1 +u.v) −(u+v).(u+v))/(1 +u.v)/(1 +u.v)
= (1 +2.u.v +u.u.v.v −u.u −2.u.v −v.v)/(1 +u.v)/(1 +u.v)
= (1 −u.u −v.v +u.u.v.v)/(1 +u.v)/(1 +u.v)
= (1 −u.u).(1 −v.v)/(1 +u.v)/(1 +u.v)

whose square root is exactly the factor we need to make the above formulae match the Lorentz transformation. I trust readers can see that this involved rather messy algebra; in preparation for showing how this works out when v = tanh(b) and u = tanh(a), let's just learn a little more about the hyperbolic functions cosh and sinh:

cosh(a).cosh(b) +sinh(a).sinh(b)
= (exp(a+b) +exp(a−b) +exp(b−a) +exp(−a−b) +exp(a+b) −exp(a−b) −exp(b−a) +exp(−b−a)) / 4

in which the middle terms from cosh(a).cosh(b) cancel those from sinh(a).sinh(b), leaving the duplicated outer terms,

= cosh(a+b)
cosh(a).sinh(b) +sinh(a).cosh(b)
= (exp(a+b) +exp(b−a) −exp(a−b) −exp(−a−b) +exp(a+b) +exp(a−b) −exp(b−a) −exp(−a−b)) / 4

in which, once more, the middle terms cancel, leaving

= sinh(a+b)

So now let's look at c.S and W in terms of c.t and x:

c.S
= c.T.cosh(a) −sinh(a).X
= (c.t.cosh(b) −sinh(b).x).cosh(a) −sinh(a).(x.cosh(b) −sinh(b).c.t)
= c.t.cosh(b).cosh(a) −sinh(b).cosh(a).x −sinh(a).x.cosh(b) +sinh(a).sinh(b).c.t
= c.t.(cosh(b).cosh(a) +sinh(a).sinh(b)) −(sinh(b).cosh(a) +cosh(b).sinh(a)).x
= c.t.cosh(a+b) −sinh(a+b).x
W
= X.cosh(a) −sinh(a).c.T
= (x.cosh(b) −sinh(b).c.t).cosh(a) −sinh(a).(c.t.cosh(b) −sinh(b).x)
= x.cosh(b).cosh(a) +sinh(a).sinh(b).x −sinh(b).c.t.cosh(a) −sinh(a).c.t.cosh(b)
= x.cosh(a+b) −sinh(a+b).c.t

Thus, rather than adding velocities, to determine the velocity of a third observer relative to a first when we know their velocities relative to a second, we must now combine hyperbolic rotations: for parallel velocities this amounts to adding the parabolic angles. Thus it makes more sense to discuss hyperbolic angles than velocities.

Given a first frame of reference, a second with velocity c.v relative to it and a third with velocity c.u relative to the second, solve tanh(b) = v, tanh(a) = u and the velocity of the third with respect to the first will be

c.tanh(a+b)
= c.sinh(a+b) / cosh(a+b)
= c.(cosh(a).sinh(b) +sinh(a).cosh(b)) / (cosh(a).cosh(b) +sinh(a).sinh(b))
= c.(tanh(b) +tanh(a))/(1 +tanh(a).tanh(b))
= c.(u+v)/(1+u.v)

exactly as found, more laboriously, above. This, then, is the addition law for velocities: at speeds tiny compared to the speed of light (i.e. the distance travelled in a nano-second must be tiny compared to the foot) the factor 1+u.v will be so close to 1 that it will be hard to detect, so the addition law will be indistinguishable from the simple addition law that was expected by a shear-based transformation of co-ordinates. Thus (again) the new rule, though different from the old, is not contradicted by the large body of evidence previously collected in support of the old.

In particular, replacing velocities with hyperbolic angles implies that we should describe the expansion of the universe by the rate of change, with distance, of hyperbolic angle or, equivalently, of the log of the Doppler shift factor. This will be the usually quoted Hubble constant, 2.3 ± .2 atto Hz, divided by the speed of light, which yields 73 pico / light year; the hyperbolic angle changes by 1 in 13.8 giga light years, give or take 10%.

See also: constant acceleration.


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