iSoul Time has three dimensions

Dual Lorentz Transformation

Victor Yakovenko has a derivation (see here) of the Lorentz Transformation (LT) in which he uses “only the equivalence of all inertial reference frames and the symmetries of space and time.” Because of the use of (spatial) reference frames and velocity, this is not completely symmetric. As we have seen, there is a dual Lorentz Transformation. Let us follow Yakovenko’s derivation but with reference timeframes and legerity (matrix forms omitted).

1) Let us consider two inertial reference timeframes P and P´. The reference timeframe P´ moves relative to P with legerity u along the t1t axis. We know that the coordinates t2 and t3 perpendicular to the legerity are the same in both reference timeframes: t2 = t2´ and t3 = t3´. So, it is sufficient to consider only transformation of the coordinates x and t from the reference timeframe P to = fx(x; t) and t´ = ft(x; t) in the reference timeframe P´.

From translational symmetry of space and time, we conclude that the functions fx(x, t) and ft(x, t) must be linear functions. Indeed, the relative distances between two events in one reference timeframe must depend only on the relative distances in another timeframe:

 t´1t´2 = ft(x1x2, t1t2),     x´12 = fx(x1x2, t1t2).          (1)

Because Eq. (1) must be valid for any two events, the functions fx(x; t) and ft(x; t) must be linear functions. Thus

t´ = At + Bx,           (2)
x´ = Ct + Dx,           (3)

where A, B, C, and D are some coefficients that depend on the legerity, u.

2) The origin of the reference timeframe P´ has the coordinate t´ = 0 and moves with legerity u relative to the reference timeframe P, so that t = ux. Substituting these values into Eq. (2), we find B = –uA. Thus, Eq. (2) has the form

t´ = A(tux);          (5)

so we need to find only three unknown functions A, C, and D of u.

3) The origin of the reference timeframe P has the coordinate t = 0 and moves with legerity –u relative to the reference timeframe P´, so that = –ut´. Substituting these values in Eqs. (2) and (3), we find D = A. Thus, Eq. (3) has the form

= Ct + Ax = A(Ft + x);          (6)

where we introduced the new variable F = C/A.

Let us change to the more common notation A = λ. Then Eqs. (5) and (6) have the form

t´ = (tux),          (7)
= (Ft + x).         (8)

Now we need to find only two unknown functions λ(u) and F(u) of u.

4) A combination of two Lorentz transformations also must be a Lorentz transformation. Let us consider a reference timeframe P´ moving relative to P with legerity u1 and a reference timeframe P´´ moving relative to P´ with legerity u2. Then

t´´ = λ(–u2) (u2),     t´´ = λ(u1) (tu1x),
x´´ = λ(u2) (F(u2) + ),     = λ(u1) (F(u1)t + x).          (10)

Substituting and from the second Eq. (10) into the first Eq. (10), we find

t´´ = λ(–u2)λ(–u1) [(1 – F(u1)u2) – (u1 + u2))],
x´´ = λ(–u2)λ(–u1) [(F(u1) + F(u2))t + (1 –  F(u2)u1)x].          (12)

For a general Lorentz transformation, the coefficients in front of t in Eq. (7) and in front of x in Eq. (8) are equal. Eqs. (12) must also satisfy this requirement:

1 – F(u1) u2 = 1 – F(u2) u1u2 / F(u2) = u1 / F(u1).          (14)

In the second Eq. (14), the left-land side depends only on u2, and the right-hand side only on u1. This equation can be satisfied only if the ratio u/F(u) is a constant b independent of legerity u, i.e.

F(u) = u/b;          (15)

Substituting Eq. (15) into Eqs. (7) and (8), we find

= u (t – ux);    = u (tu/b + x);          (16)

Now we need to find only one unknown function u, whereas the coefficient b is a fundamental constant independent of u.

5) Let us make the Lorentz transformation from the reference timeframe P to P´ and then from P´ back to P. The first transformation is performed with legerity u, and the second transformation with legality –u. The equations are similar to Eqs. (10):

t = λ(–u) ( + ux´);     x = λ(–u) (–t´u/b + ),
= λ(u) (tux);     = λ(u) (tu/b + x).          (18)

Substituting and from the first equation (18) into the second one, we find

t = λ(–u) λ(u) (1 + u2/b) x;    x = λ(–u) λ(u) (1 + u2/b) x.          (20)

Eq. (20) must be valid for any x and t, so

λ(–u) λ(u) = 1 / (1 + u2/b).          (22)

Because of the time symmetry, the function u must depend only on the absolute value of legerity u, but not on its direction, so λ(–u) = λ(u), and λ(–u) λ(u) = λ2(u) = 1 / (1 + u2/b).

6) Substituting the square root of Eq. (22) into Eqs. (16), we find the final expressions for the dual Lorentz transformation:

= λ(u) (tux),     = λ(u) (tu/b + x).          (24)

Eqs. (24) and (25) have one fundamental parameter b, which has the dimensionality of
legerity squared.

If b < 0, we can write it as b = –1/c2.          (26)

Then Eqs. (24) become the dual Lorentz transformation:

= λ(u) (tux),     = λ(u) (–tuc2 + x).          (27)

With the substitution of 1/v for u and γ for 1/λ, (27) becomes

= λ(1/v) (tx/v),     = λ(1/v) (–tc2/v + x),

and (22) becomes

λ(–1/v) λ(1/v) = 1 / (1 + c2/v2) = λ2(1/v).

It is easy to check from Eqs. (27) that, if a particle moves with velocity c in one reference timeframe, it also moves with legerity c in any other reference timeframe, i.e. if t = cx then = cx´. Thus the parameter c is the invariant speed.

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