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 (length) 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 lenticity (matrix forms omitted).

1) Let us consider two inertial reference timeframes P and P´. The reference timeframe P´ moves relative to P with lenticity u along the t₁ ≡ t axis. We know that the coordinates t₂ and t₃ perpendicular to the lenticity 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 x´ = f_x(x; t) and t´ = f_t(x; t) in the reference timeframe P´.

From translational symmetry of length and duration, we conclude that the functions f_x(x, t) and f_t(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´₁ – t´₂ = f_t(x₁ – x₂, t₁ – t₂),     x´₁ – x´₂ = f_x(x₁ – x₂, t₁ – t₂).          (1)

Because Eq. (1) must be valid for any two events, the functions f_x(x; t) and f_t(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 lenticity, u.

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

x´ = 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´ = (t – ux),          (7)
x´ = (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 lenticity u₁ and a reference timeframe P´´ moving relative to P´ with lenticity u₂. Then

t´´ = λ(–u₂) (t´ – u₂x´),     t´´ = λ(u₁) (t – u₁x),
x´´ = λ(u₂) (F(u₂)t´ + x´),     x´ = λ(u₁) (F(u₁)t + x).          (10)

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

t´´ = λ(–u₂)λ(–u₁) [(1 – F(u₁)u₂)t´ – (u₁ + u₂)x´)],
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) u₁ ⇒ u₂ / 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 lenticity u, i.e.

F(u) = u/b;          (15)

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

t´ = u (t – ux);    x´ = 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 lenticity u, and the second transformation with lenticity –u. The equations are similar to Eqs. (10):

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

Substituting x´ and t´ 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 duration symmetry, the function u must depend only on the absolute value of lenticity 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:

t´ = λ(u) (t – ux),     x´ = λ(u) (tu/b + x).          (24)

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

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

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

x´ = λ(u) (t – ux),     x´ = λ(u) (–tuc² + x).          (27)

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

x´ = λ(1/v) (t – x/v),     x´ = 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 lenticity c in one reference timeframe, it also moves with lenticity c in any other reference timeframe, i.e. if t = cx then t´ = cx´. Thus the parameter c is the invariant speed.