It is worth returning to my post on *Lorentz and Dual Lorentz transformations* in order to make the point that the Lorentz transformation is sufficient for 3D time. Superluminal speeds are not required for 3D time, contrary to what others have said, such as *here*.

What *is* required for 3D duration (with 1D stance) is the use of measures relating measured changes in duration to independent changes in length. Pace, legerity, and rapidation are needed instead of speed, velocity, and acceleration. See glossary above for definitions of terms.

The conception of length space is then simplified to a one-dimensional distance from some conventional origin point, the stance. This conception of stance is analogous to the common conception of time, as something that flows on independently of us. That is what it means to be an independent variable: it’s out of our control.

Commuters may use a conception of stance (1D space) since they are focused on the distance to and from their residence or place of employment. The shortest distance in length space may not be the shortest duration of time, which is more important. A route that takes advantage of multi-dimensional time may take the shortest time, which in this context is a dependent variable, and can be controlled to some extent.

In duration-stance (3D duration + 1D length) duration becomes like length and length becomes like duration. But length space is still composed of lengths and distances, and duration space is still made of durations.

As a reference I repeat below the previously given derivation of the Lorentz transformation for both speed and pace:

There are many expositions of a Lorentz transformation, such as *here*. It is standard to present them in terms of two reference frames and their coordinate systems in uniform relative motion along the *x*-axis. Here we take the spatial axis to be the *r*-axis, which is parallel to the spatial axis of motion. Similarly, the temporal axis is taken to be the *t*-axis, which is parallel to the temporal axis of motion.

One aspect of the exposition here is that the notation is indifferent as to the existence of other dimensions. If they exist, they are orthogonal to the direction of motion, whether spatial or temporal, and their corresponding values are the same for both frames.

The two frames are differentiated by primed and unprimed letters. Their relative speed is *v*, and their relative pace is *u* = 1/*v*. The key difference between speed and pace is their independent unit of measure: speed is measured per unit of time, whereas pace is measured per unit of stance.

A Lorentz transformation requires what I’m calling a *characteristic (modal) rate*, in units of speed or pace, which is the same for all observers within a context such as physics or a mode of travel. The characteristic speed, *c*, or pace, *k*, is a maximum that depends on the context. In the context of physics, the characteristic rate is that of light traveling in a vacuum.

The trajectory of a reference particle (or probe vehicle) that travels at the characteristic rate follows these equations in the two frames:

speed: *r* = *ct* or* r/c = t* and *r′* = *ct′*, or *r′/c* = *t′*,

pace: *kr* = *t* or *r* = *t/ k* and

*=*

*k*r′*t′*or

*r′*=

*t′/*.

*k*__Lorentz transformation
__

This starts with the Galilean transformation and includes a factor, *γ*, in the transformation equation for the direction of motion, along with the characteristic rate:

speed (–): *r′* = *γ* (*r* − *vt*) = *γr* (1 – *v/c*) = *ct′* = *γ* (*ct* – *rv/c*) = *γt* (*c* − *v*),

pace (–): *r ′* =

*γ*(

*r*–

*t/u*) =

*γr*(1 –

*k/u*) =

*t*=

*′*/k*γ*(

*t/k*–

*rk/u*) =

*γt*(1/

*k*– 1/

*u*),

with equal values for the other corresponding primed and unprimed coordinates. The inverse transformations are then:

speed (+): *r* = *γ* (*r′* + *vt′*) = *γr′* (1 + *v/c*) = *ct* = *γ* (*ct′* + *r′v/c*) = *γt′* (*c* + *v*),

pace (+): *r* = *γ* (*r ′* +

*t*) =

*′*/u*γr′*(1 +

*k/u*) =

*t/k*=

*γ*(

*t*+

*′*/k*r*) =

*′*k/u*γt′*(1/

*k*+ 1/

*u*).

Multiply each corresponding pair together to get:

speed: *rr′* = *γ²**rr′* (1 – *v²/c²*) = *c²tt′* = *γ²tt′* (*c²* − *v²*),

pace: *rr´* = *γ²rr′* (1 – * k²/u²*) =

*tt*=

*′*/k²*γ²tt′*(1/

*– 1/*

*k*²*u²*).

Dividing out *rr′* yields:

speed: 1 = *γ*^{2} (1 – *v*^{2}/*c*^{2}),

pace: 1 = *γ*^{2} (1 – *k*^{2}/*u*^{2}).

Or dividing out *tt′* yields:

speed: *c*^{2} = *γ*^{2} (*c*^{2} – *v*^{2}),

pace: 1/*ç*^{2} = *γ*^{2} (1/*k*^{2} – 1/*u*^{2}).

Either way, solving for *γ* leads to:

speed: *γ* = (1 – *v*^{2}/*c*^{2})^{–1/2},

pace: *γ* = (1 – *k*^{2}/*u*^{2})^{–1/2}.

which is the standard Lorentz transformation and applies only if |*v*| < |*c*| or |*u*| > |*k*|.