iSoul In the beginning is reality.

One-way transformations

The transformations associated with Galileo and Lorentz are two of several transformations between frames of reference. A two-way transformation assumes that the one-way speed of light is a constant, whereas a one-way transformation does not. Here are the available transformations:

Galilean transformation:  {x}' \mapsto x-vt;\; \; {t}' \mapsto t.

Dual Galilean transformation:  {x}' \mapsto x;\; \; {t}' \mapsto t-wx.

These could be combined with a selection factor κ of zero or one:

{x}' \mapsto x - \kappa vt;\; \; {t}' \mapsto t-(1-\kappa )wx.

General Lorentz transformation (see here): {x}' \mapsto \gamma (x-vt);\; \; {t}' \mapsto \gamma (t-kvx)

with \gamma =\left (1-\frac{v^{2}}{c^{2}} \right )^{-1}  and where k = 1/c² for the Lorentz transformation.

General dual Lorentz transformation:  {x}' \mapsto \lambda (x-kwt);\; \; {t}' \mapsto \lambda (t-wx)

with \lambda =\left(1-\frac{w^{2}}{\cent^{2}} \right)^{-1}.

Modify the Lorentz transformations with k1 and k2 such that k = (k1 + k2 )/2:

{x}' \mapsto \gamma (x-vt);\; \; {t}' \mapsto \gamma (t-k_{1}vx)  and  {x}' \mapsto \lambda (x-k_{2}wt);\; \; {t}' \mapsto \lambda (t-wx).

In this way, the constancy of the two-way speed of light is preserved but the appearance of instantaneous light would be allowed either in transmission or reception with k1 = 0 and k2 = 2/c² or vice versa.

Invariant intervals

Let’s begin with the space-time invariant interval r´² − ct´² = r² − ct². Then let us solve the equations:

r´ = Ar + Bt and t´ = Cr + Dt.

r´ = 0 = Ar + Btr = −tB/A = vt where v = −B/A {or} B = −Av

r´ = Ar + Bt = A(rvt)

A²(rvt)² − c²(Cr + Dt)² = r² − c²t²

A²r² − 2A²vrt + A²v²t² − C²c²r² − 2CDc²rtD²c²t² = r² − c²t²

⇒ (A² − C²c²)r² = r² {or} A² − c²C² = 1

⇒ (A²v² − D²c²)t² = −c²t² {or} D²c² − A²v² = c²

⇒ (2A²v + 2CDc²)rt = 0 {or} CDc² = −A²v

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2D light clock

The famous Michelson-Morley experiment used what could be described as a 2D light clock since their interferometer combined two light clocks at right angles. Their hypothesis was that this would show the Earth moving through the aether, but they failed to detect any motion. Einstein explained this failure as a feature of relativity. In other words, the expected difference between the two light clocks was “corrected” by relativity.

The reasoning of the Michelson-Morley experiment went like this:

Light is sent from the source and propagates with the speed of light c in the aether. It passes through the half-silvered mirror at the origin at T = 0. The reflecting mirror is at that moment at distance L (the length of the interferometer arm) and is moving with velocity v. The beam hits the mirror at time T1 and thus travels the distance cT1. At this time, the mirror has traveled the distance vT1. Thus cT1 = L + vT1 and consequently the travel time T1 = L / (cv). The same consideration applies to the backward journey, with the sign of v reversed, resulting in cT2 = LvT2 and so T2 = L / (c + v). The longitudinal travel time T|| = T1 + T2 is:

T longitudinal

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Harmonic arithmetic

This revised post follows up on harmonic addition mentioned in a previous post here.

Harmonic arithmetic is an inverse arithmetic. It is based on an automorphism that interchanges the zero with the infinite and the greater-than-one with the less-than-one: 0 ↔ ∞ and x ↔ 1/x. So zero becomes the new inaccessible number and infinity becomes the new additive unit. That is,

xy := g−1(g(x) + g(y))

xy := g−1(g(x) ∙ g(y))

etc., where g(x) = 1/x with x ≠ 0.

Regular and harmonic arithmetic are isomorphic with ordinary addition exchanged with harmonic multiplication and ordinary multiplication exchanged with harmonic division. Harmonic arithmetic is like counting down from infinity, in which an increment of one reduces the amount slightly.

Harmonic addition is defined as a power operation:

O_{-1}(a_{1},a_{2},...,a_{n})\equiv \left ( \sum_{i=1}^{n}a_{i}^{-1} \right )^{-1} ,\,\, a_{i} \neq 0

with the understanding that (1/0) → ∞ and (1/∞) → 0. An intermediate value may be zero but not a final value.

Simple harmonic addition is thus defined as:

x \oplus y = \left ( \frac{1}{x} + \frac{1}{y} \right )^{-1} \equiv \frac{xy}{x+y}; \,\, x, y \neq 0

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Lorentz in spacetime and timespace

This post builds on the previous posts here and here, and follows the approach of J.-M. Levy here.

First, consider the Lorentz transformation in spacetime along the x axis per Levy’s section III:

Let us now envision two frames in ‘standard configuration’ with having velocity v with respect to K and let x, t (resp. x´, t´) be the coordinates of event M in the two frames. Let O and be the spatial origins of the frames; O and cöıncide at time t = = 0.

Here comes the pretty argument: all we have to do is to express the relation

OM = OO´ + O´M (equation 5)

between vectors (which here reduce to oriented segments) in both frames.

In K, OM = x, OO´ = vt and O´M seen from K is /γ since is O′M as measured in . Hence a first relation is:

x = vt + /γ (equation 6)

In , OM = x/γ since x is OM as measured in K, OO′ = vt´ and O′M = . Hence a second relation:

x/γ = vt´+ x´ (equation 7)

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Light clocks with multidimensional time

A previous post on this subject is here. One reference for this post is V. A. Ugarov’s Special Theory of Relativity (Mir, 1979).

A light clock is a device with an emission-reflection-reception cycle of light that registers the current time and stance in units of cycle length and duration. Consider two identical light clocks, at first in their reference frames at rest, K, K´. Then, as the light clock in K´ moves transversely relative to K with uniform motion at velocity v (right), from K one observes the following:

transverse light clock

The illustration above shows one cycle length of the light path (i.e., wavelength), X, on the left and one cycle duration (i.e., period), T, on the right at rest in reference frames K, K´. For the reference frame K´, in motion relative to reference frame K, call the arc length of one cycle of the light path x. Call the distance between the beginning and ending points of one cycle x. For the reference frame K´ relative to reference frame K, call the arc time of one cycle of the light path t. Call the distime between the beginning and ending instants of one cycle t.

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Clock-rods

A clock-rod is a linear or planar clock with a parallel rod attached to it. A mechanical or electronic clock-rod might look like this:

clock-rod

Three clock-rods mutually perpendicular would measure length and duration in all directions.

A light clock-rod is conceptually like this:

light clock-rod

The clock and rod are parallel to each other so that parallel or perpendicular motion would change either the measurements of either the clock or the rod but not both. A complete harmonic cycle is not affected by motion:

light clock moving longitudinally

The +vt/2 increased distance of the first half-cycle is offset by the −vt/2 decreased distance of the second half-cycle. Likewise for the times +2d/v and −2d/v.

Kinematic proofs

Displacement with time: displacement s, time t, velocities v1 and v, acceleration a:

To prove: v = v1 + at

a = (vv1) / t    by definition

at = (vv1)     multiply by t

v = v1 + at

To prove: s = v1t + ½ at2

vavg = s / t          by definition

vavg = (v1 + v) / 2        by definition

s / t = (v1 + v) / 2       combining these two

s = (v1 + v) t / 2         multiply by t

s = (v1 + (v1 + at)) t / 2       from above

s = (2v1 + at) t / 2

s = v1t + ½ at2

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Space as time and time as space

Galileo parabola

Galileo used the length of uniform motion as a measure of duration, i.e., time (Dialogues Concerning Two New Sciences Tr. by Henry Crew and Alfonso de Salvio, 1914):

Accordingly we see that while the body moves from b to c with uniform speed, it also falls perpendicularly through the distance ci, and at the end of the time-interval bc finds itself at the point i. p.199

Without getting into the details of the Figure 108, notice the shift of language: “the body moves from b to c” [i.e., a length-interval], then “the time-interval bc“. Galileo uses a length interval to measure a time-interval, which is justified since the motion is “with uniform speed”.

Let there be a ball dropped out the window by a passenger on a train in uniform motion. Consider the following four scenarios, in which the length or duration of a uniform motion is measured: (1) looking down above the moving ball, measuring the length of fall; (2) looking down above the moving ball, measuring the (uniform) duration of fall; (3) looking from the side, measuring the length of motion in two dimensions; and (4) looking from the side, measuring the (uniform) duration of motion in two dimensions.

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Metric postulates for time geometry

Geometry was developed by the ancient Greeks in the language of length, but it is an abstraction that may be applied to anything that conforms to its definitions and axioms. Here we apply it to duration. We will use Brossard’s “Metric Postulates for Space Geometry” [American Mathematical Monthly, Vol. 74, No. 7, Aug.-Sep., 1967, pp. 777-788], which generalizes the plane metric geometry of Birkhoff to three dimensions. First, Brossard:

Primitive notions. Points are abstract undefined objects. Primitive terms are: point, distance, line, ray or half-line, half-bundle of rays, and angular measure. The set of all points will be denoted by the letter S and some subsets of S are called lines. The plane as well as the three-dimensional space shall not be taken as primitive terms but will be constructed.

Axioms on points, lines, and distance. The axioms on the points and on the lines are:

E1. There exist at least two points in S.

E2. A line contains at least two points.

E3. Through two distinct points there is one and only one line.

E4. There exist points not all on the same line.

A set of points is said to be collinear if this set is a subset of a line. Two sets are collinear if the union of these sets is collinear. The axioms on distance are:

D1. If A and B are points, then d(AB) is a nonnegative real number.

D2. For points A and B, d(AB) = 0 if and only if A = B.

D3. If A and B are points, then d(AB) = d(BA).

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