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 + Bt → r = −tB/A = vt where v = −B/A {or} B = −Av

⇒ r´ = Ar + Bt = A(r − vt)

⇒ A²(r − vt)² − c²(Cr + Dt)² = r² − c²t²

⇒ A²r² − 2A²vrt + A²v²t² − C²c²r² − 2CDc²rt − D²c²t² = 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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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 / (c − v). The same consideration applies to the backward journey, with the sign of v reversed, resulting in cT₂ = L − vT₂ and so T₂ = L / (c + v). The longitudinal travel time T_|| = T₁ + T₂ is:

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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 K´ 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 O´ be the spatial origins of the frames; O and O´ cöıncide at time t = 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 x´/γ since x´ is O′M as measured in K´. Hence a first relation is:

x = vt + x´/γ (equation 6)

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

x/γ = vt´+ x´ (equation 7)

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

To prove: v = v₁ + at

a = (v – v₁) / t    by definition

⇒ at = (v – v₁)     multiply by t

⇒ v = v₁ + at

To prove: s = v₁t + ½ at²

v_avg = s / t          by definition

v_avg = (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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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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