## Monday, February 22, 2021

Notes to a video lecture on http://www.unizor.com

Relativity - Transformation of Space-Time Coordinates
(notes to item #3 of Einstein's "Electrodynamics")

The following is an example of how a system of linear equations can be used to derive formulas of special theory of relativity. Albert Einstein has derived these formulas in his "Electrodynamics" in a more physical, more intuitive way. The following is pure mathematics and, as such, causes much less problems in understanding.

Assumptions:
1. Assume we have two systems of coordinates, one stationary with coordinates {X,T} (assuming for simplicity all the movements will occur in one space dimension along X-axis and one time dimension) and another with coordinates {x,t} moving along the X-axis with constant speed V.
2. Assume that at time T=0 systems coincide (i.e. X=0, t=0 and x=0).
3. Assume that the speed of something, as measured in both stationary and moving systems is the same and equal to C regardless of the direction of the movement (that "something" is the light in vacuum, but it's physical characteristics are unimportant)
4. Assume further that we are looking for linear orthogonal (i.e. preserving the distance between points and angles between vectors) transformation of coordinates from (X,T) to (x,t) that satisfies the above criteria. What would this transformation be?

Linear transformation from {X,T} system to {x,t} system should look like this:
x = pX + qT
t = rX + sT
where p, q, r and s are 4 unknown coefficients of transformation, which we are going to determine by constructing a system of 4 linear equations with them.

We should not add any constants into above transformations since {X=0,T=0} should transform into {x=0,t=0}.

A. Notice that the property of orthogonality is needed to preserve geometry (i.e. no deformation) and, therefore, to preserve the form of all physical equations of motion. As it is well known, orthogonal transformations have determinant of the matrix of coefficients equal to 1, i.e.
ps − qr = 1. An unfamiliar with this property student can study this subject separately (we at Unizor plan to include this into corresponding topic on vectors).
The above is the first equation to determine unknown coefficients. It's not linear, it is rather quadratic, but the rest of the equations will be linear and that's why we included this particular problem in the topic dedicated to linear systems.
We need three more equations to determine all the unknown coefficients.

B. Since moving system moves along X-axis with speed V, its beginning of coordinate (point x=0) must at the moment of time T be on a distance VT from the beginning of coordinates of a stationary system. Hence, if X=VT, x=0 for any T. From this and the first transformation equation x = pX + qT we derive:
0 = pVT + qT or
0 = (pV + q)T.

Since this equality is true for any T,
pV + q = 0
and, unconditionally, q = −pV.
This is the second equation for our unknown coefficients.

C. Since the speed of light C is the same in both systems {X,T} and {x,t}, an equation of its motion in the stationary system must be X = CT and in the moving system x = Ct. Therefore, if X=CT, then x=Ct. Put X=CT into equations of transformation of coordinates. We get x = pCT + qT, t = rCT + sT. Substitute these expressions into x=Ct:
pCT + qT = rC2T + sCT.
Reduce by T,
pC + q = rC2 + sC.
This is the third equation for unknown coefficients.

D. Repeat the logic of a previous paragraph for the light moving in the opposite direction with a speed −C. We get, if X = −CT, then x = −Ct. Therefore, x = −pCT + qT, t = −rCT + sT and (since x = −Ct)
−pCT + qT = rC2T − sCT.
Reduce by T,
−pC + q = rC2 − sC.
This the fourth equation for unknown coefficients.

So, this is the system of equations for 4 unknown coefficients of transformation p, q, r, s:
(a) ps − qr = 1
(b) q = −pV
(c) pC + q = rC2 + sC
(d) −pC + q = rC2 − sC

It's not exactly linear, but it has sufficient number of linear equations (all but one) to solve it using the known methodology. Let's solve this system of equations by combining the methods of substitution and elimination. We will express unknown variables q, r, s in terms of p using equations (b), (c) and (d). Then we will substitute them into (a) to get an equation for p. Solving it will allow to evaluate all other unknowns.

E. From (c) and (d), adding and subtracting these equations, we get:
2q = 2rC2, therefore q = rC2
2pC = 2sC, therefore p = s

Since from (b) q = −pV,
−pV = rC2 and r = −pV/C2.

Now all coefficients of a transformation are expressed in terms of one unknown coefficient p. To get the value of p, use the first equation (a).

F. Substituting q, r and s, expressed in terms of p, into an equation (a) ps − qr = 1, we get:
p2 − (−pV)·(−pV)/C2 = 1, therefore
p2·(1 − V2/C2) = 1 and
p = 1/√1−V2/C2 .

From this all other coefficients of a transformation matrix are derived:
q = −V/√1−V2/C2
r = −(V/C2)/√
1−V2/C2
s = 1/√
1−V2/C2

G. The final transformation matrix looks exactly like in Einstein's article on electrodynamics, but seems to be much simpler to arrive at and the derivation is strictly mathematical.
x = (1/√1−V2/C2 )·X − (V/√1−V2/C2 )·T
t = ((−V/C2)/√
1−V2/C2 )·X + (1/√1−V2/C2 )·T

Traditionally, factor V/C is replaced with Greek letter β, which results in formulas:
x = (1/√1−β2 )·X + (−V/√1−β2 )·T
t = ((−V/C2)/√
1−β2 )·X + (1/√1−β2 )·T

One more simplification is usually done by introducing Lorentz factor γ equals to 1/√1−β2 :
x = γX − γVT
= γ(X − VT)

t = −γVX/C2 + γT
= γ(T − VX/C2)

The final form of transformation of coordinates in the Special Theory of Relativity is:
x = (X − VT)/√1−(V/C)2
t = (T − VX/C2)/√1−(V/C)2