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Date: 05/22/2002 at 20:56:22
From: Alex Kamajian
Subject: Proving Terms Of 2 Polynomials Are Proportional To Each Other

An author discussing the derivation of the Lorentz transformation 
states that the expressions below are equal to zero (0).

(1) X1**2 + X2**2 + X3**2 + X4**2 = 0

(2) (X1')**2 + (X2')**2 + (X3')**2 + (X4')**2 = 0

The X's are the space-time coordinates of 'events' on an expanding 
wavefront of light as observed by 2 reference frames moving at 
constant speed relative to each other.

The assertion is made that the coordinates transform in such a way 
the these expressions are proportional to each other.

How do you demonstrate algebraically that these expressions are 
proportional to each other even when they describe 'events' that are 
not on the expanding wavefront of light (i.e. the expressions assume 
values greater than or less than zero)?

Date: 05/23/2002 at 11:05:10
From: Doctor Rick
Subject: Re: Proving Terms Of 2 Polynomials Are Proportional To Each

Hi, Alex.

Let's see if I understand the context of your question. Your 
equations are related to a quantity called the "interval" between two 
events in space-time, one at space position (0, 0, 0) and time 0, and 
the other at space position (x1, x2, x3) and time t = x4/(ic). Here c 
is the speed of light, and i is the square root of -1. Thus, in terms 
of t rather than x4, your equation (1) looks like this (using x^2, 
our standard Dr. Math notation, for x squared):

      x1^2 + x2^2 + x3^2 + (ict)^2 = 0

       x1^2 + x2^2 + x3^2 - (ct)^2  = 0

       x1^2 + x2^2 + x3^2           = (ct)^2

  sqrt(x1^2 + x2^2 + x3^2)          = |ct|

This is just the rate equation for travel at the speed of light, c: 
distance = speed * time, where the distance is calculated using the 
Pythagorean theorem. Thus the equation is indeed true for points in 
space-time on the wavefront of a light wave.

The interval is defined as

  tau = sqrt((ct)^2 - x1^2 - x2^2 - x3^2)

The special theory of relativity is based on the (observed) fact that 
the speed of light is constant in all inertial reference frames. On 
this assumption, the same equation must hold for transformed 
coordinates (x1', x2', x3', x4') in a reference frame moving at 
constant speed relative to the first reference frame. The speed of 
light c must be the same in both equations.

Now we want to consider intervals that are not zero, that is, points 
in space-time that are not on the same light wavefront. In this case 
the interval turns out to be not just proportional (I'm not sure what 
that would mean) but CONSTANT for the same two events (points in 
space-time) observed from any inertial reference frame.

This is not derived by algebra alone. At this point in the derivation 
of the Lorentz transformation, we do not know how (x1, x2, x3, x4) 
and (x1', x2', x3', x4') are related, so we cannot do algebra to 
derive the constancy of the interval. Rather, we consider the 
implications of the constancy of the speed of light on imagined 
scenarios other than two events on the same wavefront.

Consider the case of a laser and a mirror in a rocket. The mirror is 
a distance d away from the laser, facing in a direction perpendicular
to the motion of the rocket. At (0,0,0) and time 0 a pulse of light
leaves the laser. At (0,0,0) and time t' in the rocket frame, the
pulse, having reflected from the mirror, returns to the laser.

What is seen in the "ground" reference frame? The pulse leaves the 
laser at (0,0,0,0) (we'll align/synchronize our reference frames so 
this is the case). It returns to the laser at time t, but the laser 
is now at location (x1, x2, x3).

How can we relate the coordinates in the two reference frames? We 
know the speed of light is the same in both frames. In the rocket 
frame, the light traveled a total distance of 2d, so the time elapsed 
must be

  t' = 2d/c

In the ground frame, the light travels further. It takes equal times 
for the outward and return trips, so each leg takes time t/2. On the 
outward trip it travels along the hypotenuse of a right triangle 
whose sides are d (the distance perpendicular to the rocket's motion) 

  sqrt((x1/2)^2 + (x2/2)^2 + (x3/2)^2)

(the distance along the direction of the rocket's motion), since the 
rocket has moved a distance (x1/2, x2/2, x3/2) in this time. It 
travels the same distance on the return trip. Thus the total distance 

    2*sqrt(d^2 + (x1/2)^2 + (x2/2)^2 + (x3/2)^2)

  = sqrt((2d)^2 + x1^2 + x2^2 + x3^2)

and the total time elapsed is

  t = sqrt((2d)^2 + x1^2 + x2^2 + x3^2)/c

The distance d is the same in both reference frames. (Do you see why? 
Consider how you could measure it in both frames.) Solving our 
equations for t' and t to find (2d)^2, we get

  (2d)^2 = (ct')^2

  (2d)^2 = (ct)^2 - x1^2 - x2^2 - x3^2

Thus the two quantities (the interval between the same two events 
measured in two different reference frames) are equal.

Is this sufficient to answer your question? If we considered the 
interval as measured in the reference frame of a second rocket, we 
would find again

  (ct')^2 = (ct'')^2 - x1''^2 - x2''^2 - x3''^2

and therefore the intervals in the t, etc. frame and in the t'', etc. 
frame are equal. That's what you wanted proved.

As I said, it's not derived by algebra alone, but by algebraic 
analysis of a thought experiment. This thought experiment ends up 
with a positive ("timelike") interval. You might want to think up an 
experiment in which the interval is negative ("spacelike") if you're 
not convinced that the equality of intervals holds for *all* non-zero 
intervals as well as zero ("lightlike") intervals.

- Doctor Rick, The Math Forum 
Associated Topics:
College Calculus
College Physics
High School Calculus
High School Physics/Chemistry

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