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### Relativity

```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
Other

Hi, Alex.

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)
and

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
is

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.

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
http://mathforum.org/dr.math/
```
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College Physics
High School Calculus
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