(please note I don't understand the math of the Lorentz transformations or
GR yet...)
The math behind the Lorentz transformation is dead easy. Simple
algebra.
First, try very hard to mentally assume away any idea of an absolute
(or 'preferred') reference frame and ask yourself how you might start
out to describe the motion of a photon particle traveling at the speed
of light from two different non-preferred perspective frames:
x = c*t coordinate system K
and,
x' = c*t' coordinate system K'
In other words, we make no presumptions about the two reference frames
to each other, __except__ for the fact that they will each observe the
same velocity for the photon. It is as though they are completely
independent to each other.
To make this point come to home deeply and seriously, imagine that the
photon is simultaneously observable in two completely different
universes that have nothing in common to each other except that they
both interact with shared photons.
The above two equations do express the fact that x and x' have nothing
to do with each other except for the speed of light, and that t and t'
similarly have nothing to do with each other except again for the
speed of light. The only thing in common in those two equations is c,
itself. Both independent frames will see that as the same value,
whatever it turns out to be. Other than that, the two frames are
completely without any other shared referents.
Now we write these same two, but with the photon traveling in the
opposite direction, just to get all the possibilities:
x = -c*t
and,
x' = -c*t'
Then we can write:
x - c*t = 0, and also from the last case, x + c*t = 0
and,
x' - c*t' = 0, and also from the last case, x' + c*t' = 0
Of course. Nothing special about that.
But since these are zero, we can now make some statement that combines
both frames:
x - c*t = A*(x' - c*t'), and, x + c*t = B*(x' + c*t')
We don't yet know what A and B are likely to be, but that's the
problem to solve.
We can now add and subtract both of these from each other:
x - c*t = A*(x' - c*t') x - c*t = A*(x' - c*t')
+ [x + c*t = B*(x' + c*t')] - [x + c*t = B*(x' + c*t')]
----------------------------- -------------------------------
2*x = A*x'+B*x'-A*c*t'+B*c*t' 2*c*t = A*x'-B*x'-A*c*t'-B*c*t'
or,
2*x = (A+B)*x' - (A-B)*c*t' 2*c*t = (A-B)*x'-(A+B)*c*t'
All this is pretty basic algebra, accessible to most anyone I think.
At this point, to get rid of the factor 2 (the constants A and B are
just arbitrary ones, so we can create any new ones we want based on
them for convenience and use those instead), we can define the
following new derived constants:
R = (A+B)/2, and S = (A-B)/2
From that, we can re-write the above as:
x = R*x' - S*c*t' and c*t = S*x'-R*c*t'
Remember these two, we'll refer to them below.
Well, that's a start. Of course, we still haven't figured out what R
and S are. So let's do so.
At the origin of the one of the frames, we have x = 0, so
x = R*x' - S*c*t' = 0
or,
x' = t' * [S*c/R]
This appears to be in a familiar x'=v'*t' form, if we defined the
velocity v':
v' = [S*c/R]
If so, it would be the velocity for which the origin (or really, any
point) in K is moving with respect to K'.
But in perhaps a little better form, we can write that the
instantaneous change in x' is:
dx' = dt' * [S*c/R]
dx'/dt' = S*c/R
v' = dx'/dt' = S*c/R
Same thing. And v' is the relative velocity of the two systems.
Now, we also know that length of a rest-object in system K observed
from system K' must be the same as the length of a rest-object in
system K' observed from system K (due to the assumption of relativity
-- no preferred point of view.) So let's take a picture, so to speak,
at time t'=0, then
x = R*x'
Let's now look at two points in system K (x-axis) separated by a
distance of exactly 1:
x1 = R*x1'
x2 = R*x2'
and we can also say that, by definition:
x2 = x1 + 1
x2 - x1 = 1
dx = x2 - x1 = 1
But also,
R*x2' = R*x1' + 1
R*x2' - R*x1' = 1
R*(x2' - x1') = 1
x2' - x1' = 1/R
dx' = x2' - x1' = 1/R
Therefore,
dx'/dx = 1/R.
But now, if we look at the picture taken from system K at t=0 and if
we then work to eliminate t' from:
x = R*x' - S*c*t' and c*t = S*x'-R*c*t'
then we find:
c*t = S*x'-R*c*t' @ t=0
c*0 = S*x'-R*c*t'
0 = S*x'-R*c*t'
R*c*t' = S*x'
t' = S*x'/[R*c]
substituting into:
x = R*x' - S*c*t'
x = R*x' - S*c*{S*x'/[R*c]}
x = R*x' - S^2/R*x'
x = (R - S^2/R)*x'
x = (R/R)*(R - S^2/R)*x'
x = R*(1 - S^2/R^2)*x'
Now, recall that:
v' = S*c/R
So that it should be clear that:
v'^2 = S^2*c^2/R^2
v'^2/c^2 = S^2/R^2
Substituting,
x = R*(1 - S^2/R^2)*x'
x = R*(1 - v'^2/c^2)*x'
dx = R*(1 - v'^2/c^2)*dx'
dx/dx' = R*(1 - v'^2/c^2)
But we also know that, from elsewhere above:
dx'/dx = 1/R
And from the assumed relativity principle, that:
dx/dx' = dx'/dx
Therefore,
R*(1 - v'^2/c^2) = 1/R
R^2*(1 - v'^2/c^2) = 1
R^2 = 1/(1 - v'^2/c^2)
R = SQRT[1/(1 - v'^2/c^2)]
R = 1/SQRT(1 - v'^2/c^2)
From:
v' = S*c/R
we find:
S = v'*R/c
S = v'*[1/SQRT(1 - v'^2/c^2)]/c
S = (v'/c)*[1/SQRT(1 - v'^2/c^2)]
S = [1/(c/v')]*[1/SQRT(1 - v'^2/c^2)]/c
S = 1/[(c/v')*SQRT(1 - v'^2/c^2)]
S = 1/SQRT(c^2/v'^2 - 1)
So, we've solved for both R and S. Remembering,
x = R*x' - S*c*t' and c*t = S*x'-R*c*t'
And with R and S in hand, we are set to find:
x = (x' - v'*t') / SQRT(1 - v'^2/c^2)
t = (t' - x'*v'/c^2) / SQRT(1 - v'^2/c^2)
And there you have it all.
All this is based on translation and not upon systems K that rotate
with respect to K' or are under some acceleration, gravitational or
otherwise. That is left for the general theory of relativity to deal
with, where it makes the assertion that there is no difference between
acceleration and gravity, from the point of view of the observer, and
thus that inertial mass (which relates to acceleration) must be the
same as gravitational mass (which, of course, relates to gravity) in a
1:1 relationship. With that assumption in hand, all the fun begins.
Jon
