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Re: Teaching logic is urgent (the only reasonable transformations) (footnote)



Follow-up message (footnote):

* The Lorentz transformation between two frames O and O', where O' is
moving with constant speed v along the positive x-axis and the
directions of the y- and y'- axes are the same, as are the z- and
z'-axes, is:

x' = gamma (x - v t)
y' = y
z' = z
t' = gamma (t - v x/c^2), where gamma = (1 - v^2/c^2)^(-1/2)

The inverse transformation is:

x = gamma (x' + v t')
y = y'
z = z'
t = gamma (t' + v x'/c^2)

Interestingly enough, the inverse transformation can be obtained by
simultaneously replacing every primed variable by the corresponding
unprimed one, replacing every unprimed variable by the corresponding
primed one, and replacing v by -v:

In this way, x' = gamma (x - v t) becomes

x = gamma (x' - (-v) t') <-> x = gamma (x' + v t'), etc.

This shows that either reference frame is equally valid: O sees O'
moving with speed v (forward, +x-direction), O' sees O moving with speed
-v (backward, -x'-direction). Of course, the inverse transformations
can also be obtained by solving the system of equations for the desired
variables:

y' = y -> y = y'
z' = z -> z = z'

x' = gamma (x - v t) -> x' = gamma x - gamma v t (1)
t' = gamma (t - v x/c^2) -> t' = gamma t - gamma v x/c^2 (2)

To solve for t we could eliminate x from both equations, multiplying (1)
by v/c^2 and adding it to (2):

x'v/c^2 = gamma x v/c^2 - gamma v t v/c^2
t' = - gamma v x/c^2 + gamma t
-------------------------------------------------
x'v/c^2 + t' = - gamma v t v/c^2 + gamma t

-> (t' + v x'/c^2) = gamma t (1 - v^2/c^2) = gamma t gamma^-2 = t
gamma^-1

-> t = gamma (t' + v x'/c^2)

Similarly, multiplying (2) through by v and adding the result to (1)
will eliminate t from the equation and allow us to solve for x:

x' = gamma x - gamma v t
v t' = gamma v t - gamma v^2 x/c^2
--------------------------------------
x' + v t' = gamma (x - v^2 x/c^2) = gamma x (1 - v^2/c^2) = x/gamma

-> x = gamma (x' + v t')

It was claimed above that for the special cases of the observer's
world-lines, the Lorentz transformations give the desired simplification
(and thus are consistent with the Premises 2 & 3 quoted above). This
must be shown bidirectionally: i.e., to show that the Lorentz
transformation is consistent with Premise 2 (x' = 0 <-> x = vt) we
must show that (LT and x'=0 -> x=vt) and (LT and x=vt -> x'=0). This
can easily be done:

LT and x'=0 -> x' = gamma (x - v t) and x'=0
-> 0 = gamma (x - v t) -> x - v t = 0
-> x = v t

LT and x=vt -> x' = gamma (x - v t) and x - v t = 0
-> x' = gamma (0) = 0

Consistency with Premise 3 is shown in a similar manner using the
inverse transformation, since LT <-> LT^-1.

This concludes the proof that the LT is consistent with Premises 2 & 3.
It is apparent that LT is consistent with Premise 1 (x' = ax + by + cz +
dt) with the appropriate choice of constants: a = gamma, b = 0, c = 0,
d = - gamma v.

[* The Lorentz transformation between two frames O and O', where O' is
moving with constant speed v along the positive x-axis and the
directions of the y- and y'- axes are the same, as are the z- and
z'-axes, is:

x' = gamma (x - v t)
y' = y
z' = z
t' = gamma (t - v x/c^2), where gamma = (1 - v^2/c^2)^(-1/2)

The inverse transformation is:

x = gamma (x' + v t')
y = y'
z = z'
t = gamma (t' + v x'/c^2)

Interestingly enough, the inverse transformation can be obtained by
simultaneously replacing every primed variable by the corresponding
unprimed one, replacing every unprimed variable by the corresponding
primed one, and replacing v by -v:

In this way, x' = gamma (x - v t) becomes

x = gamma (x' - (-v) t') <-> x = gamma (x' + v t'), etc.

This shows that either reference frame is equally valid: O sees O'
moving with speed v (forward, +x-direction), O' sees O moving with speed
-v (backward, -x'-direction). Of course, the inverse transformations
can also be obtained by solving the system of equations for the desired
variables:

y' = y -> y = y'
z' = z -> z = z'

x' = gamma (x - v t) -> x' = gamma x - gamma v t (1)
t' = gamma (t - v x/c^2) -> t' = gamma t - gamma v x/c^2 (2)

To solve for t we could eliminate x from both equations, multiplying (1)
by v/c^2 and adding it to (2):

x'v/c^2 = gamma x v/c^2 - gamma v t v/c^2
t' = - gamma v x/c^2 + gamma t
-------------------------------------------------
x'v/c^2 + t' = - gamma v t v/c^2 + gamma t

-> (t' + v x'/c^2) = gamma t (1 - v^2/c^2) = gamma t gamma^-2 = t
gamma^-1

-> t = gamma (t' + v x'/c^2)

Similarly, multiplying (2) through by v and adding the result to (1)
will eliminate t from the equation and allow us to solve for x:

x' = gamma x - gamma v t
v t' = gamma v t - gamma v^2 x/c^2
--------------------------------------
x' + v t' = gamma (x - v^2 x/c^2) = gamma x (1 - v^2/c^2) = x/gamma

-> x = gamma (x' + v t')

It was claimed above that for the special cases of the observer's
world-lines, the Lorentz transformations give the desired simplification
(and thus are consistent with the Premises 2 & 3 quoted above). This
must be shown bidirectionally: i.e., to show that the Lorentz
transformation is consistent with Premise 2 (x' = 0 <-> x = vt) we
must show that (LT and x'=0 -> x=vt) and (LT and x=vt -> x'=0). This
can easily be done:

LT and x'=0 -> x' = gamma (x - v t) and x'=0
-> 0 = gamma (x - v t) -> x - v t = 0
-> x = v t

LT and x=vt -> x' = gamma (x - v t) and x - v t = 0
-> x' = gamma (0) = 0

Consistency with Premise 3 is shown in a similar manner using the
inverse transformation, since LT <-> LT^-1.

This concludes the proof that the LT is consistent with Premises 2 & 3.
It is apparent that LT is consistent with Premise 1 (x' = ax + by + cz +
dt) with the appropriate choice of constants: a = gamma, b = 0, c = 0,
d = - gamma v.]