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doppler's debacle
---------------
In 1842 Doppler tested a new theory about sound. From casual
observations, the pitch of a sound source approaching the listener
increased, while that of a receding source decreased. To experiment,
Doppler placed a trumpet player on a railroad car, which he hired to
run back and forth. Doppler measured the frequency by finding a
matching key on a piano at trackside. How Doppler heard the music over
the hideous noise of the train is lost in history.
The discovery of this, the Doppler effect, worked for sound, and
other mechanical waves. Given the slow speed of the train compared to
that of sound in air, the formula Doppler derived was a simple one

v[m/s] / c = (lambda'[m/s] - lambda[m/m]) / lambda[m/m]

+---------------------------------------------------------+
| DOPPLER EFFECT FOR SOUND AT LOW SPEEDS                  |
|                                                         |
| v[m/s] / c = (lambda'[m/s] - lambda[m/m]) / lambda[m/m] |
+---------------------------------------------------------+

c c is the speed of sound, not of light!, thru air. Note that we write
lambda'[m/s] and not lambda[m/s]. The wavelength here is that as
heard, observed, by the stational frame; it includes the travel time
for the sound waves to reach us.
The formula is arranged so that a recession of the source  has a
positive velocity; accession, negative. This is merely a sign
convention.
Doppler recognized three distinct scenarios for the observer and
source:

Observer still in air, trumpet moves thru air
Trumpet still in air, observer moves thru air
Both trumpet and observer move thru air

Each yielded a different formula for the frequency, or wavelength,
shift. The three cases are not equivalent or symmetrical. The one
stated above is for speeds very small compared to that of sound,
v[m/s] << c. All three equations relax to this simple form when the
speeds are small enough.

Absurd Doppler speeds
-------------------
Doppler did his work with sound and extrapolated the results to
light. There was in his day the contention about light being a
particle or a wave. Doppler took the wave side of the argument. He hit
on the idea that his wavelength shift could be applied to the stars.
Stars, he thought, have colors because their emitted wavelengths are
shifted by their motion to or from us. In fact his book about the
shift was 'On the colored light of double stars and other stars in the
heavens'!
To determine the motion of stars, he needed a reference wavelength
of a star with no radial motion. With Earth in a nearly circular orbit
around the Sun, the Sun has essentially no radial motion. Its color
represented the 'rest' wavelength for light emitted by stars. The
nature of stars was totally unknown, altho it was realized that stars
had to be sun-like luminous globes.
Doppler worked up the wavelengths of starlight from the colors.
Then he applied his sound-shift model on them. Using modern values,
Doppler came up with the following speeds.

+--------+-------+--------+----------+-------------+
| color  | lambda|v[m/s]/c| km/s vel | comments    |
+--------+-------+--------+----------+-------------+
| violet | 350nm | -0.367 | -109.091 | accession   |
| indigo | 400nm | -0.273 | - 81,878 |   |         |
| blue   | 450nm | -0.182 | - 54,545 |   |         |
| green  | 500nm | -0.091 | - 27,273 |  \|/        |
| yellow | 550nm |  0.0   |        0 | Sun at rest |
| orange | 600nm | +0.091 | + 27.273 | recession   |
| red    | 650nm | +0.182 | + 54,545 |   |         |
|far red | 700nm | +0.273 | + 81,878 |  \|/        |
+--------+-------+--------+----------+-------------+

These are crazy speeds! In physics of the early 19th century,
there was nothing wrong with such speeds. What's more, they added to
the fascination of astronomy! The figures here are worked up with the
small-speed formula. The more elaborate ones, for any of the three
cases, give similarly ridiculous results.
If we include infrared and ultraviolet radiation, which were just
discovered in the early 1800s, we can come up with speeds approaching
or exceding light. For example if the wavelength is 2,000 nanometers,
we the star is receding at, gulp, 2.636 times the speed of light!
the stars except visible light.

Fizeau's fixup
------------
Fizeau in 1848 also worked on the wavelength shift between source
and observer. He had the same early 19th century ideas about light. He
recognized that the wavelengths to compare were not those of the
star's overall color. From spectrometry, crude as it was then, he saw
that a star emits light over most colors but with certain wavelengths
missing. These are the dark lines in the star's spectrum, indicative
of various chemical elements in the star.
Fizeau compared the wavelengths of lines of a recognized element
in the star with those from a source at rest. His motion in the stars,
was of orders more modest extent. Speeds of recession or accession
were in the ones to tens of kilometers per second.
Fizeau and astronomers following him for most of the 19th century
had one hell of a time getting the measurements of wavelength from
stellar spectra. The image was examined by visual micrometry. Due to
the thick prisms in the early spectroscopes, the image was dim and the
lines were tough to discern. Spectra of only the brightest stars could
be confidently examined.
Due to this proper application of the wavelength shift, the effect
is sometimes called the Doppler-Fizeau effect (Fizeau-Doppler effect,
in France),
There was a broad disparity of speeds quoted for a given star.
Each astronomer had to independently squeeze out measurements of
wavelength shifts, each surely different from those of others. Despite
heroic valiant efforts, there was not much improvement in the
measurements.
As example, a star is receding  from ys at 600KPS, within range of
star motions in the Milky Way. The Doppler-Fizeau shift of a spectral
line of wavelength 500nm would ne

V / c = (lambda' - lambda) / lambda

where c is the lightspeed of300,000KPS.

lambda * V / c = lambda' - lambda

(lambda * V / c) + lambda = lambda'

lambda' = (lambda * V / c) + lambda
= (500nm * 60KPS / 300,000KPS) + 500nm
= 0.10nm + 500nm
= 500.10nm

An astronomer would have to discern that the line was displaced
toward longer wavelength by 1/10 nm from its home position at 500nm.
If he records that shift was 0.20nm, his speed for the star is 120KPS,
twice as great. Or if he measures the shift to be 0.05nm, the speed is
30KPS. It is tough to inspect a spectrum by eye thru the telescope and
errors like these were common.
Photography eventually could capture a permanent image of the
spectra by the 1880s. By the end of the 19th century, major projects
to photograph star spectra were underway. From the photos,
spectrograms, catalogs of stellar speeds were issued. One of the 19th
century catalogs of spectra, the Henry Draper catalog, is still in use
today (2005) for selecting candidates for extrasolar planets.

Frequency versus wavelength
-------------------------
Doppler really worked with his music waves by their frequency
rather than by wavelength. This is still done today for sound and many
parts of the electromagnetic spectrum. In the optical band we typicly
deal with wavelength, not frequency. The choice is a function of
legacy and the devices we use to explore the waves. It is no more
correct to use the one or the other, altho in many situations, the
opposite method may look peculiar at first.
Wavelength and frequency are related thru the speed of the wave.
The wave speed is the product of the frequency times wavelength.
Knowing any two of these parameters fixes the third. Wavelength is
almost always symbolized by lambda. Frequency is symbolized by either
'f' or 'nu'.

+----------------------+
| FREQUENCY-WAVELENGTH |
|                      |
| lambda * f = c       |
|                      |
| lambda = c / f       |
|                      |
| f = c / lambda       |
+----------------------+

This is a general relation which applies to any flow of items of a
given physical length. For example, a freight train with cars of 15m
length passes us at a crossing gate. 12 cars roll by per minute. How
fast is the train going? The 'wavelength' is 15 meters; frequency is
12 'cycles' per minute. The train speed is

c = lambda * f
= (15m) * (12/min)
= 180m/min
-> 10,800m/hr
= 10.8KPH.

By the physiology of human vision, wavelength is associated with
color. The spectrum has a sequence of color across its wavelength
range. The longer wavelengths are at the red end; shorter, blue
(indigo and violet) end. Astronomers speak of a shift of wavelength
toward the longer, red, end of the spectrum as a 'redshift'. A shift
toward shorter, blue, end is a 'blueshift'.

Historical blunder
----------------
Since Doppler and Fizeau, the shift of spectral lines was accepted
as the hallmark for motion of approach and reproach. Only the
component of motion along the line of sight, the radial component,
shows up by this technique. Hence, the speed is commonly called
'radial velocity'. By convention positive speeds are recession;
negative, accession.
In the 1910s, Einstein figured out that one other means of
shifting wavelengths was an expanding of space.  This expansion can be
cited as the fractional increase in scalefactor per unit time. No such
enlargement of the universe was known in the 1910s, so this cause was
largely ignored by astronomers.
Hubble in the 1920s with Humason captured spectra of galaxies.
galaxies were realized in 1923 to be other Milky Ways, but were still
commonly called 'spiral nebulae'.
Hubble & Humason correlated the distances and wavelength shifts
for dozens of galaxies. Almost all shifts were toward longer
wavelength, toward the red end of the spectrum. In 1929 Hubble
announced his results. The galaxies were rushing away from us at high
radial speeds in a direct function of distance.
Hubble assigned the spectral shifts of the galaxies to real radial
motion, speaking of radial velocity and distance as evidence for an
expanding universe. Hubble expressed the relation as an increase of
speed for unit increase in distance. This is the Hubble factor H0. It
is cited as either (km/s)/Mly or (km/s)/Mpc. This was quite normal in
his day because, in neglect of Einstein physics, the only cause of
spectral line shift was true radial velocity of the source.
Ever since then, right into the 21st century, many explanations of
the expanding universe speak of real motion of galaxies thru space.
The Hubble factor is actually the fractional increase of
scalefactor per unit time. It gives the expansion rate directly with
no need to invoke physical motion of the galaxies. Looking at the
Hubble factor, we see, for a typical value of (75km/s)/Mpc

H0 = (75km/s)/Mpc
= (75) * (km/(s.Mpc))

The units are [length]/[time*length]. A parsec is 3.08e13 kilometers.
Hence, we have

H0 = (75) * (km/(s.((1e6pc) * (3.08e13km/pc)))
= (75) * (km/(s.(3.08e19km)))
= (2.435e-18) * (km/(Km.s))
= (2.435e-18) * (1/s))
= (2.435e-18)/s
= 1/(4.107e17s)
-> 1/(~13 billion years)

This states directly that the universe expands at 2.435e-18 part every
second. If the universe started from zero size, at this rate of
growth, it would take about 13 billion years to reach its present
size.
With this simple and elegant explanation of the Hubble factor,
textbooks today -- in the 21st century! -- commonly speak of galaxies
as actually streaming thru space at hideously insane velocities. This
is on top of the erroneous use of the acoustic model for the Doppler
shift.

Reality versus Perception
-----------------------
In all the foregoing we never explained how the one platform knows
what's going on with the other. In the derivations so far we made no
explicit mention of any conversations passing between the stational
and motional frames.
In the universe there is really only one way to convey information
from place to place: electromagnetic radiation. The books on
relativity relentlessly refer to 'light' as the medium of
conversation, but we may use any wavelength of EMR. All radiation have
the one and same speed in vacuo, c.
We do have nonEMR knowledge of other places in the universe. We enjoy
rocks and dirt from the Moon, meteorites (including a bunch from Mars!),
cosmic heavy ion influx, and solar wind particles. Early in the 21st
century we could have retrieved surface samples from Mars, gas from a
comet's tail, and dust from around an asteroid.
There may be others, yet all combined the information content is
minuscule compared to the content of the EMR we receive from the
universe. So, for all practicality's sake we must communicate across
the depths of space only by EMR. We treat time in this paper as a
'clock reading' or 'unit of time measure'. Time actually applies to
ALL processes that have duration. The growth of flowers, hair, rust;
length of rainfall, erosion, combustion, mechanical movement; aging of
people, animals, structures are all dimensioned in time. Everything in
the spaceship -- not only the ticks of clocks -- is time-dilated
relative to Earth.

Observed Time
-----------
We saw that the time unit on the spaceship is experienced on Earth to
be longer than experienced on the spaceship

delT[m/s] = delT[m/m] / beta

with beta ranging between zero and one. We now let the message of each
second's tick come to Earth on a light pulse. A one pulse is sent from
the spaceship, then the next, then the next, and so on, at one second
intervals on the spaceship's clock.
The pulse travels a certain distance from the spaceship to Earth, but
the next one must travel a bit longer path because the spaceship is in
motion, say, away from us. From the stational frame the increment of
the spaceship's distance between pulses is delX[m/s] =
V[m/s]*delT[m/s]. The next pulse in travelling this extra distance
takes an increment of time longer to reach Earth. This is the distance
run divided by the speed of the pulse

delX[m/s] = V[m/s] * delT[m/s]

delT[m/s] = delX[m/s] / c
= V[m/s] * delT[m/s] / c

The perceived length of the spaceship second from the Earth's
platform is

delT'[m/s] = delT[m/s] + delT[m/s]
= delT[m/s] + V[m/s] * delT[m/s ]/ c
= delT[m/s] * (1 + V[m/s] / c)

where delT'[m/s] is the length of the time unit as made known to us by the
medium of light, as distinct from delT[m/s] which is the real length of
that unit.

delT'[m/s] = delT[m/s] * (1 + V[m/s] / c)
= delT[m/m] * (1 + V[m/s] / c) / beta

There is an alternate form of this equation that appears in other
works, but it is entirely equivalent to this here one

delT'[m/s] = delT[m/m] * (1 + V[m/s] / c) / beta
= delT[m/m] * (1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2))
= delT[m/m] * (1 + V[m/s] / c)
/ ((1 + V[m/s] / c) * (1 - V[m/s] / c))^(1/2))
= delT[m/m] * (1 + V[m/s] / c)^(1/2)
/ (1 - V[m/s] / c)^(1/2)
= delT[m/m] * (( 1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2)

+----------------------------------------------------------+
| OBSERVED TIME DILATION BETWEEN INERTIAL FRAMES           |
|                                                          |
| delT'[m/s] = delT[m/m] * (1 + V[m/s] / c) ) / beta       |
| = delT[m/m] * ((1 + V[m/s] / c) / (1 -V[m/s] / c))^(1/2) |
+----------------------------------------------------------+

Observed Time Dilation
----------------------
Consider the case of the spaceship moving away from Earth with positive
V[m/s]. The delT'[m/s] equation yields

delT'[m/s] = delT[m/m] * ((1 + (>0)) / (1 - (>0))^(1/2)
= delT[m/m] * ((>1) / (<1))^(1/2)
= delT[m/m] * (>1)^(1/2)
= delT[m/m] * (>1)
> delT[m/m]

We observe -- by the message carried by light -- the spaceship second
on Earth is longer than on the spaceship itself. Except for the size of
the difference, the sense is the same as we found for the real divergence
of time between the two platforms.
When the spaceship accedes toward us, V[m/s]<0 and

delT'[m/s] = delT[m/m] * ((1 + (<0)) / (1 - (<0))^(1/2)
= delT[m/m] * ((<1) / (>1))^(1/2)
= delT[m/m] * (<1)^(1/2)
= delT[m/m] * (<1)
< delT[m/m]

We see the observed second of the spaceship to be less, smaller, than that
on the spaceship. Time for this motional platform speeded up.
We have that by using light (EMR) we can not observe the universe as
it really is. The medium of light distorts or corrupts our perception of
the world. In reality time on the motional frame always runs slower
relative to Earth. However, we actually observe that time may be either
faster or slower according as the motional frame is acceding or receding
relative to us.
There is no getting around this distortion. It is not in any way
caused by some imperfection in our arts and science of observation. No
manner of improvement in our observation powers -- so long as those powers
live off of electromagnetic radiation as the vector of the information --
can correct this corrupted view of the universe.

EInstein and Doppler-Fizeau
-------------------------
We saw that the observed time dilation between the stational and
motional platforms is

delT'[m/s] = delT[m/m] * (1 + V[m/s] / c)) / beta
= delT[m/m] * ((1 + V[m/s] / c) / ((1 - V[m/s] / c))^(1/2)

The time unit does not have to be an external signal placed on the EMR
wave, like a radio time broadcast or television view of a clock. It
can be the time unit embedded in the wave itself. The time unit is merely
the reciprocal of the frequency of the wave, with the factor c missed out.

delT'[m/s] = 1 / f'[m/s]

1/f'[m/s] = (1 / f[m/m]) * (1 + V[m/s] / c)) / beta
= (1 / f[m/m]) * ((1 + V[m/s] / c) / ((1 - V[m/s] / c))^(1/2)

The wavelength of the wave, which is what astronomers are comfortable
with, is c/f, from lambda*f = c

c/f'[m/s] = (c/f[m/m]) * (1 + V[m/s] / c)) / beta
= (c/f[m/m]) * ((1 + V[m/s] / c) / ((1 - V[m/s] / c))^(1/2)

lambda'[m/s] = (lambda[m/m]) * (1 + V[m/s] / c)) / beta
= (lambda[m/m]) * (1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2)

lambda'[m/s] / lambda[m/m] = (1 + V[m/s] / c) / beta
= (1+V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2))

lambda'[m/s]/lambda[m/m] - 1 = ((1 + V[m/s] / c) / beta)-1

(lambda'[m/s] - lambda[m/m]) / lambda[m/m]
= (1 + V[m/s] / c) / beta) -1
= (1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2)) - 1

An other quirk of astronomy is that the lambda difference
expression is symbolized by Z and the plain lambda'/lambda, being 1
greater than Z, is called 'Z+1'. So

Z = ((1 + V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2)) - 1

Z+1 = (1 + V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2)

Many astronomy books have the alternative form

Z = ((1 + V[m/s] / c) / (1 - (V[m/s] / c)^2)^(1/2)) - 1
= ((1 + V[m/s] / c) / ((1 + V[m/s] / c)
* (1 - (V[m/s] / c})^(1/2)) - 1
= ((1 + V[m/s] / c)^(1/2)) / (1 - V[m/s] / c}^(1/2)) - 1
= (((1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2)) - 1

Z+1 = ((1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2)

+-------------------------------------------------------+
| EINSTEIN'S DOPPLER-FIZEAU EFFECT                      |
|                                                       |
| Z = ((1 + V[m/s] / c) / beta) - 1                     |
|   = (((1 + V[m/s] / c) / (1 - V[m/s] / c))^(1/2)) - 1 |
+-------------------------------------------------------+

The Low-Velocity Approximation
----------------------------
Altho the Einstein form of the Doppler-Fizeau effect is the
correct one, based on the true cause of the effect, it is easy to see
how the classical form can work so well. It turns out the speeds of
stars and other bodies in our Milky Way galaxy are small fractions of
c. So we have for V[m/s] << c

Z = ((1 + V[m/s] / c) / ((1 - (V[m/s] / c)^2)^(1/2)) - 1
= ((1 + (<<c) / c) / (1 - ((<<c) / c)^2)^(1/2)) - 1
= (1 + (<<c) / c) / (1 - (<<<<c))^(1/2)) - 1
= (1 + (<<c) / c) / (1 - 0)^(1/2)) - 1
= (1 + ((<<c) / c) / 1^(1/2)) - 1
= (1 + ((<<c) / c) / 1) - 1
= 1 + (<<c) / c - 1
= (<<c) / c
= V[m/s] / c

+-----------------------+
| LOW VELOCITY          |
| DOPPLER-FIZEAU EFFECT |
|                       |
| Z = V[m/s] / c        |
+-----------------------+

The Missing Step
--------------
As elegant as the Einstein-Doppler effect is, it presumes we know the
velocity between the source and observer and we solve for Z, or Z+1.
In astronomy we normally assess Z and we must work out V[m/s]. We need
an expression V = func(Z). Yet few textbooks ever present such a
function! We simplify V[m/s] to just V here

Z+1 = ((1 + V/c) / (1 - V/c))^(1/2)

(Z+1)^2 = (1 + V/c) / (1 - V/c)

(Z+1)^2 * (1 - V/c) = 1 + V/c

(Z+1)^2 - (Z+1)^2 * (V/c) = 1 + V/c

(Z+1)^2 = (1 + V/c) + ((Z+1)^2 * V/c)
= 1 + V/c + ((Z+1)^2 * V/c)

(Z+1)^2 - 1 = V/c + ((Z+1)^2 * V/c)
= (1 + (Z+1)^2) * V/c

((Z+1)^2 - 1) / (1 + (Z+1)^2)) = V/c

V/c = ((Z+1)^2 - 1) / ((Z+1)^ 2 +1)

+---------------------------------------+
| VELOCITY FROM EINSTEIN-DOPPLER EFFECT |
|                                       |
| V/c = ((Z+1)^2 - 1) / ((Z+1)^2 + 1)   |
+---------------------------------------+

Space Travel Near c
-----------------
One of the grandest fallacies in popular treatises on relativity
is time-dilated space travel. The spiel goes like this: A spaceship
flies to a star 100ly away at 0.8c. It takes by Earth time 125y for
the spaceship to complete the journey. But the spaceship riders feel
it took only 75y.
beta is (1-(0.8)^2)^(1/2) = 0.6. delT[m/s] = delT[m/m]/0.6 =
delT[m/m]*(1.66). A second on the spaceship is 1.66 Earth seconds. So
Earth's 125y is completed in (125y)/(1.66) = (125y)*(0.6) = 75y.
By increasing V[m/s] to 0.9c the time dilation is greater. The
trip takes only 54.5y. This means that space travel is feasible
however far the destination is merely by travelling fast enough to
dilate the travel time to some comfortable fraction of a person's
lifespan.
Not.
What's wrong?
Simply put, it's a ruse. Altho the time cited is delT[m/s], the
distance involved is delX[s/s]. That 100ly is the Earth's distance
experienced by Earth. It is not merely unfair but conniving to
commingle spacetime units from different places! The proper method of
determining the spaceship's experienced travel time is to use both the
delX[m/s] and delT[m/s].
We did not explore how distance units vary with relative motion. It
turns out that the meter on the spaceship as felt by Earth delX[m/s] is
also dilated. delX[m/s] = delX[m/m]/beta. Yes, the spaceship completes the
trip in 75y, 0.6 that experienced on Earth. But the spaceship only
travelled (100ly)*(0.6) = 60ly -- not 100! -- in that 75y. Earth feels it
finished a 100ly trip. The spaceship experiences that it has 40ly more
to go.
What's more, it'll take (40y)*(1.25) = 50y to run out that last 40ly.
So!, the spaceship eventually does travel the full 100ly, the 60ly+40ly,
but it takes 75y+50y = 125y to do so. This is exactly how long it takes on
Earth! There plain is NO saving of time by travelling near c. No matter
what inertial platform you ride, it'll take, for a give fraction of c, the
same (100)/(fraction of c) years to traverse 100ly.

----------
With the time dilation effect in Einstein's relativity system there
arises the question of getting clocks out of synch by the relative motion
between them. That is, if one inertial observer is moving across an other
inertial observer and compare their clocks, they differ. The motional
observer's clock somehow runs slower and that person ages less when
compared to the clock of the stational observer. But the motional platform
can be taken as the home observer and now it's the stational fellow that
loses time and ages slower.
To take a specific example, consider a spaceship in the motional frame
m travelling at V[m/s] = 0.8c. For this speed beta is 0.6

beta = (1 - (V[m/s] / c)^2)^(1/2)
= (1 - (0.8)^2)^(1/2)
= (1 - (0.64))^(1/2)
= (0.36)^(1/2)
= 0.6

Recall that

delT[m/s] = delT[m/m] / beta

meaning that Earth, the stational s frame, experiences the second of time
on the spaceship to be 1/beta times the second experienced on the
spaceship itself. Earth measures the spaceship time as slowed down or
dilated; its seconds are bigger.
Hence, for a given journey in Earth years the spaceship takes fewer
years and so it ages less than Earth. By picking V[m/s] close to c we can
leverage the time dilation effect so a voyage of some long time on Earth
takes on the spaceship only a few days! In this way we can set off to the
other stars how so ever far off they stand within the lifespan of a human
rider on the spaceship. In the meantime generations upon generations of
humans on Earth came and went.
This does not in fact actually happen. The spaceship in its own time
takes just as long to finish its journey as by the time kept on Earth.
There is no paradox but a severe misinterpretation.
The m platform goes out 12 lightyears, to tau Ceti. At 0.8c the trip
should take 15 years, as measured on Earth. Due to the time dilation we
experience that the spaceship took only (15y)*(0.6) = 9y. (The dilation
was derived for the unit of time; the interval, the count of units, is the
reciprocal of this dilation.)
So!, what would take 15y on Earth is accomplished in only 9y on the
spaceship, a saving of 6y. Or the motional observer ages 6y less than the
stational observer.
But we committed -- along with armies of inattentive writers on
relativity -- a colossal booboo. First off, we assumed that after our 15
years the spaceship in fact reached tau Ceti. Did it? Well, the spaceship
sees the relative speed between it and Earth as that same 0.8c, by the
symmetry of the inertial frames. So in 9 of its own years it travelled
(9y)*(0.8c) = 7.2ly. It didn't finish its trip yet!
What went wrong is that the common treatments fail to account for the
space dilation. We mixed together the time on the spaceship with the space
on the Earth. The m frame travelled 12ly in Earth space over 9y in
spaceship time. Apart from the seemingly faster-than-light speed,
(12ly)/(9y) = 1.333c, we simply can not meld the spacetime measures from
two different inertial platforms. It's like posting sales paid in French
francs and labeling the entries 'US dollars'.
Because the speed is measured the same (but in opposite directions) in
the two platforms, yet the time is dilated, the space must also be dilated
by the same factor.

V[m/s] = delX[m/s] / delT[m/s]
= delX[m/m] / delT[m/m]

delX[m/s] / delT[m/s] = delX[m/m] / delT[m/m]
= delX[m/m] / (delT[m/s] * beta)
= delX[m/m] / (delT[m/s] * beta)

delX[m/s] = delX[m/m] / beta

The spaceship's lightyear which Earth experiences is longer than the
lightyear the spaceship itself experiences. So in a given span of Earth
time the spaceship crossed fewer lightyears than on Earth. The time is 6y
less simply because the spaceship never got to tau Ceti; it has 4.8ly yet
to go.
At 0.8c it'll take (4.8ly)/(0.8c) = 6y to cover the remaining distance
and arrive at tau Ceti. This is added to the 9y already run and the whole
trip takes (9y)+(6y) = 15y.
There is no 'paradox' at all. We took the clock reading on the m frame
before it finished the trip we set it out to fulfill.
It is the pure and simple fact that from Earth the spaceship really did
traverse only 7.2ly in 9y when our clock ticked off 15y. If we wait until
the m frame completes the 12ly trip, we experience that it took after all
exactly 15y, the same as on Earth. Or on the spaceship the riders feel
that time has not slowed at all and they age just as fast as the folk on
Earth. At 0.8c it really takes 15y to get to tau Ceti. They are fully 15y
older than when they left Earth.

Simultaneous Events
-----------------
As referred to a given inertial platform, two events are simultaneous
if in that platform they occur at the same time. They occur at different
places in that platform, so their X coordinates are not the same. But they
share the same cT coordinate. In Newton physics two events which are
simultaneous for any one observer must be simultaneous for all observers.
This comes from the external and absolute nature of time in Newton
physics. In Einstein physics we must set aside the earthly concept of
absolute simultaneity.
In the motional frame two events happen at two places in the spaceship
at the same time. The X coordinates differ but the cT coordinates are the
same. delX[m/m] <> 0 and delT[m/m] = 0.

Spacetime Interval for Simultaneous Events
----------------------------------------
We start with the spacetime interval equation and set delT[m/m] to zero.

delX[m/s]^2 - c^2 * delT[m/s]^2 = delX[m/m]^2 - c^2 * delT[m/m]^2
= delX[m/m]^2 - c^2 * 0
= delX[m/m]^2

-c^2 * delT[m/s]^2 = delX[m/m]^2 - delX[m/s]^2

We saw in the Twin Paradox that delX[m/s] is dilated relative to
delX[m/m] by

delX[m/s] = delX[m/m] / beta

So

-c^2 * delT[m/s]^2 = delX[m/m]^2 - delX[m/m]^2 / beta^2
= delX[m/m]^2 - delX[m/m]^2 / beta^2
= delX[m/m]^2 * beta^2 / beta^2 - delX[m/m]^2 / beta^2
= (delX[m/m]^2 * beta^2 - delX[m/m]^2) / beta^2
= (delX[m/m]^2 - delX^2 * (V[m/s]^ 2 /c^2) - delX[m/m]^2)
/ beta^2
= (-delX[m/m]^2 * V[m/s]^2 / c^2) / beta^2

delT[m/s]^2 = (-1 / c^2) * (-delX[m/m]^2 * V[m/s]^2
/ c^2) / beta^2
= (delX[m/m]^2 * V[m/s]^2 / c^4) / beta^2
= delX[m/m]^ 2 *V[m/s]^2 / (c^4 * beta^2)

delT[m/s] = delX[m/m] * V[m/s] / (c^2 * beta)

+------------------------------------------------+
| EINSTEIN SIMULTANEITY                          |
|                                                |
| delT[m/s] = (delX[m/m] * V[m/s] / (c^2 * beta) |
+------------------------------------------------+

Platforms Can Not Agree
---------------------
So long as V[m/s] is nonzero -- there are in fact two distinct
platforms -- the right side of the above equation must be nonzero. Now in
deriving this Einstein simultaneity relation we set delT[m/m] = 0, by the
definition of simultaneity in the motional frame. Yet in the stational
frame delT[m/s] is not zero. There is a nonzero interval of time between
the stational frame's experience of the one event and the other. Or the
events in the spaceship that are occurring at the same instant in the
spaceship are experienced as occurring in sequence on the Earth.
In general events which in one platform occur together can not occur
together in any other platform. There is no absolute simultaneity under
Einstein physics.
This conclusion is even more violative against our common sense than
the time and length dilation! It does seem so perfectly natural that all
frames must agree that two simultaneous happenings for any one do occur at
the same instant for all. Yet this just is not true.
It is only the very minuscule speeds between observers we cope
with that masks this relativity of time in our everyday life.

Simultaneous in Space and Time
----------------------------
If we insist that the two events in the spaceship occur not only at the
same time but also at the same place, we have delX[m/m] = 0 as well as
delT[m/m] = 0. We then get the trivial result that in the Einstein
simultaneity equation delT[m/s] = 0. The stational frame experiences the
events to happen together. But in this case, there are not two separate
happenings. We have merely duplicated the same set of coordinates, so
there is really nothing for relativity to act on.

Lorentz Transformation
--------------------
We derived the relation between the time on the spaceship, the
motional frame, as experienced by Earth, the stational frame, in two
situations. First, we saw events in the motional platform occurring in
the same location but at two different times; we zeroed the terms with
delX[m/m]. Then we looked at two events occurring at the same time but
in two different locations by zeroing the delT[m/m] term.
Thus we obtained

delT[m/s] = delT[m/m] / beta

delT[m/s] = (V / c^2) * delX[m/m] / beta

Similar relations join the space coordinates between the two
platforms, which we did not explicitly derive

delX[m/s] = delX[m/m] / beta

delX[m/s] = (V/c) * c * delT[m/m] / beta

If we remove the constraint of holding either the location or time
of the events fixed in the spaceship, we get the more general
transformation of time and space between the two frames. These are
merely the addition of the constant-location and constant-time
formulae from above.

delT[m/s] = (delT[m/m] / beta) + ((V/c^2) * delX[m/m] / beta)
= (delT[m/m]+(V/c^2)*delX[m/m])/beta

delX[m/s] = (delX[m/m] / beta) + ((V/c) * c * delT[m/m] / beta)
= (delX[m/m] + (V/c) * c * delT[m/m]) / beta

These are the Lorentz transformation equations relating the spacetime
in the motional frame to that experienced in the stational one. They
were discovered in the 1890s by Lorentz in the era leading up to
Einstein's synthesis of relativity.

+-------------------------------------------------------+
| LORENTZ TRANSFORMATION BETWEEN TWO FRAMES             |
|                                                       |
| delT[m/s] = (delT[m/m] + (V/c^2) * delX[m/m]) / beta  |
|                                                       |
| delX[m/s] = (delX[m/m] + (V/c) * c *delT[m/m]) / beta |
+-------------------------------------------------------+

--------------------
V is the velocity of the motional frame relative to the stational
one, and is a short form for V[m/s]. beta is the Lorentz factor (1-
(V/c)^2)^1/2.
We now change the subscripts to let us introduce a second motional
frame. The m platform is now the m1 frame; the new one, m2.

delX[m1/s] = (delX[m1/m1] + (V/c) * c * delT[m1/m1]) / beta

delT[m1/s] = (delT[m1/m1] + (V/c^2) * delX[m1/m1]) / beta

Divide the first by the second

delX[m1/s] / delT[m1/s]
=  ((delX[m1/m1] + (V/c) * c * delT[m1/m1]) / beta)
/ ((delT[m1/m1] + (V/c^2) * delX[m1/m1]) / beta)

= (delX[m1 / m1] + (V/c) * c * delT[m1/m1])
/ (delT[m1/m1] + (V/c^2) * delX[m1/m1])

= (delX[m1/m1] / delT[m1/m1]) + ((V/c) * c * (delT[m1/m1]
/ delT[m1/m1])     / (delT[m1/m1] / delT[m1/m1]) + ((V/c^2)
* delX[m1/m1]   /delT[m1/m1])

= (delX[m1/m1] / delT[m1/m1]) + (V/c) * c *1)
/ (1 + (V/c^2) * delX[m1/m1] / delT[m1/m1])

= (delX[m1/m1] / delT[m1/m1]) + (V/c) * c)
/ (V/c^2) * delX[m1/m1] / delT[m1/m1])

delX[m1/m1]/delT[m1/m1] is the change of position of a point as
experienced by the m1 frame, or the speed of a point in a new motional
platform, the m2 platform, as experienced by m1. Call this V[m2/m1].
delX[m1/s]/delT[m1/s] is the speed of this same point, in the second
m2 frame, as experienced by the stational frame. Call it V[m2/s].

V[m2/s] = (V[m2/m1] + (V/c) * c * 1) / (1 + ((V/c^2) * V[m2/m1]))
= (V[m2/m1] + V) / (1 + (V * V[m2/m1] / c^2))

With more careful subscripts, we have

V[m2/s] = (V[m2/m1] + V[m1/s]) / ( 1 +(V[m1/s] * V[m2/m1] / c^2))

+--------------------------------------------------------- ---------+
| VELOCITY RELATION IN EINSTEIN PHYSICS                             |
|                                                                   |
| V[m2/s] = (V[m2/m1] + V[m1/s]) / (1 + (V[m1/s] * V[m2/m1] / c^2)) |
|                                                                   |
| V[m2/m1] = (V[m2/s] - V[m1/s]) / (1 - (V[m2/s] * V[m1/s] / c^2))  |
+-------------------------------------------------------------------+

The second expression is a rotation of the first, as explained
below. It is applicable when the speeds of the two motional platforms
are given relative to the stational frame.

Comparison with Newton
--------------------
At speeds near the speed of light, the deviation from Newton shows
up. For V[m1/s] = 0.6*c and V[m2/m1] of 0.8*c, Newton figures the
combined speed V[m2/s] to be 1.4*c. However,

V[m2/s] = (V[m2/m1] + V[m1/s]) / (1 + (V[m1/s] * V[m2/m1] / c^2))
= ((0.8) + (0.6)) / (1 + (0.8) * (0.6) / (1^2)))
= (1.4) / (1 + (0.48) / (1))
= (1.4) / (1.48)
= (0.9459)

The combined speed is still less than lightspeed. In fact, even in
the limits where both of the ingredient speeds are c, the resultant
combined speed is at most c, but never greater than c. Hence, once
again we have the imposition of a real limit on the relative speed
achievable, that of light. To see this, we try it,

V[m2/s] = (V[m2/m1] + V[m1/s]) / (1 + (V[m1/s] * V[m2/m1] / c^2))
= (c + c) / (1 + (c * c / c^2))
= 2 * c / (1 + (c^2 / c^2))
= 2 *c / (1 + 1)
= 2 * c / (2)
= c

Fizeau's Experiments
------------------
Fizeau schemed to test the aether theory by measuring the speed of
light in flowing water. In 1859 he arranged a set of tubes with water
pumped thru them and sent light beams into them. By interferometry,
which he invented, he assayed the speed of light upstream and
downstream in the water.
He found that the speed was slightly less than the Newton addition
of velocities, but derived a wrong formula for it. It worked for the
extremely low speed of the water, at most 7m/s. This was the first
indication that something peculiar was going on that compound speeds
are not adding according to the only method then known, that of
Newton.

Shift of Platforms
----------------
In the above examples we worked with 'piggybacked' platforms. That
is, an airplane was the m1 frame, a rocket fired from that plane
was the m2 frame. The speed of the rocket was added (by either Newton
or Einstein equations) to that of the plane. If we have the two
platforms m1 and m2 both referenced to the stational platform we have
a rotation of the basic equation. In Newton, the two speeds would be
subtracted to get the speed of m2 relative to m1.
Leaving out the algebra to do this rotation, we end up with

V[m2/m1] = (V[m2/s] - V[m1/s]) / (1 - (V[m2/s] * V[m1/s] / c^2))

This can be terribly confusing for the novice at relativity.
Suppose we have two spaceships running parallel with speeds of 0.9*c
and -0.7*c. (They are offset slightly so there is no actual headon
collision.) Newton gets the closing speed as (0.9*c)-(-0.7*c) =
(1.6*c).
By the Einstein formula, we have

V[m2/m1] = (V[m2/s] - V[m1/s]) / (1 - (V[m2/s] * V[m1/s] / c^2))
= ((0.9) - (-0.7)) / (1 - (0.9) * (-0.7) / (1^2)))
= (1.6) / (1-(-0.63) / (1))
= (1.6) / (1.63)
= (0.9816)

The weird thing is that from the stational frame we really do
clock the closing speed to be 1.6*c! It is from the one frame
experiencing the other that the speed is collapsed to lass than c.
All of the foregoing is so contrary to common sense that many home
astronomers valiantly try to dismiss it as some voodoo maths or
spurious illusion. Yet it is a real phaenomenon and is a direct result
of the distorted spacetime perceived by observers in relative motion. ```