On Tue, 12 Oct 2004 14:59:03 +0100, John Kennaugh wrote:

A comment on the 90 degree case, and on Einstein's use of the word
"infinity":


If it is an infinite distance then the angle between them when it was
emitted is indeterminate surely.

[ ... ]

The infinite distance is surely to avoid complications with parallax,
light coming from a fixed direction.

No, not at all. It doesn't avoid those problems at all.

The "infinite distance" isn't really supposed to be infinite, any more
than vectors are really "infinitesimal". Einstein was writing in 1905, of
course, and a lot of the definitions used in calculus on manifolds hadn't
yet been worked out. He uses "infinite" as shorthand for "the limit as
the distance becomes large". If he worked explicitly with the limits a
lot of things would be much messier.

Assume the emitter is following a path that is, at its closest
approach, a distance L from the observer, and 0 L infinity. Now draw
a line between the observer and the emitter. A photon travels from the
emitter to the observer. At the moment when the photon arrives, the line
between the observer and emitter makes a right angle with the emitter's
path ... phi = 90 degrees.

From that, one can compute the angle the line between the observer and
emitter made with the emitter's path at the moment when the photon was
emitted. That must also be the emission angle for the photon, in the
observer's frame! Call that angle theta.

Theta doesn't depend on "L". Therefore, the same relationship between the
angles will hold in Einstein's case of "infinite distance".

Let's see if I can work it out here... Call the line from the observer to
the emitter's path which makes a right angle with the path the
"perpendicular connector". The emitter's moving at v. The emitter emits
a photon which will arrive at the observer at the instant when the emitter
crosses the perpendicular connector (all this is in the observer's frame).
Then phi = 90 degrees.

Call the time of emission 0.

The distance the emitter travels between emission and reception must be
v*t, where t is the time of reception.

In that time, if C=1, the photon traveled a distance "t".

I assumed to start with that the perpendicular connector has length L.
So, we have, by Pythagoras,

L^2 + (v*t)^2 = t^2

tan(theta) = L/v*t (angle the photon is emitted at, relative to the
emitter's path, as viewed by the observer). Fiddling a bit,

(L/vt)^2 = 1/v^2 - 1 = (1 - v^2)/v^2

L/vt = +/- sqrt((1-v^2)/v^2)

L/vt = +/- 1/v*gamma

Taking the positive root,

tan(theta) = 1/(v*gamma)

So the emission angle is, indeed, independent of the distance. If the
emission angle is 90 degrees, tan(theta) = infinity, so vt=0, v=0, and
the emitter isn't moving. As v-c, gamma-infinity, and the emission
angle goes to zero.


--
I can be contacted through http://www.physicsinsights.org

Creighton Hogg wrote:

On Tue, 12 Oct 2004, EjP wrote:


John Kennaugh wrote:

Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]


His equation is correct. Paul gave a detailed answer, but the
short version is that Einstein uses a (nowadays) nonstandard
convention of measuring the angle wrt the *source*, rather
than the observer.


Okay, my question here is why is someone still using Einstein's original
paper as a reference? Why not a derivation from the books used to teach
the subject now?


Most people who go to the original paper are nuts obsessed with
finding some flaw in special relativity. They believe that
if they can identify an error in Einstein's original paper,
a century of experimental evidence will vanish. There've even
been a couple of guys who have gone on at length about the
details of the *wording* of the paper.

I don't see evidence that this poster falls into that category.
I think he was just pointing this out from a standpoint of
historical interest.

This particular "mistake" gets posted here on a fairly regular
basis.

In fairness, unlike most original works, Einstein's 1905 paper
is extremely close to a modern treatment of the subject, and
still quite readable - this particular example notwithstanding.

-E

[Usenet Monster is dropping posts all over the place. I'm reposting
from Google -- apologies if this comes through twice. I've also added
one brief comment to my original post.]

I had a comment on the 90 degree case, and on Einstein's use of the
word "infinity":

On Tue, 12 Oct 2004 14:59:03 +0100, John Kennaugh wrote:

If it is an infinite distance then the angle between them when it
was emitted is indeterminate surely.

[ ... ]

The infinite distance is surely to avoid complications with
parallax, light coming from a fixed direction.

No, not at all. It doesn't avoid those problems at all.

The "infinite distance" isn't really supposed to be infinite, any more
than vectors are really "infinitesimal". Einstein was writing in
1905, of course, and a lot of the definitions used in calculus on
manifolds hadn't yet been worked out. He uses "infinite" as shorthand
for "the limit as the distance becomes large". If he worked
explicitly with the limits a lot of things would be much messier.

Assume the emitter is following a path that is, at its closest
approach, a distance L from the observer, and 0 L infinity. Now
draw a line between the observer and the emitter. A photon travels
from the emitter to the observer. At the moment when the photon
arrives, the line between the observer and emitter makes a right angle
with the emitter's path ... phi = 90 degrees.

(NB -- At the moment when phi=90 degrees, it's 90 degrees in both the
emitter's and the observer's frame, and events which occur "at that
moment" anywhere on the line connecting them are simultaneous in both
frames. So in this case, it doesn't matter which frame you use.
That's not true for other angles.)

From that, one can compute the angle the line between the observer and
emitter made with the emitter's path at the moment when the photon was
emitted. That must also be the emission angle for the photon, in the
observer's frame! Call that angle theta.

Theta doesn't depend on "L". Therefore, the same relationship between
the angles will hold in Einstein's case of "infinite distance".

Let's see if I can work it out here... Call the line from the
observer to the emitter's path which makes a right angle with the path
the "perpendicular connector". The emitter's moving at v. The
emitter emits a photon which will arrive at the observer at the
instant when the emitter crosses the perpendicular connector (all this
is in the observer's frame). Then phi = 90 degrees.

Call the time of emission 0.

The distance the emitter travels between emission and reception must
be v*t, where t is the time of reception.

In that time, if C=1, the photon traveled a distance "t".

I assumed to start with that the perpendicular connector has length
L. So, we have, by Pythagoras,

L^2 + (v*t)^2 = t^2

tan(theta) = L/v*t (angle the photon is emitted at, relative to the
emitter's path, as viewed by the observer). Fiddling a bit,

(L/vt)^2 = 1/v^2 - 1 = (1 - v^2)/v^2

L/vt = +/- sqrt((1-v^2)/v^2)

L/vt = +/- 1/v*gamma

Taking the positive root,

tan(theta) = 1/(v*gamma)

So the emission angle is, indeed, independent of the distance. If the
emission angle is 90 degrees, tan(theta) = infinity, so vt=0, v=0, and
the emitter isn't moving. As v-c, gamma-infinity, and the emission
angle goes to zero.


--
I can be contacted through http://www.physicsinsights.org

............... ...However, in an either case, it has had to be only the
way and how to eradicate the confusion, as to arrive to distinguish among
the both sides. Therefore, because it is a systematic confusion which is
already a classical matter.
............... ...As however, to distinguish the right way as the right
side as to make the difference. And a simply as a basically, along a placing
the Special Relativity, the way it has had to be, as a simply which is out
of the gravity. And to a definitely define as a definite and absolutely
remember that the General Relativity is a strict as an absolute theory of
the Gravity, which is a systematically out of the classical mechanics, and
this is a simply what is all about, definitely as amatter a
fact!!!!!!!!!!!!!!!!!!....................

--
Ahmed Ouahi, Architect
Best Regards!

"EjP" wrote in message
...
Creighton Hogg wrote:

On Tue, 12 Oct 2004, EjP wrote:


John Kennaugh wrote:

Maybe this is something everyone knows except me (and Paul Anderson)
but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]


His equation is correct. Paul gave a detailed answer, but the
short version is that Einstein uses a (nowadays) nonstandard
convention of measuring the angle wrt the *source*, rather
than the observer.


Okay, my question here is why is someone still using Einstein's original
paper as a reference? Why not a derivation from the books used to teach
the subject now?


Most people who go to the original paper are nuts obsessed with
finding some flaw in special relativity. They believe that
if they can identify an error in Einstein's original paper,
a century of experimental evidence will vanish. There've even
been a couple of guys who have gone on at length about the
details of the *wording* of the paper.

I don't see evidence that this poster falls into that category.
I think he was just pointing this out from a standpoint of
historical interest.

This particular "mistake" gets posted here on a fairly regular
basis.

In fairness, unlike most original works, Einstein's 1905 paper
is extremely close to a modern treatment of the subject, and
still quite readable - this particular example notwithstanding.

-E

In message ,
Creighton Hogg writes


On Tue, 12 Oct 2004, EjP wrote:

John Kennaugh wrote:
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]


His equation is correct. Paul gave a detailed answer, but the
short version is that Einstein uses a (nowadays) nonstandard
convention of measuring the angle wrt the *source*, rather
than the observer.

Okay, my question here is why is someone still using Einstein's original
paper as a reference? Why not a derivation from the books used to teach
the subject now?



While I want to be sure of those parts of relativity which interest me
(i.e. the basics) I have not great desire to become an expert in
relativity and have neither the time nor the inclination to get my head
around the modern space-time geometry approach. Perhaps I have been
unlucky in the books I have looked at in that they either insult your
intelligence or jump into space time geometry. I need something in
between and Einstein's original will do me for now. In any case
Einstein's paper I can download via the WWW free. If you know a good
text book I can download let me know.

The book I have which does have a reasonable chapter on relativity does
not cover the Doppler equation. I did a search on WWW for 'Doppler
Einstein relativity' and came up with nothing useful surprisingly
enough. Maybe gave up too soon )
--
John Kennaugh
to email convert the number from hex to decimal

Androcles writes

"John Kennaugh" wrote in message
news
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer then
Cos(Fi) = 1. Einstein states that this gives equation [2] Which it does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

We know that in acoustics you get Doppler if either the source is moving
or the observer is moving or both relative to the air, the propagating
medium. We are all familiar with the diagrams/equations in the text book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in frequency
due to speed so that if you are going to produce an equation which tells
you what the frequency is when the source is moving it is necessary to
include it. Time dilation makes time intervals increase which is the same
as making frequencies reduce - frequency being the reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]


Ok, I'll interrupt here. Let me finish before interrupting.
This has been thrashed out on this newsgroup any number of times.
Given: t' = (t-vx/c^2)/sqrt(1-v^2/c^2)
Now, v is a velocity, and is the distance x divided by the time t.
So x is vt.
Substituting for x,
t' = (t-v*vt/c^2)/sqrt(1-v^2/c^2)
t' = (t- t *v^2/c^2)/sqrt(1-v^2/c^2)
t' = t(1 - v^2/c^2)/sqrt(1-v^2/c^2)
t' = t. sqrt(1 - v^2/c^2)
and look, it agrees with the moving clock (t') running slow,
sqrt(1-v^2/c^2) is less than 1.

No. Suppose t is the time between two 1s ticks. If you were right t' is
smaller then the interval between ticks is smaller - the clock is
ticking faster - time has speeded up. Look up 'dilate' = "become wider
or larger" as in eyes dilated.

The amateur to this game simply ignores x, as you'll find on numerous
websites, and writes this as
t' = (t-vx/c^2)/sqrt(1-v^2/c^2), x = 0, so
t' = t/sqrt(1-v^2/c^2)
but then the moving clock (t') is running fast,

The 'amateur' has it right. t't. A tick is stretched so the clock
appears to be going slower. It is actually easier to think in terms of
frequency.

t' = t/sqrt(1-v^2/c^2)

converts to

1/f' = 1/(f(Sqr(1-v^2/c^2)

f' = f Sqr(1-v^2/c^2)

This says the frequency of the clock ticks appears less = the clock
going slower. f' is a lower frequency than f because Sqr(1-v^2/c^2)1

[.....]
--
John Kennaugh
to email convert the number from hex to decimal

EjP writes
Creighton Hogg wrote:
On Tue, 12 Oct 2004, EjP wrote:

Okay, my question here is why is someone still using Einstein's
original paper as a reference? Why not a derivation from the books
used to teach the subject now?

Most people who go to the original paper are nuts obsessed with
finding some flaw in special relativity. They believe that
if they can identify an error in Einstein's original paper,
a century of experimental evidence will vanish. There've even
been a couple of guys who have gone on at length about the
details of the *wording* of the paper.

I don't see evidence that this poster falls into that category.

It is instructive to read the original and see just how woolly the
wording is. I certainly wouldn't give him 10/10 for it. Someone once
took exception to the wording of the version of the second postulate I
was using so I went to the original to see what that said.
What he describes as his second postulate:

" light is always propagated in empty space with a definite velocity c
which is independent of the state of motion of the emitting body"

has at least 3 meanings. I do not consider it important because it has
been 'sorted out' since. It does mean there is no definitive wording so
no one can object to mine )

"The speed of light is constant w.r.t the observer observing it".

Neither would I look for errors. If there are any I assume someone
noticed in the first decade.

I think he was just pointing this out from a standpoint of
historical interest.

This particular "mistake" gets posted here on a fairly regular
basis.

I don't think he actually defines the positive direction of v but what I
find intriguing is that simply by changing the sign of v and rearranging
you end up with what appears to be the same equation upside down
especially as my version was derived from, and appeared to be made up
of, two distinct terms describing two apparently different effects.


In fairness, unlike most original works, Einstein's 1905 paper
is extremely close to a modern treatment of the subject, and
still quite readable - this particular example notwithstanding.

I think one could do a lot worse than take his original, bring the
terminology up to date a bit and republish. For example For
"'stationary' system of co-ordinates" substitute the modern "Inertial
Frame of reference" 'stationary' was a bad word to use and the fact he
put it in quotes indicates that he was not happy with it.

Having said all that I have to say that I do not accept relativity as a
theory for many reasons but my objections do not relate to Einstein's
mathematics.

Essentially the second postulate is simply the first + Source
independence. Source independence was a well established belief when
Einstein wrote his paper. The basis of that belief was another belief
namely that the speed of light was constant w.r.t the ether so could not
be constant w.r.t the source. The first 'evidence' for source
independence was DeSitter's analysis of binary stars which came after
1905 (and was discredited in 1965 anyway) so there was no justification
in 1905 for the second postulate other than long established belief (in
the ether).

Relativity then is based on Source independence which was for 200 years
based on something we no longer believe in, included in Einstein's
theory for no logical reason, and for a further 60 years faith
maintained on evidence which has been discredited. It is hardly a
promising start and the present evidence for source independence is
hardly overwhelming and there is at least some indicating source
dependency.

Then there is the question as to why it was accepted. There were
alternatives - Walter Ritz's emission theory being the most prominent.
This was based on the idea of source dependency. This was the logical
way to go, post MMX. If you accept that MMX showed there was no ether
then you accept that a source is surrounded by nothing which can affect
the speed of light which logically means that there is nothing but the
source which the speed of light can be dependent upon. Add to that the
fact that light was made up of particles, photons, then who in his right
mind would assume that a particle ejected by a source would not have a
speed relative to the source which ejected it? One is forced to the
view that SR was accepted for political reasons rather than objectivity.
The alternative would be to accept they had been wrong for 200 years.

Then again relativity has had to ring fence itself from criticism by
declaring itself a 'principle theory'. Thus all the questions I want
answered are met with "relativity does not attempt to answer that sort
of question". A principle theory is a mathematical model and I am
interested in what it is modelling in the physical world. If light is
indeed source dependent it is not because Einstein postulated that it be
so but because the real physical process which transports real physical
energy from the source to the observer makes it so. What is that
process? It is a question of causality. What is it which causes the
speed of light to be constant w.r.t the observer observing it. The
observer? not credible. Lorentz's ether? Rejected by relativists.
Einstein's 'mobile ether'? He couldn't get further than vague
statements. What then? "The modern concept of a physical theory means
that it is not the sort of question we consider it necessary for a
theory to address." OK I cling to 'old fashioned' values which I believe
are right.

Then there is the statement that 'relativity works'. The 'fall back'
position. It has to work because physics recognises no other theory.
When the earth centred universe was the only recognised theory that
worked. While it is true that you can only disprove a theory you cannot
prove one, it is also true that a theory can only be rejected in favour
of a better one and Physics had done a first rate PR job in making sure
that few even know that there ever was an alternative.

Experiment is interpreted in terms of relativity and the pieces of the
jigsaw constructed in terms of relativity. The result is a picture where
all the pieces (apart from a few) seem to fit. The question no one can
answer is whether if Ritz's theory had been accepted whether an equally
convincing picture would have emerged. A simpler one perhaps. We do not
know because the route was never explored for political reasons and
there is far too much at stake to allow it to be explored now.

If we look at physics in action it is less than convincing. The BB
theory is a fairly recent theory compared to SR. There is not nearly so
much at stake if it is disproved yet when the facts didn't fit,
'reality' was adjusted to fit the theory by the invention of inflation,
dark matter and dark energy. The theory must be right so there must have
been a period of inflation and there must be DM and DE.

If that sort of thing has been propping up relativity for a century
there is far too much at stake now in the way of reputation to ever let
relativity fail. If you are allowed to invent the physics equivalent of
a tooth fairy, e.g. the virtual photon, which does not have to obey any
of the usual laws of physics and can therefore do whatever you want it
to do it, it is hardly difficult to make what it does 'agree fully with
experiment'. That is Harry Potter Physics. My understanding is that a
virtual photon can be everywhere at the same time, can act instantly
over any distance and can attract or repel as appropriate and of course
cannot be detected because it pops in and out of existence so quickly
there is no point in trying to prove they don't exist. Do you really
call that 'science'?

I don't think the average physicist takes string theory and its 10 or 26
dimensions seriously but where the whole thing falls apart is that it is
impossible to draw a line in the sand and say here is where serious
physics ends - beyond is speculation. If you cannot do that the whole of
physics is suspect.
--
John Kennaugh
to email convert the number from hex to decimal

"John Kennaugh" wrote in message
.uk...
Androcles writes

"John Kennaugh" wrote in message
news
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then
Cos(Fi) = 1. Einstein states that this gives equation [2] Which it does.

f' = f (1 - v/c)/Sqr(1-v^2/c^2) ----------[1a]
f' = f Sqr[(1-v/c)^2 /(1-v^2/c^2)]
= f Sqr[(1-v/c)(1-v/c)/(1-v/c)(1+v/c)]
= f Sqr[(1-v/c)/(1+v/c) ---------------- [2]

I have shown the working because the equation which interests me is [1a]

We know that in acoustics you get Doppler if either the source is
moving
or the observer is moving or both relative to the air, the propagating
medium. We are all familiar with the diagrams/equations in the text
book.

In Relativity we have two complications, one is the assumed absence of a
propagating medium and the other is time dilation. Time dilation is not
strictly Doppler shift but never the less represents a change in
frequency
due to speed so that if you are going to produce an equation which tells
you what the frequency is when the source is moving it is necessary to
include it. Time dilation makes time intervals increase which is the
same
as making frequencies reduce - frequency being the reciprocal of time.

If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]


Ok, I'll interrupt here. Let me finish before interrupting.
This has been thrashed out on this newsgroup any number of times.
Given: t' = (t-vx/c^2)/sqrt(1-v^2/c^2)
Now, v is a velocity, and is the distance x divided by the time t.
So x is vt.
Substituting for x,
t' = (t-v*vt/c^2)/sqrt(1-v^2/c^2)
t' = (t- t *v^2/c^2)/sqrt(1-v^2/c^2)
t' = t(1 - v^2/c^2)/sqrt(1-v^2/c^2)
t' = t. sqrt(1 - v^2/c^2)
and look, it agrees with the moving clock (t') running slow,
sqrt(1-v^2/c^2) is less than 1.

No.

Suppose t is the time between two 1s ticks. If you were right t' is
smaller then the interval between ticks is smaller - the clock is ticking
faster - time has speeded up. Look up 'dilate' = "become wider or larger"
as in eyes dilated.

Put some numbers in. For v = 0.866c (always convenient),
sqrt(1-v^2/c^2) = sqrt(1- 3/4) = sqrt( 1/4) = 1/2.
5 hours = 10 hours * 1/2.
The moving clock records 5 hours (t') for every 10 hours (t) of the
stationary clock, so:
t' = t.sqrt(1-v^2/c^2)

The interval between ticks of the moving clock equals two intervals of the
stationary clock and is therefore "dilated."

0----------1----------2----------3----------4---------5 time t'
0-----1----2----3----4----5----6----7-----8----9---10 time t


The amateur to this game simply ignores x, as you'll find on numerous
websites, and writes this as
t' = (t-vx/c^2)/sqrt(1-v^2/c^2), x = 0, so
t' = t/sqrt(1-v^2/c^2)
but then the moving clock (t') is running fast,

The 'amateur' has it right. t't. A tick is stretched so the clock appears
to be going slower. It is actually easier to think in terms of frequency.

t' = t/sqrt(1-v^2/c^2)

converts to

1/f' = 1/(f(Sqr(1-v^2/c^2)

f' = f Sqr(1-v^2/c^2)

This says the frequency of the clock ticks appears less = the clock going
slower. f' is a lower frequency than f because Sqr(1-v^2/c^2)1

Put some numbers in. I don't see why I should do it for you again.

[.....]

So now you've snipped after you thought you had it right
but it turns out you are totally wrong.
Well done.
Androcles.

--
John Kennaugh
to email convert the number from hex to decimal

John Kennaugh:
Maybe this is something everyone knows except me (and Paul Anderson) but
here goes anyway.

In ON THE ELECTRODYNAMICS OF MOVING BODIES Sect 7 Einstein says the
frequency of the light perceived by the observer is given by the
equation

f' = f (1 - Cos(Fi)v/c)/Sqr(1-v^2/c^2) ---------------- [1]

If the source is travelling directly to or directly from an observer
then Cos(Fi) = 1. Einstein states that this gives equation [2] Which it
does.

It's much simpler to start with the relation,

k.x - wt = k'.x' = w't',

substitute the lorentz transforms of x and t for x' and t', then solve
for k' and w'.

Androcles writes

"John Kennaugh" wrote in message
o.uk...
Androcles writes

"John Kennaugh" wrote in message
news If we ignore the Doppler component then the frequency observed by an
inertial observer due to time dilation would be.

f' = f Sqr(1-v^2/c^2) -------------- [3]


Ok, I'll interrupt here. Let me finish before interrupting.
This has been thrashed out on this newsgroup any number of times.
Given: t' = (t-vx/c^2)/sqrt(1-v^2/c^2)
Now, v is a velocity, and is the distance x divided by the time t.
So x is vt.
Substituting for x,
t' = (t-v*vt/c^2)/sqrt(1-v^2/c^2)
t' = (t- t *v^2/c^2)/sqrt(1-v^2/c^2)
t' = t(1 - v^2/c^2)/sqrt(1-v^2/c^2)
t' = t. sqrt(1 - v^2/c^2)
and look, it agrees with the moving clock (t') running slow,
sqrt(1-v^2/c^2) is less than 1.

No.

Suppose t is the time between two 1s ticks. If you were right t' is
smaller then the interval between ticks is smaller - the clock is ticking
faster - time has speeded up. Look up 'dilate' = "become wider or larger"
as in eyes dilated.

Put some numbers in. For v = 0.866c (always convenient),
sqrt(1-v^2/c^2) = sqrt(1- 3/4) = sqrt( 1/4) = 1/2.
5 hours = 10 hours * 1/2.
The moving clock records 5 hours (t') for every 10 hours (t) of the
stationary clock, so:
t' = t.sqrt(1-v^2/c^2)

The interval between ticks of the moving clock equals two intervals of the
stationary clock and is therefore "dilated."

0----------1----------2----------3----------4---------5 time t'
0-----1----2----3----4----5----6----7-----8----9---10 time t


The amateur to this game simply ignores x, as you'll find on numerous
websites, and writes this as
t' = (t-vx/c^2)/sqrt(1-v^2/c^2), x = 0, so
t' = t/sqrt(1-v^2/c^2)
but then the moving clock (t') is running fast,

The 'amateur' has it right. t't. A tick is stretched so the clock appears
to be going slower. It is actually easier to think in terms of frequency.

t' = t/sqrt(1-v^2/c^2)

converts to

1/f' = 1/(f(Sqr(1-v^2/c^2)

f' = f Sqr(1-v^2/c^2)

This says the frequency of the clock ticks appears less = the clock going
slower. f' is a lower frequency than f because Sqr(1-v^2/c^2)1

Put some numbers in. I don't see why I should do it for you again.

Had you put the numbers in to this equation rather than the other you
would have spotted your own mistake. I really don't enjoy putting you
right all the time.

You once defined a clock, quite rightly as an oscillator and a counter.
Suppose we have a 100Hz oscillator and a counter which increments every
100 cycles i.e. once per second. Suppose we now make that clock travel
at 0.866c. The equation, which you claim is wrong says the frequency
will appear to a stationary observer as

f' = f Sqr(1-v^2/c^2) = 100 x 0.5 = 50 Hz

every 100th cycle is now every 2 seconds so the clock will
increment every 2 seconds instead of every 1. It is going slower. Time
is dilated.

The problem you fell foul of is semantics. We talk about 'time' in all
sorts of different ways and it is very easy to get confused. Been there
- sorted it out. The safest way is to stick to frequency where there
isn't the semantic confusion.

In the twin paradox the moving twin appears to have been ageing for a
shorter time. Shorter implies 'smaller' you would think i.e. a smaller
number representing time. But time is measured in ticks and if the ticks
are longer (implying a larger number) then there are less of them so the
time accumulated is shorter. It is a mental minefield which is why I say
stick with frequency that only has one meaning and the correct
interpretation for time is the reciprocal of frequency. If your clock is
going slower the frequency of the ticks is lower so the period of the
ticks is longer. The correct time dilation equation should give a larger
apparent value for time.

If we take

t' = t/sqrt(1-v^2/c^2)

Which you have 'proved' to your own satisfaction is wrong and make t = 1
hour and v = 0.866c then

t' = 1/(0.5) = 2

and you would immediately claim that is wrong because 2 is more hours
than 1 and time dilation says it should be less doesn't it? No. What
that equation says is that if the chap on the space ship spends 1 hour
painting it, then the stationary observer will see it in slow motion,
time dilated, and will have to watch paint dry for 2 hours. Every 1
second tick on the space ship is slowed down and is seen by the
stationary observer to take 2 seconds.



[.....]

So now you've snipped after you thought you had it right

I had so was justified in snipping

--
John Kennaugh
to email convert the number from hex to decimal