"PD" wrote in message
oups.com...
Ah, see, and I was assuming you were out of town on pressing business.
Are you coming back to resume our discussion?

PD

Of course. It is your move.
My last post was ... hmm... let me check...

"PD" wrote in message
oups.com...
Ah, see, and I was assuming you were out of town on pressing business.
Are you coming back to resume our discussion?

PD

Of course, Its your move. My last post was...

Hmm... my mistake. I had missed you reply.
I apologise.

My indentation appears to have failed.
Androcles:
It is known that Einstein's equations --as usually understood at the
present time--when applied to moving bodies, leads to asymmetries
which
do not appear to be inherent in the phenomena. Take, for example, the
reciprocal lengths of two rods. The observable phenomenon here
depends
only on the relative motion of the rod A and the rod B, whereas
Einstein's view draws a sharp distinction between the two cases in
which
either the one or the other of these bodies is in motion. For if the
rod
A is in motion and the Rod B at rest, there arises in the
neighbourhood
of the Rod A a change in length. But if the rod A is stationary and
the
rod B in motion, a change in length arises in the neighbourhood of
the
Rod B. In the Rod A, however, we find a length change to which in
itself
there is no corresponding energy, but which gives rise--assuming
equality of relative motion in the two cases discussed--to a change
in
length of the same intensity as those produced by the motion the
former
case.


No response?

Not here. I had hoped that I addressed it above.





PD:
What I said was if a rod of fixed length is moving with speed
v in K, it will have a fixed length where it is stationary in k, not
necessarily the same fixed length as measured in K. The only thing
I'm
establishing is that a rod that isn't changing in length in one
frame
is also not changing in length in the other frame. A rod with length
1.2 (and always 1.2 -- that is, "fixed") in K may have length 0.80
(and
always 0.80 -- that is, "fixed") in k. Or it may have length 1.2
(and
always 1.2 -- that is, "fixed") in k. Of the two latter choices, we
haven't figured it out, but it's fixed -- not changing in time -- in
either case.


PD:
OK so far?

Androcles:
No. You haven't agreed on a method of synchronizing the rods yet.
PD:
Once you explain what you mean by "synchronizing the rods," we can
sort
that out.


Androcles:
I mean the same as synchronizing the clocks. I cut one rod to the
same
length as the other.
PD:
This I can only do if both the template and the synchronized rods are
at rest with respect to the cutter/observer.

Androcles:
Template? What template? I need no template. If I want to cut a rod
to
the same length as another, the master rod is the template.
I'll agree the two rods are relatively at rest when the operation is
conducted.
I can then carry the master rod to the moving frame and cut a third
rod.
I can do the same with a clock.

Androcles.
I set one clock to the same time as the other.

PD:
How do you do that if the clocks are separated by a distance?

Androcles:
Each clock is an oscillator and a counter. I'm not concerned about the
offset, as I've previously explained. The counter part of the clock
can
be anywhere. Changing the distance will change the number of
oscillations
"in flight" between the oscillator and the counter, and hence the
offset.
The issue to be addressed is any change in the oscillator, the "gain".

PD:
Einstein
proposes a solution with the emitter/mirror/receiver scheme.

Androcles:
Sure. When the clocks are at rest they can be synchronized.
No argument from me on that score. My concern isn't the offset
nor do I need to send a signal to the distant clock. I simply
count its oscillations as they arrive, or read its dial. I'm waiting
for
you to show the gain of the oscillator changes.

I don't intend to show you that. I think you *do* need more than what
you
propose, though, because your test is *merely* a gain-gauge, something
that
checks rate. Einstein's scheme sets *simultaneity* (warns about offset)
each
time the synchronization is done. Repetition of the synchronization
tells
you about gain.


Androcles:
In other words, I take out any offset. Offset can be quite useful,
New York time is offset from London time by -5 hours. It would
become a problem if the clocks (or the rods) had different gains,
though.
Why, a New York clock would eventually record a different offset
to the London clock if it had a greater gain, and the two rods would
no
longer be of the same length if I heated one of them and not the
other.
The difference in time or the difference in length is the offset.
Synchronizing rods is no different to synchronizing clocks. All we
do is remove the offset.

PD:
OK, I can work with that. That is, if we apply a synchronization
procedure to two clocks with no difference in gain, then we would find
that the synchronization equation that we use to define it is
satisfied
-- that is, the clocks are still in synchronization. However, if there
is a difference in gain, then the synchronization equation would not
be
satisfied and we would have to make an adjustment to one clock or
another to correct that. The greater the time since the last
synchronization, the less confident we would be that the clocks would
still be synchronized. Indeed, the difference in the gains in the
clocks would determine how frequently we would have to perform the
synchronization so that the agreement would be within an agreed-on
tolerance.

Androcles:
For the purpose of this debate, let us stipulate to there being no
mechanical, electrical or otherwise difference in the clocks, both
of which are perfect. Likewise, no heat expansion or similar change
in the rods.
We do not live in an ideal universe, but this is a theoretical
exercise
and we allow no tolerance whatsoever. Any difference in the clocks
that may be found is to be solely a result of calculation, not of
some practical consideration. Thus we can imagine the clock to
be many light years distant, although we cannot place it there
in practice.

PD:
OK, that's fine too. And what that means is that, once the clocks are
synchronized using Einstein's scheme, every repetition of the
synchronization will verify that the clocks are still synchronized.
(Again,
as long as the clocks are stationary in that frame.)

Androcles:
Excellent. When Stella returns to Terence, the clocks and rods are still
synchronized because they are once again stationary in the same frame.
Terence has not aged as supposed by Baez and Koks, who cannot
count to 14.
http://www.androc1es.pwp.blueyonder....oksDoppler.htm

PD:
A similar procedure would apply to rods with different gains. Note,
however that the two rods I'm synchronizing and a master rod all have
to be at rest with respect to each other to do this. I cannot apply
this procedure to a rod at rest and one flying by.

Androcles:
Agreed. All synchronization to be done at rest. Let us now declare
the clocks and the rod to be synchronized, and you may begin
your experiment.

PD:
OK, so now what we're going to do is we're going to take a rod that was
stationary in frame K, and we'll walk it backwards a few miles, and
we're
going to speed it up to speed v toward the right in frame K. It's
reached v
by the time it passes the origin of K. (It actually doesn't matter that
it
was ever stationary in the frame K, but Einstein says we'll do it that
way,
so we'll go along). After the rod has been sped up, this rod is
stationary
in the frame k. That is, the location of the emitter and receiver at one
end
of the rod are at a fixed location in k, and the location of the mirror
at
the other end of the rod is at another fixed location in k.

Now, the observer in K knows that this rod can't be used to to
synchronize
any yardsticks in K, because it's not at rest.

Moreover, the
emitter/mirror/receiver can't be used to synchronize the clocks in K,
because the system is not at rest. That doesn't really matter. The K
observer can recheck the synchronization of K's yardsticks and clocks
with
another rod and emitter/mirror/receiver stationary in K.

Androcles:
What?
Why should I go out and remeasure my front yard just because a
bus went by?
I don't see what you are getting at. Synchronization is a comparison and
correction between k and K. The moving rod IS the k-frame, and we've
agreed that we cannot synchronize when the rod is moving. What is this
"another rod" doing here anyway?
The bus was parked outside, I made chalk marks where the wheels are
the lengths are now synchronized. Now you drive the bus around
the block, come zipping by and you want me to see if the chalk marks
have moved?
NO.




PD.
As you say, we'll
assume there is no drift, so the resynching would just verify that
clocks
are still synched and yardsticks are still the same length.

However, we *can* measure a length of the moving rod in K.

Androcles:
How?

PD.
The prescription above will do it.

Androcles:
I fail to see how you can take two rods, compare them in K,
and then claim you can measure the length of the bus err..
rod flying by.


PD:
Moreover, the moving rod *can* be used as a synching system in k.

Agree so far?

Androcles:

No. Show me again. The rods were synchronized before any
motion took place. Throw that "another rod" in the trash,
we are done with it, not that we ever needed it. The wheelbase
aligns with my chalk marks, and I'm not on the bus, you are.


Androcles.




Androcles:
You are going to try to tell me that relative motion changes the gain
of the rod and the gain of the clock, are you not?

PD:
Not at all! Far from it. That would be true if length were an innate
property of the rod, because then the only thing that could change the
length of the rod would be a process that physically affected the rod.
However, that's not what's going on here.

Androcles:
I'm pleased to hear it. Do continue.


Androcles.

"Timo Nieminen" wrote in message
news:Pine.LNX.4.50.0501131152150.18995-100000@localhost...
On Thu, 13 Jan 2005, Androcles wrote:

"Timo Nieminen" wrote:

Well, the speed c-v that he objected to my use of was exactly the
speed at
which the light pulse closes with the moving mirror.

But what would you expect? Do you really think Androcles would have
admitted being wrong?

No, I will not, because I'm not wrong.
(c-v) is not equal to 1/sqrt(epsilon0 * mu0), which are constants
and the basis of Mazer's argument that Maxwell's laws apply.
Therefore c-v is not permitted in SR, which says c = (c-v)/(1-v/c)

Consider a source (stationary or moving) emitting a pulse of light.
According to SR, that pulse of light moves at c (in free space). That
is,
in any inertial reference frame, dr/dt = c, where r(t) is the
instantaneous position of the pulse of light in that reference frame.

If we have a mirror (or some other object) moving at velocity v
directly
away from the position where the pulse was emitted from, we have
dP/dt = v.

d/dt being a linear operator, d(r-P)/dt = c-v, and therefore the rate
at
which the distance between the pulse of light and the moving object
becomes smaller is |c-v|.

That's just basic Galileian kinematics. Are you really trying to claim
that Galileo was wrong?

Even if you are claiming that Galileo was wrong, that's irrelevant,
since
you claim was that SR predicts a non-null result for MMX, and SR
assumes
that Galileian kinematics work in any given reference frame.

SR does NOT explain MMX, and source dependency does.

So, even after you failed to provide any refutation of the analysis of
the
original MMX showing that SR predicts that there would be no fringe
shift
(thereby demolishing your original claim), you continue to repeat your
claim. Thank you for the demonstration of your level of intellectual
honesty.

Intellectual honesty isn't something you would know the meaning of.
SR predicts a non-null result for MMX, because SR claims the speed of
light is invariant and x' =/= y'. I see no reason to repeat the math.



Source dependency also explains cepheids as quite ordinary stars,
recurrent novae as ordinary stars, flare stars as ordinary stars
and changes the orbital parameters of eclipsing variables, which
if the source independency model is used would rip apart from
tidal forces in a mere century.
NOBODY has ever measured the speed of light from a moving
source in the vacuum of space. That is about to change.

You have an experiment ready to do? Wonderful!

Sure I do, smart-arse. Shoot the moon from the ISS with a laser and
look at the reflection with HST. Repeat continously as the
ISS approaches and recedes relative to the moon. Track position
of the HST and ISS using GPS and time the events with the
best clocks available.
I've already given you an electronic MMX, and an analysis,
you didn't say "Wonderful!" to that, Mr. Intellectual Honesty.



But, given that you don't accept de Sitter's analysis, why do you
think
the "vacuum of space" is a good enough vacuum?

I don't believe in aether as you do, Mr. Intellectual Honesty.
I do believe you are a ****ing idiot and a lying dirtbag, though.

Now **** off.
Androcles

Could I ask you to repost this reply to the thread to which it belongs?
Topic-teleportation is a terrible nuisance.

PD

Franz Heymann wrote:

"Jesse Mazer" wrote in message
...


I wrote:



Well, you're thinking about it wrong. It would be possible to


choose


units of mass such that the gravitational constant G was equal to


1,


but that wouldn't make it any less real as a physical
constant--similarly, in cgs units, the basic units of charge (emu)


Sorry, that should be esu...emu is the unit of magnetic charge in


cgs units.

You might know what you were trying to say both in the original
statement and the correction.
I doubt if anybody else would.
Comment on this garbage, for instance
"emu is the unit of magnetic charge in cgs units"


OK, magnetic moment, I confused my terms, Jeez. I think you know what I
meant.

Jesse

"PD" wrote in message
ups.com...
Could I ask you to repost this reply to the thread to which it
belongs?
Topic-teleportation is a terrible nuisance.

PD

I replied to the thread whence came the query, but since it was my
error,
sure.
Androcles.

Franz Heymann wrote:

"Jesse Mazer" wrote in message
...


Franz Heymann wrote:







even though the
experiments to determine epsilon_0 and mu_0 have nothing to do


with




light.

No experiments have ever been done to determine the values of eps0


and


mu0.
They are artefacts necessary to define a certain set of units.
Such experiments are logically impossible, for the following


reasons:


mu0 is *defined* to be precisely 4* pi*10^-7
and eps0 is *defined* to make eps0*mu0 = c^-2
(and c, in turn, has a *defined* value, but we'll let that pass)






See my other post, eps0 is no less fundamental than G. *Today* we


don't


need to define them as separate fundamental constants, since we know
that eps0 = (mu0*c^2)^-1 and vice versa, but if you know the history


of


electromagnetism you'll know that they could not have been defined


this


way before Maxwell proved that the full set of Maxwell's equations
implied that electromagnetic waves should always travel at velocity
1/squareroot(eps0*mu0),
and realized that this value was very close to
the measured value of the speed of light, implying that light was


really


an electromagnetic wave, something that no one had realized before


him.

That is pure crap, since eps0 and mu0 were invented only in the late
thirties of the previous century, by which time Maxwell was no longer
around to comment on that.

In that case I was using esp0 and mu0 as shorthand for "the permittivity
of free space" and "the permeability of free space", regardless of the
system of units--it's a lot easier to write that way. Hadn't physicists
in Maxwell's time defined the concepts of the permittivity and
permeability of free space?

I note that wikipedia uses the same abbreviation in their entry on
Maxwell's equations:

http://en.wikipedia.org/wiki/Maxwell's_equations

In the late 19th century, because of the appearance of a velocity,
1/sqrt(eps0*mu0) in the equations, Maxwell's equations were only
thought to express electromagnetism in the rest frame of the
luminiferous aether (the postulated medium for light, whose
interpretation was considerably debated).





What Maxwell showed was that the speed of EM plane waves in free space
was related to the ratio between the esu of charge and the emu of
charge, a quantity which had already been measured experimentally at
that time.


OK, I checked and you're right about this. However, if you actual
examine the equations you'll see that these two constants are
proportional to permeability and 1/permittivity. Look at this page:

http://www.ee.surrey.ac.uk/Workshop/.../unit_systems/

In the section "The laws of Ampere and Coulomb", they describe the
magnetic force between two parallel wires of length L, a distance r
apart, with currents I1 and I2 running through them:

Fm = Km I1 I2 L / r
Ampère's Force Law

where Km is a constant whose value depends upon the unit system used
to measure the currents, forces and distances. In other words, having
decided how to define distance and force, you can then choose a Km
that defines how large your unit of current will be.


Similarly, the Coulomb force between two static objects of charges Q1
and Q2, a distance r apart, is:

Fe = Ke Q1 Q2 / r^2
Coulomb's Law

Where Ke is a constant whose value depends upon the unit system used
to measure the charges, forces and distances. In other words, having
decided how to define distance and force, you can then choose a Ke
that defines how large your unit of charge will be.


What Maxwell's equations show is that you cannot define Ke and Km
independently--once you define one, you have no choice about how to
define the other. The page explains why:


To understand this we need to introduce our third equation which
expresses the fact that electric charge, Q, and current, I, are
dependent quantities -
Q = I × T (Equation USQ)

where T is the time, in seconds, for which a given current, I, flows.
Equation USQ must hold regardless of the unit system chosen; one unit
of charge being represented by the passage of one unit of current for
one second. You should now be able to see that it is possible either
to invent some convenient value for Km and then calculate Ke or else
invent some convenient value for Ke and then calculate Km. You cannot
choose both Ke and Km independently because that would violate
Equation USQ. Performing both Ampere's and Coulomb's experiments
(Weber and Kohlrausch, 1856) gives you a value for the relationship
between Km and Ke -
Ke
http://www.ee.surrey.ac.uk/Workshop/advice/coils/unit_systems/#USG=
c^2 Km / 2 (Equation USJ)

where c is the speed of light in a vacuum.


Now, in modern terms Coulomb's law is written as:

Fe = Q1*Q2 / 4pi*eps0*r^2

And the Ampere's law for two parallel wires is:

Fm = I1*I2*L*mu0 / 2pi*r

So, we can see that Ke = 1/4pi*eps0, and Km = mu0/2pi. Thus, if
Maxwell's equations say that c is equal to squareroot(2Ke/Km), this is
the same as saying it's equal to squareroot(eps0*mu0). Thus measuring
the ratio of Ke/Km is exactly equivalent to measuring the product of
eps0*mu0.

In both cases, the idea is that you are free to define units of electric
charge however you want, but depending on your choice of units you must
insert a constant (Ke or esp0) in there to get the right force.
Similarly, once you pick your units of current, you must insert another
constant (Km or mu0) in there to get the right magnetic force. As they
say above, the ratio of these constants is determined
experimentally--"Performing both Ampere's and Coulomb's experiments
(Weber and Kohlrausch, 1856) gives you a value for the relationship
between Km and Ke". This is equivalent to the statement "performing both
Ampere's and Coulomb's experiments gives you a value for the
relationship between eps0 and mu0".




Back then they were defined experimentally



That is a contradiction in terms.


I don't think so. Again, think of the modern version of Coulomb's
equation for the force between static charges:

E = Q1*Q2/4pi*eps0*r^2

Regardless of the system of units, you can observe that the force
between two charges Q1 and Q2 at a distance r is *proportional* to
Q1*Q2/r^2, but to get the correct value for the force, and to get the
units to work out right, you need to have a constant in there, just like
with Newton's gravitational law F = G*m1*m2/r^2.

But thinking about it some more, I realize that there is one difference
between this and G--we have independent ways of defining units of
"mass", which existed before it was realized that the gravitational
force between masses was proportional to Mm/R^2, but in electromagnetism
the only way for 19th-century physicists to define the amount of charge
in an object would be to measure things like the force between charged
objects. So in this sense, I guess they wouldn't have to measure eps0
like we would measure G--they could *define* the units of charge in
terms of an arbitrary choice of eps0 (or Ke). But once they do this, now
they have a fixed set of units for current (since current is just charge
moving at some velocity), so they must determine mu0 (or Km) in Ampere's
law by experiment, just like with G. Alternately, they could define the
units of current in terms of an arbitrary choice of mu0/Km, but this
would give them a fixed set of units for charge, and then they would
have to determine eps0/Ke in Coulomb's law by experiment. Either way, at
least one of the two constants has to be found experimentally, in just
the same way that G is found experimentally.

In terms of our modern understanding, we can also define units of charge
in terms of elementary charged particles like electrons. So we do have
an independent way of measuring charge, just by counting the number of
charged particles; this is another way of showing why eps0 (or
equivalently, Ke) is a real physical constant which plays exactly the
same role as G in the gravitational force equation.

Jesse

Androcles wrote:

"Jesse Mazer" wrote in message
...


I do not have a theory. I have a model, and it is the vector addition
of velocities, too simple to call a theory.
There is nothing in the description above that accounts
for the secondary maxima so commonly observed. At 6000 ly,
there MAY not be enough distance for the faster light to pass the
slower before arriving, it depends on the velocity of the star and
the orbital attitude. I have not seen the curve for T Pix.
Nor can I disprove the above, it is a matter
of plausibility. I don't see the moon sucking up Earth's oceans,
do you? What I do see is the rings of Saturn where a moon once
was, ripped apart by tidal force. Were you aware that the moon
keeps the same face toward the Earth because of tidal force?
How does a less massive star suck matter from the more massive?
It is as plausible as the moon sucking the Earth's oceans dry


from a distance of the GPS constellation.


From the description, it seems to me that it was the central dense
white dwarf which was more massive, and the orbiting star was less
massive, and also that it was in a "close orbit".



Oh, ok. The Earth sucked the water from the lunar seas, the Moon
got tired of being bullied and moved further away.
Got any more good howlers, Dr. Physicist?


No, but the earth isn't very dense, and thus is not very massive. The
earth has a density of about 5 gm/cm^3, while estimates of the density
of white dwarf stars range from 100,000 through 1,000,000,000 gm/cm^3.
If the earth was as dense as a white dwarf, and if the moon had an
atmosphere, then yes, the earth might pull some of that atmosphere away.

Try to actually think through your analogies in the future. Also, you
still didn't answer my question about your own model--if you compare a
recurrent nova at distance R with a companion star of size S with an
ordinary binary star system at distance R with a companion star of size
S, then according to your theory, wouldn't the variations in magnitude
be about the same? I predict that the actual variations in magnitude in
the two cases would be wildly different, and thus that your "machine gun
on a merry-go-round" analogy is obviously not the correct explanation
for the variations in magnitude of recurrent novas.

Jesse

On Thu, 13 Jan 2005, Androcles wrote:

"Timo Nieminen" wrote:
On Thu, 13 Jan 2005, Androcles wrote:

"Timo Nieminen" wrote:
On Wed, 12 Jan 2005, Androcles wrote:
"Timo Nieminen" wrote:
On Wed, 12 Jan 2005, Androcles wrote:

I will now give the proof.
I place two identical evacuated tubes at right angles, with
mirrors
and
with
sources and detectors of light.
When the light reaches a detector, it turns off the source.
When NO light reaches the detector, it turns the source on.
The path length can be any reasonable one.
I connect the outputs of the detectors to an up/down counter,
one
for
up
and the other for down. I calibrate the tube lengths until
there
is
no
count.
I now have MMX, electronic version and air free.

Since your original assertion was specific to MMX, I'll first
deal
with
the actual MMX. Then your modified experiment.

I rotate the apparatus 90 degrees and what happens? NO count
happens.
Acording to SR, and there being velocity as the Earth moves and
rotates
x' = (x-vt)/ sqrt(1-v^2/c^2)
y' = y.
So, x' does not equal y'.

Why would you want to analyse the experiment using a reference
frame
moving wrt the Earth?

Simple, really.
If you analyse it from a frame that is at rest, you prove nothing
at
all.
Hence we analyze from a relatively moving frame. Get your
telescope
out and watch the fringe shifts from a plane if you like. As you
look
at Michelson's light, you'll notice it is doppler-shifted. Stay at
rest
and
you won't.

Your modified version makes Doppler shifts irrelvant - you can
ignore
the
wavelength and frequency, and only the speed matters.

Exactly. Well done. To use Einstein's own words, the "tip of the
ray".
Nobody is really interested on the following wave train, but that was
all
Michelson had.

So why bring up Doppler shifts?
If you knew they were irrelevant, why introduce them into the
discussion?

Me? If I recall correctly, you were the one that was bleating about the
old interferometer. I was introducing an electronic version. That's how
it got into the discussion.

It's more accurate to say that you were the one trying to avoid an
analysis of the original experiment. Easy to see why, since you couldn't
find fault with the analysis.

You did note that the analysis of the original MMX doesn't make use of
Doppler shifts?

You are the one who introduced Doppler shifts. If you knew they were
irrelevant, why did you introduce them?

Analysing it in a frame where it at rest proves everything that is
required.

Of course it does. The tip of the ray is travelling at c relative
to the source.

So, you agree that the rest-frame analysis proves everything that is
required?

Put the experiment on a plane, and the speed of light in the plane is
c.
To an observer outside the plane on the ground, it is c+v.
Hence the speed of light is source dependent. Why are you
arguing about it, you already agreed, right?

Avoiding the question, again. Can't you give simple answer such as "yes"
or "no"?

You were the one who wrote "Of course it does" in apparent agreement.
Since this seems to contradict other things you wrote, I sought
clarification. But it looks like you aren't willing to clearly state what
you mean by what you wrote. Perhaps I should just use the literal
reading and assume you agree with me?

Here's another question I don't believe you'll answer clearly:

In your theory (such as it is, although you claimed earlier to be a
Ritzian, which would imply a rather complete theory), what is the speed of
light (relative to the mirror) reflected by a mirror from a source that
was moving at v towards the mirror, relative to tehe mirror (all in free
space):

a) c
b) c+v
c) c-v

Yes, this is irrelevant to our previous discussion, but nothing in your
last post was relevant.

Define "rest frame".

A reference frame where the MMX apparatus is at rest.

Ok, so the speed of light is source dependent. The ground is moving
beneath the plane.

Doesn't follow. Are you capable of logical thought?

To Einstein and Michelson it meant the rest of the
Universe, with the Earth in motion about to the sun.

Perhaps to Michelson. Not to Einstein.

Crazy Einstein thinks the time to get from A to B is half the time
it takes to get from A to B and back to A again.
London to New York - 7 hours against the jet stream.
New York to London - 6 hours with the jet stream

2AB/(t'A-tA) = speed, so
(tau0+tau2) / 2 = tau1
(7+6)/2 = 7, right?

Obviously I'm discussing physics with someone that belongs
in a straight -jacket.

Completely irrelevant to your claim that Einstein believed in some
absolute rest frame. I see that all you could manage was a completely
irrelevant (and wrong) diversion. The least you could try and do is make
your diversions factually correct.

Let's see, what did you manage in your last post:

Unwillingness (or inability?) to answer simple yes/no questions.
Avoiding all questions of physics.
Irrelevant diversions.
Illogic.
Insults.
Factually incorrect statements.

6 out of 7 typically characteristics of your posts! Good to see you are
maintaining consistency.

All that was missing was overt dishonesty! (But perhaps I was too
charitable in assuming that your factual error was made in ignorance of it
being an error.)

--
Timo

"Androcles" wrote in message
. uk...

"Jesse Mazer" wrote in message

Yes, this is what SR predicts, because it says the speed of light
is
the same in all directions, regardless of whose rest frame you're
in.

Take it back to the funny farm.

I say, Androclown, you seem to have found a new idiom. You are using
it up rather fast. How about eating your bananas instead?

Franz