On May 6, 4:34�pm, YBM wrote:
rbwinn wrote:

...

No, you have no answer to the use of velocities, so you claim it is a
violation of some kind or some kind of heresy against science.

You can continue to be stupid and ignorant if you like... but, please,
stop lying, especially when the proofs of your lies resides a few posts
above.

Lies? What do you claim I am lying about? I am going to continue to
use velocity of light and the Galilean transformation equations. You
scientists go ahead and do your thing with the Lorentz equations. I
really don't believe there is a distance contraction. I told you the
absolute truth about that.
Robert B. Winn

On May 6, 7:00Â*pm, " wrote:
On 6 mayo, 17:37, rbwinn wrote:





On May 6, 10:54 am, " wrote:

On 6 mayo, 12:39, rbwinn wrote:

On May 6, 7:55 am, " wrote:

On 6 mayo, 01:19, rbwinn wrote:

On May 5, 9:08�pm, The Ghost In The Machine

wrote:
In sci.physics.relativity, rbwinn

�wrote
on Mon, 5 May 2008 13:30:45 -0700 (PDT)
:

On May 5, 12:07�pm, YBM wrote:
rbwinn a �crit :

On May 5, 11:18?am, YBM wrote:
rbwinn wrote:
Einstein's own equations for
velocity of light do not work in the Lorentz equations if x or x' are
negative.
Wrong. They work.

No, they do not work. �Einstein said that x=ct, x'=ct'. �If x is
negative, then

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

cannot be used with the equation x=ct. �The velocity of light has to
be -c in the equation for t' in order for the equation to work if x is
negative. � x=(-c)t, not x=ct.

Wrong.

Let's assume that x=ct

By LT we get :

x'= gamma*(x-vt)
t'= gamma*(t-vx/c^2) � �where gamma=1/sqrt(1-v^2/c^2)

let's have a look at x'/t' (*) under the condition that x=ct :

� x'/t' = (x-vt)/(t-vx^2/c^2) = (ct-vt)/(t-vct/c^2)
� � � � = t(c-v)/( t (1 - v/c) ) = c(c-v)/(c-v)
� � � � = c

= x'=ct'

(*) the case t'=0 is trivially ok (0=c0).

You neglect the fact that if x or x' is negative in the equations
x=ct, x'=ct', then either the velocity of light has to be negative or
time has to be negative.

The substitution x^2 = c^2t^2 works equally well, yielding
x'^2 = c^2t'^2. �This is a more accurate specification of the
problem anyway, as light expands spherically from a point source.

[rest snipped]

Well, what about a photon traveling in the +x direction reflected by a
mirror back in the direction it came from? Â*I would just say it had a
velocity of -c.
I know that c^2 seems like a good idea to scientists. Â*(-c)^2=c^2
Robert B. Winn

Assume you are in a rocket in deep space, and you are not feeling any
acceleration force whatsoever. From your window you see nothing (no
stars or other objects are visible) but another rocket who appears to
be approaching you.

Now, which is the correct answer that best describe the reality (or
truth) of the situation.

a) You are moving at a constant speed approaching a static rocket.
b) The rocket you see is moving at a constant speed towards your
location.
c) Both you and the approaching rocket are moving at constant speeds
towards each other.

This is the first step in understanding relativity.

Well, it would depend on how you got into that situation, wouldn't it?
This is no different from Einstein's description of a train sitting at
the station with another train moving beside it. Â*A psassenger gets
confused about which train is moving, but that does not change
anything, does it? Â*One train is sitting still on the track, and the
other is moving. Â*The train with the slower clock would be the one
that is moving. Â*Maybe you could do the same thing with your two
spaceships. Â*The one with the slower clock would be the one in
motion. Â*If both clocks are at the same rate, they are moving toward
each other.
Robert B. Winn

You just do not get it, and it is important to understand it, if you
want to say you know how relativity works.
The history of how the ship got to its location is not relevant to the
questions (you may have been sleeping for years and just got awaken to
see a ship is approaching).
The three answers can be correct simultaneously. You in your ship can
only interpret your reality through the information you receive from
your surroundings (in this case the light informing you that another
ship is approaching). For you that is the truth: a ship is approaching
(you can not calculate that ship speed or distance just by seeing it).
You may send signals (like radar) to determine the distance and speed
of the approaching ship, and you would assure to anyone asking you
that the ship is, for instance, at 100000km approaching at 200km/sec.
But your counterpart at the other ship can say the same (by performing
similar radar experiments) that in his case he can assure your ship is
approaching his ship at 200km/sec and at a distance of 100000km.
Finally, also the scenario is compatible with both ships moving
towards the other at 100km/sec (or many other combinations of relative
speeds). The fact is that you can't choose one of the answers as the
most truthful, and you may well conclude that all the answers are
correct depending of the observer.
From these considerations, SR finds its way, after postulating:

Â* Â* * The Principle of Relativity - The laws by which the states of
physical systems undergo change are not affected, whether these
changes of state be referred to the one or the other of two systems of
inertial coordinates in uniform relative motion.
Â* Â* * The Principle of Invariant Light Speed - Light in vacuum
propagates with the speed c (a fixed constant) in terms of any system
of inertial coordinates, regardless of the state of motion of the
light source.

Miguel Rios- Hide quoted text -

- Show quoted text -

Well, I do not know exactly how scientists interpret what they are
saying, but the fact of the matter is that they claim to have a clock
running slower in a moving frame of reference. Â*My question is, What
does that do to velocity?

What you say above is wrong. No scientist or person that knows
somthing about special relativity can say that.
You have to be very careful with words here. Note that the movement
you mention has to be inertial, that is, without acceleration and with
constant speed.
Take your rocket ship. There you can define a frame of reference, let
us call it K. On the other rocket ship, one person there can define
another frame of reference and call it K'. If these two frames are,
for instance, separating at a constant relative speed (200km/sec in
the example I gave you), then you will observe from frame K (by
interchanging information signals with the other ship) that a clock in
frame K' appears to run slow with respect to your local clock. This is
because you think the other ship is going away from you.
However, as it was indicated previously, the ship with frame K' has a
different reality. For the observer in frame K' it is your ship the
one separating and for that reason that observer will see your clock
running slow.
How can this be? Well, the reason is precisely relativity. Neither you
nor the person in the other ship has any way of determine who is right
or wrong, because all the measures and tests they perform agree with
both observations, telling them both are right.

You can do the same thing with the Galilean transformation equations.
I do not personally see the need for it. How do you prove that under
the conditions you describe that both clocks are not running at the
same rate?
The Galilean transformation equations show that one frame of reference
is preferred as far as time is concerned. The other has a slower
clock. With regard to any motion, there is always a cause, even with
your space ships somewhere in outer darkness. The problem you
scientists have is your definition of scientific time. There are
other ways to tell time, as the Galiloean transformation equations
show, for instance, the way people used to do it, with the sun. I
know it bothers you folks to have time that relative, but that is just
the way it goes.

For instance, an astronaut is orbiting in a satellite. Â*Scientists on
earth compute that a clock in the satellite is slower than an
identical clock on earth. Â*The compute the velocity of the satellite
by its radius and time of orbit. Â*The astronaut in the satellite
computes his velocity by computing his radius of orbit with an
altimeter, which presumably shows the same radius used by the
scientists on earth, and the time of his orbit as shown by his clock.
His clock shows less time than the clock on earth. Â*His velocity is
faster than the velocity seen by the scientists on earth.
Â* Â* Â*Oh, no, say scientists, The velocity would be the same. Â*It is
value of pi that is different.
Â* Â* Â*No, scientists, there is no distance contraction. Â*The astronaut
in the satellite really would think he is going faster. Â*That is just
the way it is.
Robert B. Winn

Of course not. Again your use of words is not correct and poorly
represents what SR says.

I am not representing SR. I do not believe in the distance
contraction.

Robert B. Winn

On May 7, 3:57Â*am, YBM wrote:
rbwinn a écrit :

On May 6, 4:34�pm, YBM wrote:
rbwinn wrote:
No, you have no answer to the use of velocities, so you claim it is a
violation of some kind or some kind of heresy against science.
You can continue to be stupid and ignorant if you like... but, please,
stop lying, especially when the proofs of your lies resides a few posts
above.

Lies? Â*What do you claim I am lying about? Â*

Don't you read even what you write ?

you have no answer to the use of velocities

I've especially adressed this point, but you didn't read.

you claim it is a violation of some kind or some kind of heresy against science

I've never written such a claim. Your failure to grasp (or your
unwillingness to try) what is a space-time equation of movement
(in Galilean Relativity as well as in Special Relativity) is your
burden, not mine.

Well, you can try to make things complicated, but it still just comes
down to one clock running slower than another. All you have to do is
convert the time of the slower clock to the time of the faster one and
use that time in the Galilean transformation equations. Or if that
does not suit you, you can talk about relativity of simultaneity and
distance contraction with scientists.
Robert B. Winn

rbwinn a écrit :
Well, you can try to make things complicated,

If velocity as a vector is too complicated for you, you'd better give
up on physics.

Now, stop asking me any question before having reading and understanding
what I tried to explain in my previous posts.

On May 7, 7:55�am, YBM wrote:
rbwinn a �crit :

Well, you can try to make things complicated,

If velocity as a vector is too complicated for you, you'd better give
up on physics.

Now, stop asking me any question before having reading and understanding
what I tried to explain in my previous posts.

You have not explained anything in your previous posts. All you did
was show the Lorentz equations, which as anyone can see, are the
Lorentz equations. As I stated before you even posted anything, the
Lorentz equations automatically keep the velocity of light correct
because if x or x' are negative, then the velocity of light is
automatically negative, even though it is not specifically shown
because the speed of light is always shown as being squared in the
Lorentz equations. The Lorentz equations are using velocity of light
and not the two little equations that Einstein extracted, x=ct,
x'=ct'. So if w is the velocity of light, then you can substitute w
for the speed of light in the Lorentz equations and get a better idea
of what the equations mean. Or, as I said, you can continue to
discuss relativity of simultaneity and distance contraction with other
scientists. I don't care what you do.
Robert B. Winn

"YBM" wrote in message ...
Dirk Van de moortel a écrit :
YBM wrote in message

rbwinn a écrit :

for G.S, leave him alone and let him choke on it.

You're right, I give up (as I said).

BTW, did you hear about our good old friends, the Bogdanov Brothers, and
the kind of support "from Harvard" they're now trying to sell ? Look :

http://grenouille-bouillie.blogspot....li-school.html

For quite a good report of it in english.

After that, the fraudsters are about to publish (again) a book in french
on their "theories"... Let's wait for strange supporters coming on
Usenet through proxies...

I do like their grammar though :-)

Dirk Vdm