Re: The Physics Behind 'Contractions'.

On Jan 25, 4:54 pm, hw@..(Dr. Henri Wilson) wrote:
On Sun, 25 Jan 2009 00:25:51 -0800 (PST), Bruce Richmond <bsr3...@xxxxxxxxxxxx>
wrote:





On Jan 24, 6:26 pm, hw@..(Dr. Henri Wilson) wrote:
On Sat, 24 Jan 2009 07:56:37 -0800 (PST), Bruce Richmond <bsr3...@xxxxxxxxxxx>
wrote:

I have already explained to you that a particular photon's path in the moving
frame is a BROAD diagonal line.

Uh, no...  I would remember a stupid statement like that.  The line
drawn is just a representation of the path taken by the photon.  Its
width means nothing.

A line on a graph has no width. It cannot represent the path of a whole photon.
If a photon had no width or length it would have no properties at all and would
be indistinguishable from empty space.

Your argument makes as much sense as saying that the width of a line
on a road map is not drawn to scale and therefore can't represent the
path of the road.  Such a statment just makes you look stupid.

I already suspected that you had little or no knowledge of maths or physics. No
need to remind me.

Then you really are that stupid.





I have also explained to you that what moves along that diagonal is not a light
ray moving at c. It is plot of the photon's path and it obviously moves at
sqrt(c^2+v^2) in the moving frame.

It is a plot of the photon's path.  We agree on that :)  That means it
is the path traveled by the photon.  And by definition a photon is a
light particle.  So that means light traveled along that path.  

No this is where you are making your pathetic mistake. You are regarding a
photon as a little round ball that moves at c wrt everything.

Perhaps you could fill us all in on the details of the size and shape
of a photon.  Please provide a link to where you got your information,
assuming you didn't just make it up on your own.  Despite the use of
various squiggles to represent photons, wiki describes them as point
particles.  

hahahahahahhaha! What is their source of information?

Your want more. I have no problem finding many supporting my
statement. You have yet to provide one supporting yours.

You do understand what that means don't you?  Also please
provide a reference showing where light has been measured to travel at
speeds greater than c as you are claiming it does here.

Light moves at c wrt its source and c+v wrt an observer moving at -v wrt the
source.

Do you understand what "provide a reference" means.

The only light ray here is the one involving the stream of vertically aligned
photons. It moves vertically at c in both frames. The whole vertical stream
moves sideways in the moving frame....just like a light pole would.

Which would give the photons a velocity greater than c in the
stationary frame.  

Why? The beam moves verticaly at c in both frames.
Consider a line of ants crawling up a light pole. Do you really think their
speed up the pole is affected by your driving past in your car?

Their speed relative to me is. In a previous post you wrote, "It is
plot of the photon's path and it obviously moves at sqrt(c^2+v^2) in
the moving frame." A photon is a light particle, so your statement
says that light travels at a speed greater than c. If that is true
you shouldn't have any trouble providing a reference to support your
statement.

Please provide experimental evidence of light
traveling at velocities greater than c.

moron....

Name calling doesn't cut it. Provide a reference or you prove
yourself to be the stupid one.

And
unless you have experimental evidence that says otherwise I think we
should accept the multitude of experiments that confirm that light
travels at c in empty space.

I can see taht there is no point in arguing with you. You don't have the
intelligence to even try to understand what is happening here.

I have the intelligence to see that you are claiming that light
travels at speeds greater than c with no experimental evidence to back
that claim.  Sorry but I am not going to just take your word as fact.

The light from the headlights of an approaching car hits you at >c.

Says you. Provide a reference.





But since you have now agreed that the diagonal is NOT a ray of light, you must
also agree that there is no reason to believe that whatever moves diagonally
does so at speed c and not sqrt(c^2+v^2)

If A fires a photon vertically at the same instant that he and A'
coincide it will go straight up and hit B, missing B' who has moved
on.  But what if he aims to the right along a diagonal?  If he gets
the angle right he can hit B'.  In that case it would be a diagonal
ray would it not?

This is not a case of what any obswerver sees. It is a mater of plotting a path
in two different frames.

The line is plotted based on what the observers see.  

It is NOT. The diagonal line you are talking about is a plot, in the moving
frame, of the path of an infinitesimal element of the beam of light that moves
vertically in the lightclock frame.  

The diagonal is a plot of a photon's path in the stationary frame.
Try to keep things straight.

You are confused.
Are you talking about an infinitesimal line on a graph or a finite line of
thickness Lsin(v/c), where L is the length of the particular photon?

We were talking about the diagonal line in the stationary frame
describing the path of a photon in a moving light clock.

Observer A sees
the photon emitted.  Observer B' sees the photon arrive at a different
place at a later time.  The line represents the path taken by the
photon to get from A to B'.  If B' doesn't see the photon arrive we
can't plot the line since we don't know where the photon went.

Forget the observer.  As Seto would say, he will only sees a flash as he goes
past the beam.
You should plot the path in the moving frame...something you acually DID in an
earlier post.

It is a vertical line in the moving frame, and the observers in that
frame measure it to travel at c.  That is not where our argument lies
so there is no need to plot it again.

That's exactly what I have been trying to tell you.
I think you are now hopelessly confused.

And that statement make you a lying sack of *** again.

Bruce
.

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