Re: Contraction has been abolished by Special Relativity

On Sep 28, 11:27�am, Russell <rblacka...@xxxxxxxxx> wrote:
On Sep 28, 9:43�am, rbwinn <rbwi...@xxxxxxxx> wrote:





On Sep 28, 8:29 am, Russell <rblacka...@xxxxxxxxx> wrote:

I've said I can't serve as your educator in this, so I won't
promise to respond further in this thread -- nevertheless I
did read your posting (quoted below) and have a few comments
to make.

On Sep 28, 5:18 am, rbwinn <rbwi...@xxxxxxxx> wrote:
...

The difficulty encountered is shown by Einstein's train and lightning
problem. If the front of the train is x'=1, and the rear of the train
if x'=(-1) in S', the frame of reference of the train, then suppose
that lightning strikes the front and rear of the train instead of some
distance ahead of and behind the train as described by Einstein.

For the record, in Einstein's gedanken the strikes were
*at* the front and back of the train, same as in yours.

If the bolts of lightning are simultaneous from the frame of
reference of the track, then supposedly you have a little short train
as observed by an observer at x=0, the origin of S. So if the
lightning leaves marks on the front and rear of the train and marks on
the railroad track, how far apart are the marks on the railroad track?

The length of the "little short train", of course. I don't
see what the "problem" is, that you refer to in the following:

The problem is compounded further by the fact that relativity of
simultaneity says that in the frame of reference of the train, the
lightning strikes first at the front of the train, meaning that the
train travels some distance after the front mark is made on the track
before the second mark is made as seen by an observer on the train.

Correct, well done. Exactly what "problem" does this "compound"?

So according to SR, an observer by the track sees the lightning strike
both ends of a short train simultaneously, leaving marks on the track
a distance of
L sqrt(1-v^2/c^2) apart.

Yes, where L is the length of the train in S', namely 2
using the example you gave above.

An observer on the train sees the bolt of

lightning at the front of a normal length train strike first, then the
train travels some distance

Strictly speaking, it is of course the *track* that travels
in this frame.

before the other bolt of lightning

strikes, leaving two marks on the track a distance of L sqrt(1-v^2/
c^2) apart.

That's their distance apart in the track frame; but of course
in the train observer's frame the distance is contracted by
an additional factor 1/gamma, i.e. it is L(1-v^2/c^2).

Well, that is why I said the problem is compounded. �Scientists do not
even seem to agree.

What's to disagree? �That result is an easy calculation.
You yourself (a couple paragraphs above) correctly gave
the distance between the marks as L*sqrt(1-v^2/c^2) in
the track frame. �That bit of track is moving at velocity
-v in the train frame, so in the train frame that bit of
track is contracted by 1/gamma. �Hence my correct result.
It may look odd to you that the 1/gamma factor is in there
twice, but that's the correct result for your gedanken.

��The distance being contracted from the frame of

reference of the train is the length of the track. �What you just said
is the opposite of what Peter Riedt is trying to do.

Gosh, I certainly hope so! �Oh, I get it, you think Peter
Reidt is a *scientist*! �Heh heh. �That idea never occurred
to me.

��Peter is trying

to get back to no distance contraction at all. �You are trying to
compound it to agree with what the observer on the ground is said to
see, �L sqrt(1-v^2/c^2). �But I see what Peter is saying. �If the
length of the track is contracted from the frame of reference of the
train, then the train is occupying a length of track longer than the
length of the train.

I don't think that's what Peter is saying, but no matter,
let's look at it. �You use the term "length" twice in that
final clause without stating what reference frame those
lengths are measured in. �Very sloppy! �Not something a
*scientist* would say. �Your statement would be correct if
you qualified each usage as, respectively, track frame for
the first and train frame for the second. �But then the
seeming contradiction vanishes, doesn't it? �You're
comparing different things.

By the way, what you just did in your sloppy language is
called a "frame jump". �It's something that amateurs often
do, but not physicists. �How ironic that you did exactly what
you (falsely) accused physicists of doing.

��So the lightning at the front strikes first �and> the lightning at the rear of the train second, putting the marks on
the track the length of the train apart as seen from the frame of
reference of the contracted track.

� � � � � � � � � �^^^^^^^^^^?
To avoid confusing yourself, say simply "in the frame of
reference of the track". �After all, in that frame, it is the
*train* and not the track that is contracted. �And yes, in the
track frame the distance between the marks is the length of
the train in the track frame, namely, L*sqrt(1-v^2/c^2). �As
you yourself wrote above.

But scientists already said that the train was shorter, so they agree
with you.

Right. �So where is this disagreement among scientists that
you allege?

Well, for instance, I have discusswed this with Androcles, who claims
to be a scientist, and was posting here the first time I ever did, and
he definitely disagrees with you.



If the bolts of lightning are simultaneous in the frame of
reference of the train, then to an observer on the ground, the bolt of
lightning at the back of the train strikes first, etc., and the marks
on the track are further apart than the length of the train.

Yes, that's what SR says.

I decided long ago that this was all nonsense, as Peter Reidt is
trying to do, and that the correct equations were the Galilean
transformation equations.

Take care, if you "decide" something like that, that you
don't close your mind completely to (at least consideration
of) other points of view, or to well-reasoned critiques of
your own position. I note that you never addressed the
substantive point of my earlier post, that contrary to what
you claimed, the same light beam *cannot* have the same
velocity in two different frames under the Galilean
transformation. Do you *not* have an answer?

Well, yes I did have an answer. �I had equations for it. �If light is
traveling at c in both frames of reference,

STOP RIGHT THERE and explain how this is possible using
a Galilean transformation. �I don't want to consider
anything else until you do this.

�then

It is possible if a clock in S' is showing something other than t'=t.
So we call what the clock in S' is showing as measured by a photon
traveling in a given direction on the x, x' axis by the variable n'.

x'=cn'
for a photon traveling in the +x direction,

x'=(-c)n'

for a photon traveling in the -x direction.

So you might think of it this way. t'=t denotes a common measurement
of time in S and S', such as degrees of rotation of the earth, which
would be the same from either frame of reference. n' is measured by
transitions of a
cesium isotope molecule, which scientists say would be different
measured in S' than in S. So with regard to the train and two bolts
of lightning, when light from the rear of the train reaches the
observer by the track, a time of t'=t has elapsed in both frames of
reference as measured by the rotation of the earth, but a shorter time
of n' has elapsed in S' as measured by transitions of a cesium isotope
molecule in S'. When light from the front flash of lightning reaches
the origin of S', light from the rear flash of lightning is also
reaching that point in S'. When light from the front flash of
lightning reaches the origin of S, according to the rotation of the
earth, t'=t, but according to transitions of a cesium isotope molecule
in S', a greater time than t'=t has elapsed because n'=x'/c, and x' is
a greater distance than x from the points towards the front of the
train where light was emitted by the lightning in each frame of
reference. In the frame of reference of the train, light was emitted
at the front of the train. In the frame of reference of the track, it
was emitted where the lightning left a mark on the track.




� � � � � � � � w=velocity of light
� � � � � � � � x=wt
� � � � � � � � x'=wn'
Light traveling on the x axis can have a velocity of c or
-c depending on which way it is directed. �x' cannot be ct' as
Einstein defined it if using the Galilean transformation equations
because t'=t. �So we define n' as the time it takes light to travel a
distance of x'.

� � � � � � � wn'=wt-vt
� � � � � � � �n'=t(1-v/w)

So what this shows is that light has the same wavelength and frequency
in S' as it does in S. �Evidently, this makes light a little more
complex than scientists want it to be, but it is definitely traveling
at a speed of c in both frames of reference.

Might I add that IMO you seem to have learned one or two
things in the years that you have been posting here --
bravo for that -- and so, perhaps a decision taken long
ago might be profitably reexamined now. Just a thought.

Oh, I re-examine it all the time, as I just did with Peter Riedt's
idea, but I think the mathematics will show the Lorentz equations
still have marks on the track at
L sqrt(1-v^2/c^2) rather than at the length of the train as Peter is
trying to say.

Again, IMO he's not trying to say that, but anyhow I'm glad
to see you have more sense than he does.

��In any event, I do not get as emotional about this as

scientists do. �To me it is just something that I pass time doing, my
version of a video game.

That's ok by me, go for it. �For my part, I don't find lack
of emotion necessarily so desirable. �Passion is what drives
science, moreover I don't think it's your place as an amateur
to pass judgement on how scientists behave. �In any case, what
matters in the end is whether the science stands up after all
the emotional dust clears.

� � I always return to the Galilean transformation equations because
they seem to show light as energy instead of as an observer controlled
entity.

Does light have a speed or does it not? �If it does, then
you must apply the Galilean transform to *it* if you apply
it to anything else.

Well, light does have a speed, but speed is the magnitude of velocity,
and even the Lorentz equations are using velocity of light, not
speed. This is disguised by the fact that everywhere speed of light
appears in the Lorentz equations, it appears as c^2. There is a
reason for that. Wherever c appears in the Lorentz equations, it is
really w, or velocity of light, meaning that if it is talking about
light going in a -x direction, then the velocity of the light relative
to S and S' is -c. In other words, if a photon is emitted at 10 and
travels to 5 on the x axis, then its velocity is -c in both frames of
reference, just as a baseball thrown from 10 to 5 on the x axis has a
negative velocity relative to the frame of reference.
This can be shown in the Lorentz equations by trying to substitute
Einstein's two little equations in
x=ct
x'=ct'

They will not work if x and x' are negative. In that case, either c
or t has to be negative as also with c and t'. A negative time
indicates an event that occurred before t=t'=0. A negative velocity
of light indicates light traveling in the -x direction.
��The light meets at the origins of both frames of reference,

which seems to me like a very energetic thing to do.

No, the light from two flashes meets at a single point in
space at a single time, i.e. a single event. �If what *you*
said were true, the meeting point would depend on who was
looking at it -- exactly the "observer controlled entity"
you wish to avoid. �I.e. it is you who are making the
mistake.

No, observers would have nothing to do with events. If an event was
to take place when light met at the origin of S, then the event would
occur when light met at the origin of S and could only be determined
from that frame of reference, but could be calculated from any frame
of reference. In other words, if a bell rings when activated by
light from both directions, it would happen when an observer in S sees
light from both directions meet at the origin of S. The bell rings in
all frames of reference, but in frame of reference S', light from the
back of the train reaches the origin of S before the bell rings, and
light from the front of the train reaches the bell after the bell
rings. The bell rings at a time of t=.5L/c as measured by transitions
of a cesium isotope molecule in S. In S' it will ring when the origin
of S' is a distance of vt from the origin of S, regardless of what an
observer in S' thinks the light is doing.


��It shows light

to be more than just an electromagnetic wave. �My idea is founded on
lack of relativity of simultaneity, which puts the marks on the
railroad track the length of the train apart in both frames of
reference. �In the frame of reference of the track, the light from the
bolts of lightning meets halfway between the two marks on the track.
In the frame of reference of the train, the light meets at the middle
of the train. That is the way I have to call it.

You have a fundamental misunderstanding here. �The light
beams meet at a fixed event, *one* event in spacetime.
No matter who is watching this happen, all observers must
agree it is the same event, otherwise you have an observer
dependent reality (which you tell me you don't want--good!).
The Galilean transformation equations show an event dependent
reality. The bell will ring when the distance between the origin of S
and the origin of S' is vt regardless of how the observer in S' is
seeing the light. When the observer in S sees the light meet at the
origin of S, the bell rings in both frames of reference. What the
observer in S' observes depends on what can be determined concerning
his cesium clock, which we have already seen is not doing the same
thing as a cesium clock in S. It appears to me that the cesium clock
in S' does not have the same phase as a cesium clock in S.

Robert B. Winn
.

Relevant Pages

  • Re: The relativity of simultaneity
    ... which the train is moving) will reach the passenger before flash A. ... Closing velocities as perceived by the track observer ... light from different directions in the M' frame is isotropic. ... the two light rays arriving to him simultaneously by definition of the ...
    (sci.physics.relativity)
  • Re: The relativity of simultaneity
    ... which the train is moving) will reach the passenger before flash A. ... Closing velocities as perceived by the track observer ... light from different directions in the M' frame is isotropic. ... the two light rays arriving to him simultaneously by definition of the ...
    (sci.physics.relativity)
  • Re: The relativity of simultaneity
    ... which the train is moving) will reach the passenger before flash A. ... Closing velocities as perceived by the track observer ... light from different directions in the M' frame is isotropic. ... the two light rays arriving to him simultaneously by definition of the ...
    (sci.physics.relativity)
  • Re: Simultaneity of Relativity
    ... relative to the train, this is what will occur in Einstein's train ... impossible if the embankment frame of reference and the train frame of ... locations prior to the lightning strikes. ... travel from four locations to each Observer. ...
    (sci.physics.relativity)
  • Re: Simultaneity of Relativity
    ... aether to be 'entrained' by the embankment in the embankment frame of ... reference and for the aether to be 'entrained' by the train in the ... train frame of reference. ...
    (sci.physics.relativity)
Loading