Re: IRT: A New Theory of Relativity

From: kenseto (kenseto_at_erinet.com)
Date: 01/25/05 

Date: Tue, 25 Jan 2005 15:39:30 GMT

"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
news:41F5B120.90600@mail.verizon.net...
>
>
> kenseto wrote:
>
> >"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
> >news:41F597CA.5030007@mail.verizon.net...
> >
> >
> >>kenseto wrote:
> >>
> >>
> >>
> >>>"Jesse Mazer" <vze2ztqw@mail.verizon.net> wrote in message
> >>>news:41F4AEB8.8090007@mail.verizon.net...
> >>>
> >>>
> >>>
> >>>
> >>>>Are you saying that an observer at rest wrt the ether will see moving
> >>>>clocks slow down, but he *won't* see moving rulers shrink, as measured
> >>>>by his own coordinate system?
> >>>>
> >>>>
> >>>>
> >>>>
> >>>Yes the physical length of a ruler remains the same in all frames
> >>>
> >>>
> >including
> >
> >
> >>>the rest frame of the ether. However the light path length of a ruler
is
> >>>dependent of the state of absolute motion of the ruler. The higher is
the
> >>>state of absolute motion of a ruler the longer is its light path
length.
> >>>
> >>>
> >The
> >
> >
> >>>ruler at the rest frame of the ether has the shortest light path
length.
> >>>
> >>>
> >>>
> >>How do you define "light path length"? Do you just look at the amount of
> >>time the light took to get from one end to the other as measured by your
> >>own clocks, and then multiply by c?
> >>
> >>
> >
> >NO...the light path length of the observer's rod is assumed to be t*c
where
> >t is the transit time from one end of the rod to the other end. The
observer
> >will determine the light path length of an identical moving rod using IRT
as
> >follows:
> >A is the observer and B is the observed frame:
> >Lba=Laa(Faa/Fab) or Lba=Laa(Fab/Faa)
> >Laa=the light path length of the observer's rod in A's frame as measured
by
> >A.
> >Lba=the light path length of an identical rod in B's frame as predicted
by
> >A.
> >Faa=the mean frequency of a standard light source in A's frame as
measured
> >by A.
> >Fab=the mean frequency of an identical standard light source in B's frame
as
> >measured by A.
> >
>
> Why would B choose to define "light path length" by imagining what
> things would look like in A's frame?

NO....B doesn't do that. B thinks that his light path length for a meter
stick is 1 meter and he predicts that A's light path length for a meter
stick is:
Lba=Lbb(Fbb/Fba) or Lba=Lbb(Fba/Fbb)

>That's a pretty odd way to define
> measurements in different frames. Do you agree that if B uses his own
> clocks to measure the time, he will get a different time for the light
> to cross a ruler moving relative to the ether depending on which end the
> light is emitted from, assuming (as in your theory, but not in SR) that
> all observers agree about simultaneity?

NO.....My ether theory predicts that the speed of light in B's frame is also
isotropic. Therefore B will see the same transit time for light to cross a
ruler moving in the ether in all directions. The reason for this is due to
the unique structure of the ether called the E-Matrix. A description of the
E-Matrix is in the following link:
http://www.journaloftheoretics.com/Links/Papers/Seto.pdf

>

> >>No, but in SR each observer defines his time-coordinates in terms of the
> >>readings on a set of clocks which are all at rest relative to himself,
> >>and which have been "synchronized"based on the assumption that light
> >>travels at the same speed in all directions (so two clocks are defined
> >>as synchronized if a light shined at their midpoint will reach each
> >>clock at the same time).
> >>
> >>
> >
> >What is the purpose of synchronizing two clocks in the same frame?
> >
>
> So you can assign t-coordinates to events at any location in space just
> by looking at the reading of a clock at the same spatial location as the
> event. This is how coordinate systems are defined in SR.

Isn't that that's what the LT is designed to do??
I must admit that I missed this point of SR entirely. What this assume is
that all relative clocks are moving at the same intrinsic rates and thus
what A's synchronized clock read at B's location is the clock reading of B's
clock.

IRT does not assume that all clocks are running at the same intrinsic rate.
When A and B are in relative motion they are in different states of absolute
motion and thus their clocks are running at different intrisic rates. That's
why IRT have two sets of transform equations.

>
> > I don't
> >like using light to synchronize clocks. i like slow clock transport of
two
> >touching and synchronized clocks in the opposite directions. SR says that
> >such a pair of clcoks will remain synchronized.
> >
>
> Only from the point of view of an observer at rest with respect to these
> clocks. If two clocks are moving together at velocity v in my frame, and
> then one of the clocks accelerates to velocity v + s, where s is some
> very tiny velocity, then I will see the clocks become noticeably
> out-of-sync as they move a significant distance apart, even in the limit
> as s approaches 0.

IRT sees things differently. Clocks that are in the same state of absolute
motion will run at the same intrinsic rate. Two synchronized clocks A and B
started in the same frame B is accelerated and decelerated and come to rest
again wrt A. B will show a slight lower clock reading than A (depending on
how fast and how far B is moved) and this difference will remain permenant
because they are now again running at the same intrinsic rate.

>
> But yes, in the limit as their relative velocity approaches 0, "slow
> transport" will have exactly the same effect as synchronization using
> light signals, so it doesn't really matter which method you use. But if
> slow transport is your synchronization method, and if you agree that
> clocks slow down when they're moving relative to the ether, then
> observers in different frames will *not* agree about simultaneity.

The intrinsic rate of a clock is dependent on its state of absolute motion.
All clcoks in the universe are in a state of absolute motion. Two clocks A
and B in relative motion will have different intrinsic rates. This means
that A's clock cna run faster or slower compared to B's clock. That's why
IRT have two sets of eqautions.
>
> >>I don't know what you mean by "see simultaneity occur". In standard SR,
> >>you don't "see" simultaneity, simultaneity just means that two events
> >>are assigned the same time-coordinate in your reference frame (with the
> >>time-coordinate of each event determined by the reading on a clock at
> >>the same spatial location as the event, with all the clocks at rest
> >>relative to the observer and synchronized by the method I mentioned
> >>above). If you are talking about when the light from two events reaches
> >>you, that is a separate issue from simultaneity in SR--the light from
> >>two simultaneous events won't reach you at the same time unless the two
> >>events are both the same distance from you.
> >>
> >>
> >
> >Exactly....in Einstein's train gedanken the light strike
simultaneouslywhen
> >both observers are at equal distance from the strikes. The track observ
will
> >see the strikes to be simultaneous at L/c. According to the track
observer
> >the train observer will see the strikes to be simultaneous at gamma*L/c.
> >
> What do you mean by the phrase "see the strikes to be simultaneous at
> X"? Each observer either will see the light from both strikes reach him
> at the same moment or he won't.

That's not true. Different observers will have different light path lengths
for an identical physical distance. In the case of the train gedanken, the
train observer have a higher state of absolute motion than the track
observer and thus the light path length in the train is longer than the
light path length in the track and thus the train observer will see the
strikes to be simultaneous at a later time.

>If the train observer and the track
> observer are both at the same position when the light from both strikes
> reaches them, they will both *see* the light from both strikes
> simultaneously. However, the train observer will say the strikes weren't
> really simultaneous, because he was heading towards one strike and away
> from the other,

This assertion is bogus and it violates the isotropy of the speed of light
in the train.

>so for the light from both to reach him at the same
> moment, he will have to say that the strike he was heading away from
> happened before the strike he was heading towards (if he assumes light
> travels at the same speed in all directions in his own rest frame, as in
> SR).

This reasoning assumes that you know how light moves from the source to the
target. IOW, you assumed that you know the velocity and position of the
leading edge of the light ray (the first photon). This is a direct violation
of the uncertainty principle.
>
> But simultaneity is a lot easier to think about if you assume both
> observers have a network of clocks at rest relative to themselves, and
> synchronized using light signals (or slow transport, if you prefer), and
> they assign time-coordinates to events based on the reading on the clock
> in their network which was at the same location as the event when it
> happened. This is the standard way that Einstein assumed coordinates
> should be assigned to events. If you have a different way to assign
> time-coordinates to events, please outline it.

This is based on the faulty assumption that the intrinsic rate of the train
and the track clocks are running at the same intrinsic rate.

Ken Seto

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