twin paradox experiment done in lab

Spine less Brian Quincy Hutchings" QncyMI at
netscape.net is Al Gore's recycled Dingleberry of
AGW Relativity, who asked:
||||Brian wrote:||| "do I have to kiss the dingleberries?"
Brian was originally Lyndon LaRouche's roach, that
morphed into "Spudnick", son of "Mr. Potato head",
rasterspace", "tensegriboy" & is now a brain-fossil in
"1treePetrifiedForestLane"
& is no longer able to realize what's going on and so:
||| Brian asks: " do I have to kiss the Dingleberries?"
||| Brian says: Too bad, I can't just go "backwards"
||| Brian says: in time, and kill the mofo Einstein in the
||| Brian says: patent office.
||||Brian says: E=mcc is "just" an elaboration of KE=mvv
||||Brian says: I want to believe in wormholes
||||Brian says: that are absolute F and FS.
||||Brian says: Your format is extremely tiresome;
||||Brian says: it's no longer worth replying to such
||||Brian says: dingbats as yourself, Poraatt,
||||Brian says: Kobbee-doo, Neinstein.
||||Brian says: have a nice _____, all of you.

hanson wrote:
Then DO what you preach, Brian.
Have some backbone & STOP responding to my posts.
I told you that for the *** 12th time*** now.
"have a nice ___," too, Brian, you "God-am ____, fool"
ahahahaha.. ahahahanson

if you are going to misrepresent a dead guy,
you should at least post a handwritten caricature.

all of you Einsteinmaniacs are essentially floundred
in those two "problems" that I mention,
which are merely a couple of literary conventions
in the "field" of physics education.

Mental patient "Rob Greason"
is the spineless Brian Quincy Hutchings" QncyMI at
netscape.net and is Al Gore's recycled Dingleberry of
AGW Relativity, who asked:
||||Brian wrote:||| "do I have to kiss the dingleberries?"
Brian was originally Lyndon LaRouche's roach, that
morphed into "Spudnick", son of "Mr. Potato head",
"rasterspace", "tensegriboy" & is now a brain-fossil in
"1treePetrifiedForestLane"
showing now as "Rob Greason"
& is no longer able to realize what's going on and so:
||| Brian asks: " do I have to kiss the Dingleberries?"
||| Brian says: Too bad, I can't just go "backwards"
||| Brian says: in time, and kill the mofo Einstein in the
||| Brian says: patent office.
||||Brian says: E=mcc is "just" an elaboration of KE=mvv
||||Brian says: I want to believe in wormholes
||||Brian says: that are absolute F and FS.

and now buzzwording as ever, but never capable of
producing a simple equation or a numerical solution
Brian "Rubs Grease on" and declares himself as a
being a " caricature".

AHAHAHA... ahahahaha.. ahahahanson


--- Posted via news://freenews.netfront.net/ - Complaints to ---

you don't dispute any thing; nice.

On Jul 19, 3:23 pm, PD wrote:

The transformation equations tell you, in the case of a common origin at
a particular location and a particular time between two coordinate
systems, what the relationship is between the *current* time and space
coordinates (according to a local and stationary clock and ruler system)
in one reference frame and the *current* time and space coordinates
(according to a local and stationary clock and ruler system) in another
reference system, for any selected event. You can repeat that process
for another selected event.

This is a myth and absolutely not true. The Lorentz, the Voigt, and
the Galilean transforms all describe the relationship between how one
observer observes the observed and how another observer observes the
same observed. These observations do not have to be done at the
global level (sharing the same common origin). They only apply to
local level, (dx’ = dx – v dt) instead of (x’ = x – v t). srgu

The transformation equations do NOT "project" out in space or time to
something not local.

Once again, these transforms involve three parties --- two observers
observing the same observed. These transforms all describe how the
observations of these two observers are related in observing the same
observed.

it isn't a difficult problem in terms of perspective;
if you, et us say a fourth observer, get far-enough
away from the other three observers/observees,
then you can see the relative merits ... well,
especially if they are "going" in a plane
that is normal to your line of sight,
otherwise use dopplergangers.

Once again, these transforms involve three parties --- two observers
observing the same observed. *These transforms all describe how the
observations of these two observers are related in observing the same
observed.

PD wrote:
On 7/20/2011 11:27 AM, Koobee Wublee wrote:

The transformation equations tell you, in the case of a common origin at
a particular location and a particular time between two coordinate
systems, what the relationship is between the *current* time and space
coordinates (according to a local and stationary clock and ruler system)
in one reference frame and the *current* time and space coordinates
(according to a local and stationary clock and ruler system) in another
reference system, for any selected event. You can repeat that process
for another selected event.

This is a myth and absolutely not true.

I didn't say the Lorentz transform was the only transform that does
this. I said that the Lorentz transform does the things described, which
it does.

One CAN do a Lorentz transformation where there is an offset between
origins at some given time, but this simply introduces constants, and so
adds nothing.

Otherwise, what you say below is fine.

The Lorentz, the Voigt, and
the Galilean transforms all describe the relationship between how one
observer observes the observed and how another observer observes the
same observed. These observations do not have to be done at the
global level (sharing the same common origin). They only apply to
local level, (dx’ = dx – v dt) instead of (x’ = x – v t).srgu

Once again, these transforms involve three parties --- two observers
observing the same observed. These transforms all describe how the
observations of these two observers are related in observing the same
observed.

You still have not said anything that makes sense. Which version of
the Lorentz transform below has a better chance of representing
reality?

**** Global Version where it is a pure coordinate transform of linear
rectangular coordinate system

** t’ = (t – v x / c^2) / sqrt(1 – v^2 / c^2)
** x’ = (x – v t) / sqrt(1 – v^2 / c^2)
** y’ = y
** z’ = z

Or

**** Geometry Transform where it transforms a point in spacetime to
another (there is no such offset in this version)

** dt’ = (dt – v dx / c^2) / sqrt(1 – v^2 / c^2)
** dx’ = (dx – v dt) / sqrt(1 – v^2 / c^2)
** dy’ = dy
** dz’ = dz

not sure what you mean; I believe that it was Dirac,
who showed the relativistic properties of matter "at rest,
even."

**** *Global Version where it is a pure coordinate transform of linear
rectangular coordinate system

**** *Geometry Transform where it transforms a point in spacetime to
another (there is no such offset in this version)

On Jul 20, 2:45 pm, PD wrote:
On 7/20/2011 2:21 PM, Koobee Wublee wrote:

You still have not said anything that makes sense. Which version of
the Lorentz transform below has a better chance of representing
reality?

**** Global Version where it is a pure coordinate transform of linear
rectangular coordinate system

** t’ = (t – v x / c^2) / sqrt(1 – v^2 / c^2)
** x’ = (x – v t) / sqrt(1 – v^2 / c^2)
** y’ = y
** z’ = z

Note there is no constant offset in the above.

Or

**** Geometry Transform where it transforms a point in spacetime to
another (there is no such offset in this version)

** dt’ = (dt – v dx / c^2) / sqrt(1 – v^2 / c^2)
** dx’ = (dx – v dt) / sqrt(1 – v^2 / c^2)
** dy’ = dy
** dz’ = dz

Do you not know these two transforms presented above represent very
different meanings?

the "transform" to motion alone the x-axis is supposed
to be done "without loss of generality," although
it represents a significant loss to orienteering, if
you *never* do it "in general position" ...
Kama Sutra #777 (the missionary position).

** *t’ = (t – v x / c^2) / sqrt(1 – v^2 / c^2)
** *x’ = (x – v t) / sqrt(1 – v^2 / c^2)
** *y’ = y
** *z’ = z

** *dt’ = (dt – v dx / c^2) / sqrt(1 – v^2 / c^2)
** *dx’ = (dx – v dt) / sqrt(1 – v^2 / c^2)
** *dy’ = dy
** *dz’ = dz
Do you not know these two transforms presented above represent very
different meanings?