LORENTZ TRANSFORMATION IN ONE LINE AT .... LEVEL
Alexander Abian
Starting with a very appealing physics (Minkowski's) equality
(1) x^2 - (ct)^2 = (x')^2 - (ct')^2
and applying it at the origin of the Primed -system, i.e. where x'= 0
we have
(2) x^2 = c^2 (t^2 - (t')^2)
THE CRUCIAL ONE LINE is to recognize the utmost elementary (grammar
school) fact that distance x is the product of velocity v and time t) i.e.,
(3) x^2 = (vt)^2
from (2) and (3) we have
/------------
(t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
which is the Lorentz transformation of time
Except for (2) and (3) which together use ONLY ONE LINE the remainings
are grammar school algebra.
P.S. I can give also ONE LINE grammar school proof of formula for the
relativistic addition of velocities and one LINE grammar school proof
of E = mc^2. If interested post a polite request - no e-mails.
With love, Alexander Abian
--
-------------------------------------------------------------------------
ABIAN TIME-MASS EQUIVALENCE FORMULA T = A m^2 in Abian units.
ALTER EARTH'S ORBIT AND TILT TO STOP GLOBAL DISASTERS AND EPIDEMICS.
JOLT THE MOON TO JOLT THE EARTH INTO A SANER ORBIT.ALTER THE SOLAR SYSTEM.
REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT TO CREATE A BORN AGAIN EARTH(1990)
THERE WAS A BIG SUCK AND DILUTION OF PRIMEVAL MASS INTO THE VOID OF SPACE
ryk...@my-deja.com
ab...@iastate.edu (Alexander Abian) wrote:
> -------------
>
> Starting with a very appealing physics (Minkowski's) equality
>
> (1) x^2 - (ct)^2 = (x')^2 - (ct')^2
>
> and applying it at the origin of the Primed -system, i.e. where x'=
0
> we have
>
> (2) x^2 = c^2 (t^2 - (t')^2)
>
> THE CRUCIAL ONE LINE is to recognize the utmost elementary (grammar
> school) fact that distance x is the product of velocity v and time t)
i.e.,
>
> (3) x^2 = (vt)^2
>
> from (2) and (3) we have
> /------------
> (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
>
> which is the Lorentz transformation of time
>
> Except for (2) and (3) which together use ONLY ONE LINE the
remainings
> are grammar school algebra.
>
> P.S. I can give also ONE LINE grammar school proof of formula for the
> relativistic addition of velocities and one LINE grammar school proof
> of E = mc^2. If interested post a polite request - no e-mails.
> With love, Alexander Abian
Rod: To bad you can't use the transforms .
The classic light clock senerio is flawed .
Do you have a better gamma derivation
showing time dilation ?
--
Rod Ryker...
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
Alan Horowitz
your paper on "Equations of the best fitting lines" was good.
I couldn't get it here - I had to write to Kuala Lampur.
--
Alan Horowitz al...@widomaker.com
Wayne Throop
: from (2) and (3) we have
: (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
:
: which is the Lorentz transformation of time
Actually, it isn't. That's only the form for time dilation.
It's not the lorentz transform. The two are distinct.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Alexander Abian
In <9315...@sheol.org> thr...@sheol.org (Wayne Throop) writes:
>: ab...@iastate.edu (Alexander Abian) wrote
>: from (2) and (3) we have
>: /-------------
>: (t')^2 = (1 - (v^2/t^2)) t^2 or t' = \/ 1 - (v^2/c^2) t
>:
>: which is the Lorentz transformation of time
>Actually, it isn't. That's only the form for time dilation.
>It's not the lorentz transform. The two are distinct.
Abian answers:
Actually IT IS. Because by grammar school algebra it is trivial to show
/------------- /------------
t'= \/ 1 - (v^2/c^2) t if and only if t'= (t - (vx/c^2))/ \/1 - (v^2/c^2)
(for any reply, please post, no e-mail)
Wayne Throop
::: from (2) and (3) we have
::: (t')^2 = (1 - (v^2/t^2)) t^2 or t' = \/ 1 - (v^2/c^2) t
:::
::: which is the Lorentz transformation of time
:: thr...@sheol.org (Wayne Throop)
:: Actually, it isn't. That's only the form for time dilation.
: ab...@iastate.edu (Alexander Abian)
: Actually IT IS. Because by grammar school algebra it is trivial to show
:
: /------------- /------------
: t'= \/ 1 - (v^2/c^2) t if and only if t'= (t - (vx/c^2))/ \/1 - (v^2/c^2)
OK. Go ahead and show it. Hint: you will find that t'=sqrt(1-(v/c)^2)*t
only where x=vt, but not in general. And that was my point; the lorentz
transform is a more general statement than time dilation; the two are
no co-equal; they aren't the same thing.
Francis Rey
Francis Rey writes
A. Abian wrote
-------------
Starting with a very appealing physics (Minkowski's) equality
(1) x^2 - (ct)^2 = (x')^2 - (ct')^2
and applying it at the origin of the Primed -system, i.e. where x'= 0
we have
(2) x^2 = c^2 (t^2 - (t')^2)
THE CRUCIAL ONE LINE is to recognize the utmost elementary (grammar
school) fact that distance x is the product of velocity v and time t) i.e.,
(3) x^2 = (vt)^2
from (2) and (3) we have
/------------
(t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
which is the Lorentz transformation of time
Except for (2) and (3) which together use ONLY ONE LINE the remainings
are grammar school algebra.
P.S. I can give also ONE LINE grammar school proof of formula for the
relativistic addition of velocities and one LINE grammar school proof
of E = mc^2. If interested post a polite request - no e-mails.
With love, Alexander Abian
-----------
Francis Rey
Why use so much words and so much line for your demonstration ?
One letter is enough !!
ZERO : 0
You have forgotten something somewhere, the equations are:
(1) x^2 - (ct)^2 = (x')^2 - (ct')^2 = 0
From : x = ct and x' = ct'
the starting conditions of the building ( wrong) of the Lorentz formulae.
If you don' belive read the good relativist text books. And if your math is
pretty light read the Appendix 1, of the Einstein's book "Relativity" ,
where the derivation of the Lorentz Formulae is Simple , as Einstein said.
Simple and wrong, but that is not the actual problem.
--
francis Rey
The fact much to learn don't teach intelligence
Heraclite of Ephesus.
franc...@wanadoo.fr
http://perso.wanadoo.fr/francis.rey/
http://perso.infonie.fr/rascasse
Francis Rey
Waynr Throop wrote
: ab...@iastate.edu (Alexander Abian)
: from (2) and (3) we have
: /------------
: (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
:
: which is the Lorentz transformation of time
Actually, it isn't. That's only the form for time dilation.
It's not the lorentz transform. The two are distinct.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Francis Rey
Very good since Throop is able to say the two are distinct he can be able
to show us why they are distinct and where comes the form of time
dilatation from.
Daryl McCullough
>
>Francis Rey writes
>
> Waynr Throop wrote
>: ab...@iastate.edu (Alexander Abian)
>: from (2) and (3) we have
>: /------------
>: (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
>:
>: which is the Lorentz transformation of time
>
>Actually, it isn't. That's only the form for time dilation.
>It's not the lorentz transform. The two are distinct.
>
>
>Wayne Throop thr...@sheol.org http://sheol.org/throopw
>
>Francis Rey
>Very good since Throop is able to say the two are distinct he can be able
>to show us why they are distinct and where comes the form of time
>dilatation from.
The Lorentz transformations (for only one spatial coordinate,
to make it simple) are:
t' = gamma(t - v/c^2 x)
x' = gamma(x - vt)
This means that if, in the unprimed coordinate system,
an event has coordinates (x,t), then in the primed coordinate
system, the same event will have coordinates (x',t').
The rate of a clock at rest in the primed coordinate system,
as measured in the unprimed coordinate system, is given by:
Change in t' gamma ((Change in t) - v/c^2 (Change in x))
rate = ----------- = -------------------------------------------
Change in t Change in t
If the clock is at rest in the primed coordinate system,
then it is moving at speed v in the unprimed coordinate
system. In that case, (Change in x) = v (Change in t). So
gamma ((Change in t) - v/c^2 * v (Change in t))
rate = -----------------------------------------------
Change in t
= gamma ( 1 - v^2/c^2)
= 1/gamma
So, the clock in the primed coordinate system is running at
a rate of 1/gamma, as measured in the primed coordinate system.
Daryl McCullough
CoGenTex, Inc.
Ithaca, NY
Wayne Throop
::: from (2) and (3) we have
::: /------------
::: (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
:::
::: which is the Lorentz transformation of time
:: thr...@sheol.org (Wayne Throop)
:: Actually, it isn't. That's only the form for time dilation. It's
:: not the lorentz transform. The two are distinct.
: "Francis Rey" <franc...@wanadoo.fr>
: ery good since Throop is able to say the two are distinct he can be
: able to show us why they are distinct and where comes the form of time
: dilatation from.
Been there. Done that. The form for time dilation comes from plugging
the position of the moving clock into the transform. But you can't recover
the transform from the form for time dilation alone.
Rey is the fellow who thinks you can only plug lightlike lines
into the lorentz transforms. So you see how poorly he understands,
despite all this having been explained to him multiple times before.
ryk...@my-deja.com
thr...@sheol.org (Wayne Throop) wrote:
> Rey is the fellow who thinks you can only plug lightlike lines
> into the lorentz transforms. So you see how poorly he understands,
> despite all this having been explained to him multiple times before.
Rod: You use gamma after I have proven you cannot .
Gamma : derived by a thought experiment that
cannot be physically recreated , and wouldn't
work if you could .
--
Rod Ryker...
It is reasoning and
faith that bind truth .
Wayne Throop
: You use gamma after I have proven you cannot .
You've proven no such thing. Not even close.
: Gamma : derived by a thought experiment that cannot be physically
: recreated , and wouldn't work if you could .
What ARE you raving about?
ryk...@my-deja.com
thr...@sheol.org (Wayne Throop) wrote:
> : ryk...@my-deja.com
> : You use gamma after I have proven you cannot .
>
> You've proven no such thing. Not even close.
Rod: Oh , and by what thread do you refer ?
> : Gamma : derived by a thought experiment that cannot be physically
> : recreated , and wouldn't work if you could .
>
> What ARE you raving about?
Rod: Raving ? You're ranting . Obviously you know , from what
you wrote above . Otherwise , what are you raving about ?
Daryl McCullough
>>Gamma : derived by a thought experiment that cannot be physically
>>recreated , and wouldn't work if you could .
>
>What ARE you raving about?
He's talking about his confused idea about light clocks. His
point is this: if you start off with a light clock "at rest",
and then accelerate it, the light pulse will escape from the
clock. Let's let one mirror be parallel to the x-z plane at
location y = -L/2. Let the other mirror be parallel to the
x-y plane at location y = +L/2. Let a light pulse be bouncing
up and down in the y-direction. Now, accelerate the mirrors
in the x-direction.
Now, each time the light pulse bounces up and down, the
mirror moves forward a bit. Before long, the light pulse
escapes from the mirrors, and your light clock doesn't
work any more.
What Ryker doesn't realize is that just about *any* kind
of mechanical device will fail to work correctly during
a period of acceleration. After accelerating the light
clock, it is necessary to re-aim the light pulse so that
it is once again bouncing back and forth between the two
moving mirrors. At this point, the path of the light
pulse will be straight up and down, as measured in the
rest frame of the light clock, but will be a zig-zag
pattern, as measured in the original rest frame.
Tom Roberts
> thr...@sheol.org (Wayne Throop) says...
> >ryk...@my-deja.com:
> >>Gamma : derived by a thought experiment that cannot be physically
> >>recreated , and wouldn't work if you could .
> >What ARE you raving about?
>
> What Ryker doesn't realize is that just about *any* kind
> of mechanical device will fail to work correctly during
> a period of acceleration. After accelerating the light
> clock, it is necessary to re-aim the light pulse so that
> it is once again bouncing back and forth between the two
> moving mirrors.
Lasers are light clocks in precisely this sense. They are mounted
in tactical aircraft and operate even in the highly non-inertial
system of a maneuvering jet fighter. All one need do is engineer the
system to account for possible accelerations (e.g. make the mirrors
appropriately spherical rather than planar -- most lasers do this
anyway to enhance their optical properties).
Of course the ~10 g acceleration of a fighter it still very small in
relativistic terms (~ 0.1 lightyear in appropriate units).
Tom Roberts tjro...@lucent.com
ryk...@my-deja.com
da...@cogentex.com (Daryl McCullough) wrote:
> thr...@sheol.org (Wayne Throop) says...
>
> >ryk...@my-deja.com:
> >>Gamma : derived by a thought experiment that cannot be physically
> >>recreated , and wouldn't work if you could .
> >
> >What ARE you raving about?
>
> point is this: if you start off with a light clock "at rest",
> and then accelerate it, the light pulse will escape from the
> clock. Let's let one mirror be parallel to the x-z plane at
> location y = -L/2. Let the other mirror be parallel to the
> x-y plane at location y = +L/2. Let a light pulse be bouncing
> up and down in the y-direction. Now, accelerate the mirrors
> in the x-direction.
>
> Now, each time the light pulse bounces up and down, the
> mirror moves forward a bit. Before long, the light pulse
> escapes from the mirrors, and your light clock doesn't
> work any more.
>
> of mechanical device will fail to work correctly during
> a period of acceleration. After accelerating the light
> clock, it is necessary to re-aim the light pulse so that
> it is once again bouncing back and forth between the two
> pulse will be straight up and down, as measured in the
> rest frame of the light clock, but will be a zig-zag
> pattern, as measured in the original rest frame.
>
> Daryl McCullough
Rod: Yes , you understand ! My rantings .
But fail , to recieve the whole argument .
Acceleration is but one mistake wrt the argument SR makes .
A LCA moving at a steady velocity will also have the same effect
on the photon . The photon has no momentum wrt the LCA , and if
it did this momentum would quickly drop off WHEN NOT AFFIXED
(touching) THE MIRROR ! The photon is independent not affixed
to the moving LCA . All I ask is that you honestly reason this
in your head with your knowledge . :)
Wayne Throop
::: recreated , and wouldn't work if you could .
:: What ARE you raving about?
: da...@cogentex.com (Daryl McCullough)
: He's talking about his confused idea about light clocks. His point is
: this: if you start off with a light clock "at rest", and then
: accelerate it, the light pulse will escape from the clock.
Thanks.
: What Ryker doesn't realize is that just about *any* kind of mechanical
: device will fail to work correctly during a period of acceleration.
Well, more to the point in the specific gedanken by which gamma may be
illustrated, we start our light clocks ticking after any periods of
acceleration. Hence the "re-aiming" Daryl mentions occurs "for free",
because light moving down an aiming tube (or reflecting off a mirror, or
going through a lens, or however it is aimed) will be aimed in the exact
diagonal direction to keep it with the light clock, without adjusting
anything at all on the clock itself; the two clocks are identical.
Further, consider a light clock which operates by firing a laser at a
mirror either when triggered "by hand", or when a spot of light strikes
a photocell beside the laser. Such a clock would operate well under
quite high accelerations; just so long as the acceleration doesn't take
the photocell completely out of the laser's path in a single light
bounce. The light's path would be consistently vertical in the
instantaneously comoving frame of the clock, and, of course, more and
more diagonal in the original rest frame.
Ryker also doesn't seem to realize that the specific example
used to ilustrate the derivation of gamma isn't how SR "gets" that factor.
It's ONLY an illustrative example.
Indeed, way back when ryker1 first asked "where does gamma come from",
I summarized the light clock example, stated it was only one way to
look at it, and that Einstein used bouncing light in quite a different
way to derive the lorentz transforms. Ryker1's argument is based on
the silly idea that if he shows that one idealized scenario faulty,
he's eliminated all derivations of gamma.
Wayne Throop
:: He's talking about his confused idea about light clocks. His point
:: is this: if you start off with a light clock "at rest", and then
:: accelerate it, the light pulse will escape from the clock.
: ryk...@my-deja.com
: you understand ! My rantings . But fail , to recieve the whole
: argument . Acceleration is but one mistake wrt the argument SR makes .
What mistake? There's no acceleration in the light clock scenario.
: A LCA moving at a steady velocity will also have the same effect on
: the photon .
Wrong. You haven't thought it out. Think of how you aim your photon
to get it to bounce in the first place, once the two clocks are
in uniform relative motion. See
http://sheol.org/throopw/transverse-light-bounce.gif
http://sheol.org/throopw/transverse-light-bounce2.gif
: The photon has no momentum wrt the LCA , and if it did this momentum
: would quickly drop off WHEN NOT AFFIXED (touching) THE MIRROR ! The
: photon is independent not affixed to the moving LCA .
The fact that the photon moves in a straight line through the
aiming aparatus, and isn't "pushed" by the sides of any aiming tube,
nor any part of the LCA, is exactly why the light clock DOES work.
: All I ask is that you honestly reason this in your head with your
: knowledge . :)
Good advice. Please consider taking it yourself.
Jeremy Boden
<tjro...@lucent.com> writes
...
>Of course the ~10 g acceleration of a fighter it still very small in
>relativistic terms (~ 0.1 lightyear in appropriate units).
>
to say?
--
Jeremy Boden
Bennett Standeven
On Thu, 15 Jul 1999 ryk...@my-deja.com wrote:
> In article <9320...@sheol.org>,
> thr...@sheol.org (Wayne Throop) wrote:
> > : ryk...@my-deja.com
> > : You use gamma after I have proven you cannot .
> >
> > You've proven no such thing. Not even close.
>
> Rod: Oh , and by what thread do you refer ?
>
Which thread did you _not_ prove that one cannot use gamma? How about all
of them?
ryk...@my-deja.com
> ::: recreated , and wouldn't work if you could .
>
> :: What ARE you raving about?
>
> : He's talking about his confused idea about light clocks. His point
is
> : this: if you start off with a light clock "at rest", and then
> : accelerate it, the light pulse will escape from the clock.
>
>
> : What Ryker doesn't realize is that just about *any* kind of
mechanical
> : device will fail to work correctly during a period of acceleration.
>
> Well, more to the point in the specific gedanken by which gamma may be
> illustrated, we start our light clocks ticking after any periods of
> acceleration. Hence the "re-aiming" Daryl mentions occurs "for free",
> because light moving down an aiming tube (or reflecting off a mirror,
or
> going through a lens, or however it is aimed) will be aimed in the
exact
> diagonal direction to keep it with the light clock, without adjusting
> anything at all on the clock itself; the two clocks are identical.
Rod: By means of momentum no doubt . Right Wayne ?
Which analogy will you use ? The tube ? The lens ...
Pick one , explain in detail and I'll show error .
> Further, consider a light clock which operates by firing a laser at a
> mirror either when triggered "by hand", or when a spot of light
strikes
> a photocell beside the laser. Such a clock would operate well under
> quite high accelerations; just so long as the acceleration doesn't
take
> the photocell completely out of the laser's path in a single light
> bounce. The light's path would be consistently vertical in the
> instantaneously comoving frame of the clock, and, of course, more and
> more diagonal in the original rest frame.
Rod: To broad . Concentrate on specifics . Is the light a pulse ?
How long in length . What interval if any , distance between
mirrors etc .
> Ryker also doesn't seem to realize that the specific example
> used to ilustrate the derivation of gamma isn't how SR "gets" that
factor.
> It's ONLY an illustrative example.
> Indeed, way back when ryker1 first asked "where does gamma come from",
> I summarized the light clock example, stated it was only one way to
> look at it, and that Einstein used bouncing light in quite a different
> way to derive the lorentz transforms. Ryker1's argument is based on
> the silly idea that if he shows that one idealized scenario faulty,
> he's eliminated all derivations of gamma.
Rod: Gamma is based on what original Einstien
or other , during his time , derivation wrt SR time dilation .
BTW only gamma wrt SR time dilation is what I refer to .
> Wayne Throop thr...@sheol.org http://sheol.org/throopw
Francis Rey
Francis Rey writes
Daryl wrote
Francis says...
>
>Francis Rey writes
>
> Waynr Throop wrote
>: ab...@iastate.edu (Alexander Abian)
>: from (2) and (3) we have
>: /------------
>: (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
>:
>: which is the Lorentz transformation of time
>
>Actually, it isn't. That's only the form for time dilation.
>It's not the lorentz transform. The two are distinct.
>
>
>Wayne Throop thr...@sheol.org http://sheol.org/throopw
>
>Francis Rey
>Very good since Throop is able to say the two are distinct he can be able
>to show us why they are distinct and where comes the form of time
>dilatation from.
Daryl
The Lorentz transformations (for only one spatial coordinate,
to make it simple) are:
t' = gamma(t - v/c^2 x)
x' = gamma(x - vt)
Francis Rey:
Wrong beginning: Lorentz transforms never had been written for more than
one coordinate, being given that it concern only a light ray. An unique and
independant free light ray, and not a spherical wave or wave front. Quite
normal thing since it describe every radius of a spherical emitted light
wave.
Daryl:
This means that if, in the unprimed coordinate system,
an event has coordinates (x,t), then in the primed coordinate
system, the same event will have coordinates (x',t').
Francis
Where that "event " is coming from ? Not any "event", a perfectly
identified fact : a light ray .
Daryl
The rate of a clock at rest in the primed coordinate system,
as measured in the unprimed coordinate system, is given by:
Change in t' gamma ((Change in t) - v/c^2 (Change in x))
rate = ----------- = -------------------------------------------
Change in t Change in t
Francis Rey:
We ignore totally the rate of any clock in the constructioin of Lorentz
formulae.
That rate arrives at an ackward moment . And makes no sense.
Are you not able to write dt or dt' for differentials ?
dt' / dt never has been a rate of anything ?
Why a dt' is giving a dx ? What links t' or t , to x or x' ?
Obviously because: x = c t and x' = ct'
the former relations of Lorentz.
Daryl
If the clock is at rest in the primed coordinate system,
then it is moving at speed v in the unprimed coordinate
system. In that case, (Change in x) = v (Change in t). So
gamma ((Change in t) - v/c^2 * v (Change in t))
rate = -----------------------------------------------
Change in t
= gamma ( 1 - v^2/c^2)
= 1/gamma
Francis
We don't care about a clock at rest on the moving system, since the Lorentz
formulae are not devoted to a clock but to the time of a light ray
travelling from an at rest frame to, and in, a moving system.
Your :
(Change in x) = v (Change in t)
is very childish .
We are old and we understand dx = v dt .
Which means merely : x = v t
If x = v t
then your x' becomes:
x' = gamma( x - vt ) = 0
As x' = c t'
that means t' = 0 , or c = 0
t' = 0 means :
no event in the moving system.
c = 0 means:
the problem is not related to a light ray then the Lorentz formulae can't
be used.
If you argue x' is not equal to c t' you have to say to what "event" it
belongs.
t' = 0 means also:
t' = gamma(t - v/c^2 x) = 0
and t = 0
It seems your "event " would better be named a "non-event".
But then we are not in the problem of light with x = c t and x' = c
t' !!
Then it does not exist any c in the formulae !
A clock has right to move at velocity v , but in that case the time t is
not even necessarily the time displayed by the clock itself, it can be any
time of any other clock. We don't care.
Daryl
So, the clock in the primed coordinate system is running at
a rate of 1/gamma, as measured in the primed coordinate system.
Francis
A perfectly wrong statement. The clock is moving alone and sad, and it runs
what the physicist has adjusted to be dispalyed.
That reasoning, and derivation, is not a personal reasoning of Daryl, it is
the wrong reasoning of Einstein himself.
If you permit I can recall the Wayne Throop affirmation at starting of this
post:
<<That's only the form for time dilation.
It's not the lorentz transform. The two are distinct. >>
Where is the distinction in that Daryl derivation? If we can yet speak
about these formulae.
Conclusion:
It never existed in SR any time dilatation or any length contraction.
At best it would be a time dilatation according to the velocity v in the
alone case where v = 0 . Put v = 0 in the gamma coefficient .
Complement: it does no more exist any length contraction in SR, except for
the case where v = 0 !!!
Sorry.
Francis Rey
Francis Rey writes
Wayne Throop wrote
::: ab...@iastate.edu (Alexander Abian)
::: from (2) and (3) we have
::: /------------
::: (t')^2 = (1 - v^2/t^2) t^2 or t' = \/ (1- v^2/c^2) t
:::
::: which is the Lorentz transformation of time
:: thr...@sheol.org (Wayne Throop)
:: Actually, it isn't. That's only the form for time dilation. It's
:: not the lorentz transform. The two are distinct.
: "Francis Rey" <franc...@wanadoo.fr>
: ery good since Throop is able to say the two are distinct he can be
: able to show us why they are distinct and where comes the form of time
: dilatation from.
Been there. Done that. The form for time dilation comes from plugging
the position of the moving clock into the transform. But you can't recover
the transform from the form for time dilation alone.
Rey is the fellow who thinks you can only plug lightlike lines
into the lorentz transforms. So you see how poorly he understands,
despite all this having been explained to him multiple times before.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Francis Rey
Wayne is the fellow who believes he is physicist when he makes a mish-mash
with mathematics and physics never understood.
Physics is not to "plug something " into any formulae. The proof he writes
""likelight lines"" when the problem is about real light lines .
It is very curious not to seen when light is in question.
We must recognize that Wayne is very much dreaming: he never has explained
that Lorentz formulae can be used in any way for any moving body. He has
said, as he does now, but to say is not an explanation. Parrots say .
A proof of his misunderstanding is he names "trnasform" the Lorentz
formulae, when they are merely formula which describes a real physical
phenomenon.
I don't recall the to day post to Daryl Mc Cullough on the same thread,
about time dilatation. That will permit Wayne to do as if he had not been
aware, then to continue to believe.
Tom Roberts
> <tjro...@lucent.com> writes
> >Of course the ~10 g acceleration of a fighter it still very small in
> >relativistic terms (~ 0.1 lightyear in appropriate units).
> 10g is an acceleration; a light year is a distance. What are you trying
> to say?
I'm trying to say just what I said: in appropriate units a 10 g
acceleration is ~0.1 lightyear -- OOPS: I meant ~ 0.1 inverse-lightyear.
The appropriate units are ones in which c=1 and both distance and time
are measured in lightyears (equivalent to measuring time in years because
c=1). In such units acceleration is (length)^-1. I also happen to
remember that the gravitational acceleration at the surface of the earth
(1 g) is approximately 1 lightyear in these units.
1/a is the distance at which a constantly-accelerated observer has an
event horizon....
Tom Roberts tjro...@lucent.com
Wayne Throop
:: moving down an aiming tube (or reflecting off a mirror, or going
:: through a lens, or however it is aimed) will be aimed in the exact
:: diagonal direction to keep it with the light clock, without adjusting
:: anything at all on the clock itself; the two clocks are identical.
: By means of momentum no doubt . Right Wayne ?
Wrong, "Rod".
: Which analogy will you use ? The tube ? The lens ... Pick one ,
: explain in detail and I'll show error .
I've already refered you to the relevant diagrams and explanations.
See http://sheol.org/throopw/transverse-light-bounce.gif
and http://sheol.org/throopw/transverse-light-bounce2.gif
:: Further, consider a light clock which operates by firing a laser at a
:: mirror either when triggered "by hand", or when a spot of light
:: strikes a photocell beside the laser. Such a clock would operate
:: well under quite high accelerations; just so long as the acceleration
:: doesn't take the photocell completely out of the laser's path in a
:: single light bounce. The light's path would be consistently vertical
:: in the instantaneously comoving frame of the clock, and, of course,
:: more and more diagonal in the original rest frame.
: To broad . Concentrate on specifics . Is the light a pulse ? How
: long in length . What interval if any , distance between mirrors etc
: .
1) Yes. 2) Negligably short. 3) There's only one mirror,
and it's many, many, many times as far away from the laser/lightcell
assembly as the pulse is long.
: Gamma is based on what original Einstien or other , during his time ,
: derivation wrt SR time dilation .
Meaningless gibberish.
Please translate it into English if you want me to comment.
ryk...@my-deja.com
thr...@sheol.org (Wayne Throop) wrote:
> :: the "re-aiming" Daryl mentions occurs "for free", because light
> :: moving down an aiming tube (or reflecting off a mirror, or going
> :: through a lens, or however it is aimed) will be aimed in the exact
> :: diagonal direction to keep it with the light clock, without
adjusting
> :: anything at all on the clock itself; the two clocks are identical.
Rod: ? What ? There is no re-aiming . How can there be ?
The mirrors are inline verticle , flat and paralell . The
apparent photon path , is diagonal which some call an aberation .
Is gamma based on aberation ? The photon moves verticle .
The assembly moves perpendicular wrt the photon's path .
These two seperate axis's put together , is what makes the
path of the photon appear to move in a diagonal path .
Rod: Lets say you walk in place on a tread mill for 1 hour
and the tread mill registers 3 miles traveled - 3 mph .
Now we try this again , only we put your tread mill on
a trailer traveling 100 mph perpendicular to the direction
you are walking . Your speedometer reads 3 mph.
the trailer's reads 100mph . After 1 hour Wayne says he
needs a rest . He claims to have walked sqrt(a^2+b^2)
Don't like the tread mill ? Use a 3 mile long bridge
on wheels .
> : ryk...@my-deja.com
> : By means of momentum no doubt . Right Wayne ?
>
> Wrong, "Rod".
Rod: Does the momentum of a light clock assembly moving
at v have any effect on a photon as to keep it comoving with a
light clock assembly ?
> : Which analogy will you use ? The tube ? The lens ... Pick one ,
> : explain in detail and I'll show error .
>
> I've already refered you to the relevant diagrams and explanations.
> See http://sheol.org/throopw/transverse-light-bounce.gif
> and http://sheol.org/throopw/transverse-light-bounce2.gif
Rod: Regarding your first bounce diagram , well , you
can't reason . Your reasoning is flawed . The aiming
tube you describe is null . It serves no aiming purpose
at all . No reasoning skills at all . Get the money back
for the book you copied this from . You must be smarter
than this ?
Rod: The second bounce diagram regards the ether . Which is
off subject .
> :: Further, consider a light clock which operates by firing a laser at
a
> :: mirror either when triggered "by hand", or when a spot of light
> :: strikes a photocell beside the laser. Such a clock would operate
> :: well under quite high accelerations; just so long as the
acceleration
> :: doesn't take the photocell completely out of the laser's path in a
> :: single light bounce. The light's path would be consistently
vertical
> :: in the instantaneously comoving frame of the clock, and, of course,
> :: more and more diagonal in the original rest frame.
>
> : To broad . Concentrate on specifics . Is the light a pulse ? How
> : long in length . What interval if any , distance between mirrors
etc
> : .
>
> 1) Yes. 2) Negligably short. 3) There's only one mirror,
> and it's many, many, many times as far away from the laser/lightcell
> assembly as the pulse is long.
Rod: The light will miss the mirror unless the mirror is very long .
Note that this example provided by you was not concieved by
Einstein or someone else at the time gamma was derived .
Which should tell you that gamma needs to be re evaluated .
Einstein made a lucky guess based upon a bogus light clock ?
Rod: How did Einstein derive gamma . If it wasn't Einstein ,
who and how did this person derive gamma . Please be specific .
Pictures available upon request .
> Wayne Throop thr...@sheol.org http://sheol.org/throopw
Gerry Quinn
>Rod: The light will miss the mirror unless the mirror is very long .
It may help you to consider a light bulb instead of a laser. The wave
equations of photons emitted from the bulb will interfere constructively
only along paths between the bulb and the photocell which go
'diagonally' between the mirrors. So those are the photons you will
measure, even if you don't 'aim' anything at all.
- Gerry Quinn
ryk...@my-deja.com
Rod: Below is a response I wrote to someone else
on another thread . Take time and study it carefully .
The point is that photon B is traveling diagonally .
The importance is strictly on the difference between the
two photon's path wrt their LCA's . Photon B WILL NOT
reach the top mirror in B's LCA at time 1 in B's frame . However ,
A in A's frame will see photon A hit A's top mirror at time 1 . This
makes the LCA's NOT identical . It takes longer for the B photon
to travel to reach the top mirror of the B LCA in B's frame .
If A checks his LCA against a non LC at A's position , both of
these clocks will show the same time . However , wrt the B LCA ,
a non LC WILL NOT show the same time in B's frame .
Graph it out and see for yourself .
ryk...@my-deja.com
ger...@indigo.ie (Gerry Quinn) wrote:
> In article <7mojkl$vf5$1...@nnrp1.deja.com>, ryk...@my-deja.com wrote:
>
> >Rod: The light will miss the mirror unless the mirror is very long .
>
> It may help you to consider a light bulb instead of a laser. The wave
> equations of photons emitted from the bulb will interfere
constructively
> only along paths between the bulb and the photocell which go
> 'diagonally' between the mirrors. So those are the photons you will
> measure, even if you don't 'aim' anything at all.
>
> - Gerry Quinn
Rod: Now you create another dilema here .
A stationary LCA will see it's photon oscillate in time
wrt a non light clock in it's own frame .
However , the moving LCA will see it's photon oscillate
more slowly wrt a non light clock in the same frame .
The reason being , is that the diagonally traveling photon
natuarally travels a greater distance , therefore it takes
more time for it to bounce between the mirrors . The LCA's
are not identical wrt the photons perspective . And above ,
I show the SIGNIFICANCE of this dilema .
Do you have another example ?
Next ...
Gerry Quinn
>In article <okXm3.476$Ua3...@news.indigo.ie>,
> ger...@indigo.ie (Gerry Quinn) wrote:
>> In article <7mojkl$vf5$1...@nnrp1.deja.com>, ryk...@my-deja.com wrote:
>>
>> >Rod: The light will miss the mirror unless the mirror is very long .
>>
>> It may help you to consider a light bulb instead of a laser. The wave
>> equations of photons emitted from the bulb will interfere
>constructively
>> only along paths between the bulb and the photocell which go
>> 'diagonally' between the mirrors. So those are the photons you will
>> measure, even if you don't 'aim' anything at all.
>>
>> - Gerry Quinn
>
>
>Rod: Now you create another dilema here .
>A stationary LCA will see it's photon oscillate in time
>wrt a non light clock in it's own frame .
>However , the moving LCA will see it's photon oscillate
>more slowly wrt a non light clock in the same frame .
>The reason being , is that the diagonally traveling photon
>natuarally travels a greater distance , therefore it takes
>more time for it to bounce between the mirrors . The LCA's
>are not identical wrt the photons perspective . And above ,
>I show the SIGNIFICANCE of this dilema .
>Do you have another example ?
>
>Next ...
I am not interested in your argument about mirrors. I was just tired of
seeing nonsense about the photons missing them, and I hoped that my post
would at least move matters onwards in that regard.
The point you do not seem to understand is that all clocks that depend
on the electromagnetic force for their construction _ARE_ light clocks.
Every atom is in a sense a system of mirrors reflecting an
electromagnetic field that controls the movement of its constituents.
There are other non-electromagnetic clocks such as 'muon clocks'. The
only difficulty for ether theories is to explain why they agree with
light clocks.
- Gerry Quinn
ryk...@my-deja.com
Rod: Yes you are , because you know my argument is
correct re this subject . Einstein derived gamma
with a bogus thought experiment . But what a relief ,
he made a (lucky guess) , since current data shows SR
experiments to be accurate ? .
> I was just tired of
> seeing nonsense about the photons missing them,
> and I hoped that my post
> would at least move matters onwards in that regard.
Rod: The WHOLE thought experiment Einstein used to derive
gamma is nonsense .
BTW , you post awfully late in order to move things along .
> The point you do not seem to understand is that all clocks that depend
> on the electromagnetic force for their construction _ARE_ light
clocks.
> Every atom is in a sense a system of mirrors reflecting an
> electromagnetic field that controls the movement of its constituents.
Rod: *My* point is , gamma was derived from a bogus thought-
experiment . If ONE takes the time to READ , one will clearly
see I'm correct .
> There are other non-electromagnetic clocks such as 'muon clocks'. The
> only difficulty for ether theories is to explain why they agree with
> light clocks.
Rod: Ether ? Me ? I hold (no) single theory as absolute at this
time . Although , upon reading new material presented by Hemetis
I may quickly change my mind .
> - Gerry Quinn