All you need to know about SR
Richard
According to Dirk Van de moortel:
As seen in Moe's frame where time t is used:
http://users.pandora.be/vdmoortel/dirk/Stuff/MoeLarryCurly.gif
<Total proper time Larry>
= Int{ 0 to T/2; 1/glout dt } + Int{ T/2 to T; 1/glin dt }
<Total proper time Curly>
= Int{ 0 to T/2; 1/gcout dt } + Int{ T/2 to T; 1/gcin dt }
where the gamma values are given by
glout = 1/sqrt[ 1 - v^2/c^2 ] Larry outbound
glin = 1/sqrt[ 1 - (-v)^2/c^2 ] Larry inbound
gcout = 1/sqrt[ 1 - (-v)^2/c^2 ] Curly outbound
gcin = 1/sqrt[ 1 - v^2/c^2 ] Curly inbound
and you see that
glout = glin = gcout = gcin
so finally
<Total proper time Larry> = <Total proper time Curly>
= 1/g * Int{ 0 to T; dt }
= 1/g * T
= <Total proper time Moe> / g
where
g = 1/sqrt[ 1 - v^2/c^2 ]
___________________________________
Larry and Curly have accellerated equally away from Moe's rest frame in
opposite directions, thus Dirk concludes, nay 'proves' above that from
Moe's rest frame Larry's and Curly's clocks will have ticked equally wrt
Moe upon thier symetric return to Moes frame. Moreover Moe's clock will
have ticked a greater number of times than Larry's and Curly's clocks.
Now let's suppose that Moe, Larry, and Curly were previously all at rest
in Shem's frame, which is identical to Curly's outboud frame above.
Let's also suppose that when Moe, Larry, and Curly departed from Shem's
rest frame to enter Moe's rest frame above, that another party
accellerated equally in the opposite direction to Moe, Larry, and Curly,
wrt Shem's frame in a symetric manner just as in the case above:
Thus we have an identical system as above, that is, accoding to Dirk
Moe's clock must be ticking slower than Shem's clock.
So after Larry and Curly accellerate away from Moe, in Dirks argument
above, then we have:
1) Curly's clock is ticking slower than Moe's clock, wrt Moe.
2) Moe's clock is ticking slower than Shem's clock, wrt Shem
3) Curly's clock is ticking at the same rate as Shem's clock, wrt Shem
If this last sequence is sustained for a very large interval, then the
durations of the acclerations do not enter into the outcome. For all
practical purposes Shem's and Curly's clocks remained at rest wrt each
other throught the total sequence.
Thus according to Shem, Curly's clock is ticking faster than Moe's
clock, but according to Moe Curly's clock is ticking slower than Moes'
clock, these ticking rates are reversed. Shem and Moe cannot agree upon
the relative ticking rates of these clocks. OTOH, there will only be one
objective outcome, which is independent of perceptions of it, thus
either Shem or Moe, or both, are morons for thinking that the Lorentz
transform is invariant wrt frames of reference, and that it has anything
valid to say about the ticking rates of clocks.
No amount of mathematical torture will cure this mishap, SR is
internally inconsistent. OTOH if K is always taken as the frame of the
medium, then no such contradictions arise. Time dilation can only be
logically maintained when it is a function related to some absolute
field such as the gravitational field of a planet, Sun, Galaxy, etc.
Thus the very real need for a 'general' theory of relativity to
eliminate the contradictions of SR, or IOW to undo the mistake that was
SR. It yet remains to be shown how even Einstien's version of the
'general' theory will resolve the pelucid contradiction illustrated
above.
Richard Perry
For an understanding of the root source of the virus known as "Special
Relativity", consult any text on Classical Electromagnetism. Then see
why the approach of the Classical model is lacking: View the following
correction to Classical Electromagnetism, note the use of the
mathematically and conceptually correct method of providing invariance.
Richard
According to Dirk Van de moortel:
As seen in Moe's frame where time t is used:
http://users.pandora.be/vdmoortel/dirk/Stuff/MoeLarryCurly.gif
<Total proper time Larry>
= Int{ 0 to T/2; 1/glout dt } + Int{ T/2 to T; 1/glin dt }
<Total proper time Curly>
= Int{ 0 to T/2; 1/gcout dt } + Int{ T/2 to T; 1/gcin dt }
where the gamma values are given by
glout = 1/sqrt[ 1 - v^2/c^2 ] Larry outbound
glin = 1/sqrt[ 1 - (-v)^2/c^2 ] Larry inbound
gcout = 1/sqrt[ 1 - (-v)^2/c^2 ] Curly outbound
gcin = 1/sqrt[ 1 - v^2/c^2 ] Curly inbound
and you see that
glout = glin = gcout = gcin
so finally
<Total proper time Larry> = <Total proper time Curly>
= 1/g * Int{ 0 to T; dt }
= 1/g * T
= <Total proper time Moe> / g
where
g = 1/sqrt[ 1 - v^2/c^2 ]
___________________________________
Larry and Curly have accelerated equally away from Moe's rest frame in
opposite directions, thus Dirk concludes, nay 'proves' above that from
Moe's rest frame Larry's and Curly's clocks will have ticked equally wrt
Moe upon their symmetric return to Moe's frame. Moreover Moe's clock
will
have ticked a greater number of times than Larry's and Curly's clocks.
Now let's suppose that Moe, Larry, and Curly were previously all at rest
in Shem's frame, which is identical to Curly's outbound frame above.
Let's also suppose that when Moe, Larry, and Curly departed from Shem's
rest frame to enter Moe's rest frame above, that another party
accelerated equally in the opposite direction to Moe, Larry, and Curly,
wrt Shem's frame in a symmetric manner just as in the case above:
Thus we have an identical system as above, that is, according to Dirk
Moe's clock must be ticking slower than Shem's clock.
So after Larry and Curly accelerate away from Moe, in Dirk's argument
above, then we have:
1) Curly's clock is ticking slower than Moe's clock, wrt Moe.
2) Moe's clock is ticking slower than Shem's clock, wrt Shem
3) Curly's clock is ticking at the same rate as Shem's clock, wrt Shem
If this last sequence is sustained for a very large interval, then the
durations of the accelerations do not enter into the outcome. For all
practical purposes Shem's and Curly's clocks remained at rest wrt each
other throughout the total sequence.
Thus according to Shem, Curly's clock is ticking faster than Moe's
clock, but according to Moe Curly's clock is ticking slower than Moe's
clock, these ticking rates are reversed. Shem and Moe cannot agree upon
the relative ticking rates of these clocks. OTOH, there will only be one
objective outcome, which is independent of perceptions of it, thus
either Shem or Moe, or both, are morons for thinking that the Lorentz
transform is invariant wrt frames of reference, and that it has anything
valid to say about the ticking rates of clocks.
No amount of mathematical torture will cure this mishap, SR is
internally inconsistent. OTOH if K is always taken as the frame of the
medium, then no such contradictions arise. Time dilation can only be
logically maintained when it is a function related to some absolute
field such as the gravitational field of a planet, Sun, Galaxy, etc.
Thus the very real need for a 'general' theory of relativity to
eliminate the contradictions of SR, or IOW to undo the mistake that was
SR. It yet remains to be shown how even Einstein's version of the
'general' theory will resolve the pellucid contradiction illustrated
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3F4DADA7...@yahoo.com...
>
> According to Dirk Van de moortel:
>
> As seen in Moe's frame where time t is used:
> http://users.pandora.be/vdmoortel/dirk/Stuff/MoeLarryCurly.gif
> <Total proper time Larry>
> = Int{ 0 to T/2; 1/glout dt } + Int{ T/2 to T; 1/glin dt }
> <Total proper time Curly>
> = Int{ 0 to T/2; 1/gcout dt } + Int{ T/2 to T; 1/gcin dt }
> where the gamma values are given by
> glout = 1/sqrt[ 1 - v^2/c^2 ] Larry outbound
> glin = 1/sqrt[ 1 - (-v)^2/c^2 ] Larry inbound
> gcout = 1/sqrt[ 1 - (-v)^2/c^2 ] Curly outbound
> gcin = 1/sqrt[ 1 - v^2/c^2 ] Curly inbound
> and you see that
> glout = glin = gcout = gcin
> so finally
> <Total proper time Larry> = <Total proper time Curly>
> = 1/g * Int{ 0 to T; dt }
> = 1/g * T
> = <Total proper time Moe> / g
> where
> g = 1/sqrt[ 1 - v^2/c^2 ]
You forgot to include my conclusion:
| HTH but I'm sure it won't. Never mind, many engineers
| have the struggle of their life with it. Yet, bridges do get
| built, don't they?
Since you don't understand this, I will clarify:
If you can build a bridge without understanding relativity,
then relativity must be wrong, right?
> ___________________________________
>
> Larry and Curly have accelerated equally away from Moe's rest frame in
> opposite directions, thus Dirk concludes, nay 'proves' above that from
> Moe's rest frame Larry's and Curly's clocks will have ticked equally wrt
> Moe upon their symmetric return to Moe's frame. Moreover Moe's clock
> will
> have ticked a greater number of times than Larry's and Curly's clocks.
Splash already :-)
I have *not* proven what you say here, because what
you say is total rubbish, showing that you don't know
what you are talking about.
I have proven above that Larry's and Curly's clocks will
have ticked equally upon their symmetric return to Moe.
Moreover Moe's clock will have ticked a greater number
of times than Larry's and Curly's clocks.
Read it carefully and spot the differences with your
nonsense.
Then try again, but keep in mind that linear algebra and
analytical geometry garantee that my proof is sufficient.
We don't have to check the calculations in other frames
since the Lorentz transformation is linear.
No amount of words rooted in your pathetically malicious
self-inflicted misunderstanding will make that go away.
Do the fucking math like you asked me on
http://groups.google.com/groups?&threadm=3F43B5A6...@yahoo.com
| "I want to see your fucking math, no irrelevant BS,
| just the fucking math."
You got the fucking math.
Now you give me your fucking math.
And - please - make it slightly more intelligent than
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowdyDoo.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LorentzPerry.html
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AAB.html
Dirk Vdm
Mathew Orman
Hallo Richard,
DVM is trying to tell you that one must be one of those elite 15 people in
the entire population of humans on the Earth
in order to fully understand and appreciate SR.
But obviously you are not!
So you just have to put up with being a normal logical down to the Earth
man who's mind cannot be easy brainwashed :)
Sincerely,
Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com
Richard
Dirk isn't brainwashed, he's just fucking dumb;)
Fucking SR Morons, eh!
>
> Sincerely,
>
> Mathew Orman
> www.ultra-faster-than-light.com
> www.radio-faster-than-light.com
Richard Perry
Titan Point
> You got the fucking math.
> Now you give me your fucking math.
> And - please - make it slightly more intelligent than
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/HowdyDoo.html
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LorentzPerry.html
> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AAB.html
>
> Dirk Vdm
Still collecting winners for the "Immortal Fumbles", Dirk? 'Tis a strange
but noble hobby...
Richard
He still has to collect his own fumbles and post them. The above reply
is one of those. The dipshit doesn't know that he already provided the
fucking math.
Richard Perry
Titan Point
He has provided the fucking math but better still, you've provided the
fucking fumble and entertainment for us all.
Mathew Orman
"Titan Point" <titanpoi...@myrealbox.com> wrote in message
news:pan.2003.08.29....@myrealbox.com...
Here is the tough one for you and Dirk:
In reality motion velocity is expressed as m/m so it is a dimensionless
number.
Or is it?
Robert J. Kolker
Mathew Orman wrote:
>
> In reality motion velocity is expressed as m/m so it is a dimensionless
> number.
> Or is it?
>
Velocity can be expressed as distance per unit time. Or it can be
expressed as a fraction of the speed of light in a specified direction.
Speed as a fraction of the speed of light appears to be dimensionless,
but it isn't.
Bob Kolker
Tom Potter
"Robert J. Kolker" <bobk...@attbi.com> wrote in message
news:bioou7$bhnpt$1...@ID-76471.news.uni-berlin.de...
Bob raises a good point!
As he indicates, and this is very important,
"Speed as a fraction of the speed of light appears to be dimensionless,
but it isn't."
When you divide speed by C you end up with a
seemingless dimensionless number,
which is really a tangent function,
as I graphically show in the physics tutorial
which can be downloaded from my web site.
And the tutorial goes on top show that
Special Relativity is basically the law of tangents.
--
Tom Potter http://tompotter.us
Sam Wormley
0.8 c is velocity with unit, but the 0.8 is unitless.
Special Relativity is NOT basically the law of tangents.
Special Relativity
http://scienceworld.wolfram.com/physics/SpecialRelativity.html
Law of Tangents
http://mathworld.wolfram.com/LawofTangents.html
Tom Potter
"Sam Wormley" <swor...@mchsi.com> wrote in message
news:3F500CB5...@mchsi.com...
Sam, I don't, and didn't, expect you to comprehend this fact,
as you have a serious mind lock,
and are an evangelist for the holy algorithm.
I posted this information so that the folks who don't
have locks on their minds,
can take a look at my presentation,
and decide for themselves.
Addition of tangents:
tan(X+Y) = tan(X) + tan(Y) / 1 + tan(X) * tan(Y)
Addition of velocities ( "C" set to one (1)):
velocity(X+Y) = velocity(X)+velocity(Y) / 1 + velocity(X) * velocity(Y)
Sam Wormley
> >
> > Special Relativity is NOT basically the law of tangents.
> >
> > Special Relativity
> > http://scienceworld.wolfram.com/physics/SpecialRelativity.html
> >
> > Law of Tangents
> > http://mathworld.wolfram.com/LawofTangents.html
>
> Sam, I don't, and didn't, expect you to comprehend this fact,
> as you have a serious mind lock,
> and are an evangelist for the holy algorithm.
>
> I posted this information so that the folks who don't
> have locks on their minds,
> can take a look at my presentation,
> and decide for themselves.
>
> Addition of tangents:
> tan(X+Y) = tan(X) + tan(Y) / 1 + tan(X) * tan(Y)
>
> Addition of velocities ( "C" set to one (1)):
> velocity(X+Y) = velocity(X)+velocity(Y) / 1 + velocity(X) * velocity(Y)
>
> --
> Tom Potter http://tompotter.us
Velocity != Tangent
Dirk Van de moortel
"Tom Potter" <t...@hotsheet.com> wrote in message news:bip9fu$b7djo$3...@ID-188019.news.uni-berlin.de...
[snip]
> Sam, I don't, and didn't, expect you to comprehend this fact,
> as you have a serious mind lock,
> and are an evangelist for the holy algorithm.
>
> I posted this information so that the folks who don't
> have locks on their minds,
> can take a look at my presentation,
> and decide for themselves.
>
> Addition of tangents:
> tan(X+Y) = tan(X) + tan(Y) / 1 + tan(X) * tan(Y)
tan(X+Y) = [ tan(X) + tan(Y) ] / [ 1 - tan(X) * tan(Y) ]
Dirk Vdm
Tom Potter
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:LuZ3b.8435$tg7.4...@phobos.telenet-ops.be...
Addition of velocities ( "C" set to one (1)):
velocity(X+Y) = [ velocity(X)+velocity(Y)] / [ 1 + velocity(X) *
velocity(Y) ]
As you can see from this example,
and from the graphic development in the physics tutorial
that can be downloaded from my web site,
[velocity / C ] is a tangent function,
and Special Relativity is basically the law of tangents.
In other words, velocities are basically tangents,
and they add according to the law of tangents.
Dirk Van de moortel
"Tom Potter" <t...@hotsheet.com> wrote in message news:biqac3$buste$3...@ID-188019.news.uni-berlin.de...
They don't.
If you insist on using your notation
velocity(X+Y) =
[ velocity(X)+velocity(Y)] / [ 1 + velocity(X) * velocity(Y) ]
then that is okay, provided X and Y refer to the so-called "rapidity"
and the velocity function is the *hyperbolic* tangent:
v = velocity(X) = Tanh(X) or c*Tanh(X/c) if we don't take c=1
u = velocity(Y) = Tanh(Y) or c*Tanh(Y/c) if we don't take c=1
*Only* then you can say in a mathematically correct way:
velocity(X+Y)
= Tanh(X+Y)
= [ Tanh(X) + Tanh(Y) ] / [ 1 + Tanh(X)*Tanh(Y) ]
= [ velocity(X)+velocity(Y)] / [ 1 + velocity(X) * velocity(Y) ]
So in this case the notation you used is okay, provided of course
you use the hyperbolic tangent and not the trigonometric tangent.
(see also
http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html )
If you are not prepared to do this, you immediately get in
trouble with your notation.
Of course there *is* a connection between velocity and
the trigonometric tangent, but that has nothing to do with
addition of velocities or with relativity. That is purely
elementary calculus and ditto geometry.
Dirk Vdm
The Ghost In The Machine
<swor...@mchsi.com>
wrote
on Sat, 30 Aug 2003 04:44:10 GMT
<3F502B95...@mchsi.com>:
Maybe not, but Tom raises an interesting point, although unfortunately
makes a sign error as well, as
tan(a+b) = (tan(a)+tan(b)) / (1 - tan(a)*tan(b))
However, if one looks at the problem abstractedly,
v' = (v + w) / (1 + vw/c^2)
If we state that c = 1 (which can be done by an appropriate
choice of units, we get)
v' = (v + w) / (1 + vw)
which is awfully close to
tanh(a + b) = (tanh(a) + tanh(b)) / (1 + tanh(a) * tanh(b))
So a velocity could be construed as a form of hyperbolic tangent.
Whether it's useful I for one don't know; at this point it's
merely an odd sort of transformation.
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Dirk Van de moortel
"The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:t0k621-...@lexi2.athghost7038suus.net...
If you haven't (or hadn't) already, see my second reply to TP
on this thread.
Dirk Vdm
Bilge
>If we state that c = 1 (which can be done by an appropriate
>choice of units, we get)
>
>v' = (v + w) / (1 + vw)
>
>which is awfully close to
>
>tanh(a + b) = (tanh(a) + tanh(b)) / (1 + tanh(a) * tanh(b))
>
>So a velocity could be construed as a form of hyperbolic tangent.
>Whether it's useful I for one don't know; at this point it's
>merely an odd sort of transformation.
It should look similar. The derivation for the lorentz transforms
gives you both rotations and boosts. Does,
x' = \gamma x - \gamma\beta ct
ct' = \gamma ct - \gamma\beta x
resemble:
x' = x cosh(A) - ct\beta sinh(A) = cosh(A)[ x - ct tanh(A)]
ct' = ct cosh(A) - x\beta sinh(A) = cosh(A)[ct - x tanh(A)]
and look similar to:
x' = x cos(B) + y sin(B)
y' = y cos(B) - x sin(B)
-----------
\beta = tanh(A)
\gamma = cosh(A)
\gamma\beta = sinh(A)
The Ghost In The Machine
<dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
on Sat, 30 Aug 2003 16:15:14 GMT
<m444b.9261$Mb7.4...@phobos.telenet-ops.be>:
>
> "The Ghost In The Machine" <ew...@sirius.athghost7038suus.net> wrote in message news:t0k621-...@lexi2.athghost7038suus.net...
[snip for brevity]
>> So a velocity could be construed as a form of hyperbolic tangent.
>> Whether it's useful I for one don't know; at this point it's
>> merely an odd sort of transformation.
>
> If you haven't (or hadn't) already, see my second reply to TP
> on this thread.
Blame it on "post delay". :-) I use leafnode.
>
> Dirk Vdm
Bilge
>resemble:
>
> x' = x cosh(A) - ct\beta sinh(A) = cosh(A)[ x - ct tanh(A)]
> ct' = ct cosh(A) - x\beta sinh(A) = cosh(A)[ct - x tanh(A)]
typo alert: sinh(A) not \beta \sinh(A)