Error in the Aether debate
Richard
Both parties to the ongoing aether debate have made a common error in
judgment, i.e. that transforming from any frame K' to any other frame
K'' is as valid as transforming from the absolute frame k to either of
these.
The math is incorrect. I have already provided a very simple proof in a
related thread.
We have for the transformation from K to K' and K to K'' respectively:
t = t' gamma' where v' is the speed of K' wrt K
t = t'' gamma'' where v'' is the speed of K'' wrt K
Speed is the key word here, it doesn't matter what direction the clock
is moving wrt K, thus you add v' and v'' as vectors. In fact you cannot
add them at all unless you know an algebraic trick that I am unaware of.
By substitution from the above we get this 'correct' relationship
between the clocks in frames K' and K'':
t' = t''gamma''/gamma'
which is decidedly not the Lorentz transformation equation.
The Lorentz equation is valid only for transformations to or from the K
frame (aether frame).
And the above correct relationship between these clocks is at odds with
SR's
t' = t''gamma''', or is it perhaps
t'' = t'gamma''' ?
Who here can decide which of the above two equations applies?
It simply doesn't matter, they are both wrong.
Expanding the latter, what both sides seem to believe to be correct, we
get:
t' = t''/sqrt(1 - (v'- v'')^2/c^2)
If you can derive this by simplifying the above correct relationship,
which gives
t' = t'' sqrt[(c^2 - (v')^2)/(c^2 - (v'')^2)]
then I'd be interested in your methods.
As you can see, there is an objective difference in ticking rates
between the K' and K'' clocks, and moreover they are 'equal' when
v' = v'', even though the clocks may be moving in opposite directions,
i.e. even when their relative velocity is 2v'.
Unfortunately, and this is the problem that plaques SR, we must
arbitrarily decide which of these clocks is ticking faster than the
other, not a problem using the correct equations that I just provided
you with.
Consider the analogy to sound, and maybe even cracking a book on sound
wave doppler and related, you'll find that these same equations will be
derived when considering a sound clock whose waves are propagating in
the transverse direction between the reflectors. The longitudinal
propagation need not even be considered for the purpose of clock rate
comparisons. Nor does it change the above correct equations when it 'is'
considered, because the length contractions are also quite absolute.
Again they are the same when v' = v''. SR is insanity, and it would
seem, ironically, that is, if the reports of his statements are
reliable, that Lorentz began the insanity with complete lack of
mathematical skills, or else a failure to understand the difference
between a scalar and a vector.
There are only a small handful of people engaged in this debate that
understand what I just posted above, you know who you are, and thank you
for existing, I don't feel so alone in a world full of morons as long as
you are around;)
Richard Perry
Richard
Both parties to the ongoing aether debate have made a common error in
judgment, i.e. that transforming from any frame K' to any other frame
K'' is as valid as transforming from the absolute frame k to either of
these.
The math is incorrect. I have already provided a very simple proof in a
related thread.
We have for the transformation from K to K' and K to K'' respectively:
t = t' gamma' where v' is the speed of K' wrt K
t = t'' gamma'' where v'' is the speed of K'' wrt K
Speed is the key word here, it doesn't matter what direction the clock
is moving wrt K, thus you 'cannot' add v' and v'' as vectors. In fact
Uncle Al
>
> Both parties to the ongoing aether debate have made a common error in
> judgment, i.e. that transforming from any frame K' to any other frame
> K'' is as valid as transforming from the absolute frame k to either of
> these.
1) Aether must be locally at rest - Michelson-Moreley,
Phys. Rev. Lett. 88(1) 010401 (2002)
Phys. Rev. Lett. 90 060403 (2003)
Phys. Rev. Lett. 42(9) 549 (1979)
Phys. Bull. 21 255 (1970)
2) Aether cannot be locally at rest - Hipparcos star displacements,
http://www.astro.lu.se/~dainis/HTML/ASTROMET.html
middle
<http://www.oso.chalmers.se/~michael/astrobiologi-2002/
j.1468-4004.2001.42512.x.pdf>
http://ilrs.gsfc.nasa.gov/docs/iers_1996_conventions.pdf
http://arXiv.org/abs/astro-ph/0302522
3) Therefore, there is no aether.
4) If empirical reality says you are an idiot, you are an empirical
idiot.
http://w0rli.home.att.net/youare.swf
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Richard
Uncle Al wrote:
>
> Richard wrote:
> >
> > Both parties to the ongoing aether debate have made a common error in
> > judgment, i.e. that transforming from any frame K' to any other frame
> > K'' is as valid as transforming from the absolute frame k to either of
> > these.
> [snip]
>
> 1) Aether must be locally at rest - Michelson-Moreley,
> Phys. Rev. Lett. 88(1) 010401 (2002)
> Phys. Rev. Lett. 90 060403 (2003)
> Phys. Rev. Lett. 42(9) 549 (1979)
> Phys. Bull. 21 255 (1970)
The MM null results alone only determined that the MM was incapable of
determining motion wrt the aether. Similarly two way doppler, even wrt
classical sound waves, is incapable of being used to determine motion
wrt the medium. There are however other forms of experiment that can by
performed to determine motion wrt the medium. I've already suggested
three.
>
> 2) Aether cannot be locally at rest - Hipparcos star displacements,
> http://www.astro.lu.se/~dainis/HTML/ASTROMET.html
> middle
> <http://www.oso.chalmers.se/~michael/astrobiologi-2002/
> j.1468-4004.2001.42512.x.pdf>
> http://ilrs.gsfc.nasa.gov/docs/iers_1996_conventions.pdf
> http://arXiv.org/abs/astro-ph/0302522
>
> 3) Therefore, there is no aether.
Bullshit, stars and other matter 'are' the aether.
>
> 4) If empirical reality says you are an idiot, you are an empirical
> idiot.
> http://w0rli.home.att.net/youare.swf
I'm waiting for that empirical evidence, I haven't seen any to date.
OTOH there does exist empirical evidence in support of the correct
relativistic equations that I posted, namely Hafele-Keating. It was the
clock in motion wrt Earth (which 'is' the local aether) that ticked
slower. The Fizeau data also requires an Earth centered local aether.
Get past the original aether as an independent medium spew, and
understand that all matter is (not is the source of, but 'is') the
aether. This is exactly Einstein's gravitational aether. It exists,
motion wrt it is measurable, and it'll take a bit more than an
unqualified statement from you to negate that fact. I OTOH, unlike you,
am not guessing or parroting, I know beyond doubt that I am correct.
I've done the math for myself, I've posted the math, and it is
immaculately correct. Every challenge I've laid to discredit the math
has been glossed over. You know why that is Al? Because its fucking
right asshole!
Richard Perry
Paul Cardinale
> Both parties to the ongoing aether debate have made a common error in
> judgment, i.e. that transforming from any frame K' to any other frame
> K'' is as valid as transforming from the absolute frame k to either of
> these.
>
> The math is incorrect. I have already provided a very simple proof in a
> related thread.
>
> We have for the transformation from K to K' and K to K'' respectively:
>
> t = t' gamma' where v' is the speed of K' wrt K
> t = t'' gamma'' where v'' is the speed of K'' wrt K
>
That's not the LT, you moron.
[remaining excrement snipped]
Paul Cardinale
Ed Keane III
>
>
> I've done the math for myself, I've posted the math, and it is
> immaculately correct. Every challenge I've laid to discredit the math
> has been glossed over. You know why that is Al? Because its fucking
> right asshole!
>
You do not have a right asshole. The hole that shits in the toilet
is your ass hole. The other hole is your mouth. Do not take it so
personal when someone describes what you are saying. It is
just their opinion.
Richard
It is equivalent to the LT, thus it 'is' the LT.
If not then you have a problem, because your clocks will be
simultaneously ticking at two distinct ratios. They are equivalent
though, so not a problem. You are however just fucking dumb, and that is
an incurable condition.
Richard Perry
Paul Cardinale
> Paul Cardinale wrote:
> >
> > Richard <no_mail...@yahoo.com> wrote in message news:<3F4048BE...@yahoo.com>...
> > > Both parties to the ongoing aether debate have made a common error in
> > > judgment, i.e. that transforming from any frame K' to any other frame
> > > K'' is as valid as transforming from the absolute frame k to either of
> > > these.
> > >
> > > The math is incorrect. I have already provided a very simple proof in a
> > > related thread.
> > >
> > > We have for the transformation from K to K' and K to K'' respectively:
> > >
> > > t = t' gamma' where v' is the speed of K' wrt K
> > > t = t'' gamma'' where v'' is the speed of K'' wrt K
> > >
> >
> > That's not the LT, you moron.
> >
> > [remaining excrement snipped]
> >
> > Paul Cardinale
>
> It is equivalent to the LT, thus it 'is' the LT.
>
No it isn't you imbecile. You are completely clueless about the LT.
And your too stupid and too arogant to realize that you are ignorant.
If you would bother to look it up (but you won't because you're a lazy
dishonest troll), you would see that you are a ridiculous fool.
Paul Cardinale
Richard
Just for you dumbass:
t' = (t - vx/c^2)gamma'
t' = (t - tvv/c^2)gamma'
t' = t(1 - v^2/c^2)gamma'
t' = t gamma'/gamma'^2
t' = t/gamma'
t = t'gamma'
Therefore, in conclusion, you are just fucking dumb.
Richard Perry
Paul Cardinale
That is not correct, you pathetic idiot.
You assumed that v = x/t, which is wrong. v is dx/dt. There is a
difference between x/t and dx/dt, and that difference matters.
[crap snipped]
>
> Therefore, in conclusion, you are just fucking dumb.
>
Look in a mirror when you write that.
Paul Cardinale
Richard
You're pathetic.
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3F4DA0D9...@yahoo.com...
hm, I missed this thread before :-)
Retrieving context from google
http://groups.google.com/groups?&threadm=3F404943...@yahoo.com
|||||||||||||||||||||||||||||||||
| "Both parties to the ongoing aether debate have made a
| common error in judgment, i.e. that transforming from
| any frame K' to any other frame K'' is as valid as
| transforming from the absolute frame k to either of
| these.
|
| The math is incorrect. I have already provided a very
| simple proof in a related thread.
|
| We have for the transformation from K to K' and K
| to K'' respectively:
|
| t = t' gamma' where v' is the speed of K' wrt K
| t = t'' gamma'' where v'' is the speed of K'' wrt K
|
| Speed is the key word here, it doesn't matter what
| direction the clock is moving wrt K, thus you 'cannot'
| add v' and v'' as vectors. In fact you cannot add them
| at all unless you know an algebraic trick that I am
| unaware of.
|
| By substitution from the above we get this 'correct'
| relationship between the clocks in frames K' and K'':
|
| t' = t''gamma''/gamma'
|
| which is decidedly not the Lorentz transformation
| equation.
|||||||||||||||||||||||||||||||||
That equation:
t = t' gamma' where v' is the speed of K' wrt K
is valid for events with x'=0 (or v'=0) only, since
t = gamma' ( t' + v' x' /c^2 )
Likewise that equation:
t = t" gamma" where v" is the speed of K" wrt K
is valid for events with x"=0 (or v"=0) only, since
t = gamma" ( t" + v" x" /c^2 )
Let's forget about the silly v'=0 and v"=0, okay?
So, if you are going to combine these two equations to
t' = t''gamma''/gamma' ,
then you have an equation that is only valid for events
that satisfy
x'=0 and x"=0
together.
Using the Lorentz transformation between K' and K"
(supposing that K" has velocity v w.r.t. K'):
x" = gamma (x'-vt') ,
we see that we must also have:
t'=0
and thus
t=0
and thus
t"=0 .
Finally with
x = gamma'( x' + v't' ) ,
we have
x=0
Conclusion: your equations are valid if and only if
x = x' = x" = t = t' = t" = 0
or
v' = v" = v = 0
How interesting.
Reminds me of Marcel Luttgens.
Dirk Vdm
Titan Point
>
> Conclusion: your equations are valid if and only if
> x = x' = x" = t = t' = t" = 0
> or
> v' = v" = v = 0
> How interesting.
> Reminds me of Marcel Luttgens.
>
> Dirk Vdm
You mean he proved nothing? ;-)
Richard
Again:
t' = (t - vx/c^2)gamma'
t' = (t - tvv/c^2)gamma'
t' = t(1 - v^2/c^2)gamma'
t' = t gamma'/gamma'^2
t' = t/gamma'
t = t'gamma'
_________________________________
Thus by substitution:
t = (t - vx/c^2)gamma'
gamma' = t/(t - vx/c^2)
gamma' = 1/(1-vx/tc^2)
And thus
(1-vx/tc^2) = sqrt(1 - v^2/c^2)
Squaring both sides
1 - (vx/tc^2)^2 = 1 - v^2/c^2
and
(vx/tc^2)^2 = v^2/c^2
vx/tc^2 = v/c
x/tc = 1
t = x/c
or ct = x
Thus the velocity defined by the two events is c.
Thus the relationship
t = t'gamma'
which is commonly cited, and derived by Einstein himself, applies only
to the light clock, which was the device he used to derive the equation
in the first place. It is valid only when applied to the propagation of
light at c. No other clock will tick at the same rate wrt the same
observer.
Also, Dirk's math was incorrect. Dirk is just fucking dumb, so you'll
have to forgive him. Moreover I've addressed this issue before, and
again in a related thread, viz. that the relative ticking rate of two
clocks depends upon frame of reference, and can even be reversed
depending upon your arbitrary starting point. Thus Dirk's misuse of
Lorentz (which is just standard SR) is simply moronic. See the thread
"All you need to know about SR".
BTW, you're just fucking dumb too.
Richard Perry
Richard
successive locations of one of the frames itself wrt the other, and of
course that observer himself will always measure his change in x to be
zero, but not so for his time. Setting x to zero in the fourth equation
of the transform
t' = (t - vx/c^2)gamma'
gives:
t' = t gamma'
OTOH, this isn't the relative ticking rates of the clocks moving in
tandem with frames K and K', it is rather the relative respective
measurements of the ticking rate of the 'same' clock.
Thus the ticking ratio of the respective clocks in frames K and K' is
the inverse of this, or
t = t' gamma'.
And I have simply taken a shortcut to Einstein's derivation of this
equation.
OTOH, I still have the Larry, Curly, Moe, and "Shem", thought experiment
to fall back on, that is, it is still true that Dirk is just fucking
dumb:)
Richard Perry
Richard
Richard wrote:
>
> Ok, I see what I did here, I set the two events as the respective
> successive locations of one of the frames itself wrt the other, and of
> course that observer himself will always measure his change in x to be
> zero, but not so for his time. Setting x to zero in the fourth equation
> of the transform
>
> t' = (t - vx/c^2)gamma'
>
> gives:
>
> t' = t gamma'
>
> OTOH, this isn't the relative ticking rates of the clocks moving in
> tandem with frames K and K', it is rather the relative respective
> measurements of the ticking rate of the 'same' clock.
>
> Thus the ticking ratio of the respective clocks in frames K and K' is
> the inverse of this, or
>
> t = t' gamma'.
>
> And I have simply taken a shortcut to Einstein's derivation of this
> equation.
>
> OTOH, I still have the Larry, Curly, Moe, and "Shem", thought experiment
> to fall back on, that is, it is still true that Dirk is just fucking
> dumb:)
>
> Richard Perry
Hell, I'm still off, what I did above (setting x to zero) 'did' give me
a relative ticking rate of the clocks, it just happens to be
contradictory to the result of setting x' to zero. Thus strictly in
accord with my complaint that 'which clock is the faster ticking clock'
depends entirely upon your arbitrary starting point. Or IOW, which
frame do we designate as K and which as K'?
Russell Blackadar
[after some painful flailing]
> Hell, I'm still off, what I did above (setting x to zero) 'did' give me
> a relative ticking rate of the clocks, it just happens to be
> contradictory to the result of setting x' to zero.
x' of *what*? Some clock?
*Of course* you can't put a clock in a position such that
its x and x' are both zero for successive ticks, unless v=0.
(But x and x' *can* both be zero at a single event.)
At least you seem to be slowly coming to awareness that you
have a few things still to learn in this business. That's
an encouraging sign.
I suggest you stop posting equations until you have figured
out what each symbol means; in particular for your x's and t's,
primed and unprimed, what events they are supposed to be
coordinates for, and what objects (if any) are supposed to
be present at those events. You might be surprised to find
that things will begin to make sense after all.
Thus strictly in
> accord with my complaint that 'which clock is the faster ticking clock'
> depends entirely upon your arbitrary starting point. Or IOW, which
> frame do we designate as K and which as K'?
Either one. (But stick to the choice, once you have made
it.)
You are assuming that the question "which clock is faster"
has an answer that is independent of frame. But experiments
have found that there is no answer of that kind; ticking
rate of a clock, as measured by the usual procedure in some
frame, is a phenomenon that depends on that frame. The thing
that you "complain" about in SR is precisely what puts it
in accord with experiment. Essentially, you are complaining
about how the universe works and blaming SR for it. That's
a natural first reaction, I suppose, but there's no reason
you can't learn to outgrow it.
Richard
Hafele-Keating! You cannot arbitrarily chose a frame asshole, one clock
objectively ticked slower than the other. As for your rant about x and
x' set at zero, you're a fucking moron.
x' = (x - vt)gamma
Bear with me now,
This is called the first equation of the lorentz transform.
x' is the K' frame space interval between two events.
x is the K frame space interval between the same two events.
If the two events are the respective locations of the K' frame itself
wrt K and K' respectively, then x' can indeed be zero, it is zero from
the frame K'. OTOH x is zero from the frame K.
Now before you prove yourself a greater idiot by stating that these are
not intervals but positions, let me remind you that when you assign a
real number to x you get an interval whether you like it or not, that
interval is |x - 0|, i.e. you have set one end of the interval at the
origin of the coordinate system. IOW you will get the same result from
either form
x' = (x - vt)gamma
or
dx' = (dx - vdt)gamma
Thus they are equivalent interpretations.
According to K' the space interval is always zero when that interval is
that defined by his respective positions over some time interval.
Likewise for K. When x or x' are set to zero, then it follows that
x/t = v, or x'/t' = v
I'll be more than happy to explain further if you are having trouble
with this math.
I don't deal well with condescending morons, because that's an oxymoron,
which BTW you are. If you think that I'm confused, or mathematically
inept, then perhaps you would care to examine the electromagnetic theory
linked below, and tell me which part of it is incorrect. See if you can
show either mathematically or empirically that it is incorrect. When you
finally realize that I am your better, then go fuck yourself.
Richard Perry
http://www.cswnet.com/~rper
Richard
Hafele-Keating! You cannot arbitrarily chose a frame asshole, one clock
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3F4ECBC2...@yahoo.com...
What a string of fumbles...
It might help if you first try to understand what *I* gave you,
and if you ask some polite questions about it. That is the way
to learn something.
And if you *really* think you still have Larry, Curly, Moe and
"Shem", I'll give the math to you, and I'll add a "Flurk" as well.
I will not work with c=1 because that might be too confusing
for you.
Refer to the new version of
http://users.pandora.be/vdmoortel/dirk/Stuff/MoeLarryCurly.gif
<Autopilot>
---------------------------------
1) Put Shem's frame to coincide with Curly's outbound frame
and let him use coordinates (x",t") and calculate the proper
times.
Velocities w.r.t. Shem:
Moe: velocity v
Curly outbound: velocity 0
Larry outbound: velocity 2v/(1+v^2/c^2) := V
Curly inbound: velocity 2v/(1+v^2/c^2) := V
Larry inbound: velocity 0
Remark:
sqrt[1-V^2/c^2] = ... = (1-v^2/c^2)/(1+v^2/c^2)
Verify this!
Lorentz transformation between Moe and Shem:
x" = g(x+vt)
t" = g(t+vx/c^2)
g = 1/sqrt(1-v^2/c^2)
Moe uses coordinates (x,t).
Departure event:
(x,t) = (0,0)
so t" = 0
Curly's turnaround event:
(x,t) = (-vT/2,T/2)
so t" = gT/2(1-v^2/c^2)
Larry's turnaround event:
(x,t) = (vT/2,T/2)
so t" = gT/2(1+v^2/c^2)
Reunion event:
(x,t) = (0,T)
so t" = gT
<Total proper time Moe>
= Int{ 0 to gT; sqrt[1-v^2/c^2] dt" }
= T
<Total proper time Curly>
= Int{ 0 to gT/2(1-v^2/c^2); sqrt[1-0^2/c^2] dt" }
+ Int{ gT/2(1-v^2/c^2) to gT; sqrt[1-V^2/c^2] dt" }
= gT/2(1-v^2/c^2)
+ sqrt[1-V^2/c^2](gT-gT/2(1-v^2/c^2))
= ...
= T/g (Verify this!)
<Total proper time Larry>
= Int{ 0 to gT/2(1+v^2/c^2); sqrt[1-V^2/c^2] dt" }
+ Int{ gT/2(1+v^2/c^2) to gT; sqrt[1-0^2] dt" }
= sqrt[1-V^2/c^2]*gT/2(1+v^2/c^2)
+ (gT-gT/2(1+v^2/c^2))
= ...
= T/g (Verify this!)
so again we find
<Total proper time Larry> = <Total proper time Curly>
= <Total proper time Moe> / g
---------------------------------
2) Put Flurk's frame to coincide with Larry's outbound frame
and let him use coordinates (x',t') and calculate the proper
times.
Velocities w.r.t. Flurk:
Moe: velocity -v
Curly outbound: velocity -2v/(1+v^2/c^2) := -V
Larry outbound: velocity 0
Curly inbound: velocity 0
Larry inbound: velocity -2v/(1+v^2/c^2) := -V
Remark:
sqrt[1-V^2/c^2] = ... = (1-v^2/c^2)/(1+v^2/c^2)
You already have verified this.
Lorentz transformation between Moe and Flurk:
x' = g(x-vt)
t' = g(t-vx/c^2)
g = 1/sqrt(1-v^2/c^2)
Moe uses coordinates (x,t).
Departure event:
(x,t) = (0,0)
so t' = 0
Curly's turnaround event:
(x,t) = (-vT/2,T/2)
so t' = gT/2(1+v^2/c^2)
Larry's turnaround event:
(x,t) = (vT/2,T/2)
so t' = gT/2(1-v^2/c^2)
Reunion event:
(x,t) = (0,T)
so t' = gT
<Total proper time Moe>
= Int{ 0 to gT; sqrt[1-v^2/c^2] dt' }
= T
<Total proper time Curly>
= Int{ 0 to gT/2(1+v^2/c^2); sqrt[1-V^2/c^2] dt' }
+ Int{ gT/2(1+v^2/c^2) to gT; sqrt[1-0^2/c^2] dt' }
= sqrt[1-V^2/c^2]*gT/2(1+v^2/c^2)
+ (gT-gT/2(1+v^2/c^2))
= ...
= T/g (Verify this!)
<Total proper time Larry>
= Int{ 0 to gT/2(1-v^2/c^2); sqrt[1-0^2/c^2] dt' }
+ Int{ gT/2(1-v^2/c^2) to gT; sqrt[1-V^2/c^2] dt' }
= gT/2(1-v^2/c^2)
+ sqrt[1-V^2/c^2]*(gT-gT/2(1-v^2/c^2))
= ...
= T/g (Verify this!)
so again we find
<Total proper time Larry> = <Total proper time Curly>
= <Total proper time Moe> / g
---------------------------------
</Autopilot>
Dirk Vdm
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3F4F6953...@yahoo.com...
You *really* have no idea what you are talking about :-)))
Just listen to yourself:
| "If the two events are the respective locations of the
| K' frame itself wrt K and K' respectively,..."
Can you actually read this without laughing out load?
You are even worse than Marcel Luttgens has ever been.
Do you know what an event is?
Do you know what a frame is?
Do you know what coordinates are?
Do you know what transformations are?
but let's go on...
> Now before you prove yourself a greater idiot by stating that these are
> not intervals but positions, let me remind you that when you assign a
> real number to x you get an interval whether you like it or not, that
> interval is |x - 0|, i.e. you have set one end of the interval at the
> origin of the coordinate system. IOW you will get the same result from
> either form
>
> x' = (x - vt)gamma
>
> or
>
> dx' = (dx - vdt)gamma
>
> Thus they are equivalent interpretations.
>
> According to K' the space interval is always zero when that interval is
> that defined by his respective positions over some time interval.
> Likewise for K. When x or x' are set to zero, then it follows that
>
> x/t = v, or x'/t' = v
>
> I'll be more than happy to explain further if you are having trouble
> with this math.
>
> I don't deal well with condescending morons, because that's an oxymoron,
> which BTW you are. If you think that I'm confused, or mathematically
> inept, then perhaps you would care to examine the electromagnetic theory
> linked below, and tell me which part of it is incorrect. See if you can
> show either mathematically or empirically that it is incorrect. When you
> finally realize that I am your better, then go fuck yourself.
Yes, and when
t = t' gamma' and t = t" gamma"
are to be valid together, then
x = x' = x" = t = t' = t" = 0 or v' = v" = v = 0
and vice versa.
And there is nothing you can do about it.
Go ahead, try.
Do the math and give me a counter-example.
Dirk Vdm
Richard
Dirk Van de moortel wrote:
>
It is the four-vector position of a particle (point)
> Do you know what a frame is?
An inertial reference coordinate system in constant uniform motion.
> Do you know what coordinates are?
>Do you know what transformations are?
Fuck you. What aren't you following?
K' is initially at {x_1,y_1,z_1,t_1},
K' moves to {{x_2,y_2,z_2,t_2}
These are two space-time events, and taken together they define a
space-time interval.
The lorentz transform addresses differences in intervals of two events
wrt different frames of reference. Nothing bars the motion of the origin
of one of the frames itself from describing those intervals, and when
this is done, the clock ticking ratio wrt the two frames, which was
derived by Einstein by a only slightly different argument, falls right
out of the transform. Are you really that fucking dumb that you're
having a problem with this?
We both need a break Dirk, you have just spewed a bigger load of crap
than I could ever manage.
I assume that you're referring to my absolute medium interpretation of
the transform here, where t is the frame of the medium.
In that case gammas are not equal here, v' =/= v''
With the premises:
t = t' gamma' and t = t" gamma"
then direct substitution gives:
t' gamma' = t'' gamma''
expanding this gives
t'/sqrt(1 - (v'/c)^2) = t/sqrt(1 - (v''/c)^2)
And t' = t'' only when v' = v'' (verify this for yourself)
Richard Perry
Richard
Dirk Van de moortel wrote:
>
I've printed it off. When I've had time to detail your errors I'll post
them to a new thread, should be lot's of fun;)
Richard Perry
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3F4FA2C8...@yahoo.com...
>
>
> Dirk Van de moortel wrote:
> >
> > "Richard" <no_mail...@yahoo.com> wrote in message news:3F4F6953...@yahoo.com...
> > >
[snip]
> > Do you know what an event is?
>
> It is the four-vector position of a particle (point)
>
> > Do you know what a frame is?
>
> An inertial reference coordinate system in constant uniform motion.
>
> > Do you know what coordinates are?
>
> >Do you know what transformations are?
>
> Fuck you. What aren't you following?
>
> K' is initially at {x_1,y_1,z_1,t_1},
> K' moves to {{x_2,y_2,z_2,t_2}
> These are two space-time events, and taken together they define a
> space-time interval.
> The lorentz transform addresses differences in intervals of two events
> wrt different frames of reference. Nothing bars the motion of the origin
> of one of the frames itself from describing those intervals, and when
> this is done, the clock ticking ratio wrt the two frames, which was
> derived by Einstein by a only slightly different argument, falls right
> out of the transform. Are you really that fucking dumb that you're
> having a problem with this?
No problem. It makes me wonder why on earth you
manage to write something as silly as
| "If the two events are the respective locations of the
| K' frame itself wrt K and K' respectively,..."
>
> >
I don't need a break.
> I assume that you're referring to my absolute medium interpretation of
> the transform here, where t is the frame of the medium.
"t is the frame of the medium"?
Nonsense.
>
> In that case gammas are not equal here, v' =/= v''
>
> With the premises:
>
> t = t' gamma' and t = t" gamma"
>
> then direct substitution gives:
>
> t' gamma' = t'' gamma''
>
> expanding this gives
>
> t'/sqrt(1 - (v'/c)^2) = t/sqrt(1 - (v''/c)^2)
>
> And t' = t'' only when v' = v'' (verify this for yourself)
Unless v'=0 and/or v"=0
from t = t' gamma' follows x'=0 (Lorentz K-K')
from t = t" gamma" follows x"=0 (Lorentz K-K")
from x' = x" = 0 follows t' = t" = 0 (Lorentz K'-K" and K"-K')
from x'=0 and t'=0 follows x = t = 0 (Lorentz K-K')
So x = x' = x" = t = t' = t" = 0.
No way out.
Dirk Vdm
Dirk Van de moortel
"Richard" <no_mail...@yahoo.com> wrote in message news:3F4FA342...@yahoo.com...
>
[snip]
>
> I've printed it off. When I've had time to detail your errors I'll post
> them to a new thread, should be lot's of fun;)
Good. I have deliberately made one error.
Let's see if you can find it.
Dirk Vdm
Paul Cardinale
> Titan Point wrote:
> >
> > On Thu, 28 Aug 2003 11:29:43 +0200, Dirk Van de moortel wrote:
> >
> > >
> > > Conclusion: your equations are valid if and only if
> > > x = x' = x" = t = t' = t" = 0
> > > or
> > > v' = v" = v = 0
> > > How interesting.
> > > Reminds me of Marcel Luttgens.
> > >
> > > Dirk Vdm
> >
> > You mean he proved nothing? ;-)
>
> Again:
>
> t' = (t - vx/c^2)gamma'
>
> t' = (t - tvv/c^2)gamma'
Hey puke-brain, you must be dumber than a pile of dog crap if you think you
can pass that off on anybody.
v is the relative velocity of frame K' with respect to K.
x is the position, in K, of an event.
t is the time, in frame, of that event.
v IS NOT x/t. x/t is not the velocity of anything.
The has been pointed out to you before,
but YOU, RICHARD PERRY, have the learning abilities of a dead worm.
You are stupid, arogant, willfully ignorant, dishonest, and vulgar.
In short, you are a pathetic exuse for a human being.
Paul Cardinale
Starblade Darksquall
> Both parties to the ongoing aether debate have made a common error in
> judgment, i.e. that transforming from any frame K' to any other frame
> K'' is as valid as transforming from the absolute frame k to either of
> these.
>
> The math is incorrect. I have already provided a very simple proof in a
> related thread.
>
> We have for the transformation from K to K' and K to K'' respectively:
>
> t = t' gamma' where v' is the speed of K' wrt K
> t = t'' gamma'' where v'' is the speed of K'' wrt K
>
> add them at all unless you know an algebraic trick that I am unaware of.
>
> By substitution from the above we get this 'correct' relationship
>
> t' = t''gamma''/gamma'
>
>
> Richard Perry
>
> http://www.cswnet.com/~rper
Well, obviously the math from aether theorums is different from the
math of SR. SR uses lorentz transformations, and those DO commute
properly. However in aether mathematics, the transformations do not
commute between reference frames, as you have shown, so if we are not
at rest wrt the aether then we DO have to boost to the aether frame
and then to the new frame which is also not at rest wrt the aether.
SR does not have this problem. This makes the math of SR simpler. Even
though it would seem to make it complex since it uses second order
tensors, it actually is easier since you do not have to worry about
your movement relative to the aether, since in SR, there is no aether.
And, as far as I can tell, all the experimental evidence points to the
correctness of SR, and the incorrectness of aether theories in
general.
(...Starblade Riven Darksquall...)
Bill Vajk
> Well, obviously the math from aether theorums is different from the
snip
> And, as far as I can tell, all the experimental evidence points to the
> correctness of SR, and the incorrectness of aether theories in
> general.
There's the first problem, the lack of a hook to hang one's hat
on, that is to say, no single aether theory to discuss. Einstein
sure thought there's something missing and he thought it might
be some form of aether which he was unable to define.
To support your penultimate statement further, care to present a
list of Einstein's "other" grievous errors?
Dirk Van de moortel
"Starblade Darksquall" <Starb...@Yahoo.com> wrote in message news:4aa861fb.03083...@posting.google.com...
[snip]
> > Richard Perry
> >
> > http://www.cswnet.com/~rper
>
> Well, obviously the math from aether theorums is different from the
> math of SR.
Nah, they have the same math.
Even Perry has the same math.
The only problem is that he does not understand
what the symbols stand for :-)
Dirk Vdm
Cimsocretic01
by a team of quantum of the fact that the polarization coupling of "paired
photons" travels at avelocity of at least 4 times that of light. This would not
be possible if there were not an absolute velocity reference frame, ie:- the
Aether.
Starblade Darksquall
This does not contradict SR. This happens even if it would make
something in the 'future' cause something in the 'past'. However,
since no messages can be sent, this does not contradict causality or
SR.
(...Starblade Riven Darksquall...)