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Asymptotical error bars in SR prove its incompleteness



 
 
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  #21  
Old May 13th 08 posted to sci.physics.relativity
Eric Gisse
external usenet poster
 
Posts: 15,984
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 12:31*am, Albertito wrote:
[snip]

Simple question: What do you hope to accomplish by writing things that
are purposefully experimentally indistinguishable from SR when you
have no underlying theory to derive them from?
Ads
  #22  
Old May 13th 08 posted to sci.physics.relativity
PD
external usenet poster
 
Posts: 16,603
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 3:31*am, Albertito wrote:
On May 12, 11:21 pm, PD wrote:



On May 12, 6:56 am, Albertito wrote:


On May 11, 10:07 pm, PD wrote:


On May 9, 6:44 am, Albertito wrote:


Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c *below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is


* * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ....


Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't *still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors


* * * * * * * *w = u + v.


That sum can't still be experimentally distinguished from the sum
of rapidities


* * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c).


In addition, we must also conclude that the relativistic Doppler


* * * * * * * *f' = Exp(-r) f ,
* * * * * * * *where r = arctanh(v/c),


can't still be experimentally distinguished from this one


* * * * * * * *f' = Exp(-v/c) f


There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.


PD


Interpretation under my model and that provided
by SR can't still be experimentally distinguished.


Consider this sum of betas,


* * v/c = v_p/c + v_mu'/c ,
* * with
* * v_p, pion speed at the moment of its decay wrt muon
* * * * * * detector.
* * v_mu', *the relative speed of a muon with respect
* * * * * * * * *to the point where the pion has decayed


Consider this sum of rapidities,


*arctanh(w/c) = *arctanh(v_p/c) *+ *arctanh(v_mu'/c)


In both sums, neither v_p nor v_m_u' exceed c.


SR assumes the final speed of the muon in the detector
must be


* * * * w = c tanh(arctanh(v_p/c) *+ *arctanh(v_mu'/c)),


* * * * but it needs also the assumption there is time dilation
* * * * of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2),
* * * * in order to fit the prediction to the observation.


In my model, the final speed of the muon in the detector
is
* * * * * * *v *= v_p + v_mu',
* * * * * * without the assumption of any time dilation at all.


If it still is not clear to you, I can repeat,


* * * * * * *v/c =(v_p + v_mu')/c


And, if you do the math, you find out the left-hand-side of this
formula should be greater than 1, given the amount of kinetic energy
released in pions decaying to muons. Go ahead, plug in the numbers.


What's the problem with finding out the left-hand-side
of this formula is greater than 1?


Well, for one thing, it doesn't happen. As I mentioned to you, the
time of flight of muons is *measured*. We know what the speed is. It's
not greater than 1.

That's only a problem
if you use the relativistic kinetic energy.


No, sir. It's a problem with measurement.

Of course,
v/c = (v_p + v_mu')/c *is greater than 1, because v_p/c
is very close to 1, and v_mu'/c can be greater than
(1 - v_p/c).


  #23  
Old May 13th 08 posted to sci.physics.relativity
Albertito
external usenet poster
 
Posts: 609
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 10:04 am, PD wrote:
On May 13, 3:31 am, Albertito wrote:



On May 12, 11:21 pm, PD wrote:


On May 12, 6:56 am, Albertito wrote:


On May 11, 10:07 pm, PD wrote:


On May 9, 6:44 am, Albertito wrote:


Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is


r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...


Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors


w = u + v.


That sum can't still be experimentally distinguished from the sum
of rapidities


arctanh(w/c) = arctanh(u/c) + arctanh(v/c).


In addition, we must also conclude that the relativistic Doppler


f' = Exp(-r) f ,
where r = arctanh(v/c),


can't still be experimentally distinguished from this one


f' = Exp(-v/c) f


There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.


PD


Interpretation under my model and that provided
by SR can't still be experimentally distinguished.


Consider this sum of betas,


v/c = v_p/c + v_mu'/c ,
with
v_p, pion speed at the moment of its decay wrt muon
detector.
v_mu', the relative speed of a muon with respect
to the point where the pion has decayed


Consider this sum of rapidities,


arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)


In both sums, neither v_p nor v_m_u' exceed c.


SR assumes the final speed of the muon in the detector
must be


w = c tanh(arctanh(v_p/c) + arctanh(v_mu'/c)),


but it needs also the assumption there is time dilation
of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2),
in order to fit the prediction to the observation.


In my model, the final speed of the muon in the detector
is
v = v_p + v_mu',
without the assumption of any time dilation at all.


If it still is not clear to you, I can repeat,


v/c =(v_p + v_mu')/c


And, if you do the math, you find out the left-hand-side of this
formula should be greater than 1, given the amount of kinetic energy
released in pions decaying to muons. Go ahead, plug in the numbers.


What's the problem with finding out the left-hand-side
of this formula is greater than 1?


Well, for one thing, it doesn't happen. As I mentioned to you, the
time of flight of muons is *measured*. We know what the speed is. It's
not greater than 1.

That's only a problem
if you use the relativistic kinetic energy.


No, sir. It's a problem with measurement.

Of course,
v/c = (v_p + v_mu')/c is greater than 1, because v_p/c
is very close to 1, and v_mu'/c can be greater than
(1 - v_p/c).


You can't keep track of single muons to measure their
'time of flight'. All you can do is to perform statistics and
extrapolate to single muons. You can't say a muon called
A was emitted by a pion called P at time t=0, and then A
was detected at distance R at time t'. The extrapolation for
a single muon you perfom is pure nonsense. That's the
reason why we find out in particle physics there are three
odd generations of charged leptons, and nobody can guess
why there are actually those three generations and why they
exhibit the rest masses they exhibit. SR is unable to account for
the mass of muons and tau leptons with respect to the mass of
electrons. Should we call that impossibility a fraud?. If you avoid
SR, and try to explain why muon and tau lepton have the rest
masses they have wrt that of an electron, you will be able to find
out those three generations of charged leptons, are just exact
ratios derived from multiples of c.
  #24  
Old May 13th 08 posted to sci.physics.relativity
Eric Gisse
external usenet poster
 
Posts: 15,984
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 1:53*am, Albertito wrote:
On May 13, 10:04 am, PD wrote:



On May 13, 3:31 am, Albertito wrote:


On May 12, 11:21 pm, PD wrote:


On May 12, 6:56 am, Albertito wrote:


On May 11, 10:07 pm, PD wrote:


On May 9, 6:44 am, Albertito wrote:


Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c *below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is


* * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...


Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't *still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors


* * * * * * * *w = u + v.


That sum can't still be experimentally distinguished from the sum
of rapidities


* * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c).


In addition, we must also conclude that the relativistic Doppler


* * * * * * * *f' = Exp(-r) f ,
* * * * * * * *where r = arctanh(v/c),


can't still be experimentally distinguished from this one


* * * * * * * *f' = Exp(-v/c) f


There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.


PD


Interpretation under my model and that provided
by SR can't still be experimentally distinguished.


Consider this sum of betas,


* * v/c = v_p/c + v_mu'/c ,
* * with
* * v_p, pion speed at the moment of its decay wrt muon
* * * * * * detector.
* * v_mu', *the relative speed of a muon with respect
* * * * * * * * *to the point where the pion has decayed


Consider this sum of rapidities,


*arctanh(w/c) = *arctanh(v_p/c) *+ *arctanh(v_mu'/c)


In both sums, neither v_p nor v_m_u' exceed c.


SR assumes the final speed of the muon in the detector
must be


* * * * w = c tanh(arctanh(v_p/c) *+ *arctanh(v_mu'/c)),


* * * * but it needs also the assumption there is time dilation
* * * * of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2),
* * * * in order to fit the prediction to the observation.


In my model, the final speed of the muon in the detector
is
* * * * * * *v *= v_p + v_mu',
* * * * * * without the assumption of any time dilation at all.


If it still is not clear to you, I can repeat,


* * * * * * *v/c =(v_p + v_mu')/c


And, if you do the math, you find out the left-hand-side of this
formula should be greater than 1, given the amount of kinetic energy
released in pions decaying to muons. Go ahead, plug in the numbers.


What's the problem with finding out the left-hand-side
of this formula is greater than 1?


Well, for one thing, it doesn't happen. As I mentioned to you, the
time of flight of muons is *measured*. We know what the speed is. It's
not greater than 1.


That's only a problem
if you use the relativistic kinetic energy.


No, sir. It's a problem with measurement.


Of course,
v/c = (v_p + v_mu')/c *is greater than 1, because v_p/c
is very close to 1, and v_mu'/c can be greater than
(1 - v_p/c).


You can't keep track of single muons to measure their
'time of flight'. All you can do is to perform statistics and
extrapolate to single muons. You can't say a muon called
A was emitted by a pion called P at time t=0, and then A
was detected at distance R at time t'. The extrapolation for
a single muon you perfom is pure nonsense. That's the
reason why we find out in particle physics there are three
odd generations of charged leptons, and nobody can guess
why there are actually those three generations and why they
exhibit the rest masses they exhibit. SR is unable to account for
the mass of muons and tau leptons with respect to the mass of
electrons. Should we call that impossibility a fraud?. If you avoid
SR, and try to explain why muon and tau lepton have the rest
masses they have wrt that of an electron, you will be able to find
out *those three generations of charged leptons, are just exact
ratios derived from multiples of c.


I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist.
  #25  
Old May 13th 08 posted to sci.physics.relativity
Albertito
external usenet poster
 
Posts: 609
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 10:03 am, Eric Gisse wrote:
On May 13, 12:31 am, Albertito wrote:
[snip]

Simple question: What do you hope to accomplish by writing things that
are purposefully experimentally indistinguishable from SR when you
have no underlying theory to derive them from?


Is the second postulate of SR a first principle?
Yes? Is it a first principle to claim the speed of
light does not depend on the speed of the source,
but then we will have to deal with time dilations and
length contractions, in order to be able explain
phenomena? Well, if it is so that c' = c, is a first
principle from which SR can be derived, then
c' = c Exp(-v/c^2) is also a first principle from which
my model can be derived, and time dilations and
length contractions will not be required.

  #26  
Old May 13th 08 posted to sci.physics.relativity
PD
external usenet poster
 
Posts: 16,603
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 4:53*am, Albertito wrote:
On May 13, 10:04 am, PD wrote:



On May 13, 3:31 am, Albertito wrote:


On May 12, 11:21 pm, PD wrote:


On May 12, 6:56 am, Albertito wrote:


On May 11, 10:07 pm, PD wrote:


On May 9, 6:44 am, Albertito wrote:


Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c *below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is


* * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...


Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't *still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors


* * * * * * * *w = u + v.


That sum can't still be experimentally distinguished from the sum
of rapidities


* * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c).


In addition, we must also conclude that the relativistic Doppler


* * * * * * * *f' = Exp(-r) f ,
* * * * * * * *where r = arctanh(v/c),


can't still be experimentally distinguished from this one


* * * * * * * *f' = Exp(-v/c) f


There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.


PD


Interpretation under my model and that provided
by SR can't still be experimentally distinguished.


Consider this sum of betas,


* * v/c = v_p/c + v_mu'/c ,
* * with
* * v_p, pion speed at the moment of its decay wrt muon
* * * * * * detector.
* * v_mu', *the relative speed of a muon with respect
* * * * * * * * *to the point where the pion has decayed


Consider this sum of rapidities,


*arctanh(w/c) = *arctanh(v_p/c) *+ *arctanh(v_mu'/c)


In both sums, neither v_p nor v_m_u' exceed c.


SR assumes the final speed of the muon in the detector
must be


* * * * w = c tanh(arctanh(v_p/c) *+ *arctanh(v_mu'/c)),


* * * * but it needs also the assumption there is time dilation
* * * * of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2),
* * * * in order to fit the prediction to the observation.


In my model, the final speed of the muon in the detector
is
* * * * * * *v *= v_p + v_mu',
* * * * * * without the assumption of any time dilation at all.


If it still is not clear to you, I can repeat,


* * * * * * *v/c =(v_p + v_mu')/c


And, if you do the math, you find out the left-hand-side of this
formula should be greater than 1, given the amount of kinetic energy
released in pions decaying to muons. Go ahead, plug in the numbers.


What's the problem with finding out the left-hand-side
of this formula is greater than 1?


Well, for one thing, it doesn't happen. As I mentioned to you, the
time of flight of muons is *measured*. We know what the speed is. It's
not greater than 1.


That's only a problem
if you use the relativistic kinetic energy.


No, sir. It's a problem with measurement.


Of course,
v/c = (v_p + v_mu')/c *is greater than 1, because v_p/c
is very close to 1, and v_mu'/c can be greater than
(1 - v_p/c).


You can't keep track of single muons to measure their
'time of flight'. All you can do is to perform statistics and
extrapolate to single muons. You can't say a muon called
A was emitted by a pion called P at time t=0, and then A
was detected at distance R at time t'.


Of course you can. You didn't read what I wrote. You time a muon in
flight the same way you could a car on the highway: you time its
crossings on successive, spaced gates. (Note you don't have to follow
the car all the way from the garage to do so.) You make a scintillator
paddle triplet, with each paddle separated by some appreciable
difference (say, 20m), and then the signals from each of the
scintillators to an o'scope or a TDC by an equal length cable. 20 m
would be about 60 ns if the speed were as high as c, trivial to
measure by either of the devices mentioned. The presence of the MIP
signal in all three paddles assures that you are seeing the passage of
the muon. You measure the distance, and you measure the time it takes
the muon to cover that distance. You have a direct measurement of the
velocity. It does not exceed c. *Measured*.

It might do you good to go into the laboratory to see for yourself,
rather than just jacking around with piddly little ideas.

PD

  #27  
Old May 13th 08 posted to sci.physics.relativity
Eric Gisse
external usenet poster
 
Posts: 15,984
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 2:04*am, Albertito wrote:
On May 13, 10:03 am, Eric Gisse wrote:

On May 13, 12:31 am, Albertito wrote:
[snip]


Simple question: What do you hope to accomplish by writing things that
are purposefully experimentally indistinguishable from SR when you
have no underlying theory to derive them from?


Is the second postulate of SR a first principle?


Yes...

Yes? Is it a first principle to claim the speed of
light does not depend on the speed of the source,


Yes...

but then we will have to deal with time dilations and
length contractions, in order to be able explain
phenomena?


No. These are /predictions/, not first principles. You might want to
actually study a little science first.

Well, if it is so that c' = c, *is a first
principle from which SR can be derived, then
c' = c Exp(-v/c^2) is also a first principle from which
my model can be derived, and time dilations and
length contractions will not be required.


Except there is zero evidence for your model, and you haven't actually
derived anything meaningful yet. You write down equations you think
your model has, but you don't ever show where they come from. In fact,
the equations you write are always arbitrary and only picked because
they match some SR equation to a few orders in a power series
expansion.

Why should anyone care what you have to say at all when that is all
you ever do?


  #28  
Old May 13th 08 posted to sci.physics.relativity
Albertito
external usenet poster
 
Posts: 609
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 11:24 am, PD wrote:
[snipped]
Of course you can. You didn't read what I wrote. You time a muon in
flight the same way you could a car on the highway: you time its
crossings on successive, spaced gates. (Note you don't have to follow
the car all the way from the garage to do so.) You make a scintillator
paddle triplet, with each paddle separated by some appreciable
difference (say, 20m), and then the signals from each of the
scintillators to an o'scope or a TDC by an equal length cable. 20 m
would be about 60 ns if the speed were as high as c, trivial to
measure by either of the devices mentioned. The presence of the MIP
signal in all three paddles assures that you are seeing the passage of
the muon. You measure the distance, and you measure the time it takes
the muon to cover that distance. You have a direct measurement of the
velocity. It does not exceed c. *Measured*.

It might do you good to go into the laboratory to see for yourself,
rather than just jacking around with piddly little ideas.

PD


That only proves those experiments are biased by
relativistic formulas. Time dilation is not an observable
by definition, it is just an interpretation to explain an
observable. You can't label single muons to keep track
of them, the same way you can't label single electrons.
Perform a double-slit experiment to keep track of a single
muon. Can you clarify the path a muon can follow in that
double-slit experiment? If you can't clarify a muon path
why are so sure a muon has travelled a distance R in time
t'? Didn't you know that intermediate measurements alter
the experimental results.
  #29  
Old May 13th 08 posted to sci.physics.relativity
Albertito
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Posts: 609
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 11:48 am, Eric Gisse wrote:
On May 13, 2:04 am, Albertito wrote:

On May 13, 10:03 am, Eric Gisse wrote:


On May 13, 12:31 am, Albertito wrote:
[snip]


Simple question: What do you hope to accomplish by writing things that
are purposefully experimentally indistinguishable from SR when you
have no underlying theory to derive them from?


Is the second postulate of SR a first principle?


Yes...

Yes? Is it a first principle to claim the speed of
light does not depend on the speed of the source,


Yes...

but then we will have to deal with time dilations and
length contractions, in order to be able explain
phenomena?


No. These are /predictions/, not first principles. You might want to
actually study a little science first.

Well, if it is so that c' = c, is a first
principle from which SR can be derived, then
c' = c Exp(-v/c^2) is also a first principle from which
my model can be derived, and time dilations and
length contractions will not be required.


Except there is zero evidence for your model, and you haven't actually
derived anything meaningful yet. You write down equations you think
your model has, but you don't ever show where they come from. In fact,
the equations you write are always arbitrary and only picked because
they match some SR equation to a few orders in a power series
expansion.

Why should anyone care what you have to say at all when that is all
you ever do?


Sorry, I meant c'= c Exp(-v/c), hehe :-)
BTW, just one question: If I can derive the prediction for
the ratios of charged lepton masses (issue that noboby
still can resolve), will you become a renegade of SR,
and become a faithful supporter of my model?

  #30  
Old May 13th 08 posted to sci.physics.relativity
Eric Gisse
external usenet poster
 
Posts: 15,984
Default Asymptotical error bars in SR prove its incompleteness

On May 13, 3:11*am, Albertito wrote:
On May 13, 11:48 am, Eric Gisse wrote:



On May 13, 2:04 am, Albertito wrote:


On May 13, 10:03 am, Eric Gisse wrote:


On May 13, 12:31 am, Albertito wrote:
[snip]


Simple question: What do you hope to accomplish by writing things that
are purposefully experimentally indistinguishable from SR when you
have no underlying theory to derive them from?


Is the second postulate of SR a first principle?


Yes...


Yes? Is it a first principle to claim the speed of
light does not depend on the speed of the source,


Yes...


but then we will have to deal with time dilations and
length contractions, in order to be able explain
phenomena?


No. These are /predictions/, not first principles. You might want to
actually study a little science first.


Well, if it is so that c' = c, *is a first
principle from which SR can be derived, then
c' = c Exp(-v/c^2) is also a first principle from which
my model can be derived, and time dilations and
length contractions will not be required.


Except there is zero evidence for your model, and you haven't actually
derived anything meaningful yet. You write down equations you think
your model has, but you don't ever show where they come from. In fact,
the equations you write are always arbitrary and only picked because
they match some SR equation to a few orders in a power series
expansion.


Why should anyone care what you have to say at all when that is all
you ever do?


Sorry, I meant c'= c Exp(-v/c), hehe :-)
BTW, just one question: If I can derive the prediction for
the ratios of charged lepton masses (issue that noboby
still can resolve), will you become a renegade of SR,
and become a faithful supporter of my model?


SR isn't a theory of interactions or particles. Try to understand the
theory you seek to replace.
 




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