Re: Stefanovich's theory is not Poincare invariant. [Archive]

Igor Khavkine

Sep2-05, 10:58 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-09-01, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n>> No. I will only discuss special relativity and E&M in the generally\n>> accepted formulation, not a strawman theory that you concoct.\n>\n> Then we have nothing to discuss. I prefer discussing my\n> "strawman theory".\n\nThat\'s fine by me. I\'ve changed the title accordingly. I can\'t say I\ndisagree with it.\n\n>> Now, to continue, I want to hear only one thing: whether you understood\n>> everything I\'ve explained so far. Again, I emphasize that only\n>> understanding, not agreement, is required. If you still don\'t understand\n>> something, I\'m open for questions.\n>\n> Sure, I understand special relativity and classical\n> electrodynamics well. I don\'t have any questions there.\n> Only disagreements at some critical points.\n\nI have yet to see you apply your understanding of SR and E&M correctly.\nThis makes me doubt that you understand them very well. If you don\'t\nwant to rectify this situation, that\'s your choice. From what I\'ve seen,\nyou disagree with some of SR\'s postulates (you demand proof, yet forget\nthat postulates are not proven). That\'s fine, except that when you claim\nto draw physically unreasonable consequences from SR, you must specify\nthat SR stands for Stefanovich\'s Relativity, which is distinct from\nSpecial Relativity, since key postulates are dropped.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-09-01, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

>> No. I will only discuss special relativity and E&M in the generally
>> accepted formulation, not a strawman theory that you concoct.
>
> Then we have nothing to discuss. I prefer discussing my
> "strawman theory".

That's fine by me. I've changed the title accordingly. I can't say I
disagree with it.

>> Now, to continue, I want to hear only one thing: whether you understood
>> everything I've explained so far. Again, I emphasize that only
>> understanding, not agreement, is required. If you still don't understand
>> something, I'm open for questions.
>
> Sure, I understand special relativity and classical
> electrodynamics well. I don't have any questions there.
> Only disagreements at some critical points.

I have yet to see you apply your understanding of SR and E&M correctly.
This makes me doubt that you understand them very well. If you don't
want to rectify this situation, that's your choice. From what I've seen,
you disagree with some of SR's postulates (you demand proof, yet forget
that postulates are not proven). That's fine, except that when you claim
to draw physically unreasonable consequences from SR, you must specify
that SR stands for Stefanovich's Relativity, which is distinct from
Special Relativity, since key postulates are dropped.

Igor

Eugene Stefanovich

Sep2-05, 02:29 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n> From what I\'ve seen,\n> you disagree with some of SR\'s postulates\n\nYes, I disagree with the postulate of Einstein\'s special relativity\n(ESR) which claims that observables in all (even interacting) physical\nsystems transform by universal linear tensor formulas.\n\n> (you demand proof, yet forget\n> that postulates are not proven).\n\nI understand that this is a postulate, and therefore it doesn\'t require\nproof in the framework of ESR.\n\n> That\'s fine, except that when you claim\n> to draw physically unreasonable consequences from SR, you must specify\n> that SR stands for Stefanovich\'s Relativity, which is distinct from\n> Special Relativity, since key postulates are dropped.\n\nYou got it exactly right. I drop the above postulate, and therefore\nmy theory (I would suggest the name \'SSR\') is different from the\nconventional ESR. Then, it is not a surprise that physical predictions\nof the two theories are different. For example, SSR predicts violations\nof the Einstein\'s time dilation formula in decays of moving particles.\nSSR also allows faster-than-light propagation of interactions which is\nstrictly forbidden by ESR.\n\nYou say that these are "physically unreasonable consequences",\nbut so far there was no direct experimental proof that these\npredictions are wrong.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> From what I've seen,
> you disagree with some of SR's postulates

Yes, I disagree with the postulate of Einstein's special relativity
(ESR) which claims that observables in all (even interacting) physical
systems transform by universal linear tensor formulas.

> (you demand proof, yet forget
> that postulates are not proven).

I understand that this is a postulate, and therefore it doesn't require
proof in the framework of ESR.

> That's fine, except that when you claim
> to draw physically unreasonable consequences from SR, you must specify
> that SR stands for Stefanovich's Relativity, which is distinct from
> Special Relativity, since key postulates are dropped.

You got it exactly right. I drop the above postulate, and therefore
my theory (I would suggest the name 'SSR') is different from the
conventional ESR. Then, it is not a surprise that physical predictions
of the two theories are different. For example, SSR predicts violations
of the Einstein's time dilation formula in decays of moving particles.
SSR also allows faster-than-light propagation of interactions which is
strictly forbidden by ESR.

You say that these are "physically unreasonable consequences",
but so far there was no direct experimental proof that these
predictions are wrong.

Eugene.

Eugene Stefanovich

Sep2-05, 02:29 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n>> I prefer discussing my\n>>"strawman theory".\n>\n>\n> That\'s fine by me. I\'ve changed the title accordingly. I can\'t say I\n> disagree with it.\n\nI strongly disagree with your title. In RQD the Poincare commutation\nrelations are satisfied in all systems, including interacting ones.\nTake a look at subsection 11.1.4 of my book\nwhere I reproduce Weinberg\'s proof of Poincare commutators in QED.\n\nWhat surprises me a lot is that everybody seem to understand\n(after Dirac\'s 1949 paper on "relativistic forms of dynamics")\nthat at least some of the generators of the Poincare group\n(in addition to the time translation generator - the Hamiltonian)\nmust contain interactions. However, nobody realizes that these\ninteractions require modifications in the way physical observables\nare transformed with respect to space translations, rotations,\nor boosts.\n\nFor example, people seriously discuss Dirac\'s point form\nof dynamics in which space translations are interaction-dependent.\nBut this dependence implies that the well-known simple relationship\nbetween observations in two reference frames shifted wrt each other\nmust be broken. There is a good experimental evidence against such a\npossibility. In my view, this evidence completely excludes the point\nform dynamics as a viable candidate.\n\nThe only form of dynamics which has a chance to be consistent with\nexperiment is the "instant form". In this case, space translations\nand rotations are interaction-free, as expected. However, the\ntransformations of observables wrt boosts are non-trivial.\nThe interaction corrections to the Lorentz transformation formulas\nare small, but they should be observable, in principle.\n\nNobody questions the fact that interaction terms in the Hamiltonian\nlead to observable interaction effects in the time evolution.\nThen, why you resist the idea that interaction terms in the boost\noperators must lead to observable interaction effects in the\nboost transformations?\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
>> I prefer discussing my
>>"strawman theory".
>
>
> That's fine by me. I've changed the title accordingly. I can't say I
> disagree with it.

I strongly disagree with your title. In RQD the Poincare commutation
relations are satisfied in all systems, including interacting ones.
Take a look at subsection 11.1.4 of my book
where I reproduce Weinberg's proof of Poincare commutators in QED.

What surprises me a lot is that everybody seem to understand
(after Dirac's 1949 paper on "relativistic forms of dynamics")
that at least some of the generators of the Poincare group
(in addition to the time translation generator - the Hamiltonian)
must contain interactions. However, nobody realizes that these
interactions require modifications in the way physical observables
are transformed with respect to space translations, rotations,
or boosts.

For example, people seriously discuss Dirac's point form
of dynamics in which space translations are interaction-dependent.
But this dependence implies that the well-known simple relationship
between observations in two reference frames shifted wrt each other
must be broken. There is a good experimental evidence against such a
possibility. In my view, this evidence completely excludes the point
form dynamics as a viable candidate.

The only form of dynamics which has a chance to be consistent with
experiment is the "instant form". In this case, space translations
and rotations are interaction-free, as expected. However, the
transformations of observables wrt boosts are non-trivial.
The interaction corrections to the Lorentz transformation formulas
are small, but they should be observable, in principle.

Nobody questions the fact that interaction terms in the Hamiltonian
lead to observable interaction effects in the time evolution.
Then, why you resist the idea that interaction terms in the boost
operators must lead to observable interaction effects in the
boost transformations?

Eugene.

stefanbanev@yahoo.com

Sep2-05, 05:12 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Dear Eugene,\n\nIt is impossible to reason person when it gets personal. He definitely\nhas no intention to understand and discuss constructively your ideas.\nYour questioning of SR assumptions is perfectly reasonable. Theory can\nnot be correct or wrong it can only possess more or less predictive\npower; so, it would be far more productive to think some "simple"\nexperiment to make the point rather trying persuading opponent who has\nno intention even consider your assumptions.\n\nRegards,\nStefan\n\n\nEugene Stefanovich wrote:\n> Igor Khavkine wrote:\n> > From what I\'ve seen,\n> > you disagree with some of SR\'s postulates\n>\n> Yes, I disagree with the postulate of Einstein\'s special relativity\n> (ESR) which claims that observables in all (even interacting) physical\n> systems transform by universal linear tensor formulas.\n>\n> > (you demand proof, yet forget\n> > that postulates are not proven).\n>\n> I understand that this is a postulate, and therefore it doesn\'t require\n> proof in the framework of ESR.\n>\n> > That\'s fine, except that when you claim\n> > to draw physically unreasonable consequences from SR, you must specify\n> > that SR stands for Stefanovich\'s Relativity, which is distinct from\n> > Special Relativity, since key postulates are dropped.\n>\n> You got it exactly right. I drop the above postulate, and therefore\n> my theory (I would suggest the name \'SSR\') is different from the\n> conventional ESR. Then, it is not a surprise that physical predictions\n> of the two theories are different. For example, SSR predicts violations\n> of the Einstein\'s time dilation formula in decays of moving particles.\n> SSR also allows faster-than-light propagation of interactions which is\n> strictly forbidden by ESR.\n>\n> You say that these are "physically unreasonable consequences",\n> but so far there was no direct experimental proof that these\n> predictions are wrong.\n>\n> Eugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Dear Eugene,

It is impossible to reason person when it gets personal. He definitely
has no intention to understand and discuss constructively your ideas.
Your questioning of SR assumptions is perfectly reasonable. Theory can
not be correct or wrong it can only possess more or less predictive
power; so, it would be far more productive to think some "simple"
experiment to make the point rather trying persuading opponent who has
no intention even consider your assumptions.

Regards,
Stefan

Eugene Stefanovich wrote:
> Igor Khavkine wrote:
> > From what I've seen,
> > you disagree with some of SR's postulates
>
> Yes, I disagree with the postulate of Einstein's special relativity
> (ESR) which claims that observables in all (even interacting) physical
> systems transform by universal linear tensor formulas.
>
> > (you demand proof, yet forget
> > that postulates are not proven).
>
> I understand that this is a postulate, and therefore it doesn't require
> proof in the framework of ESR.
>
> > That's fine, except that when you claim
> > to draw physically unreasonable consequences from SR, you must specify
> > that SR stands for Stefanovich's Relativity, which is distinct from
> > Special Relativity, since key postulates are dropped.
>
> You got it exactly right. I drop the above postulate, and therefore
> my theory (I would suggest the name 'SSR') is different from the
> conventional ESR. Then, it is not a surprise that physical predictions
> of the two theories are different. For example, SSR predicts violations
> of the Einstein's time dilation formula in decays of moving particles.
> SSR also allows faster-than-light propagation of interactions which is
> strictly forbidden by ESR.
>
> You say that these are "physically unreasonable consequences",
> but so far there was no direct experimental proof that these
> predictions are wrong.
>
> Eugene.

Igor Khavkine

Sep3-05, 03:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-09-02, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n> > From what I\'ve seen,\n>> you disagree with some of SR\'s postulates\n>\n> Yes, I disagree with the postulate of Einstein\'s special relativity\n> (ESR) which claims that observables in all (even interacting) physical\n> systems transform by universal linear tensor formulas.\n\nNo it does not, at least not in the sense that you seem to want to\nassign to that statement.\n\n> You got it exactly right. I drop the above postulate, and therefore\n> my theory (I would suggest the name \'SSR\') is different from the\n> conventional ESR. Then, it is not a surprise that physical predictions\n> of the two theories are different. For example, SSR predicts violations\n> of the Einstein\'s time dilation formula in decays of moving particles.\n> SSR also allows faster-than-light propagation of interactions which is\n> strictly forbidden by ESR.\n>\n> You say that these are "physically unreasonable consequences",\n> but so far there was no direct experimental proof that these\n> predictions are wrong.\n\nYou, unfortunately again, misunderstand my meaning. The statements you\nmake about (Einsten\'s) special relativity are incorrect, such as the\nstatement you made at the top of this post. You base these statements on\nwhat would appear to be logical deductions from a certain set of\npostultes.\n\nWhat are these postulates? I don\'t know. But I do know that they are not\nthe same Einstein\'s. How do I know this? Because the conclusions you\nseem to draw from them are false for special relativity. The only\npossibilities are that either your logical deductions are flawed or that\nyou are not talking about special relativity. Since you insist that your\nlogic is sound, I can only conclude the latter.\n\nBTW, I meant that the conclusions you draw from this unknown theory,\nwhich for lack of imagination I\'ve called Stephanovich\'s theory, are\njudged "physically unreasonable" not by me, but by yourself.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-09-02, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:
> > From what I've seen,
>> you disagree with some of SR's postulates
>
> Yes, I disagree with the postulate of Einstein's special relativity
> (ESR) which claims that observables in all (even interacting) physical
> systems transform by universal linear tensor formulas.

No it does not, at least not in the sense that you seem to want to
assign to that statement.

> You got it exactly right. I drop the above postulate, and therefore
> my theory (I would suggest the name 'SSR') is different from the
> conventional ESR. Then, it is not a surprise that physical predictions
> of the two theories are different. For example, SSR predicts violations
> of the Einstein's time dilation formula in decays of moving particles.
> SSR also allows faster-than-light propagation of interactions which is
> strictly forbidden by ESR.
>
> You say that these are "physically unreasonable consequences",
> but so far there was no direct experimental proof that these
> predictions are wrong.

You, unfortunately again, misunderstand my meaning. The statements you
make about (Einsten's) special relativity are incorrect, such as the
statement you made at the top of this post. You base these statements on
what would appear to be logical deductions from a certain set of
postultes.

What are these postulates? I don't know. But I do know that they are not
the same Einstein's. How do I know this? Because the conclusions you
seem to draw from them are false for special relativity. The only
possibilities are that either your logical deductions are flawed or that
you are not talking about special relativity. Since you insist that your
logic is sound, I can only conclude the latter.

BTW, I meant that the conclusions you draw from this unknown theory,
which for lack of imagination I've called Stephanovich's theory, are
judged "physically unreasonable" not by me, but by yourself.

Igor

Eugene Stefanovich

Sep7-05, 01:25 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nstefanbanev@yahoo.com wrote:\n> Dear Eugene,\n>\n> It is impossible to reason person when it gets personal. He definitely\n> has no intention to understand and discuss constructively your ideas.\n> Your questioning of SR assumptions is perfectly reasonable. Theory can\n> not be correct or wrong it can only possess more or less predictive\n> power; so, it would be far more productive to think some "simple"\n> experiment to make the point rather trying persuading opponent who has\n> no intention even consider your assumptions.\n\nAgree entirely. The experimental predictions of the proposed theory\n(Relativistic Quantum Dynamics, or RQD)\nhave been discussed on this NG a number of times. The predicted\ndeviations from standard approaches are rather small, so their\nunambiguous observation would require a significant experimental effort.\nI just wanted to reiterate here two areas in which such experiments have\na chance for success:\n\n1. Deviations from Einstein\'s time dilation formula in decays (and\noscillations) of moving particles. See:\n\nE. V. Stefanovich, Quantum effects in relativistic decays,\nInt. J. Theor. Phys. 35 (1996), 2539.\nhttp://www.geocities.com/meopemuk/IJTPpaper.pdf\n\nL. A. Khalfin, Quantum theory of unstable particles and relativity,\nPreprint of Steklov Mathematical Institute, St. Petersburg Department,\nPDMI-6/1997 (1997), http://www.pdmi.ras.ru/preprint/1997/97-06.html\n\nM.I. Shirokov, Decay law of moving unstable particle, Int.\nJ. Theor. Phys. 43} (2004), 1541.\nhttp://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf\n\nM.I. Shirokov, Evolution in time of moving unstable systems,\nhttp://arxiv.org/abs/quant-ph/0508087\n\n2. Superluminal propagation of electromagnetic interactions.\nI suggest that this effect was already seen in numerous experiments\nusually quoted under the name "photon tunneling". In particular,\nsuch experiments were conducted in the last 15 years by G. Nimtz,\nG.C. Giakos and T.K.Ishii, R.Y. Chiao, A. Ranfagni and D. Mugnai,\nK. Wynne, L.J. Wang and many others. The experimental evidence for\nsuperluminal group velocities is overwhelming. A consistent\ntheoretical explanation is still lacking.\n\nYou can find more detailed discussion of these experiments and some more\nreferences in my book "Relativistic Quantum Dynamics",\nhttp://arxiv.org/abs/physics/0504062\n\nRegards.\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>stefanbanev@yahoo.com wrote:
> Dear Eugene,
>
> It is impossible to reason person when it gets personal. He definitely
> has no intention to understand and discuss constructively your ideas.
> Your questioning of SR assumptions is perfectly reasonable. Theory can
> not be correct or wrong it can only possess more or less predictive
> power; so, it would be far more productive to think some "simple"
> experiment to make the point rather trying persuading opponent who has
> no intention even consider your assumptions.

Agree entirely. The experimental predictions of the proposed theory
(Relativistic Quantum Dynamics, or RQD)
have been discussed on this NG a number of times. The predicted
deviations from standard approaches are rather small, so their
unambiguous observation would require a significant experimental effort.
I just wanted to reiterate here two areas in which such experiments have
a chance for success:

1. Deviations from Einstein's time dilation formula in decays (and
oscillations) of moving particles. See:

E. V. Stefanovich, Quantum effects in relativistic decays,
\Int. J. Theor. Phys. 35 (1996), 2539.
http://www.geocities.com/meopemuk/IJTPpaper.pdf

L. A. Khalfin, Quantum theory of unstable particles and relativity,
Preprint of Steklov Mathematical Institute, St. Petersburg Department,
PDMI-6/1997 (1997), http://www.pdmi.ras.ru/preprint/1997/97-06.html

M.I. Shirokov, Decay law of moving unstable particle, \Int.
J. Theor. Phys. 43} (2004), 1541.
http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf

M.I. Shirokov, Evolution in time of moving unstable systems,
http://arxiv.org/abs/http://www.arxiv.org/abs/quant-ph/0508087

2. Superluminal propagation of electromagnetic interactions.
I suggest that this effect was already seen in numerous experiments
usually quoted under the name "photon tunneling". In particular,
such experiments were conducted in the last 15 years by G. Nimtz,
G.C. Giakos and T.K.Ishii, R.Y. Chiao, A. Ranfagni and D. Mugnai,
K. Wynne, L.J. Wang and many others. The experimental evidence for
superluminal group velocities is overwhelming. A consistent
theoretical explanation is still lacking.

You can find more detailed discussion of these experiments and some more
references in my book "Relativistic Quantum Dynamics",
http://arxiv.org/abs/http://www.arxiv.org/abs/physics/0504062

Regards.
Eugene.

Ken S. Tucker

Sep7-05, 01:25 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi Stefan...\n\nstefanbanev@yahoo.com wrote:\n> Dear Eugene,\n>\n> It is impossible to reason person when it gets personal. He definitely\n> has no intention to understand and discuss constructively your ideas.\n> Your questioning of SR assumptions is perfectly reasonable. Theory can\n> not be correct or wrong it can only possess more or less predictive\n> power; so, it would be far more productive to think some "simple"\n> experiment to make the point rather trying persuading opponent who has\n> no intention even consider your assumptions.\n>\n> Regards,\n> Stefan\n\nEinstein\'s SR (ESR) was replaced in 1916 by GR,\nso there really is no ESR. GR does everything\nSR did and more, and place the Theory of Relativity\non the firm Principle of Relativity stating,\n\n*no absolute motion exists*,\n\nwhich in turn means there is\n\n*no preferred Frame of Reference*,\n\nwhich in turn means,\n\n*physical laws are equally valid in all FoR\'s*.\n\nFor Eugene to say that he finds deviation in\nSR means he\'s questioning inertial FoR\'s in\nGR, hence GR itself. That\'s fine, what is\nwrong with GR?\n\nRegards\nKen S. Tucker\n\n> Eugene Stefanovich wrote:\n> > Igor Khavkine wrote:\n> > > From what I\'ve seen,\n> > > you disagree with some of SR\'s postulates\n> >\n> > Yes, I disagree with the postulate of Einstein\'s special relativity\n> > (ESR) which claims that observables in all (even interacting) physical\n> > systems transform by universal linear tensor formulas.\n> >\n> > > (you demand proof, yet forget\n> > > that postulates are not proven).\n> >\n> > I understand that this is a postulate, and therefore it doesn\'t require\n> > proof in the framework of ESR.\n> >\n> > > That\'s fine, except that when you claim\n> > > to draw physically unreasonable consequences from SR, you must specify\n> > > that SR stands for Stefanovich\'s Relativity, which is distinct from\n> > > Special Relativity, since key postulates are dropped.\n> >\n> > You got it exactly right. I drop the above postulate, and therefore\n> > my theory (I would suggest the name \'SSR\') is different from the\n> > conventional ESR. Then, it is not a surprise that physical predictions\n> > of the two theories are different. For example, SSR predicts violations\n> > of the Einstein\'s time dilation formula in decays of moving particles.\n> > SSR also allows faster-than-light propagation of interactions which is\n> > strictly forbidden by ESR.\n> >\n> > You say that these are "physically unreasonable consequences",\n> > but so far there was no direct experimental proof that these\n> > predictions are wrong.\n> >\n> > Eugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi Stefan...

stefanbanev@yahoo.com wrote:
> Dear Eugene,
>
> It is impossible to reason person when it gets personal. He definitely
> has no intention to understand and discuss constructively your ideas.
> Your questioning of SR assumptions is perfectly reasonable. Theory can
> not be correct or wrong it can only possess more or less predictive
> power; so, it would be far more productive to think some "simple"
> experiment to make the point rather trying persuading opponent who has
> no intention even consider your assumptions.
>
> Regards,
> Stefan

Einstein's SR (ESR) was replaced in 1916 by GR,
so there really is no ESR. GR does everything
SR did and more, and place the Theory of Relativity
on the firm Principle of Relativity stating,

*no absolute motion exists*,

which in turn means there is

*no preferred Frame of Reference*,

which in turn means,

*physical laws are equally valid in all FoR's*.

For Eugene to say that he finds deviation in
SR means he's questioning inertial FoR's in
GR, hence GR itself. That's fine, what is
wrong with GR?

Regards
Ken S. Tucker

> Eugene Stefanovich wrote:
> > Igor Khavkine wrote:
> > > From what I've seen,
> > > you disagree with some of SR's postulates
> >
> > Yes, I disagree with the postulate of Einstein's special relativity
> > (ESR) which claims that observables in all (even interacting) physical
> > systems transform by universal linear tensor formulas.
> >
> > > (you demand proof, yet forget
> > > that postulates are not proven).
> >
> > I understand that this is a postulate, and therefore it doesn't require
> > proof in the framework of ESR.
> >
> > > That's fine, except that when you claim
> > > to draw physically unreasonable consequences from SR, you must specify
> > > that SR stands for Stefanovich's Relativity, which is distinct from
> > > Special Relativity, since key postulates are dropped.
> >
> > You got it exactly right. I drop the above postulate, and therefore
> > my theory (I would suggest the name 'SSR') is different from the
> > conventional ESR. Then, it is not a surprise that physical predictions
> > of the two theories are different. For example, SSR predicts violations
> > of the Einstein's time dilation formula in decays of moving particles.
> > SSR also allows faster-than-light propagation of interactions which is
> > strictly forbidden by ESR.
> >
> > You say that these are "physically unreasonable consequences",
> > but so far there was no direct experimental proof that these
> > predictions are wrong.
> >
> > Eugene.

Eugene Stefanovich

Sep7-05, 01:26 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n\n>>Yes, I disagree with the postulate of Einstein\'s special relativity\n>>(ESR) which claims that observables in all (even interacting) physical\n>>systems transform by universal linear tensor formulas.\n>\n>\n> No it does not, at least not in the sense that you seem to want to\n> assign to that statement.\n\nI don\'t know what sense you alluded to. In ESR, space-time coordinates\nof a particle (x,t) form a 4-vector. This means that under boosts they\ntransform as\n\nx\' = x \\cosh \\theta - ct \\sinh \\theta (1)\nt\' = t \\cosh \\theta - x/c \\sinh \\theta (2)\n\nwhere v = c \\tanh \\theta is the velocity of the moving reference frame.\n\nIn ESR, momentum and energy\nof a particle (p,e) form a 4-vector. This means that under boosts they\ntransform as\n\np\' = p \\cosh \\theta - e/c \\sinh \\theta (3)\ne\' = e \\cosh \\theta - cp \\sinh \\theta (4)\n\nThese transformation laws are supposed to be independent on whether the\nparticle is free or it interacts with other particles.\nDid I get it right? If not, then I must have misunderstood Einstein\'s\nspecial relativity completely. Though, I doubt that.\n\n> You, unfortunately again, misunderstand my meaning. The statements you\n> make about (Einsten\'s) special relativity are incorrect, such as the\n> statement you made at the top of this post.\n\nDo you agree that equations (1) - (4) are assumed to be exact in\nEinstein\'s theory independent on interactions? If, as you say, my\nstatements are incorrect, then what are the correct transformation\nlaws in ESR?\n\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>>Yes, I disagree with the postulate of Einstein's special relativity
>>(ESR) which claims that observables in all (even interacting) physical
>>systems transform by universal linear tensor formulas.
>
>
> No it does not, at least not in the sense that you seem to want to
> assign to that statement.

I don't know what sense you alluded to. In ESR, space-time coordinates
of a particle (x,t) form a 4-vector. This means that under boosts they
transform as

x' = x \cosh \theta - ct \sinh \theta (1)
t' = t \cosh \theta - x/c \sinh \theta (2)

where v = c \tanh \theta is the velocity of the moving reference frame.

In ESR, momentum and energy
of a particle (p,e) form a 4-vector. This means that under boosts they
transform as

p' = p \cosh \theta - e/c \sinh \theta (3)
e' = e \cosh \theta - cp \sinh \theta (4)

These transformation laws are supposed to be independent on whether the
particle is free or it interacts with other particles.
Did I get it right? If not, then I must have misunderstood Einstein's
special relativity completely. Though, I doubt that.

> You, unfortunately again, misunderstand my meaning. The statements you
> make about (Einsten's) special relativity are incorrect, such as the
> statement you made at the top of this post.

Do you agree that equations (1) - (4) are assumed to be exact in
Einstein's theory independent on interactions? If, as you say, my
statements are incorrect, then what are the correct transformation
laws in ESR?

Eugene.

stefanbanev@yahoo.com

Sep7-05, 02:11 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi Ken,\n\nMy understanding of Stefanovich\'s approach is that nothing wrong with\nSR and GR; SR is a good approximation of GR for near flat space-time.\nStefanovich\'s point is that SR is a perfect toll for passive observer\nto predict the outcome of experiment but it is no so perfect do predict\noutcome of experiment if observer is involved in interaction. He has\nnot said the same about GR but I think the statement above is correct\nfor GR as well. GR is a predictive theory for passive observer but it\nmay not be so good to describe observer event interaction. I am not\nwell qualified to discuss the technical details; physics is not my area\nof expertise however I can see a general picture.\n\nRegards,\nStefan\n\n\n> For Eugene to say that he finds deviation in\n> SR means he\'s questioning inertial FoR\'s in\n> GR, hence GR itself. That\'s fine, what is\n> wrong with GR?\n\n\n\n\n> Hi Stefan...\n>\n> stefanbanev@yahoo.com wrote:\n> > Dear Eugene,\n> >\n> > It is impossible to reason person when it gets personal. He definitely\n> > has no intention to understand and discuss constructively your ideas.\n> > Your questioning of SR assumptions is perfectly reasonable. Theory can\n> > not be correct or wrong it can only possess more or less predictive\n> > power; so, it would be far more productive to think some "simple"\n> > experiment to make the point rather trying persuading opponent who has\n> > no intention even consider your assumptions.\n> >\n> > Regards,\n> > Stefan\n>\n> Einstein\'s SR (ESR) was replaced in 1916 by GR,\n> so there really is no ESR. GR does everything\n> SR did and more, and place the Theory of Relativity\n> on the firm Principle of Relativity stating,\n>\n> *no absolute motion exists*,\n>\n> which in turn means there is\n>\n> *no preferred Frame of Reference*,\n>\n> which in turn means,\n>\n> *physical laws are equally valid in all FoR\'s*.\n>\n> For Eugene to say that he finds deviation in\n> SR means he\'s questioning inertial FoR\'s in\n> GR, hence GR itself. That\'s fine, what is\n> wrong with GR?\n>\n> Regards\n> Ken S. Tucker\n>\n> > Eugene Stefanovich wrote:\n> > > Igor Khavkine wrote:\n> > > > From what I\'ve seen,\n> > > > you disagree with some of SR\'s postulates\n> > >\n> > > Yes, I disagree with the postulate of Einstein\'s special relativity\n> > > (ESR) which claims that observables in all (even interacting) physical\n> > > systems transform by universal linear tensor formulas.\n> > >\n> > > > (you demand proof, yet forget\n> > > > that postulates are not proven).\n> > >\n> > > I understand that this is a postulate, and therefore it doesn\'t require\n> > > proof in the framework of ESR.\n> > >\n> > > > That\'s fine, except that when you claim\n> > > > to draw physically unreasonable consequences from SR, you must specify\n> > > > that SR stands for Stefanovich\'s Relativity, which is distinct from\n> > > > Special Relativity, since key postulates are dropped.\n> > >\n> > > You got it exactly right. I drop the above postulate, and therefore\n> > > my theory (I would suggest the name \'SSR\') is different from the\n> > > conventional ESR. Then, it is not a surprise that physical predictions\n> > > of the two theories are different. For example, SSR predicts violations\n> > > of the Einstein\'s time dilation formula in decays of moving particles.\n> > > SSR also allows faster-than-light propagation of interactions which is\n> > > strictly forbidden by ESR.\n> > >\n> > > You say that these are "physically unreasonable consequences",\n> > > but so far there was no direct experimental proof that these\n> > > predictions are wrong.\n> > >\n> > > Eugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi Ken,

My understanding of Stefanovich's approach is that nothing wrong with
SR and GR; SR is a good approximation of GR for near flat space-time.
Stefanovich's point is that SR is a perfect toll for passive observer
to predict the outcome of experiment but it is no so perfect do predict
outcome of experiment if observer is involved in interaction. He has
not said the same about GR but I think the statement above is correct
for GR as well. GR is a predictive theory for passive observer but it
may not be so good to describe observer event interaction. I am not
well qualified to discuss the technical details; physics is not my area
of expertise however I can see a general picture.

Regards,
Stefan

> For Eugene to say that he finds deviation in
> SR means he's questioning inertial FoR's in
> GR, hence GR itself. That's fine, what is
> wrong with GR?

> Hi Stefan...
>
> stefanbanev@yahoo.com wrote:
> > Dear Eugene,
> >
> > It is impossible to reason person when it gets personal. He definitely
> > has no intention to understand and discuss constructively your ideas.
> > Your questioning of SR assumptions is perfectly reasonable. Theory can
> > not be correct or wrong it can only possess more or less predictive
> > power; so, it would be far more productive to think some "simple"
> > experiment to make the point rather trying persuading opponent who has
> > no intention even consider your assumptions.
> >
> > Regards,
> > Stefan
>
> Einstein's SR (ESR) was replaced in 1916 by GR,
> so there really is no ESR. GR does everything
> SR did and more, and place the Theory of Relativity
> on the firm Principle of Relativity stating,
>
> *no absolute motion exists*,
>
> which in turn means there is
>
> *no preferred Frame of Reference*,
>
> which in turn means,
>
> *physical laws are equally valid in all FoR's*.
>
> For Eugene to say that he finds deviation in
> SR means he's questioning inertial FoR's in
> GR, hence GR itself. That's fine, what is
> wrong with GR?
>
> Regards
> Ken S. Tucker
>
> > Eugene Stefanovich wrote:
> > > Igor Khavkine wrote:
> > > > From what I've seen,
> > > > you disagree with some of SR's postulates
> > >
> > > Yes, I disagree with the postulate of Einstein's special relativity
> > > (ESR) which claims that observables in all (even interacting) physical
> > > systems transform by universal linear tensor formulas.
> > >
> > > > (you demand proof, yet forget
> > > > that postulates are not proven).
> > >
> > > I understand that this is a postulate, and therefore it doesn't require
> > > proof in the framework of ESR.
> > >
> > > > That's fine, except that when you claim
> > > > to draw physically unreasonable consequences from SR, you must specify
> > > > that SR stands for Stefanovich's Relativity, which is distinct from
> > > > Special Relativity, since key postulates are dropped.
> > >
> > > You got it exactly right. I drop the above postulate, and therefore
> > > my theory (I would suggest the name 'SSR') is different from the
> > > conventional ESR. Then, it is not a surprise that physical predictions
> > > of the two theories are different. For example, SSR predicts violations
> > > of the Einstein's time dilation formula in decays of moving particles.
> > > SSR also allows faster-than-light propagation of interactions which is
> > > strictly forbidden by ESR.
> > >
> > > You say that these are "physically unreasonable consequences",
> > > but so far there was no direct experimental proof that these
> > > predictions are wrong.
> > >
> > > Eugene.

Eugene Stefanovich

Sep7-05, 06:30 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ken S. Tucker wrote:\n\n> *no absolute motion exists*,\n\nAgree completely\n\n> which in turn means there is\n>\n> *no preferred Frame of Reference*,\n\nExactly true.\n\n\n> which in turn means,\n>\n> *physical laws are equally valid in all FoR\'s*.\n\nYes, you are right.\n\n>\n> For Eugene to say that he finds deviation in\n> SR means he\'s questioning inertial FoR\'s in\n> GR, hence GR itself. That\'s fine, what is\n> wrong with GR?\n\nI do not question the principle of relativity.\nI agree completely that the speed of light is an invariant.\nMy point is that there is no (and there can be no) proof\nthat these two statements imply the universal validity of\ntensor transformation laws for all observables in all physical\nsystems. The problem with SR (and GR) is that this\nuniversal validity is tacitly assumed. However, a more careful\nconsideration of this issue reveals that this assumption is\nwrong.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ken S. Tucker wrote:

> *no absolute motion exists*,

Agree completely

> which in turn means there is
>
> *no preferred Frame of Reference*,

Exactly true.

> which in turn means,
>
> *physical laws are equally valid in all FoR's*.

Yes, you are right.

>
> For Eugene to say that he finds deviation in
> SR means he's questioning inertial FoR's in
> GR, hence GR itself. That's fine, what is
> wrong with GR?

I do not question the principle of relativity.
I agree completely that the speed of light is an invariant.
My point is that there is no (and there can be no) proof
that these two statements imply the universal validity of
tensor transformation laws for all observables in all physical
systems. The problem with SR (and GR) is that this
universal validity is tacitly assumed. However, a more careful
consideration of this issue reveals that this assumption is
wrong.

Eugene.

Eugene Stefanovich

Sep7-05, 06:31 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>stefanbanev@yahoo.com wrote:\n> Hi Ken,\n>\n> My understanding of Stefanovich\'s approach is that nothing wrong with\n> SR and GR; SR is a good approximation of GR for near flat space-time.\n> Stefanovich\'s point is that SR is a perfect toll for passive observer\n> to predict the outcome of experiment but it is no so perfect do predict\n> outcome of experiment if observer is involved in interaction. He has\n> not said the same about GR but I think the statement above is correct\n> for GR as well. GR is a predictive theory for passive observer but it\n> may not be so good to describe observer event interaction. I am not\n> well qualified to discuss the technical details; physics is not my area\n> of expertise however I can see a general picture.\n\n1. I am not talking about gravity and GR, only about the special\nprinciple of relativity that connects observations in different\ninertial frames of reference.\n2. The interactions I am talking about are not between the observer\nand the physical system. They are between different parts of\nthe physical system.\n\nFor example, take a system of two classical charges interacting\nwith each other. At time t they are described by positions\nx_1(t), x_2(t) and\nmomenta p_1(t), p_2(t). My approach establishes that transformations\nof these observables to the moving frame of reference\n[x\'_1(t\'), x\'_2(t\'), p\'_1(t\'), p\'_2(t\'), where t\' is the time\nmeasured by the moving observer] are NOT\ngiven by linear tensor Lorentz formulas (as mandated in SR).\nThe exact transformations depend on the state of this system and\non the strength of interaction between the particles.\nSo, something IS wrong with SR.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>stefanbanev@yahoo.com wrote:
> Hi Ken,
>
> My understanding of Stefanovich's approach is that nothing wrong with
> SR and GR; SR is a good approximation of GR for near flat space-time.
> Stefanovich's point is that SR is a perfect toll for passive observer
> to predict the outcome of experiment but it is no so perfect do predict
> outcome of experiment if observer is involved in interaction. He has
> not said the same about GR but I think the statement above is correct
> for GR as well. GR is a predictive theory for passive observer but it
> may not be so good to describe observer event interaction. I am not
> well qualified to discuss the technical details; physics is not my area
> of expertise however I can see a general picture.

1. I am not talking about gravity and GR, only about the special
principle of relativity that connects observations in different
inertial frames of reference.
2. The interactions I am talking about are not between the observer
and the physical system. They are between different parts of
the physical system.

For example, take a system of two classical charges interacting
with each other. At time t they are described by positions
x_1(t), x_2(t) and
momenta p_1(t), p_2(t). My approach establishes that transformations
of these observables to the moving frame of reference
[x'_1(t'), x'_2(t'), p'_1(t'), p'_2(t'), where t' is the time
measured by the moving observer] are NOT
given by linear tensor Lorentz formulas (as mandated in SR).
The exact transformations depend on the state of this system and
on the strength of interaction between the particles.
So, something IS wrong with SR.

Eugene.

Eugene Stefanovich

Sep7-05, 06:31 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> stefanbanev@yahoo.com wrote:\n\n\n>>Theory can\n>>not be correct or wrong it can only possess more or less predictive\n>>power; so, it would be far more productive to think some "simple"\n>>experiment to make the point rather ...\n\nThis is a good point, and I agree to do exactly that.\n\nAs a topic for discussion I would like to suggest the decay law\nof moving unstable particles. This problem has a number of advantages\nwhich makes it an ideal candidate for discussion:\n\n1. Einstein\'s theory makes an exact prediction: the decay law\nof a moving particle is EXACTLY \\gamma times slower than for\nthe particle at rest.\n\n2. The quantum relativistic treatment of this problem is rather\nsimple. The number of particles is limited (the unstable\nparticle and decay products), so one does not\nneed to resort to quantum fields and questionable\nrenormalization tricks. The relevant Hilbert space of the\nsystem consists of just two sectors of the Fock space.\nThe problem can solved exactly, even analytically.\nThe solution does not depend on any assumptions about\nthe details of the interaction.\n\n3. The obtained solution fundamentally disagrees with Einstein\'s\nformula in 1. Though, numerically, the differences are very small\nfor common unstable systems.\n\n4. This result was obtained not by just one suspicious crackpot,\nbut also by two distinguished professors independently.\nSee references below.\n\n\n> E. V. Stefanovich, Quantum effects in relativistic decays,\n> Int. J. Theor. Phys. 35 (1996), 2539.\n> http://www.geocities.com/meopemuk/IJTPpaper.pdf\n>\n> L. A. Khalfin, Quantum theory of unstable particles and relativity,\n> Preprint of Steklov Mathematical Institute, St. Petersburg Department,\n> PDMI-6/1997 (1997), http://www.pdmi.ras.ru/preprint/1997/97-06.html\n>\n> M.I. Shirokov, Decay law of moving unstable particle, Int.\n> J. Theor. Phys. 43} (2004), 1541.\n> http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf\n>\n> M.I. Shirokov, Evolution in time of moving unstable systems,\n> http://arxiv.org/abs/quant-ph/0508087\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> stefanbanev@yahoo.com wrote:

>>Theory can
>>not be correct or wrong it can only possess more or less predictive
>>power; so, it would be far more productive to think some "simple"
>>experiment to make the point rather ...

This is a good point, and I agree to do exactly that.

As a topic for discussion I would like to suggest the decay law
of moving unstable particles. This problem has a number of advantages
which makes it an ideal candidate for discussion:

1. Einstein's theory makes an exact prediction: the decay law
of a moving particle is EXACTLY \gamma times slower than for
the particle at rest.

2. The quantum relativistic treatment of this problem is rather
simple. The number of particles is limited (the unstable
particle and decay products), so one does not
need to resort to quantum fields and questionable
renormalization tricks. The relevant Hilbert space of the
system consists of just two sectors of the Fock space.
The problem can solved exactly, even analytically.
The solution does not depend on any assumptions about
the details of the interaction.

3. The obtained solution fundamentally disagrees with Einstein's
formula in 1. Though, numerically, the differences are very small
for common unstable systems.

4. This result was obtained not by just one suspicious crackpot,
but also by two distinguished professors independently.
See references below.

> E. V. Stefanovich, Quantum effects in relativistic decays,
> \Int. J. Theor. Phys. 35 (1996), 2539.
> http://www.geocities.com/meopemuk/IJTPpaper.pdf
>
> L. A. Khalfin, Quantum theory of unstable particles and relativity,
> Preprint of Steklov Mathematical Institute, St. Petersburg Department,
> PDMI-6/1997 (1997), http://www.pdmi.ras.ru/preprint/1997/97-06.html
>
> M.I. Shirokov, Decay law of moving unstable particle, \Int.
> J. Theor. Phys. 43} (2004), 1541.
> http://www.jinr.ru/publish/Preprints/2004/004(E2-2004-4).pdf
>
> M.I. Shirokov, Evolution in time of moving unstable systems,
> http://arxiv.org/abs/http://www.arxiv.org/abs/quant-ph/0508087

Eugene.

Igor Khavkine

Sep8-05, 12:36 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-09-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n>\n>>>Yes, I disagree with the postulate of Einstein\'s special relativity\n>>>(ESR) which claims that observables in all (even interacting) physical\n>>>systems transform by universal linear tensor formulas.\n>>\n>> No it does not, at least not in the sense that you seem to want to\n>> assign to that statement.\n>\n> I don\'t know what sense you alluded to. In ESR, space-time coordinates\n> of a particle (x,t) form a 4-vector. This means that under boosts they\n> transform as\n>\n> x\' = x \\cosh \\theta - ct \\sinh \\theta (1)\n> t\' = t \\cosh \\theta - x/c \\sinh \\theta (2)\n\nNope, that\'s what the coordinates of space-time events transform like,\nnot particle positions coordinates.\n\n> where v = c \\tanh \\theta is the velocity of the moving reference frame.\n>\n> In ESR, momentum and energy\n> of a particle (p,e) form a 4-vector. This means that under boosts they\n> transform as\n>\n> p\' = p \\cosh \\theta - e/c \\sinh \\theta (3)\n> e\' = e \\cosh \\theta - cp \\sinh \\theta (4)\n>\n> These transformation laws are supposed to be independent on whether the\n> particle is free or it interacts with other particles.\n\nNope, that\'s not how the canonical energy and momentum of a particle\ntransform when formulated in Hamiltonian form.\n\nThis is what I meant when I said "not in the sense...".\n\n> Did I get it right? If not, then I must have misunderstood Einstein\'s\n> special relativity completely. Though, I doubt that.\n\nNo, unfortunately, you still do not have it right. Based on previous\ndiscussion, I\'m inclined to believe that you have indeed completely\nmisunderstood Einstein\'s special relativity and Hamiltonian mechanics.\n\n>> You, unfortunately again, misunderstand my meaning. The statements you\n>> make about (Einsten\'s) special relativity are incorrect, such as the\n>> statement you made at the top of this post.\n>\n> Do you agree that equations (1) - (4) are assumed to be exact in\n> Einstein\'s theory independent on interactions?\n\nNo, they are not true in Einstein\'s theory. They may be true, in\nStefanovich\'s theory, which is why I retitled this thread. Although, I\ncan\'t say for certain, since I don\'t know what that theory is exactly.\n\n> If, as you say, my\n> statements are incorrect, then what are the correct transformation\n> laws in ESR?\n\nI was in the middle of deriving the correct predictions of Einstein\'s\ntheory earlier in this thread, when it seemed like you ran out of\npatience. Would you like me to continue now? Like I said, I will do this\nkeeping all the postulates of the standard formulation of relativity.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-09-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:
>
>>>Yes, I disagree with the postulate of Einstein's special relativity
>>>(ESR) which claims that observables in all (even interacting) physical
>>>systems transform by universal linear tensor formulas.
>>
>> No it does not, at least not in the sense that you seem to want to
>> assign to that statement.
>
> I don't know what sense you alluded to. In ESR, space-time coordinates
> of a particle (x,t) form a 4-vector. This means that under boosts they
> transform as
>
> x' = x \cosh \theta - ct \sinh \theta (1)
> t' = t \cosh \theta - x/c \sinh \theta (2)

Nope, that's what the coordinates of space-time events transform like,
not particle positions coordinates.

> where v = c \tanh \theta is the velocity of the moving reference frame.
>
> In ESR, momentum and energy
> of a particle (p,e) form a 4-vector. This means that under boosts they
> transform as
>
> p' = p \cosh \theta - e/c \sinh \theta (3)
> e' = e \cosh \theta - cp \sinh \theta (4)
>
> These transformation laws are supposed to be independent on whether the
> particle is free or it interacts with other particles.

Nope, that's not how the canonical energy and momentum of a particle
transform when formulated in Hamiltonian form.

This is what I meant when I said "not in the sense...".

> Did I get it right? If not, then I must have misunderstood Einstein's
> special relativity completely. Though, I doubt that.

No, unfortunately, you still do not have it right. Based on previous
discussion, I'm inclined to believe that you have indeed completely
misunderstood Einstein's special relativity and Hamiltonian mechanics.

>> You, unfortunately again, misunderstand my meaning. The statements you
>> make about (Einsten's) special relativity are incorrect, such as the
>> statement you made at the top of this post.
>
> Do you agree that equations (1) - (4) are assumed to be exact in
> Einstein's theory independent on interactions?

No, they are not true in Einstein's theory. They may be true, in
Stefanovich's theory, which is why I retitled this thread. Although, I
can't say for certain, since I don't know what that theory is exactly.

> If, as you say, my
> statements are incorrect, then what are the correct transformation
> laws in ESR?

I was in the middle of deriving the correct predictions of Einstein's
theory earlier in this thread, when it seemed like you ran out of
patience. Would you like me to continue now? Like I said, I will do this
keeping all the postulates of the standard formulation of relativity.

Igor

Eugene Stefanovich

Sep8-05, 04:59 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n>> In ESR, space-time coordinates\n>>of a particle (x,t) form a 4-vector. This means that under boosts they\n>>transform as\n>>\n>>x\' = x \\cosh \\theta - ct \\sinh \\theta (1)\n>>t\' = t \\cosh \\theta - x/c \\sinh \\theta (2)\n>\n>\n> Nope, that\'s what the coordinates of space-time events transform like,\n> not particle positions coordinates.\n\nI don\'t see what is the difference. Suppose you have two particles,\nand the event is defined as\nintersection of their trajectories. Then the space\ncoordinates of this event coincide with the particles\' position\nat the instant of collision. So, "event coordinates" and\n"particle positions coordinates" are the same things. They should have\nthe same transformations.\n\n\n>>If, as you say, my\n>>statements are incorrect, then what are the correct transformation\n>>laws in ESR?\n>\n>\n> I was in the middle of deriving the correct predictions of Einstein\'s\n> theory earlier in this thread, when it seemed like you ran out of\n> patience. Would you like me to continue now? Like I said, I will do this\n> keeping all the postulates of the standard formulation of relativity.\n\n\nOK, I promise to be patient this time.\nPlease explain me what are Einstein\'s\ntransformation laws for space-time coordinates of events and for\nparticle positions.\nWhich postulates they are based on, and how they are derived?\n\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
>> In ESR, space-time coordinates
>>of a particle (x,t) form a 4-vector. This means that under boosts they
>>transform as
>>
>>x' = x \cosh \theta - ct \sinh \theta (1)
>>t' = t \cosh \theta - x/c \sinh \theta (2)
>
>
> Nope, that's what the coordinates of space-time events transform like,
> not particle positions coordinates.

I don't see what is the difference. Suppose you have two particles,
and the event is defined as
intersection of their trajectories. Then the space
coordinates of this event coincide with the particles' position
at the instant of collision. So, "event coordinates" and
"particle positions coordinates" are the same things. They should have
the same transformations.

>>If, as you say, my
>>statements are incorrect, then what are the correct transformation
>>laws in ESR?
>
>
> I was in the middle of deriving the correct predictions of Einstein's
> theory earlier in this thread, when it seemed like you ran out of
> patience. Would you like me to continue now? Like I said, I will do this
> keeping all the postulates of the standard formulation of relativity.

OK, I promise to be patient this time.
Please explain me what are Einstein's
transformation laws for space-time coordinates of events and for
particle positions.
Which postulates they are based on, and how they are derived?

Eugene.

Eugene Stefanovich

Sep9-05, 10:41 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:slrndi0g09.vrl.igor.kh@corum.multive rse.ca...\n\n> > In ESR, momentum and energy\n> > of a particle (p,e) form a 4-vector. This means that under boosts they\n> > transform as\n> >\n> > p\' = p \\cosh \\theta - e/c \\sinh \\theta (3)\n> > e\' = e \\cosh \\theta - cp \\sinh \\theta (4)\n> >\n> > These transformation laws are supposed to be independent on whether the\n> > particle is free or it interacts with other particles.\n>\n> Nope, that\'s not how the canonical energy and momentum of a particle\n> transform when formulated in Hamiltonian form.\n\nNow you got me confused. Are there some other non-canonical\nparticle momentum and energy? Are they (also?) measurable?\nHow do they transform?\n\nEugene\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:slrndi0g09.vrl.igor.kh@corum.multiverse.ca...

> > In ESR, momentum and energy
> > of a particle (p,e) form a 4-vector. This means that under boosts they
> > transform as
> >
> > p' = p \cosh \theta - e/c \sinh \theta (3)
> > e' = e \cosh \theta - cp \sinh \theta (4)
> >
> > These transformation laws are supposed to be independent on whether the
> > particle is free or it interacts with other particles.
>
> Nope, that's not how the canonical energy and momentum of a particle
> transform when formulated in Hamiltonian form.

Now you got me confused. Are there some other non-canonical
particle momentum and energy? Are they (also?) measurable?
How do they transform?

Eugene

Igor Khavkine

Sep9-05, 02:41 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Now you got me confused. Are there some other non-canonical\n> particle momentum and energy? Are they (also?) measurable?\n> How do they transform?\n\nCanonical positions and momenta are coordinates on the phase space on\nwhich the Hamiltonian formulation of mechanics is defined. In a\nrelativistic theory, the Hamiltonian formulation takes a little care to\nset up.\n\nThe energy and momentum of a given particle can also be defined simply\nas a function of the embedding of its world-line in space-time. All of\nthese quantities transform by the rules dictated by special relativity\nand the Minkowski geometry of space-time. However, since you already\nmisunderstand what that means, the above is unlikely to tell you\nanything new. So, stay tuned to this thread for details.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Now you got me confused. Are there some other non-canonical
> particle momentum and energy? Are they (also?) measurable?
> How do they transform?

Canonical positions and momenta are coordinates on the phase space on
which the Hamiltonian formulation of mechanics is defined. In a
relativistic theory, the Hamiltonian formulation takes a little care to
set up.

The energy and momentum of a given particle can also be defined simply
as a function of the embedding of its world-line in space-time. All of
these quantities transform by the rules dictated by special relativity
and the Minkowski geometry of space-time. However, since you already
misunderstand what that means, the above is unlikely to tell you
anything new. So, stay tuned to this thread for details.

Igor

Igor Khavkine

Sep10-05, 09:57 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-09-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> For example, take a system of two classical charges interacting\n> with each other. At time t they are described by positions\n> x_1(t), x_2(t) and\n> momenta p_1(t), p_2(t). My approach establishes that transformations\n> of these observables to the moving frame of reference\n> [x\'_1(t\'), x\'_2(t\'), p\'_1(t\'), p\'_2(t\'), where t\' is the time\n> measured by the moving observer] are NOT\n> given by linear tensor Lorentz formulas (as mandated in SR).\n\nThe above is not mandated by SR, contrary to what you believe. Well, it\nwould be if SR stood for \'Stefanovich\'s Relativity\'. But it is not in\nEinstein\'s theory.\n\n> The exact transformations depend on the state of this system and\n> on the strength of interaction between the particles.\n\nThe same thing happens in SR.\n\n> So, something IS wrong with SR.\n\nNope, nothing wrong there.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-09-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> For example, take a system of two classical charges interacting
> with each other. At time t they are described by positions
> x_1(t), x_2(t) and
> momenta p_1(t), p_2(t). My approach establishes that transformations
> of these observables to the moving frame of reference
> [x'_1(t'), x'_2(t'), p'_1(t'), p'_2(t'), where t' is the time
> measured by the moving observer] are NOT
> given by linear tensor Lorentz formulas (as mandated in SR).

The above is not mandated by SR, contrary to what you believe. Well, it
would be if SR stood for 'Stefanovich's Relativity'. But it is not in
Einstein's theory.

> The exact transformations depend on the state of this system and
> on the strength of interaction between the particles.

The same thing happens in SR.

> So, something IS wrong with SR.

Nope, nothing wrong there.

Igor

Ralph Hartley

Sep10-05, 10:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> In ESR, space-time coordinates of a particle (x,t) form a 4-vector.\n\nThe space-time coordinates of an *event* form a 4-vector.\n\nA particle does not have unique space-time coordinates, it has a *path*\nin space-time. It can be parameterized more than one way. For example\n(x(t),t(t)) and (x\'(t),t\'(t)).\n\nNotation note: (x(t),t(t)) = (x(t),t) and (x\'(t),t\'(t)) are both\nunprimed coordinates, they use a different parameterization of the curve\nso that (x(t),t(t)) = (x\'(t\'),t\'(t\')). The corresponding primed\ncoordinates are (x(t)\',t(t)\') and (x\'(t)\',t\'(t)\') respectively so that\n(x(t)\',t(t)\') = (x\'(t\')\',t\'(t\')\'). Similarly for (p,e).\n\nSo when you speak of how the coordinates transform, you have a choice.\nYou can transform the coordinates at a fixed event, or at a fixed time\ncoordinate. Both are OK, but you have to be consistent.\n\n> under boosts they transform as\n>\n> x\' = x \\cosh \\theta - ct \\sinh \\theta (1)\n> t\' = t \\cosh \\theta - x/c \\sinh \\theta (2)\n>\n> where v = c \\tanh \\theta is the velocity of the moving reference frame.\n\nOk, this is how the coordinates of an event transform. So it transforms\n(x(t),t(t)) to (x(t)\',t(t)\'), which is not to be confused with\n(x\'(t)\',t\'(t)\'), a different, but related, pair of functions.\n\n> In ESR, momentum and energy\n> of a particle (p,e) form a 4-vector. This means that under boosts they\n> transform as\n>\n> p\' = p \\cosh \\theta - e/c \\sinh \\theta (3)\n> e\' = e \\cosh \\theta - cp \\sinh \\theta (4)\n\nSince p and e are not assumed to be constants, it matters which (p,e)\nyou transform. For the formula to be correct it needs to apply to a\nfixed event, not a fixed time coordinate.\n\n(3) and (4) map (p(t),e(t)) to (p(t)\',e(t)\') and also map (p\'(t),e\'(t))\nto (p\'(t)\',e\'(t)\'). They do *not* transform (p(t),e(t)) to\n(p\'(t)\',e\'(t)\'). Those are *differnt* vector valued functions.\n\nThis would all be fine *if* you were consistent about it, but in an\nearlier post\n\nEugene Stefanovich wrote:\n> In your approach (please correct me if I am wrong)\n> Lorentz transformations\n> connect properties of _the same_ event from the point of view of\n> different observers. I.e., you interpret Boost[v]p as momentum of the\n> particle at point A on the world-line.\n\nThat is what (3) and (4) do. Interpreting them to do something else is\nwrong.\n\nIf you want Boost[v]p to mean something else, then (3) and (4) are wrong.\n\nYou can\'t just cahnge the definition and not the equations.\n\n> The problem with your definition\n> is that it does not include the (obviously) dynamical character of\n> time translation. I.e., by applying the same rules to Time[t] you must\n> get\n>\n> Time[t] p = p [(A)]\n>\n> because independent on the value of t, you should get the momentum\n> at the same worldline point A. However, in my definition\n>\n> Time[t] p = p + t L [(B)]\n> where L is the Lorentz force.\n\nBut (3) and (4) are the transformation for a fixed event. if you use\nthem you *must* use (A) if you want to be consistent.\n\n(B) maps p(t) to p\'(t)\' but (3),(4) maps (p(t),e(t)) to (p(t)\',e(t)\')\nwhich is not the same thing at all.\n\nIf you wish to use (B) you need the formula to go from (p(t),e(t)) to\n(p\'(t)\',e\'(t)\'). SR does let you write such a fourmula, but it is not\n(3) and (4).\n\nIf you use two different parameterizations of a curve in the same\nequation, of course you will find an inconsistency.\n\n> Did I get it right?\n\nIf you were intending to write the transformation rules for a fixed\nevent, then yes (and (A) is correct). Otherwise no.\n\n> If not, then I must have misunderstood Einstein\'s special relativity\n> completely.\n\nApparently.\n\nRalph Hartley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> In ESR, space-time coordinates of a particle (x,t) form a 4-vector.

The space-time coordinates of an *event* form a 4-vector.

A particle does not have unique space-time coordinates, it has a *path*
in space-time. It can be parameterized more than one way. For example
(x(t),t(t)) and (x'(t),t'(t)).

Notation note: (x(t),t(t)) = (x(t),t) and (x'(t),t'(t)) are both
unprimed coordinates, they use a different parameterization of the curve
so that (x(t),t(t)) = (x'(t'),t'(t')). The corresponding primed
coordinates are (x(t)',t(t)') and (x'(t)',t'(t)') respectively so that
(x(t)',t(t)') = (x'(t')',t'(t')'). Similarly for (p,e).

So when you speak of how the coordinates transform, you have a choice.
You can transform the coordinates at a fixed event, or at a fixed time
coordinate. Both are OK, but you have to be consistent.

> under boosts they transform as
>
> x' = x \cosh \theta - ct \sinh \theta (1)
> t' = t \cosh \theta - x/c \sinh \theta (2)
>
> where v = c \tanh \theta is the velocity of the moving reference frame.

Ok, this is how the coordinates of an event transform. So it transforms
(x(t),t(t)) to (x(t)',t(t)'), which is not to be confused with
(x'(t)',t'(t)'), a different, but related, pair of functions.

> In ESR, momentum and energy
> of a particle (p,e) form a 4-vector. This means that under boosts they
> transform as
>
> p' = p \cosh \theta - e/c \sinh \theta (3)
> e' = e \cosh \theta - cp \sinh \theta (4)

Since p and e are not assumed to be constants, it matters which (p,e)
you transform. For the formula to be correct it needs to apply to a
fixed event, not a fixed time coordinate.

(3) and (4) map (p(t),e(t)) to (p(t)',e(t)') and also map (p'(t),e'(t))
to (p'(t)',e'(t)'). They do *not* transform (p(t),e(t)) to
(p'(t)',e'(t)'). Those are *differnt* vector valued functions.

This would all be fine *if* you were consistent about it, but in an
earlier post

Eugene Stefanovich wrote:
> In your approach (please correct me if I am wrong)
> Lorentz transformations
> connect properties of _the same_ event from the point of view of
> different observers. I.e., you interpret Boost[v]p as momentum of the
> particle at point A on the world-line.

That is what (3) and (4) do. Interpreting them to do something else is
wrong.

If you want Boost[v]p to mean something else, then (3) and (4) are wrong.

You can't just cahnge the definition and not the equations.

> The problem with your definition
> is that it does not include the (obviously) dynamical character of
> time translation. I.e., by applying the same rules to Time[t] you must
> get
>
> Time[t] p = p [(A)]
>
> because independent on the value of t, you should get the momentum
> at the same worldline point A. However, in my definition
>
> Time[t] p = p + t L [(B)]
> where L is the Lorentz force.

But (3) and (4) are the transformation for a fixed event. if you use
them you *must* use (A) if you want to be consistent.

(B) maps p(t) to p'(t)' but (3),(4) maps (p(t),e(t)) to (p(t)',e(t)')
which is not the same thing at all.

If you wish to use (B) you need the formula to go from (p(t),e(t)) to
(p'(t)',e'(t)'). SR does let you write such a fourmula, but it is not
(3) and (4).

If you use two different parameterizations of a curve in the same
equation, of course you will find an inconsistency.

> Did I get it right?

If you were intending to write the transformation rules for a fixed
event, then yes (and (A) is correct). Otherwise no.

> If not, then I must have misunderstood Einstein's special relativity
> completely.

Apparently.

Ralph Hartley

Eugene Stefanovich

Sep11-05, 01:25 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Ralph Hartley" <hartley@aic.nrl.navy.mil> wrote in message\nnews:dfplil\\$pbt\\$1@ra.nrl.navy.mil...\ n\n> > In ESR, momentum and energy\n> > of a particle (p,e) form a 4-vector. This means that under boosts they\n> > transform as\n> >\n> > p\' = p \\cosh \\theta - e/c \\sinh \\theta (3)\n> > e\' = e \\cosh \\theta - cp \\sinh \\theta (4)\n>\n> Since p and e are not assumed to be constants, it matters which (p,e)\n> you transform. For the formula to be correct it needs to apply to a\n> fixed event, not a fixed time coordinate.\n>\n> (3) and (4) map (p(t),e(t)) to (p(t)\',e(t)\') and also map (p\'(t),e\'(t))\n> to (p\'(t)\',e\'(t)\'). They do *not* transform (p(t),e(t)) to\n> (p\'(t)\',e\'(t)\'). Those are *differnt* vector valued functions.\n>\n> This would all be fine *if* you were consistent about it, but in an\n> earlier post\n>\n> Eugene Stefanovich wrote:\n> > In your approach (please correct me if I am wrong)\n> > Lorentz transformations\n> > connect properties of _the same_ event from the point of view of\n> > different observers. I.e., you interpret Boost[v]p as momentum of the\n> > particle at point A on the world-line.\n>\n> That is what (3) and (4) do. Interpreting them to do something else is\n> wrong.\n>\n> If you want Boost[v]p to mean something else, then (3) and (4) are wrong.\n\nLet me first agree with you. In eq. (3), momentum p and energy e are\nmeasured\nby the observer K at rest at time t=0 by his clock. Momentum p\'\nis measured by the moving observer K\' at time t\' by his clock whre t\' is\nrelated\nto t by the Lorentz formula\n\nt\' = t \\cosh \\theta - x/c \\sinh \\theta (2)\n\nwhere we can set t=0 and x is the position of the particle measured by K at\ntime t=0.\n\nOn the other hand,\n\nBoost[v]p (3\')\n\nis the momentum of the particle measured by observer K\' at time t\'=0 by his\nclock\nTherefore, in order to obtain p\' in (3) one should apply an additional time\ntranslation\nto eq. (3\'). The boost and time translation transformations in my original\npost\nshould have been written as\n\nBoost[v]p = p + vM (A)\nTime[t] p = p + t L (B)\n\nwhere both L and M are complex expressions depending on the interaction,\npositions/momenta of all particles, and the field configuration. L and M\nshould be\nsuch that the Poincare algebra commutators are satisfied between all\ngenerators.\nI agree that my "proof" of the Poincare-non-invariance of Maxwell\'s theory\nwas not correct.\n\nHowever, I am not ready to give up my major claim yet. The claim is that\neq. (3) is not accurate in interacting theories.\n\nAs I understand, your point is that in any interacting theory eq. (3) should\nbe obtained,\nif we first transform momentum to the moving frame (as in eq. (A)), and then\nperform a time translation to the time t\' in (2). I haven\'t seen such a\ncalculation in\nMaxwell\'s theory, and I doubt it will actually yield the Lorentz formula\n(3).\n\nMy conjecture is that eq. (3) is valid only in non-interacting theories. I\nadmit that I\ndo not have a proof for that, but I have a couple of arguments that make\nthis conjecture\nplausible:\n\n1. For particles-only theories this conjecture is proven in the\nCurrie-Jordan-\nSudarshan theorem. (Though, Igor Khavkine is right that this proof is not\nvalid for particles+fields theories).\n\n2. There is a thoroughly studied physical system (the decay law of a moving\nunstable particle) in which interactions L and M can be rigorously\nconstructed,\nand deviations from the Lorentz transformations formulas (i.e., the\nEinstein\'s\ntime dilation law) can be explicitly calculated. See references in another\npost in this\nthread.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Ralph Hartley" <hartley@aic.nrl.navy.mil> wrote in message
news:dfplil$pbt$1@ra.nrl.navy.mil...

> > In ESR, momentum and energy
> > of a particle (p,e) form a 4-vector. This means that under boosts they
> > transform as
> >
> > p' = p \cosh \theta - e/c \sinh \theta (3)
> > e' = e \cosh \theta - cp \sinh \theta (4)
>
> Since p and e are not assumed to be constants, it matters which (p,e)
> you transform. For the formula to be correct it needs to apply to a
> fixed event, not a fixed time coordinate.
>
> (3) and (4) map (p(t),e(t)) to (p(t)',e(t)') and also map (p'(t),e'(t))
> to (p'(t)',e'(t)'). They do *not* transform (p(t),e(t)) to
> (p'(t)',e'(t)'). Those are *differnt* vector valued functions.
>
> This would all be fine *if* you were consistent about it, but in an
> earlier post
>
> Eugene Stefanovich wrote:
> > In your approach (please correct me if I am wrong)
> > Lorentz transformations
> > connect properties of _the same_ event from the point of view of
> > different observers. I.e., you interpret Boost[v]p as momentum of the
> > particle at point A on the world-line.
>
> That is what (3) and (4) do. Interpreting them to do something else is
> wrong.
>
> If you want Boost[v]p to mean something else, then (3) and (4) are wrong.

Let me first agree with you. In eq. (3), momentum p and energy e are
measured
by the observer K at rest at time t=0 by his clock. Momentum p'
is measured by the moving observer K' at time t' by his clock whre t' is
related
to t by the Lorentz formula

t' = t \cosh \theta - x/c \sinh \theta (2)

where we can set t=0 and x is the position of the particle measured by K at
time t=0.

On the other hand,

Boost[v]p (3')

is the momentum of the particle measured by observer K' at time t'=0 by his
clock
Therefore, in order to obtain p' in (3) one should apply an additional time
translation
to eq. (3'). The boost and time translation transformations in my original
post
should have been written as

Boost[v]p = p + vM (A)
Time[t] p = p + t L (B)

where both L and M are complex expressions depending on the interaction,
positions/momenta of all particles, and the field configuration. L and M
should be
such that the Poincare algebra commutators are satisfied between all
generators.
I agree that my "proof" of the Poincare-non-invariance of Maxwell's theory
was not correct.

However, I am not ready to give up my major claim yet. The claim is that
eq. (3) is not accurate in interacting theories.

As I understand, your point is that in any interacting theory eq. (3) should
be obtained,
if we first transform momentum to the moving frame (as in eq. (A)), and then
perform a time translation to the time t' in (2). I haven't seen such a
calculation in
Maxwell's theory, and I doubt it will actually yield the Lorentz formula
(3).

My conjecture is that eq. (3) is valid only in non-interacting theories. I
admit that I
do not have a proof for that, but I have a couple of arguments that make
this conjecture
plausible:

1. For particles-only theories this conjecture is proven in the
Currie-Jordan-
Sudarshan theorem. (Though, Igor Khavkine is right that this proof is not
valid for particles+fields theories).

2. There is a thoroughly studied physical system (the decay law of a moving
unstable particle) in which interactions L and M can be rigorously
constructed,
and deviations from the Lorentz transformations formulas (i.e., the
Einstein's
time dilation law) can be explicitly calculated. See references in another
post in this
thread.

Eugene.

Eugene Stefanovich

Sep11-05, 03:33 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:slrndi0f8d.vrl.igor.kh@corum.multive rse.ca...\n> On 2005-09-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n> > For example, take a system of two classical charges interacting\n> > with each other. At time t they are described by positions\n> > x_1(t), x_2(t) and\n> > momenta p_1(t), p_2(t). My approach establishes that transformations\n> > of these observables to the moving frame of reference\n> > [x\'_1(t\'), x\'_2(t\'), p\'_1(t\'), p\'_2(t\'), where t\' is the time\n> > measured by the moving observer] are NOT\n> > given by linear tensor Lorentz formulas (as mandated in SR).\n>\n> The above is not mandated by SR, contrary to what you believe. Well, it\n> would be if SR stood for \'Stefanovich\'s Relativity\'. But it is not in\n> Einstein\'s theory.\n\nThen please tell me what are the transformations of observables in\nEinstein\'s theory.\n\n> > The exact transformations depend on the state of this system and\n> > on the strength of interaction between the particles.\n>\n> The same thing happens in SR.\n\nYes, I am listening.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:slrndi0f8d.vrl.igor.kh@corum.multiverse.ca...
> On 2005-09-07, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
> > For example, take a system of two classical charges interacting
> > with each other. At time t they are described by positions
> > x_1(t), x_2(t) and
> > momenta p_1(t), p_2(t). My approach establishes that transformations
> > of these observables to the moving frame of reference
> > [x'_1(t'), x'_2(t'), p'_1(t'), p'_2(t'), where t' is the time
> > measured by the moving observer] are NOT
> > given by linear tensor Lorentz formulas (as mandated in SR).
>
> The above is not mandated by SR, contrary to what you believe. Well, it
> would be if SR stood for 'Stefanovich's Relativity'. But it is not in
> Einstein's theory.

Then please tell me what are the transformations of observables in
Einstein's theory.

> > The exact transformations depend on the state of this system and
> > on the strength of interaction between the particles.
>
> The same thing happens in SR.

Yes, I am listening.

Eugene.