View Full Version : wavelength of a peaceful electron
vernonner3voltazim
Oct11-04, 01:06 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\nLet me start by defining a particle to be "at peace"\nif it is not doing much interacting with other ordinary\nparticles. The definition might sometimes correspond\nwith that of "at rest", especially when the particle\'s\nvelocity is minimized, but "at rest" has a problem, in\nthat with the Uncertainty Principle operating in the\nbackground, no particle can ever really be "at rest".\n\nThus, helium atoms at Absolute Zero may be "at peace",\nbut because of "zero point energy" (courtesy of the\nUncertainty Principle), they are not actually "at\nrest" and so Absolutely Cold liquid helium does not\ncoalesce into the solid state at ordinary pressure.\n\nNow I can ask you folks a Question:\nConsider an electron that has minimal velocity and\nhappens to be at peace. If it was truly at rest its\nwavelength (courtesy of the wave/particle duality)\nwould be spread widely, even to an indefinite extent,\nbut since it cannot actually be at rest, what is its\n(by preceding explanation: maximal) wavelength?\n\nThanks in advance!\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>Let me start by defining a particle to be "at peace"
if it is not doing much interacting with other ordinary
particles. The definition might sometimes correspond
with that of "at rest", especially when the particle's
velocity is minimized, but "at rest" has a problem, in
that with the Uncertainty Principle operating in the
background, no particle can ever really be "at rest".
Thus, helium atoms at Absolute Zero may be "at peace",
but because of "zero point energy" (courtesy of the
Uncertainty Principle), they are not actually "at
rest" and so Absolutely Cold liquid helium does not
coalesce into the solid state at ordinary pressure.
Now I can ask you folks a Question:
Consider an electron that has minimal velocity and
happens to be at peace. If it was truly at rest its
wavelength (courtesy of the wave/particle duality)
would be spread widely, even to an indefinite extent,
but since it cannot actually be at rest, what is its
(by preceding explanation: maximal) wavelength?
Thanks in advance!
<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\nvnemitz@pinn.net (vernonner3voltazim) wrote in message news:<42336979.0410110906.dfae1dd@posting.google.c om>...\n> Let me start by defining a particle to be "at peace"\n> if it is not doing much interacting with other ordinary\n> particles. The definition might sometimes correspond\n> with that of "at rest", especially when the particle\'s\n> velocity is minimized, but "at rest" has a problem, in\n> that with the Uncertainty Principle operating in the\n> background, no particle can ever really be "at rest".\n>\n> Thus, helium atoms at Absolute Zero may be "at peace",\n> but because of "zero point energy" (courtesy of the\n> Uncertainty Principle), they are not actually "at\n> rest" and so Absolutely Cold liquid helium does not\n> coalesce into the solid state at ordinary pressure.\n>\n> Now I can ask you folks a Question:\n> Consider an electron that has minimal velocity and\n> happens to be at peace. If it was truly at rest its\n> wavelength (courtesy of the wave/particle duality)\n> would be spread widely, even to an indefinite extent,\n> but since it cannot actually be at rest, what is its\n> (by preceding explanation: maximal) wavelength?\n>\n> Thanks in advance!\n\n\nDepends on which wavelength you are referring to. An electron has a\nCompton wavelength equal to hbar/mc, which is a relativistic invariant\nas well as a constant, being only dependent on particle mass. Thus\neveryone agrees on its value. It will also have a de Broglie\nwavelength given by hbar divided by its momentum, which will appear\ndifferently in different Lorentz frames.\n\nIn answer to your question, there is really no limit to particle\nmomentum under the uncertainty principle. What this principle\nactually states is that there will be a relationship between intrinsic\nuncertainties in position and momentum, but not in the definition of\nthese quantities themselves. Since the electron has a finite mass, it\nhas a well-defined rest frame. So we can still talk about an\nelectron at rest, where it would have an infinite de Broglie\nwavelength.\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>vnemitz@pinn.net (vernonner3voltazim) wrote in message news:<42336979.0410110906.dfae1dd@posting.google.com>...
> Let me start by defining a particle to be "at peace"
> if it is not doing much interacting with other ordinary
> particles. The definition might sometimes correspond
> with that of "at rest", especially when the particle's
> velocity is minimized, but "at rest" has a problem, in
> that with the Uncertainty Principle operating in the
> background, no particle can ever really be "at rest".
>
> Thus, helium atoms at Absolute Zero may be "at peace",
> but because of "zero point energy" (courtesy of the
> Uncertainty Principle), they are not actually "at
> rest" and so Absolutely Cold liquid helium does not
> coalesce into the solid state at ordinary pressure.
>
> Now I can ask you folks a Question:
> Consider an electron that has minimal velocity and
> happens to be at peace. If it was truly at rest its
> wavelength (courtesy of the wave/particle duality)
> would be spread widely, even to an indefinite extent,
> but since it cannot actually be at rest, what is its
> (by preceding explanation: maximal) wavelength?
>
> Thanks in advance!
Depends on which wavelength you are referring to. An electron has a
Compton wavelength equal to \hbar/mc, which is a relativistic invariant
as well as a constant, being only dependent on particle mass. Thus
everyone agrees on its value. It will also have a de Broglie
wavelength given by \hbar divided by its momentum, which will appear
differently in different Lorentz frames.
In answer to your question, there is really no limit to particle
momentum under the uncertainty principle. What this principle
actually states is that there will be a relationship between intrinsic
uncertainties in position and momentum, but not in the definition of
these quantities themselves. Since the electron has a finite mass, it
has a well-defined rest frame. So we can still talk about an
electron at rest, where it would have an infinite de Broglie
wavelength.
Uncle Al
Oct12-04, 10:50 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>\nvernonner3voltazim wrote:\n\n[snip]\n\n> Now I can ask you folks a Question:\n> Consider an electron that has minimal velocity and\n> happens to be at peace. If it was truly at rest its\n> wavelength (courtesy of the wave/particle duality)\n> would be spread widely, even to an indefinite extent,\n> but since it cannot actually be at rest, what is its\n> (by preceding explanation: maximal) wavelength?\n\nWhat does "velocity" mean? The velocity observed depends on the\nrelative velocities of inertial observers. Formulate your question so\nit is frame-independent.\n\n--\nUncle Al\nhttp://www.mazepath.com/uncleal/\n(Toxic URL! Unsafe for children and most mammals)\nhttp://www.mazepath.com/uncleal/qz.pdf\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>vernonner3voltazim wrote:
[snip]
> Now I can ask you folks a Question:
> Consider an electron that has minimal velocity and
> happens to be at peace. If it was truly at rest its
> wavelength (courtesy of the wave/particle duality)
> would be spread widely, even to an indefinite extent,
> but since it cannot actually be at rest, what is its
> (by preceding explanation: maximal) wavelength?
What does "velocity" mean? The velocity observed depends on the
relative velocities of inertial observers. Formulate your question so
it is frame-independent.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
vernonner3voltazim
Oct13-04, 02:13 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\n> thoovler@excite.com (Igor) wrote:\n> > vnemitz@pinn.net (vernonner3voltazim) wrote:\n> >\n> > Now I can ask you folks a Question:\n> > Consider an electron that has minimal velocity and\n> > happens to be at peace. If it was truly at rest its\n> > wavelength (courtesy of the wave/particle duality)\n> > would be spread widely, even to an indefinite extent,\n> > but since it cannot actually be at rest, what is its\n> > (by preceding explanation: maximal) wavelength?\n>\n> Depends on which wavelength you are referring to. An\n> electron has a Compton wavelength equal to hbar/mc,\n> which is a relativistic invariant as well as a constant,\n> being only dependent on particle mass. Thus everyone\n> agrees on its value. It will also have a de Broglie\n> wavelength given by hbar divided by its momentum, which\n> will appear differently in different Lorentz frames.\n\nThanks! I think that tells me much of what I wanted\nto know. To Uncle Al, who wanted a particular Lorentz\nframe, I say, "relative to the observer".\n\n> In answer to your question, there is really no limit\n> to particle momentum under the uncertainty principle.\n> What this principle actually states is that there will\n> be a relationship between intrinsic uncertainties in\n> position and momentum, but not in the definition of\n> these quantities themselves.\n\nI think there can be some legitimate disagreement\nabout that. Please recall the Energy/Time formulation\nof the Uncertainty Principle, which is used to allow\nvirtual particles to spontaneously appear from out of\nnowhere on "borrowed" energy. I think that because\nthe Energy/Time and Position/Momentum formulations\nare equivalent, then a particle can borrow momentum\nand an equivalent non-resting position in the same\nway as it can borrow energy for a limited time.\n\n> Since the electron has a finite mass, it has a\n> well-defined rest frame. So we can still talk\n> about an electron at rest, where it would have\n> an infinite de Broglie wavelength.\n\nThat would be consistent with the other part of\nwhat you wrote. However, because of the borrowing-\nequivalence that I described, we should conclude\nthat an electron can Never be truly considered to\nbe At Rest. And so my original Question here is\nto find out, since I assume it\'s always borrowing\na little momentum, what its maximal de Broglie\nwavelength would likely be. In a way, this is\nnot terribly unlike asking, "What is the average\nvirtual-energy content of the vacuum?" Because\nSOME virtual particles are always present,\nexisting on borrowed energy. So, SOME borrowed\nmomentum should be expected to be present, for\nany given particle.\n\nThanks, Again!\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>thoovler@excite.com (Igor) wrote:
> > vnemitz@pinn.net (vernonner3voltazim) wrote:
> >
> > Now I can ask you folks a Question:
> > Consider an electron that has minimal velocity and
> > happens to be at peace. If it was truly at rest its
> > wavelength (courtesy of the wave/particle duality)
> > would be spread widely, even to an indefinite extent,
> > but since it cannot actually be at rest, what is its
> > (by preceding explanation: maximal) wavelength?
>
> Depends on which wavelength you are referring to. An
> electron has a Compton wavelength equal to \hbar/mc,
> which is a relativistic invariant as well as a constant,
> being only dependent on particle mass. Thus everyone
> agrees on its value. It will also have a de Broglie
> wavelength given by \hbar divided by its momentum, which
> will appear differently in different Lorentz frames.
Thanks! I think that tells me much of what I wanted
to know. To Uncle Al, who wanted a particular Lorentz
frame, I say, "relative to the observer".
> In answer to your question, there is really no limit
> to particle momentum under the uncertainty principle.
> What this principle actually states is that there will
> be a relationship between intrinsic uncertainties in
> position and momentum, but not in the definition of
> these quantities themselves.
I think there can be some legitimate disagreement
about that. Please recall the Energy/Time formulation
of the Uncertainty Principle, which is used to allow
virtual particles to spontaneously appear from out of
nowhere on "borrowed" energy. I think that because
the Energy/Time and Position/Momentum formulations
are equivalent, then a particle can borrow momentum
and an equivalent non-resting position in the same
way as it can borrow energy for a limited time.
> Since the electron has a finite mass, it has a
> well-defined rest frame. So we can still talk
> about an electron at rest, where it would have
> an infinite de Broglie wavelength.
That would be consistent with the other part of
what you wrote. However, because of the borrowing-
equivalence that I described, we should conclude
that an electron can Never be truly considered to
be At Rest. And so my original Question here is
to find out, since I assume it's always borrowing
a little momentum, what its maximal de Broglie
wavelength would likely be. In a way, this is
not terribly unlike asking, "What is the average
virtual-energy content of the vacuum?" Because
SOME virtual particles are always present,
existing on borrowed energy. So, SOME borrowed
momentum should be expected to be present, for
any given particle.
Thanks, Again!
<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\nvnemitz@pinn.net (vernonner3voltazim) wrote in message news:<42336979.0410130942.84631c8@posting.google.c om>...\n> > thoovler@excite.com (Igor) wrote:\n> > > vnemitz@pinn.net (vernonner3voltazim) wrote:\n> > >\n> > > Now I can ask you folks a Question:\n> > > Consider an electron that has minimal velocity and\n> > > happens to be at peace. If it was truly at rest its\n> > > wavelength (courtesy of the wave/particle duality)\n> > > would be spread widely, even to an indefinite extent,\n> > > but since it cannot actually be at rest, what is its\n> > > (by preceding explanation: maximal) wavelength?\n> >\n> > Depends on which wavelength you are referring to. An\n> > electron has a Compton wavelength equal to hbar/mc,\n> > which is a relativistic invariant as well as a constant,\n> > being only dependent on particle mass. Thus everyone\n> > agrees on its value. It will also have a de Broglie\n> > wavelength given by hbar divided by its momentum, which\n> > will appear differently in different Lorentz frames.\n>\n> Thanks! I think that tells me much of what I wanted\n> to know. To Uncle Al, who wanted a particular Lorentz\n> frame, I say, "relative to the observer".\n>\n> > In answer to your question, there is really no limit\n> > to particle momentum under the uncertainty principle.\n> > What this principle actually states is that there will\n> > be a relationship between intrinsic uncertainties in\n> > position and momentum, but not in the definition of\n> > these quantities themselves.\n>\n> I think there can be some legitimate disagreement\n> about that. Please recall the Energy/Time formulation\n> of the Uncertainty Principle, which is used to allow\n> virtual particles to spontaneously appear from out of\n> nowhere on "borrowed" energy. I think that because\n> the Energy/Time and Position/Momentum formulations\n> are equivalent, then a particle can borrow momentum\n> and an equivalent non-resting position in the same\n> way as it can borrow energy for a limited time.\n>\n> > Since the electron has a finite mass, it has a\n> > well-defined rest frame. So we can still talk\n> > about an electron at rest, where it would have\n> > an infinite de Broglie wavelength.\n>\n> That would be consistent with the other part of\n> what you wrote. However, because of the borrowing-\n> equivalence that I described, we should conclude\n> that an electron can Never be truly considered to\n> be At Rest. And so my original Question here is\n> to find out, since I assume it\'s always borrowing\n> a little momentum, what its maximal de Broglie\n> wavelength would likely be. In a way, this is\n> not terribly unlike asking, "What is the average\n> virtual-energy content of the vacuum?" Because\n> SOME virtual particles are always present,\n> existing on borrowed energy. So, SOME borrowed\n> momentum should be expected to be present, for\n> any given particle.\n>\n> Thanks, Again!\n\nYou bring up some interesting points, but still relativity tells us\nthat any massive particle has a rest frame. Even while all this\nso-called borrowing is going on, and the electron is moving all over\nthe place relative to an external observer, it will still see itself\nas being at rest relative to everything else around it. So my\noriginal point still holds for the electron itself.\n\nYou\'ll notice that in QM, things like wavepackets are always defined\nwrt an outside observer. The square of the amplitude of the wave\nfunction determines the probability density of locating the particle\nin a particular region. However, an observer in the particle\'s rest\nframe will always have a 100% probability of locating the particle, so\nthe traditional wave function model would seem to break down in that\ncase.\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>vnemitz@pinn.net (vernonner3voltazim) wrote in message news:<42336979.0410130942.84631c8@posting.google.com>...
> > thoovler@excite.com (Igor) wrote:
> > > vnemitz@pinn.net (vernonner3voltazim) wrote:
> > >
> > > Now I can ask you folks a Question:
> > > Consider an electron that has minimal velocity and
> > > happens to be at peace. If it was truly at rest its
> > > wavelength (courtesy of the wave/particle duality)
> > > would be spread widely, even to an indefinite extent,
> > > but since it cannot actually be at rest, what is its
> > > (by preceding explanation: maximal) wavelength?
> >
> > Depends on which wavelength you are referring to. An
> > electron has a Compton wavelength equal to \hbar/mc,
> > which is a relativistic invariant as well as a constant,
> > being only dependent on particle mass. Thus everyone
> > agrees on its value. It will also have a de Broglie
> > wavelength given by \hbar divided by its momentum, which
> > will appear differently in different Lorentz frames.
>
> Thanks! I think that tells me much of what I wanted
> to know. To Uncle Al, who wanted a particular Lorentz
> frame, I say, "relative to the observer".
>
> > In answer to your question, there is really no limit
> > to particle momentum under the uncertainty principle.
> > What this principle actually states is that there will
> > be a relationship between intrinsic uncertainties in
> > position and momentum, but not in the definition of
> > these quantities themselves.
>
> I think there can be some legitimate disagreement
> about that. Please recall the Energy/Time formulation
> of the Uncertainty Principle, which is used to allow
> virtual particles to spontaneously appear from out of
> nowhere on "borrowed" energy. I think that because
> the Energy/Time and Position/Momentum formulations
> are equivalent, then a particle can borrow momentum
> and an equivalent non-resting position in the same
> way as it can borrow energy for a limited time.
>
> > Since the electron has a finite mass, it has a
> > well-defined rest frame. So we can still talk
> > about an electron at rest, where it would have
> > an infinite de Broglie wavelength.
>
> That would be consistent with the other part of
> what you wrote. However, because of the borrowing-
> equivalence that I described, we should conclude
> that an electron can Never be truly considered to
> be At Rest. And so my original Question here is
> to find out, since I assume it's always borrowing
> a little momentum, what its maximal de Broglie
> wavelength would likely be. In a way, this is
> not terribly unlike asking, "What is the average
> virtual-energy content of the vacuum?" Because
> SOME virtual particles are always present,
> existing on borrowed energy. So, SOME borrowed
> momentum should be expected to be present, for
> any given particle.
>
> Thanks, Again!
You bring up some interesting points, but still relativity tells us
that any massive particle has a rest frame. Even while all this
so-called borrowing is going on, and the electron is moving all over
the place relative to an external observer, it will still see itself
as being at rest relative to everything else around it. So my
original point still holds for the electron itself.
You'll notice that in QM, things like wavepackets are always defined
wrt an outside observer. The square of the amplitude of the wave
function determines the probability density of locating the particle
in a particular region. However, an observer in the particle's rest
frame will always have a 100% probability of locating the particle, so
the traditional wave function model would seem to break down in that
case.
vernonner3voltazim
Oct19-04, 03:29 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\n> thoovler@excite.com (Igor) wrote:\n> > vnemitz@pinn.net (vernonner3voltazim) wrote:\n> >\n> > However, because of the borrowing-\n> > equivalence that I described, we should conclude\n> > that an electron can Never be truly considered to\n> > be At Rest. And so my original Question here is\n> > to find out, since I assume it\'s always borrowing\n> > a little momentum, what its maximal de Broglie\n> > wavelength would likely be. In a way, this is\n> > not terribly unlike asking, "What is the average\n> > virtual-energy content of the vacuum?" Because\n> > SOME virtual particles are always present,\n> > existing on borrowed energy. So, SOME borrowed\n> > momentum should be expected to be present, for\n> > any given particle.\n> >\n> You bring up some interesting points, but still\n> relativity tells us that any massive particle has\n> a rest frame.\n> >\nWell, OK, but we both know that Relativity (at least\nthe General version) has incompatibilities with QM.\nWhat you wrote at the end looks to me like another\nincompatibility.\n> >\n> Even while all this so-called\n> borrowing is going on, and the electron is moving\n> all over the place relative to an external observer,\n> it will still see itself as being at rest relative\n> to everything else around it. So my original point\n> still holds for the electron itself.\n> >\nMaybe. On the other hand, from a pure QM perspective,\nevery particle is going to be seeing every OTHER\nparticle jumping all around every supposed "resting"\nposition. I don\'t think any observer-particle can\ntell if it is itself also doing a lot of jumping.\nThat\'s why it might "THINK" itself is at rest....\n> >\n> You\'ll notice that in QM, things like wavepackets\n> are always defined wrt an outside observer. The\n> square of the amplitude of the wave function\n> determines the probability density of locating\n> the particle in a particular region.\n> >\nYes, and that relates to why I asked the main Question\nin this Thread. If a particle could truly be at rest\n(no borrowing), then its de Broglie wavelength will\nbe infinitely large, and its "particular region" might\nencompass the entire Universe. On the other hand,\nconsider the special case of liquid helium at Absolute\nZero: Even AT that temperature it will remain liquid,\ncourtesy of "zero point motion" -- energy/momentum\nborrowed via Uncertainty, that is. I think it\nreasonable to conclude that ALL particles AT Absolute\nZero (the MOST "at-rest" state allowed, right?) are\ngoing to be experiencing zero point motion. Thus,\nWITH borrowing, there is some maximal de Broglie\nwavelength for any particle, and its "particular\nregion" should never be just as limited. For an\nelectron, that\'s what I was asking about here.\n> >\n> However, an\n> observer in the particle\'s rest frame will always\n> have a 100% probability of locating the particle,\n> so the traditional wave function model would seem\n> to break down in that case.\n> >\n"Incompatible" theories....\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>thoovler@excite.com (Igor) wrote:
> > vnemitz@pinn.net (vernonner3voltazim) wrote:
> >
> > However, because of the borrowing-
> > equivalence that I described, we should conclude
> > that an electron can Never be truly considered to
> > be At Rest. And so my original Question here is
> > to find out, since I assume it's always borrowing
> > a little momentum, what its maximal de Broglie
> > wavelength would likely be. In a way, this is
> > not terribly unlike asking, "What is the average
> > virtual-energy content of the vacuum?" Because
> > SOME virtual particles are always present,
> > existing on borrowed energy. So, SOME borrowed
> > momentum should be expected to be present, for
> > any given particle.
> >
> You bring up some interesting points, but still
> relativity tells us that any massive particle has
> a rest frame.
> >
Well, OK, but we both know that Relativity (at least
the General version) has incompatibilities with QM.
What you wrote at the end looks to me like another
incompatibility.
> >
> Even while all this so-called
> borrowing is going on, and the electron is moving
> all over the place relative to an external observer,
> it will still see itself as being at rest relative
> to everything else around it. So my original point
> still holds for the electron itself.
> >
Maybe. On the other hand, from a pure QM perspective,
every particle is going to be seeing every OTHER
particle jumping all around every supposed "resting"
position. I don't think any observer-particle can
tell if it is itself also doing a lot of jumping.
That's why it might "THINK" itself is at rest....
> >
> You'll notice that in QM, things like wavepackets
> are always defined wrt an outside observer. The
> square of the amplitude of the wave function
> determines the probability density of locating
> the particle in a particular region.
> >
Yes, and that relates to why I asked the main Question
in this Thread. If a particle could truly be at rest
(no borrowing), then its de Broglie wavelength will
be infinitely large, and its "particular region" might
encompass the entire Universe. On the other hand,
consider the special case of liquid helium at Absolute
Zero: Even AT that temperature it will remain liquid,
courtesy of "zero point motion" -- energy/momentum
borrowed via Uncertainty, that is. I think it
reasonable to conclude that ALL particles AT Absolute
Zero (the MOST "at-rest" state allowed, right?) are
going to be experiencing zero point motion. Thus,
WITH borrowing, there is some maximal de Broglie
wavelength for any particle, and its "particular
region" should never be just as limited. For an
electron, that's what I was asking about here.
> >
> However, an
> observer in the particle's rest frame will always
> have a 100% probability of locating the particle,
> so the traditional wave function model would seem
> to break down in that case.
> >
"Incompatible" theories....
<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 <thoovler@excite.com> writes\n>In answer to your question, there is really no limit to particle\n>momentum under the uncertainty principle. What this principle\n>actually states is that there will be a relationship between intrinsic\n>uncertainties in position and momentum, but not in the definition of\n>these quantities themselves. Since the electron has a finite mass, it\n>has a well-defined rest frame. So we can still talk about an\n>electron at rest, where it would have an infinite de Broglie\n>wavelength.\n\nYes, but that doesn\'t mean we can observe such a thing.\n\nClearly in a universe that contained just one particle (an electron),\nposition would be meaningless.\n\nIf we observe an electron then, surely, zero point energy would click in\nand the momentum of the electron would, could, never be precisely zero.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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 <thoovler@excite.com> writes
>In answer to your question, there is really no limit to particle
>momentum under the uncertainty principle. What this principle
>actually states is that there will be a relationship between intrinsic
>uncertainties in position and momentum, but not in the definition of
>these quantities themselves. Since the electron has a finite mass, it
>has a well-defined rest frame. So we can still talk about an
>electron at rest, where it would have an infinite de Broglie
>wavelength.
Yes, but that doesn't mean we can observe such a thing.
Clearly in a universe that contained just one particle (an electron),
position would be meaningless.
If we observe an electron then, surely, zero point energy would click in
and the momentum of the electron would, could, never be precisely zero.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
<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>\nIgor <thoovler@excite.com> writes\n\n>You bring up some interesting points, but still relativity tells us\n>that any massive particle has a rest frame. Even while all this\n>so-called borrowing is going on, and the electron is moving all over\n>the place relative to an external observer, it will still see itself\n>as being at rest relative to everything else around it. So my\n>original point still holds for the electron itself.\n\nYes, but that isn\'t answering the question.\n\n>You\'ll notice that in QM, things like wavepackets are always defined\n>wrt an outside observer. The square of the amplitude of the wave\n>function determines the probability density of locating the particle\n>in a particular region.\n\nThat was the OP\'s point. When an electron (presumably sitting in the\nvacuum) has the lowest energy state it can be seen to reach, what is its\ndistribution in space.\n\n>However, an observer in the particle\'s rest\n>frame will always have a 100% probability of locating the particle, so\n>the traditional wave function model would seem to break down in that\n>case.\n\nNo, because both observer and electron live in a vacuum with a positive\nfixed point energy. So, by definition, the observer and the electron can\nnever share the exact same rest frame.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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 <thoovler@excite.com> writes
>You bring up some interesting points, but still relativity tells us
>that any massive particle has a rest frame. Even while all this
>so-called borrowing is going on, and the electron is moving all over
>the place relative to an external observer, it will still see itself
>as being at rest relative to everything else around it. So my
>original point still holds for the electron itself.
Yes, but that isn't answering the question.
>You'll notice that in QM, things like wavepackets are always defined
>wrt an outside observer. The square of the amplitude of the wave
>function determines the probability density of locating the particle
>in a particular region.
That was the OP's point. When an electron (presumably sitting in the
vacuum) has the lowest energy state it can be seen to reach, what is its
distribution in space.
>However, an observer in the particle's rest
>frame will always have a 100% probability of locating the particle, so
>the traditional wave function model would seem to break down in that
>case.
No, because both observer and electron live in a vacuum with a positive
fixed point energy. So, by definition, the observer and the electron can
never share the exact same rest frame.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
<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\nOz <oz@farmeroz.port995.com> wrote in message news:<t53UWuAlk6cBFw5S@farmeroz.port995.com>...\n> Igor <thoovler@excite.com> writes\n>\n> >You bring up some interesting points, but still relativity tells us\n> >that any massive particle has a rest frame. Even while all this\n> >so-called borrowing is going on, and the electron is moving all over\n> >the place relative to an external observer, it will still see itself\n> >as being at rest relative to everything else around it. So my\n> >original point still holds for the electron itself.\n>\n> Yes, but that isn\'t answering the question.\n>\n> >You\'ll notice that in QM, things like wavepackets are always defined\n> >wrt an outside observer. The square of the amplitude of the wave\n> >function determines the probability density of locating the particle\n> >in a particular region.\n>\n> That was the OP\'s point. When an electron (presumably sitting in the\n> vacuum) has the lowest energy state it can be seen to reach, what is its\n> distribution in space.\n>\n> >However, an observer in the particle\'s rest\n> >frame will always have a 100% probability of locating the particle, so\n> >the traditional wave function model would seem to break down in that\n> >case.\n>\n> No, because both observer and electron live in a vacuum with a positive\n> fixed point energy. So, by definition, the observer and the electron can\n> never share the exact same rest frame.\n\n\nBut I\'m not referring to an "external" observer. I\'m talking about\nthe electron\'s rest frame itself. So I don\'t understand your\nobjection.\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>Oz <oz@farmeroz.port995.com> wrote in message news:<t53UWuAlk6cBFw5S@farmeroz.port995.com>...
> Igor <thoovler@excite.com> writes
>
> >You bring up some interesting points, but still relativity tells us
> >that any massive particle has a rest frame. Even while all this
> >so-called borrowing is going on, and the electron is moving all over
> >the place relative to an external observer, it will still see itself
> >as being at rest relative to everything else around it. So my
> >original point still holds for the electron itself.
>
> Yes, but that isn't answering the question.
>
> >You'll notice that in QM, things like wavepackets are always defined
> >wrt an outside observer. The square of the amplitude of the wave
> >function determines the probability density of locating the particle
> >in a particular region.
>
> That was the OP's point. When an electron (presumably sitting in the
> vacuum) has the lowest energy state it can be seen to reach, what is its
> distribution in space.
>
> >However, an observer in the particle's rest
> >frame will always have a 100% probability of locating the particle, so
> >the traditional wave function model would seem to break down in that
> >case.
>
> No, because both observer and electron live in a vacuum with a positive
> fixed point energy. So, by definition, the observer and the electron can
> never share the exact same rest frame.
But I'm not referring to an "external" observer. I'm talking about
the electron's rest frame itself. So I don't understand your
objection.
<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 <thoovler@excite.com> writes\n\n>But I\'m not referring to an "external" observer. I\'m talking about\n>the electron\'s rest frame itself. So I don\'t understand your\n>objection.\n\nIts meaningless to describe the \'position\' of a particle in a universe\nthat contains only that one particle. Come to that describing it\'s \'rest\nframe\' has no meaning either.\n\nDistance is relative.\n\nIt only makes sense to describe the position and even the rest frame\nwith respect to another particle. That, of necessity, demands one\nconsider zero point energy.\n\nHence the OP\'s question.\n\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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 <thoovler@excite.com> writes
>But I'm not referring to an "external" observer. I'm talking about
>the electron's rest frame itself. So I don't understand your
>objection.
Its meaningless to describe the 'position' of a particle in a universe
that contains only that one particle. Come to that describing it's 'rest
frame' has no meaning either.
Distance is relative.
It only makes sense to describe the position and even the rest frame
with respect to another particle. That, of necessity, demands one
consider zero point energy.
Hence the OP's question.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
vernonner3voltazim
Oct21-04, 02:47 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\n>thoovler@excite.com (Igor) wrote:\n> >Oz <oz@farmeroz.port995.com> wrote:\n> > >Igor <thoovler@excite.com> writes\n> >\n> > > However, an observer in the particle\'s rest\n> > > frame will always have a 100% probability of\n> > > locating the particle, so the traditional\n> > > wave function model would seem to break down\n> > > in that case.\n> >\n> > No, because both observer and electron live in\n> > a vacuum with a positive fixed point energy.\n> > So, by definition, the observer and the electron\n> > can never share the exact same rest frame.\n>\n> But I\'m not referring to an "external" observer.\n> I\'m talking about the electron\'s rest frame itself.\n> So I don\'t understand your objection.\n\nThis business of switching reference frames for no\nreason needs to stop. From my original post is:\n"Consider an electron that has minimal velocity and\nhappens to be at peace." And I specifically\nreferred to a "preceding explanation" in that post,\nin which WE are doing the observing/considering,\nnot the electron. Which means I\'m asking about the\nelectron from the observer\'s point of view.\nI don\'t care if in its own reference frame whether\nit or anything else is always at Perfect Rest. In\nthe Real World, we most often describe how things\nlook RELATIVE TO US (and they are never still...).\n\nSo, what IS the de Broglie wavelength of a peaceful\nelectron? Thanks!\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>>thoovler@excite.com (Igor) wrote:
> >Oz <oz@farmeroz.port995.com> wrote:
> > >Igor <thoovler@excite.com> writes
> >
> > > However, an observer in the particle's rest
> > > frame will always have a 100% probability of
> > > locating the particle, so the traditional
> > > wave function model would seem to break down
> > > in that case.
> >
> > No, because both observer and electron live in
> > a vacuum with a positive fixed point energy.
> > So, by definition, the observer and the electron
> > can never share the exact same rest frame.
>
> But I'm not referring to an "external" observer.
> I'm talking about the electron's rest frame itself.
> So I don't understand your objection.
This business of switching reference frames for no
reason needs to stop. From my original post is:
"Consider an electron that has minimal velocity and
happens to be at peace." And I specifically
referred to a "preceding explanation" in that post,
in which WE are doing the observing/considering,
not the electron. Which means I'm asking about the
electron from the observer's point of view.
I don't care if in its own reference frame whether
it or anything else is always at Perfect Rest. In
the Real World, we most often describe how things
look RELATIVE TO US (and they are never still...).
So, what IS the de Broglie wavelength of a peaceful
electron? Thanks!
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