View Full Version : [SOLVED] Defining physical variables
Phil Gardner
Apr21-04, 03:25 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>Any variable that we use to describe or model physical phenomena can\nbe defined in one or both of two ways:\n\n(1) An operational definition (an OD) - a set of procedures by which\nthe absolute value of the variable can be measured;\n\n(2) A functional definition (an FD) - an equation which defines the\nvariable as a function of other variables, all of which are defined by\nODs.\n\nIf someone writes an FD for a variable that can be defined by an OD\nthis FD is clearly a testable hypothesis that can be disproved by\nexperiment.\n\nIf however someone writes an FD for a variable that cannot be defined\nby an OD this FD is clearly an untestable hypothesis.\n\nTo take one specific example, a widely accepted FD for the total\nenergy, E, of a particle is E^2 = m^2 c^4 + p^2 c^2. I cannot think\nof an OD for E. Can anyone?\n\nPhil Gardner\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>Any variable that we use to describe or model physical phenomena can
be defined in one or both of two ways:
(1) An operational definition (an OD) - a set of procedures by which
the absolute value of the variable can be measured;
(2) A functional definition (an FD) - an equation which defines the
variable as a function of other variables, all of which are defined by
ODs.
If someone writes an FD for a variable that can be defined by an OD
this FD is clearly a testable hypothesis that can be disproved by
experiment.
If however someone writes an FD for a variable that cannot be defined
by an OD this FD is clearly an untestable hypothesis.
To take one specific example, a widely accepted FD for the total
energy, E, of a particle is E^2 = m^2 c^4 + p^2 c^2. I cannot think
of an OD for E. Can anyone?
Phil Gardner
Doug Sweetser
Apr22-04, 03:25 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>Hello:\n\nIs anyone else bothered by this expression?\n\n> E^2 = m^2 c^4 + p^2 c^2.\n\nI consistently think of E and p as components of a 4-vector, that when\ngiven a metric can be used to calculate the Lorentz invariant scalar m.\nFor that reason, I always write:\n\nm^2 c^4 = E^2 - p^2 c^2\n\nThe lefthand side is the Lorentz scalar, the right a contraction of a\n4-vector. The first expression doesn\'t translate well into tensor\nlingo.\n\ndoug\nquaternions.com\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>Hello:
Is anyone else bothered by this expression?
> E^2 = m^2 c^4 + p^2 c^2.
I consistently think of E and p as components of a 4-vector, that when
given a metric can be used to calculate the Lorentz invariant scalar m.
For that reason, I always write:
m^2 c^4 = E^2 - p^2 c^2
The lefthand side is the Lorentz scalar, the right a contraction of a
4-vector. The first expression doesn't translate well into tensor
lingo.
doug
quaternions.com
Mark Palenik
Apr23-04, 03:24 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"Doug Sweetser" <sweetser@alum.mit.edu> wrote in message\nnews:c67g7a\\$bck\\$1@pcls4.std.com...\n> Hello:\n>\n> Is anyone else bothered by this expression?\n>\n> > E^2 = m^2 c^4 + p^2 c^2.\n>\n> I consistently think of E and p as components of a 4-vector, that when\n> given a metric can be used to calculate the Lorentz invariant scalar m.\n> For that reason, I always write:\n>\n> m^2 c^4 = E^2 - p^2 c^2\n>\n> The lefthand side is the Lorentz scalar, the right a contraction of a\n> 4-vector. The first expression doesn\'t translate well into tensor\n> lingo.\n>\n\nYeah, but if you want to calculate the energy, it\'s more useful to put it in\nthe first form. Presumably, we know m.\n\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>"Doug Sweetser" <sweetser@alum.mit.edu> wrote in message
news:c67g7a$bck$1@pcls4.std.com...
> Hello:
>
> Is anyone else bothered by this expression?
>
> > E^2 = m^2 c^4 + p^2 c^2.
>
> I consistently think of E and p as components of a 4-vector, that when
> given a metric can be used to calculate the Lorentz invariant scalar m.
> For that reason, I always write:
>
> m^2 c^4 = E^2 - p^2 c^2
>
> The lefthand side is the Lorentz scalar, the right a contraction of a
> 4-vector. The first expression doesn't translate well into tensor
> lingo.
>
Yeah, but if you want to calculate the energy, it's more useful to put it in
the first form. Presumably, we know m.
Siew-Ann Cheong
Apr24-04, 11:16 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>Phil Gardner <pej_dg@dodo.com.au> wrote:\n: Any variable that we use to describe or model physical phenomena can\n: be defined in one or both of two ways:\n:\n: (1) An operational definition (an OD) - a set of procedures by which\n: the absolute value of the variable can be measured;\n:\n: (2) A functional definition (an FD) - an equation which defines the\n: variable as a function of other variables, all of which are defined by\n: ODs.\n:\n: If someone writes an FD for a variable that can be defined by an OD\n: this FD is clearly a testable hypothesis that can be disproved by\n: experiment.\n:\n: If however someone writes an FD for a variable that cannot be defined\n: by an OD this FD is clearly an untestable hypothesis.\n:\n: To take one specific example, a widely accepted FD for the total\n: energy, E, of a particle is E^2 = m^2 c^4 + p^2 c^2. I cannot think\n: of an OD for E. Can anyone?\n\nm and c can both be defined operationally.\n\nThe kinetic energy T = (\\gamma - 1) m c^2 can be defined operationally.\n\nOne then have E = T + m c^2.\n\nIs that what you sought?\n\n\nSiew-Ann Cheong\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>Phil Gardner <pej_dg@dodo.com.au> wrote:
: Any variable that we use to describe or model physical phenomena can
: be defined in one or both of two ways:
:
: (1) An operational definition (an OD) - a set of procedures by which
: the absolute value of the variable can be measured;
:
: (2) A functional definition (an FD) - an equation which defines the
: variable as a function of other variables, all of which are defined by
: ODs.
:
: If someone writes an FD for a variable that can be defined by an OD
: this FD is clearly a testable hypothesis that can be disproved by
: experiment.
:
: If however someone writes an FD for a variable that cannot be defined
: by an OD this FD is clearly an untestable hypothesis.
:
: To take one specific example, a widely accepted FD for the total
: energy, E, of a particle is E^2 = m^2 c^4 + p^2 c^2. I cannot think
: of an OD for E. Can anyone?
m and c can both be defined operationally.
The kinetic energy T = (\gamma - 1) m c^2 can be defined operationally.
One then have E = T + m c^2.
Is that what you sought?
Siew-Ann Cheong
FrediFizzx
Apr24-04, 11:18 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>"Doug Sweetser" <sweetser@alum.mit.edu> wrote in message\nnews:c67g7a\\$bck\\$1@pcls4.std.com...\n| Hello:\n|\n| Is anyone else bothered by this expression?\n|\n| > E^2 = m^2 c^4 + p^2 c^2.\n\nIt bothers the heck out of me. I think it should be written,\n\n|E^2| = |m^2| c^4 + p^2 c^2.\n\nOr your way with,\n\n|m^2| c^4 = |E^2| - p^2 c^2.\n\nSo there can\'t be such things as negative mass or energy. But why should\nthe absolute value signs appear?\n\nFrediFizzx\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>"Doug Sweetser" <sweetser@alum.mit.edu> wrote in message
news:c67g7a$bck$1@pcls4.std.com...
| Hello:
|
| Is anyone else bothered by this expression?
|
| > E^2 = m^2 c^4 + p^2 c^2.
It bothers the heck out of me. I think it should be written,
|E^2| = |m^2| c^4 + p^2 c^2.
Or your way with,
|m^2| c^4 = |E^2| - p^2 c^2.
So there can't be such things as negative mass or energy. But why should
the absolute value signs appear?
FrediFizzx
akhaneles@yahoo.com
Apr24-04, 08:16 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>pej_dg@dodo.com.au (Phil Gardner) wrote in message news:<ea961a86.0404210336.4a10957d@posting.google. com>...\n> Any variable that we use to describe or model physical phenomena can\n> be defined in one or both of two ways:\n>\n> (1) An operational definition (an OD) - a set of procedures by which\n> the absolute value of the variable can be measured;\n>\n> (2) A functional definition (an FD) - an equation which defines the\n> variable as a function of other variables, all of which are defined by\n> ODs.\n>\n> If someone writes an FD for a variable that can be defined by an OD\n> this FD is clearly a testable hypothesis that can be disproved by\n> experiment.\n>\n> If however someone writes an FD for a variable that cannot be defined\n> by an OD this FD is clearly an untestable hypothesis.\n>\n> To take one specific example, a widely accepted FD for the total\n> energy, E, of a particle is E^2 = m^2 c^4 + p^2 c^2. I cannot think\n> of an OD for E. Can anyone?\n>\n> Phil Gardner\n\nSomething is weird here. If you look at two extreme cases:\nA. Photon: E^2 = p^2 c^2 - (vector)\nB. Massive particle at rest: E^2 = m^2 c^4 - (scalar)\n\nThe tautology (B) between (scalar) energy and mass for particle at\nrest is understandable. Do we have the same kind of tautology (A)\nbetween energy and (vector) momentum of photon?\n\nWhat would be the recipe to choose between vector and scalar rules for\nenergy/momentum components of photons?\nE = Ex + Ey + Ez = (Px + Py + Pz) c\nE = sqrt(Px^2 + Py^2 + Pz^2) c\n\nAlex\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>pej_dg@dodo.com.au (Phil Gardner) wrote in message news:<ea961a86.0404210336.4a10957d@posting.google.com>...
> Any variable that we use to describe or model physical phenomena can
> be defined in one or both of two ways:
>
> (1) An operational definition (an OD) - a set of procedures by which
> the absolute value of the variable can be measured;
>
> (2) A functional definition (an FD) - an equation which defines the
> variable as a function of other variables, all of which are defined by
> ODs.
>
> If someone writes an FD for a variable that can be defined by an OD
> this FD is clearly a testable hypothesis that can be disproved by
> experiment.
>
> If however someone writes an FD for a variable that cannot be defined
> by an OD this FD is clearly an untestable hypothesis.
>
> To take one specific example, a widely accepted FD for the total
> energy, E, of a particle is E^2 = m^2 c^4 + p^2 c^2. I cannot think
> of an OD for E. Can anyone?
>
> Phil Gardner
Something is weird here. If you look at two extreme cases:
A. Photon: E^2 = p^2 c^2 - (vector)
B. Massive particle at rest: E^2 = m^2 c^4 - (scalar)
The tautology (B) between (scalar) energy and mass for particle at
rest is understandable. Do we have the same kind of tautology (A)
between energy and (vector) momentum of photon?
What would be the recipe to choose between vector and scalar rules for
energy/momentum components of photons?
E = Ex + Ey + Ez = (Px + Py + Pz) cE = \sqrt(Px^2 + Py^2 + Pz^2) c
Alex
Doug Sweetser
Apr28-04, 01:49 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>Hello FrediFizzx:\n\nThe reason signs are tricky is the Minkowski pseudo-metric. It is that\npseudo that creates these issues, whether for events in spacetime or\nthe relationships between mass, energy and momentum (both have the same\nunderlying geometry).\n\nRecently I have come up with a simple way to manage data so that\nMinkowski\'s tensor behaves like a metric (if anyone has a reference to\nthis approach, please tell me). For any collection of data, should t\n>|R/c|, or E > |Pc|, graph as usual. This is for points in that past\nand future lightcone. For this set, the Minkowski metric is positive\ndefinite, as a metric should be :-)\n\nFor the data in the opposite case, where t < |R/c|, or E < |Pc|, I think\nthose points should be plotted on a complete different graph, one where\nthe axes are imaginary. One cannot get to an event that is spacelike\nseparated, nor travel an imaginary distance R, so this imaginary\ntangent space is consistent with the definition of spacelike\nseparation, "You can\'t get there from here." For this set of data, the\nMinkowski metric is again positive definite.\n\nYou can have negative energy. It is known as potential energy. What\ncauses problems in quantum field theory is E < |Pc| because it\nindicates negative probabilities. That issue evaporates if such data\ngets plotted on a graph with imaginary axes.\n\ndoug\nquaternions.com\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>Hello FrediFizzx:
The reason signs are tricky is the Minkowski pseudo-metric. It is that
pseudo that creates these issues, whether for events in spacetime or
the relationships between mass, energy and momentum (both have the same
underlying geometry).
Recently I have come up with a simple way to manage data so that
Minkowski's tensor behaves like a metric (if anyone has a reference to
this approach, please tell me). For any collection of data, should t
>|R/c|, or E > |Pc|, graph as usual. This is for points in that past
and future lightcone. For this set, the Minkowski metric is positive
definite, as a metric should be :-)
For the data in the opposite case, where t < |R/c|, or E < |Pc|, I think
those points should be plotted on a complete different graph, one where
the axes are imaginary. One cannot get to an event that is spacelike
separated, nor travel an imaginary distance R, so this imaginary
tangent space is consistent with the definition of spacelike
separation, "You can't get there from here." For this set of data, the
Minkowski metric is again positive definite.
You can have negative energy. It is known as potential energy. What
causes problems in quantum field theory is E < |Pc| because it
indicates negative probabilities. That issue evaporates if such data
gets plotted on a graph with imaginary axes.
doug
quaternions.com
John Baez
May29-04, 11:53 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>In article <c6aeeo\\$9p3vr\\$1@ID-185976.news.uni-berlin.de>,\nFrediFizzx <fredifizzx@hotmail.com> wrote:\n\n>"Doug Sweetser" <sweetser@alum.mit.edu> wrote in message\n>news:c67g7a\\$bck\\$1@pcls4.std.com...\n \n>| Is anyone else bothered by this expression?\n>|\n>| > E^2 = m^2 c^4 + p^2 c^2.\n\nNo... what with the war in Iraq and so on, this is pretty\nfar down on my list of worries.\n\n>I consistently think of E and p as components of a 4-vector, that when\n>given a metric can be used to calculate the Lorentz invariant scalar m.\n>For that reason, I always write:\n>\n> m^2 c^4 = E^2 - p^2 c^2\n\nFine, you\'ve got a point - but there\'s a trick called subtraction\nthat shows this is equivalen to the equation you don\'t like. :-)\n\n>It bothers the heck out of me. I think it should be written,\n>\n> |E^2| = |m^2| c^4 + p^2 c^2.\n\nThis would give the wrong answer for tachyons.\n\n>So there can\'t be such things as negative mass or energy.\n\nWhat appears in these equations is not mass and energy, but\ntheir squares, so negative mass or energy don\'t affect anything\nhere - they\'re irrelevant.\n\nTachyons have negative mass^2, and your formula would be wrong for\nthose. And if you don\'t care about tachyons, then m^2 and p^2\nwill always be positive for you, so why bother sticking in absolute\nvalues?\n\n>But why should the absolute value signs appear?\n\nWhy indeed? The answer is: they shouldn\'t.\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>In article <c6aeeo$9p3vr$1@ID-185976.news.uni-berlin.de>,
FrediFizzx <fredifizzx@hotmail.com> wrote:
>"Doug Sweetser" <sweetser@alum.mit.edu> wrote in message
>news:c67g7a$bck$1@pcls4.std.com...
>| Is anyone else bothered by this expression?
>|
>| > E^2 = m^2 c^4 + p^2 c^2.
No... what with the war in Iraq and so on, this is pretty
far down on my list of worries.
>I consistently think of E and p as components of a 4-vector, that when
>given a metric can be used to calculate the Lorentz invariant scalar m.
>For that reason, I always write:
>
> m^2 c^4 = E^2 - p^2 c^2
Fine, you've got a point - but there's a trick called subtraction
that shows this is equivalen to the equation you don't like. :-)
>It bothers the heck out of me. I think it should be written,
>
> |E^2| = |m^2| c^4 + p^2 c^2.
This would give the wrong answer for tachyons.
>So there can't be such things as negative mass or energy.
What appears in these equations is not mass and energy, but
their squares, so negative mass or energy don't affect anything
here - they're irrelevant.
Tachyons have negative mass^2, and your formula would be wrong for
those. And if you don't care about tachyons, then m^2 and p^2
will always be positive for you, so why bother sticking in absolute
values?
>But why should the absolute value signs appear?
Why indeed? The answer is: they shouldn't.
FrediFizzx
May31-04, 04:22 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"John Baez" <baez@galaxy.ucr.edu> wrote in message\nnews:c91145\\$or4\\$1@glue.ucr.edu...\n| In article <c6aeeo\\$9p3vr\\$1@ID-185976.news.uni-berlin.de>,\n| FrediFizzx <fredifizzx@hotmail.com> wrote:\n\n| >It bothers the heck out of me. I think it should be written,\n| >\n| > |E^2| = |m^2| c^4 + p^2 c^2.\n|\n| This would give the wrong answer for tachyons.\n|\n| >So there can\'t be such things as negative mass or energy.\n|\n| What appears in these equations is not mass and energy, but\n| their squares, so negative mass or energy don\'t affect anything\n| here - they\'re irrelevant.\n|\n| Tachyons have negative mass^2, and your formula would be wrong for\n| those. And if you don\'t care about tachyons, then m^2 and p^2\n| will always be positive for you, so why bother sticking in absolute\n| values?\n\nYes, I don\'t much care for tachyons. I don\'t much care for negative mass or\nenergy either unless someone could show me negative length.\n\n| >But why should the absolute value signs appear?\n|\n| Why indeed? The answer is: they shouldn\'t.\n\nSo even if solving for E or m, we just throw out the negative solutions if\nwe believe they don\'t represent something physical? I suppose that is\nsimple enough. Just doesn\'t seem "pure" though.\n\nFrediFizzx\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>"John Baez" <baez@galaxy.ucr.edu> wrote in message
news:c91145$or4$1@glue.ucr.edu...
| In article <c6aeeo$9p3vr$1@ID-185976.news.uni-berlin.de>,
| FrediFizzx <fredifizzx@hotmail.com> wrote:
| >It bothers the heck out of me. I think it should be written,
| >
| > |E^2| = |m^2| c^4 + p^2 c^2.
|
| This would give the wrong answer for tachyons.
|
| >So there can't be such things as negative mass or energy.
|
| What appears in these equations is not mass and energy, but
| their squares, so negative mass or energy don't affect anything
| here - they're irrelevant.
|
| Tachyons have negative mass^2, and your formula would be wrong for
| those. And if you don't care about tachyons, then m^2 and p^2
| will always be positive for you, so why bother sticking in absolute
| values?
Yes, I don't much care for tachyons. I don't much care for negative mass or
energy either unless someone could show me negative length.
| >But why should the absolute value signs appear?
|
| Why indeed? The answer is: they shouldn't.
So even if solving for E or m, we just throw out the negative solutions if
we believe they don't represent something physical? I suppose that is
simple enough. Just doesn't seem "pure" though.
FrediFizzx
Frank Hellmann
Jun1-04, 02:12 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>"FrediFizzx" <fredifizzx@hotmail.com> wrote in message news:<2hvsu4FggoocU1@uni-berlin.de>...\n\n> So even if solving for E or m, we just throw out the negative solutions if\n> we believe they don\'t represent something physical? I suppose that is\n> simple enough. Just doesn\'t seem "pure" though.\n>\n\nYou can see (somewhat heuristically, if anybody want\'s to expand I\'d\nbe glad) that the negative energy solutions can\'t interact with the\npositive energy sollutions due to the \'mass gap\'. In classical\nmechanics every interaction is smooth and smoothness prevents a\nparticle to go from +m to -m immidiately\n\nIf you do QFT, E is really a frequency and the Energy Expectation\nvalue is always well defined and positive, negative frequency parts of\nthe field correspond to antiparticles.\n\n(Remember Diracs negative energy electron \'sea\', that too introduced\nanti particles to interpretate the negative energy bits)\n\n---\nfrank\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>"FrediFizzx" <fredifizzx@hotmail.com> wrote in message news:<2hvsu4FggoocU1@uni-berlin.de>...
> So even if solving for E or m, we just throw out the negative solutions if
> we believe they don't represent something physical? I suppose that is
> simple enough. Just doesn't seem "pure" though.
>
You can see (somewhat heuristically, if anybody want's to expand I'd
be glad) that the negative energy solutions can't interact with the
positive energy sollutions due to the 'mass gap'. In classical
mechanics every interaction is smooth and smoothness prevents a
particle to go from +m to -m immidiately
If you do QFT, E is really a frequency and the Energy Expectation
value is always well defined and positive, negative frequency parts of
the field correspond to antiparticles.
(Remember Diracs negative energy electron 'sea', that too introduced
anti particles to interpretate the negative energy bits)
---
frank
FrediFizzx
Jun4-04, 02:48 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>"Frank Hellmann" <C.i.m@gmx.net> wrote in message\nnews:e7f834be.0406010624.4c5d83b2@posting .google.com...\n| "FrediFizzx" <fredifizzx@hotmail.com> wrote in message\nnews:<2hvsu4FggoocU1@uni-berlin.de>...\n|\n| > So even if solving for E or m, we just throw out the negative solutions\nif\n| > we believe they don\'t represent something physical? I suppose that is\n| > simple enough. Just doesn\'t seem "pure" though.\n| >\n|\n| You can see (somewhat heuristically, if anybody want\'s to expand I\'d\n| be glad) that the negative energy solutions can\'t interact with the\n| positive energy sollutions due to the \'mass gap\'. In classical\n| mechanics every interaction is smooth and smoothness prevents a\n| particle to go from +m to -m immidiately\n\nWell, my big problem with the above is that there necessarily isn\'t a gap\nfrom m to -m since I don\'t see how there can be -m in the first place.\nNor -m*c^2.\n\n| If you do QFT, E is really a frequency and the Energy Expectation\n| value is always well defined and positive, negative frequency parts of\n| the field correspond to antiparticles.\n\nThis is somewhat better in using frequency as long as negative frequency\nmaybe represents an opposite direction relative to what we are calling the\npositive frequency. But it seems there has to be more than just this.\n\n| (Remember Diracs negative energy electron \'sea\', that too introduced\n| anti particles to interpretate the negative energy bits)\n\nI suppose I should explain my problem more. In reading Griffiths\'\n"Introduction to Elementary Particles", he sort of left a bad taste in my\nmind about how he did away with the negative energy sea in the solutions to\nthe Dirac equation. Basically what it looks like, is he just flipped the\nsign to negative on the four momentum for the two anti-particle solutions so\nthey have positive energy. ???\n\nFrediFizzx\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>"Frank Hellmann" <C.i.m@gmx.net> wrote in message
news:e7f834be.0406010624.4c5d83b2@posting.google.c om...
| "FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:<2hvsu4FggoocU1@uni-berlin.de>...
|
| > So even if solving for E or m, we just throw out the negative solutions
if
| > we believe they don't represent something physical? I suppose that is
| > simple enough. Just doesn't seem "pure" though.
| >
|
| You can see (somewhat heuristically, if anybody want's to expand I'd
| be glad) that the negative energy solutions can't interact with the
| positive energy sollutions due to the 'mass gap'. In classical
| mechanics every interaction is smooth and smoothness prevents a
| particle to go from +m to -m immidiately
Well, my big problem with the above is that there necessarily isn't a gap
from m to -m since I don't see how there can be -m in the first place.
Nor -m*c^2.
| If you do QFT, E is really a frequency and the Energy Expectation
| value is always well defined and positive, negative frequency parts of
| the field correspond to antiparticles.
This is somewhat better in using frequency as long as negative frequency
maybe represents an opposite direction relative to what we are calling the
positive frequency. But it seems there has to be more than just this.
| (Remember Diracs negative energy electron 'sea', that too introduced
| anti particles to interpretate the negative energy bits)
I suppose I should explain my problem more. In reading Griffiths'
"Introduction to Elementary Particles", he sort of left a bad taste in my
mind about how he did away with the negative energy sea in the solutions to
the Dirac equation. Basically what it looks like, is he just flipped the
sign to negative on the four momentum for the two anti-particle solutions so
they have positive energy. ???
FrediFizzx
Frank Hellmann
Jun6-04, 04:27 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>"FrediFizzx" <fredifizzx@hotmail.com> wrote in message news:<2i53vbFivsevU1@uni-berlin.de>...\n> "Frank Hellmann" <C.i.m@gmx.net> wrote in message\n> news:e7f834be.0406010624.4c5d83b2@posting.google.c om...\n> | "FrediFizzx" <fredifizzx@hotmail.com> wrote in message\n> news:<2hvsu4FggoocU1@uni-berlin.de>...\n> |\n> | > So even if solving for E or m, we just throw out the negative solutions\n> if\n> | > we believe they don\'t represent something physical? I suppose that is\n> | > simple enough. Just doesn\'t seem "pure" though.\n> | >\n> |\n> | You can see (somewhat heuristically, if anybody want\'s to expand I\'d\n> | be glad) that the negative energy solutions can\'t interact with the\n> | positive energy sollutions due to the \'mass gap\'. In classical\n> | mechanics every interaction is smooth and smoothness prevents a\n> | particle to go from +m to -m immidiately\n>\n> Well, my big problem with the above is that there necessarily isn\'t a gap\n> from m to -m since I don\'t see how there can be -m in the first place.\n> Nor -m*c^2.\n>\n\nWell I guess it depends on what you want to start from. If you start\nfrom the Energy equation E^2 = m^2 + p^2 which really just means that\nthe energy momentum four vector has length m, then obviously you could\nget positive and negative Energy sollutions to that. (Which does not\nimply negative mass since you get it from considering the negative\nroot of E^2 = something larger then zero)\n\nActually even in Newtonian mechanics, why not have negative mass? What\nprevents it. That\'s the question to ask. Landau and Lifschitz argue\nthat there would be no minimum of the action but just a maximum, but\nall we ever use is the extremum, so this is not very satisfactory.\nGoing from Newton it seems odd that the change in momentum should be\nopposite the force acting on it, however aparently it does lead to a\nconsistent theory.\n\n(and in the spr archives you\'ll find lots of talk about negative mass\ngravity)\n\nOf course just because a theory is consistent does not mean it\'s\nrealized in nature, however if one part of the theory is and the other\nisn\'t then we would of course like some theoretical reason for why\nthat is the case.\n\n\n> | If you do QFT, E is really a frequency and the Energy Expectation\n> | value is always well defined and positive, negative frequency parts of\n> | the field correspond to antiparticles.\n>\n> This is somewhat better in using frequency as long as negative frequency\n> maybe represents an opposite direction relative to what we are calling the\n> positive frequency. But it seems there has to be more than just this.\n>\n\nPositrons can be viewed as electrons goiung backwards in time, and\nthat\'s where their arrows in Feynman diagrams come from.\n\n> | (Remember Diracs negative energy electron \'sea\', that too introduced\n> | anti particles to interpretate the negative energy bits)\n>\n> I suppose I should explain my problem more. In reading Griffiths\'\n> "Introduction to Elementary Particles", he sort of left a bad taste in my\n> mind about how he did away with the negative energy sea in the solutions to\n> the Dirac equation. Basically what it looks like, is he just flipped the\n> sign to negative on the four momentum for the two anti-particle solutions so\n> they have positive energy. ???\n\nUh... I\'ve never looked in detail at the Dirac sea thing, basically\nthe thing is that it\'s wrong, QFT is the right answer to these\nquestions (the whole thing doesn\'t work for bosons anyway...).\n\nI guess a good analogy would be a hole conductor. If you have a\nnegative particle missing then the positively charged "hole"\npropagates just like a positive particle.\nThe absence of a negative energy sollution looks just like a positive\nenergy sollution.\n\nI personally would suggest putting these questions of until you study\nQFT which was the theory that resolved them.\n\n---\nfrank.\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>"FrediFizzx" <fredifizzx@hotmail.com> wrote in message news:<2i53vbFivsevU1@uni-berlin.de>...
> "Frank Hellmann" <C.i.m@gmx.net> wrote in message
> news:e7f834be.0406010624.4c5d83b2@posting.google.c om...
> | "FrediFizzx" <fredifizzx@hotmail.com> wrote in message
> news:<2hvsu4FggoocU1@uni-berlin.de>...
> |
> | > So even if solving for E or m, we just throw out the negative solutions
> if
> | > we believe they don't represent something physical? I suppose that is
> | > simple enough. Just doesn't seem "pure" though.
> | >
> |
> | You can see (somewhat heuristically, if anybody want's to expand I'd
> | be glad) that the negative energy solutions can't interact with the
> | positive energy sollutions due to the 'mass gap'. In classical
> | mechanics every interaction is smooth and smoothness prevents a
> | particle to go from +m to -m immidiately
>
> Well, my big problem with the above is that there necessarily isn't a gap
> from m to -m since I don't see how there can be -m in the first place.
> Nor -m*c^2.
>
Well I guess it depends on what you want to start from. If you start
from the Energy equation E^2 = m^2 + p^2 which really just means that
the energy momentum four vector has length m, then obviously you could
get positive and negative Energy sollutions to that. (Which does not
imply negative mass since you get it from considering the negative
root of E^2 = something larger then zero)
Actually even in Newtonian mechanics, why not have negative mass? What
prevents it. That's the question to ask. Landau and Lifschitz argue
that there would be no minimum of the action but just a maximum, but
all we ever use is the extremum, so this is not very satisfactory.
Going from Newton it seems odd that the change in momentum should be
opposite the force acting on it, however aparently it does lead to a
consistent theory.
(and in the spr archives you'll find lots of talk about negative mass
gravity)
Of course just because a theory is consistent does not mean it's
realized in nature, however if one part of the theory is and the other
isn't then we would of course like some theoretical reason for why
that is the case.
> | If you do QFT, E is really a frequency and the Energy Expectation
> | value is always well defined and positive, negative frequency parts of
> | the field correspond to antiparticles.
>
> This is somewhat better in using frequency as long as negative frequency
> maybe represents an opposite direction relative to what we are calling the
> positive frequency. But it seems there has to be more than just this.
>
Positrons can be viewed as electrons goiung backwards in time, and
that's where their arrows in Feynman diagrams come from.
> | (Remember Diracs negative energy electron 'sea', that too introduced
> | anti particles to interpretate the negative energy bits)
>
> I suppose I should explain my problem more. In reading Griffiths'
> "Introduction to Elementary Particles", he sort of left a bad taste in my
> mind about how he did away with the negative energy sea in the solutions to
> the Dirac equation. Basically what it looks like, is he just flipped the
> sign to negative on the four momentum for the two anti-particle solutions so
> they have positive energy. ???
Uh... I've never looked in detail at the Dirac sea thing, basically
the thing is that it's wrong, QFT is the right answer to these
questions (the whole thing doesn't work for bosons anyway...).
I guess a good analogy would be a hole conductor. If you have a
negative particle missing then the positively charged "hole"
propagates just like a positive particle.
The absence of a negative energy sollution looks just like a positive
energy sollution.
I personally would suggest putting these questions of until you study
QFT which was the theory that resolved them.
---
frank.
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