View Full Version : particles of opposite mass
John Baez
Apr15-04, 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>In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,\nOz <oz@farmeroz.port995.com> wrote:\n\n>John Baez <baez@galaxy.ucr.edu> writes:\n\n>>As long as general relativity applies:\n>>\n>>A positive-mass body will curve spacetime in a way that bends geodesics\n>>"towards" it, so it will *attract* other bodies regardless of the sign\n>>of their mass.\n>>\n>>A negative-mass body will curve spacetime in a way that bends geodesics\n>>"away from" it, so it will *repel* other bodies regardless of the sign\n>>of their mass.\n\nIn short:\n\na positive-mass body attracts EVERYTHING;\na negative-mass body repels EVERYTHING.\n\n>That strikes me as very reasonable. Of course we must be careful to\n>distinguish between a positive and negative inertia, too. In this sort\n>of scenario I don\'t think we can assume mass and inertia will\n>necessarily be either the same, or a different, sign.\n\nI\'m assuming general relativity holds. Given that, the equivalence\nprinciple says mass and inertia are the same. If we don\'t assume\ngeneral relativity holds, all bets are off - we just have to do the\nexperiment.\n\n>Fortunately in GR\n>when following a geodesic, there is no acceleration so this can be\n>conveniently swept under the carpet.\n\nYes: that\'s a more elegant way of saying the equivalence principle holds.\n\n>>Now you\'ve got all the necessary knowledge to take a crack at this:\n\n>Oh .. my .. god! He never changes! Straight into homework.\n\nHeh - but this is an easy one, just for old time\'s sake.\n\n>>PUZZLE:\n>>\n>> Figure out what happens if you have two planets near each\n>> other: Earth and Anti-Earth, the first with positive mass, the\n>> second with an "equal but opposite" negative mass.\n\n>I expect we will have the \'accelerate across the universe\' scenario...\n>\n>This needs some thought. I trust you are not expecting me to solve an\n>equivalent of schild metric for this scenario?\n\nNo, I don\'t expect miracles - just a little logic!\n\n>I assume embedded in an otherwise empty flat spacetime. For convenience\n>I will consider the masses as point particles.\n\nGood.\n\n>Now what?\n\nNow solve the problem.\n\n>Well, there will be a point halfway between the two which will be\n>locally flat. Eh? No, that can\'t be right. A test particle on the\n>repulsive body will fall straight down and hit the attractive one, since\n>it will be repelled by the repulsive and attracted by the attractive.\n\nHmm, you\'ve certainly managed to make it more complicated by\nintroducing this unnecessary "test particle". Now *I\'m* confused!\n\n>So if both bodies were dust then the repulsive one would expand and the\n>attractive one would collapse.\n\nYou assumed they were points a minute ago, so there\'s no\nneed to worry about what would happen if they were made of dust -\nthough you\'re perfectly right about what *would* happen!\n\n>If they were solid enough to resist\n>gravitational forces then they clearly would accelerate across the\n>universe, trailing their gravitational fields behind them.\n\nRight! Excellent!\n\nThe positive mass Earth attracts the negative mass Anti-Earth.\nThe negative mass Anti-Earth repels the positive mass Anti-Earth.\n\nSince they have "equal and opposite mass", they both accelerate\nin the same direction at the same rate.\n\nSo, the Anti-Earth chases the Earth faster and faster, approaching\nthe speed of light... but never catches it.\n\nAnd energy is conserved, since the total kinetic energy is zero\nno matter how fast they\'re going!\n\n>If they were orbiting each other as well, then they would have a\n>complex circular path (probably).\n\nOh??\n\nThis is fun to think about, but I\'m highly dubious of this idea of\nparticles of opposite mass "orbiting" each other. Do you see why?\n\n>What if they were different sized masses?\n\nThis is even *more* fun.\n\n>Well a -m particle would orbit a large +m particle, but presumably in\n>its immediate vicinity space would be less curved.\n\nYou can do all these problems with Newtonian gravity as long as\nnothing goes too fast and none of your point masses get too close.\n\nYou should do them this way before worrying about fancy "spacetime\ncurvature" effects.\n\n>I think this means it\n>has a slightly larger orbit. The two bodies will orbit round a centre of\n>mass that will be outside the line between them. This will be a patch of\n>flat spacetime. For an infinitely small orbiting mass, the only patch of\n>flat spacetime (not at inf) will be the saddle on the major body,\n>clearly a -ve mass will push this further away from the -ve particle.\n\nI\'m not sure what you mean here - let\'s keep things simple and\nNewtonian for a while; we\'ll have enough fun that way.\n\nIn the Newtonian approximation, the center of mass of our two particles\nwill move along a straight line at constant speed. This is conservation\nof momentum, so it holds no matter what the signs of the masses - under\nour default assumption that GR still works.\n\nBut: what\'s the center of mass of a positive mass particle and a\nnegative mass particle?\n\n>As their masses tend to being equal and opposite then this patch will\n>recede to infinity and we get the \'follow my leader\' scenario again.\n>\n>My head hurts ....\n\nYeah, it\'s tough. The math works just as well when you change\nthe signs in these problems. The hard part, but the fun part,\nis to solve them using "intuition".\n\n>>By the way, it currently seems like I\'ll be in Oxford this July 7-9,\n>>to speak at the Workshop on Gerbes: Recent Developments and Future\n>>Perspectives, at Oxford, organized by Nuno Reis. So, maybe we can\n>>get together while I\'m there.\n\n>Should be fine.\n>I can\'t contact you, but you can contact me using reply-to of this post.\n\nOkay, I\'ll contact you shortly after I arrive in Cambridge on July 1st.\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>In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,
Oz <oz@farmeroz.port995.com> wrote:
>John Baez <baez@galaxy.ucr.edu> writes:
>>As long as general relativity applies:
>>
>>A positive-mass body will curve spacetime in a way that bends geodesics
>>"towards" it, so it will *attract* other bodies regardless of the sign
>>of their mass.
>>
>>A negative-mass body will curve spacetime in a way that bends geodesics
>>"away from" it, so it will *repel* other bodies regardless of the sign
>>of their mass.
In short:
a positive-mass body attracts EVERYTHING;
a negative-mass body repels EVERYTHING.
>That strikes me as very reasonable. Of course we must be careful to
>distinguish between a positive and negative inertia, too. In this sort
>of scenario I don't think we can assume mass and inertia will
>necessarily be either the same, or a different, sign.
I'm assuming general relativity holds. Given that, the equivalence
principle says mass and inertia are the same. If we don't assume
general relativity holds, all bets are off - we just have to do the
experiment.
>Fortunately in GR
>when following a geodesic, there is no acceleration so this can be
>conveniently swept under the carpet.
Yes: that's a more elegant way of saying the equivalence principle holds.
>>Now you've got all the necessary knowledge to take a crack at this:
>Oh .. my .. god! He never changes! Straight into homework.
Heh - but this is an easy one, just for old time's sake.
>>PUZZLE:
>>
>> Figure out what happens if you have two planets near each
>> other: Earth and Anti-Earth, the first with positive mass, the
>> second with an "equal but opposite" negative mass.
>I expect we will have the 'accelerate across the universe' scenario...
>
>This needs some thought. I trust you are not expecting me to solve an
>equivalent of schild metric for this scenario?
No, I don't expect miracles - just a little logic!
>I assume embedded in an otherwise empty flat spacetime. For convenience
>I will consider the masses as point particles.
Good.
>Now what?
Now solve the problem.
>Well, there will be a point halfway between the two which will be
>locally flat. Eh? No, that can't be right. A test particle on the
>repulsive body will fall straight down and hit the attractive one, since
>it will be repelled by the repulsive and attracted by the attractive.
Hmm, you've certainly managed to make it more complicated by
introducing this unnecessary "test particle". Now *I'm* confused!
>So if both bodies were dust then the repulsive one would expand and the
>attractive one would collapse.
You assumed they were points a minute ago, so there's no
need to worry about what would happen if they were made of dust -
though you're perfectly right about what *would* happen!
>If they were solid enough to resist
>gravitational forces then they clearly would accelerate across the
>universe, trailing their gravitational fields behind them.
Right! Excellent!
The positive mass Earth attracts the negative mass Anti-Earth.
The negative mass Anti-Earth repels the positive mass Anti-Earth.
Since they have "equal and opposite mass", they both accelerate
in the same direction at the same rate.
So, the Anti-Earth chases the Earth faster and faster, approaching
the speed of light... but never catches it.
And energy is conserved, since the total kinetic energy is zero
no matter how fast they're going!
>If they were orbiting each other as well, then they would have a
>complex circular path (probably).
Oh??
This is fun to think about, but I'm highly dubious of this idea of
particles of opposite mass "orbiting" each other. Do you see why?
>What if they were different sized masses?
This is even *more* fun.
>Well a -m particle would orbit a large +m particle, but presumably in
>its immediate vicinity space would be less curved.
You can do all these problems with Newtonian gravity as long as
nothing goes too fast and none of your point masses get too close.
You should do them this way before worrying about fancy "spacetime
curvature" effects.
>I think this means it
>has a slightly larger orbit. The two bodies will orbit round a centre of
>mass that will be outside the line between them. This will be a patch of
>flat spacetime. For an infinitely small orbiting mass, the only patch of
>flat spacetime (not at inf) will be the saddle on the major body,
>clearly a -ve mass will push this further away from the -ve particle.
I'm not sure what you mean here - let's keep things simple and
Newtonian for a while; we'll have enough fun that way.
In the Newtonian approximation, the center of mass of our two particles
will move along a straight line at constant speed. This is conservation
of momentum, so it holds no matter what the signs of the masses - under
our default assumption that GR still works.
But: what's the center of mass of a positive mass particle and a
negative mass particle?
>As their masses tend to being equal and opposite then this patch will
>recede to infinity and we get the 'follow my leader' scenario again.
>
>My head hurts ....
Yeah, it's tough. The math works just as well when you change
the signs in these problems. The hard part, but the fun part,
is to solve them using "intuition".
>>By the way, it currently seems like I'll be in Oxford this July 7-9,
>>to speak at the Workshop on Gerbes: Recent Developments and Future
>>Perspectives, at Oxford, organized by Nuno Reis. So, maybe we can
>>get together while I'm there.
>Should be fine.
>I can't contact you, but you can contact me using reply-to of this post.
Okay, I'll contact you shortly after I arrive in Cambridge on July 1st.
alistair
Apr16-04, 02:28 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>Figure out what happens if you have two planets near each\n>> other: Earth and Anti-Earth, the first with positive mass, the\n>> second with an "equal but opposite" negative mass.\n\nIf you had a universe made of just two large masses, one negative mass\nand the other positive,the two masses would oscillate towards and away\nfrom one another perpetually (unless they started out static at\nmaximum separation, in which case they would keep at a fixed\ndistance).\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>Figure out what happens if you have two planets near each
>> other: Earth and Anti-Earth, the first with positive mass, the
>> second with an "equal but opposite" negative mass.
If you had a universe made of just two large masses, one negative mass
and the other positive,the two masses would oscillate towards and away
from one another perpetually (unless they started out static at
maximum separation, in which case they would keep at a fixed
distance).
Ken S. Tucker
Apr17-04, 05:08 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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c5fcj8\\$6ua\\$1@glue.ucr.edu>...\n\nSorry to interrupt, this is fun, in view of symmetry.\n\n>In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,\n>Oz <oz@farmeroz.port995.com> wrote:\n>>John Baez <baez@galaxy.ucr.edu> writes:\n>>>As long as general relativity applies:\n\n>>So if both bodies were dust then the repulsive one would expand and the\n>>attractive one would collapse.\n>\n>You assumed they were points a minute ago, so there\'s no\n>need to worry about what would happen if they were made of dust -\n>though you\'re perfectly right about what *would* happen!\n\nUsing Old Newton\'s Force = - G (M) (m) /r^2 the universe\nwould behave the same if one used (-M) and (-m) in Newtons,\nso I think there is no easy way to decide if mass/energy is positive\nor negative. So I think a negative energy "dust cloud" would\ncondense as a positive energy cloud.\nTo satisfy GR, we should presume a photon, born\nfrom negative energy would possess negative energy,\nand deflect in the the negative mass universe as it would\npresuming positive mass. (?)\nIOW\'s could we do an experiment to determine the\npolarization of the scalar "mass"?\n\nKen S. Tucker\nPS: snippable, there is interesting symmetry in the\n+/- mass universe. But if you really want a repulsive\ndust cloud you would need (i = sqrt(-1))\n\nF\' = - G(Mi)(mi)/r^2 = + G(M)(m)/r^2\n\n(last term is repulsive because of the +)\n\nand it looks like that universe would be equal to\nours if the "arrow of time" were to reverse to\nconvert F\' to F.\nkst\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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c5fcj8$6ua$1@glue.ucr.edu>...
Sorry to interrupt, this is fun, in view of symmetry.
>In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,
>Oz <oz@farmeroz.port995.com> wrote:
>>John Baez <baez@galaxy.ucr.edu> writes:
>>>As long as general relativity applies:
>>So if both bodies were dust then the repulsive one would expand and the
>>attractive one would collapse.
>
>You assumed they were points a minute ago, so there's no
>need to worry about what would happen if they were made of dust -
>though you're perfectly right about what *would* happen!
Using Old Newton's Force = - G (M) (m) /r^2 the universe
would behave the same if one used (-M) and (-m) in Newtons,
so I think there is no easy way to decide if mass/energy is positive
or negative. So I think a negative energy "dust cloud" would
condense as a positive energy cloud.
To satisfy GR, we should presume a photon, born
from negative energy would possess negative energy,
and deflect in the the negative mass universe as it would
presuming positive mass. (?)
IOW's could we do an experiment to determine the
polarization of the scalar "mass"?
Ken S. Tucker
PS: snippable, there is interesting symmetry in the
+/- mass universe. But if you really want a repulsive
dust cloud you would need (i = \sqrt(-1))F' = - G(Mi)(mi)/r^2 = + G(M)(m)/r^2
(last term is repulsive because of the +)
and it looks like that universe would be equal to
ours if the "arrow of time" were to reverse to
convert F' to F.
kst
Charles Francis
Apr18-04, 03:50 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>In article <c5fcj8\\$6ua\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\nwrites\n>In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,\n>Oz <oz@farmeroz.port995.com> wrote:\n>\n>>John Baez <baez@galaxy.ucr.edu> writes:\n>\n>>>As long as general relativity applies:\n>>>\n>>>A positive-mass body will curve spacetime in a way that bends geodesics\n>>>"towards" it, so it will *attract* other bodies regardless of the sign\n>>>of their mass.\n>>>\n>>>A negative-mass body will curve spacetime in a way that bends geodesics\n>>>"away from" it, so it will *repel* other bodies regardless of the sign\n>>>of their mass.\n\nHuh, I\'ve missed something. Mass is generally a magnitude, hence\npositive.\n>\n>In short:\n>\n>a positive-mass body attracts EVERYTHING;\n>a negative-mass body repels EVERYTHING.\n\nWell, if you say so.\n\n>\n>>>PUZZLE:\n>>>\n>>> Figure out what happens if you have two planets near each\n>>> other: Earth and Anti-Earth, the first with positive mass, the\n>>> second with an "equal but opposite" negative mass.\n\n>\n>>If they were solid enough to resist\n>>gravitational forces then they clearly would accelerate across the\n>>universe, trailing their gravitational fields behind them.\n>\n>Right! Excellent!\n\nIs it?\n\n>The positive mass Earth attracts the negative mass Anti-Earth.\n>The negative mass Anti-Earth repels the positive mass Anti-Earth.\n>\nBut since active gravitational mass is normally the same as passive\ngravitational mass the positive mass Earth should attract the negative\nmass anti-Earth negatively. I.e. it repels it, so we have the opposite\nof em, like masses attract, unlike repel.\n\n>negative mass particle?\n>\n>>As their masses tend to being equal and opposite then this patch will\n>>recede to infinity and we get the \'follow my leader\' scenario again.\n>>\n>>My head hurts ....\n>\n>Yeah, it\'s tough. The math works just as well when you change\n>the signs in these problems. The hard part, but the fun part,\n>is to solve them using "intuition".\n\nCertainly math which isn\'t formalised intuition is no fun. And not much\nuse either in my book.\n\n\nRegards\n\n--\nCharles Francis\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 <c5fcj8$6ua$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
writes
>In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,
>Oz <oz@farmeroz.port995.com> wrote:
>
>>John Baez <baez@galaxy.ucr.edu> writes:
>
>>>As long as general relativity applies:
>>>
>>>A positive-mass body will curve spacetime in a way that bends geodesics
>>>"towards" it, so it will *attract* other bodies regardless of the sign
>>>of their mass.
>>>
>>>A negative-mass body will curve spacetime in a way that bends geodesics
>>>"away from" it, so it will *repel* other bodies regardless of the sign
>>>of their mass.
Huh, I've missed something. Mass is generally a magnitude, hence
positive.
>
>In short:
>
>a positive-mass body attracts EVERYTHING;
>a negative-mass body repels EVERYTHING.
Well, if you say so.
>
>>>PUZZLE:
>>>
>>> Figure out what happens if you have two planets near each
>>> other: Earth and Anti-Earth, the first with positive mass, the
>>> second with an "equal but opposite" negative mass.
>
>>If they were solid enough to resist
>>gravitational forces then they clearly would accelerate across the
>>universe, trailing their gravitational fields behind them.
>
>Right! Excellent!
Is it?
>The positive mass Earth attracts the negative mass Anti-Earth.
>The negative mass Anti-Earth repels the positive mass Anti-Earth.
>
But since active gravitational mass is normally the same as passive
gravitational mass the positive mass Earth should attract the negative
mass anti-Earth negatively. I.e. it repels it, so we have the opposite
of em, like masses attract, unlike repel.
>negative mass particle?
>
>>As their masses tend to being equal and opposite then this patch will
>>recede to infinity and we get the 'follow my leader' scenario again.
>>
>>My head hurts ....
>
>Yeah, it's tough. The math works just as well when you change
>the signs in these problems. The hard part, but the fun part,
>is to solve them using "intuition".
Certainly math which isn't formalised intuition is no fun. And not much
use either in my book.
Regards
--
Charles Francis
Arnold Neumaier
Apr19-04, 01:29 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>John Baez wrote:\n> In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,\n> Oz <oz@farmeroz.port995.com> wrote:\n>\n>\n>>John Baez <baez@galaxy.ucr.edu> writes:\n>\n>\n>>>As long as general relativity applies:\n>>>\n>>>A positive-mass body will curve spacetime in a way that bends geodesics\n>>>"towards" it, so it will *attract* other bodies regardless of the sign\n>>>of their mass.\n>>>\n>>>A negative-mass body will curve spacetime in a way that bends geodesics\n>>>"away from" it, so it will *repel* other bodies regardless of the sign\n>>>of their mass.\n>\n>\n> In short:\n>\n> a positive-mass body attracts EVERYTHING;\n> a negative-mass body repels EVERYTHING.\n\nThis sounds paradoxical - it would mean a positive-mass body attracts\na negative-mass body, while the latter repels the former.\nProbably they are chasing after each other???\n\nBut shouldn\'t we have actio = reactio? So:\nDo they get closer to each other or farther apart if initially they\nare at rest with respect to each other?\n\nTo which extent is the general relativistic 2-body problem solved?\n\n\nArnold Neumaier\n\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>John Baez wrote:
> In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,
> Oz <oz@farmeroz.port995.com> wrote:
>
>
>>John Baez <baez@galaxy.ucr.edu> writes:
>
>
>>>As long as general relativity applies:
>>>
>>>A positive-mass body will curve spacetime in a way that bends geodesics
>>>"towards" it, so it will *attract* other bodies regardless of the sign
>>>of their mass.
>>>
>>>A negative-mass body will curve spacetime in a way that bends geodesics
>>>"away from" it, so it will *repel* other bodies regardless of the sign
>>>of their mass.
>
>
> In short:
>
> a positive-mass body attracts EVERYTHING;
> a negative-mass body repels EVERYTHING.
This sounds paradoxical - it would mean a positive-mass body attracts
a negative-mass body, while the latter repels the former.
Probably they are chasing after each other???
But shouldn't we have actio = reactio? So:
Do they get closer to each other or farther apart if initially they
are at rest with respect to each other?
To which extent is the general relativistic 2-body problem solved?
Arnold Neumaier
<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 17/04/2004 10:08 am, Ken S. Tucker at dynamics@vianet.on.ca wrote:\n\n> baez@galaxy.ucr.edu (John Baez) wrote in message\n> news:<c5fcj8\\$6ua\\$1@glue.ucr.edu>...\n>\n> Sorry to interrupt, this is fun, in view of symmetry.\n>\n>> In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,\n>> Oz <oz@farmeroz.port995.com> wrote:\n>>> John Baez <baez@galaxy.ucr.edu> writes:\n>>>> As long as general relativity applies:\n>\n>>> So if both bodies were dust then the repulsive one would expand and the\n>>> attractive one would collapse.\n>>\n>> You assumed they were points a minute ago, so there\'s no\n>> need to worry about what would happen if they were made of dust -\n>> though you\'re perfectly right about what *would* happen!\n>\n> Using Old Newton\'s Force = - G (M) (m) /r^2 the universe\n> would behave the same if one used (-M) and (-m) in Newtons,\n\nNo! The _force_ would be the same, but the acceleration would be different.\nThere\'s only one m in F=ma.\n\nTim\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>on 17/04/2004 10:08 am, Ken S. Tucker at dynamics@vianet.on.ca wrote:
> baez@galaxy.ucr.edu (John Baez) wrote in message
> news:<c5fcj8$6ua$1@glue.ucr.edu>...
>
> Sorry to interrupt, this is fun, in view of symmetry.
>
>> In article <ZM6JLBCV7PdAFwAA@btopenworld.com>,
>> Oz <oz@farmeroz.port995.com> wrote:
>>> John Baez <baez@galaxy.ucr.edu> writes:
>>>> As long as general relativity applies:
>
>>> So if both bodies were dust then the repulsive one would expand and the
>>> attractive one would collapse.
>>
>> You assumed they were points a minute ago, so there's no
>> need to worry about what would happen if they were made of dust -
>> though you're perfectly right about what *would* happen!
>
> Using Old Newton's Force = - G (M) (m) /r^2 the universe
> would behave the same if one used (-M) and (-m) in Newtons,
No! The _force_ would be the same, but the acceleration would be different.
There's only one m in F=ma.
Tim
<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 wrote:\n>>>Figure out what happens if you have two planets near each\n>>>other: Earth and Anti-Earth, the first with positive mass, the\n>>>second with an "equal but opposite" negative mass.\n>.....[cut].....\n>\n> The positive mass Earth attracts the negative mass Anti-Earth.\n> The negative mass Anti-Earth repels the positive mass Anti-Earth.\n>\n> Since they have "equal and opposite mass", they both accelerate\n> in the same direction at the same rate.\n>\n> So, the Anti-Earth chases the Earth faster and faster, approaching\n> the speed of light... but never catches it.\n>\n> And energy is conserved, since the total kinetic energy is zero\n> no matter how fast they\'re going!\n>\n>\n>....[cut]\n>\n>\n> You can do all these problems with Newtonian gravity as long as\n> nothing goes too fast and none of your point masses get too close.\n>\n> You should do them this way before worrying about fancy "spacetime\n> curvature" effects.\n>\n\n\n\nOK, hang on here! I tried this with Newton\'s Universal Law of\nGravitation, and did not get the result you claim.\n\nF_12=-GM_1M_2/r^2*hat{r}_12\n\nwhere F_12 is the force particle 1 exerts on particle 2, M_1 is the mass\nof particle 1, M_2 is the mass of particle 2, and hat{r}_12 is the unit\nvector r2-r1/|r2-r1| pointing from r1 to r2\n\n\ni have a mass 1 of +m at x=0 and mass 2 -m at x=r. for the force\nparticle 1 exerts on particle 2, hat{r}_12=hat{x}, and F12=Gm^2/r^2\n\non the other hand, for the force particle 2 exerts on particle one, the\nunit vector hat{r}_21=-hat{x}, so i get the negative of the previous\nresult: F_21=-Gm^2/r^2\n\nin other words, the two particles don\'t go in the same direction, they\ngo in opposite directions (as Newton\'s third law requires). The\ndifference between this case and the normal positive mass case is that\nthe two particles repel, instead of attract. Like masses attract,\nopposite masses repel. It\'s Coulomb\'s Law with an extra minus sign\nthrown in.\n\n[Moderator\'s note: It\'s incorrect to say that the two particles "go"\nin opposite directions. The forces are in opposite directions, but\nsince the inertial masses have opposite signs, the accelerations are\nin the same direction. -TB]\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 wrote:
>>>Figure out what happens if you have two planets near each
>>>other: Earth and Anti-Earth, the first with positive mass, the
>>>second with an "equal but opposite" negative mass.
>.....[cut].....
>
> The positive mass Earth attracts the negative mass Anti-Earth.
> The negative mass Anti-Earth repels the positive mass Anti-Earth.
>
> Since they have "equal and opposite mass", they both accelerate
> in the same direction at the same rate.
>
> So, the Anti-Earth chases the Earth faster and faster, approaching
> the speed of light... but never catches it.
>
> And energy is conserved, since the total kinetic energy is zero
> no matter how fast they're going!
>
>
>....[cut]
>
>
> You can do all these problems with Newtonian gravity as long as
> nothing goes too fast and none of your point masses get too close.
>
> You should do them this way before worrying about fancy "spacetime
> curvature" effects.
>
OK, hang on here! I tried this with Newton's Universal Law of
Gravitation, and did not get the result you claim.
F_{12}=-GM_1M_2/r^2*hat{r}_12
where F_{12} is the force particle 1 exerts on particle 2, M_1 is the mass
of particle 1, M_2 is the mass of particle 2, and hat{r}_12 is the unit
vector r2-r1/|r2-r1| pointing from r1 to r2
i have a mass 1 of +m at x=0 and mass 2 -m at x=r. for the force
particle 1 exerts on particle 2, hat{r}_12=hat{x}, and F12=Gm^2/r^2
on the other hand, for the force particle 2 exerts on particle one, the
unit vector hat{r}_21=-hat{x}, so i get the negative of the previous
result: F_{21}=-Gm^2/r^2
in other words, the two particles don't go in the same direction, they
go in opposite directions (as Newton's third law requires). The
difference between this case and the normal positive mass case is that
the two particles repel, instead of attract. Like masses attract,
opposite masses repel. It's Coulomb's Law with an extra minus sign
thrown in.
[Moderator's note: It's incorrect to say that the two particles "go"
in opposite directions. The forces are in opposite directions, but
since the inertial masses have opposite signs, the accelerations are
in the same direction. -TB]
Italo Vecchi
Apr20-04, 02:33 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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c5fcj8\\$6ua\\$1@glue.ucr.edu>...\n....\n>\n > In short:\n>\n> a positive-mass body attracts EVERYTHING;\n> a negative-mass body repels EVERYTHING.\n>\n\nHuh. So in a negative mass world masses repel each other, as do\npositive and negative charges (positive force on negative mass), while\nsame sign charges attract each other(negative force on negative mass).\nIt\'s like a movie playing backwards.\n\nIV\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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c5fcj8$6ua$1@glue.ucr.edu>...
....
>
> In short:
>
> a positive-mass body attracts EVERYTHING;
> a negative-mass body repels EVERYTHING.
>
Huh. So in a negative mass world masses repel each other, as do
positive and negative charges (positive force on negative mass), while
same sign charges attract each other(negative force on negative mass).
It's like a movie playing backwards.
IV
John Baez
Apr20-04, 02:35 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 <BCA6E161.81A4Tim@timsilverman.demon.co.uk>,\nTi m S <Tim@timsilverman.demon.co.uk> wrote:\n\n>on 17/04/2004 10:08 am, Ken S. Tucker at dynamics@vianet.on.ca wrote:\n\n>> Using Old Newton\'s Force = - G (M) (m) /r^2 the universe\n>> would behave the same if one used (-M) and (-m) in Newtons,\n>> I think there is no easy way to decide if mass/energy is positive\n>> or negative. So I think a negative energy "dust cloud" would\n>> condense as a positive energy cloud.\n\nWhoops! Actually a negative mass dust cloud would *expand*, not collapse\nlike a positive mass one does. The reason is...\n\n>The _force_ would be the same, but the acceleration would be different.\n>There\'s only one m in F=ma.\n\nRight!\n\nIf you have two negative masses, they REPEL each other,\nsince we get two minus signs from the masses in F = GMm/r^2,\nbut a third when we work out the acceleration using F = ma.\n\nTriple negatives seem tricky, so it may help to remember\nwhat I wrote earlier in this thread:\n\na positive-mass body attracts EVERYTHING;\na negative-mass body repels EVERYTHING.\n\nAnd again, if we use GR this is just the equivalence principle in action:\n\nA positive-mass body will curve spacetime in a way that bends geodesics\n"towards" it, so it will *attract* other bodies regardless of the sign\nof their mass.\n\nA negative-mass body will curve spacetime in a way that bends geodesics\n"away from" it, so it will *repel* other bodies regardless of the sign\nof their mass.\n\nNext puzzle... I may have asked Oz this already:\n\nPUZZLE: in Newtonian gravity, how does a small negative mass\n"orbit" a big positive mass? What curve does it trace out?\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 <BCA6E161.81A4Tim@timsilverman.demon.co.uk>,
Tim S <Tim@timsilverman.demon.co.uk> wrote:
>on 17/04/2004 10:08 am, Ken S. Tucker at dynamics@vianet.on.ca wrote:
>> Using Old Newton's Force = - G (M) (m) /r^2 the universe
>> would behave the same if one used (-M) and (-m) in Newtons,
>> I think there is no easy way to decide if mass/energy is positive
>> or negative. So I think a negative energy "dust cloud" would
>> condense as a positive energy cloud.
Whoops! Actually a negative mass dust cloud would *expand*, not collapse
like a positive mass one does. The reason is...
>The _force_ would be the same, but the acceleration would be different.
>There's only one m in F=ma.
Right!
If you have two negative masses, they REPEL each other,
since we get two minus signs from the masses in F = GMm/r^2,
but a third when we work out the acceleration using F = ma.
Triple negatives seem tricky, so it may help to remember
what I wrote earlier in this thread:
a positive-mass body attracts EVERYTHING;
a negative-mass body repels EVERYTHING.
And again, if we use GR this is just the equivalence principle in action:
A positive-mass body will curve spacetime in a way that bends geodesics
"towards" it, so it will *attract* other bodies regardless of the sign
of their mass.
A negative-mass body will curve spacetime in a way that bends geodesics
"away from" it, so it will *repel* other bodies regardless of the sign
of their mass.
Next puzzle... I may have asked Oz this already:
PUZZLE: in Newtonian gravity, how does a small negative mass
"orbit" a big positive mass? What curve does it trace out?
John Baez
Apr20-04, 02:35 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 <c612a9\\$d0e\\$1@lfa222122.richmond.edu>,\nArno ld Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n\n>John Baez wrote:\n\n>> a positive-mass body attracts EVERYTHING;\n>> a negative-mass body repels EVERYTHING.\n\n>This sounds paradoxical - it would mean a positive-mass body attracts\n>a negative-mass body, while the latter repels the former.\n>Probably they are chasing after each other???\n\nRight. It **sounds** paradoxical - that\'s why I\'m talking about it!\nBut it\'s not; it\'s just weird.\n\nOf course, the fact that we never see such runaway motions is evidence\nthat there aren\'t negative masses!\n\nQuite generally, anything with unbounded negative energy runs the risk\nof causing motions with speeds that approach infinity. The simplest\nexample is the negative *potential* energy of two positive point\nmasses in Newtonian mechanics: they can speed up indefinitely as\nthey fall into each other. But negative masses give negative\n*kinetic* energy, and that causes other funny effects.\n\n>But shouldn\'t we have actio = reactio? So:\n>Do they get closer to each other or farther apart if initially they\n>are at rest with respect to each other?\n\nThey stay the same distance apart, at least in Newtonian mechanics,\nwhere the question is completely unambiguous because we don\'t need\nto worry about Lorentz transforms.\n\n>To which extent is the general relativistic 2-body problem solved?\n\nIt\'s a huge open problem which needs to be solved by supercomputers\nfor people to analyze the gravitational wave data that may eventually\ncome out of LIGO. People are really struggling to solve it.\n\nBut everything I\'ve been discussing becomes easy to understand\nin the Newtonian limit.\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 <c612a9$d0e$1@lfa222122.richmond.edu>,
Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>John Baez wrote:
>> a positive-mass body attracts EVERYTHING;
>> a negative-mass body repels EVERYTHING.
>This sounds paradoxical - it would mean a positive-mass body attracts
>a negative-mass body, while the latter repels the former.
>Probably they are chasing after each other???
Right. It **sounds** paradoxical - that's why I'm talking about it!
But it's not; it's just weird.
Of course, the fact that we never see such runaway motions is evidence
that there aren't negative masses!
Quite generally, anything with unbounded negative energy runs the risk
of causing motions with speeds that approach infinity. The simplest
example is the negative *potential* energy of two positive point
masses in Newtonian mechanics: they can speed up indefinitely as
they fall into each other. But negative masses give negative
*kinetic* energy, and that causes other funny effects.
>But shouldn't we have actio = reactio? So:
>Do they get closer to each other or farther apart if initially they
>are at rest with respect to each other?
They stay the same distance apart, at least in Newtonian mechanics,
where the question is completely unambiguous because we don't need
to worry about Lorentz transforms.
>To which extent is the general relativistic 2-body problem solved?
It's a huge open problem which needs to be solved by supercomputers
for people to analyze the gravitational wave data that may eventually
come out of LIGO. People are really struggling to solve it.
But everything I've been discussing becomes easy to understand
in the Newtonian limit.
<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> writes\n>PUZZLE: in Newtonian gravity, how does a small negative mass\n>"orbit" a big positive mass? What curve does it trace out?\n\nI think I already answered this. Although I used my usual randomly\ntalking around the problem method (which does generate some \'feel\' for\nwhat is going on) the answer is simple.\n\nIn essence the large positive mass will dominate and result in\nattraction. The small negative mass will orbit in a newtonian manner.\n\nHowever the common centre will move from being between the two masses,\nto one on ... ohh ascii art is easier:\n\nO = large +ve mass\n- = small -ve mass (sketch #1)\n+ = small +ve mass (sketch #2)\nx = centre of rotation\n\nviewed perp to axis of orbit\n\nx O - sketch #1 -ve mass\n\nO x + sketch #2 +ve mass\n\nAdjust appropriately using normal newtonian methods for elliptical\norbits.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\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> writes
>PUZZLE: in Newtonian gravity, how does a small negative mass
>"orbit" a big positive mass? What curve does it trace out?
I think I already answered this. Although I used my usual randomly
talking around the problem method (which does generate some 'feel' for
what is going on) the answer is simple.
In essence the large positive mass will dominate and result in
attraction. The small negative mass will orbit in a newtonian manner.
However the common centre will move from being between the two masses,
to one on ... ohh ascii art is easier:
O = large +ve mass
- = small -ve mass (sketch #1)
+ = small +ve mass (sketch #2)
x = centre of rotation
viewed perp to axis of orbit
x O - sketch #1 -ve mass
O x + sketch #2 +ve mass
Adjust appropriately using normal newtonian methods for elliptical
orbits.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<
Richard Saam
Apr21-04, 04:23 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>John Baez wrote:\n\n>In article <BCA6E161.81A4Tim@timsilverman.demon.co.uk>,\n>T im S <Tim@timsilverman.demon.co.uk> wrote:\n>\n>\n>\n>>on 17/04/2004 10:08 am, Ken S. Tucker at dynamics@vianet.on.ca wrote:\n>>\n>>\n>\n>\n>Whoops! Actually a negative mass dust cloud would *expand*, not collapse\n>like a positive mass one does. The reason is...\n>\n>\n>\n>>The _force_ would be the same, but the acceleration would be different.\n>>There\'s only one m in F=ma.\n>>\n>>\n>\n>Right!\n>\n>\n\nWhat are the implications for an expanding universe? Are there negative\nmass dust clouds here and there?\n\nRichard Saam\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 wrote:
>In article <BCA6E161.81A4Tim@timsilverman.demon.co.uk>,
>Tim S <Tim@timsilverman.demon.co.uk> wrote:
>
>
>
>>on 17/04/2004 10:08 am, Ken S. Tucker at dynamics@vianet.on.ca wrote:
>>
>>
>
>
>Whoops! Actually a negative mass dust cloud would *expand*, not collapse
>like a positive mass one does. The reason is...
>
>
>
>>The _force_ would be the same, but the acceleration would be different.
>>There's only one m in F=ma.
>>
>>
>
>Right!
>
>
What are the implications for an expanding universe? Are there negative
mass dust clouds here and there?
Richard Saam
Ken S. Tucker
Apr21-04, 04:24 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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c62ciq\\$h9g\\$1@glue.ucr.edu>...\n>In article <c612a9\\$d0e\\$1@lfa222122.richmond.edu>,\n>Arnol d Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>>John Baez wrote:\n>>> a positive-mass body attracts EVERYTHING;\n>>> a negative-mass body repels EVERYTHING.\n>>This sounds paradoxical - it would mean a positive-mass body attracts\n>>a negative-mass body, while the latter repels the former.\n>>Probably they are chasing after each other???\n>\n>Right. It **sounds** paradoxical - that\'s why I\'m talking about it!\n>But it\'s not; it\'s just weird.\n>\n>Of course, the fact that we never see such runaway motions is evidence\n>that there aren\'t negative masses!\n>\n>Quite generally, anything with unbounded negative energy runs the risk\n>of causing motions with speeds that approach infinity. The simplest\n>example is the negative *potential* energy of two positive point\n>masses in Newtonian mechanics: they can speed up indefinitely as\n>they fall into each other. But negative masses give negative\n>*kinetic* energy, and that causes other funny effects.\n>\n>>But shouldn\'t we have actio = reactio? So:\n>>Do they get closer to each other or farther apart if initially they\n>>are at rest with respect to each other?\n>\n>They stay the same distance apart, at least in Newtonian mechanics,\n>where the question is completely unambiguous because we don\'t need\n>to worry about Lorentz transforms.\n>\n>>To which extent is the general relativistic 2-body problem solved?\n>\n>It\'s a huge open problem which needs to be solved by supercomputers\n>for people to analyze the gravitational wave data that may eventually\n>come out of LIGO. People are really struggling to solve it.\n>\n>But everything I\'ve been discussing becomes easy to understand\n>in the Newtonian limit.\n\nHi Dr. Baez et al, I commited a stupid error above (maybe).\n((in Ken Tucker posts)).\n\nThe quantities Mass, Time, Length are all relative,\nbut, "c", "h", "q" are invariants, (speed of light, Planck\'s\nconstant, fundamental charge) and I think in any\nuniverse, whether positive or negative matter,\nwe should find c=h=1 and q^2 =1.\n\nFor example, using the units of "h" = energy*time\nand energy =mass*c^2, then 1*h = mass*time =1.\n\nHere we should denote the common varibles M,R,T\nas being those (Mass, Radius,Time) we use in our familiar\nuniverse.\nHence in a unverse of negative matter, using "h=1"\nwe should find,\nh = m*t =1\n\nwhere m = -M and t=-T\n\nLet\'s look at acceleration, given in OUR universe\nby A= R/T^2, that becomes a= r/t^2 in the negative\nuniverse, because t^2 =T^2 and r=-R thus,\n\nA = - a.\n\nHence in the negitive mass universe, Newton suggests\n\nA = M/R^2 in our universe\n\nis transformed too,\n\na = m/r^2 in the negative universe.\n\nI don\'t want to make a mistake, so I\'ll transform\nthis way,\n\nm=-M , A= -a and r^2 = R^2,,,\n\nto get,\n\nM*A = m*a = F = f.\n\nSorry, about the detail, I was just trying to compare\nthe polarity of the scalar "m", with our usual "M".\nRegards\nKen S. Tucker\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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c62ciq$h9g$1@glue.ucr.edu>...
>In article <c612a9$d0e$1@lfa222122.richmond.edu>,
>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>>John Baez wrote:
>>> a positive-mass body attracts EVERYTHING;
>>> a negative-mass body repels EVERYTHING.
>>This sounds paradoxical - it would mean a positive-mass body attracts
>>a negative-mass body, while the latter repels the former.
>>Probably they are chasing after each other???
>
>Right. It **sounds** paradoxical - that's why I'm talking about it!
>But it's not; it's just weird.
>
>Of course, the fact that we never see such runaway motions is evidence
>that there aren't negative masses!
>
>Quite generally, anything with unbounded negative energy runs the risk
>of causing motions with speeds that approach infinity. The simplest
>example is the negative *potential* energy of two positive point
>masses in Newtonian mechanics: they can speed up indefinitely as
>they fall into each other. But negative masses give negative
>*kinetic* energy, and that causes other funny effects.
>
>>But shouldn't we have actio = reactio? So:
>>Do they get closer to each other or farther apart if initially they
>>are at rest with respect to each other?
>
>They stay the same distance apart, at least in Newtonian mechanics,
>where the question is completely unambiguous because we don't need
>to worry about Lorentz transforms.
>
>>To which extent is the general relativistic 2-body problem solved?
>
>It's a huge open problem which needs to be solved by supercomputers
>for people to analyze the gravitational wave data that may eventually
>come out of LIGO. People are really struggling to solve it.
>
>But everything I've been discussing becomes easy to understand
>in the Newtonian limit.
Hi Dr. Baez et al, I commited a stupid error above (maybe).
((in Ken Tucker posts)).
The quantities Mass, Time, Length are all relative,
but, "c", "h", "q" are invariants, (speed of light, Planck's
constant, fundamental charge) and I think in any
universe, whether positive or negative matter,
we should find c=h=1 and q^2 =1.
For example, using the units of "h" = energy*time
and energy =mass*c^2, then 1*h = mass*time =1.
Here we should denote the common varibles M,R,T
as being those (Mass, Radius,Time) we use in our familiar
universe.
Hence in a unverse of negative matter, using "h=1"
we should find,
h = m*t =1
where m = -M and t=-T
Let's look at acceleration, given in OUR universe
by A= R/T^2, that becomes a= r/t^2 in the negative
universe, because t^2 =T^2 and r=-R thus,
A = - a[/itex].
Hence in the negitive mass universe, Newton suggests
A = M/R^2 in our universe
is transformed too,
a = m/r^2 in the negative universe.
I don't want to make a mistake, so I'll transform
this way,
m=-M , A= -a and [itex]r^2 = R^2,,,
to get,
M*A = m*a = F = f.
Sorry, about the detail, I was just trying to compare
the polarity of the scalar "m", with our usual "M".
Regards
Ken S. Tucker
<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 20/04/2004 7:35 am, John Baez at baez@galaxy.ucr.edu wrote:\n\n>\n> Next puzzle... I may have asked Oz this already:\n>\n> PUZZLE: in Newtonian gravity, how does a small negative mass\n> "orbit" a big positive mass? What curve does it trace out?\n\nThat\'s easy: 2-body orbits are conic sections, so it must be a hyperbola.\nWith the big body at the focus of the other branch.\n\nTim\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>on 20/04/2004 7:35 am, John Baez at baez@galaxy.ucr.edu wrote:
>
> Next puzzle... I may have asked Oz this already:
>
> PUZZLE: in Newtonian gravity, how does a small negative mass
> "orbit" a big positive mass? What curve does it trace out?
That's easy: 2-body orbits are conic sections, so it must be a hyperbola.
With the big body at the focus of the other branch.
Tim
robert bristow-johnson
Apr22-04, 04:17 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>In article c62ciq\\$h9g\\$1@glue.ucr.edu, John Baez at baez@galaxy.ucr.edu wrote\non 04/20/2004 02:35:\n\n> In article <c612a9\\$d0e\\$1@lfa222122.richmond.edu>,\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>> John Baez wrote:\n>\n>>> a positive-mass body attracts EVERYTHING;\n>>> a negative-mass body repels EVERYTHING.\n>\n>> This sounds paradoxical - it would mean a positive-mass body attracts\n>> a negative-mass body, while the latter repels the former.\n>> Probably they are chasing after each other???\n>\n> Right. It **sounds** paradoxical - that\'s why I\'m talking about it!\n> But it\'s not; it\'s just weird.\n>\n> Of course, the fact that we never see such runaway motions is evidence\n> that there aren\'t negative masses!\n\nokay, if that is the case, then from what principle does your "a\npositive-mass body attracts EVERYTHING; a negative-mass body repels\nEVERYTHING" statement come from? from a Newtonian POV it seems nonsensical\nbecause of Newton\'s 3rd law. if it were like static E&M except for a sign\nchange (like signed masses attract, unlike signed masses repel), that could\nmake sense since both would be attracting or repelling each other.\n\nso where (from what parent theory) does that principle come from, John?\n\nr b-j\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>In article c62ciq$h9g$1@glue.ucr.edu, John Baez at baez@galaxy.ucr.edu wrote
on 04/20/2004 02:35:
> In article <c612a9$d0e$1@lfa222122.richmond.edu>,
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>> John Baez wrote:
>
>>> a positive-mass body attracts EVERYTHING;
>>> a negative-mass body repels EVERYTHING.
>
>> This sounds paradoxical - it would mean a positive-mass body attracts
>> a negative-mass body, while the latter repels the former.
>> Probably they are chasing after each other???
>
> Right. It **sounds** paradoxical - that's why I'm talking about it!
> But it's not; it's just weird.
>
> Of course, the fact that we never see such runaway motions is evidence
> that there aren't negative masses!
okay, if that is the case, then from what principle does your "a
positive-mass body attracts EVERYTHING; a negative-mass body repels
EVERYTHING" statement come from? from a Newtonian POV it seems nonsensical
because of Newton's 3rd law. if it were like static E&M except for a sign
change (like signed masses attract, unlike signed masses repel), that could
make sense since both would be attracting or repelling each other.
so where (from what parent theory) does that principle come from, John?
r b-j
John Baez
Apr22-04, 04:29 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <kabhc.8641\\$um3.210686@bgtnsc04-news.ops.worldnet.att.net>,\nRichard Saam <rdsaam@att.net> wrote:\n\n>What are the implications for an expanding universe? Are there negative\n>mass dust clouds here and there?\n\nNobody has ever seen them, and I doubt they exist. For one,\nas we\'ve seen here, negative mass stuff would do incredibly weird\nthings - so it probably doesn\'t exist, or we\'d notice! For two,\na negative-mass dust cloud would automatically spread apart and\ndissolve under the effects of gravity! There\'d be no reason for\nit to form in the first place.\n\nThe physics of negative mass particles is mainly good for\nstretching our brains and having a little fun.\n\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>In article <kabhc.8641$um3.210686@bgtnsc04-news.ops.worldnet.att.net>,
Richard Saam <rdsaam@att.net> wrote:
>What are the implications for an expanding universe? Are there negative
>mass dust clouds here and there?
Nobody has ever seen them, and I doubt they exist. For one,
as we've seen here, negative mass stuff would do incredibly weird
things - so it probably doesn't exist, or we'd notice! For two,
a negative-mass dust cloud would automatically spread apart and
dissolve under the effects of gravity! There'd be no reason for
it to form in the first place.
The physics of negative mass particles is mainly good for
stretching our brains and having a little fun.
Aaron Bergman
Apr23-04, 04:13 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>In article <c68tqt\\$84l\\$1@glue.ucr.edu>, John Baez wrote:\n>\n> The physics of negative mass particles is mainly good for\n> stretching our brains and having a little fun.\n\nCan you write down a lagrangian for a negative mass particle?\n\nAaron\n--\nAaron Bergman\n<http://zippy.ph.utexas.edu/~abergman/>\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 <c68tqt$84l$1@glue.ucr.edu>, John Baez wrote:
>
> The physics of negative mass particles is mainly good for
> stretching our brains and having a little fun.
Can you write down a lagrangian for a negative mass particle?
Aaron
--
Aaron Bergman
<http://zippy.ph.utexas.edu/~abergman/>
Stephen Riley
Apr23-04, 04:22 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In message <c62ciq\\$h9g\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\nwrites\n\n>They stay the same distance apart, at least in Newtonian mechanics,\n>where the question is completely unambiguous because we don\'t need\n>to worry about Lorentz transforms.\n\nSo looking at this in 1 dimension, their separation remains constant but\nthey accelerate away together at a constant rate in the direction of the\npositive mass, forever. This assumes equal magnitude masses with each\ninitially at rest relative to the other?\n\nBut, staying in 1D for simplicity, what if one body were less massive\nthan the other, even slightly? If for example there\'s a large positive\nmass and a smaller negative mass; the same force acts between them so\nthere\'s a net attraction and they do eventually collide and don\'t\naccelerate forever. They annihilate, or collide and come back where they\nstarted. If instead the larger mass were negative, acceleration between\nthem (repulsion in this case) is proportional to the inverse square of\nthe distance, so while they do accelerate forever this tends to zero. So\nthere are three extremes, in the first case bodies collide, in the\nsecond acceleration approaches zero, and in the third case, where masses\nare equal, acceleration is constant. But the third case is unstable in\nthat if things do not start out with each body still wrt to the other\none of the other two cases results, from conservation of energy, because\nthe bodies continue to separate or approach at this initial relative\nspeed. So we probably wouldn\'t see runaway acceleration?\n\nIn the first case they collide - the collision formulas still work since\nthe denominator only goes to zero for equal (and opposite signed)\nmasses. But what if one mass was hurled at another opposite signed mass\nof equal magnitude? Another force would be required to stop equal (and\nopposite) masses hitting each other otherwise the collision formulas\nblow up, which presumably means energy and/or momentum wouldn\'t be\nconserved. Any self respecting universe allow that surely :) If in the\nunlikely event that any of the above is correct, what would that force\nlook like?\n\n--\nStephen Riley\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 message <c62ciq$h9g$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
writes
>They stay the same distance apart, at least in Newtonian mechanics,
>where the question is completely unambiguous because we don't need
>to worry about Lorentz transforms.
So looking at this in 1 dimension, their separation remains constant but
they accelerate away together at a constant rate in the direction of the
positive mass, forever. This assumes equal magnitude masses with each
initially at rest relative to the other?
But, staying in 1D for simplicity, what if one body were less massive
than the other, even slightly? If for example there's a large positive
mass and a smaller negative mass; the same force acts between them so
there's a net attraction and they do eventually collide and don't
accelerate forever. They annihilate, or collide and come back where they
started. If instead the larger mass were negative, acceleration between
them (repulsion in this case) is proportional to the inverse square of
the distance, so while they do accelerate forever this tends to zero. So
there are three extremes, in the first case bodies collide, in the
second acceleration approaches zero, and in the third case, where masses
are equal, acceleration is constant. But the third case is unstable in
that if things do not start out with each body still wrt to the other
one of the other two cases results, from conservation of energy, because
the bodies continue to separate or approach at this initial relative
speed. So we probably wouldn't see runaway acceleration?
In the first case they collide - the collision formulas still work since
the denominator only goes to zero for equal (and opposite signed)
masses. But what if one mass was hurled at another opposite signed mass
of equal magnitude? Another force would be required to stop equal (and
opposite) masses hitting each other otherwise the collision formulas
blow up, which presumably means energy and/or momentum wouldn't be
conserved. Any self respecting universe allow that surely :) If in the
unlikely event that any of the above is correct, what would that force
look like?
--
Stephen Riley
<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>> Next puzzle... I may have asked Oz this already:\n>\n> PUZZLE: in Newtonian gravity, how does a small negative mass\n> "orbit" a big positive mass? What curve does it trace out?\n\nI\'ll bite : It must still be an elipse (one minus sign that appears in\nboth equations hence a remains unchanged). Presumeably the point about\nwhich both planets mutually orbit is on the opposite side of the larger\nmass from usual though ?\n\n--\nBoo\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>> Next puzzle... I may have asked Oz this already:
>
> PUZZLE: in Newtonian gravity, how does a small negative mass
> "orbit" a big positive mass? What curve does it trace out?
I'll bite : It must still be an elipse (one minus sign that appears in
both equations hence a remains unchanged). Presumeably the point about
which both planets mutually orbit is on the opposite side of the larger
mass from usual though ?
--
Boo
<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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c68tqt\\$84l\\$1@glue.ucr.edu>...\n> In article <kabhc.8641\\$um3.210686@bgtnsc04-news.ops.worldnet.att.net>,\n> Richard Saam <rdsaam@att.net> wrote:\n>\n> >What are the implications for an expanding universe? Are there negative\n> >mass dust clouds here and there?\n>\n> Nobody has ever seen them, and I doubt they exist. For one,\n> as we\'ve seen here, negative mass stuff would do incredibly weird\n> things - so it probably doesn\'t exist, or we\'d notice! For two,\n> a negative-mass dust cloud would automatically spread apart and\n> dissolve under the effects of gravity! There\'d be no reason for\n> it to form in the first place.\n>\n> The physics of negative mass particles is mainly good for\n> stretching our brains and having a little fun.\n\nSo that\'s very strong evidence against the existence of wormholes\nsince wormholes would require negative mass to exist.\n\nDavid\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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c68tqt$84l$1@glue.ucr.edu>...
> In article <kabhc.8641$um3.210686@bgtnsc04-news.ops.worldnet.att.net>,
> Richard Saam <rdsaam@att.net> wrote:
>
> >What are the implications for an expanding universe? Are there negative
> >mass dust clouds here and there?
>
> Nobody has ever seen them, and I doubt they exist. For one,
> as we've seen here, negative mass stuff would do incredibly weird
> things - so it probably doesn't exist, or we'd notice! For two,
> a negative-mass dust cloud would automatically spread apart and
> dissolve under the effects of gravity! There'd be no reason for
> it to form in the first place.
>
> The physics of negative mass particles is mainly good for
> stretching our brains and having a little fun.
So that's very strong evidence against the existence of wormholes
since wormholes would require negative mass to exist.
David
ebunn@lfa221051.richmond.edu
Apr24-04, 12:19 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>In article <BCAC39B8.A921%rbj@surfglobal.net>,\nrobert bristow-johnson <rbj@surfglobal.net> wrote:\n\n>okay, if that is the case, then from what principle does your "a\n>positive-mass body attracts EVERYTHING; a negative-mass body repels\n>EVERYTHING" statement come from?\n\nI won\'t pretend to speak for John B., but personally, I\'d say that this\nprinciple comes from\n\n1. The equivalence principle\n2. The fact that a positive-mass object attracts another positive-mass\nobject\n3. Newton\'s 3rd Law.\n\nI know you don\'t like this, because you say\n\n>from a Newtonian POV it seems nonsensical\n>because of Newton\'s 3rd law.\n\nI\'m not quite sure why you think that, though. Perhaps there\'s just a\nsemantic issue here? In the sentence "a negative-mass body repels\neverything," the word "repels" doesn\'t mean "exerts a force that\npoints away from itself"; it means "causes the other body to\naccelerate away from it."\n\nIn a world consisting only of positive-mass objects, those two\nare equivalent, but with negative masses, they\'re not.\n\nLet\'s write down all the directions in an explicit case. Say\nwe have a positive mass at the origin, and a negative mass\nto the right of it (on the positive x axis). Then the directions\nof all the relevant quantities are\n\nForce on positive-mass object: Left\nForce on negative-mass object: Right\nAcceleration of positive-mass object: Left\nAcceleration of negative-mass object: Left\n\nNote that there\'s no problem with Newton\'s 3rd Law: the forces\nare equal and opposite. The positive-mass object attracts\nthe negative-mass object (that is, it causes the negative-mass\nobject to accelerate towards it). The negative-mass object\nrepels the positive-mass object (that is, it causes the positive-mass\nobject to accelerate away from it).\n\nGiven the assumptions above, there\'s no other way it could be. At the\nrisk of belaboring the obvious, here\'s why. (In case it\'s not clear,\nin every sentence below, the word "it" refers to the subject of the\nsentence. That is, "A exerts a force on B that points away from it"\nmeans "A exerts a force on B that points away from A.")\n\nA. A positive mass causes another positive mass to accelerate towards\nit.\n\nB. A positive mass causes anything to accelerate towards it.\n(Equivalence Principle: everything accelerates the same way under a\ngravitational force.) In particular, a positive mass causes a\nnegative mass to accelerate towards it.\n\nC. A positive mass exerts a force on a negative mass that points away\nfrom it. This comes from the definition of mass. If m is negative,\nthen F = ma implies that F and a point in opposite directions.\n\nD. A negative mass exerts a force on a positive mass that points away\nfrom it. (Newton\'s 3rd Law)\n\nE. A negative mass causes a positive mass to accelerate away from it.\n(F = ma)\n\nF. A negative mass causes anything to accelerate away from it.\n(Equivalence Principle)\n\nStatements B and F are the ones that can also be phrased\n\n>"a positive-mass body attracts EVERYTHING; a negative-mass body\n>repels EVERYTHING"\n\n(with or without the shouting).\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\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 <BCAC39B8.A921%rbj@surfglobal.net>,
robert bristow-johnson <rbj@surfglobal.net> wrote:
>okay, if that is the case, then from what principle does your "a
>positive-mass body attracts EVERYTHING; a negative-mass body repels
>EVERYTHING" statement come from?
I won't pretend to speak for John B., but personally, I'd say that this
principle comes from
1. The equivalence principle
2. The fact that a positive-mass object attracts another positive-mass
object
3. Newton's 3rd Law.
I know you don't like this, because you say
>from a Newtonian POV it seems nonsensical
>because of Newton's 3rd law.
I'm not quite sure why you think that, though. Perhaps there's just a
semantic issue here? In the sentence "a negative-mass body repels
everything," the word "repels" doesn't mean "exerts a force that
points away from itself"; it means "causes the other body to
accelerate away from it."
In a world consisting only of positive-mass objects, those two
are equivalent, but with negative masses, they're not.
Let's write down all the directions in an explicit case. Say
we have a positive mass at the origin, and a negative mass
to the right of it (on the positive x axis). Then the directions
of all the relevant quantities are
Force on positive-mass object: Left
Force on negative-mass object: Right
Acceleration of positive-mass object: Left
Acceleration of negative-mass object: Left
Note that there's no problem with Newton's 3rd Law: the forces
are equal and opposite. The positive-mass object attracts
the negative-mass object (that is, it causes the negative-mass
object to accelerate towards it). The negative-mass object
repels the positive-mass object (that is, it causes the positive-mass
object to accelerate away from it).
Given the assumptions above, there's no other way it could be. At the
risk of belaboring the obvious, here's why. (In case it's not clear,
in every sentence below, the word "it" refers to the subject of the
sentence. That is, "A exerts a force on B that points away from it"
means "A exerts a force on B that points away from A.")
A. A positive mass causes another positive mass to accelerate towards
it.
B. A positive mass causes anything to accelerate towards it.
(Equivalence Principle: everything accelerates the same way under a
gravitational force.) In particular, a positive mass causes a
negative mass to accelerate towards it.
C. A positive mass exerts a force on a negative mass that points away
from it. This comes from the definition of mass. If m is negative,
then F = ma implies that F and a point in opposite directions.
D. A negative mass exerts a force on a positive mass that points away
from it. (Newton's 3rd Law)
E. A negative mass causes a positive mass to accelerate away from it.
(F = ma)
F. A negative mass causes anything to accelerate away from it.
(Equivalence Principle)
Statements B and F are the ones that can also be phrased
>"a positive-mass body attracts EVERYTHING; a negative-mass body
>repels EVERYTHING"
(with or without the shouting).
-Ted
--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]
John Baez
Apr24-04, 12:21 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>In article <BCAC39B8.A921%rbj@surfglobal.net>,\nrobert bristow-johnson <rbj@surfglobal.net> wrote:\n\n>>> John Baez wrote:\n\n>>>> a positive-mass body attracts EVERYTHING;\n>>>> a negative-mass body repels EVERYTHING.\n\n>from what principle does your "a\n>positive-mass body attracts EVERYTHING; a negative-mass body repels\n>EVERYTHING" statement come from?\n\nIn Newtonian mechanics it comes from\n\nF = Gmm\'/r^2\n\nand\n\nF = ma\n\nIn general relativity it comes from Einstein\'s equations, which\nsay:\n\nThe rate at which a small initially comoving ball of freely falling\ntest particles begins to shrink is proportional to its volume times:\nthe energy density at the center of the ball, plus the pressure in\nthe x direction at that point, plus the pressure in the y direction,\nplus the pressure in the z direction.\n\nA blob with negative mass and negligible pressure would thus repel\nfreely falling particles in its vicinity, regardless of their mass.\n\nI believe you can also see this sort of thing by taking the Schwarzschild\nsolution, sticking in a negative value of m, and working out the geodesics\nin this metric.\n\n>from a Newtonian POV it seems nonsensical because of Newton\'s 3rd law.\n\nEh? Conservation of momentum *demands* that if a positive-mass\nparticle attracts a negative-mass one, the negative-mass particle\nrepels the positive-mass one. It has nothing to say about whether\ntwo negative-mass particles will attract or repel each other, but\n\nF = Gmm\'/r^2\n\nand\n\nF = ma\n\nsay they repel each other.\n\n>if it were like static E&M except for a sign\n>change (like signed masses attract, unlike signed masses repel), that could\n>make sense since both would be attracting or repelling each other.\n\nNewtonian gravity is *not* like electrostatics, since m shows up twice\nin\n\nF = Gmm\'/r^2\n\nand\n\nF = ma\n\nbut only once in\n\nF = Gqq\'/r^2\n\nand\n\nF = ma.\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 <BCAC39B8.A921%rbj@surfglobal.net>,
robert bristow-johnson <rbj@surfglobal.net> wrote:
>>> John Baez wrote:
>>>> a positive-mass body attracts EVERYTHING;
>>>> a negative-mass body repels EVERYTHING.
>from what principle does your "a
>positive-mass body attracts EVERYTHING; a negative-mass body repels
>EVERYTHING" statement come from?
In Newtonian mechanics it comes from
F = Gmm'/r^2
and
F = ma
In general relativity it comes from Einstein's equations, which
say:
The rate at which a small initially comoving ball of freely falling
test particles begins to shrink is proportional to its volume times:
the energy density at the center of the ball, plus the pressure in
the x direction at that point, plus the pressure in the y direction,
plus the pressure in the z direction.
A blob with negative mass and negligible pressure would thus repel
freely falling particles in its vicinity, regardless of their mass.
I believe you can also see this sort of thing by taking the Schwarzschild
solution, sticking in a negative value of m, and working out the geodesics
in this metric.
>from a Newtonian POV it seems nonsensical because of Newton's 3rd law.
Eh? Conservation of momentum *demands* that if a positive-mass
particle attracts a negative-mass one, the negative-mass particle
repels the positive-mass one. It has nothing to say about whether
two negative-mass particles will attract or repel each other, but
F = Gmm'/r^2
and
F = ma
say they repel each other.
>if it were like static E&M except for a sign
>change (like signed masses attract, unlike signed masses repel), that could
>make sense since both would be attracting or repelling each other.
Newtonian gravity is *not* like electrostatics, since m shows up twice
in
F = Gmm'/r^2
and
F = ma
but only once in
F = Gqq'/r^2
and
F = ma.
John Baez
Apr24-04, 12:21 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>In article <slrnc8gcri.21a.abergman@cardinal2.Stanford.EDU>,\ nAaron Bergman <abergman@physics.utexas.edu> wrote:\n\n>In article <c68tqt\\$84l\\$1@glue.ucr.edu>, John Baez wrote:\n\n>> The physics of negative mass particles is mainly good for\n>> stretching our brains and having a little fun.\n\n>Can you write down a lagrangian for a negative mass particle?\n\nAh, now this is getting really interesting! If you can\nwrite down one for a positive mass particle, I can switch\nthe sign and get one for a negative mass particle. But\nsometimes you can\'t do this!\n\nAre we talking Newtonian classical point particles, special-relativistic\nclassical point particles, general relativistic classical point\nparticles, quantum mechanics, or quantum field theory? All the\nvariations are interesting, but I\'ll just think about two.\n\nIf we\'re talking Newtonian classical point particles, the\nLagrangian is\n\nmv^2/2 - V(q)\n\nThe sign of m here really affects things: if we make it negative,\nwe get particles that accelerate the opposite way than we\'re used to!\n\nIf we\'re talking about a massive spin-0 quantum field, the\nLagrangian doesn\'t say anything about whether the particle\nin question has positive or negative mass, because all that\nshows up in the formula is m^2.\n\nIn this case, we can consistently quantize this Lagrangian in two\ndifferent ways. One way gives particles with positive rest\nmass; the other gives particles with negative rest mass!\n\nDepending on which we choose, we get two unitarily inequivalent\nrepresentations of the Poincare group on the single-particle Hilbert\nspace. Either choice is equally good, since they are equivalent\nby an *antiunitary* operator. The trouble starts when we try\nto cook up theories that allow positive-mass particles to interact\nwith negative-mass ones.\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 <slrnc8gcri.21a.abergman@cardinal2.Stanford.EDU>,
Aaron Bergman <abergman@physics.utexas.edu> wrote:
>In article <c68tqt$84l$1@glue.ucr.edu>, John Baez wrote:
>> The physics of negative mass particles is mainly good for
>> stretching our brains and having a little fun.
>Can you write down a lagrangian for a negative mass particle?
Ah, now this is getting really interesting! If you can
write down one for a positive mass particle, I can switch
the sign and get one for a negative mass particle. But
sometimes you can't do this!
Are we talking Newtonian classical point particles, special-relativistic
classical point particles, general relativistic classical point
particles, quantum mechanics, or quantum field theory? All the
variations are interesting, but I'll just think about two.
If we're talking Newtonian classical point particles, the
Lagrangian is
mv^2/2 - V(q)
The sign of m here really affects things: if we make it negative,
we get particles that accelerate the opposite way than we're used to!
If we're talking about a massive spin-0 quantum field, the
Lagrangian doesn't say anything about whether the particle
in question has positive or negative mass, because all that
shows up in the formula is m^2.
In this case, we can consistently quantize this Lagrangian in two
different ways. One way gives particles with positive rest
mass; the other gives particles with negative rest mass!
Depending on which we choose, we get two unitarily inequivalent
representations of the Poincare group on the single-particle Hilbert
space. Either choice is equally good, since they are equivalent
by an *antiunitary* operator. The trouble starts when we try
to cook up theories that allow positive-mass particles to interact
with negative-mass ones.
Danny Ross Lunsford
Apr24-04, 09:16 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>John Baez wrote:\n\n> The physics of negative mass particles is mainly good for\n> stretching our brains and having a little fun.\n\nHas the work on "ponderomotive theory" of Einstein, Infeld, and Hoffmann\nbeen done for negative mass test particles?\n\n-drl\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 wrote:
> The physics of negative mass particles is mainly good for
> stretching our brains and having a little fun.
Has the work on "ponderomotive theory" of Einstein, Infeld, and Hoffmann
been done for negative mass test particles?
-drl
Danny Ross Lunsford
Apr27-04, 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>John Baez wrote:\n\n> Ah, now this is getting really interesting! If you can\n> write down one for a positive mass particle, I can switch\n> the sign and get one for a negative mass particle. But\n> sometimes you can\'t do this!\n\nWhat about a mass that is positive *and* negative?\n\nHere\'s a relativistic wave equation:\n\n(ym dm - y5 m) psi = 0\n\nTo see that it is formally equivalent to the usual equation, make a\ntransformation of basis\n\nym -> y\'m = iy5 ym\n\nso\n\n{y\'m, y\'n} = -{y5 ym, y5 yn} = 2 gmn\n\nA suitable S leading from one basis to the other is\n\nS = 1/root2(1 + iy5) S-1 = 1/root2(1 - iy5)\n\nS ym S-1 = 1/2(1 + iy5) ym (1 - iy5)\n= 1/2(ym + i[y5,ym] - ym)\n= iy5 ym\n\nThe transformed spinor is\n\npsi\' = 1/root2(1 + iy5) psi\n\nand\n\n(i y5 y\'m dm - m y5) psi\' = 0\n\nor\n\n(y\'m dm + im) psi\' = 0\n\nGeometrically, however, the "pseudo" Dirac equation is a very different\nanimal. In order to be fully Lorentz invariant the mass has to change\nsign under spacetime parity. If you really "believe" that antimatter is\nnegative mass, then this is "correct" equation!\n\n-drl\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 wrote:
> Ah, now this is getting really interesting! If you can
> write down one for a positive mass particle, I can switch
> the sign and get one for a negative mass particle. But
> sometimes you can't do this!
What about a mass that is positive *and* negative?
Here's a relativistic wave equation:
(ym dm - y5 m) \psi =
To see that it is formally equivalent to the usual equation, make a
transformation of basis
ym ->[/itex] y'm = iy5 ym
so
{y'm, y'n} = -{y5 ym, y5 yn} = 2 gmn
A suitable S leading from one basis to the other is
S = 1/root2(1 + iy5) S-1 = 1/root2(1 - iy5)
S ym S-1 = 1/2(1 + iy5) ym (1 - iy5)
= 1/2(ym + i[y5,ym] - ym)
= iy5 ym
The transformed spinor is
\psi' = 1/root2(1 + iy5) [itex]\psi
and
(i y5 y'm dm - m y5) \psi' =
or
(y'm dm + im) \psi' =
Geometrically, however, the "pseudo" Dirac equation is a very different
animal. In order to be fully Lorentz invariant the mass has to change
sign under spacetime parity. If you really "believe" that antimatter is
negative mass, then this is "correct" equation!
-drl
Aaron Denney
Apr27-04, 02:49 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 2004-04-22, John Baez <baez@galaxy.ucr.edu> wrote:\n> Nobody has ever seen them, and I doubt they exist. For one,\n> as we\'ve seen here, negative mass stuff would do incredibly weird\n> things - so it probably doesn\'t exist, or we\'d notice! For two,\n> a negative-mass dust cloud would automatically spread apart and\n> dissolve under the effects of gravity! There\'d be no reason for\n> it to form in the first place.\n\nWhat about charged negative-mass particles? If they have like charges,\nshouldn\'t they glom together rapidly?\n\n--\nAaron Denney\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>On 2004-04-22, John Baez <baez@galaxy.ucr.edu> wrote:
> Nobody has ever seen them, and I doubt they exist. For one,
> as we've seen here, negative mass stuff would do incredibly weird
> things - so it probably doesn't exist, or we'd notice! For two,
> a negative-mass dust cloud would automatically spread apart and
> dissolve under the effects of gravity! There'd be no reason for
> it to form in the first place.
What about charged negative-mass particles? If they have like charges,
shouldn't they glom together rapidly?
--
Aaron Denney
-><-
Charles Francis
Apr27-04, 02:54 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>In message <c62cq7\\$hev\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\nwrites\n>In article <UYfIiCF6rWgAFw5c@clef.demon.co.uk>,\n>Charles Francis <charles@clef.demon.co.uk> wrote:\n>\n>> In article <c5fcj8\\$6ua\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\n>>writes\n>\n>>>a positive-mass body attracts EVERYTHING;\n>>>a negative-mass body repels EVERYTHING.\n>\n>>Well, if you say so.\n>\n>Perhaps "if" I say so, but not simply "because" I do. :-)\n>\n>>>>>PUZZLE:\n>>>>>\n>>>>> Figure out what happens if you have two planets near each\n>>>>> other: Earth and Anti-Earth, the first with positive mass, the\n>>>>> second with an "equal but opposite" negative mass.\n>\n>>>>If they were solid enough to resist\n>>>>gravitational forces then they clearly would accelerate across the\n>>>>universe, trailing their gravitational fields behind them.\n>\n>>>Right! Excellent!\n>\n>>Is it?\n>\n>Yes - because then, the fact that this crazy behavior is not\n>seen is evidence that there are no negative-mass objects.\n>\nNo, it just means we live in a matter universe, and that antimatter\nscooting it would rapidly get destroyed. In any case absence of\nobservation is not necessarily evidence of absence. I would rather see a\ntheoretical demonstration. As I recall D\'Inverno shows the equivalence\nof inertial mass with active and passive gravitational mass, and since\nwe know negative mass is time reversed and manifests as positive mass,\nthat ought to be enough.\n\n--\nCharles Francis\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 message <c62cq7$hev$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
writes
>In article <UYfIiCF6rWgAFw5c@clef.demon.co.uk>,
>Charles Francis <charles@clef.demon.co.uk> wrote:
>
>> In article <c5fcj8$6ua$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
>>writes
>
>>>a positive-mass body attracts EVERYTHING;
>>>a negative-mass body repels EVERYTHING.
>
>>Well, if you say so.
>
>Perhaps "if" I say so, but not simply "because" I do. :-)
>
>>>>>PUZZLE:
>>>>>
>>>>> Figure out what happens if you have two planets near each
>>>>> other: Earth and Anti-Earth, the first with positive mass, the
>>>>> second with an "equal but opposite" negative mass.
>
>>>>If they were solid enough to resist
>>>>gravitational forces then they clearly would accelerate across the
>>>>universe, trailing their gravitational fields behind them.
>
>>>Right! Excellent!
>
>>Is it?
>
>Yes - because then, the fact that this crazy behavior is not
>seen is evidence that there are no negative-mass objects.
>
No, it just means we live in a matter universe, and that antimatter
scooting it would rapidly get destroyed. In any case absence of
observation is not necessarily evidence of absence. I would rather see a
theoretical demonstration. As I recall D'Inverno shows the equivalence
of inertial mass with active and passive gravitational mass, and since
we know negative mass is time reversed and manifests as positive mass,
that ought to be enough.
--
Charles Francis
Charles Francis
Apr27-04, 02:54 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In message <c6cg9f\\$mg2\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\nwrites\n>\n>>if it were like static E&M except for a sign\n>>change (like signed masses attract, unlike signed masses repel), that could\n>>make sense since both would be attracting or repelling each other.\n>\n>Newtonian gravity is *not* like electrostatics, since m shows up twice\n>in\n>\n>F = Gmm\'/r^2\n>\n>and\n>\n>F = ma\n\nI make that thrice!\n\n>but only once in\n>\n>F = Gqq\'/r^2\n>\n>and\n>\n>F = ma.\n>\nThe only trouble is we already know that the transformation m -> -m\ntakes matter to antimatter and t to -t, so there is an extra minus of\nwhich you take no account, and as a result negative inertial mass\nmanifests as positive.\n\n\n\n--\nCharles Francis\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 message <c6cg9f$mg2$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
writes
>
>>if it were like static E&M except for a sign
>>change (like signed masses attract, unlike signed masses repel), that could
>>make sense since both would be attracting or repelling each other.
>
>Newtonian gravity is *not* like electrostatics, since m shows up twice
>in
>
>F = Gmm'/r^2
>
>and
>
>F = ma
I make that thrice!
>but only once in
>
>F = Gqq'/r^2
>
>and
>
>F = ma.
>
The only trouble is we already know that the transformation m -> -m
takes matter to antimatter and t to -t, so there is an extra minus of
which you take no account, and as a result negative inertial mass
manifests as positive.
--
Charles Francis
Kefka G
Apr27-04, 02:59 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>rbj@surfglobal.net wrote:\n\n>>> John Baez wrote:\n>>\n>>>> a positive-mass body attracts EVERYTHING;\n>>>> a negative-mass body repels EVERYTHING.\n>>\n>>> This sounds paradoxical - it would mean a positive-mass body attracts\n>>> a negative-mass body, while the latter repels the former.\n>>> Probably they are chasing after each other???\n>>\n>> Right. It **sounds** paradoxical - that\'s why I\'m talking about it!\n>> But it\'s not; it\'s just weird.\n>>\n>> Of course, the fact that we never see such runaway motions is evidence\n>> that there aren\'t negative masses!\n>\n>okay, if that is the case, then from what principle does your "a\n>positive-mass body attracts EVERYTHING; a negative-mass body repels\n>EVERYTHING" statement come from? from a Newtonian POV it seems nonsensical\n>because of Newton\'s 3rd law. if it were like static E&M except for a sign\n>change (like signed masses attract, unlike signed masses repel), that could\n>make sense since both would be attracting or repelling each other.\n>\n>so where (from what parent theory) does that principle come from, John?\n>\n\nI\'m not John, but I think confusion regarding Newton\'s third law arises because\nof the following - the third law is, indeed, satisfied, it\'s just that a\npositive force on a negative mass object induces a negative acceleration. In\nother words, a "push" is effectively a "pull" when we\'re dealing with negative\nmass. Yes, the resulting physics is somewhat pathological, but it is good to\nnote that in classical E/M, we have much the same problem - when we assume\npoint charges exist, we\'re forced to assume that the "bare" mass of a particle\nis negative so that we end up with a finite total mass. This leads to the well\nknown runaway solutions when we calculate the self force of a point charge.\nThese end up conserving energy precisely because of the negative bare mass -\nthe increase of energy in the fields is balanced by a decrease of energy in the\nparticle. We generally choose to ignore these solutions, preferring the\nacausal ones (see Rohrlich\'s 1965 book for a better explanation).\n\n-Eric\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>rbj@surfglobal.net wrote:
>>> John Baez wrote:
>>
>>>> a positive-mass body attracts EVERYTHING;
>>>> a negative-mass body repels EVERYTHING.
>>
>>> This sounds paradoxical - it would mean a positive-mass body attracts
>>> a negative-mass body, while the latter repels the former.
>>> Probably they are chasing after each other???
>>
>> Right. It **sounds** paradoxical - that's why I'm talking about it!
>> But it's not; it's just weird.
>>
>> Of course, the fact that we never see such runaway motions is evidence
>> that there aren't negative masses!
>
>okay, if that is the case, then from what principle does your "a
>positive-mass body attracts EVERYTHING; a negative-mass body repels
>EVERYTHING" statement come from? from a Newtonian POV it seems nonsensical
>because of Newton's 3rd law. if it were like static E&M except for a sign
>change (like signed masses attract, unlike signed masses repel), that could
>make sense since both would be attracting or repelling each other.
>
>so where (from what parent theory) does that principle come from, John?
>
I'm not John, but I think confusion regarding Newton's third law arises because
of the following - the third law is, indeed, satisfied, it's just that a
positive force on a negative mass object induces a negative acceleration. In
other words, a "push" is effectively a "pull" when we're dealing with negative
mass. Yes, the resulting physics is somewhat pathological, but it is good to
note that in classical E/M, we have much the same problem - when we assume
point charges exist, we're forced to assume that the "bare" mass of a particle
is negative so that we end up with a finite total mass. This leads to the well
known runaway solutions when we calculate the self force of a point charge.
These end up conserving energy precisely because of the negative bare mass -
the increase of energy in the fields is balanced by a decrease of energy in the
particle. We generally choose to ignore these solutions, preferring the
acausal ones (see Rohrlich's 1965 book for a better explanation).
-Eric
Aaron Bergman
Apr28-04, 02:45 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 <c6chsp\\$n0k\\$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)\nwrote:\n\n> Depending on which we choose, we get two unitarily inequivalent\n> representations of the Poincare group on the single-particle Hilbert\n> space. Either choice is equally good, since they are equivalent\n> by an *antiunitary* operator. The trouble starts when we try\n> to cook up theories that allow positive-mass particles to interact\n> with negative-mass ones.\n\nI don\'t see offhand how you\'re going to tell the difference between the\nordinary theory and the \'negative mass\' theory run backwards in time.\n\nAaron\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 <c6chsp$n0k$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)
wrote:
> Depending on which we choose, we get two unitarily inequivalent
> representations of the Poincare group on the single-particle Hilbert
> space. Either choice is equally good, since they are equivalent
> by an *antiunitary* operator. The trouble starts when we try
> to cook up theories that allow positive-mass particles to interact
> with negative-mass ones.
I don't see offhand how you're going to tell the difference between the
ordinary theory and the 'negative mass' theory run backwards in time.
Aaron
Ken S. Tucker
Apr28-04, 02:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c6chsp\\$n0k\\$1@glue.ucr.edu>...\n>In article <slrnc8gcri.21a.abergman@cardinal2.Stanford.EDU>,\ n>Aaron Bergman <abergman@physics.utexas.edu> wrote:\n>\n>>In article <c68tqt\\$84l\\$1@glue.ucr.edu>, John Baez wrote:\n>>> The physics of negative mass particles is mainly good for\n>>> stretching our brains and having a little fun.\n>>Can you write down a lagrangian for a negative mass particle?\n>\n>Ah, now this is getting really interesting! If you can\n>write down one for a positive mass particle, I can switch\n>the sign and get one for a negative mass particle. But\n>sometimes you can\'t do this!\n>\n>Are we talking Newtonian classical point particles, special-relativistic\n>classical point particles, general relativistic classical point\n>particles, quantum mechanics, or quantum field theory? All the\n>variations are interesting, but I\'ll just think about two.\n>\n>If we\'re talking Newtonian classical point particles, the\n>Lagrangian is\n>\n>mv^2/2 - V(q)\n>\n>The sign of m here really affects things: if we make it negative,\n>we get particles that accelerate the opposite way than we\'re used to!\n>\n>If we\'re talking about a massive spin-0 quantum field, the\n>Lagrangian doesn\'t say anything about whether the particle\n>in question has positive or negative mass, because all that\n>shows up in the formula is m^2.\n>\n>In this case, we can consistently quantize this Lagrangian in two\n>different ways. One way gives particles with positive rest\n>mass; the other gives particles with negative rest mass!\n>\n>Depending on which we choose, we get two unitarily inequivalent\n>representations of the Poincare group on the single-particle Hilbert\n>space. Either choice is equally good, since they are equivalent\n>by an *antiunitary* operator. The trouble starts when we try\n>to cook up theories that allow positive-mass particles to interact\n>with negative-mass ones.\n\nPardon my knee-jerk, let\'s look at the geodesics\nof any material particle, (from the standpoint of GR),\nwith any scalar polarity, (m, -m, sqrt(-1)*m)...).\n\nAn easy definition of a material particle in classical\ngeodesic motion is one where an accelometer on the\ngeodesic would read zero, aka free-fall.\n\nSuppose we were to construct our accelometer using\n*negative matter*, we should expect that accelometer\nto read zero, as does an accelometer constructed from\npositive matter. This is in respect of Equivalence.\n\nMy point is, the material of the construction is\nunimportant, since they will both read zero.\n\nHence, I think, GR would work independant of the\nrelative mass scalar polarity.\n\nInteresting, Ken S. Tucker\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>baez@galaxy.ucr.edu (John Baez) wrote in message news:<c6chsp$n0k$1@glue.ucr.edu>...
>In article <slrnc8gcri.21a.abergman@cardinal2.Stanford.EDU>,
>Aaron Bergman <abergman@physics.utexas.edu> wrote:
>
>>In article <c68tqt$84l$1@glue.ucr.edu>, John Baez wrote:
>>> The physics of negative mass particles is mainly good for
>>> stretching our brains and having a little fun.
>>Can you write down a lagrangian for a negative mass particle?
>
>Ah, now this is getting really interesting! If you can
>write down one for a positive mass particle, I can switch
>the sign and get one for a negative mass particle. But
>sometimes you can't do this!
>
>Are we talking Newtonian classical point particles, special-relativistic
>classical point particles, general relativistic classical point
>particles, quantum mechanics, or quantum field theory? All the
>variations are interesting, but I'll just think about two.
>
>If we're talking Newtonian classical point particles, the
>Lagrangian is
>
>mv^2/2 - V(q)
>
>The sign of m here really affects things: if we make it negative,
>we get particles that accelerate the opposite way than we're used to!
>
>If we're talking about a massive spin-0 quantum field, the
>Lagrangian doesn't say anything about whether the particle
>in question has positive or negative mass, because all that
>shows up in the formula is m^2.
>
>In this case, we can consistently quantize this Lagrangian in two
>different ways. One way gives particles with positive rest
>mass; the other gives particles with negative rest mass!
>
>Depending on which we choose, we get two unitarily inequivalent
>representations of the Poincare group on the single-particle Hilbert
>space. Either choice is equally good, since they are equivalent
>by an *antiunitary* operator. The trouble starts when we try
>to cook up theories that allow positive-mass particles to interact
>with negative-mass ones.
Pardon my knee-jerk, let's look at the geodesics
of any material particle, (from the standpoint of GR),
with any scalar polarity, (m, -m, \sqrt(-1)*m)...).
An easy definition of a material particle in classical
geodesic motion is one where an accelometer on the
geodesic would read zero, aka free-fall.
Suppose we were to construct our accelometer using
*negative matter*, we should expect that accelometer
to read zero, as does an accelometer constructed from
positive matter. This is in respect of Equivalence.
My point is, the material of the construction is
unimportant, since they will both read zero.
Hence, I think, GR would work independant of the
relative mass scalar polarity.
Interesting, Ken S. Tucker
alistair
Apr30-04, 11:31 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>Is negative matter associated with gravitons which have negative\nenergy?\nAnd if so, do these gravitons carry momentum in the same direction\nthat they are moving?\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>Is negative matter associated with gravitons which have negative
energy?
And if so, do these gravitons carry momentum in the same direction
that they are moving?
Gordon D. Pusch
May13-04, 02:00 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>NNTP-Posting-Host: perimeterinstitute.ca\nDate: 13 May 2004 06:25:09 -0400\nX-Trace: news.sentex.net 1084443909 perimeterinstitute.ca (13 May 2004 06:25:09 -0400)\nLines: 27\nPath: news.easynews.com!core-easynews!newsfeed1.easynews.com!easynews.com!easyn ews!sjc1.usenetserver.com!news.usenetserver.com!cy clone.bc.net!newsfeed2.telusplanet.net!newsfeed.te lus.net!news.sentex.net!not-for-mail\nXref: core-easynews sci.physics.research:56147\nX-Received-Date: Thu, 13 May 2004 03:24:48 MST (news.easynews.com)\n\n\nTim S <Tim@timsilverman.demon.co.uk> writes:\n\n> on 20/04/2004 7:35 am, John Baez at baez@galaxy.ucr.edu wrote:\n>\n>>\n>> Next puzzle... I may have asked Oz this already:\n>>\n>> PUZZLE: in Newtonian gravity, how does a small negative mass\n>> "orbit" a big positive mass? What curve does it trace out?\n>\n> That\'s easy: 2-body orbits are conic sections, so it must be a hyperbola.\n> With the big body at the focus of the other branch.\n\n....Actually, that would be what you\'d get if a small mass was "orbiting" a\nlarge negative one.\n\nFor a small negative mass orbiting a large positive one, the orbit of the\nsmall negative mass would be a normal ellipse around the postive mass body;\nwhat would be a bit odd is that the positive mass body would be _between_\nthe small negative mass and the center of mass of the combined system,\nwhich would still be at one of the foci of the ellipse...\n\n\n-- Gordon D. Pusch\n\nperl -e \'\\$_ = "gdpusch\\@NO.xnet.SPAM.com\\n"; s/NO\\.//; s/SPAM\\.//; print;\'\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>NNTP-Posting-Host: perimeterinstitute.ca
Date: 13 May 2004 06:25:09 -0400
X-Trace: news.sentex.net 1084443909 perimeterinstitute.ca (13 May 2004 06:25:09 -0400)
Lines: 27
Path: news.easynews.com!core-easynews!newsfeed1.easynews.com!easynews.com!easyn ews!sjc1.usenetserver.com!news.usenetserver.com!cy clone.bc.net!newsfeed2.telusplanet.net!newsfeed.te lus.net!news.sentex.net!not-for-mail
Xref: core-easynews sci.physics.research:56147
X-Received-Date: Thu, 13 May 2004 03:24:48 MST (news.easynews.com)
Tim S <Tim@timsilverman.demon.co.uk> writes:
> on 20/04/2004 7:35 am, John Baez at baez@galaxy.ucr.edu wrote:
>
>>
>> Next puzzle... I may have asked Oz this already:
>>
>> PUZZLE: in Newtonian gravity, how does a small negative mass
>> "orbit" a big positive mass? What curve does it trace out?
>
> That's easy: 2-body orbits are conic sections, so it must be a hyperbola.
> With the big body at the focus of the other branch.
....Actually, that would be what you'd get if a small mass was "orbiting" a
large negative one.
For a small negative mass orbiting a large positive one, the orbit of the
small negative mass would be a normal ellipse around the postive mass body;
what would be a bit odd is that the positive mass body would be _between_
the small negative mass and the center of mass of the combined system,
which would still be at one of the foci of the ellipse...
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
John Baez
May17-04, 03:03 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <abergman-ED510D.20563024042004@localhost>,\nAaron Bergman <abergman@physics.utexas.edu> wrote:\n\n>In article <c6chsp\\$n0k\\$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)\n>wrote:\n\n>> Depending on which we choose, we get two unitarily inequivalent\n>> representations of the Poincare group on the single-particle Hilbert\n>> space. Either choice is equally good, since they are equivalent\n>> by an *antiunitary* operator.\n\n>I don\'t see offhand how you\'re going to tell the difference between the\n>ordinary theory and the \'negative mass\' theory run backwards in time.\n\nRight: time reversal is the antiunitary operator I was referring to\nabove.\n\nHowever, there is a real difference between the theory of two species\nof equal-mass particles, and the theory of two species of particles\nof opposite mass! In the latter case the spectrum of the Hamiltonian\nis the whole real line, while in the former the spectrum lies in a\nhalf-line. This is for free particles. But....\n\n>> The trouble starts when we try\n>> to cook up theories that allow positive-mass particles to interact\n>> with negative-mass ones.\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 <abergman-ED510D.20563024042004@localhost>,
Aaron Bergman <abergman@physics.utexas.edu> wrote:
>In article <c6chsp$n0k$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)
>wrote:
>> Depending on which we choose, we get two unitarily inequivalent
>> representations of the Poincare group on the single-particle Hilbert
>> space. Either choice is equally good, since they are equivalent
>> by an *antiunitary* operator.
>I don't see offhand how you're going to tell the difference between the
>ordinary theory and the 'negative mass' theory run backwards in time.
Right: time reversal is the antiunitary operator I was referring to
above.
However, there is a real difference between the theory of two species
of equal-mass particles, and the theory of two species of particles
of opposite mass! In the latter case the spectrum of the Hamiltonian
is the whole real line, while in the former the spectrum lies in a
half-line. This is for free particles. But....
>> The trouble starts when we try
>> to cook up theories that allow positive-mass particles to interact
>> with negative-mass ones.
John Baez
May17-04, 06:05 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <c6ma97\\$tom\\$1@lfa222122.richmond.edu>,\nCharle s Francis <charles@clef.demon.co.uk> wrote:\n\n>In message <c6cg9f\\$mg2\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\n>writes\n\n>>Newtonian gravity is *not* like electrostatics, since m shows up twice\n>>in\n>>\n>>F = Gmm\'/r^2\n>>\n>>and\n>>\n>>F = ma\n\n>I make that thrice!\n\nThe concept of mass shows up thrice, but m shows up twice:\n\nThus, when you reverse the sign of just the mass m, the particle whose\nmass was m feels the same acceleration.\n\nBut, if you switch the signs of *all* masses in a gravitational\nproblem, all accelerations switch sign.\n\nBoth the previous two sentences would work the other way if we\nwere doing electrostatics and said "charge" and "q" instead of\n"mass" and "m":\n\nWhen you reverse the sign of just the charge q, the particle whose\ncharge was q feels the opposite acceleration.\n\nBut: if you switch the signs of *all* charges in a electrostatics\nproblem, all accelerations stay the same.\n\nHas someone pointed out yet that this is related to the fact that\nphotons have odd spin while gravitons have even spin?\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>In article <c6ma97$tom$1@lfa222122.richmond.edu>,
Charles Francis <charles@clef.demon.co.uk> wrote:
>In message <c6cg9f$mg2$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
>writes
>>Newtonian gravity is *not* like electrostatics, since m shows up twice
>>in
>>
>>F = Gmm'/r^2
>>
>>and
>>
>>F = ma
>I make that thrice!
The concept of mass shows up thrice, but m shows up twice:
Thus, when you reverse the sign of just the mass m, the particle whose
mass was m feels the same acceleration.
But, if you switch the signs of *all* masses in a gravitational
problem, all accelerations switch sign.
Both the previous two sentences would work the other way if we
were doing electrostatics and said "charge" and "q" instead of
"mass" and "m":
When you reverse the sign of just the charge q, the particle whose
charge was q feels the opposite acceleration.
But: if you switch the signs of *all* charges in a electrostatics
problem, all accelerations stay the same.
Has someone pointed out yet that this is related to the fact that
photons have odd spin while gravitons have even spin?
John Baez
May17-04, 06:05 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIn article <c6ma8i\\$tol\\$1@lfa222122.richmond.edu>,\nCharle s Francis <charles@clef.demon.co.uk> wrote:\n\n>since we know negative mass is time reversed and manifests\n>as positive mass,\n\nIn all my discussion of this subject I\'ve been talking about\nnegative mass that acts like negative mass - NOT negative mass\nthat through some other sign-switching trick acts like positive\nmass.\n\nEveryone else is free to play other games with other rules,\nbut I\'m just discussing the effect of switching the sign of "m"\nin all my favorite physics formulas.\n\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>In article <c6ma8i$tol$1@lfa222122.richmond.edu>,
Charles Francis <charles@clef.demon.co.uk> wrote:
>since we know negative mass is time reversed and manifests
>as positive mass,
In all my discussion of this subject I've been talking about
negative mass that acts like negative mass - NOT negative mass
that through some other sign-switching trick acts like positive
mass.
Everyone else is free to play other games with other rules,
but I'm just discussing the effect of switching the sign of "m"
in all my favorite physics formulas.
Aaron Bergman
May20-04, 12:51 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>In article <c89154\\$7qb\\$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)\nwrote:\n\n> >> The trouble starts when we try\n> >> to cook up theories that allow positive-mass particles to interact\n> >> with negative-mass ones.\n\nIf they don\'t interact, they might as well be in completely different\nuniverses.\n\nAaron\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 <c89154$7qb$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)
wrote:
> >> The trouble starts when we try
> >> to cook up theories that allow positive-mass particles to interact
> >> with negative-mass ones.
If they don't interact, they might as well be in completely different
universes.
Aaron
Charles Francis
May20-04, 04:41 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>In message <c891to\\$7vh\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>\nwrites\n>\n>\n>In article <c6ma8i\\$tol\\$1@lfa222122.richmond.edu>,\n>Charl es Francis <charles@clef.demon.co.uk> wrote:\n>\n>>since we know negative mass is time reversed and manifests\n>>as positive mass,\n>\n>In all my discussion of this subject I\'ve been talking about\n>negative mass that acts like negative mass - NOT negative mass\n>that through some other sign-switching trick acts like positive\n>mass.\n>\n>Everyone else is free to play other games with other rules,\n>but I\'m just discussing the effect of switching the sign of "m"\n>in all my favorite physics formulas.\n\nAnd it has been a fun game. I have been teasing you just a touch. But\nthere is a little more than fun to it. Ignoring gravity the maths of\nantimatter is, I think pretty solid, but it is not instantly obvious\nwhether negative inertial mass manifesting as positive mass will yield a\nnegative or a positive gravitational mass. I think the game shows that\nin practice gravitational mass must always be positive, so when we are\nlooking for quantum gravity theories that is something we should require\nof them.\n\n--\nCharles Francis\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 message <c891to$7vh$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu>
writes
>
>
>In article <c6ma8i$tol$1@lfa222122.richmond.edu>,
>Charles Francis <charles@clef.demon.co.uk> wrote:
>
>>since we know negative mass is time reversed and manifests
>>as positive mass,
>
>In all my discussion of this subject I've been talking about
>negative mass that acts like negative mass - NOT negative mass
>that through some other sign-switching trick acts like positive
>mass.
>
>Everyone else is free to play other games with other rules,
>but I'm just discussing the effect of switching the sign of "m"
>in all my favorite physics formulas.
And it has been a fun game. I have been teasing you just a touch. But
there is a little more than fun to it. Ignoring gravity the maths of
antimatter is, I think pretty solid, but it is not instantly obvious
whether negative inertial mass manifesting as positive mass will yield a
negative or a positive gravitational mass. I think the game shows that
in practice gravitational mass must always be positive, so when we are
looking for quantum gravity theories that is something we should require
of them.
--
Charles Francis
<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 13/05/2004 11:25 am, Gordon D. Pusch at\ng_d_pusch_remove_underscores@xnet.com wrote:\n\n>\n> Tim S <Tim@timsilverman.demon.co.uk> writes:\n>\n>> on 20/04/2004 7:35 am, John Baez at baez@galaxy.ucr.edu wrote:\n>>\n>>>\n>>> Next puzzle... I may have asked Oz this already:\n>>>\n>>> PUZZLE: in Newtonian gravity, how does a small negative mass\n>>> "orbit" a big positive mass? What curve does it trace out?\n>>\n>> That\'s easy: 2-body orbits are conic sections, so it must be a hyperbola.\n>> With the big body at the focus of the other branch.\n>\n> ...Actually, that would be what you\'d get if a small mass was "orbiting" a\n> large negative one.\n>\n> For a small negative mass orbiting a large positive one, the orbit of the\n> small negative mass would be a normal ellipse around the postive mass body;\n> what would be a bit odd is that the positive mass body would be _between_\n> the small negative mass and the center of mass of the combined system,\n> which would still be at one of the foci of the ellipse...\n\nYeah, ten minutes after I\'d sent that I realised I\'d misread the question,\nbut my correction seems to have got lost in the moderation process (maybe\njust as well, since there was a mistake in that too...)\n\nI think I\'ve got it all sussed, though. If the absolute value of the\npositive mass is larger than the absolute value of the negative mass, we\nhave the following situation:\n\nTotal energy negative: unbound, hyperbolic paths\nTotal energy zero: marginally unbound, parabolic paths\nTotal energy positive: bound, elliptical orbits.\n\n(The energy conditions are the other way round from when both masses are\npositive. But this makes sense: a large positive mass means it has a small\nkinetic energy, so the energy is dominated by the negative kinetic energy of\nthe negative-mass body, and the positive potential energy. These have the\nopposite sign from normal).\n\nThe centre of mass is, as you say, not between but beyond the masses. In\nthis case, it\'s beyond the positive mass.\n\nCM ... Pos Neg\n\nIf the absolute value of the positive mass is smaller than the absolute\nvalue of the negative mass, then:\n\nThe total energy must be positive (because the potential energy is positive\nand the kinetic energy of the positive mass outweighs that of the negative\nmass).\nThe paths are always hyperbolic.\nThe centre of mass is beyond the negative mass.\n\nPos Neg ... CM\n\nIf the absolute value of the positive mass is equal to the absolute\nvalue of the negative mass, then the situation is more complicated,\nbecause there is no centre of mass frame (the centre of mass is off at\nprojective infinity).\n\nThe simple thing about this situation is that the relative acceleration\nof the two bodies is zero (their relative velocity is constant). If\ntheir relative velocity is zero, their distance, and hence acceleration,\nis constant, and they follow parabolas. If their relative velocity is\nnon-zero, one might expect hyperbolas, but oddly enough it appears that\nthe actual path is a catenary. Not sure if this is right but the\nderivation looks OK to me...\n\nTim\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 13/05/2004 11:25 am, Gordon D. Pusch at
g_{d_pusch_remove_underscores}@xnet.com wrote:
>
> Tim S <Tim@timsilverman.demon.co.uk> writes:
>
>> on 20/04/2004 7:35 am, John Baez at baez@galaxy.ucr.edu wrote:
>>
>>>
>>> Next puzzle... I may have asked Oz this already:
>>>
>>> PUZZLE: in Newtonian gravity, how does a small negative mass
>>> "orbit" a big positive mass? What curve does it trace out?
>>
>> That's easy: 2-body orbits are conic sections, so it must be a hyperbola.
>> With the big body at the focus of the other branch.
>
> ...Actually, that would be what you'd get if a small mass was "orbiting" a
> large negative one.
>
> For a small negative mass orbiting a large positive one, the orbit of the
> small negative mass would be a normal ellipse around the postive mass body;
> what would be a bit odd is that the positive mass body would be _between_
> the small negative mass and the center of mass of the combined system,
> which would still be at one of the foci of the ellipse...
Yeah, ten minutes after I'd sent that I realised I'd misread the question,
but my correction seems to have got lost in the moderation process (maybe
just as well, since there was a mistake in that too...)
I think I've got it all sussed, though. If the absolute value of the
positive mass is larger than the absolute value of the negative mass, we
have the following situation:
Total energy negative: unbound, hyperbolic paths
Total energy zero: marginally unbound, parabolic paths
Total energy positive: bound, elliptical orbits.
(The energy conditions are the other way round from when both masses are
positive. But this makes sense: a large positive mass means it has a small
kinetic energy, so the energy is dominated by the negative kinetic energy of
the negative-mass body, and the positive potential energy. These have the
opposite sign from normal).
The centre of mass is, as you say, not between but beyond the masses. In
this case, it's beyond the positive mass.
CM ... Pos Neg
If the absolute value of the positive mass is smaller than the absolute
value of the negative mass, then:
The total energy must be positive (because the potential energy is positive
and the kinetic energy of the positive mass outweighs that of the negative
mass).
The paths are always hyperbolic.
The centre of mass is beyond the negative mass.
Pos Neg ... CM
If the absolute value of the positive mass is equal to the absolute
value of the negative mass, then the situation is more complicated,
because there is no centre of mass frame (the centre of mass is off at
projective infinity).
The simple thing about this situation is that the relative acceleration
of the two bodies is zero (their relative velocity is constant). If
their relative velocity is zero, their distance, and hence acceleration,
is constant, and they follow parabolas. If their relative velocity is
non-zero, one might expect hyperbolas, but oddly enough it appears that
the actual path is a catenary. Not sure if this is right but the
derivation looks OK to me...
Tim
Richard Saam
May25-04, 11:01 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\n\nTim S wrote:\n\n>\n>\n>I think I\'ve got it all sussed, though. If the absolute value of the\n>positive mass is larger than the absolute value of the negative mass, we\n>have the following situation:\n>\n>Total energy negative: unbound, hyperbolic paths\n>Total energy zero: marginally unbound, parabolic paths\n>Total energy positive: bound, elliptical orbits.\n>\nJust a crazy thought.\n\nFree energy is the criterion for the direction of chemical change.\n\nConventionally a negative free energy for a given reaction means the\nreaction thermodynamically should go forward.\n\nThough not conventional, perhaps free mass could be considered\nequivalent to free energy through the identity E = m c^2. A negative\nfree energy would be equivalent to negative mass.\n\nThe free mass equivalent to free energy would be extremely small for\nlaboratory chemical reactions but perhaps on a cosmic scale, the numbers\nwould be significant.\n\nThere perhaps is something intuitive between:\n\nenergy negative: unbound, hyperbolic paths - negative free energy\nenergy positive: bound, elliptical orbits. - positive free energy\n\nThe equilibrium position:\n\nenergy zero: marginally unbound, parabolic paths - zero free energy\n\ndoes not sound good.\n\nRichard Saam\n\n\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>Tim S wrote:
>
>
>I think I've got it all sussed, though. If the absolute value of the
>positive mass is larger than the absolute value of the negative mass, we
>have the following situation:
>
>Total energy negative: unbound, hyperbolic paths
>Total energy zero: marginally unbound, parabolic paths
>Total energy positive: bound, elliptical orbits.
>
Just a crazy thought.
Free energy is the criterion for the direction of chemical change.
Conventionally a negative free energy for a given reaction means the
reaction thermodynamically should go forward.
Though not conventional, perhaps free mass could be considered
equivalent to free energy through the identity E = m c^2. A negative
free energy would be equivalent to negative mass.
The free mass equivalent to free energy would be extremely small for
laboratory chemical reactions but perhaps on a cosmic scale, the numbers
would be significant.
There perhaps is something intuitive between:
energy negative: unbound, hyperbolic paths - negative free energy
energy positive: bound, elliptical orbits. - positive free energy
The equilibrium position:
energy zero: marginally unbound, parabolic paths - zero free energy
does not sound good.
Richard Saam
>
>
>
John Baez
May26-04, 06:12 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>\nIn article <abergman-AA90CF.03115019052004@localhost>,\nAaron Bergman <abergman@physics.utexas.edu> wrote:\n\n>In article <c89154\\$7qb\\$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)\n>wrote:\n\n>> >> The trouble starts when we try\n>> >> to cook up theories that allow positive-mass particles to interact\n>> >> with negative-mass ones.\n\n>If they don\'t interact, they might as well be in completely different\n>universes.\n\nRight. So, in short, no truly interesting quantum field theories\nare known that have both positive-mass and negative-mass particles.\nSo in quantum field theory, we might as well assume (by convention)\nthat all particles have positive mass. And this is what we do.\n\nPlease don\'t think I was trying to make this whole subject seem more\ninteresting than it actually is! It gets a little more interesting\nwhen you take gravity into account, since the interaction of positive-\nand negative-mass particles via Newtonian gravity is sufficiently\ncounterintuitive to trick people into making some sign errors.\n\n....................................... ..............................\n\nWe should declare instead candidly that we dwell on mathematics\nand affirm its statements for the sake of its intellectual beauty,\nwhich betokens the reality of its conceptions and the truth of\nits assertions. For if this passion were extinct, we would cease\nto understand mathematics; its conceptions would dissolve and its\nproofs carry no conviction. - Michael Polyani\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>In article <abergman-AA90CF.03115019052004@localhost>,
Aaron Bergman <abergman@physics.utexas.edu> wrote:
>In article <c89154$7qb$1@glue.ucr.edu>, baez@galaxy.ucr.edu (John Baez)
>wrote:
>> >> The trouble starts when we try
>> >> to cook up theories that allow positive-mass particles to interact
>> >> with negative-mass ones.
>If they don't interact, they might as well be in completely different
>universes.
Right. So, in short, no truly interesting quantum field theories
are known that have both positive-mass and negative-mass particles.
So in quantum field theory, we might as well assume (by convention)
that all particles have positive mass. And this is what we do.
Please don't think I was trying to make this whole subject seem more
interesting than it actually is! It gets a little more interesting
when you take gravity into account, since the interaction of positive-
and negative-mass particles via Newtonian gravity is sufficiently
counterintuitive to trick people into making some sign errors.
.................................................. ...................
We should declare instead candidly that we dwell on mathematics
and affirm its statements for the sake of its intellectual beauty,
which betokens the reality of its conceptions and the truth of
its assertions. For if this passion were extinct, we would cease
to understand mathematics; its conceptions would dissolve and its
proofs carry no conviction. - Michael Polyani
Daryl McCullough
May29-04, 12:55 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 says...\n\n>Right. So, in short, no truly interesting quantum field theories\n>are known that have both positive-mass and negative-mass particles.\n>So in quantum field theory, we might as well assume (by convention)\n>that all particles have positive mass. And this is what we do.\n>\n>Please don\'t think I was trying to make this whole subject seem more\n>interesting than it actually is! It gets a little more interesting\n>when you take gravity into account, since the interaction of positive-\n>and negative-mass particles via Newtonian gravity is sufficiently\n>counterintuitive to trick people into making some sign errors.\n\nWhen we get to relativity, though, it is not clear to me if there\nis a meaningful distinction between "negative mass" and "negative\nenergy". The quantity "m" never occurs by itself in the equations\nof SR, only multiplied by gamma: For example, the Lorentz force\nequation is\n\nd/dt (gamma m v) = q (E + v x B)\n\nSo, I think that relativistically, there would be no physical\ndifference between (1) letting m be negative, while gamma is\npositive, and (2) letting m be positive, while gamma is negative.\nThe important quantity is gamma m, which is the particle\'s energy.\nSo a negative mass is equivalent to choosing the opposite sign\nfor energy.\n\nI believe that this extends to gravity, as well---that what\nis important is not m but the energy.\n\nSo it seems to me that trying to work out the consequences of\nnegative mass in a relativistic context gets us back to Dirac\'s\noriginal idea of a positron as a "hole" in a sea of negative-energy\nelectrons. Dirac of course was only considering the electromagnetic\nproperties of such holes, not the gravitational effects.\n\n--\nDaryl McCullough\nIthaca, NY\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 says...
>Right. So, in short, no truly interesting quantum field theories
>are known that have both positive-mass and negative-mass particles.
>So in quantum field theory, we might as well assume (by convention)
>that all particles have positive mass. And this is what we do.
>
>Please don't think I was trying to make this whole subject seem more
>interesting than it actually is! It gets a little more interesting
>when you take gravity into account, since the interaction of positive-
>and negative-mass particles via Newtonian gravity is sufficiently
>counterintuitive to trick people into making some sign errors.
When we get to relativity, though, it is not clear to me if there
is a meaningful distinction between "negative mass" and "negative
energy". The quantity "m" never occurs by itself in the equations
of SR, only multiplied by \gamma: For example, the Lorentz force
equation is
d/dt (\gamma m v) = q (E + v x B)
So, I think that relativistically, there would be no physical
difference between (1) letting m be negative, while \gamma is
positive, and (2) letting m be positive, while \gamma is negative.
The important quantity is \gamma m, which is the particle's energy.
So a negative mass is equivalent to choosing the opposite sign
for energy.
I believe that this extends to gravity, as well---that what
is important is not m but the energy.
So it seems to me that trying to work out the consequences of
negative mass in a relativistic context gets us back to Dirac's
original idea of a positron as a "hole" in a sea of negative-energy
electrons. Dirac of course was only considering the electromagnetic
properties of such holes, not the gravitational effects.
--
Daryl McCullough
Ithaca, NY
<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 wrote:\n\n> Right. So, in short, no truly interesting quantum field theories\n> are known that have both positive-mass and negative-mass particles.\n> So in quantum field theory, we might as well assume (by convention)\n> that all particles have positive mass. And this is what we do.\n\n\n\nThat\'s not necessarily completely true though is it? If the sign of the\nmass indicates that the asymmetry that exists between the two classes of\nparticles is due to the fact that the anti-particle exists in a negative\nenergy state, by way of -rho and negative mass, until enough vacuum\nenergy is condensed over an isolated area to achieve positive curvature.\n\nParticle theory says that for every fermion type there is another\nfermion type that has exactly the same mass, and negative mass and\ndensity particles appear naively to explain this without jumping to the\nconclusion that particle theory is necessarily wrong because our\nobservations don\'t seem to support this predicted symmetry.\n\nThe symmetry is then maintained if particles that are created from the\nenergy of the vacuum, have negative mass and density before they are\ncondensed into positive mass and density virtual particles, which can\nthen be converted into real particles, given enough energy.\n\nNegative energy and density is then **Generally** maintained by the\nnegative pressure component, so both virtual and real that are created\nwill increase negative pressure via further rarefaction of the vacuum.\n\nRight, John???! Please tell me that I\'m not off my rocker on this one!\n\nSnip this if I AM wrong, but the above will affect vacuum expansion\nwhile G will remain constant, because the increase in mass energy which\noccurs by way of condensation of vacuum energy, will immediately be\noffset by the described increase in negative pressure which necessarily\noccurs if negative mass particles have negative density... until they\ndon\'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>John Baez wrote:
> Right. So, in short, no truly interesting quantum field theories
> are known that have both positive-mass and negative-mass particles.
> So in quantum field theory, we might as well assume (by convention)
> that all particles have positive mass. And this is what we do.
That's not necessarily completely true though is it? If the sign of the
mass indicates that the asymmetry that exists between the two classes of
particles is due to the fact that the anti-particle exists in a negative
energy state, by way of -\rho and negative mass, until enough vacuum
energy is condensed over an isolated area to achieve positive curvature.
Particle theory says that for every fermion type there is another
fermion type that has exactly the same mass, and negative mass and
density particles appear naively to explain this without jumping to the
conclusion that particle theory is necessarily wrong because our
observations don't seem to support this predicted symmetry.
The symmetry is then maintained if particles that are created from the
energy of the vacuum, have negative mass and density before they are
condensed into positive mass and density virtual particles, which can
then be converted into real particles, given enough energy.
Negative energy and density is then **Generally** maintained by the
negative pressure component, so both virtual and real that are created
will increase negative pressure via further rarefaction of the vacuum.
Right, John???! Please tell me that I'm not off my rocker on this one!
Snip this if I AM wrong, but the above will affect vacuum expansion
while G will remain constant, because the increase in mass energy which
occurs by way of condensation of vacuum energy, will immediately be
offset by the described increase in negative pressure which necessarily
occurs if negative mass particles have negative density... until they
don't!
<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>Daryl McCullough wrote:\n>\n> John Baez says...\n>\n> >Right. So, in short, no truly interesting quantum field theories\n> >are known that have both positive-mass and negative-mass particles.\n> >So in quantum field theory, we might as well assume (by convention)\n> >that all particles have positive mass. And this is what we do.\n> >\n> >Please don\'t think I was trying to make this whole subject seem more\n> >interesting than it actually is! It gets a little more interesting\n> >when you take gravity into account, since the interaction of positive-\n> >and negative-mass particles via Newtonian gravity is sufficiently\n> >counterintuitive to trick people into making some sign errors.\n>\n> When we get to relativity, though, it is not clear to me if there\n> is a meaningful distinction between "negative mass" and "negative\n> energy". The quantity "m" never occurs by itself in the equations\n> of SR, only multiplied by gamma: For example, the Lorentz force\n> equation is\n>\n> d/dt (gamma m v) = q (E + v x B)\n>\n> So, I think that relativistically, there would be no physical\n> difference between (1) letting m be negative, while gamma is\n> positive, and (2) letting m be positive, while gamma is negative.\n> The important quantity is gamma m, which is the particle\'s energy.\n> So a negative mass is equivalent to choosing the opposite sign\n> for energy.\n>\n> I believe that this extends to gravity, as well---that what\n> is important is not m but the energy.\n>\n> So it seems to me that trying to work out the consequences of\n> negative mass in a relativistic context gets us back to Dirac\'s\n> original idea of a positron as a "hole" in a sea of negative-energy\n> electrons. Dirac of course was only considering the electromagnetic\n> properties of such holes, not the gravitational effects.\n>\n> --\n> Daryl McCullough\n> Ithaca, NY\n\n\nNot exactly "back to Dirac\'s original idea", because, in this case, both\nthe electron as well as the anti-electron will leave holes in the\nvacuum.\n\nThe observed anti-electron has the same gravitational properties as an\nelectron, and so, (unlike dirac\'s hole theory), there will be a\ncontribution -e for each occupied state of positive energy and a\ncontribution -e for each unoccupied state of negative energy, because\nnegative pressure increases in proportion to the holes that both\nparticles leave in the "sea".\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>Daryl McCullough wrote:
>
> John Baez says...
>
> >Right. So, in short, no truly interesting quantum field theories
> >are known that have both positive-mass and negative-mass particles.
> >So in quantum field theory, we might as well assume (by convention)
> >that all particles have positive mass. And this is what we do.
> >
> >Please don't think I was trying to make this whole subject seem more
> >interesting than it actually is! It gets a little more interesting
> >when you take gravity into account, since the interaction of positive-
> >and negative-mass particles via Newtonian gravity is sufficiently
> >counterintuitive to trick people into making some sign errors.
>
> When we get to relativity, though, it is not clear to me if there
> is a meaningful distinction between "negative mass" and "negative
> energy". The quantity "m" never occurs by itself in the equations
> of SR, only multiplied by \gamma: For example, the Lorentz force
> equation is
>
> d/dt (\gamma m v) = q (E + v x B)
>
> So, I think that relativistically, there would be no physical
> difference between (1) letting m be negative, while \gamma is
> positive, and (2) letting m be positive, while \gamma is negative.
> The important quantity is \gamma m, which is the particle's energy.
> So a negative mass is equivalent to choosing the opposite sign
> for energy.
>
> I believe that this extends to gravity, as well---that what
> is important is not m but the energy.
>
> So it seems to me that trying to work out the consequences of
> negative mass in a relativistic context gets us back to Dirac's
> original idea of a positron as a "hole" in a sea of negative-energy
> electrons. Dirac of course was only considering the electromagnetic
> properties of such holes, not the gravitational effects.
>
> --
> Daryl McCullough
> Ithaca, NY
Not exactly "back to Dirac's original idea", because, in this case, both
the electron as well as the anti-electron will leave holes in the
vacuum.
The observed anti-electron has the same gravitational properties as an
electron, and so, (unlike dirac's hole theory), there will be a
contribution -e for each occupied state of positive energy and a
contribution -e for each unoccupied state of negative energy, because
negative pressure increases in proportion to the holes that both
particles leave in the "sea".
<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>Negative (effective) mass is a perfectly standard experimental\nrealisable in silicon. Try typing "negative effective mass" into\nGoogle, and take the first couple of sites you see (not the crank\nones).\nwww.ph.unimelb.edu.au/ tcmp/publications/andy/ICPS.pdf.pdf\nwww.iop.org/EJ/abstract/0256-307X/19/10/336\nwww-ee.eng.buffalo.edu/ faculty/mitin/Papers/paper159.pdf\n\nIf you want to know whether a particle of negative mass falls up or\ndown, why not measure one and find out like these guys did (negative\neffective mass neutron in silicon)?\nhttp://www1.physik.tu-muenchen.de/lehrstuehle/E21/Projects/Gravity/gravity.html\n\nNegative mass is measured to fall upwards. Its passive gravitational\nmass equals its negative inertial mass. In fact, its passive\ngravitational mass also equals its non-free-space inertial mass when\npositive too. At least one person (J. Baez) has claimed that a\npositive mass attracts everything, independent of the sign of the mass\n- well, that is experimentally incorrect. BTW, don\'t tell me this is\njust the same as a balloon, which has negative effective weight, but\nstill positive effective mass. If you push a balloon, it moves away\nfrom you. If you transfer right-going momentum to a NEM electron, it\nmoves leftwards (experimentally).\n\nNot sure where that leaves the arguments about geodesics and GR :)\n\nAs to whether its active gravitational mass also equals its passive\ngravitational mass, that too should only be decided by experiment,\nwhich has not been carried out, although I would certainly let my\nexpectations be guided by the equivalence principle prediction that it\nshould be.\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>Negative (effective) mass is a perfectly standard experimental
realisable in silicon. Try typing "negative effective mass" into
Google, and take the first couple of sites you see (not the crank
ones).
www.ph.unimelb.edu.au/ tcmp/publications/andy/ICPS.pdf.pdf
www.iop.org/EJ/abstract/0256-307X/19/10/336
www-ee.eng.buffalo.edu/ faculty/mitin/Papers/paper159.pdf
If you want to know whether a particle of negative mass falls up or
down, why not measure one and find out like these guys did (negative
effective mass neutron in silicon)?
http://www1.physik.tu-muenchen.de/lehrstuehle/E21/Projects/Gravity/gravity.html
Negative mass is measured to fall upwards. Its passive gravitational
mass equals its negative inertial mass. In fact, its passive
gravitational mass also equals its non-free-space inertial mass when
positive too. At least one person (J. Baez) has claimed that a
positive mass attracts everything, independent of the sign of the mass
- well, that is experimentally incorrect. BTW, don't tell me this is
just the same as a balloon, which has negative effective weight, but
still positive effective mass. If you push a balloon, it moves away
from you. If you transfer right-going momentum to a NEM electron, it
moves leftwards (experimentally).
Not sure where that leaves the arguments about geodesics and GR :)
As to whether its active gravitational mass also equals its passive
gravitational mass, that too should only be decided by experiment,
which has not been carried out, although I would certainly let my
expectations be guided by the equivalence principle prediction that it
should be.
<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>For some reason everyone seems to be treating negative mass as a toy\nmodel only. But in semiconductors (silicon, GaAs/AlGaAs etc), negative\neffective mass is a perfectly standard concept, all it means is that\ndE/dk is negative. Try google "negative effective mass". I don\'t see\nwhy a particle dressed by lattice interactions is any less "real" than\na particle dressed by interactions with its virtual charge cloud, like\nlow-energy EM-coupling running.\n\nwww-ee.eng.buffalo.edu/ faculty/mitin/Papers/paper142.pdf\nwww.ph.unimelb.edu.au/ tcmp/publications/andy/ICPS.pdf.pdf\nwww.iop.org/EJ/abstract/0256-307X/19/10/336\n\nMoreover, someone has actually done the experiment on such a beast,\nand shown that it falls upwards, in contradiction to J. Baez (and\nothers) predictions that a positive mass attracts everything (positive\nof negative) from a geodesic argument\n\nwww1.physik.tu-muenchen.de/lehrstuehle/\nE21/Projects/Gravity/gravity.html\n\nPlease note that this is not like a helium balloon falling upwards. A\nballoon has negative "effective weight" (in air), but still has\npositive inertial mass. If I exert a rightwards force on a balloon, it\ngains a rightwards velocity. Otherwise, a neutrally buoyant balloon\nwould have zero effective inertial mass, and we know that is not true\n(because it would have to move at c)\nBut an electron near the top of the band has a genuine negative\ninertial mass. A rightwards force on the electron causes it to gain\nvelocity in the leftwards direction.\n\nSo what we know is that: negative inertial mass => negative passive\ngravitational mass. We don\'t know about active gravitational mass,\nalthough the equivalence principle would lead us to suspect that it\nwould be negative.\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>For some reason everyone seems to be treating negative mass as a toy
model only. But in semiconductors (silicon, GaAs/AlGaAs etc), negative
effective mass is a perfectly standard concept, all it means is that
dE/dk is negative. Try google "negative effective mass". I don't see
why a particle dressed by lattice interactions is any less "real" than
a particle dressed by interactions with its virtual charge cloud, like
low-energy EM-coupling running.
www-ee.eng.buffalo.edu/ faculty/mitin/Papers/paper142.pdf
www.ph.unimelb.edu.au/ tcmp/publications/andy/ICPS.pdf.pdf
www.iop.org/EJ/abstract/0256-307X/19/10/336
Moreover, someone has actually done the experiment on such a beast,
and shown that it falls upwards, in contradiction to J. Baez (and
others) predictions that a positive mass attracts everything (positive
of negative) from a geodesic argument
www1.physik.tu-muenchen.de/lehrstuehle/
E21/Projects/Gravity/gravity.html
Please note that this is not like a helium balloon falling upwards. A
balloon has negative "effective weight" (in air), but still has
positive inertial mass. If I exert a rightwards force on a balloon, it
gains a rightwards velocity. Otherwise, a neutrally buoyant balloon
would have zero effective inertial mass, and we know that is not true
(because it would have to move at c)
But an electron near the top of the band has a genuine negative
inertial mass. A rightwards force on the electron causes it to gain
velocity in the leftwards direction.
So what we know is that: negative inertial mass => negative passive
gravitational mass. We don't know about active gravitational mass,
although the equivalence principle would lead us to suspect that it
would be negative.
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