island
Apr14-04, 03:18 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Danny Ross Lunsford wrote:\n\n> John Baez wrote:\n\n> >>I am unclear about this though. Will a large antimatter body repel\n> >>ordinary matter or attract it, similarly for antimatter. Its all those\n> >>double and triple negatives that confuse the heck out of me.\n\n> > Right, they\'re confusing - and I never worked them out myself until we\n> > discussed this a couple of times here on sci.physics.research. But now\n> > I know how it goes. 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> This is consistent with taking the other sign for 2M in the\n> Schwarzschild solution. I suppose that was done.\n\n> > Now you\'ve got all the necessary knowledge to take a crack at this:\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> > (We\'ve already discussed *everything* here. We\'ve even been through\n> > a discussion before about how "equal and opposite" is a slightly stupid\n> > thing to say - but we all know what it means.)\n\n> Without looking up the answer, if it\'s going to be realistic then the\n> two have to be capable of erasing each other into some kind of\n> radiation. So they must be capable of forming some odd topogical\n> relation. This is like a magnetic pole in the vicinity of an electric one.\n\n\nThat assumes that an \'antiplanet\' has the same characteristics as an\nantiparticle, but antiparticles don\'t have the characteristics of\nnegative mass.\n\nA negative mass object produces negative pressure because, like John\nsaid... "a negative-mass body will curve spacetime in a way that bends\ngeodesics "away from" it"... which means that negative mass produces the\nsame effects as a positive cosmological constant.\n\n\n~\n\nQuoting from the Sci.Astro faqs:\n\nhttp://www.astro.ucla.edu/~wright/cosmo_constant.html\n"The magnitude of the negative pressure needed for energy conservation\nis easily found to be P = -u = -rho*c2 where P is the pressure, u is the\nvacuum energy density, and rho is the equivalent mass density using E =\nm*c2.\n\nBut in General Relativity, pressure has weight, which means that the\ngravitational acceleration at the edge of a uniform density sphere is\nnot given by\n\ng = GM/R2 = (4*pi/3)*G*rho*R\n\nbut is rather given by\ng = (4*pi/3)*G*(rho+3P/c2)*R\n\nNow Einstein wanted a static model, which means that g = 0, but he also\nwanted to have some matter, so rho > 0, and thus he needed P < 0. In\nfact, by setting\nrho(vacuum) = 0.5*rho(matter)\n\nhe had a total density of 1.5*rho(matter) and a total pressure of\n-0.5*rho(matter)*c2 since the pressure from ordinary matter is\nessentially zero (compared to rho*c2). Thus rho+3P/c2=0 and the\ngravitational acceleration was zero,\ng = (4*pi/3)*G*(rho(matter)-2*rho(vacuum))*R = 0\n\nallowing a static Universe."\n/quote\n\n\nThat\'s the reason why we get all those weird, contrdictory answers when\nwe try to posit an antimass particle into our world, because there\n\'ain\'t no such animal\', because an antiparticle doesn\'t have -rho.\n\nBoth, Positrons and Electrons, are produced at the event horizon of a\nBlack Hole from virtual particle pairs. As with electric charge, this\nmeans that the *normal* distribution of negative energy electrons does\nnot contribute to pair creation. Only *departures* from the normal\ndistribution in a vacuum will isolate enough vacuum energy to produce\nvirtual particle pairs. These pairs can be converted into real\nparticles if enough energy is introduced, but they do not have -rho if\nthey represent localized departures from the norm.\n\nGeneral relativity tells us that gravitation is essentially curvature\ndue to the energy contained in a region and pair production changes this\nenergy to the positve mass of particle pairs, so the \'departure\' is\nmaintained in this manner. These departures cannot produce negative\ncurvature, so they cannot have negative mass, because the energy density\nof these particles does *not* represent the background density.\n\nThe anti-electron has the same gravitational properties as an electron,\nand the electron has a greater chance for survival, (thus maintaining\nthe departure, *indefinitely*), since it might be a long time before it\nmeets an antiparticle if its counterpart antiparticle gets sucked into\nthe black hole.\n\nThere will be a contribution -e for each occupied state of positive\nenergy and a contribution -e for each unoccupied state of negative\nenergy, because negative pressure increases in proportion to the hole\nthat the departures represent.\n\n\nIn other words, *both* particles leave "holes", not just one.\n\n\n\nMore from the faq:\n\n-Einstein\'s Greatest Blunder\n"However, there is a basic flaw in this Einstein static model: it is\nunstable - like a pencil balanced on its point. For imagine that the\nUniverse grew slightly: say by 1 part per million in size. Then the\nvacuum energy density stays the same, but the matter energy density goes\ndown by 3 parts per million. This gives a net negative gravitational\nacceleration, which makes the Universe grow even more! If instead the\nUniverse shrank slightly, one gets a net positive gravitational\nacceleration, which makes it shrink more! Any small deviation gets\nmagnified, and the model is fundamentally flawed."\n\n\nThat\'s not correct if the increase in mass-energy is offset by the\nincrease in negative pressure that results from the "departure", because\nthe vacuum expands naturally, as a function of rarefaction that results\nfrom pair production, so the number of particles in the universe always\nequals the square of the ratio of the electric and the gravitational\nforce between two electrons, as the number of particles in the universe\nincreases, while G remains constant.\n\nTension between ordinary matter and the vacuum increases when you\nincrease mass energy, while at the same time increasing negative\npressure by way of particle pair production.\n\n\n"In addition to this flaw of instability, the static model\'s premise of\na static Universe was shown by Hubble to be incorrect. This led Einstein\nto refer to the cosmological constant as his greatest blunder, and to\ndrop it from his equations. But it still exists as a possibility -- a\ncoefficient that should be determined from observations or fundamental\ntheory."\n\n\nThere is no instability if vacuum expansion is offset by an increase in\nmass energy, as previuously described.\n\n\n-The Quantum Expectation\n"The equations of quantum field theory describing interacting particles\nand anti-particles of mass M are very hard to solve exactly. With a\nlarge amount of mathematical work it is possible to prove that the\nground state of this system has an energy that is less than infinity.\nBut there is no obvious reason why the energy of this ground state\nshould be zero. One expects roughly one particle in every volume equal\nto the Compton wavelength of the particle cubed, which gives a vacuum\ndensity of\n\nrho(vacuum) = M4c3/h3 = 1013 [M/proton mass]4 gm/cc\n\nFor the highest reasonable elementary particle mass, the Planck mass of\n20 micrograms, this density is more than 1091 gm/cc. So there must be a\nsuppression mechanism at work now that reduces the vacuum energy density\nby at least 120 orders of magnitude."\n\n\nOne particle in every volume equal to the Compton wavelength of the\nparticle cubed\'... describes the "depature", *not* the normal\ndistribution, which, a rough guess would put at about 120 orders of\nmagnitude greater.\n\n\n\n"It never hurts to poke around in the basement"\n-drl\n\n"I think it\'ll be something that we\'ve all missed"\n-John Baez\n\n"Such a variation lies outside ordinary general relativity, but can be\nincorporated by a fairly simple modification of the theory"\n-Steve Carlip\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>Danny Ross Lunsford wrote:
> John Baez wrote:
> >>I am unclear about this though. Will a large antimatter body repel
> >>ordinary matter or attract it, similarly for antimatter. Its all those
> >>double and triple negatives that confuse the heck out of me.
> > Right, they're confusing - and I never worked them out myself until we
> > discussed this a couple of times here on sci.physics.research. But now
> > I know how it goes. 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.
> This is consistent with taking the other sign for 2M in the
> Schwarzschild solution. I suppose that was done.
> > Now you've got all the necessary knowledge to take a crack at this:
> > 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.
> > (We've already discussed *everything* here. We've even been through
> > a discussion before about how "equal and opposite" is a slightly stupid
> > thing to say - but we all know what it means.)
> Without looking up the answer, if it's going to be realistic then the
> two have to be capable of erasing each other into some kind of
> radiation. So they must be capable of forming some odd topogical
> relation. This is like a magnetic pole in the vicinity of an electric one.
That assumes that an 'antiplanet' has the same characteristics as an
antiparticle, but antiparticles don't have the characteristics of
negative mass.
A negative mass object produces negative pressure because, like John
said... "a negative-mass body will curve spacetime in a way that bends
geodesics "away from" it"... which means that negative mass produces the
same effects as a positive cosmological constant.
~
Quoting from the Sci.Astro faqs:
http://www.astro.ucla.edu/~wright/cosmo_constant.html
"The magnitude of the negative pressure needed for energy conservation
is easily found to be P = -u = -\rho*c2 where P is the pressure, u is the
vacuum energy density, and \rho is the equivalent mass density using E =
m*c2.
But in General Relativity, pressure has weight, which means that the
gravitational acceleration at the edge of a uniform density sphere is
not given by
g = GM/R2 = (4*\pi/3)*G*\rho*R
but is rather given by
g = (4*\pi/3)*G*(\rho+3P/c2)*R
Now Einstein wanted a static model, which means that g = 0, but he also
wanted to have some matter, so \rho > 0, and thus he needed P < . In
fact, by setting
\rho(vacuum) = .5*\rho(matter)
he had a total density of 1.5*\rho(matter) and a total pressure of
-0.5*\rho(matter)*c2 since the pressure from ordinary matter is
essentially zero (compared to \rho*c2). Thus \rho+3P/c2=0 and the
gravitational acceleration was zero,
g = (4*\pi/3)*G*(\rho(matter)-2*\rho(vacuum))*R =
allowing a static Universe."
/quote
That's the reason why we get all those weird, contrdictory answers when
we try to posit an antimass particle into our world, because there
'ain't no such animal', because an antiparticle doesn't have -\rho.
Both, Positrons and Electrons, are produced at the event horizon of a
Black Hole from virtual particle pairs. As with electric charge, this
means that the *normal* distribution of negative energy electrons does
not contribute to pair creation. Only *departures* from the normal
distribution in a vacuum will isolate enough vacuum energy to produce
virtual particle pairs. These pairs can be converted into real
particles if enough energy is introduced, but they do not have -\rho if
they represent localized departures from the norm.
General relativity tells us that gravitation is essentially curvature
due to the energy contained in a region and pair production changes this
energy to the positve mass of particle pairs, so the 'departure' is
maintained in this manner. These departures cannot produce negative
curvature, so they cannot have negative mass, because the energy density
of these particles does *not* represent the background density.
The anti-electron has the same gravitational properties as an electron,
and the electron has a greater chance for survival, (thus maintaining
the departure, *indefinitely*), since it might be a long time before it
meets an antiparticle if its counterpart antiparticle gets sucked into
the black hole.
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 hole
that the departures represent.
In other words, *both* particles leave "holes", not just one.
More from the faq:
-Einstein's Greatest Blunder
"However, there is a basic flaw in this Einstein static model: it is
unstable - like a pencil balanced on its point. For imagine that the
Universe grew slightly: say by 1 part per million in size. Then the
vacuum energy density stays the same, but the matter energy density goes
down by 3 parts per million. This gives a net negative gravitational
acceleration, which makes the Universe grow even more! If instead the
Universe shrank slightly, one gets a net positive gravitational
acceleration, which makes it shrink more! Any small deviation gets
magnified, and the model is fundamentally flawed."
That's not correct if the increase in mass-energy is offset by the
increase in negative pressure that results from the "departure", because
the vacuum expands naturally, as a function of rarefaction that results
from pair production, so the number of particles in the universe always
equals the square of the ratio of the electric and the gravitational
force between two electrons, as the number of particles in the universe
increases, while G remains constant.
Tension between ordinary matter and the vacuum increases when you
increase mass energy, while at the same time increasing negative
pressure by way of particle pair production.
"In addition to this flaw of instability, the static model's premise of
a static Universe was shown by Hubble to be incorrect. This led Einstein
to refer to the cosmological constant as his greatest blunder, and to
drop it from his equations. But it still exists as a possibility -- a
coefficient that should be determined from observations or fundamental
theory."
There is no instability if vacuum expansion is offset by an increase in
mass energy, as previuously described.
-The[/itex] Quantum Expectation
"The equations of quantum field theory describing interacting particles
and anti-particles of mass M are very hard to solve exactly. With a
large amount of mathematical work it is possible to prove that the
ground state of this system has an energy that is less than infinity.
But there is no obvious reason why the energy of this ground state
should be zero. One expects roughly one particle in every volume equal
to the Compton wavelength of the particle cubed, which gives a vacuum
density of
\rho(vacuum) = M4c3/h3 = 1013 [M/proton mass]4 [itex]gm/cc
For the highest reasonable elementary particle mass, the Planck mass of
20 micrograms, this density is more than 1091 gm/cc. So there must be a
suppression mechanism at work now that reduces the vacuum energy density
by at least 120 orders of magnitude."
One particle in every volume equal to the Compton wavelength of the
particle cubed'... describes the "depature", *not* the normal
distribution, which, a rough guess would put at about 120 orders of
magnitude greater.
"It never hurts to poke around in the basement"
-drl
"I think it'll be something that we've all missed"
-John Baez
"Such a variation lies outside ordinary general relativity, but can be
incorporated by a fairly simple modification of the theory"
-Steve Carlip
> John Baez wrote:
> >>I am unclear about this though. Will a large antimatter body repel
> >>ordinary matter or attract it, similarly for antimatter. Its all those
> >>double and triple negatives that confuse the heck out of me.
> > Right, they're confusing - and I never worked them out myself until we
> > discussed this a couple of times here on sci.physics.research. But now
> > I know how it goes. 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.
> This is consistent with taking the other sign for 2M in the
> Schwarzschild solution. I suppose that was done.
> > Now you've got all the necessary knowledge to take a crack at this:
> > 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.
> > (We've already discussed *everything* here. We've even been through
> > a discussion before about how "equal and opposite" is a slightly stupid
> > thing to say - but we all know what it means.)
> Without looking up the answer, if it's going to be realistic then the
> two have to be capable of erasing each other into some kind of
> radiation. So they must be capable of forming some odd topogical
> relation. This is like a magnetic pole in the vicinity of an electric one.
That assumes that an 'antiplanet' has the same characteristics as an
antiparticle, but antiparticles don't have the characteristics of
negative mass.
A negative mass object produces negative pressure because, like John
said... "a negative-mass body will curve spacetime in a way that bends
geodesics "away from" it"... which means that negative mass produces the
same effects as a positive cosmological constant.
~
Quoting from the Sci.Astro faqs:
http://www.astro.ucla.edu/~wright/cosmo_constant.html
"The magnitude of the negative pressure needed for energy conservation
is easily found to be P = -u = -\rho*c2 where P is the pressure, u is the
vacuum energy density, and \rho is the equivalent mass density using E =
m*c2.
But in General Relativity, pressure has weight, which means that the
gravitational acceleration at the edge of a uniform density sphere is
not given by
g = GM/R2 = (4*\pi/3)*G*\rho*R
but is rather given by
g = (4*\pi/3)*G*(\rho+3P/c2)*R
Now Einstein wanted a static model, which means that g = 0, but he also
wanted to have some matter, so \rho > 0, and thus he needed P < . In
fact, by setting
\rho(vacuum) = .5*\rho(matter)
he had a total density of 1.5*\rho(matter) and a total pressure of
-0.5*\rho(matter)*c2 since the pressure from ordinary matter is
essentially zero (compared to \rho*c2). Thus \rho+3P/c2=0 and the
gravitational acceleration was zero,
g = (4*\pi/3)*G*(\rho(matter)-2*\rho(vacuum))*R =
allowing a static Universe."
/quote
That's the reason why we get all those weird, contrdictory answers when
we try to posit an antimass particle into our world, because there
'ain't no such animal', because an antiparticle doesn't have -\rho.
Both, Positrons and Electrons, are produced at the event horizon of a
Black Hole from virtual particle pairs. As with electric charge, this
means that the *normal* distribution of negative energy electrons does
not contribute to pair creation. Only *departures* from the normal
distribution in a vacuum will isolate enough vacuum energy to produce
virtual particle pairs. These pairs can be converted into real
particles if enough energy is introduced, but they do not have -\rho if
they represent localized departures from the norm.
General relativity tells us that gravitation is essentially curvature
due to the energy contained in a region and pair production changes this
energy to the positve mass of particle pairs, so the 'departure' is
maintained in this manner. These departures cannot produce negative
curvature, so they cannot have negative mass, because the energy density
of these particles does *not* represent the background density.
The anti-electron has the same gravitational properties as an electron,
and the electron has a greater chance for survival, (thus maintaining
the departure, *indefinitely*), since it might be a long time before it
meets an antiparticle if its counterpart antiparticle gets sucked into
the black hole.
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 hole
that the departures represent.
In other words, *both* particles leave "holes", not just one.
More from the faq:
-Einstein's Greatest Blunder
"However, there is a basic flaw in this Einstein static model: it is
unstable - like a pencil balanced on its point. For imagine that the
Universe grew slightly: say by 1 part per million in size. Then the
vacuum energy density stays the same, but the matter energy density goes
down by 3 parts per million. This gives a net negative gravitational
acceleration, which makes the Universe grow even more! If instead the
Universe shrank slightly, one gets a net positive gravitational
acceleration, which makes it shrink more! Any small deviation gets
magnified, and the model is fundamentally flawed."
That's not correct if the increase in mass-energy is offset by the
increase in negative pressure that results from the "departure", because
the vacuum expands naturally, as a function of rarefaction that results
from pair production, so the number of particles in the universe always
equals the square of the ratio of the electric and the gravitational
force between two electrons, as the number of particles in the universe
increases, while G remains constant.
Tension between ordinary matter and the vacuum increases when you
increase mass energy, while at the same time increasing negative
pressure by way of particle pair production.
"In addition to this flaw of instability, the static model's premise of
a static Universe was shown by Hubble to be incorrect. This led Einstein
to refer to the cosmological constant as his greatest blunder, and to
drop it from his equations. But it still exists as a possibility -- a
coefficient that should be determined from observations or fundamental
theory."
There is no instability if vacuum expansion is offset by an increase in
mass energy, as previuously described.
-The[/itex] Quantum Expectation
"The equations of quantum field theory describing interacting particles
and anti-particles of mass M are very hard to solve exactly. With a
large amount of mathematical work it is possible to prove that the
ground state of this system has an energy that is less than infinity.
But there is no obvious reason why the energy of this ground state
should be zero. One expects roughly one particle in every volume equal
to the Compton wavelength of the particle cubed, which gives a vacuum
density of
\rho(vacuum) = M4c3/h3 = 1013 [M/proton mass]4 [itex]gm/cc
For the highest reasonable elementary particle mass, the Planck mass of
20 micrograms, this density is more than 1091 gm/cc. So there must be a
suppression mechanism at work now that reduces the vacuum energy density
by at least 120 orders of magnitude."
One particle in every volume equal to the Compton wavelength of the
particle cubed'... describes the "depature", *not* the normal
distribution, which, a rough guess would put at about 120 orders of
magnitude greater.
"It never hurts to poke around in the basement"
-drl
"I think it'll be something that we've all missed"
-John Baez
"Such a variation lies outside ordinary general relativity, but can be
incorporated by a fairly simple modification of the theory"
-Steve Carlip