Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
On Oct 17, 2008, at 10:09 PM, Michel Jullian wrote: Ah, yes, this is what I was missing. Even though I still believe the surface charge density is uniform on _most_ of the thin conducting disk surface (as it is on the plates of a parallel plate capacitor), I now realize the non-uniform charges at the periphery must make up for non-infinity of the radius. So I now think Robin and you are correct, there must be a radial component of the field for a uniform disk of charge (or mass). Many thanks for patiently enlightening me. Actually it was Robin who was correct all along. I changed views after he patiently came up with a proof so clear I couldn't misunderstand it. However, I still hold that in close proximity to a thin x-y planar disk having planar mass density rho the z axis gravitational field is given by gravimagnetic theory to be: g = a rho/(2 * epsilon_0_g) where a is a unit vector normal to and directed toward the plane, i.e. in the z axis, for positive rho, away for negative rho, where rho is mass density (say in i kg/m^2), and epsilon_0_g is given by: epsilon_0_g = 1/(4 Pi G) = 1.192299(31)x10^9 kg s^2/m^3 If the mass is in small free to move chunks though this is a great understatement of the complexity because many instabilities are feasible, though I think it is this force that eventually gives planetary rings their flat shape. I think galaxies can be a lot more complicated due to spin induced Lorentz forces bending arms into 3D curled spirals, etc. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
2008/10/17 Horace Heffner <[EMAIL PROTECTED]>: > > On Oct 16, 2008, at 11:17 PM, Michel Jullian wrote: > >> 2008/10/16 Horace Heffner <[EMAIL PROTECTED]>: >>> >>> On Oct 16, 2008, at 7:17 AM, Michel Jullian wrote: >>> But if you get closer and closer to a finite disk of charge, whether on-axis or off-axis, it will look more and more like an infinite sheet of charge, because the 1/r^2 law makes the effect of the most remote charges rapidly negligible compared to that of the closest ones right under you. >>> >>> Exactly what I thought intuitively. Not true though, due to >>> superposition. >>> Suppose you have a virus located on the center of a surface of a cm >>> square >>> plane segment of an insulator having a uniformly distributed charge of >>> 10^-9 >>> C. The field is normal to the surface where the virus is located. You >>> now >>> bring a 1 C charge within a cm of the virus. By superposition, the field >>> lines will be about parallel to the surface where the virus is located, >>> not >>> normal to it. >> >> Well maybe, but we were considering a uniform distribution. > > Actually not, and that was the point of Robin's very clear proof. In the > vicinity of the measuring point the field is uniform. However, for a > measuring point not in the center of the disk there is a large mass of > unbalanced charge that is all to one side of the measuring point. Since > this unbalanced is all to one side, it can be replaced with the full charge > located at its center of charge. The fact it is spread uniformly over its > (distant) surface becomes irrelevant. Finite charge distributions can not > be treated in the same manner as infinite charge distributions. > > > >> So the field _right above the surface_, where the disk looks infinite, will indeed be perpendicular to the surface. BTW you said you knew this to be the case close to a conductor (freely moving charges) didn't you, well now imagine we suddenly freeze the charges on that conductor, will the field above it change its direction? Michel >>> >>> The problem is that on a finite planar conductor the charges are not >>> uniformly distributed. They are only uniformly distributed on a sphere. >>> The charges are distributed toward the edges. On a real surface the >>> field >>> gradients are largest at protrusions where the surface curvature is >>> maximized convexly. >> >> ... >> >> Yes, but at only a few mm away from the edge of a metal disk the >> surface charge density becomes uniform to a very good approximation >> (provided it is isolated of course). > > The charge distribution on a thin conducting disc is far from uniform > though. The thinner the disk the stronger the field on the periphery. In a > conductor it is the buildup of charge toward the periphery that permits the > field lines normal to the surface. Ah, yes, this is what I was missing. Even though I still believe the surface charge density is uniform on _most_ of the thin conducting disk surface (as it is on the plates of a parallel plate capacitor), I now realize the non-uniform charges at the periphery must make up for non-infinity of the radius. So I now think Robin and you are correct, there must be a radial component of the field for a uniform disk of charge (or mass). Many thanks for patiently enlightening me. Cheers, Michel > > > >> >> A couple resources on the subject: >> >> http://www.goiit.com/posts/show/126825.htm >> >> "A large plate with uniform charge density. >> Consider the plate below, which we will assume is infininte ( a good >> assumption if you are close to the plate) and has a charge per unit >> area, s. " > > This is merely a repeating of the same mistake. This assumption depends on > the degree to which the plate is uniform, the measuring point being central, > and the fact the plate is a conductor. Robin shows a clear lack of field > orthogonality in the case of measuring points not on the center of a > uniformly charged insulating disk. > > >> http://www2.truman.edu/~edis/courses/186/lab3.html >> >> "Electric field lines and equipotential lines each behave in a certain >> way in the vicinity of a source---any distribution of charges which >> give rise to an electric field---and in the vicinity of a conducting > > > Note the use of the word "conducting" above. That is critical. > > >> surface: >> - Near a source, the equipotential lines are parallel to the surface >> of the charge distribution, and the electric field lines are >> perpendicular to the surface of the charge distribution, >> - Near a conducting surface, the equipotential lines are parallel to >> the conducting surface, and the electric field lines are perpendicular >> to the conducting surface." >> >> Cheers, >> >> Michel > > Consider again Robin's proof. > > On Oct 14, 2008, at 7:54 PM, Robin van Spaandonk wrote: > >> [snip] >> Consider the attached diagram. >> >> With the exception of "C" (for Center), all letters label int
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
On Oct 16, 2008, at 11:17 PM, Michel Jullian wrote: 2008/10/16 Horace Heffner <[EMAIL PROTECTED]>: On Oct 16, 2008, at 7:17 AM, Michel Jullian wrote: But if you get closer and closer to a finite disk of charge, whether on-axis or off-axis, it will look more and more like an infinite sheet of charge, because the 1/r^2 law makes the effect of the most remote charges rapidly negligible compared to that of the closest ones right under you. Exactly what I thought intuitively. Not true though, due to superposition. Suppose you have a virus located on the center of a surface of a cm square plane segment of an insulator having a uniformly distributed charge of 10^-9 C. The field is normal to the surface where the virus is located. You now bring a 1 C charge within a cm of the virus. By superposition, the field lines will be about parallel to the surface where the virus is located, not normal to it. Well maybe, but we were considering a uniform distribution. Actually not, and that was the point of Robin's very clear proof. In the vicinity of the measuring point the field is uniform. However, for a measuring point not in the center of the disk there is a large mass of unbalanced charge that is all to one side of the measuring point. Since this unbalanced is all to one side, it can be replaced with the full charge located at its center of charge. The fact it is spread uniformly over its (distant) surface becomes irrelevant. Finite charge distributions can not be treated in the same manner as infinite charge distributions. So the field _right above the surface_, where the disk looks infinite, will indeed be perpendicular to the surface. BTW you said you knew this to be the case close to a conductor (freely moving charges) didn't you, well now imagine we suddenly freeze the charges on that conductor, will the field above it change its direction? Michel The problem is that on a finite planar conductor the charges are not uniformly distributed. They are only uniformly distributed on a sphere. The charges are distributed toward the edges. On a real surface the field gradients are largest at protrusions where the surface curvature is maximized convexly. ... Yes, but at only a few mm away from the edge of a metal disk the surface charge density becomes uniform to a very good approximation (provided it is isolated of course). The charge distribution on a thin conducting disc is far from uniform though. The thinner the disk the stronger the field on the periphery. In a conductor it is the buildup of charge toward the periphery that permits the field lines normal to the surface. A couple resources on the subject: http://www.goiit.com/posts/show/126825.htm "A large plate with uniform charge density. Consider the plate below, which we will assume is infininte ( a good assumption if you are close to the plate) and has a charge per unit area, s. " This is merely a repeating of the same mistake. This assumption depends on the degree to which the plate is uniform, the measuring point being central, and the fact the plate is a conductor. Robin shows a clear lack of field orthogonality in the case of measuring points not on the center of a uniformly charged insulating disk. http://www2.truman.edu/~edis/courses/186/lab3.html "Electric field lines and equipotential lines each behave in a certain way in the vicinity of a source---any distribution of charges which give rise to an electric field---and in the vicinity of a conducting Note the use of the word "conducting" above. That is critical. surface: - Near a source, the equipotential lines are parallel to the surface of the charge distribution, and the electric field lines are perpendicular to the surface of the charge distribution, - Near a conducting surface, the equipotential lines are parallel to the conducting surface, and the electric field lines are perpendicular to the conducting surface." Cheers, Michel Consider again Robin's proof. On Oct 14, 2008, at 7:54 PM, Robin van Spaandonk wrote: [snip] Consider the attached diagram. With the exception of "C" (for Center), all letters label intersections. The line segment "DF" is perpendicular to the radial line segment "BC". Let there be a test mass at "A". We examine the component of the gravitational forces within the plane for the moment. The arc segment "DEF" is a mirror image of "DBF" about the line segment "DF". The forces acting on A within the plane due to the two segments "DEFAD" and "DBFAD" exactly cancel, because these two regions have the same area (uniform thickness of the disc is assumed). The rest of the mass of the disc, excluding these two segments, is all to the left of A. Hence there is a net force acting on A, pulling it to the left. This remains valid if A is outside the plane of the disk. It only ceases to be true when A is exactly on the axis of the disc, at which poin
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
2008/10/16 Horace Heffner <[EMAIL PROTECTED]>: > > On Oct 16, 2008, at 7:17 AM, Michel Jullian wrote: > >> But if you get closer and closer to a finite disk of charge, whether >> on-axis or off-axis, it will look more and more like an infinite >> sheet of charge, because the 1/r^2 law makes the effect of the most >> remote charges rapidly negligible compared to that of the closest ones >> right under you. > > Exactly what I thought intuitively. Not true though, due to superposition. > Suppose you have a virus located on the center of a surface of a cm square > plane segment of an insulator having a uniformly distributed charge of 10^-9 > C. The field is normal to the surface where the virus is located. You now > bring a 1 C charge within a cm of the virus. By superposition, the field > lines will be about parallel to the surface where the virus is located, not > normal to it. Well maybe, but we were considering a uniform distribution. >> So the field _right above the surface_, where the disk looks infinite, >> will indeed be perpendicular to the surface. BTW you said you knew >> this to be the case close to a conductor (freely moving charges) >> didn't you, well now imagine we suddenly freeze the charges on that >> conductor, will the field above it change its direction? >> >> Michel > > The problem is that on a finite planar conductor the charges are not > uniformly distributed. They are only uniformly distributed on a sphere. > The charges are distributed toward the edges. On a real surface the field > gradients are largest at protrusions where the surface curvature is > maximized convexly. ... Yes, but at only a few mm away from the edge of a metal disk the surface charge density becomes uniform to a very good approximation (provided it is isolated of course). A couple resources on the subject: http://www.goiit.com/posts/show/126825.htm "A large plate with uniform charge density. Consider the plate below, which we will assume is infininte ( a good assumption if you are close to the plate) and has a charge per unit area, s. " http://www2.truman.edu/~edis/courses/186/lab3.html "Electric field lines and equipotential lines each behave in a certain way in the vicinity of a source---any distribution of charges which give rise to an electric field---and in the vicinity of a conducting surface: - Near a source, the equipotential lines are parallel to the surface of the charge distribution, and the electric field lines are perpendicular to the surface of the charge distribution, - Near a conducting surface, the equipotential lines are parallel to the conducting surface, and the electric field lines are perpendicular to the conducting surface." Cheers, Michel
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
On Oct 16, 2008, at 7:17 AM, Michel Jullian wrote: But if you get closer and closer to a finite disk of charge, whether on-axis or off-axis, it will look more and more like an infinite sheet of charge, because the 1/r^2 law makes the effect of the most remote charges rapidly negligible compared to that of the closest ones right under you. Exactly what I thought intuitively. Not true though, due to superposition. Suppose you have a virus located on the center of a surface of a cm square plane segment of an insulator having a uniformly distributed charge of 10^-9 C. The field is normal to the surface where the virus is located. You now bring a 1 C charge within a cm of the virus. By superposition, the field lines will be about parallel to the surface where the virus is located, not normal to it. So the field _right above the surface_, where the disk looks infinite, will indeed be perpendicular to the surface. BTW you said you knew this to be the case close to a conductor (freely moving charges) didn't you, well now imagine we suddenly freeze the charges on that conductor, will the field above it change its direction? Michel The problem is that on a finite planar conductor the charges are not uniformly distributed. They are only uniformly distributed on a sphere. The charges are distributed toward the edges. On a real surface the field gradients are largest at protrusions where the surface curvature is maximized convexly. The infinite plane allows the assumption of uniform charge density only at the expense of having to assume the availability of an infinite charge. That's mine opinion this hour anyway. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
But if you get closer and closer to a finite disk of charge, whether on-axis or off-axis, it will look more and more like an infinite sheet of charge, because the 1/r^2 law makes the effect of the most remote charges rapidly negligible compared to that of the closest ones right under you. So the field _right above the surface_, where the disk looks infinite, will indeed be perpendicular to the surface. BTW you said you knew this to be the case close to a conductor (freely moving charges) didn't you, well now imagine we suddenly freeze the charges on that conductor, will the field above it change its direction? Michel 2008/10/16 Horace Heffner <[EMAIL PROTECTED]>: > > On Oct 15, 2008, at 11:54 PM, Michel Jullian wrote: > >> 2008/10/15 Horace Heffner <[EMAIL PROTECTED]>: >> ... >>> >>> Agreed! It appears I am mistaken about the field lines near the plane of >>> a >>> finite 2D disc. I was confused by thinking I knew the field lines at a >>> charged surface become normal to the surface as you approach the surface >>> (in >>> the limit). This only applies to conductors, where the charges are free >>> to >>> redistribute to make this so. >> >> ... >> >> Nope, no such redistribution required, only uniform charge >> distribution. Perpendicularity of E field to uniform sheet of charge >> easily shown using Gauss law: >> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1 >> Infinity of the sheet of charge is not required either, provided one >> is much closer to the sheet than to its edge (Gauss box much thinner >> than wide => flux through the sides can be neglected). >> >> Michel > > Well, this is indeed the assumption I was going on, but it now seems to me > to be wrong. It is certainly easy to prove for a conductor's static surface > charge that the field is normal to the surface. Assume the field is not > normal and has an x component to the field vector. If there is an x > component to the field vector the surface charge will move, there will be a > current, until there is no such vector. This denies the assumption that the > surface charge was static. QED > > It certainly is true for the infinite plane case and uniform charge > distribution rho that the field is uniform near the surface. However, Robin > provides pretty convincing proof that in the finite surface case on an > insulator in the x-y plane the field can have an x or y component close to > the surface. > > Suppose we take a much more simple case than the disc - a finite rectangular > plane segment. Such a surface can be examined as a the integral of a series > of line charges, so we are now down to the much more simple finite uniform > line charge case analyzed here: > > http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potlin.html#c1 > > http://tinyurl.com/4pn59j > > The potential V at point x located distance a from one end and b from the > other, and d from the line charge of uniform density lambda, is given as: > > > V = k lambda ln[ (b + (b^2 + d^2)^(1/2)) / (-a + (a^2+d^2)^(1/2)) ] > > where k = 1/(4 Pi epsilon_0). This is not symmetric in a and b, so the > potential varies as a and b change, leaving a non-normal field except where > a = b. The integral of a bunch of vectors slanted in a given way will also > be slanted in a given way, so the rectangular plane segment of uniform will > not have normal field except in the center. > > Note that the reference you give appears to assume an "infinite sheet of > charge" and thus assumes a normal field at the surface. > >> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1 > > > Best regards, > > Horace Heffner > http://www.mtaonline.net/~hheffner/ > > > > >
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On Oct 15, 2008, at 11:54 PM, Michel Jullian wrote: 2008/10/15 Horace Heffner <[EMAIL PROTECTED]>: ... Agreed! It appears I am mistaken about the field lines near the plane of a finite 2D disc. I was confused by thinking I knew the field lines at a charged surface become normal to the surface as you approach the surface (in the limit). This only applies to conductors, where the charges are free to redistribute to make this so. ... Nope, no such redistribution required, only uniform charge distribution. Perpendicularity of E field to uniform sheet of charge easily shown using Gauss law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1 Infinity of the sheet of charge is not required either, provided one is much closer to the sheet than to its edge (Gauss box much thinner than wide => flux through the sides can be neglected). Michel Well, this is indeed the assumption I was going on, but it now seems to me to be wrong. It is certainly easy to prove for a conductor's static surface charge that the field is normal to the surface. Assume the field is not normal and has an x component to the field vector. If there is an x component to the field vector the surface charge will move, there will be a current, until there is no such vector. This denies the assumption that the surface charge was static. QED It certainly is true for the infinite plane case and uniform charge distribution rho that the field is uniform near the surface. However, Robin provides pretty convincing proof that in the finite surface case on an insulator in the x-y plane the field can have an x or y component close to the surface. Suppose we take a much more simple case than the disc - a finite rectangular plane segment. Such a surface can be examined as a the integral of a series of line charges, so we are now down to the much more simple finite uniform line charge case analyzed here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potlin.html#c1 http://tinyurl.com/4pn59j The potential V at point x located distance a from one end and b from the other, and d from the line charge of uniform density lambda, is given as: V = k lambda ln[ (b + (b^2 + d^2)^(1/2)) / (-a + (a^2+d^2)^(1/2)) ] where k = 1/(4 Pi epsilon_0). This is not symmetric in a and b, so the potential varies as a and b change, leaving a non-normal field except where a = b. The integral of a bunch of vectors slanted in a given way will also be slanted in a given way, so the rectangular plane segment of uniform will not have normal field except in the center. Note that the reference you give appears to assume an "infinite sheet of charge" and thus assumes a normal field at the surface. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1 Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
2008/10/15 Horace Heffner <[EMAIL PROTECTED]>: ... > Agreed! It appears I am mistaken about the field lines near the plane of a > finite 2D disc. I was confused by thinking I knew the field lines at a > charged surface become normal to the surface as you approach the surface (in > the limit). This only applies to conductors, where the charges are free to > redistribute to make this so. ... Nope, no such redistribution required, only uniform charge distribution. Perpendicularity of E field to uniform sheet of charge easily shown using Gauss law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1 Infinity of the sheet of charge is not required either, provided one is much closer to the sheet than to its edge (Gauss box much thinner than wide => flux through the sides can be neglected). Michel
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... bobbing parabolas ... Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
I wrote: " ... matter with some z axis velocity and a stable circular orbit will essentially sustain simple harmonic motion in the z axis ... ". That should say: " ... matter with some z axis velocity and a stable circular orbit will essentially sustain oscillating in the z axis ... ". The motion is not simple harmonic because the z axis field is constant, not increasing with distance. This makes for a pretty weird orbit shape - not one that is merely an inclined ellipse. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
On Oct 14, 2008, at 7:54 PM, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Mon, 13 Oct 2008 02:08:35 -0800: Hi, [snip] I disagree. You are ignoring the 1/r^2 nature of gravity or electrostatic charge. The field near a line charge is 1/r normal to the line. The field near a plane charge is uniform and normal to the plane. The closer you get to a finite line or plane segment the closer it approximates an infinite line or plane. [snip] Consider the attached diagram. With the exception of "C" (for Center), all letters label intersections. The line segment "DF" is perpendicular to the radial line segment "BC". Let there be a test mass at "A". We examine the component of the gravitational forces within the plane for the moment. The arc segment "DEF" is a mirror image of "DBF" about the line segment "DF". The forces acting on A within the plane due to the two segments "DEFAD" and "DBFAD" exactly cancel, because these two regions have the same area (uniform thickness of the disc is assumed). The rest of the mass of the disc, excluding these two segments, is all to the left of A. Hence there is a net force acting on A, pulling it to the left. This remains valid if A is outside the plane of the disk. It only ceases to be true when A is exactly on the axis of the disc, at which point the two segments each comprise half the area of the disc. Of course, the attractive force exerted by the mass of the disc also has a component normal to the plane, and the combination of the two vectors (in the plane and normal to the plane), produces the total force acting on the test mass. Regards, Robin van Spaandonk <[EMAIL PROTECTED]> Agreed! It appears I am mistaken about the field lines near the plane of a finite 2D disc. I was confused by thinking I knew the field lines at a charged surface become normal to the surface as you approach the surface (in the limit). This only applies to conductors, where the charges are free to redistribute to make this so. I think you have indeed shown for a finite disc with a uniform mass density rho that some force exists along the AC line, i.e the x axis for radii not zero. What I have shown is that in close proximity to a mass x-y planar disk having planar mass density rho the z axis gravitational field is given by gravimagnetic theory to be: g = a rho/(2 * epsilon_0_g) where a is a unit vector normal to and directed toward the plane, i.e. in the z axis, for positive rho, away for negative rho, where rho is mass density (say in i kg/m^2), and epsilon_0_g is given by: epsilon_0_g = 1/(4 Pi G) = 1.192299(31)x10^9 kg s^2/m^3 However it is also true a galactic disc that starts out with uniform density but some angular velocity would change that density in order to achieve rotational equilibrium, a balance of the radial forces. It is also true that matter/gas in a disk should tend to locally coalesce, so the dynamics are complex, a lot more complex than for a simple charged surface. Ignoring that complexity, I think we can see that matter with some z axis velocity and a stable circular orbit will essentially sustain simple harmonic motion in the z axis, which provides the prospect that some matter in decaying orbits will have a polar angle of approach on a central black hole. That z axis harmonic motion will convert to an inclined elliptical orbit as the orbit decays and approaches a central black hole and the field nearby the hole increasingly approximates a 1/r^2 radial field. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
In reply to Horace Heffner's message of Mon, 13 Oct 2008 02:08:35 -0800: Hi, [snip] >I disagree. You are ignoring the 1/r^2 nature of gravity or >electrostatic charge. > >The field near a line charge is 1/r normal to the line. The field >near a plane charge is uniform and normal to the plane. The closer >you get to a finite line or plane segment the closer it approximates >an infinite line or plane. > [snip] Consider the attached diagram. With the exception of "C" (for Center), all letters label intersections. The line segment "DF" is perpendicular to the radial line segment "BC". Let there be a test mass at "A". We examine the component of the gravitational forces within the plane for the moment. The arc segment "DEF" is a mirror image of "DBF" about the line segment "DF". The forces acting on A within the plane due to the two segments "DEFAD" and "DBFAD" exactly cancel, because these two regions have the same area (uniform thickness of the disc is assumed). The rest of the mass of the disc, excluding these two segments, is all to the left of A. Hence there is a net force acting on A, pulling it to the left. This remains valid if A is outside the plane of the disk. It only ceases to be true when A is exactly on the axis of the disc, at which point the two segments each comprise half the area of the disc. Of course, the attractive force exerted by the mass of the disc also has a component normal to the plane, and the combination of the two vectors (in the plane and normal to the plane), produces the total force acting on the test mass. Regards, Robin van Spaandonk <[EMAIL PROTECTED]> <>
