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
2008/10/16 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 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 point the two segments each
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
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif - segements.gif
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
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/
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
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]segements.gif 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
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
... bobbing parabolas ... Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
On Oct 13, 2008, at 3:59 AM, Michel Jullian wrote: Agreed. I think the misunderstanding arises from Robin talking about the field _inside_ the disk, and the two of us talking about the field _near_ the disk, i.e. that felt by matter ejected out of the disk thickness. Michel This is true. I am talking about an idealized situation, a planar charge (mass) of uniform density rho, and the field immediately surrounding it and generated by it. However, a thin disk of approximately uniform density approximates this. Matter passing through the z axis of an x-y plane disc of finite thickness will see a reducing z axis field which reaches a minimum of zero at z=0. Matter that remains close to a thin disk, except when passing through the disc, will see an approximately uniform z axis field that produces periodic motion about the plane of the disk. In the case of a thin but finitely thick disc of uniform density the local fields near the disk overwhelm the more remotely generated fields, even out a radial distance that is a fairly good percentage of the disc diameter away from the center of the disc. Such a disc then approximates an infinite plane. The closer you are to the center of a plane segment without being within the plane the closer it approximates an infinite plane. Except at the periphery, the boundary, the closer you are to a thin disk without being within the disk the closer it approximates an infinite plane. The radial field of a thin uniform disc, just outside the disk, but not out near the radial periphery of the disk, is small in comparison to the z axis field. It is the material that moves in the z axis with respect to the disk that would end up in a polar jet by a slingshot scenario, and which would approach a central black hole from a polar direction. Material approaching a black hole in the plane of its spin would not end up in a jet via a sligshot mechanism, unless perturbed by material having a z axis component, but maybe could by a compression scenario. That's my impression anyway. It is coincidental perhaps the solar system is currently passing through the plane of the Milky Way, though the Milky way has an arm structure and is thus not a planar disk. We are in a galaxy colliding with the Milky Way, so the mechanics of our future motion is complex and possibly chaotic. A near pass with another star or stars could send us in most any direction. This is not a comforting thought. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
On Oct 14, 2008, at 3:13 AM, Michel Jullian wrote: I agree on all points except your coincidental remark that We are in a galaxy colliding with the Milky Way, isn't the Milky Way our galaxy (as etymology indicates) any more? Michel We are a member of the Sagittarius Dwarf Galaxy, which is now colliding with the Milky Way. Our solar system takes a polar route over the top of the Milky Way. See: http://viewzone.com/milkyway.html We are from another galaxy in the process of joining with the Milky Way. The Milky Way is actually not our parent galaxy. The mystery of why the Milky Way has always been sideways in the night sky has never been answered -- until now. This first full-sky map of Sagittarius shows its extensive interaction with the Milky Way, Majewski said. Both stars and star clusters now in the outer parts of the Milky Way have been 'stolen' from Sagittarius as the gravitational forces of the Milky Way nibbled away at its dwarf companion. This one vivid example shows that the Milky Way grows by eating its smaller neighbors. The study's map of M giants depicts 2 billion years of Sagittarius stripping by the Milky Way, and suggests that Sagittarius has reached a critical phase in what had been a slow dance of death. After slow, continuous gnawing by the Milky Way, Sagittarius has been whittled down to the point that it cannot hold itself together much longer, said 2MASS Science Team member and study co-author Martin Weinberg of the University of Massachusetts. We are seeing Sagittarius at the very end of its life as an intact system. We are now also a member of the milky way, but ... we are the aliens. The above article describes what I think to be secondary effects from numerous meteor hits and gas accumulation effects, exactly what one would expect from an incoming (at a much higher relative velocity than expected due to intergalactic interaction) Nemesis cloud, laced with a few big chunks. The Nemesis cloud is possibly the Milky Way itself. I almost had it right: http://mtaonline.net/~hheffner/Nemesis.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
This theory seems to be a hoax based on out of context extracts from real scientific papers. It was debunked here: http://blogs.discovermagazine.com/badastronomy/2007/06/27/is-the-sun-from-another-galaxy/ Note the above debunking is not devoid of flaws either, e.g. it asserts that the solar system orbits in the galactic plane, whereas it is well known that it bobs up and down significantly around that ideal orbit. This bobbing motion is believed to have caused most mass extinctions BTW (the galactic plane we cross twice per bobbing period being very crowded with putative colliders/perturbators). Michel 2008/10/14 Horace Heffner [EMAIL PROTECTED]: On Oct 14, 2008, at 3:13 AM, Michel Jullian wrote: I agree on all points except your coincidental remark that We are in a galaxy colliding with the Milky Way, isn't the Milky Way our galaxy (as etymology indicates) any more? Michel We are a member of the Sagittarius Dwarf Galaxy, which is now colliding with the Milky Way. Our solar system takes a polar route over the top of the Milky Way. See: http://viewzone.com/milkyway.html We are from another galaxy in the process of joining with the Milky Way. The Milky Way is actually not our parent galaxy. The mystery of why the Milky Way has always been sideways in the night sky has never been answered -- until now. This first full-sky map of Sagittarius shows its extensive interaction with the Milky Way, Majewski said. Both stars and star clusters now in the outer parts of the Milky Way have been 'stolen' from Sagittarius as the gravitational forces of the Milky Way nibbled away at its dwarf companion. This one vivid example shows that the Milky Way grows by eating its smaller neighbors. The study's map of M giants depicts 2 billion years of Sagittarius stripping by the Milky Way, and suggests that Sagittarius has reached a critical phase in what had been a slow dance of death. After slow, continuous gnawing by the Milky Way, Sagittarius has been whittled down to the point that it cannot hold itself together much longer, said 2MASS Science Team member and study co-author Martin Weinberg of the University of Massachusetts. We are seeing Sagittarius at the very end of its life as an intact system. We are now also a member of the milky way, but ... we are the aliens. The above article describes what I think to be secondary effects from numerous meteor hits and gas accumulation effects, exactly what one would expect from an incoming (at a much higher relative velocity than expected due to intergalactic interaction) Nemesis cloud, laced with a few big chunks. The Nemesis cloud is possibly the Milky Way itself. I almost had it right: http://mtaonline.net/~hheffner/Nemesis.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
On Oct 14, 2008, at 1:36 PM, Michel Jullian wrote: This theory seems to be a hoax based on out of context extracts from real scientific papers. It was debunked here: http://blogs.discovermagazine.com/badastronomy/2007/06/27/is-the- sun-from-another-galaxy/ Interesting! Thanks for the reference. I didn't know there was a dispute at all. I originally got the reference to the article from: http://www.sciencedaily.com/ which is fairly reliable. Curious the dispute seems to be centered as much on the cause of global warming as on the underlying astronomy. Also of interest may be the fact that gravimagnetism, assuming it exists, tends to ensure that the spins of long existing bodies which are subject to tidal effects will necessarily tend to align with and reinforce the local galactic gravimagnetic field. Note the above debunking is not devoid of flaws either, e.g. it asserts that the solar system orbits in the galactic plane, whereas it is well known that it bobs up and down significantly around that ideal orbit. This bobbing motion is believed to have caused most mass extinctions BTW (the galactic plane we cross twice per bobbing period being very crowded with putative colliders/perturbators). Michel Yes, and I believe we now are in the galactic plane, crossing the plane, of the Milky Way at this time. There seems to be some degree of doubt as to the velocity of the solar system. Here is yet another article which might be controversial: http://redshift.vif.com/JournalFiles/Pre2001/V03NO2PDF/V03N2MON.PDF http://tinyurl.com/3ufm69 It gives: v_o = 359 ± 180 km/s in the direction of right ascension alpha_o = 8.7 ± 3.5h and declination delta_o= –1.1 ± 10.0°. It is a doubtful article because it suggests an absolute velocity can be determined by muon decay anisotropy, i.e. by the cosmic ray muon gammas. Perhaps they just mean absolute relative to some normative source of cosmic rays within the Milky Way. The table at the end of the article seems to me to show a wide range of directions though. In any case the absolute velocity of the solar system is complicated by the apparently absolute velocity of the Milky Way. See: http://www.astro.ucla.edu/~wright/CMB-dipole-history.html http://tinyurl.com/4t3ssp The following apparently reliable source gives the coordinates for the center of the Milky Way as Right Ascension 21:12.0, Declination +48:19: http://seds.org/messier/more/mw.html http://tinyurl.com/52lqzd The Milky Way and neighboring galaxies are thought to be moving in the general direction of the Great Attractor: Right Ascension: 243 53 12, Declination: 64 S 55: http://www.philipsedgwick.com/Galactic/GreatAttractor.htm http://tinyurl.com/5xauj It might take some work to sort all this out. 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] attachment: segements.gif
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
The BH being a relatively small object, and there being near-continuous collisions in the accretion disk, it seems to me that matter from the disk attracted to the BH and missing it can make their closest approach from basically all directions (in 3D, not just 2D), and therefore get slingshot-ejected in all directions. Hence my hypothesis that only that which is ejected fastest and closest to the polar direction, a small minority, does not fall back on the disk (escape depending only on the near field in the central area of the disk as Horace pointed out, not on the far field which we all agree is not perpendicular to the disk). Michel 2008/10/13 Robin van Spaandonk [EMAIL PROTECTED]: In reply to Horace Heffner's message of Sun, 12 Oct 2008 15:19:12 -0800: Hi, [snip] My initial point was that Michel's explanation of jet formation was unlikely to be correct IMO, because there is little or no matter ejected at an angle between that of the disc and that of the jet. His explanation made use of the supposition that the gravitational field of the disc was perpendicular to it, and I was pointing out that that wasn't so. In short, I still don't see how the slingshot effect can provide an adequate explanation for the jets. The only comment I made about your theory, was to point out that the disc is not infinite. On Oct 12, 2008, at 1:24 PM, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Sat, 11 Oct 2008 17:49:52 -0800: Hi, [snip] This is because the electric field about an infinite plane of uniform charge is given by: E = a rho/(2 * epsilon_0) so it is just a matter of applying the gravimagnetic isomorphism to obtain the result. In both formulations rho includes the sign of [snip] however in reality, the plane is not infinite. In fact if you look at real galactic jets, the jet usually extends much farther out into space than the diameter of the accretion disc. Sure, but that is probably irrelevant to the mechanism which creates the near light speed jets. Such a mechanism must occur very close to the black hole. Once the near light speed jets are formed there the effect of the BH or disk at great distance is likely moot, true? In any case, a model of jets which includes negative mass charge creation by black holes seems to me to make much more sense. BTW, congrats on the All Ordinaries being up 3% at the moment. A propitious sign for all markets Monday I hope. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Regards, Robin van Spaandonk [EMAIL PROTECTED]
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
Agreed. I think the misunderstanding arises from Robin talking about the field _inside_ the disk, and the two of us talking about the field _near_ the disk, i.e. that felt by matter ejected out of the disk thickness. Michel 2008/10/13 Horace Heffner [EMAIL PROTECTED]: 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.
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
On Oct 13, 2008, at 2:02 AM, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Mon, 13 Oct 2008 01:31:05 -0800: Hi, [snip] But it is so for a very thin disc, therefore a very thin disc can not exist in the vicinity of the black hole. A thin disc's field is not a 1/r^2 field, nor even a 1/r field, but rather a uniform field directed at the disc. Actually, it is precisely the opposite. The gravitational field of the disc is only perpendicular to the surface for an infinitely *thick* disk, because then the centre of gravity (halfway down the length of what has become a column), is at an angle which approaches 90 degrees to the plane of the disc. 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. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
In reply to Michel Jullian's message of Mon, 13 Oct 2008 08:35:02 +0200: Hi, [snip] The BH being a relatively small object, and there being near-continuous collisions in the accretion disk, it seems to me that matter from the disk attracted to the BH and missing it can make their closest approach from basically all directions (in 3D, not just 2D), and therefore get slingshot-ejected in all directions. Agreed. Hence my hypothesis that only that which is ejected fastest and closest to the polar direction, a small minority, does not fall back on the disk Why? What is special about the polar direction? I can agree with the fastest, but not with the direction. In fact if the slingshot effect were responsible, then I would expect to see most matter primarily ejected in the plane of the accretion disc, with progressively less ejected as the ejection angle with the disc increases, and the *least* ejected in the polar directions. Now you might easily argue that when matter is ejected within the disc, it usually gets thermalized (to borrow a term), and soon just once again becomes part of the disc. However this doesn't explain why the jets are so strongly collimated, and so narrow, and why they are *maximal* perpendicular to the disc. What might explain it is if the jets comprise fast charged particles and the whole thing is an incredibly powerful magnet, such that the particles are forced to circulate around the magnetic field lines (which I think Horace says in his theory, though I only skimmed it, so I could have misunderstood). BTW if this is true, then they should also be incredibly strong emitters of cyclotron radiation (though probably not coherent). If one thinks of the empty space around the jets as a huge invisible magnetic doughnut, with a very small hole, then the jets escape out through the holes. At least that's how I could envisage it happening. [snip] Regards, Robin van Spaandonk [EMAIL PROTECTED]
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
In reply to Horace Heffner's message of Mon, 13 Oct 2008 01:31:05 -0800: Hi, [snip] But it is so for a very thin disc, therefore a very thin disc can not exist in the vicinity of the black hole. A thin disc's field is not a 1/r^2 field, nor even a 1/r field, but rather a uniform field directed at the disc. Actually, it is precisely the opposite. The gravitational field of the disc is only perpendicular to the surface for an infinitely *thick* disk, because then the centre of gravity (halfway down the length of what has become a column), is at an angle which approaches 90 degrees to the plane of the disc. Regards, Robin van Spaandonk [EMAIL PROTECTED]
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
On Oct 12, 2008, at 7:11 PM, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Sun, 12 Oct 2008 15:19:12 -0800: Hi, [snip] My initial point was that Michel's explanation of jet formation was unlikely to be correct IMO, because there is little or no matter ejected at an angle between that of the disc and that of the jet. His explanation made use of the supposition that the gravitational field of the disc was perpendicular to it, and I was pointing out that that wasn't so. But it is so for a very thin disc, therefore a very thin disc can not exist in the vicinity of the black hole. A thin disc's field is not a 1/r^2 field, nor even a 1/r field, but rather a uniform field directed at the disc. This gives rise to rather extended z axis excursions for even slight z axis velocities. The field of the disc itself is directed toward the x-y plane of the disc. As material moves toward the comparatively tiny black hole this should give rise to a bulge in the disc and a considerable percentage of material arriving at the disc with large z axis velocity components. However, material doesn't tend to arrive as particles or a gas. It tends to arrive in the form of stars or black holes. In the case of stars the accretion disc is very small and clearly very thin. We would thus not expect accretion into large black holes to be an explanation for jets that last millions of years, because hundreds of stars might be involved in that kind of time frame. Galactic centers are densely populated. In short, I still don't see how the slingshot effect can provide an adequate explanation for the jets. On this I think you and I are agreed. A very narrow jet would not be logical from the sligshot effect alone. The only comment I made about your theory, was to point out that the disc is not infinite. True, and it is not infinitesimally thin, either condition of which is required for a true uniform gravitational field in close proximity to the disk and disk center. However, even for an approximately thin disc, the central field is far from a 1/r^2 field, thus we see central galactic bulges. A large accretion disc will eventually impart a large angular momentum to a black hole. Black holes created by accretion of a binary star, should have an initially large angular momentum. Under any quantum theory of gravity, including mine, a fast spinning black hole should have a powerful gravimagnetic field. Such a field would indeed tend to focus a beam into jets, oriented along the spin axis of the black hole, regardless the axis of the accretion disk of the moment. This is because the gravimagnetic Lorentz force cancels, redirects, tangential motion, while leaving the z axis motion alone. The z axis velocity, as with all velocity components, I think is greatly increased however by compression and heating of the accreting disc, so maybe the initial z axis velocity component is irrelevant. Of further interest is the deduction in my theory that virtual particles carry no gravitational charge. Therefore, the black hole, essentially representing a single nucleus comprised of all neutrons, magnetically aligned, will project a powerful magnetic field beyond its Swartzchild radius, with a strength corresponding to its mass. This magnetic field will assist ionization of the incoming accretion disc, development of a powerful equatorial current of counter revolving electrons and nuclei, and the formation of polar jets, again through application of the Lorentz force. One interesting thing about this magnetic model is that microwaves should tend to be issued in the plane of the black hole spin more than in the jet direction. Here again, even under this assumption, the ejection velocity of material in the jets should not be very uniform, even though the jet would be very narrow. The only scenario I can see whereby the jets would take on a uniform velocity near c is the case where negative mass matter, dark matter coincidentally also having dark energy, originating within the black hole and interacting with the jet matter, accelerates the jet material to near light speed, and to an energy (velocity) spectrum corresponding to the mass of the black hole. This is the scenario which is consistent with all aspects of my theory, and which notably does not even require an accretion disk for formation of the jets. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
In reply to Horace Heffner's message of Sat, 11 Oct 2008 17:49:52 -0800: Hi, [snip] This is because the electric field about an infinite plane of uniform charge is given by: E = a rho/(2 * epsilon_0) so it is just a matter of applying the gravimagnetic isomorphism to obtain the result. In both formulations rho includes the sign of [snip] however in reality, the plane is not infinite. In fact if you look at real galactic jets, the jet usually extends much farther out into space than the diameter of the accretion disc. Therefore, consider a point e.g. 10% in from the edge of an accretion disk and some distance away from it. An inscribed circle in the plane of the accretion disk centered on the normal projection of the point onto the plane thereof, and with a radius of 10% of that of the accretion disc will have perpendicular gravity vector components that cancel one another, while the parallel components (toward the disc) all reinforce one another. IOW if that small (non-concentric) circle were all there were, then the point mass would indeed experience an attractive force normal to the disc. However it isn't all there is. The rest of the accretion disc is there too, and it is largely to one side of the small virtual disk, hence its gravitational component will shift the direction of the overall vector toward the centre of the accretion disk. (and that's without taking the mass of the black hole itself into account). (The virtual disc is inside the real one, has a smaller radius, and it's outer edge just touches the outer edge of the real disc - see attached gif file). Regards, Robin van Spaandonk [EMAIL PROTECTED] attachment: discs.gif
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
On Oct 12, 2008, at 1:24 PM, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Sat, 11 Oct 2008 17:49:52 -0800: Hi, [snip] This is because the electric field about an infinite plane of uniform charge is given by: E = a rho/(2 * epsilon_0) so it is just a matter of applying the gravimagnetic isomorphism to obtain the result. In both formulations rho includes the sign of [snip] however in reality, the plane is not infinite. In fact if you look at real galactic jets, the jet usually extends much farther out into space than the diameter of the accretion disc. Sure, but that is probably irrelevant to the mechanism which creates the near light speed jets. Such a mechanism must occur very close to the black hole. Once the near light speed jets are formed there the effect of the BH or disk at great distance is likely moot, true? In any case, a model of jets which includes negative mass charge creation by black holes seems to me to make much more sense. BTW, congrats on the All Ordinaries being up 3% at the moment. A propitious sign for all markets Monday I hope. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Black Holes from Newtonian Gravity? - discs.gif
In reply to Horace Heffner's message of Sun, 12 Oct 2008 15:19:12 -0800: Hi, [snip] My initial point was that Michel's explanation of jet formation was unlikely to be correct IMO, because there is little or no matter ejected at an angle between that of the disc and that of the jet. His explanation made use of the supposition that the gravitational field of the disc was perpendicular to it, and I was pointing out that that wasn't so. In short, I still don't see how the slingshot effect can provide an adequate explanation for the jets. The only comment I made about your theory, was to point out that the disc is not infinite. On Oct 12, 2008, at 1:24 PM, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Sat, 11 Oct 2008 17:49:52 -0800: Hi, [snip] This is because the electric field about an infinite plane of uniform charge is given by: E = a rho/(2 * epsilon_0) so it is just a matter of applying the gravimagnetic isomorphism to obtain the result. In both formulations rho includes the sign of [snip] however in reality, the plane is not infinite. In fact if you look at real galactic jets, the jet usually extends much farther out into space than the diameter of the accretion disc. Sure, but that is probably irrelevant to the mechanism which creates the near light speed jets. Such a mechanism must occur very close to the black hole. Once the near light speed jets are formed there the effect of the BH or disk at great distance is likely moot, true? In any case, a model of jets which includes negative mass charge creation by black holes seems to me to make much more sense. BTW, congrats on the All Ordinaries being up 3% at the moment. A propitious sign for all markets Monday I hope. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Regards, Robin van Spaandonk [EMAIL PROTECTED]
