|
|
Post by Progenitor A on Apr 6, 2011 12:09:00 GMT 1
I am saying that 1 + (-1)= 0; ie a positive mass plus a negative mass is equivalent to no mass,ie the hole. Good! So we do not have a negative mass at the hole, we have zero mass and your model of negative gravity does not apply |
|
|
|
Post by speakertoanimals on Apr 6, 2011 12:24:40 GMT 1
Except that is NOT in the plane indicated by the spinning top. Hence no contradiction. Remember the 'horizontal plane'.......... Plus if you MEANT stuff to do with the Lagrange points, WHY didn't you bother to call them that (amd provide a reference with a diagram) which would have made the situation clearer to everyone. Yes it is. We have net effect of shell is zero everywhere inside. When we remove a plug of matter, the NET effect is the same as just being repelled by a negative mass at the position of the plug, which is a damn lot easier to compute than that net effect of the shell minus the plug. In algebraic terms, we have A ( effect of shell minus plus plug) plus B (effect of positive mass at plug) is zero. A + B = 0 Hence when we REMOVE the plug, we just have A, which is the same as -B. Hence to get -B (minus the effect of the plug), we just change the sign of the mass of the plug. Hence being repelled by the negative mass of the plug is the same as being attracted by the positive mass of shell-minus-plug, as long as we are within the hollow sphere (which of course is the only place where the equation A+B = 0 holds). Wrong. By symmetry, there will be zero gravitational force at the exact centre of the ring, but within the ring (using the conventional meaning of within) and in the plane of the ring, the force is towards the nearest point on the ring, and AWAY from the centre: www.mathpages.com/home/kmath402/kmath402.htmGauss law considers CLOSED surfaces, and the surface BOUNDED by a circle is NOT closed. Because a closed surface doesn't have a boundary. Or perhaps NM just can't see the difference between closed and ENclosed. |
|
|
|
Post by Progenitor A on Apr 6, 2011 12:43:44 GMT 1
Except that is NOT in the plane indicated by the spinning top. Hence no contradiction. Remember the 'horizontal plane'.......... Plus if you MEANT stuff to do with the Lagrange points, WHY didn't you bother to call them that (amd provide a reference with a diagram) which would have made the situation clearer to everyone. This person repeatedly contradicts herself without being aware! Yes it is. We have net effect of shell is zero everywhere inside. When we remove a plug of matter, the NET effect is the same as just being repelled by a negative mass at the position of the plug, which is a damn lot easier to compute than that net effect of the shell minus the plug. In algebraic terms, we have A ( effect of shell minus plus plug) plus B (effect of positive mass at plug) is zero. A + B = 0 Hence when we REMOVE the plug, we just have A, which is the same as -B. Hence to get -B (minus the effect of the plug), we just change the sign of the mass of the plug. Hence being repelled by the negative mass of the plug is the same as being attracted by the positive mass of shell-minus-plug, as long as we are within the hollow sphere (which of course is the only place where the equation A+B = 0 holds). Why anyone would want to contradict this (unless their ONLY aim to to try to sew confusion) is beyond me. A divergent repulsive force is not the same as a convergent attractive force. And if this person cared to think about what you she is saying she would see that the consequences are absurd. If anyone else is interested in examining the absurdities of this model then I will oblige She had better consult I think with eamonshute ( who thought up this idea) as he says (rightly) that a negative mass at the hole position will cause the ball beareng to hit the opposite point on th sphere only if it happens to be lying on the diameter joining the hole and the opposite point. Otherwise it will diverge (that's what diverging fields do) According to our resident 'physicist' it will always be attracted to that point she says here When inside it, the net effect is a repulsion from the hole, hence the ball bearing will tend to be attracted by the rest of the sphere towards the point opposite the hole. Again she does not recognise her contradictions! I am afraid that this person is no physicist! ;D Wrong. By symmetry, there will be zero gravitational force at the exact centre of the ring, but within the ring (using the conventional meaning of within) and in the plane of the ring, the force is towards the nearest point on the ring, and AWAY from the centre: www.mathpages.com/home/kmath402/kmath402.htmWell I concede that point, having read the link. There is a radial force. I am surprised Gauss law considers CLOSED surfaces, and the surface BOUNDED by a circle is NOT closed. Because a closed surface doesn't have a boundary! Or perhaps NM just can't see the difference between closed and ENclosed. Sheer uncomprehending gobbledegook! Regurgitated misunderstood Google ! |
|
|
|
Post by speakertoanimals on Apr 6, 2011 12:52:17 GMT 1
Why bother 'intuiting', as I pointed out, it was pointless! Plus a solution probably IS possible, you need (as I think I pointed out) to move to prolate spheroidal coordinates: en.wikipedia.org/wiki/Prolate_spheroidal_coordinates and a whole mess of stuff using Legendre polynomials and Legendre functions..................The point, you CAN'T deduce the answer by using simple reasoning (such as Gauss law). nor can you deduce it by making incorrect assumptions about the gravitational field within a ring.............. |
|
|
|
Post by Progenitor A on Apr 6, 2011 13:13:26 GMT 1
Right Carnyx, modification to the gravitational field within a rugby ball
I honestly thought, that by extension, a ring would have a net zero gravitational field.
I have been shown that that is wong,
There is a radial component, but the radial field at the centre is zero
That makes the only area that is free of a gravitational field as the plane of the central circle forming the fattest part of the ball - I think
Any thoughts?
|
|
|
|
Post by Joanne Byers on Apr 6, 2011 14:15:04 GMT 1
Nay Missus and STA stop abusing eachother or you will bring Proboards down on us for harassment. It's childish. Refrain, otherwise I will suspend you, and Carnyx, too.
I've already spent more time than I care to editing your contributions last night.
|
|
|
|
Post by speakertoanimals on Apr 6, 2011 14:20:29 GMT 1
Who said the NET effect of sphere minus plug is a CONVERGING attractive force? It isn't, it is the diverging repulsive force from the negative mass plug. you just ASSUMED the convergent part (although I grant you I may have been a bit foggy on this myself earlier). Plus this DOESN'T alter the algebraic validity of the vector equation I sketched -- the logic is unassailable. Ah, seems you don't know the difference between closed and enclosed then..................... All that I said was that a closed surface lying wholly inside the hollow sphere contains no mass -- obvious really! A circle drawn on the inside of the shell isn't a surface. It is the boundary of a surface, but that surface isn't closed, hence not what I was talking about. Stop playing silly huggers (O come one, what the heck is happening here, you already can't write L I E without it changing, which is a teeny problem since LYES WITHIN is a common phrase in these discussions!) Alleluia! Do you also concede that your rugby ball argument is hence also wrong.............. As regards sphere with plug -- the force is repulsion AWAY from the removed plug. Hence if starts off-axis wrt plug, will be repelled until it hits the innner surface, then still pushed until it is as far away as it can get from the removed plug -- hence the point diametrically opposite the removed plug. Except in the real case it would probably oscillate about this point, like a ball rolling around the bottom of a bowl. Hence a useful (if slightly slapdash) way to think of it would be ball bearing attracted to point opposite plug, even if that does give the incorrect view of a convergent force, rather thah the complete description as sketched above. Seems you don't know what you want NM -- if I give a quick and dirty answer, I get snipped, if I give a full and complete answer, I get snipped. I think you just don't like me, and would rather gouge your own eyeballs out rather than admit I'm sometimes right (apart from the once in a month turn-around like the toroidal one above! Was it perhaps because it WASN'T from wikipedia, which you seem to think is the root of all evil...................). So, rugby ball case is I think exactly solvable, in terms of all that Legendre shit. I'm not going to do it, because Legendre functions are messy buggers! People anyway seem more interested in using Prolate spheroidal coordinates for other problems, and I could only find ONE paper that tried to solve for the interior field of a prolate spheroidal shell -- but that was in the weak-field approximation to general relativity, and in the case of a rotating spheroid, designed to be intermediate between the case of a sphere (answer known), and an infinite rotating cylinder (answer known). The classical answer is probably hidden in that somewhere, with a little work, but since that would involve deciphering a whole load of notation, and then extracting the classical, non-rotating limit, I don't have time! Classical and Quantum Gravity Volume 14 Number 2 B V Ivanov 1997 Class. Quantum Grav. 14 509 Gravitational effects in a rotating prolate spheroid **************STOP PRESS!****************** Finally, I managed to find it! And the surprising result is Newtons theorem, which states that the gravitational attraction at ANY point inside a homoeoidal ellipsoid is zero (look it up, I had to!). This pdf lists it clearly. brd4.braude.ac.il/~karp/Calgary.pdfListed as non-trivial compared to sphere case -- probably means only way to prove it is doing the damn integral, which isn't exactly illuminating! I'll admit, I'm damn surprised -- may be well-known amongst maths community, but most physics texts don't get much beyond sphere case, and I'd never come across Newtons Theorem at graduate level either. In fact, I might almost say (censorship and daft software permitting), well, bugger me sideways, I'd never heard of that! I'll admit that homoeoidal is a bit of a mouthful, silly mathematicians! |
|
|
|
Post by Progenitor A on Apr 6, 2011 14:35:40 GMT 1
Who said the NET effect of sphere minus plug is a CONVERGING attractive force? It isn't, it is the diverging repulsive force from the negative mass plug. you just ASSUMED the convergent part (although I grant you I may have been a bit foggy on this myself earlier). Plus this DOESN'T alter the algebraic validity of the vector equation I sketched -- the logic is unassailable. Who said that gravity converges to a point or plane? Try Newton ONLY Negative Gravity diverges and - listen closely - THAT DOES NOT EXISTYour 'model' is a lunatic contrivance Absolute and utter balderdash!!! There is no negative gravity
The ball bearing does not follow the curve of negative gravity I do not believe I am having this discussion! Stop playing silly huggers. Don't worry, I will I am tiring of this utter childish nonsense As regards sphere with plug -- the force is repulsion AWAY from the removed plug. Hence if starts off-axis wrt plug, will be repelled until it hits the innner surface, then still pushed until it is as far away as it can get from the removed plug -- hence the point diametrically opposite the removed plug. Except in the real case it would probably oscillate about this point, like a ball rolling around the bottom of a bowl. Hence a useful (if slightly slapdash) way to think of it would be ball bearing attracted to point opposite plug, even if that does give the incorrect view of a convergent force, rather thah the complete description as sketched above. Absolute astonishing utterly childish tosh! Seems you don't know what you want NM -- if I give a quick and dirty answer, I get snipped, if I give a full and complete answer, I get snipped. I think you just don't like me, and would rather gouge your own eyeballs out rather than admit I'm sometimes right (apart from the once in a month turn-around like the toroidal one above! Was it perhaps because it WASN'T from wikipedia, which you seem to think is the root of all evil...................) Whether or not I like you is irrelevant. You simply speak nonsense Byr bye! I have had enough of you Note that Eamon is not supporting your childish argument |
|
|
|
Post by eamonnshute on Apr 6, 2011 15:07:27 GMT 1
Note that Eamon is not supporting your childish argument You are wrong, STA is right. |
|
|
|
Post by speakertoanimals on Apr 6, 2011 16:05:15 GMT 1
Except the very torus example you supposedly agreed with does precisely that! The field inside the torus is radially OUTWARD, hence for two particles starting off at slightly different positions, but not at the centre, their paths DIVERGE under gravity as they are each attracted to the nearest point on the torus.
Gravity from a POINT mass is strictly converging, but arrange enough point masses, and you can get various shapes, including the diverging repulsion from a point in the case of the punctured sphere.
Lets go back again to the supposed lunatic contrivance. It is actually a perfectly standard construction, possible since the net effect of a whole load of point masses is just computed by summing the effects of each of them (or integrating in the continuous limit). Hence all we are doing in this argument is rather than using A + B + C... to compute the result, we instead start from a known result (complete spherical shell), and in effect subtract to obtain the result we want.
Why? Because if A is complete spherical shell (zero field inside), B is punctured shell (what we want), and C is field due to plug that later gets removed, then the linear, additive property of gravity means we can say:
(punctured sphere) + (plug) = (complete shell)
Or, B+C = A. We know A (identically zero), hence we can just rearrange the equation to show that B = -C. We know how to compute C, we hence just negate it to get what we want.
Negative of a converging field is a diverging one.
If you think this is wrong, then it means you are doubting the linear, additive property of gravity, which then means you can't compute ANYTHING since most situations involve more than 1 point mass. All integrals or sums disallowed, if we followed this logic...................
Clearly correct, and a damn shame that some people seem more interested in this pretend argument that Newtons Third Theorem, which was news to me, is actually quite weird and freaky, and I'll be damned if I can gain any intuition as to WHY it is true (and I don't think anyone else can either, else so much time would noty have been spent historically trying to deduce the external fields for elliptical bodies!).
I just wonder how Newton proved it originally, but it probably involved some totally incomprehensible geometric proof, and I don't have a copy of the Principia to hand.................................
See Googlebooks, Galactic dynamics
By James Binney, Scott Tremaine, page 51. Messy stuff, and of course some pages missing from preview, but seems that a proof involves the usual messy integrals.
Interesting to astrophysicsts, since MOST bodies (planets, stars, galaxies), are closer to ellipsoidal than spherical.
|
|
|
|
Post by carnyx on Apr 6, 2011 22:57:24 GMT 1
@sta
So as a direct result of idle musings on a chatboard.. you finally got to learn something new .. Who'd a thunk it !
You say;
...my italics for emphasis ...
Then you go on to say;
But you might like this intuitive recipe;
Assuming that an ellipse is really a tilted circle, then tilt the ellipse back so it looks like a circle. Then you can do the analysis as a 2D case, you can translate to a sphere, and prove the point that way.
Now, if it becomes "obvious" that spherical and ovoid shells have the same property of zero gravitational pull at all internal points, then can the same thing be shown for any shape of closed shell with a single wall, for example a box, or canister?
Then with regard to the toroid puzzle, I thought it refered to a ball inside the tube ... and that the ball must be attracted to the inside of the tube by the effect of the mass of the tube on the opposite side of the toroid!
Now I understand the toroid puzzle involves a ball within the inner space, I see that exact centre will be an unstable null-point. But as the ball moves towards the inner tube wall it becomes attracted, we have a situation where the sum of the 1/D^2 forces do not equal zero.
The real reason is that there is more mass in the outer wall of the tube of the torus than there is on the inner wall. And as mass has density, we are dealing with cubic relationship. So rather than being governed by 1/D^2, there is also a ^3 relationship which gives a nonlinearity, and so for differeing linear distances we will get an inequality of forces.
Now, I would ask posters to imagine what would happen inside a spherical shell that was enclosed within a larger spherical shell. Would the inner ball move towards the inner shell wall, as it does for a toroid?
And here is a problem. If you make the shells fit together so well, they effectively become one thick shell, will the ball be attracted to the wall?
Then if it does, how thin does the shell wall have to be, to get back to zero gravity?
|
|
|
|
Post by speakertoanimals on Apr 7, 2011 1:43:10 GMT 1
Carnyx, Newton and various mathematicians since then have been pondering the case of ellipsoidal distributions of mass, and none of them have come up with a EASY way to solve the problem! As I pointed out with the symmetry case, sphere or cylinder are relatively straightforward, and you can argue what the solution is without doing the maths. You can't do the same for an ellipsoid. Tilting circles and 2D examples don't cut the mustard, that's just the way it is.
People are stll writing research papers on that one. If there was a simple argument, I think it would have been found.
Nope, not a ball inside an inner tube, a ball inside the hole of a ring doughnut, and the limit where the torus becomes a ring. Again, another example where the only way to solve it is by doing the maths.
Except we already know -- zero field within a sphere, doesn't matter if you put another sphere inside that, answer remains the same, since gravitational effects just a sum of vectors, and zero plus zero still gives zero. Doesn't matter if you add shell upon shell, until you get a sphere with a hole in the middle, answer still zero.
I had a look at the Principia , and commentary says that what Newton was trying to do (having solved simple case of the sphere), was look at the gravitational field of non-spherical bodies, and easier example after a sphere is the prolate spheroid. But what is really amazing is that Newton invented not just universal law of gravitation (or discovered rather1!), but also the mathematical techniques to solve such probelms (the calculus). For anyone else, just doing ONE of those things would have been a great achievement. But its the breadth of what Newton did that is amazing, gravitation, motion, planetary orbits, fluids, optics, the whole shebang.
|
|
|
|
Post by carnyx on Apr 7, 2011 8:34:55 GMT 1
@sta,
Sorrry, old thing .. such a tired 'non-argument from authority' is just stalling.Try again.
Then we have the rest of you post ..starting with;
Clearly, you completely ignored the i most salient part of the post, which is quoted below .. but drifted off into a kind of OT fugue.
STA, I would ask you again to comment on this, but I now realise that you really cannot play this game properly. You lack the basics, Rule 1 of which is to 'answer the mail' ( i.e. stick to the point).
I'm sorry to have to say this but having reviewed the considerable number of attempts at civilised intercourse that I have tried, using all kinds of gambits, and also observed the considerable efforts of others .. I have come to the conclusion that it is fruitless and sterile.
There is no fun to be had with you, you see? You have only the one unstable social game ... that of the death-spiral .... and that acrid monotony makes you a bore.
It would not be so bad if any of your insights were original, but we all know that they are other men's flowers, plucked from the net or other sources, and to which all of us now have access. So even your attempts at argument from the authority of gnostic academicism is tattered, and we can all see that it is your childish fury and rage that motivates your every attempt at civilised intercouse.
Madam, I leave you to your solo tarantella.
|
|
|
|
Post by speakertoanimals on Apr 7, 2011 10:28:20 GMT 1
Except if you aren't capable of understanding the argument, you won't be able to see why the ellipsoid problem is harder than the sphere or the infinite cylinder.
If you don't want to take into account all those mathematicians that have tried and failed, or the explanation as to why the ellipsoid problem is harder, fair enough, but you'll be wasting your time! not argument from authority, but understanding what efforts have been made by others, and WHY they came to the conclusion they did. Refusing to look at that is just choosing to remain ignorant of what has been done, hence you won't be able to understand what has been shown can't be done, and why.
You seem to have very rigid rules as to what people can and can't talk about -- I see no reason to respond to everything you may care to say, not if I don't consider it interesting, as long as I stay on topic (which I did). You aren't the sole arbiter and director of the conversation you know!
|
|
|
|
Post by carnyx on Apr 7, 2011 12:09:24 GMT 1
STA .. you are once demostrating your failure to grasp what was actually stated.
Try again.
...Or, carry on dancing your angry little solo jig.
|
|
