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Newton's Sphere Theorem Argument is false.

  1. Jun 17, 2005 #1
    In his Principia Newton claimed that a hollow sphere exerts no force on a particle floating inside (Propositions 70 to 74 of Book I). His argument went as follows:

    (1) Imagine a test-particle at the vertex of a cone of fixed acuteness. You can picture the cone as a flashlight beam coming from the particle. The cone makes a circle on a flat sheet of uniform density and thickness. Moving the sheet further away makes a larger disk on the sheet. The size of the disk increases as the square of the distance (d^2) and so does the mass, while the gravitational force of the disk on the test-particle decreases as the square of the distance (d^2), cancelling out any change. That is, at any given distance, the cone traces the size disk needed to keep the attractive force constant.

    (images can be provided if I am allowed)

    (2) Now imagine the particle is between two parallel sheets: You can picture two equal opposing cones beaming out from your particle like a lighthouse beam. (Equal cones can be made by rotating one line passing through the particle around another line-axis through it.) Either sheet or the particle can be moved independantly, and the force from the disks will remain equal and balanced, as long as the sheets stay parallel.

    (3) Similarly, opposing cones will cut out opposing spherical caps on the surface of a hollow sphere if we put our particle and its cones inside. The arbitrary position of the particle and the varying distance of the caps is compensated by their size and mass. Newton proposed constructing more cone pairs, covering the entire surface and showing that the forces inside balance.

    (4) Someone might object that the spherical caps marked by the cones aren't really flat disks. Newton responds by shrinking the cones to very small (infinitesimals) so that we can make the caps as flat as we wish.

    (5) Another might ask how you can cover the sphere surface completely with circles. Again, Newton argues that spaces between disks can be filled with smaller circles and so on. We can iterate the procedure as many times as we wish to closely cover the surface area.

    The task is to disprove Newton's argument in a few simple steps using high-school level mathematics, in a way that anyone can understand.
    To prevent any misunderstanding, this is a physics theorem meant to be applied to physical problems. This requires more than just some simple math. It also requires a physicist's ability to intelligently apply and interpret the physical significance of the theorem. A little of column A, and a little of column B.

    Now for the mathematics:

    (1) Not all points in the sphere are equal.

    Firstly, all points inside the shell are not equal. There are two distinct sets of points inside a sphere. Set 1 contains only one point, the origin, or geometric centre (GC), equidistant from all points on the surface. Set 2 contains all the other points. All the other key geometric features follow from this.

    Let P be a point anywhere inside a sphere, but not at the centre. Only lines passing through a sphere's centre will pierce the surface perpendicularly. All other lines pierce the surface at some other angle. So a line passing through P must also pass through the centre to be perpendicular at the surface. Only one line passes through both P and the centre, and is perpendicular at the surface.

    (2) Most of Newton’s Cones Cut Sphere on an Angle

    Similarly, since a cone-axis is a line, only one cone-axis passing through P is perpendicular at the surface. Only cones formed on this axis will make perpendicular disks or spherical caps. Cones on some other cone-axis will cut tilted ellipses or spherical caps.

    Newton’s scheme is to cover the surface using cone-pairs. Each pair must have a different axis, so only one pair can be perpendicular at the surface if P is not at the centre. All the other cone-pairs pierce the surface on an angle, and make caps or disks which are tilted relative to P.

    (3) Tilted Disks Pull Off-Centre

    A point-mass is only pulled directly toward the centre of a uniform disk when the point-mass lies in the same plane as the disk or when the point-mass lies on the axis of the disk. That is, whenever a disk has any other tilt relative to the point-mass, there is a residual force toward the nearest disk edge, pulling the point-mass off-course from the disk centre. This is because the Centre of Mass theorem fails in close proximity or for significant spreads in distribution of mass, since it is only an approximation.

    (4) Tilted Disk Pairs on Sphere Don’t Balance Out

    Could disks balance their forces in spite of tilt? Yes, but only between two (infinite) uniform parallel planes, where the tilts cancel. Then masses between the planes can indeed experience zero net force. Newton's claim for hollow spheres actually turns out to be true (in theory) for parallel planes

    We can understand intuitively that even though the direction of force is off-center because of tilt, we can exactly counter that with an equal and opposite tilt on the opposing disk, without having to correct the direction change from the tilt. But on the sphere this is impossible: The disks actually double the error. This is why Newton's thought experiment works with parallel planes, but not with spheres.

    Newton himself posed that making the cones very small (infinitesimal) would eliminate any accuracies, (one problem is the fact that spherical caps aren't really flat disks at any size), but Newton didn't foresee the problem of tilt.

    - - - - - - - - - - - - - - - - -
    Tilt indeed vanishes or becomes insignificant if the disks are made very small relative to their distance from the test-particle. However this is just a kind of sleight of hand;

    (a) The problem isn't tilt per se, but the actual distribution of mass, in this case manifested as tilt with larger disks.

    (b) But just as a curve can be chopped into infinitesimally small straight lines, while it's curvature remains constant, this has no effect on the actual distribution of mass, which is fixed. We can measure things many different ways, but this doesn't move the mass.

    The information concerning distribution of mass can be contained by the tilt, or translated into other forms, but problem of the imbalance of forces remains, and Newton's argument is a failure.

    This section shows that Newton's argument is wrong. If the Sphere Theorem is correct after all, it must be correct for reasons other than those that Newton gave.
    Now lets show why the Sphere Theorem is false in the real physical situation:

    The problem here is that simply applying the result of the integration to the sphere is not an accurate description of the physical case.

    To get right at the issue, since mass is in reality distributed in clumps, and is not a continuum, the Sphere Theorem is only a valid or useful approximation for very large uniform spheres at a macro-level, involving millions of atoms.

    It is not in serious dispute that the bulk of the mass of atoms reside in the nucleus and this is demonstrated both mechanically and vis the gravitational field by the scattering matrix. Most scientists agree that this experiment has already been done to death.

    (1) If we were to make a sphere out of a thin layer of gold atoms for instance, the actual gravitational potential field would not at all reflect the result of an integration of the continuum model proposed by Newton's Sphere Theorem, and so the integral is inapplicable to this problem. The *REAL* field would look more like a golf-ball, and there would be no flat field inside the sphere. All particles floating inside would accelerate outward toward the inside surface due to imbalance of forces, no matter how carefully the sphere was constructed.

    (2) One might think, "So what? The Sphere Theorem fails at the molecular level. Big deal." But this is not the case at all. The failure is independant of size entirely. It is not tied to physical size, but to the coarseness of the quanization of the mass distribution. This would also be true for charge distribution as in both classical and relativistic electrostatics.

    (3) You could also have a 3-meter diameter aluminium sphere which would for all intents and purposes would be a continuum. However, once the static charge on it dropped below a few thousand excess electrons or hole charges, even though these charges would spread out as evenly as possible due to repulsion, the electric field would be as lumpy as gravitational potential of the gold-leaf ball we referred to above.

    This thought experiment is all that is required for any reasonable person to see that the Sphere Theorem is an approximation similar to the Center of Mass theorem, and it fails miserably in many physical situations. you don't have to be a rocket scientist to see this.

    The Sphere Theorem is not accurate, practical or useful. QED.
    Last edited: Jun 17, 2005
  2. jcsd
  3. Jun 17, 2005 #2
    It is impossible to disprove Newton's result, because it is correct. Using high school calculus it is easy enough to integrate the new g-field in a sphere and show that it is zero. This is actually what Newton did himself, but because the calculus was relatively new he prefered to publish a geometric proof.

    I would agree with you that the argument is not very rigorous and that it left me unconvinced at the time I studied it.
  4. Jun 17, 2005 #3


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    Keep it on sciforums, Rogue Physicist. It's not welcome here.

    - Warren
  5. Jun 17, 2005 #4
    r u sure his argument is false?? i also dont think that the particle will experience any force at all considering the sphere to be free of any other forces or as an isolated body with the particle inside.the forces will cancel as per me....correct if its wrong and then i will give my explanation abt this ans...
  6. Jun 17, 2005 #5
    I have edited the original post since the Admin moved it here. Take a another look above.
  7. Jun 17, 2005 #6
    a short description to your long long ans...

    its an isolated sphere so external forces =0 and inside forces will cancel each other...this can be proved also by equating forces BUT a short simple example is our Universe which is a sphere of infinite radius...does an astronaut floating in space feel any acceleration when its NOT withing the field od gravitation of any astronomical object????????? NO he/she doesnt!!! I think this is a very short example but if u still are not satisfied, i may come up with the long description....and if the theory has this point that the spherical hollow disc can be constructed with infinitesimally small circles then its NOT correct A two dimentional object can never extend to a three dimentional one!!!but i dont think this is there ..have u inserted this yourself??????????

  8. Jun 17, 2005 #7


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    So, to make it clear: What happens to a particle inside a hollow cylinder?
  9. Jun 17, 2005 #8
    Clever question. In the real physical case, since the circle is made of atoms with discretely localized masses spread far apart, there is no uniform field here, even for a single slice of the cylinder. If you plot the field potential in polar coordinates from the center outward, you will see that the unevenness of the field increases as you leave the centre. Since the force is attractive, even the Geometric Centre (GC) is unstable, and placing a particle there would be like balancing a marble on a hilltop. eventually, even just a few degrees above absolute zero there would be a breaking of symmetry and the test-particle would accelerate toward the nearest inside surface. The geodesic trajectory would resemble an interesting type of scattering matrix, as the particle flew to the cylinder side, and with a high probability passed through.
  10. Jun 17, 2005 #9

    Tom Mattson

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    So you're on about Newton because his argument isn't rigorous enough for you? Well then why not apply Gauss' law for gravitation to the uniform spherical shell? You will certainly get a zero force inside the shell.

    Yes, one might. In fact, this one does.

    Still: So what? In many macroscopic situations it is perfectly acceptable to ignore the coarseness of the mass distribution.

    Puh-leeze. It's as accurate, practical, and useful as the assumption of continuous mass distributions.
  11. Jun 17, 2005 #10


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    Did you understand that his "argument" is claiming that what you are saying here is false?
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