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Zero Point: Solving for it and what it is.

  1. Aug 10, 2004 #1
    Zero Point

    What it is:

    Zero Point is the point of equivalence of the forces acting on each other of two objects(Gravity and EM). This is where the forces are equal if they were acting on a peice of bread, for instance. If the masses/charges are equal, the Zero Point of the force is directly in the middle, but if it is lopsided, the Zero Point is closer to the lesser mass/charge, due to the inverse square law.

    Solving for it:

    The basic equations to solve for Zero Point, where:

    [tex]M[/tex] is the greater mass.

    [tex]Q[/tex] is the greater charge.

    [tex]q[/tex] is the lesser charge.

    [tex]m[/tex] is the lesser mass(of course, none of this is needed when the masses are equal)

    [tex]d[/tex] is the total distance in between the two bodies.

    [tex]r_1[/tex] is the distance of M(or Q) from the Zero Point.

    [tex]r_2[/tex] is the distance of m(or q) from the Zero Point.



    [tex]r_1=\frac{Md - Mr_1}{m}[/tex]

    [tex]r_2=\frac{md - mr_2}{M}[/tex]


    [tex]r_1=\frac{Qd - Qr_1}{q}[/tex]

    [tex]r_2=\frac{qd - qr_2}{Q}[/tex]

    To solve for the Mass(or Charge)-




    Last edited: Aug 15, 2004
  2. jcsd
  3. Aug 10, 2004 #2
  4. Aug 12, 2004 #3
    C'mon! Someone criticize this or tell me why you can't!
  5. Aug 12, 2004 #4
    What you did are just linear combinations of weighted distances similar to adding vectors when they are collinear. And you also take into account that linear coefficients can also be greater than unity which is not normalized. You can get results that are infinite and therefore cannot be part of a solution to your equations.
  6. Aug 12, 2004 #5
    Please, explain it to we who do not have enough knowledge to find your words explaining of your point!
  7. Aug 12, 2004 #6


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    I get different answers. Bear in mind that [tex]r_1 + r_2 = d[/tex].
    Using the inverse square law and your notation the ratio of mass [or charge], I get a 'zero point' distance of
    [tex]M = m\frac{r_1^2}{r_2^2}[/tex]

    Your solution results in
    [tex]M=m(\frac{d}{r_2} - 1)[/tex]
    Substituting for d
    [tex]M=m(\frac{r_1 + r_2}{r_2} - 1)[/tex] which simplifies to
    Last edited: Aug 12, 2004
  8. Aug 12, 2004 #7


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    Sorry, I just got hung up on this:

  9. Aug 13, 2004 #8
    For normalization of a quantity, the dividend can never be greater than the divisor. This is to say that the quantity is a fraction. And if the quotient is unity then the dividend is equal to the divisor. So to avoid getting infinity in finding a solution when adding fractional numbers, the sum is always unity. When all the fractional numbers (dimensionless quantities such as ratios of masses and distances) do not add to unity then the result is holism. This philosophy says that the whole is greater than the sum of its parts.

    Since the total mass of universe cannot be detected as if some mass is missing, the total mass of the universe is a holistic quantity. Since the total size of the universe cannot be determined which is limited by the visible universe, the total distance of the universe is also a holistic quantity.
    Last edited: Aug 13, 2004
  10. Aug 13, 2004 #9
    I don't know...Just an example-:)
  11. Aug 13, 2004 #10
    OK. I am assuming that holistic is wholistic...whatever.

    I understand that, though I cannot get that from the original statement.
  12. Aug 14, 2004 #11
    Is there a theory that says anything about the total charge of the universe?
  13. Aug 14, 2004 #12


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    Yes. Gauss's Law. Think global charge neutrality. The Laplace equation solution for the total charge of the universe is zero.
  14. Aug 14, 2004 #13
    Excellent answer! Is it not remarkable that we do not have an equivalent Gauss's Law for the magnetic field, as we cannot have isolated poles? Both the gravitational and the electric field are conservative fields, but the former is a non conservative field, so in a certain sense it is right to say that space is pervaded by a sea of magnetic field, the origin of the so-called Zero-point-energy?, conservative fields on other hand seem to be derived fields.


  15. Aug 14, 2004 #14


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    Spoken like a true EE. I think the magnetic moment eventually submits to the same solution. Try using gauge invariance. As I recall, that was how unification of the EM field with the weak nuclear force was achieved.
  16. Aug 15, 2004 #15
    But what is a magnetic moment? Its origin is not in an inherent magnetic field, I mean, in the electron? And oh yes, the breaking of parity in the weak nuclear force, but does not the magnetic field and its even parity played a central role on it?
    Why should we have waited parity was not broken when the imposed field in the system had an even parity?
    Just some questionings about parity
  17. Aug 15, 2004 #16
    Should we consider a generalized Gauss's Law, where antimatter and matter, white gravity bodies and black gravity bodies are included, so we can give reason of what is observed at great astronomical distances and maintain the steady state of the whole gravitational cosmic system?
    I mean the total gravitational charge!
  18. Aug 15, 2004 #17
    Oh, yes the chemistry of nuclear interactions. My point and concern at this moment is not precisely that chemistry, let the experts on such a field finish their work!
    My point has to do with a way of "seeing" the relation between classical concepts and non classical ones:
    - can those concepts with which the normal engineer work everyday be put in the same context as those non classical concepts that began with QM?
    - Does not EE has worked succesfully for more than 100 years with those non classical concepts by using complex numbers?
    - Is it possible to aply that same methodology to QM, to the Lorentz transformation group, to the pendulum and even the gravitational fields?
    - can this be done with that methodology used to explain the chemistry of the nuclear interactions, I mean, can all those equations mentioned above be put under that same roof?

    Just some questionings
  19. Aug 15, 2004 #18
    Why is the total gravity charge always attractive?
  20. Aug 15, 2004 #19
    As far as I know there are just two kinds of physical fields, I mean not theoretical fields:
    - non conservative fields such as the magnetic field that do not have associated with it a Gauss's Law, having then the chance to have the two kinds of forces:attractive and repulsive: the inherent magnetic center has in it the two polarities
    - conservative or derived fields for which Gauss's Law is valid, where we find sort of odd parity; its lines of force go from one center to another center, so just one type of force we have because of this. In the mean while in this kind of field seems to prevail the second law of thermodynamics, it seems not to be the case(?) with the former, as space is surrounded by a sea of a magnetic field, some physicists have been thinking in a Zero-Point-Energy, sort of free energy? But are not both types of field complementary to each other? The latter being used for applications in the mean while the former is the original field?

    This is the way I see those things
    Best regards
  21. Aug 16, 2004 #20
    Thanks. I have no question at the moment. But when you say Gauss' law do you mean the existence of a divergence of a vector field while scalar field never can have a divergence?
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