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Force and Spacetime Equivalence

  1. Apr 22, 2004 #1
    The Principle of Equivalence between inertial mass and gravitational mass used by Einstein to formulate his general theory of relativity is a prelude to another equivalence principle.

    This is the equivalence between a fundamental force and the geometry of spacetime.

    The forces used in this discussion will only be the classical forces of gravity and electromagnetism. The weak and strong force will not be discussed at this time.
  2. jcsd
  3. Apr 22, 2004 #2
    The following classical forces are denoted by their respective symbols. [itex]F_E[/itex] for the electric force. Magnetic force is denoted by [itex] F_B[/itex], The gravitational force is [itex] F_G [/itex], the inertial force is [itex] F_I [/itex].

    [tex] \vec{F_E} = q \vec{E} [/tex]

    [tex] \vec{F_B} = q \vec{v} \times \vec{B} [/tex]

    [tex] F_G = G \frac {m_1 m_2}{r^2} [/tex]

    [tex] \vec{F_I} = m \vec{a} [/tex]
  4. Apr 22, 2004 #3
    The equivalence between [itex]F_I[/itex] and [itex] F_G [/itex] is what Einstein did. But it can shown that it is also possible to establish the equivalence between [itex] F_G [/itex] and [itex]F_E [/itex] with [itex] F_B[/itex].
  5. Apr 22, 2004 #4
    There are two distinct relationships that are equally important and must be taken for the study of their geometrical structures in spacetime. These can be done without the use of the calculus of tensor.

    [tex] F^{-}_G = F_E - F_B [/tex]


    [tex] F^{+}_G = F_B - F_E [/tex]
    Last edited: Apr 23, 2004
  6. Apr 22, 2004 #5
    Because of the minus sign between the electric force and magnetic force, the spacetime geometry can appear to be hyperbolic in contrast to Euclidean, spherical or elliptic geometry. But the true geometry is none of the above.
  7. Apr 22, 2004 #6
    The spacetime geometry is that of a doubly twisted Moebius strip and split or tear thru the middle creating two loops that are linked together. With the principle of directional invariance, two distinct strips can be created that are not topologically equivalent. That is to say they cannot be transformed into each other.
  8. Apr 23, 2004 #7
    If we are restricted to just one dimensionality of these Moebius strips, we can form two types of Hopf links that are not topologically transformable into each other. One of these two Hopf links can be the seat of the force of gravity, [itex] F^{-}_G [/itex] and the other the force of antigravity, [itex] F^{+}_G [/itex].
  9. Apr 23, 2004 #8
    If the electric force dominates the magnetic force of the vacuum then

    [tex] F^{-}_G = F_E - F_B [/tex]

    if the magentic force dominates the electric force of the vacuum then

    [tex] F^{+}_G = F_B - F_E [/tex]
  10. Apr 23, 2004 #9
    If infinitesimal lengths of r's are introduced together with the forces of electricity and magnetism of the pure vacuum, the true quantum of square of energy, E can be formulated.

    [tex] E^2 = r_E \times F_E \cdot r_B \times F_B [/tex]
  11. Apr 23, 2004 #10
    By switching the position of r and F in one of the two outer products, it is also valid for the square of energy to be given by:

    [tex] E^2 = F_E \times r_E \cdot r_B \times F_B [/tex]

    When these are expanded by Lagrange's identity the following squares of E are valid.

    [tex] E^2 = (r_E \cdot r_B)(F_E \cdot F_B) - (r_E \cdot F_B)(r_B \cdot F_E) [/tex]


    [tex] E^2 = (r_E \cdot F_B)(r_B \cdot F_E) - (r_E \cdot r_B)(F_E \cdot F_B) [/tex]

    No matter how many switches of the positions of r and F, these are the only two possible forms of the square of E. One represents positive (real) polarity of energy and the other represents negative (imaginary) polarity of energy depending on whether

    [tex] (r_E \cdot r_B)(F_E \cdot F_B) = 0 [/tex]
  12. Apr 23, 2004 #11
    If r is four dimensional then [itex] r_E \cdot r_B [/itex] can be interpreted as the spacetime interval.

    if the electric force and the magnetic force are orthogonal then [itex] F_E \cdot F_B [/itex] vanishes.

    Although [itex] r_E [/itex] and [itex] F_E [/itex] must always be orthogonal (same thing with the B's), the angle between [itex] r_E [/itex] and [itex] F_B [/itex] varies from 0 to 360 degrees.
  13. Apr 23, 2004 #12

    The equivalence of force and spacetime occurs at either of two independent conditions:

    1. When [itex] r_E \cdot r_B = 0 [/itex].

    2. When [itex] F_E \cdot F_B = 0 [/itex].

    or both.
  14. Apr 24, 2004 #13
    The two true quanta of square of energy will now be denoted by:

    [itex] H^{+}[/itex] and [itex] H^{-}[/itex]. These are symbols for two types of Hadamard matrices. With these matrices and their generalization to higher order, the geometrical structure of positive and negative electricity can be described. And the experimental values of the mass can also be described. This description and by the use of abstract algebraic rings and the Abelian group of matrix addition operator and semigroup of matrix multiplication, makes a Theory of Quantization for one dimensional Space (TQS). This theory, TQS, at the least, can removed the mystery behind the mass ratio of an electron to that of the mass of the proton as determined by experiments to be 1836. TQS predicts a value 0f 1832 which error is less than 1 percent of the experimental number.
  15. Apr 25, 2004 #14
    The central domain of Einstein's special and general theories of relativity is to give the justification to why

    [tex] r_E \cdot r_B = 0 [/tex]

    The central domain of quantum mechanics is to give the justification to why

    [tex] F_E \cdot F_B = 0 [/tex]

    And finally the task of quantum gravity is to justify both conditions.
  16. Apr 25, 2004 #15
    When both conditions are satisfied then the r's are equivalence to the wave functions of quantum mechanics and the F's are equivalence to the linear momenta and the square of energy is given by:

    [tex] E^2 = \psi_i \times \phi_i \cdot \psi_j \times \phi_j [/tex]
  17. Apr 26, 2004 #16
    The question now is to ask: "can some certainties be recovered" from the square of energy? Or becoming even more uncertain?

    Heisenberg's uncertainty principle is given by the products of conjugate variables of position and momentum or energy and time.

    [tex] \Delta \psi_i \Delta \phi_i \geq \frac{h}{2 \pi}[/tex]

    [tex] \Delta E \Delta t \geq \frac{h}{2 \pi} [/tex]
  18. Apr 26, 2004 #17
    But the square of the uncertainty should become more certain.

    [tex] \Delta E^2 \Delta t^2 \leq \frac {h}{2 \pi} [/tex]
  19. Apr 26, 2004 #18
    The physical meaning of the difference of the square of time must now be clarified.

    What is [itex] \Delta t^2 [/itex] ?

    Does this implies there are two directions of time?
  20. Apr 26, 2004 #19
    If [itex] \Delta E^2 = \Delta F^2 \Delta \psi^2 [/itex] then [itex] \Delta \psi^2 \Delta t^2 [/itex] is the spacetime interval.
  21. Apr 26, 2004 #20
    This shows that the physical meaning [itex] \Delta t^2 [/itex] is equivalent to the square of the linear momentum, [itex] \Delta \phi^2 [/itex] when mass is still not clearly defined.
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