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Can Quantum Jumps Same as Dimension Jumps ?

  1. Jul 26, 2004 #1
    Can Quantum Jumps Same as Dimension Jumps ???

    By going back to Bohr's stationary orbits of electrons, The idea of quantum jumps was born.

    We can hypothesize that matter such as electrons are 4D objects and photons are 3D objects and the continuous space between orbits is a 2D continuum. So what happens when one electron jumps from one orbits to another orbit is that the 4D electron transforms to 3D energy and the 3D energy transforms to a 2D continuum then 2D continuum back to 3D energy and then back to a 4D electron at a lower or higher orbit whichever the case in question. And transformation of 2D continuous space (S) is given by:

    [tex] S = cE [/tex]

    where c is the speed of light in vacuum and it acts as dimension lowering constant for energy (E).

    And by dimensional analysis, the 2D continuous space is proportional to an arbitrary surface area with the proportionality constant as force per unit of time. Futhermore force per unit time is proportional to the time derivative of acceleration with mass as the constant of proportionality.

    [tex] \frac {Force}{time} = m \frac {\partial a}{\partial t}[/tex]
     
    Last edited: Jul 26, 2004
  2. jcsd
  3. Jul 26, 2004 #2
    If the surface area changes with time then the gradient of continuous space is given by

    [tex] \nabla S = m \frac {\partial \vec{a}}{\partial t} \frac{\partial Area}{\partial t}[/tex]

    Futhermore for volume as a function of time, V(t) and given density [itex] \rho [/itex], the gradient of continuous space is given by

    [tex] \nabla S = \frac {c^4 \rho_T}{\vec{r}} V(t) [/tex]
     
    Last edited: Jul 27, 2004
  4. Jul 26, 2004 #3
    If we can define the ratio of instantaneous density to total density of the universe as a function of time given by

    [tex]\frac{1}{\rho_T} \int_{0}^{\infty} \rho_i (t) dt = 1 [/tex]

    then the instantaneous density is the sum of kinetic density and potential density given by

    [tex] \rho_i = \rho_k + \rho_p [/tex]

    If the total mass of the universe is the sum of total potential mass and total kinetic mass

    [tex] m_T = m_k + m_p [/tex]

    then instantaneous kinetic and potential volumes can be related to the total volume given by

    [tex] \frac{m_T}{m_p} - \frac{V_k}{V_p} \left ( \frac{V_T - V_p}{V_k - V_T} \right ) = 1 [/tex]
     
    Last edited: Jul 26, 2004
  5. Jul 26, 2004 #4
    [tex] \frac{m_T}{m_p} \geq 1 [/tex]

    and

    [tex] \frac{V_k}{V_p} \left ( \frac{V_T - V_p}{V_k - V_T} \right ) < 1 [/tex]

    if

    [tex] \frac{m_T}{m_p} = 1 [/tex]

    then

    [tex] \frac{V_k}{V_p} \left ( \frac{V_T - V_p}{V_k - V_T} \right ) = 0 [/tex]
     
  6. Jul 26, 2004 #5
    [itex]V_k [/itex] is the same as the instantaneous volume of radiation. [itex] V_p [/itex] is the instantaneous volume of matter. [itex] V_T [/itex] is the total volume of the universe at each epoch.
     
  7. Jul 28, 2004 #6
    The gradient of continuous space becomes a function of instantaneous density and volume.

    [tex]\nabla S = \frac{c^4}{\vec{r}} \int_{0}^{\infty} \rho_i (t) V(t) dt [/tex]

    From this, an integral force exist.

    [tex] F = \frac{c^3}{\vec{r} \times \vec{r}} \int_{0}^{\infty} \rho_i (t) V(t) dt [/tex]
     
  8. Jul 28, 2004 #7
    This force is infinitely large because the term [itex] \vec{r} \times \vec{r} = 0 [/itex]. But if the r's are orthogonal and comparable to Planck length then the force is just simply large. If orthogonal r's is very large then the force is small.
     
  9. Jul 28, 2004 #8
    Orthogonality seems to be a necessary condition for the force to exist. This implies that orthogonal forces must exist and the extremely high magnitude of this force must be able to quantize spacetime at the local infinitesimal region of a spacetime continuum.
     
  10. Jul 28, 2004 #9
    For low velocity and small mass and together with the invariance, [itex] \vec{a} \cdot \vec{r} = c^2 [/itex], this force becomes Newton's 2nd law of motion.
     
  11. Jul 28, 2004 #10
    In post #7, the r's can become large only if the individual r's can be added collinearly together.

    [tex] \vec{r} = \int_{0}^{\infty} \vec{r}_i [/tex]
     
  12. Jul 28, 2004 #11
    Collinearity implies that [itex] \vec{r} = \alpha \vec{r}_i[/itex] where [itex] \alpha[/itex] is an integer.
     
  13. Jul 28, 2004 #12
    When the linear momentum is zero and when both density and volume are functions of energy and time, the square of mass is given by

    [tex] m^2 = \int \int \frac{\partial \rho^2}{\partial t} \frac{\partial V^2}{\partial t} dt dt [/tex]
     
  14. Jul 28, 2004 #13
    Furthermore, the square of mass is a new constant of nature given by

    [tex] m^2 = \left( \frac{h}{ac} \right)^2 [/tex]

    where h is Planck's constant, a is Planck length and c is the speed of light.
     
  15. Jul 28, 2004 #14
    Implication is the existence of real positive and negative root for mass given by

    [tex] \pm \frac{h}{ac} [/tex]

    without the use of complex number.
     
  16. Jul 28, 2004 #15
    The use of these roots is to constraint the value of

    [tex] \frac {V_k}{V_p} \left( \frac{V_T - V_p}{V_k - V_T} \right) [/tex]

    or

    [tex] \frac {V_k}{V_p} \left( \frac{V_p - V_T}{V_T - V_k} \right) [/tex]

    so that the expressions are always less than unity.
     
  17. Jul 31, 2004 #16
    Why is the quantum of mass using [itex] \pm \frac{h}{ac}[/itex] numerically equal to the Planck mass using [itex] \sqrt{\frac{hc}{G}}[/itex]?

    If Planck's constant was determined first together with the speed of light and knowing the value of the Planck length, the value of the gravitational constant can be found by

    [tex] G = \frac{a^2 c^3}{h}[/tex]
     
  18. Jul 31, 2004 #17
    In a planetary system where one of the earthlike planets is in a perpetual cloud cover, the inhabitants will never be able to have the opportunity of seeing the starry night sky. In this planet, there could be no Galileo nor Kepler and then nor Newton before Maxwell's and Planck's discoveries. Yet the scientists of this planet can derive the laws of electromagnetism and also quantum mechanics first and then discover the law of universal gravitation afterward.

    Although Einstein's theories of relativity can clinch the final constancy of the speed of light, these are not necessary for the first order determination of the gravitational constant.
     
    Last edited: Jul 31, 2004
  19. Jul 31, 2004 #18
    Quite clever example of different paradigms and how they will guide our quests in different ways, but I was wondering if you did choose those examples to be discovered first(Maxwell equations without the displacement current concept and Plank's discoveries) because in the last analysis they have to do with more fundamental laws of nature than those of gravitation?

    Regards

    EP
     
    Last edited: Jul 31, 2004
  20. Jul 31, 2004 #19
    Epsilon Pi,

    Will respond to your reply at the earliest possible time.
     
  21. Aug 1, 2004 #20
    I think all physical constants which are independent on other constants should appear in physical laws that are not depended on the sequential order of their discovery. For example, Planck's constant is depended on Boltzmann's constant therefore Boltzmann's constant must necessarily come first. Maxwell's theory of electromagnetism and quantum theory are in sequential order. One must follow the other since the later one uses some constants found in the earlier theory.
     
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