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  1. #1
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    Lagrangian square energy

    The Lagrangian energy is defined as the difference of kinetic and potential energies. On the other hand, the Hamiltonian energy is the sum of kinetic and potential energies. If K is the kinetic energy and P is the potential energy then the Lagrangian L is L=K-P and the Hamiltonian H is H=K+P. The product of L and H can be defined as the Lagrangian square energy ℒ²=LH=(K-P)(K+P)=K²-P². This nomenclature or naming convention is justified by the fact for the existence of a minuend, a subtrahend, and a difference. These can either be the first power or the second power of the physical observables. Incidentally, the renormalization process of QED also uses similar nomenclatures. In this case, both minuend and subtrahend are about infinities while the difference is finite and non-vanishing. This nonzero value allows all gauge theories to be renormalized. Likewise, all gauge theories also depend on these nomenclatures but uniquely the differences remain the same for every chosen gauge.

    However, if then ℒ²=LH is separately defined as ℒ²=LH=L-H or alternatively ℒ²=LH=H-L by direct substitution ℒ²=LH=±2P. Moreover, ℒ²=LH is non-Abelian iff H-L≠L-H. The right hand side expression, ±2P, is clearly independent of the kinetic energy. If both K²-P² and ±2P are true then ±2P=K²-P² giving the polynomial energy equation: K²±2P-P²=0 or separately as two conics of hyperbola K²+2P-P²=0 and K²-2P-P²=0. Although their real domains exist, the vertices, the foci, the transverse axes, and the centers can solely be defined properly on an imaginary axis. The centers are (��,0) and (-��,0). The vertices are (0,0), (2��,0), (-2��,0), and (0,0). The surprisingly unique property is that all the foci are located at (0,0) thus making the distance constant of the hyperbola always equal to zero. Fundamentally, zero distance defines a spinor as a vector of zero length in any spacetime dimension. Furthermore, the double integral of the Lagrangian square energy with respect to two time derivatives (∬ℒ²��������) can be used to define the double least action principle of physical reality. If one action is monotonically increasing then the other is monotonically decreasing, vice versa.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Graybeard (12-28-2010)

  3. #2
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    Re: Lagrangian square energy

    Dear Antonio ..... I can never understand everything you say. So for the 'man in the street' can the concept of the 'Hamiltonian' be described as the amplitude of the kinetic and potential of the Quantum particle.

    The spread of possible values ... much like quantifying the spread of a shotgun shell and providing a value for any given slice along the spreading cone ?


    greg
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    AntonioLao (12-29-2010)

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    Re: Lagrangian square energy

    Greg, describing the Hamiltonian as amplitude is properly used for the interaction of waves, specifically the wavefunctions of quantum mechanics as solutions to the Schroedinger's equation. Here, I'm using it in a simplified but generalized form without introducing any particular generalized coordinate system. However, in a conservative system (e.g. any of the isolated or closed thermodynamics systems) where the total energy is a constant, the Hamiltonian is simply the sum of kinetic energy and potential energy such that if the kinetic energy increases the potential energy must decrease, vice versa, in order for the total energy to remain a constant.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

  6. #4
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    Re: Lagrangian square energy

    Quote Originally Posted by AntonioLao View Post
    Greg, describing the Hamiltonian as amplitude is properly used for the interaction of waves, specifically the wavefunctions of quantum mechanics as solutions to the Schroedinger's equation.
    As the wavefunction can be described as the sum of the amplitudes ... correct ? .... then are these amplitudes calculated for any time/space plane of the particle such as the area of the spread of a shotgun charge at any given point in time/space ??

    Is this what is called the Hamiltonian ?

    Quote Originally Posted by AntonioLao View Post
    Here, I'm using it in a simplified but generalized form without introducing any particular generalized coordinate system. However, in a conservative system (e.g. any of the isolated or closed thermodynamics systems) where the total energy is a constant, the Hamiltonian is simply the sum of kinetic energy and potential energy such that if the kinetic energy increases the potential energy must decrease, vice versa, in order for the total energy to remain a constant.
    But is the hamiltonian calculated for each possible moment of the particle. Why would you calculate the Hamiltonian unless you had two given parameters, one for time and the other for space. What purpose is it used for ?

    Or alternatively .... CLICK ??

    greg
    Last edited by Graybeard; 12-29-2010 at 11:21 PM.
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  7. #5
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    Re: Lagrangian square energy

    Quote Originally Posted by Graybeard
    the wavefunction can be described as the sum of the amplitudes
    That is correct. This is true because the wavefunctions are operated by linear operators together with the validity of the principle of superposition. Surprisingly, the square of the amplitude is interpreted as the probability for finding the physical location of the quantized particle. The Hamiltonian calculated for each particle is simply its energy content but which according to relativistic quantum mechanics or usually called quantum field theory (QFT) can increase without limit such that QFT are always plagued by infinities in their calculations. Fortunately, these infinities can be removed by the physical processes called first and second renormalization for mass and electric charge of every given particle using the concept of gauge symmetry. Nevertheless, there is still no gauge theory that can renormalize the infinities of gravity at zero distances.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Graybeard (12-30-2010)

 

 

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