Singularitarian
Banned
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- Jul 14, 2009
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Ajay Sharma is a teacher of physics, who sprung to fame due to of course, the media. When changing something in physics is considered a big thing, it's usually someone tampering with the master equations of physics history, one most notably for its popularity is the [latex]E=Mc^2[/latex] conservation equation. The equation in this form before Sharma's contribution was already flawed, known to many scientists, that the conservation of energy is best seen as an approximation rather than an equivalance. However, Sharma developed an ingenious way to reconcile the error. His paper can be found here:
http://merlin.fic.uni.lodz.pl/concepts/2006_4/2006_4_351.pdf
Using a conversion factor given as [latex]A[/latex], we use it in his equations to describe the ''feasibility'' or the general extent in which energy and mass are convertible. So it is described for this reason as a dimensionless coefficient, which is quite different from any popular universal constant like [latex]G[/latex]. Upon readin his work, i came across some derivations he had made where the coefficient could be made to express positive and negative solutions due to either a mass increase or an energy decrease. In his equations, he gives:
[latex]\int dE=Ac^2 \int dM[/latex] [1]
from here he defines the initial and final masses, which can be found in the link given, but little time here to go over it. He comes to the conclusion which are by character linear equations as:
[latex]E_f - E_i=Ac^2(M_f - M_i)[/latex] [2]
It is from here, i derive some relationships. His equation funnily looked similar to density equations in a gravitational field i had created some time ago, which gave me the basis to formulate the relationships, which give a very neat, Impulse relationship with density.
We start by assuming that for [latex]v*^2=\pm \sqrt{2gh}[/latex] then we can equate the following linear system:
[latex]Ac^2 (F\Delta x)=Ac^2(\frac{M_f v*^2}{2}- \frac{M_i v^2}{2})[/latex] [3]
can be given the form:
[latex]Ac^2 \Box \phi(\frac{M_f v*^2}{2g}- \frac{M_i v^2}{2g})=Ac^2 (- \frac{1}{2}dMgt^2 \rho)[/latex] [4]
Using Nordstrom notation. The solution inside the brackets are close to that of relating it to the kinetic energy, since the simpler form:
[latex]Ac^2 (F\Delta x)=Ac^2(\frac{M_f v*^2}{2}- \frac{M_i v^2}{2})[/latex] [5]
Actually simplifies to: [latex]Ac^2W=Ac^2 \Delta K_e[/latex]. The relationship between work and kinetic energy within the brackets are now defined in this alteration of using Sharma's Coefficient. More interestingly, is that it is within a gravitational field, a certain distance from the gravitational source. Because energy can be gained or lost, the correct equation to use from now on would be of the form:
[latex]\int dE= \pm Ac^2 \int dM[/latex] [6]
Moving on, equation [5] if found to have the relationship with force given as:
[latex]\pm Ac^2 \int -\frac{1}{2} Mg t^2 \rho=\Box \phi \frac{1}{2}Fdt^2[/latex] [7]
This relationship on the left hand side yields the said relationhip to the impulse and density:
[latex]\Box \phi \frac{1}{2}Fdt^2=\frac{1}{2} \vec{I} \rho (dt)[/latex] [8]
Going back to the origins of equation [3], we have:
[latex]Ac^2(F\Delta x)=Ac^2(\frac{M_fv*^2}{2} - \frac{M_iv^2}{2})[/latex]
In a Hamiltonian of the classical energy-conservation equation, the mass is best to be considered as:
[latex]E=\pm Mc^2[/latex]
In relativity, we have the identity [latex]E^2=M^2+p^2[/latex] and accoring to the rule above, it is better expressed as [latex]E=\pm \sqrt{M^2+p^2}[/latex]. Sharma's equation would make this expressed now as:
[latex]\int dEv^2=Ac^2 \pm \sqrt{\int dM^2 + p^2}[/latex]
which would imply a particle with a vanishingly small amount of mass, i would calculate possibly close to the order of [latex]10^{-50}kg[/latex] to be about right; this is true in a more simpler form:
[latex]E=\sqrt{M^2+p^2(c)}[/latex]
For the negative and postive relationships of matter, the equation [latex]-\int dM^2=Ac^2(\frac{\int dE p}{c^4})[/latex] reduces to
[latex]\int dEv^2= \pm(Ac^2 \sqrt{M^2+p^2})[/latex]
http://merlin.fic.uni.lodz.pl/concepts/2006_4/2006_4_351.pdf
Using a conversion factor given as [latex]A[/latex], we use it in his equations to describe the ''feasibility'' or the general extent in which energy and mass are convertible. So it is described for this reason as a dimensionless coefficient, which is quite different from any popular universal constant like [latex]G[/latex]. Upon readin his work, i came across some derivations he had made where the coefficient could be made to express positive and negative solutions due to either a mass increase or an energy decrease. In his equations, he gives:
[latex]\int dE=Ac^2 \int dM[/latex] [1]
from here he defines the initial and final masses, which can be found in the link given, but little time here to go over it. He comes to the conclusion which are by character linear equations as:
[latex]E_f - E_i=Ac^2(M_f - M_i)[/latex] [2]
It is from here, i derive some relationships. His equation funnily looked similar to density equations in a gravitational field i had created some time ago, which gave me the basis to formulate the relationships, which give a very neat, Impulse relationship with density.
We start by assuming that for [latex]v*^2=\pm \sqrt{2gh}[/latex] then we can equate the following linear system:
[latex]Ac^2 (F\Delta x)=Ac^2(\frac{M_f v*^2}{2}- \frac{M_i v^2}{2})[/latex] [3]
can be given the form:
[latex]Ac^2 \Box \phi(\frac{M_f v*^2}{2g}- \frac{M_i v^2}{2g})=Ac^2 (- \frac{1}{2}dMgt^2 \rho)[/latex] [4]
Using Nordstrom notation. The solution inside the brackets are close to that of relating it to the kinetic energy, since the simpler form:
[latex]Ac^2 (F\Delta x)=Ac^2(\frac{M_f v*^2}{2}- \frac{M_i v^2}{2})[/latex] [5]
Actually simplifies to: [latex]Ac^2W=Ac^2 \Delta K_e[/latex]. The relationship between work and kinetic energy within the brackets are now defined in this alteration of using Sharma's Coefficient. More interestingly, is that it is within a gravitational field, a certain distance from the gravitational source. Because energy can be gained or lost, the correct equation to use from now on would be of the form:
[latex]\int dE= \pm Ac^2 \int dM[/latex] [6]
Moving on, equation [5] if found to have the relationship with force given as:
[latex]\pm Ac^2 \int -\frac{1}{2} Mg t^2 \rho=\Box \phi \frac{1}{2}Fdt^2[/latex] [7]
This relationship on the left hand side yields the said relationhip to the impulse and density:
[latex]\Box \phi \frac{1}{2}Fdt^2=\frac{1}{2} \vec{I} \rho (dt)[/latex] [8]
Going back to the origins of equation [3], we have:
[latex]Ac^2(F\Delta x)=Ac^2(\frac{M_fv*^2}{2} - \frac{M_iv^2}{2})[/latex]
In a Hamiltonian of the classical energy-conservation equation, the mass is best to be considered as:
[latex]E=\pm Mc^2[/latex]
In relativity, we have the identity [latex]E^2=M^2+p^2[/latex] and accoring to the rule above, it is better expressed as [latex]E=\pm \sqrt{M^2+p^2}[/latex]. Sharma's equation would make this expressed now as:
[latex]\int dEv^2=Ac^2 \pm \sqrt{\int dM^2 + p^2}[/latex]
which would imply a particle with a vanishingly small amount of mass, i would calculate possibly close to the order of [latex]10^{-50}kg[/latex] to be about right; this is true in a more simpler form:
[latex]E=\sqrt{M^2+p^2(c)}[/latex]
For the negative and postive relationships of matter, the equation [latex]-\int dM^2=Ac^2(\frac{\int dE p}{c^4})[/latex] reduces to
[latex]\int dEv^2= \pm(Ac^2 \sqrt{M^2+p^2})[/latex]
