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General relativity, dark energy and the cosmological constant

  1. Mar 13, 2012 #1
    I see where someone listed a theory that they had https://www.physicsforums.com/showthread.php?t=49230
    so going to attempt to do the same.
    I listed my theory at http://www.bautforum.com/showthread.php/129309-Negative-Mass-Interpretation-of-General-Relativity but they are very quiet and I would appreciate some healthy debate.

    I listed my paper at http://www.vixra.org/pdf/1203.0025v1.pdf

    Comments appreciated.

    This theory states that the differential geometry of General Relativity is better physically interpreted through the equation

    [tex]R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R=G_{\mu \nu}=g_{\mu \nu}\Lambda-\Pi_{\mu \nu}[/tex]

    where [tex]G_{\mu \nu}[/tex] is Einstein's normal curvature tensor, [tex]g_{\mu \nu}\Lambda[/tex] is the cosmological constant term and [tex]\Pi_{\mu \nu}[/tex] represents the curvature tensor of negative (or residual) mass-energy.

    The first benefit is that if one were to interpret [tex]g_{\mu \nu}\Lambda[/tex] as the potential energy of all quantum harmonic oscillator states, then [tex]G_{\mu \nu}[/tex] would represent the "occupied" states and [tex]\Pi_{\mu \nu}[/tex] the unoccupied. Thus the cosmological constant would appear to be very close to zero (probably showing an error in the equation on large scales that [tex]\Lambda[/tex] isn't exactly constant over those scales.

    Both Einstein's tensor and the negative mass-energy tensor should both be capable of the exact same transformations.

    Locally, there would be no way to distinguish one from the other experimentally, as they simplify down to equivalent Newtonian potentials (I will post my derivation here if requested, it is long but algebraically trivial, can be found http://www.vixra.org/pdf/1203.0025v1.pdf ) of

    [tex]\vec{g}=-\nabla\Phi=-\frac{GM}{r^{2}}\hat{\vec{r}}=-\frac{\Lambda_{\mathrm{vac}} c^{2}r}{6}\hat{\vec{r}}+\frac{G\rho_{\mathrm{res}}V}{r^{2}}\hat{\vec{r}}[/tex]

    for a spherical mass [tex]\textit{M}[/tex] where the "res" subscript denotes residual or remaining negative mass-energy. As a visual aid, the gradients of the top and bottom Euclidean two dimensional scalar fields are equivalent (multiply both by a -1):
    See attached image

    The empirical evidence that suggests the negative mass interpretation is actually the more correct interpretation would be the accelerating expansion detected in 1998. On large scales, the Newtonian potential equations lose their meaning since as baryonic mass is reduced, the unit vector loses its meaning and the equation would be considered as repulsive.

    If the Newtonian equation were correct, there should be a point where for a test mass of 1kg, [tex]\vec{g}=0[/tex]. Namely

    [tex]r=(\frac{6G\rho_{\mathrm{res}}V}{\Lambda_{\mathrm{vac}} c^{2}})^{1/3}[/tex]

    where greater than this, gravity would be repulsive.

    Attached Files:

  2. jcsd
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