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Scalar Gravitational Theory with Variable Rest Mass

  1. Bad theory badly developed

    0 vote(s)
  2. Bad theory ok development

    0 vote(s)
  3. Good theory poorly developed

    0 vote(s)
  4. Good theory ok development

    0 vote(s)
  1. Nov 26, 2006 #1
    In this paper we will present the mechanical dynamics of a gravitational system resulting from a specific, rest mass, scalar potential relation, that is equivalent in predicting orbital and photon motion to that of General Relativity in the weak field solutions. The weak solutions of General Relativity do not appear to be contradicted by this development, and in this range the physical difference may not be measurable. The strong field solutions will be significantly different, however since, in this scalar relation, the rest mass goes to zero at at Scwarzschild boundary. The consequences of the mass dependence gravitational potential results, for large masses, not in the prediction of black holes, but rather mass to Gamma ray converters.

    Attached Files:

  2. jcsd
  3. Nov 28, 2007 #2
    Revised version

    Due to the helpful input of several readers this paper has gone through a series of revisions, the latest of which is located at:
    http://www.arxdtf.org/css/grav2.pdf [Broken]
    The math in this paper has been gone through multiple times by myself and others, but errors are not inconceivable. The most notable differences in this paper with GR are in the predicted properties of black holes, and should be the most likely avenue of success or crash for the paper.
    Last edited by a moderator: May 3, 2017
  4. Nov 28, 2007 #3
    Technical does not mean sensible


    Equation 1 looks like nonsense to me. In standard GR, we have a few well defined tensors.

    The Riemann curvature tensor:
    [tex]R^{\alpha}_{\beta \mu \nu}[/tex]

    The Ricci tensor, a contract of the Riemann curvature tensor:
    [tex]R_{\mu \nu}[/tex]

    The Ricci scalar, a contract of the Ricci tensor:

    The metric tensor
    [tex]g_{\mu \nu}[/tex]

    And the stress energy tensor:
    [tex]T_{\mu \nu}[/tex]

    One starts with the Hilbert action:

    [tex]S = \int \sqrt{-g} d^4 x R[/tex]

    and by varying this action with respect to the metric tensor, one gets three terms. One of these is zero (I think it is the boundary of a boundary term based on a theorem by Gauss). The result is the Einstein field equations:

    [tex]R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R= 8 \pi T_{\mu \nu}[/tex]

    As I vaguely understand it, should you choose not to work with the Ricci scalar, it can in a way be moved over to the other side of the equation.

    Anyway, that is a brief description of the road to the Einstein field equations.

    Here is equation 1, taken directly from the paper:

    [tex]R^t_t = 8 \pi (T^t_t - T)[/tex]

    By Einstein's summation convension, the left hand side should just be the Ricci scalar, and the right hand side should be zero. The Ricci scalar can equal zero, but that is not too interesting. At this point I stop reading.

  5. Nov 29, 2007 #4


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    Gold Member

    I found this a bit difficult to follow. Do you use Newton's definition of the grav field to calculate [tex]\alpha[/tex] ? What is [tex]\mu[/tex] ?

    (my emphasis) Conservation of energy should be guaranteed, not assumed.

    You do not mention inertial vs gravitational mass. I think you might be saying that the mass which generates the field is different from the mass that 'feels' the field.

    I don't think it's right, but it is interesting to see another theory of gravity.
  6. Nov 30, 2007 #5
    Apologies and thanks for the 1/2 typo which has been corrected.
    [tex]R^t_t = 8 \pi (T^t_t - 1/2T)[/tex]
    This historical reference to negative Komar mass is on well worked out physics available in most texts, and I didn't think the notatio needed to be that precise, I just used the short notation used by Ansorg in the paper cited.
    My apologizes if it triggered misgivings.
  7. Dec 1, 2007 #6
    [tex]\mu[/tex] is just the gravitationla radius Gm/c^2

    The presumption is that particle energy
    [tex] mc ^{2} = m_{\o}c^{2} + G M m/r = m _{\o}c^{2} +1/2mv}^{2} =constant [/tex]
    , is not the normal assumption. The normal assumption is that gravitational energy is being contributed by the gravitational field as the particle decends. The present presumption is that the total particle energy is constant, with no contribution by the field. The source of the kinetic energy being extracted from the loss of the rest mass
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