Anamitra Palit on the mathematics of General Relativity

Discussion in 'Pseudoscience' started by Anamitra Palit, Dec 18, 2020.

  1. Anamitra Palit Registered Senior Member

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    102
    Moderator note:

    The first couple of pages of this thread merge together approproximately 6 threads started by Anamitra Palit, all concerning various mathematical aspects of General Relativity. In general, this poster seems to be claiming that GR is mathematically flawed, inconsistent, or produces trivial results that do not correctly describe the mathematics of space and time.

    ----

    Certain unusual/conflicting features follow from the analysis of the transformation of the covariant derivative:
    https://drive.google.com/file/d/1K0Z_zljtOTaxmDketEY2JDxthKmadnpE/view?usp=sharing
     
    Last edited by a moderator: Jan 26, 2021
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  3. QuarkHead Remedial Math Student Valued Senior Member

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    Look neither I nor, I hope, will anyone else open an URL kink that looks highly suspect. If you have something to say, then say it.

    However, the covariant derivative in differential geometry is defined to be invariant under any and all coordinate transformations - that is its whole reason for existence.
     
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  5. Anamitra Palit Registered Senior Member

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    QuarkHead I don't understand what you mean by 'an URL kink that looks highly suspect'
    Equation (3) in the paper brings out a discrepancy in the theory
     
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  7. James R Just this guy, you know? Staff Member

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    Please post the summary of the problem you have identified.
     
  8. Anamitra Palit Registered Senior Member

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    @ James R:The paper is a short mathematical one.It is the mathematics that has to be understood to get to its objective....to have a clear view of it. In the initial posting it has been stated:"Certain unusual/conflicting features follow from the analysis of the transformation of the covariant derivative". The covariant derivative of a rank one contravariant tensor is a mixed tensor of rank two.Applying the transformation rules of a rank two mixed tensor we have arrived at an equation[equation (3)] which expresses a contradiction in the theory.
     
  9. Anamitra Palit Registered Senior Member

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    102
    I tried out the basic equations in Latex. These equations were obtained by using Mathpix software. But I could not obtain the correct preview.But on posting the equations came up correctly . The detailed description would be posted shortly
    \begin{equation}
    \begin{aligned}
    &\frac{\partial \bar{A}^{\mu}}{\partial \bar{x} \rho}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}}\left[\frac{\partial A^{\alpha}}{\partial x^{\beta}}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma}\right]\\
    &\frac{\partial \bar{A}^{\mu}}{\partial \bar{x}^{\rho}} d \bar{x}^{\rho}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}}\left[\frac{\partial A^{\alpha}}{\partial x^{\beta}} d \bar{x}^{\rho}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} d \bar{x}^{\rho}\right]\\
    &d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial \bar{x} \rho}\left[\frac{\partial A^{\alpha}}{\partial x^{\beta}} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} d x^{k}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} d x^{k}\right]\\
    &d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} \frac{\partial A^{\alpha}}{\partial x^{\beta}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} \frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}\\
    &d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\delta^{\beta} k \frac{\partial A^{\alpha}}{\partial x^{\beta}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} \delta_{k}^{\beta} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}
    \end{aligned}
    \end{equation}

    \begin{equation}

    \begin{array}{l}

    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial A^{\alpha}}{\partial x^{k}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \\

    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\mu}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial A^{\alpha}}{\partial x^{k}} d x^{k}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \\

    \qquad \bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \quad(2) \mid

    \end{array}


    Equation (2) holds for any manifold.In particular we consider the flat space time manifold.
    In the flat space time context the Christoffel symbols[all of them] are zero only in the Cartesian system but non zero[all not zero] in the others. On the right side of equation (2)we consider Cartesian coordinates in flat space time: Γ=0 . On the left side we consider some other coordinate system manifold being the same that is flat space time.

    Therefore from ( 2 ) we obtain:
    $$
    \bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=0(3)
    $$
    But $\bar{\Gamma}_{\rho \sigma}^{\mu} \neq 0,$ and the field $\bar{A}^{\sigma}$ is arbitrary! The possibility of equation (3) materializing comes into
    question.
    3
    We may obtain the Christoffel symbols for flat pace time in the spherical system by applying $\mathrm{M}=0$ to the Schwarzschild Christoffel symbols $^{[2]} .$ We have six non vanishing Christoffel symbols for $\mathrm{M}=0$
    $$
    \Gamma_{\theta \theta}^{r}=-r, \Gamma_{\varphi \varphi}^{r}=-r \sin ^{2} \theta, \Gamma_{r \theta}^{\theta}=\frac{1}{r}, \Gamma_{\varphi \varphi}^{\theta}=-\cos \theta \sin \theta, \Gamma_{r \varphi}^{\varphi}=1 / r, \Gamma_{\theta \varphi}^{\varphi}=\cot \theta
    $$
    Direct Verification[flat space time, spherical]:
    Following the usual technique ${ }^{[3]}$
    $$
    \begin{array}{c}
    \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2} g^{\alpha s}\left[\frac{\partial g_{s \beta}}{\partial x^{\gamma}}+\frac{\partial g_{s \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{s}}\right] \\
    g_{\alpha k} \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2} g_{\alpha k} g^{\alpha s}\left[\frac{\partial g_{s \beta}}{\partial x^{\gamma}}+\frac{\partial g_{s \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{s}}\right] \\
    g_{\alpha k} \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2} \delta_{k}^{s}\left[\frac{\partial g_{s \beta}}{\partial x^{\gamma}}+\frac{\partial g_{s \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{s}}\right]
    \end{array}
    $$
     
    Last edited: Dec 26, 2020
  10. Anamitra Palit Registered Senior Member

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  11. Anamitra Palit Registered Senior Member

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    102
    Transformation of the Covariant Derivative
    Anamitra Palit
    Physicist,freelancer
    P154 Motijheel Avenue,Flat C4,Kolkata 700074,India
    palit.anamitra@gmail.com
    Cell No:+ 919163892336
    Abstract
    The article considers the transformation of the covariant derivative of a rank one contravariant tensor to bring out a conflicting aspect of the theory in that we arrive at an impossible equation pointing to conflict in the theory.
    Introduction
    The covariant derivative of a rank one contravariant tensor is a mixed tensor of rank two. Its transformation leads to an impossible equation to bring out a contradiction in the theory.
    Calculations
    We consider the transformation of the covariant derivative[1] of the rank one contravariant tensor[which is a mixed tensor of rank two ]
    \begin{equation}

    \begin{aligned}

    &\frac{\partial \bar{A}^{\mu}}{\partial \bar{x} \rho}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}}\left[\frac{\partial A^{\alpha}}{\partial x^{\beta}}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma}\right]\\

    &\frac{\partial \bar{A}^{\mu}}{\partial \bar{x}^{\rho}} d \bar{x}^{\rho}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}}\left[\frac{\partial A^{\alpha}}{\partial x^{\beta}} d \bar{x}^{\rho}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} d \bar{x}^{\rho}\right]\\

    &d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial \bar{x} \rho}\left[\frac{\partial A^{\alpha}}{\partial x^{\beta}} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} d x^{k}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} d x^{k}\right]\\

    &d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} \frac{\partial A^{\alpha}}{\partial x^{\beta}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\rho}}{\partial x^{k}} \frac{\partial x^{\beta}}{\partial \bar{x}^{\rho}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}\\

    &d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\delta^{\beta} k \frac{\partial A^{\alpha}}{\partial x^{\beta}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{\beta \gamma}^{\alpha} A^{\gamma} \delta_{k}^{\beta} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}

    \end{aligned}

    \end{equation}

    \begin{equation}
    \begin{array}{l}
    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial A^{\alpha}}{\partial x^{k}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \\
    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial A^{\alpha}}{\partial x^{k}} d x^{k}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \\
    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\mu}=d \bar{A}^{\mu}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}
    \end{array}
    \end{equation}

    \begin{equation}
    \bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}
    \end{equation}
    -----(2)
    Equation (2) holds for any manifold.In particular we consider the flat space time manifold.
    In the flat space time context the Christoffel symbols[all of them] are zero only in the Cartesian system but non zero[all not zero] in the others. On the right side of equation (2)we consider Cartesian coordinates in flat space time:
    \begin{equation}
    \Gamma_{k \gamma}^{\alpha}=0
    \end{equation}.
    On the left side we consider some other coordinate system, manifold remaining the same that is flat space time.
    Therefore from (2) we obtain:
    \begin{equation}
    \bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=0
    \end{equation}
    ---------------(3)

    But
    \begin{equation}
    \bar{\Gamma}_{\rho \sigma}^{\mu} \neq 0
    \end{equation}
    ,and the field̅
    \begin{equation}
    \bar{A}^{\sigma}
    \end{equation}
    is arbitrary! The possibility of equation (3) materializing comes into question.
    We may obtain the Christoffel symbols for flat pace time in the spherical system by applying M=0 to the Schwarzschild Christoffel symbols[2]. We have six non vanishing Christoffel symbols for M=0
    \begin{equation}
    \Gamma_{\theta \theta}^{r}=-r, \Gamma_{\varphi \varphi}^{r}=-r \sin ^{2} \theta, \Gamma_{r \theta}^{\theta}=\frac{1}{r}, \Gamma_{\varphi \varphi}^{\theta}=-\cos \theta \sin \theta, \Gamma_{r \varphi}^{\varphi}=1 / r, \Gamma_{\theta \varphi}^{\varphi}=\cot \theta
    \end{equation}

    Direct Verification[flat space time, spherical]:
    Following the usual technique[3],
    \begin{equation}
    \begin{array}{c}
    \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2} g^{\alpha s}\left[\frac{\partial g_{s \beta}}{\partial x^{\gamma}}+\frac{\partial g_{s \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{s}}\right] \\
    g_{\alpha k} \Gamma_{\beta r}^{\alpha}=\frac{1}{2} g_{\alpha k} g^{\alpha s}\left[\frac{\partial g_{s \beta}}{\partial x^{\gamma}}+\frac{\partial g_{s \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{s}}\right] \\
    g_{\alpha k} \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2} \delta_{k}^{s}\left[\frac{\partial g_{s \beta}}{\partial x^{\gamma}}+\frac{\partial g_{s \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{s}}\right] \\
    g_{\alpha k} \Gamma_{\beta \gamma}^{\alpha}=\frac{1}{2}\left[\frac{\partial g_{k \beta}}{\partial x^{\gamma}}+\frac{\partial g_{k \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{k}}\right]
    \end{array}
    \end{equation}

    [No summation on k]

    In the orthogonal system the only surviving term on the left side is
    \begin{equation}
    g_{k k} \Gamma_{\beta r}^{k}
    \end{equation}
    with no summation on k.

    We have,
    \begin{equation}
    \begin{array}{l}
    g_{k k} \Gamma_{\beta \gamma}^{k}=\frac{1}{2}\left[\frac{\partial g_{k \beta}}{\partial x^{\gamma}}+\frac{\partial g_{k \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta r}}{\partial x^{k}}\right] \\
    \Gamma_{\beta \gamma}^{k}=\frac{1}{2 g_{k k}}\left[\frac{\partial g_{k \beta}}{\partial x^{\gamma}}+\frac{\partial g_{k \gamma}}{\partial x^{\beta}}-\frac{\partial g_{\beta \gamma}}{\partial x^{k}}\right]
    \end{array}
    \end{equation}

    As for an example we may have,
    \begin{equation}
    \begin{array}{c}
    \Gamma_{r \varphi}^{\varphi}=\frac{1}{2 g_{\varphi \varphi}}\left[\frac{\partial g_{\varphi r}}{\partial \varphi}+\frac{\partial g_{\varphi \varphi}}{\partial r}-\frac{\partial g_{r \varphi}}{\partial \varphi}\right] \\
    \Gamma_{r \varphi}^{\varphi}=\frac{1}{2 g_{\varphi \varphi}} \frac{\partial g_{\varphi \varphi}}{\partial r}=\frac{1}{-2 r^{2} \sin ^{2} \theta} \frac{\partial\left(-r^{2} \sin ^{2} \theta\right)}{\partial r}=1 / r
    \end{array}
    \end{equation}
    The other Christoffel symbols may be verified in a similar manner.[Covariant derivative reduces to partial derivative in the fat space time context only in the Cartesian system]
    Conclusions
    As stated at the outset we have arrived at an impossible equation starting from the transformation of the covariant derivative of a contravariant tensor. This points to difficulties in the basic theory.

    References
    1. Spiegel M R,Vector Analysis and an Introduction to Tensor Analysis, Schaum’s Outline Series, MacGraw Hill Book Company,Singapore,1974,Chapter 8,Tensor Analysis, problem 52,p197-198.
    2. Hartle J. B.,Gravity ,Pearson Education Inc,2003,Appendix B,p570
     
    Last edited: Dec 26, 2020
  12. Anamitra Palit Registered Senior Member

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    102
    In the last post we consider equations prior to (2)
    \begin{equation}
    \begin{array}{l}
    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial A^{\alpha}}{\partial x^{k}} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \\
    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\rho}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} \frac{\partial A^{\alpha}}{\partial x^{k}} d x^{k}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k} \\
    d \bar{A}^{\mu}+\bar{\Gamma}_{\rho \sigma}^{\mu} \bar{A}^{\sigma} d \bar{x}^{\mu}=d \bar{A}^{\mu}+\Gamma_{k \gamma}^{\alpha} A^{\gamma} \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}} d x^{k}
    \end{array}
    \end{equation}
    We should take note of the fact
    \begin{equation}
    \frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}}\frac{\partial A^{\alpha}}{\partial x^k} dx^{k}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}}d A^{\alpha}=d\bar{A}^{\mu}\end{equation}
     
    Last edited: Dec 26, 2020
  13. Anamitra Palit Registered Senior Member

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    The article on the link investigates the metric coefficients in the light of the Field Equations of General Relativity. We drive results connecting the metric coefficients(exclusively). They follow from the consistent application of mathematics on the existing theory of General Relativity is quite starling in view of the fact that these results impose a heavy constraint on the theory.

    Using the(insert) link facility a google drive file has been provided:
    https://drive.google.com/file/d/1O5zOIHjd0OnjwO_NnYkmX_akejPLy6Ii/view?usp=sharing

    I tried to put out the material in latex . A large part of the file was written in Latex. But the preview did not come. Assistance requested...
     
  14. Anamitra Palit Registered Senior Member

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    102
    Last edited: Jan 11, 2021
  15. Anamitra Palit Registered Senior Member

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    102
    The Equations representing the Energy momentum transformations or the Lorentz transformations for that matter are not independent;

    https://vixra.org/abs/2101.0050
     
  16. Anamitra Palit Registered Senior Member

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    102
    We deduce in the paper the following results
    v1.v2>=c^2
    [v1 and v2 are four velocities]
    p1.p2>=m1m2c^2
    [p1 and p2 are four velocities]

    https://drive.google.com/file/d/1yTw0x5uFs1zaT9bd1n6E7jgMvpA56k6W/view?usp=sharing

    By applying the reversed Cauchy Schwarz Inequality we may arrive directly at the same results
    Let's consider the reversed Cauchy Schwarz inequality.

    c^2 t1 t2-x1 x2 -y1 y2-z1 z2>=Sqrt[c^2t1^2-x1^2-y1^2-z1^2]Sqrt[c^2t2^2-x2^2-y2^2-z2^2]

    The equality sign holds when (t1,x1,y1,z1) and (t2,x2,y2,z2) are identical vectors

    Replacing x^i b y dx^i we obtain

    c^2 dt1 dt2-dx1dx2 -dy1 dy2-dz1 dz2>=Sqrt[c^2dt1^2-dx1^2-dy1^2-dz1^2]Sqrt[c^2dt2^2-dx2^2-dy2^2-dz2^2]
    A paper on the Reversed Cauchy Schwarz Inequality:
    https://drive.google.com/file/d/1z69d0OO4WRK6CthRln0s5LBJWbDS290f/view?usp=sharing

    Wikipedia Link on the Cauchy Schwarz Inequality:
    https://en.wikipedia.org/wiki/Minkowski_space#Norm_and_reversed_Cauchy_inequality

    Dividing both sides by dtau^2 we obtain

    Four dot product v1.v2>=c^2
    Multiplying both sides by m1m2[m1 and m2 being rest masses] we obtain,
    m1v1.mv2>=m1 m2c^2
    or,p1.p2>=m1m2 c^2
     
  17. James R Just this guy, you know? Staff Member

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    And so? What's the point?
     
  18. James R Just this guy, you know? Staff Member

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    Why don't you submit your papers to peer reviewed journals, for publication?
     
  19. Anamitra Palit Registered Senior Member

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    102
    The formulas v1.v2>=c^2,
    p1.p2 >m1m2c^2 are important just like v,v=c^2

    But alas..
    c^2dtau^2=c^2dt^2-dx^2-dy^2-dz^2

    c^2=c^2[dt/dtau]^2-[dx/dtau]^2-[dy/dtau]^2-[dz/dtau]^2

    c^2=c^2v_t^2-v_x^2-v_y^2-v_z^2 (1)

    v_i are proper speeds and as such they can exceed the speed of light without hurting or violating Special Relativity

    For two proper velocities v1 and v2at the same point of the manifold.Since tensors are additive we have

    c^2=c^2(v1_t+v2_t)^2-(v1_x+v2_x)^2-(v1_y+v2_y)^2-(v1_z+v2_z)^2(2)

    or,c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2+c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2

    or, c^2=c^2+c^2+2v1.v2

    v1.v2=-c^2 (3)

    By calculations we have arrived at an untenable result.

    An analogous result may be obtained in the General Relativity context
     
  20. Anamitra Palit Registered Senior Member

    Messages:
    102
    (in continuation)
    Indeed v1,v2 and v=v1+v2 all satisfy(1) and hence heir existence is certified by Special relativity or even by General Relativity for that matter.

    An analogous result may be obtained in the General Relativity context

    c^2dtau^2=c^2 g_tt dt^2-g_xx dx^2-g_yy dy^2-g_zz dz^2 (4)

    We consider transformations g_tt dt^2=dT^2, g_xxdx^2=dX^2, g_yydy^2=dY^2,g_zzdz^2=dZ^2 (5)

    Local or even transformations over infinitesimally small regions would suffice.

    Equations (4) and (5) combined gives us the flat space time metric[mathematical form of it]

    c^2dtau^2=c^2 dT^2- dX^2- dY^2- dZ^2 (6)

    All conclusions we made earlier follow.

    Incidentally, there is one point to take note of:the Lorentz transformations follow from (6) in a unique manner [Reference; Steve Wienberg,Gravitation and Cosmology,Chapter 2:Special Relativity]
     
  21. Anamitra Palit Registered Senior Member

    Messages:
    102
    In relation to the last post

    We may always choose the eight unknowns unknowns:v1_i and v2_j with each i and j=1,2,3,4 , in such a manner that the next three equations hold

    c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2 (1)

    c^2=c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2 (2)

    and c^2=c^2(v_1+v2_t)^2_-(v1_x+v2_x)^2-(v1_y+v2_y)^2-(v1_z+v2_z)^2 (3)
    Equations (1),(2) and (3) are all certified by the relation
    c^2=c^2v_t^2-v_x^2-v-y^2-v_z^2 which is equivalent to the Lorentz transformations
    as stated earlier.
    Equations (1),(2) and (3) are all certified by the relation
    c^2=c^2v_t^2-v_x^2-v-y^2-v_z^2 which is equivalent to the Lorentz transformations
    as stated earlier.
    The three equations finally lead will lead to 2v1 .v2<=-c^2
     
    Last edited: Jan 12, 2021
  22. exchemist Valued Senior Member

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    What is it you want to discuss?
     
  23. Michael 345 New year. PRESENT is 71 years old Valued Senior Member

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    The Inner Mind?

    I sense a link

    Please Register or Log in to view the hidden image!

     

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