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Light Ray deflection in Nordstrom's theory of gravity...

  1. Sep 25, 2015 #1
    Hi, firstly I wanted to say thankyou for this forum, its provided me with a lot of help over the years and now it has come to that dreaded question that can not seem to poke at at all. I'm in my 4th year and taking General Relativity for the first time, with a hap-dash differential geometry "intro" in the middle of it. Whilst I have been working at GR to get the intuition down-packed, I must admit I am lacking the physical interpretation of many of these equations.

    Okay so :

    1. The problem statement, all variables and given/known data

    Nordstrom's theory of gravity has a metric

    $$\tilde{g}_{\mu \nu} = \exp( -2 \Phi / c^{2}) \eta_{\mu \nu}$$
    where ##\eta_{\mu \nu}## is the Minkowski metric and ##\Phi## is a scalar field. By considering geodesic reparameterisations, show that light rays in Nordstrom's theory follow the same lines in the flat space, i.e they are not deflected by gravity, whereas massive particles are deflected.

    In the previous section of the question we are asked to prove the corresponding Christoffel symbols ## \Gamma^{\lambda}_{\mu \nu}## and ##\tilde{\Gamma}^{\lambda}_{\mu \nu}## between ##g_{\mu \nu}## and ##\tilde{g}_{\mu \nu}##. Where

    $$ \tilde{g}_{\mu \nu} = \Omega^{2} g_{\mu \nu}$$

    now, clearly for Nordstrom's theory of gravity ## \Omega = \exp(- \Phi / c^{2})##

    The relation I arrived at was :

    $$ \tilde{\Gamma}^{\alpha}_{\mu \beta} = \Gamma^{\alpha}_{\mu \beta} + C^{\alpha}_{\mu \beta}$$


    $$ C^{\alpha}_{\mu \beta} = \frac{1}{\Omega} \left[ \partial_{\mu}(\Omega) \delta^{\alpha}_{\beta} + \partial_{\beta}(\Omega) \delta^{\alpha}_{\mu} - \partial_{\lambda}(\Omega) g^{\alpha \lambda} g_{\mu \beta} \right] $$

    2. Relevant equations
    So, the question asks us, that via Geodesic reparameterisations, show that the light rays in Nordstrom's theory follow the same lines as in flat space.

    One looks to the geodesic equation (perhaps niavely)

    $$ \frac{d^{2} x^{\alpha}}{d \lambda^{2}} + \tilde{\Gamma}^{\alpha}_{\mu\beta} \frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\mu}}{\partial \lambda}=0$$

    3. The attempt at a solution
    The Previous sections have had aspects of my attempt in them, though I will dot point my solution up until I lose the scent.

    1) State metric

    $$ \tilde{g}_{\mu \nu} = \Omega^{2} \eta_{\mu \nu}$$

    2) The connection

    $$ \tilde{\Gamma}^{\alpha}_{\mu \beta} = \Gamma^{\alpha}_{\mu \beta} + C^{\alpha}_{\mu \beta}$$

    The connection ##\Gamma^{\alpha}_{\mu \beta}## for flat space time, is zero. Thus we neglect it.

    3) The scalar field for Nordstroms theory

    $$\Omega = \exp(- \Phi / c^{2}) $$

    4) The geodesics Equation

    $$ \frac{d^{2} x^{\alpha}}{d \lambda^{2}} + \tilde{\Gamma}^{\alpha}_{\mu\beta} \frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\mu}}{\partial \lambda} =0$$

    $$ \frac{d^{2} x^{\alpha}}{d \lambda^{2}}+\left( \Gamma^{\alpha}_{\mu \beta}+C^{\alpha}_{\mu \beta}\right) \frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\mu}}{\partial \lambda} =0$$

    5) Simplification

    $$ \frac{d^{2} x^{\alpha}}{d \lambda^{2}}+C^{\alpha}_{\mu \beta}\frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\mu}}{\partial \lambda} =0$$

    6) Detirmination of ## C^{\alpha}_{\mu \beta} ## for Nordstrom's metric.

    $$ C^{\alpha}_{\mu \beta} = \frac{1}{\Omega} \left[ \partial_{\mu}(\Omega) \delta^{\alpha}_{\beta} + \partial_{\beta}(\Omega) \delta^{\alpha}_{\mu} - \partial_{\lambda}(\Omega) g^{\alpha \lambda} g_{\mu \beta} \right] $$


    $$ \Omega = \exp(- \Phi / c^{2}) $$

    we find ## C^{\alpha}_{\mu \beta}## to be :

    $$C^{\alpha}_{\mu \beta} = \left[ - \frac{\partial_{\mu} \Phi }{c^{2}} \delta^{\alpha}_{\beta} - \frac{\partial_{\beta} \Phi}{c^{2}} \delta^{\alpha}_{\mu} + \frac{\partial_{\lambda} \Phi}{c^{2}} g^{\alpha \lambda} g_{\mu \beta} \right] $$

    7) Substitution into the second term in the geodesic equation :

    $$ C^{\alpha}_{\mu \beta} \frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\beta}}{\partial \lambda} = - -\frac{ \partial_{\mu} \Phi}{c^{2}} \frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\alpha}}{\partial \lambda} - \frac{ \partial_{\beta} \Phi}{c^{2}} \frac{\partial x^{\alpha}}{\partial \lambda} \frac{\partial x^{\beta}}{\partial \lambda} + \frac{\partial_{\lambda}\Phi}{c^{2}} \frac{\partial x_{\beta}}{\partial \lambda} \frac{\partial x^{\beta}}{\partial \lambda} g^{\alpha \lambda} $$

    we then group terms and use ## \frac{\partial x^{\mu}}{\partial \lambda} = p^{\mu} ##.

    $$ C^{\alpha}_{\mu \beta} \frac{\partial x^{\mu}}{\partial \lambda} \frac{\partial x^{\beta}}{\partial \lambda} = -\frac{2 \partial_{\mu} \Phi}{c^{2}} p^{\mu} p^{\alpha} + \frac{\partial_{\lambda}\Phi}{c^{2}} p^{2} g^{\alpha \lambda} $$

    Giving the Geodesic equation :

    $$ \frac{\partial^{2} x^{\alpha}}{\partial \lambda^{2}} -\frac{2 \partial_{\mu} \Phi}{c^{2}} p^{\mu} p^{\alpha} + \frac{\partial_{\lambda}\Phi}{c^{2}} p^{2} g^{\alpha \lambda} = 0 $$

    8) ## p^{2} = m^{2}## For the light ray :

    $$ \frac{\partial^{2} x^{\alpha}}{\partial \lambda^{2}} -\frac{2 \partial_{\mu} \Phi}{c^{2}} p^{\mu} p^{\alpha} + \underbrace{\frac{\partial_{\lambda}\Phi}{c^{2}} p^{2} g^{\alpha \lambda}}_{=0} = 0 $$

    $$ \frac{\partial p^{\alpha}}{\partial \lambda} -\frac{2 \partial_{\mu} \Phi}{c^{2}} p^{\mu} p^{\alpha} = 0 $$
    Now at this point the concept of geodesic parameterization comes in :

    I know:

    $$\frac{d^{2} x^{\mu}}{d \alpha^{2}} + \Gamma^{\mu}_{\rho \sigma} \frac{d x^{\rho}}{d \alpha} \frac{d x^{\sigma}}{d \alpha} = f(\alpha) \frac{d x^{\mu}}{d \alpha} $$

    for some parameter ##\alpha(\lambda)##, where ##f(\lambda)## is related to the affine parameter via :

    $$ f(\alpha) = - \left( \frac{d^{2} \alpha}{d \lambda^{2}}\right) \left( \frac{d \alpha}{d \lambda}\right)^{-2}$$

    I had the thought that perhaps

    $$-\frac{2 \partial_{\mu} \Phi}{c^{2}} p^{\mu} = f(\alpha) $$.

    but then I'm not entirely sure what this gets me.

    So to summarise this question. Given that I have arrived at the point I have, what steps must I take to

    1) use geodesic parameterization and

    2) show that the deflection of the light ray is zero. (as in, how will i know that the deflection is zero).

    I feel like I am almost there - I must be missing something obvious.

    I hope I have been clear in my description of the question and working out,

    Thanks for your guidance
  2. jcsd
  3. Sep 25, 2015 #2
    Hi J, I note the lack of response to your issue. There seems to be a lot of editorializing in your question and wonder if you could restate the question so as to allow a response. For example you force a respondent to accept the use of geodesic parameterization, meaning that you should be able to identify an effective degree of freedom of your geo "object" Perhaps you should consider that Nordstrom's Theory of Gravity is not a 'object' at all but is akin to a 'process' and approach your work from that vantage point. Parameterization would still work on a process or system as well as upon an object.
  4. Oct 4, 2015 #3
    Hi JOliver,

    I can't answer your question, but this link might help.

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