- #1

JOliver

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Okay so :

1. Homework Statement

1. Homework Statement

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 :

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

Where

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

## Homework 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)

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

## 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.

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

Recalling

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

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