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

• JOliver
In summary, the conversation discusses a question about Nordstrom's theory of gravity and its metric, which involves a scalar field and geodesic reparameterizations. The question asks to show that light rays in this theory follow the same lines as in flat space, while massive particles are deflected. The conversation also touches on the concept of geodesic parameterization and the need to restate the question in order to receive a response. The idea of Nordstrom's theory being a process rather than an object is also suggested.
JOliver
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. 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,

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.

Hi JOliver,

Cheers.

1. How does Nordstrom's theory of gravity explain the deflection of light rays?

Nordstrom's theory posits that gravity is not a result of the curvature of space-time, as described by Einstein's General Theory of Relativity, but rather a modification of the law of inertia. In this theory, the motion of objects is affected by the inertia of the surrounding space. This leads to the deflection of light rays as they pass through regions of varying inertia, similar to how a ray of light would bend when passing through a medium with varying refractive index.

2. How does Nordstrom's theory differ from Einstein's theory of General Relativity?

Nordstrom's theory differs from Einstein's in two major ways. Firstly, it proposes a modification of the law of inertia, while General Relativity maintains the classical notion of inertia. Secondly, Nordstrom's theory does not consider the curvature of space-time as the source of gravity, whereas General Relativity does.

3. Can Nordstrom's theory explain the bending of light in strong gravitational fields?

Yes, Nordstrom's theory does account for the bending of light in strong gravitational fields, such as those near massive objects like stars and galaxies. This is because the theory still considers the effects of inertia on the motion of objects, and in strong gravitational fields, inertia plays a significant role in determining the path of light rays.

4. Are there any experimental observations that support Nordstrom's theory of gravity?

No, there are currently no experimental observations that support Nordstrom's theory of gravity. In fact, the theory has largely been abandoned in favor of Einstein's General Relativity due to its inability to fully explain various phenomena, such as the precession of Mercury's orbit and the bending of starlight near the Sun.

5. What are the potential implications of Nordstrom's theory of gravity if it were proven to be true?

If Nordstrom's theory were proven to be true, it would revolutionize our understanding of gravity and the universe as a whole. It would require a major paradigm shift in the field of physics, and would likely lead to a better understanding of the relationship between gravity and other fundamental forces, such as electromagnetism and the strong and weak nuclear forces.

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