# Detailing why a scalar gravity theory predicts no bending of light [closed]

I want to understand in technical detail why a particular scalar theory for gravity predicts no bending of light. It is left as a question, either in "Gravitation" by Misner, Thorne, and Wheeler, question 7.1, parts A and E, or in a paper by Thirring (Ann. Phys., 16:96-117, 1961). I am not in graduate school for physics (and so this is not a homework problem for me, just self-study). I am getting stuck at the end of 1.A so cannot proceed to 1.E. Here is the problem

Exercise 7.1. Scalar Gravitational Field, $\Phi$

A. Consider the variation principle $\delta I = 0$, where $$I = - m \int e^{\phi} \sqrt{ -\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \lambda}\frac{d z^{\beta}}{d \lambda} }~d \lambda,$$ Here $m = (\rm{rest~mass})$ and $\lambda = (\rm{parametrized~world~line})$ for a test particle in the scalar gravitational field $\Phi$. By varying the particle's world line, derive differential equations governing the particle's motion. Write them using the particle's proper time as the path parameter, $$d \tau = \sqrt{-\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \lambda}\frac{d z^{\beta}}{d \lambda}}~d \lambda,$$ so that $u^{\alpha}=d z^{\alpha}/d \tau$ satisfies $u^{\alpha}u^{\beta}\eta_{\alpha \beta}=-1$.

The scalar gravity field $e^{\phi}$ is along for the ride in part A. It is simple enough to take 4 derivates of the action with respect to the 4 parametrized velocities $d z^{\alpha}/d \lambda$. I remain confused as to how to "[Re?]write them using the particle's proper time as the path parameter".

For completeness, I will include part E which will be my next hurdle after A is understood.

[B, C, and D involve the field equations and perihelion precision.]

E. Pass to the limit of a zero rest-mass particle in the equations of motion of part A. Do this by using a parameter $\lambda$ different from proper time, so chosen that $k^{\mu} = d x^{\mu}/d \lambda$ is the energy-momentum vector, and by taking the limit $m \rightarrow 0$ with $k^0 = \gamma m = E$ remaining finite (so $u^0 = \gamma \rightarrow \infty$). Use these equations to show that the quantities $q^{\mu} = k^{\mu} e^{\Phi}$ are constants of motion, and from this deduce that there is no bending of light by the Sun in this scalar theory.

1. Start with the action: $$I = - m \int e^{\phi} \sqrt{ -\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \lambda}\frac{d z^{\beta}}{d \lambda} }~d \lambda$$
2. Replace $\lambda$ with $\tau$ $$I = - m \int e^{\phi} \sqrt{ -\eta_{\alpha \beta}\frac{d z^{\alpha}}{d \tau}\frac{d z^{\beta}}{d \tau} }~d \tau$$
3. Write the Lagrangian as: $$\mathcal{L} = - m e^{\Phi} \sqrt{-u^{\mu} u_{\mu}}$$
4. Use Euler-Lagrange to calculate the equations of motion: $$0 = m e^{\Phi} \frac{d}{d \tau} \frac{u_{\mu}}{\sqrt{-u^{\nu} u_{\nu}}}$$ Because we are working with a worldline, the contraction of $u^{\mu}$ will create the ratio of an interval over an interval, good old unity. To dodge the imatinary bullet, it must be negative unity. I am not clear about the two minus signs in the action, but I will not lose sleep over it.

I did a survey of answers to this question on the Internet. While it is good to read many of them, my favorite was "Gravitational Bending of Light" by Don Edwards. The problem is described using Huygen's principle for deflection. One looks at the ratio of changes in space over those in time. Since those are the same, there is no deflection. That is not the way MTW requested the answer, but it is less abstract to me and more grounded in physics.

## closed as unclear what you're asking by tpg2114♦, user10851, Emilio Pisanty, Dan, Waffle's Crazy PeanutNov 8 '13 at 3:17

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• There is no question... – jinawee Nov 7 '13 at 18:06
• I don't work with the calculus of variations much at all. The question is problem 7.1 (Derive differential equations ect.) – sweetser Nov 7 '13 at 18:45
• Goldstein's "Classical Mechanics" has useful discussions of variational principles (Chapter 2 in the 2nd ed.) and their application to relativistic physics (Sections 7-8 and 7-9). Section 7-9 in particular gives you a feel for the calculations involved in the MTW exercise. One drawback: Goldstein 2 still uses the antiquated imaginary "time" component (ict) in four-vectors. I don't know about the 3rd ed. – Art Brown Nov 10 '13 at 7:32
• Thanks for the specific reference @art-brown. I just ordered a used version of Goldstein from Amazon. The ict thing will be trivial to deal with. In MTW right above this question, they say "(b) Become familiar with the results of the other exercise (7.2 or 7.1) by discussing it with someone who has worked it in detail." A goal was to have such a discussion here. – sweetser Nov 11 '13 at 12:53
• Again, my reopen vote is taken out of the queue. – Abhimanyu Pallavi Sudhir Nov 27 '13 at 16:21

Start by varying the action in the case where $\phi=0$, i.e., no gravity, and learn how to derive the equations of motion for a particle of mass $m$. You should be in a better position to approach this problem then.