A Limited Toy Model for Classical Gravity

Start the year off with a new toy model for gravity. It is a toy, so let's just play with it and see what we learn. Feel free in the comments section to vote if I should keep this toy or return it for a refund.

At a minimum, the toy must do three things. First, it must explain why I weigh a little over 150 pounds here on Earth. Second, it must explain why light bends in both time and space when flying by the edge of the Sun. Third, it needs to explain why I don't weigh over 300 pounds here on Earth, feeling only half the effects that light apparently does.

[Point of clarification: this is also the maximum accomplishment of this toy model. In particular, critical issues such as gauge symmetry and the spin of the mediating force will not be addressed. The toy is limited to the classical force domain.]

The toy model is a variation on EM, one I will call the double O. Henry because two of the moves are forced and the ending is a surprise. There are three terms that make up the action: the matter, the interaction, and the field actions.

$S_{O.H.2}=S_{matter}+S_{interaction}+S_{fields}$

There is nothing to play with for the matter term, so it stays the same as it ever was:

$S_{matter}= \int{ dt \;(-\frac{m}{\gamma})} \approx \int{dt \; (-m + \frac{1}{2} m v^2)}$

If the interaction term were kept identical to what appears in EM, then the force equation would be identical, and the physics would be identical. To make the physics different, I am forced to change a little, choosing the sign of the dot product term:

$S_{interaction}= \int{ dt \;(-q \phi - q (V \cdot A) })$

The sign on the field term must flip relative to the first interaction term to go from Gauss's law in EM to Newton's law of gravity at least in the static case. Here is the double O. Henry action [corrected sign of the matter term]:

$S_{O.H.2}= \int{ dt \left( (-m + \frac{1}{2} m v^2) - q \phi - q (V \cdot A) + \int{ \int{ \int{dx \, dy\, dz\;(B^2 - E^2) }}} \right) }$

The action is set. Everything flows from here.

Really this is just EM with a few sign flips. Count the number of sign changes, and then, based on EM, the rest pops out.

For the field equations, the first field equation has one sign flip, but the other three have two changes, so no net change. Write the equations in the Lorenz gauge:

$\begin{align*} \rho &= -\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} + \nabla^2 \phi \\ J_i &= \frac{1}{c^2}\frac{\partial^2 A_i}{\partial t^2} - \nabla^2 A_i \end{align*}$

The solutions of interest for the Earth and the Sun are static, the same year after year. Use the old reliable charge over distance:

$\\ \phi = 1 - \frac{G M}{c^2 R} \\ \\ A_i = 1 + \frac{G M}{c^2 R}$

I have a few fetishes, like using dimensionless quantities, making sure all potential energy densities are positive, and having solutions that look similar.

The Lorenz force equation gets a sign flip for anything with an A:

$\vec{F} = m c^2 \left(- \vec{\nabla} \phi + \frac{1}{c} \frac{\partial \vec{A}}{\partial t} - \frac{\vec{v}}{c} \times (\vec{\nabla} \times \vec{A}) \right)$

When I jump on the scale, the only term that matters is the gradient of phi [clarification: I removed the Rhats as they just added an unnecessary layer of complexity]:

$\begin{align*} \vec{F} &= - m c^2 \vec{\nabla} \phi \\ &= -m c^2 \nabla \left(1 - \frac{G M}{c^2 R} \right) \\ &= - \frac{G M m}{R^2} \end{align*}$

Good old Newton's law of gravity. My scale will do its thing.

Now consider a photon flying by the Sun. The field is static, which is too bad, it looks like we cannot use our solutions for A. The time derivative of A is zero, case closed. Next.

Actually, we better find a way to use our solutions for A, they are solutions to the field equations after all! Remember the chain rule:

$\frac{\partial A_i}{\partial t} = \frac{\partial x_i}{\partial t} \frac{\partial A_i}{\partial x_i}$

The force equation for a static field should be rewritten like so:

$\vec{F} = m c^2 \left(- \vec{\nabla} \phi + \frac{\vec{v}}{c} \nabla A - \frac{\vec{v}}{c} \times (\vec{\nabla} \times \vec{A}) \right)$

Now plug in the solutions for phi and A into the first two terms of the force:

$\begin{align*} \vec{F} &= m c^2 (-\vec{\nabla} \phi + \frac{\vec{v}}{c} \nabla A) \\ \frac{d (mv)}{dt}&= m c^2 \left( \nabla \left( - 1 + \frac{G M}{c^2 R} \right) + \frac{c}{c}\nabla \left(1 + \frac{G M}{c^2 R} \right) \right)\\ m \frac{d v}{d t} + v \frac{d m}{d t} &= - 2 \frac{G M m}{R^2} \end{align*}$

There are two equal contributions, one from the usual gradient of phi, the other from the chain rule applied to the time derivative of A. There are a number of issues I feel insecure about. First is that the small m is the stand-in for light whose mass is zero. I should probably take a limit as m approaches zero, but how to handle the rocket term (dm/dt) is not clear.

[After more reflection, I decided any references to an Rhat or vhat vector were not justified by the calculation, so have been removed.]

This explains why I don't weigh 300 pounds on the scale. My velocity relative to the Earth is zero, so the v/c term is zero. I am 100% the gradient of phi.

On Christmas day, I had fun with the family, but I was also thinking about these solutions for phi and A, exactly as written here. I thought it would be cool if I could find the solutions under Schwarzschild's tree. I fired up Mathematica long enough to ask the following two questions: for the Schwarzschild metric written in isotropic coordinates, what is the Taylor series expansion for the square root of the g₀₀ term, and likewise for the square root of the g_RR term?

$\begin{align*} \sqrt{g_{00}} &= \frac{1-\frac{G M}{2 c^2 R}}{1+\frac{G M}{2 c^2 R}} \approx 1 - \frac{G M}{c^2 R} \\ \\ \sqrt{g_{RR}} &= (1+\frac{G M}{2 c^2 R})^2 \approx 1 + \frac{G M}{c^2 R} \end{align*}$

Bingo, bingo, a nice gift to myself.

Doug

Snarky puzzle:
Find a new constant velocity solution to the double O. Henry gravitational force equation. If you presume the same chain rule game, show the analytic solution has a 1/R - not 1/R² - dependence. Do not share your answer with anyone looking for dark matter. Do share with people building models of stars in motion in galaxies. Don't expect them to be happy as much new programming will be required to test it.

MIT IAP classes, "Animating New Physics," Jan. 18-20, 3-5pm if a member of the MIT community and in Cambridge, MA. Otherwise, will webcast via Google+, http://gplus.to/sweetser

Next Monday/Tuesday: First Snarky Puzzle Answers of 2012

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