A list of eleven steps to analyzing an action will be presented. I could have added one more, but then people might think I was giving up on my physics addiction. The method will be applied to Newton's theory of gravity in this blog. The Maxwell action will be up next. Only if something acceptable comes through the set of tools I construct will I do the third blog that would make the Newtonian action more like Maxwell.

Once an action is written down, the rest is calculation. To provide structure to the discussion, I have compiled eleven steps to look at. One could add more or use less. There is nothing sacrosanct about this eleven step implementation for analysis of actions. With these caveats in mind, these are the eleven steps.

1. Define the action: matter, interactions, fields
2. The static form
3. The action under rotations
4. The action under boosts
5. Gauge transformations of fields
6. Derive the force equations
7. Derive the field equations
8. Solutions to the field equations
9. Solutions to the force equations
10. The experimental data
11. Spin

The structure will help pinpoint issues with Newton's theory of gravity. The steps will show why Maxwell's work still shines. Should there be a new proposal, any areas of concern can be clearly marked.

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I will be doing a blogging communications experiment using the above tag. As a continuous text, it may appear that I always have the same level of confidence. That is not the case. These tags can serve to point out areas I am insecure about, such as spin. We will have to see how this goes.
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Step G1: Define the action: matter, interactions, fields
Note: Assume quaternion products throughout.

\begin{align*} S_{Newton ~G} &= S_{matter} + S_{interaction} + S_{fields} \\ & \approx \int{ dt \left( \sum_{particles} \left( -\frac{\sqrt{((E, \vec{P})^2+ ((E, \vec{P})^2)^*)/2} }{\gamma} - m \phi \right) + \int{ \int{ \int{ d x ~ d y ~ d z~ \left(-\rho \phi - \frac{1}{2}(\nabla \phi)^2 \right) } } } \right) } \\ & \approx \int{ dt \left( \sum_{particles} \left( \frac{1}{2} \frac{m v^2}{c^2} - m \phi \right) + \int{ \int{ \int{ d x ~ d y ~ d z~ \left( - \rho \phi - \frac{1}{2}(\nabla \phi)^2 \right) } } } \right) } \end{align*}

[corrected. All the terms evaluate to a mass m. The mass density must be integrated over a volume to yield a mass. Less obvious is the square of the derivative of the potential must have the same units. One can say that as the number of particles becomes infinite, the volume integration of the mass density is a way to write the same thing. The reason I write both is because I don't know how to tell Mathematica how to do this. It is much simpler to write sum m and integrate over a volume rho.]

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The middle line has an unusual look, which is why I tagged this discussion. I recall in a comment David Halliday saying the matter term was some sort of square root of a contraction. When I thought about how could I demonstrate the mass m is invariant under a boost in Mathematica, I realized I needed to write the mass in this longer way. While writing the simple mass out this long way is odd, it is more precise.
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Step G2: Newtonian gravity's static form - same as the above

The fact that Newtonian gravity is so successful with the static case underlies its fatal flaw: any changes must propagate instantly which is in violation of special relativity.

Step G3: The action under 3D rotations - invariant

The matter and interaction terms are scalars, so there is no spatial arm to twist. The field strength, Del phi, will rotate, but its dot product will be unchanged by a spatial rotation.

Step G4: The action under boosts - [Correction: not] invariant

To understand the matter term, one needs to know the following about gamma:

$\gamma = \frac{\partial t}{\partial \tau}$

Gamma is a measure of how time changes with respect to the invariant interval. Take a closer look at the matter action:

\begin{align*} S_{matter}&=\int{ dt \left(\sum_{particles} \left( -\frac{\sqrt{((E, \vec{P})^2+ ((E, \vec{P})^2)^*)/2}}{\gamma} \right) \right)} \\ &=\int{dt \left(\sum_{particles} -m \frac{\partial \tau}{\partial t} \right)}\\ &=\int{d \tau \left(\sum_{particles} -m \right)} \end{align*}
[Clarification. In the force derivation, this is not the matter term used. The matter term used is 1/2 mv2/c2
which is not invariant under boosts.]

The interaction action is invariant because both terms are invariant under a boost.

The field would be called a "pure quaternion", the sort Maxwell also used, that have a zero in the first slot. The zero plus the other three values transform like a 4-vector. A boost would move the pure quaternion to a new position, moving it along an invariant hyperbola. The square results in a Lorentz invariant for the field action.

[Correction: The Laplacian takes derivatives with respect to space only, and thus is manifestly not covariant. A switch to the 4-vector is a fundamental change of the action, and would not constitute an accurate representation of Newton's theory of gravity. An illegal math operation led me astray. I was well aware that Newton's theory was not invariant under a Lorentz transformation, and therefore the action should also share that property. Now where the action breaks Lorentz symmetry jumps off the page - in only one place, the field term.]

Step G5: Gauge transformations of fields

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[Clarifications: There is a global gauge symmetry. A constant field k can be added to the potential:

$\\ \phi \rightarrow \phi' = \phi + k \\ \\ \vec{\nabla} \phi \rightarrow \vec{\nabla} \phi'=\vec{\nabla} \phi +\vec{\nabla} k =\vec{\nabla} \phi$

There does not appear to be a local gauge transformation, one that depends on x, y, and z. Both general relativity and EM have local gauge symmetries. It might be the instantaneous propagation of changes bar forming a gauge theory (gauge as in measure), but I am unsure of that point.

For completeness, I have keep my initial idea which is hardly worth your time.

I was unable to find any online references in the Newtonian gravity context.
I don't think this line of reasoning is right, but it is all I was able to come up with. Use a gauge field that is the gradient of a harmonic function, $\vec{\nabla} f$, a fancy way to say it is a solution to LaPlace's equation:

$\nabla \cdot \nabla f = 0$

The gradient of a harmonic function could be added to phi without altering the result:

$\inline \\ \phi \rightarrow \phi' = \phi + \vec{\nabla f} \\ \\ \vec{\nabla} \phi \rightarrow \vec{\nabla} \phi' = \vec{\nabla} \phi + \nabal \cdot \nabla f = \vec{\nabla} \phi$

This looks incorrect to me because phi is invariant under a boost as discussed in step G4, but the gauge field would be altered by a boost. An example of a harmonic function is a charge/distance for a point source, essentially the answer to the field equations themselves.
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Step G6: Derive the force equations

The matter and interaction actions are the key.

$\inline S_G \approx \int{ dt \left( \sum_{particles} \left( \frac{1}{2} m \frac{v^2}{c^2} - m \phi \right) \right) }$

Vary this with respect to position and velocity to generate the force equation:

$\frac{d ~ m\vec{v}}{d t} = \vec{F} = -m c^2\vec{\nabla} \phi$

Do like charges attract or repel? It is too early to say. Solutions to the field equations must be found, then plugged in.

Step G7: Derive the field equations

The interaction and field actions are the vital players here:

$S_G = \int{ dt \left( \sum_{particles} \left( - m \phi \right) - (\nabla \phi)^2 \right)}$

This time the potentials and changes in the potential are varied. If instead of a collection of point masses, one has a continuous mass distribution, then the sum over particles would become an integral. Apply Euler-Lagrange to generate the field equations:

$\rho = \nabla^2 \phi \quad eq.~ G7$

Step G8: Find solutions to the field equations

A first guess at a solution might be charge over distance. That does solve the homogeneous problem where there is no mass density. To figure out the sign of the charge, one asks about the charge near the origin:

$\phi = \nabla^2 \frac{\rho}{R^2 + a^2}$
[clarification: this is one of many ways of investigating a point singularity. One looks into this solution as "a" goes to zero. This is done in Jackson's Classical Electrodynamics, section 1.7 Poisson and Laplace equations.]

The Laplacian of this near phi is negative. Oops. Many authors put a minus sign in front of the potential and move on. This minus sign creates it own, more subtle issue about how to avoid negative energy potentials. The solution is to include a huge positive constant which is effectively the rest energy of the test particle (see the article "Why Maxwell Couldn't Explain Gravity"  for a more detailed discussion):

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

The constants make the charge/distance value tiny, so this will be a positive potential energy. We don't detect phi, only changes in phi, so the constant drops out.

Step G9: Solutions into the force equation

[clarification: the action has units of mass, so the c2 comes from the dimensionless velocity. Bring that over to the other side to take the standard form];

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

Like charges attract for gravity, as expected. One unusual thing about the implementation details as done here is that the gravitational constant G comes in from the dimensionless potential. Nice.

Step 10: Solutions and experimental data

Any experiment with light and gravity gets half the answer right, half wrong, so gets marked as wrong in the grade book of science. What is right is how time changes. What cannot be accounted for is how space changes due to gravity. Newton's theory has one scalar potential to do one job right. Since gravity works on both time and space, Newtonian gravity is destine to come up short.

Step 11: Spin
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While I feel I can say some relevant things about spin and EM by looking at the action, I don't feel I can do that about Newtonian gravity theory. Another nail in the coffin is not needed, the static nature and disagreements with experiment.
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As a form of penance, and to aid the next blogs, I created a Mathematica notebook in support of the work written out here [update: a new version is up as of Friday, Nov. 18]. Two things of note from that process. The first was I was not skilled enough in the Mathematica arts to write one functional for the EulerEquations[] both the force and field equations. The former depends on position and velocity, while the latter depend on the field and its derivatives. Getting the syntax exactly right is the kind of thing to pay their tech support folks to do (but I dropped that service a while ago, too expensive).

There is a factor of 3 that confuses me in the field equation solutions. If I look at the solution for rho/(R2 + a2)1/2, the contributions from x, y, and z add up for a factor of 3. I see the factor of 3 in Jackson, second edition, section 1.7, Poisson and Laplace Equations. Dirac delta functions are tricky beasts.

[Additional observation. I was not able to use the terms with the integral directly to get the field equations by applying Euler-Lagrange. The reason was amusing. Instead of the triple integral of second derivatives, I got double integrals of first derivatives. That was the fundamental rule of calculus at work.]

Doug

Snarky puzzle: Take Newton's gravitational theory and make it consistent with special relativity, nothing more. Oops, that is a totally unfair question. You have to be as good as  Gupta, Thirring, Feynman, Weinberg, or Deser to take that one technical requirement and end up at general relativity.

Google+ hangout: 11:00-11:45pm Eastern time, Tuesday-Friday. http://gplus.to/sweetser

This could be an efficient way to exchange a few ideas. If you have a question or two, hangout.

Next Monday/Tuesday: Analyzing Actions: EM (2/3?)