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In The Quaternion Group and Modern Physics by P.R. Girard, the quaternion form of the general relativistic equation of motion is derived from

$du'/ds = (d a / d s ) u {a_c}^* + a u ( d {a_c}^* / d s ) + a (du/ds ) {a_c}^*$

as

$du/ds = - [ ( a_c \frac{da}{ds})u - u ( a_c \frac{da}{ds})^*]$.

where $u$ is a "minquat" of the form $(ct,ix,iy,iz)$ and a is an arbitrary function satisfying $a{a_c}=1$

This is said to "correspond" to the equation of motion of GR. It is an incredibly elegant formulation of the equation.

I've tried applying this formula to $a=\cos(s)-i\sin(s) \hat{i}$ and, unless I've made some errors-and I probably did, this results in $x(1+\cos(2s))+ ict(1+\sin(2s))$ for the first two terms.

1.What is the physical significance of these terms?

2.Does this really represent the GR equations of motion?

*3. Does the set of arbitrary functions satisfying $a{a_c}=1$ represent a general relativity symmetry group?

4.What other functions satisfying the condition $a{a_c}=1$ should I use? (I've also used $e^{i\theta}$ and this switched $x$ and $ct$ which is interesting, but-again-what does it actually represent?)

5.Can I use this equation to derive any standard results? I would like to see the centrifugal and other fictitious forces come out of it maybe.

I've found that the Runge Lenz vector is used to calculate orbits in Newtonian Gravity and GR, but I'm baffled by the introduction of this vector in this case.

Can it be translated into a standard formulation of the equations of motion?

Any information or advice for tackling this would be appreciated.

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I just learned about the quaternion group from your question. Thank you! I skimmed through the paper, sounds intriguing (why no body taught us this before?) I'm printing the paper and will study it now. Sorry if my comment is irrelevant to your question. –  stupidity Jun 8 '12 at 12:42
    
It is fascinating. This is the best presentation on the relationship between quaternions and physics I've found. (Other than the section on spinors by Wheeler in Gravitation.) The equation I've mentioned is so elegant, it is hard to believe it captures the equations of motion of GR. –  MadScientist Jun 8 '12 at 14:46
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1 Answer 1

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This equation is the geodesic equation in the quaternionic coordinates, where $u$ is the four velocity vector.

I'll give you a derivation of this equation in the more familiar local Lorentz formalism, and then establish the equivalence.

Given a frame of vielbeins $e_a = e^{\mu}_a \frac{\partial}{\partial x^{\mu}}$. The geodesic equation is simply the statement that the four velocity $ v = v^a e_a$is an invariant:

$dv= 0 $

Writing this equation in components, we obtain:

$ dv^a + \omega^a_b v^b = 0 $

where: $e^b$ are the inverse vielbeins and $\omega^a_b = \omega^a_{\mu b} dx^{\mu}$ is the spin connection one form defined by:

$ de_b = \omega^b_c e_c $

The geodesic equation can be written in the usual coordinate representation by substituting:

$v^a = v^{\mu} e_{\mu}^a$

and using the torsion free condition:

$\partial_{ \mu} e_{\nu}^a + \omega_{\mu b}^a e_{\nu}^a - \Gamma_{\mu \nu}^{\lambda} e_{\lambda}^a = 0$

Now, having established that the invarianve of the four velocity vector gives the geodesic equation, there is still an other representation of the same equation along the following principle:

There exists a local Lorentz transformation (may be singular), which transforms to the particle rest frame i.e., there exists

$v\prime^a = \Lambda^a_b(x) v^a$

such that

$ dv\prime^a = 0$ (componentwise, this equation just means that the components of the velocity in the rest frame are constants).

Using the above two equations, we see that locally:

$\omega^a_b = d\Lambda_a^c \Lambda_c^b$

Substuituting this equation in the first version of the geodesic equation, we obtain:

$ dv^a + d\Lambda_a^c \Lambda_c^b v^b = 0 $

This equation is just the vectorial version of the geodesic equation from the article.

The right hand side is composed of a finite local Lorentz transformation followed by an infinitesimal one as in the quaternionic representation.

In summary, this form of the geodesic equation is based on the following two principles:

  1. In general relativity, thefour velocity is an invariant.

  2. There exists a local Lorentz frame in which all velocity components are constants.

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Thank You. I need to review frame vielbiens to really understand this, but it looks accurate. –  MadScientist Jun 18 '12 at 13:30
    
After studying this, I'll attempt to apply this equation directly to some traditional calculations. This should be very interesting. –  MadScientist Jun 18 '12 at 13:42
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Please notice that the term containing the derivative of the local Lorentz transformation (in both vectorial and quaternionic representations) is only seemingly linear in the velocity components . The derivative of the local Lorentz transformation itself is linear in the velocity components, thus this term is quadratic in the velocity as in the standard form of the geodesic equation. –  David Bar Moshe Jun 18 '12 at 14:19
    
@Moshe Thanks. That is why the equation appears incorrect. –  MadScientist Jun 19 '12 at 14:31
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