I start with an attempt to communicate the issue of why gravity was thought perhaps to be a consequence of relativistic transformations.

Relativity is the mathematical transformation between two different geometric coordinate systems. Back in Newton's day, such transformations were quite straight forward as Euclidean coordinate systems were assumed applicable to reality. If the origin of a Euclidean coordinate system (coordinate system “b”) was at point (x

_{0},y

_{0},z

_{0}) in the original coordinate system (coordinate system “a”) then any point in coordinate system “a”, say point (x

_{a},y

_{a},z

_{a}) was simply represented by the point (x

_{a}-x

_{0},y

_{a}-y

_{0},z

_{a}-z

_{0}) in coordinate system “b”. This transformation was exactly the same even if the point being referred to as (x

_{0},y

_{0},z

_{0}) was moving in any arbitrary manner.

I would like to point out that, as opposed to Einstein's space time continuum, my paradigm results in a representation via a Euclidean coordinate system. The required transformation between any two different Euclidean coordinate systems are exactly what was introduced above. The difference between my system and Euclid's original system is that tau axis. Momentum quantization along that tau axis (mass) introduces some very subtle consequences when it comes to physical measurements such that the simple transformation above does not yield the measurements as taken by a person at rest in the moving system: i.e., special relativity is a necessary part of such a transformation in order to compensate for the effects of momentum quantization along the tau axis. Note that the transformation required by an observer in the original coordinate system to transform physical phenomena from one to the other has nothing to do with measurements made by the other observer; special relativity has to do with the apparent form of the transformed laws. Essentially, the “principal of relativity” as introduced by Galileo was the statement that the laws of physics were mathematically the same in any two coordinate systems so long as the only movement of one coordinate system relative to the other was at a constant velocity: i.e., as seen by the original rest observer, there should be no change in the laws of physics when transformed to another frame moving at a constant velocity. This is not quite the same as Einstein's paradigm; he sees the the laws themselves transforming, quite a different thing.

The form of the laws of physics was exactly the issue behind Newton's concept of an “inertial frame”. Newton's fundamental law, “things at rest tend to stay at rest and things in motion tend to stay in motion” is actually a special case of the more general (and simpler) inductive conclusion, “if something has been changing in a specific way up to now, it will most probably continue to change in the same manner in the future”. The value of my representation of that statement is that it actually has nothing to do with reality and everything to do with one's expectations. Newton further defined “a force” as the thing which changed that state and came up with the relationship [math]\vec{F}=m\vec{a}[/math] as a definition of how that rate of change changed. Acceleration is, after all, a change in the rate of change.

The reason I brought the above up is that I wanted to make it clear that Newton's [math]\vec{F}=m\vec{a}[/math] has to do with how things change and is not at all limited to things moving through space so to speak. The equation applies to geometric coordinate systems for the simple reason that coordinate systems are mental constructs created to display collections of information; as such, they are particularly valuable when it comes to displaying extremely large amounts of information. “A picture is worth a thousand words” is a rather extreme understatement of that fact. Essentially, Newton defined an “inertial frame” to be a coordinate system where [math]\vec{F}=m\vec{a}[/math] is a valid expression: i.e., the statement is “true” because “force” is actually defined by that very expression, so long as you are using an “inertial” frame of reference. You have to be very careful here because definitions such as that can tend to lead to circular reasoning if not handled properly.

That idea together with differential calculus created an extremely powerful mathematical method of predicting the dynamic behavior of objects (an object being any suitable stable defined collection of information). Since acceleration is the time derivative of velocity (where velocity is the time derivative of position in that “inertial frame”) force can be thought of as the time derivative of momentum (momentum being given by [math]m\vec{v}[/math]). If m is a constant, the expression [math]\vec{F}=\frac{d}{dt}(m \vec{v})[/math] is identical to [math]\vec{F}=m\vec{a}[/math] and if m is allowed to change, that fact simply allows a rather simple mechanism to handle cases where identical forces cause different accelerations. As a consequence, m ends up being little more than a parameter allowing a more versatile definition of force.

What made Newton's expression so powerful was the fact that, when it came to what people commonly call “physical objects” that more versatile definition of force was quite limited. That parameter “m” (which we call mass) could be assigned to a physical object and essentially remained constant forever. As I said these equations are defined to be true in an inertial frame (which was essentially the same collection of Galilean frames asserted in his “principal of relativity”). Furthermore, the Euclidean transformations brought up in the first paragraph above made it quite easy to transform Newton's mathematical solutions to any frame moving in any arbitrary manner desired. Thus it is that Newton seemed to have solved the entire field of dynamic behavior of physical objects.

That brings up the interesting question, “what does the dynamic behavior look if one is using a non-inertial frame”. Clearly, a non-inertial frame is a frame which is accelerating relative to an inertial frame. It should be clear to the reader that any object at rest in any inertial frame (which then, by definition, has no forces accelerating it) will appear to be accelerating if its position is represented by coordinates in a non-inertial frame representation. It should also be quite clear that it is not really accelerating at all, it is only the reference frame which is actually changing (accelerating). The apparent motion of any force free physical object who's position is being represented via the non-inertial frame will be an exact mirror image of that frames acceleration.

That fact brings up a very important issue; if one has two objects of different mass at rest in an inertial frame, their acceleration, as seen in the accelerating frame, will be exactly the same (that fact follows directly from the fact that the acceleration is nothing more than the mirror image of the frames acceleration). From the perspective of Newton's definition of force, [math]\vec{F}=m\vec{a}[/math], the apparent force on such objects (when seen from the perspective of an accelerating frame) must be exactly proportional to the mass of the object of interest. When I was a student, such “apparent” forces (those which seemed to exist only because one was working in a non-inertial frame), were called “pseudo forces”; a term, according to Qfwfq, which is apparently no longer in common use. The most common examples of such forces are centrifugal and Coriolis forces.

From the above (the fact that all “pseudo forces” are proportional to the mass of the object being accelerated by that force), the physics community began to consider any case where the force was proportional to the mass of the object to be a possible example of a “pseudo force”: i.e., such behavior suggested the reference frame being used was not a proper inertial frame. Since the force of gravity appeared to be absolutely and always directly proportional to the mass of the object being accelerated, quite a number of theoreticians concluded that gravity should be explainable as a fictional force due to the researcher's failure to use the proper inertial frame of reference. What many of them tried to discover was, what was the “proper inertial frame” one should be using: i.e., what sort of geometry would make gravity a fictitious force. This interest lead to much work in the field of mathematical transformations between strange and unusual geometric frames of reference (which turned out to be an interesting field all on its own).

No such proper inertial frame of reference was ever found and, sometime in the middle of the eighteenth century, Pierre Louis Maupertuis proved, mathematically, that no such frame of reference could possibly exist. His proof pretty well stopped any efforts to continue the search; however, the work of generating transformations between strange and unusual geometric frames had already evolved into a field of its own and much valuable work stands quite independent of any interest in gravitational effects.

Astonishing methods of solving a great many very difficult dynamic mechanical problems can be traced directly to the early interest in such transformations. Hamiltonian mechanics was originally an extension of Lagrangian mechanics which was essentially a completely general mechanism (based upon Maupertuis' concept of action) for transforming Newtonian mechanics to an absolutely general reference frame represented by any set of abstract coordinates (q

_{1},q

_{2}, ... q

_{n}). Hamilton extended Lagrange's work into what has come to be called Hamiltonian mechanics. And finally, analysis of Hamiltonian mechanics can be shown to be the underlying impetus which initially pushed physics towards modern quantum mechanics.

But that is all essentially beside the point here as our interest is in gravitational forces and the real impact of coordinate transformations upon that issue. It was exactly that issue which made Einstein's Relativity into an incontestable truth of the twentieth century. Einstein's general relativity yielded gravity as a consequence of geometric effects. Not exactly as a consequence of a geometric transformation (as were centrifugal and Coriolis forces) but rather through the idea that there was only one valid frame of reference under which correct physics could be generated (actually, in many ways, quite analogous to Newton's “inertial” frame). According to Adler, Bazin and Schiffer, (in their text book,

*Introduction to General Relativity*, McGraw-Hill Co., New York, 1965, p. 7.) Einstein

**proved**that "a reduction of gravitational theory to geodesic motion in an appropriate geometry could be carried out

**only**in the four-dimensional space-time continuum of [Einstein's] relativity theory". It was, in fact, this astonishing achievement which made Einstein's approach to relativity so overwhelming a part of accepted modern theory.

The question then arises, how is it that Einstein's theory appears to circumvent Maupertuis' proof? The answer revolves around the principal of “least action” he invented as a means of calculating paths consistent with Newtonian physics (essentially minimizing the energy with respect to the path). Maupertuis showed that the problem was a consequence of the fact that, when it came to gravitational paths,

**different velocities led to different paths**: i.e., two different objects behavior could not be reduced to geodesic motion in the same reference frame, something which must be true in the proper inertial frame. The inclusion of time in Einstein's space-time continuum allows this critical variation to be achieved; however, it turns out that this is exactly that same issue which creates the critical problems when it comes to “quantizing the gravitational field”. Thus it is that there are very real problems bringing quantum mechanics into Einstein's General Relativity theory. To date all attempts, that I am aware of, have resulted in failure.

Compare this to my presentation which is totally consistent with quantum mechanics from the very beginning. Beyond that, in my presentation, mass is defined to be momentum in the tau direction of a four dimensional Euclidean geometry. As a consequence, that hypothetical (see as unexaminable) tau dimension can be simply scaled to make the velocity of every elemental entity through that four dimensional geometry look exactly the same. That final fact totally circumvents Maupertuis' proof. It seems certainly reasonable to once again look at the consequences of general transformations and perhaps find that “non-inertial” geometry which yields gravity as a pseudo force. So let us proceed to examine some aspects of that circumstance.

To begin with, my fundamental equation does indeed require a specific frame of reference: that frame of reference being at rest with respect to the entire universe. In that particular frame, (remember, that frame is a standard Euclidean frame) any object (any collection of fundamental elements forming a stable pattern) which can be seen as essentially

**not interacting with the rest of the universe**(i.e., those interactions may be ignored and we are looking at a “free” object) will obey Newton's laws of motion in the absence of a force (it will not accelerate in any way). That is, the only forces which appear in such a circumstance are those pseudo forces we want to examine in this analysis: i.e., apparent forces which are entirely due to the fact that our coordinate system is not at rest with respect to the universe.

True, we have created a situation which obviously circumvents Maupertuis' proof but, since energy is now not a function of velocity, we have also made the original formation of his principal of action into an unusable procedure (his relationship related velocities and we now have no relationship to minimize); however, there is another attack (actually the same attack but somewhat subtly different). If gravity is to be a mere pseudo force, we can use the fact that our model must reproduce the exactly the same classical pseudo forces produced by Newton mechanics. This is true as these forces are no more than a direct consequence of expressing the path in a non-inertial frame or, in my case, a reference frame not at rest with respect to the universe.

It is interesting to look at centrifugal force as a well understood Newtonian pseudo force. From the perspective of an observer at the center of the rotation, a string exerting a force equal to the centrifugal force will appear to maintain the object under the influence of that pseudo force at rest: i.e., a rock at the end of a string swinging in a circle will appear to be at rest in a reference frame rotating with that rock. We can then see the object as a test probe into the force field describing that specific pseudo force. Since my total interest is in explaining gravity as a pseudo force, I want to examine this force as seen from the perspective of being m times the negative gradient of a gravitational potential (which is the typical way of defining a gravitational potential). In this case [math]\vec{F}=-m\vec{\nabla}\Phi [/math] where [math]\Phi(\vec{x})[/math] is the gravitational potential. Any freshman physics text will provide an excellent derivation of centrifugal force. The result is the quite simple form, [math]\;\vec{F}=mr\omega^2\hat{r}\;[/math] where omega is the angular velocity and [math]\hat{r}[/math] is a unit vector in the radial direction. It follows that the analogous gravitational representation implies that the required [math]\Phi(\vec{x})[/math] (which, by the way, must by symmetry be a radial function) must obey the relationship

[math]-\frac{\partial}{\partial r}\Phi(r)=r\omega^2[/math].

This fact directly implies that [math]-\Phi=\frac{1}{2}(r\omega)^2=\frac{1}{2}|\vec{v}|^2[/math] or, multiplying by [math]\frac{2}{c^2}[/math], one can conclude that

[math]\frac{2}{c^2}\Phi(r)=-\left(\frac{|\vec{v}|}{c}\right)^2[/math].

In other words, the gravitational potential (as seen in the frame where the object being observed appears to be at rest) seems to be directly related to the actual velocity as seen from the correct frame (the frame at rest with the universe). This result is very interesting. As the observed object is actually moving in the correct frame, we should expect a clock (or any temporal physical process moving with that object) to proceed in accordance with special relativity. This implies that the correct relativistic transformation of the instantaneous time differentials should be given by

[math]dt'=dt\sqrt{1-\left(\frac{|\vec{v}|}{c}\right)^2}\equiv dt\sqrt{1+\frac{2\Phi}{c^2}}[/math]

which happens to be exactly the standard gravitational red shift. This implies that any geometry which yields gravity as a pseudo force must also yield the standard gravitational red shift; or, alternately, gravitational red shift is not really a valid test of Einstein's general theory of relativity. This really isn't very enlightening as the gravitational red shift can be shown to be required by conservation of energy, but it does nonetheless imply that the above analysis is valid.

More importantly, the above suggests an attack towards determining the geometry which will yield gravity as a pseudo force in our four dimensional Euclidean geometry. I have already shown how static structures appear as three dimensional objects in this geometry so let us examine what is commonly called “a gravitational well”. The gravitational well consists of a vertical hole where there is a gravitational field in the vertical direction. If an experimenter in a gravitational well sets up a clock via a light pulse traveling back and forth between two horizontally displaced mirrors, since we can establish horizontal measure (since we are currently using a Euclidean geometry, simple vertical lines carry those measures to different heights in the hole) and his clock must run slow, we must see the apparent velocity of light to be

[math]c'=c\sqrt{1+\frac{2\Phi}{c^2}}[/math]

(at least for small gravitational effects). The gravitational red shift (which we are using here) is an instantaneous differential effect whereas the expression above concerns the horizontal path of the light in the clock. To see that problem, consider the same experiment done in an accelerating elevator as seen from a rest observer (at rest with the universe; still ignoring everything except the objects explicitly under examination). From the rest observers perspective, the direction of the light cannot be parallel to the floor of the elevator; if it were it would miss the mirror on the other side of the elevator. It must be already have a component of its velocity in the direction the elevator is accelerating.

Secondly, the faces of the mirrors cannot be perpendicular to the floor of the elevator; it they were, the light would miss the original mirror on the return (the component of the elevator vertical velocity has changed, the vertical component of the light has not). It must now have an additional component of its velocity in the direction the elevator is accelerating. As a result, both mirrors must lean slightly towards the outside such as to assure that the reflected light always picks up that required change in velocity to keep up with the elevator. A little thought about the situation should be enough to prove to the reader that, at the instant immediately prior to reflection (as seen by the rest observer), the path of the light must be parallel to the floor of the elevator and immediately after reflection the path must be towards the position the other mirror will be in when the light gets to the other side.

So, with reference to the floor of the elevator, the light will initially rise as it will be going up faster than the elevator. The acceleration of the elevator will eventually bring this apparent rise rate to zero and then let the elevator catch up so that the light can reflect off the opposite mirror. Thus it is that the observer in the elevator (who will regard the floor as level) will see the light following a curved line. The pseudo force on the light seems to cause the light to follow a curved path. We can again note that curvature of the light in a gravitational field is not a test of Einstein's theory, any theory which yields gravity as a pseudo force must also require light to curve in a gravitational field.

Thus it is that the clock we have proposed will run even slower than what was predicted a few paragraphs ago. The error is clearly a function of the physical size of the clock. This effect can be eliminated by defining our clock's size to be small enough to make the error negligible. I point this out so that we can discuss further ramifications of those instantaneous relativistic effects.

Since, in my four dimensional geometry, it has already been shown that clocks actually measure tau, the above also implies any lower object will appear to proceed in the tau direction at a slower rate proportional to exactly that same factor: i.e., the fundamental velocity of any object in my geometry will appear to be slower in a lower gravitational potential. This immediately suggests that we should be using Fermat's principle to establish the metric which will yield the proper geodesics: i.e., we should consider the phenomena of refraction.

As an aside, it is quite clear that the proper adjustment to our geometry which will yield the standard concept of gravity as a pseudo force is a change in the presumed measure of that geometry: i.e., instead of seeing the speed of light as slower in a gravitational field we could just as well see the speed as unchanged and the distances as increased. After all, once time is defined, distances are reckoned via the speed of light. Though that satisfies the original goal expressed above, the idea of refraction (the speed of light being slowed in a gravitational field) is a much simpler expression of the solution. It is certainly most convenient method of finding the proper geodesics. In fact, there is a very simple view of the situation which will yield exactly that result.

With regard to the issue of refraction, my fundamental equation is a wave equation with Dirac delta function interactions. Clearly, in the absence of interactions, the probability wave representing an event will proceed at a fixed velocity. Any specific delta function interaction can be seen as an impact changing the direction and energy of that probability wave. What is important here is the fact that interaction will depend upon the distance between the two elements connected by that delta function interaction: i.e., the hypothetical element (which must be a boson) must carry the momentum and energy being transfered and the transfer must be consistent with the Heisenberg uncertainty principal: i.e., the uncertainty in momentum is directly related to the uncertainty in position. This implies that the further apart the interacting fermions are, the less momentum transfered must be (see virtual particle exchange).

Any physical object (any structure stable enough to be thought of as an object) must have internal forces maintaining that structure. Any interaction with another distant object must be via the virtual particle exchange I just commented about. Thus it is that one would expect the fundamental element of that physical object interacting via that delta function would have its momentum altered, not the whole object; however, that alteration would create a discrepancy in the structure of the object under discussion. Since that object must have internal forces maintaining its structure, it is to be expected that those internal interactions (which are also mediated by delta function exchange forces) will bring the trajectory of that interacting fundamental element essentially back to its original path (at least on average).

Thus it is that the path of that fundamental element can be seen as crooked as compared to its path in the absence of that distant object. Of issue is the fact that, if the influence of the distant object is ignored, the influenced element will inexplicitly appear to be proceeding at a slightly slower velocity than it would if the distant object didn't exist. What is important here is that this effect decreases as the distance from the distant object increases. That means that the net effect is to yield a very slight change in the speed of the elements which make up that object as one moves across the object. The net effect of such an interaction is to refract the wave function of the object under examination.

If the distant object and the object under observation are not moving with respect to one another (they are moving parallel to one another in the tau direction), the net effect of that refraction is to curve the paths of the two objects towards one another: i.e., there will be an apparent attraction between them. It is also evident that, since the mass of the source object (the source of these bosons external to the object of interest) is proportional to the total momentum of that object, one should expect the apparent density (as seen from the object of interest) should be proportional to its mass: i.e., one should expect the exchange forces to be proportional to mass.

Clearly the interaction just discussed arises from differential effects in the basic interactions thus it will amount to a force considerably less than the underlying force standing behind that differential effect. Thus it is that the two forces I have already discussed (the forces due to massless boson exchange: shown to yield electromagnetic effects and the forces due to massive boson exchange: shown to yield fundamental nuclear forces) will end up being split into four forces. Differential effects will yield a correction to both basic forces which correspond quite well with the forces observed in nature. The differential effect on massless boson exchange yields what appears to be a very weak gravitational force (weak when compared to the underlying electromagnetic effects) and the differential effect on massive boson exchange yields what appears to be a very weak nuclear force (weak compared to the underlying nuclear force). What is interesting is that the “weak nuclear force” can be shown to violate parity symmetry whereas the “weak electrical force” (gravity) does not. This is a direct consequence of the fact that the nuclear exchange bosons are massive.

But let us get back to gravitational effects. It is my position that gravity is a direct consequence of the fact that the presence of an object with momentum in the tau direction yields a secondary differential effect which causes the speed of elements through my four dimensional Euclidean space to slow: i.e., refraction of the wave solution to my fundamental equation occurs. In order to check my proposition, I need to be able to work out the geodesic paths implied by such a notion. To begin with, my assertion is that the four dimensional velocity of all elements can be seen as slowed in the presence of a gravitational potential. Thus it is that the “index of refraction” of a gravitational field is given by the instantaneous value of [math]n=\frac{c}{c'}[/math] evaluated a specific point in that field (where c is the velocity of light far removed from any massive influence). Geodesic paths can be obtained by minimizing nds where “ds” is the distance differential along that geodesic path in my four dimensional Euclidean geometry.

In order to accomplish that result, we need to use what is called “the calculus of variations”. We want the variation of the path integral along the geodesic to vanish: i.e.,

[math]\delta \int^{P_2} _{P_1}nds = 0\quad\quad where \quad\quad n=\frac{1}{\sqrt{1+\frac{2}{c^2}\Phi(\vec{x})}}[/math].

Thankfully this is a problem already solved by the physics community. The vanishing of the path integral occurs when the function being integrated satisfies the Euler-Lagrange equation.

In order to simplify the problem, I would like to turn to a spherically symmetric case. Essentially a point source with a very large momentum in the tau direction and negligible momentum in the other three dimensions. This is in order to look at geodesic paths in the vicinity of a single very massive object where the other properties of the object can be ignored. In such a case, I can write [math]\Phi(\vec{x})[/math] in a very simple form consistent with standard notation:

[math]\Phi(\vec{x})=-\frac{\kappa M}{r}[/math]

where [math]\kappa[/math] is the proper proportionality constant to yield the correct potential generated by the mass M. This expression can be further simplified by changing to cylindrical coordinates (omit z as by symmetry we need only consider motion in the x-y plane). Thus it is that the differential of the path along which our metric is to be minimized is given by [math]ds=\sqrt{(d\tau)^2+(dr)^2+r^2(d\theta)^2}[/math]. It follows that the metric to be minimized to yield gravity consistent with a pseudo force is [math]dl=nds[/math] which can be written

[math]dl={\left[1-\frac{2\kappa M}{c^2r}\right]^{-\frac{1}{2}}}\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}[/math]

Using the standard Euler-Lagrange notation, the path integral over which the variation is to vanish (“J” in equation #1) is given by

[math] J =\int f(t,y,\dot{y})dt=\int^{P_2}_{P_1}\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt[/math]

Since “t” in my reference frame is actually a path length measure (essentially a reference as to where one is on the referenced path) I have changed the “ds” into “dt” thus converting the original expression given for ds, [math]\left[(d\tau)^2+(dr)^2+r^2(d\theta)^2\right]^{\frac{1}{2}}[/math], into [math]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}dt[/math]. Following the standard Euler-Lagrange attack, if the expression

[math]f(t,y,\dot{y})=\frac{1}{\sqrt{1-\frac{2\kappa M}{c^2r}}}\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/math]

satisfies the Euler-Lagrange equation, the variation of the integral over the implied path will vanish. In this case we have three independent variables to deal with (note that “y”, in this case, stands for the variable of interest) which leads to three independent differential equations. However, we have one additional constraint which simplifies the problem considerably: [math](\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2[/math]. The partials of f with respect to tau and theta vanish leading to the fact that the first integrals of the Euler-Lagrange equation for those variables are trivial. Substituting c for [math]\left[(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2\right]^{\frac{1}{2}}[/math], one has the following two first integrals:

[math] \int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\tau}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\frac{\dot{\tau}}{c}=\;[/math]

a constant [math]=\;\frac{1}{cl}[/math]

and

[math]\int \frac{d}{dt}\left(\frac{\partial}{\partial \dot{\theta}}\left\{\left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}\sqrt{(\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2}\right\}\right)dt = \left[ \frac{1}{1-\frac{2\kappa M}{c^2r}}\right]^{\frac{1}{2}}r^2\dot{\theta}\frac{1}{c}=[/math]

a constant [math]=\;\frac{h}{cl}[/math].

I have chosen to represent the two constants as I have, in terms of the constants

*l*and

*h*, because this notation will bring my solution into exactly the same form as the standard Schwarzschild solution to the Einsteinian field equations (with one very important difference).

From this solution we can presume that

[math]\dot{\tau}=\frac{1}{l}\sqrt{1-\frac{2\kappa M}{c^2r}}\quad\quad and \quad\quad \dot{\theta}=\frac{h}{r^2l}\sqrt{1-\frac{2\kappa M}{c^2r}}[/math]

Using [math](\dot{\tau})^2+(\dot{r})^2+r^2(\dot{\theta})^2=c^2[/math], we can write

[math]\frac{1}{l^2}\left(1-\frac{2\kappa M}{c^2r}\right)+(\dot{r})^2+\frac{h^2}{r^2l^2}\left(1-\frac{2\kappa M}{c^2r}\right)=c^2.[/math]

This differential equation (relating r and t) can be transformed to the actual differential equation of interest (that would be r versus [math]\theta[/math]) through the following well known replacement

Since [math]\quad r'=\frac{dr}{d\theta}=\frac{\dot{r}}{\dot{\theta}}, \quad \dot{r}=\dot{\theta}r'=\frac{h}{r^2l}r'\sqrt{1-\frac{2\kappa M}{c^2 r}},[/math]

and the equation for the geodesic can be directly written as

[math]\frac{1}{l^2}\left(1-\frac{2\kappa M}{c^2 r}\right)+\frac{h^2}{r^4l^2}(r')^2\left(1-\frac{2\kappa M}{c^2 r}\right)+\frac{h^2}{r^2l^2}\left(1-\frac{2\kappa M}{c^2 r}\right)= c^2[/math]

which can be rearranged to yield:

[math]\left(1-\frac{2\kappa M}{c^2 r}\right)=c^2l^2 -\frac{h^2}{r^4}(r')^2-\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)+\left[\frac{h^2}{r^4}(r')^2\left(\frac{2\kappa M}{c^2r}\right)\right],[/math]

which, except for the final term in square brackets, is precisely the Schwarzschild solution to Einstein's field equations for a spherically symmetric case (see Adler, Bazin and Schiffer,

*Introduction to General Relativity*, McGraw-Hill Co., New York, 1965, p. 180.).

I have certainly shown gravitational forces can be reduced to geodesics in my geometry by virtue of the fact that I have just done so. The fact that my result is not exactly the same as that obtained from Einstein's theory is not too troubling. It is possible that I have made a subtle deductive error in the above as none of my work has ever been checked by anyone competent to follow my reasoning; however, in the absence of an error, my result must be correct as it is deduced and not theorized as Einstein's solution is.

In order to comprehend exactly what the impact of that extra term is, I would like to compare the Schwarzschild solution to to the classical Newtonian result. Note that, if one differentiates the classical elliptical solution to the Newtonian problem, [math]\frac{1}{r}=\frac{1}{r_0}(1-\epsilon \;cos (\theta))[/math], and chooses the appropriate constants; the Newtonian solution can be written:

[math]\left(1-\frac{2\kappa M}{c^2 r}\right)= c^2l^2-\frac{h^2}{r^4}(r')^2-\frac{h^2}{r^2}.[/math]

It may then be seen that the Schwarzschild solution amounts to a small adjustment to the energy contribution of the angular momentum term of Newton's solution (the most important consequence being the experimentally verified precession of the orbit of Mercury).

[math]-\frac{h^2}{r^2}\quad\Rightarrow\quad -\frac{h^2}{r^2}\left(1-\frac{2\kappa M}{c^2r} \right).[/math]

In the same vein, my "error" (if indeed it is an error) amounts to an equally small adjustment to the energy contribution of the radial momentum term. It should be noted that both Einstein's change to Newton's solution and my change to Einstein's solution are not only small but they also both contain an additional factor of one over r relative to their respective base terms. Whereas Einsteins theory makes no changes whatsoever in the radial term of Newton's equation,

[math]-\frac{h^2}{r^4}(r')^2\quad\Rightarrow\quad -\frac{h^2}{r^4}(r')^2,[/math]

my solution makes a very small change to that term:

[math]-\frac{h^2}{r^4}(r')^2\quad\Rightarrow\quad -\frac{h^2}{r^4}(r')^2\left(1-\frac{2\kappa M}{c^2r} \right).[/math]

Since the energy attributable to the radial momentum of Mercury is very small compared to the energy attributable to its orbital angular momentum, the perhelic shift of Mercury is certainly not a test of the validity of Einstein's theory as, within the accuracy of experiment, I obtain exactly the same result. The only test which could possibly remain is the deflection of star light. In that regard, please notice that in the above deduction, I selected my constants

*l*and

*h*so as to reproduce the exact form of Schwarzschild's solution. By doing so I forced my two constants of first integration to be related.

Under normal circumstances, one would simply write the first integrals as equal to an arbitrary constant: i.e., one would ordinarily write,

[math]\frac{\partial f}{\partial \dot{\tau}}= \left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}\frac{\dot{\tau}}{c}=\frac{A_1}{c} \quad and \quad \frac{\partial f}{\partial \dot{\theta}}= \left(1-\frac{2\kappa M}{c^2r}\right)^{-\frac{1}{2}}r^2\frac{\dot{\theta}}{c}=\frac{A_2}{c}[/math]

In this case, one would obtain a differential equation for the geodesic of the form

[math]A^2_1\left(1-\frac{2\kappa M}{c^2 r}\right)=c^2 -\frac{A^2_2}{r^4}(r')^2-\frac{A^2_2}{r^2}\left(1-\frac{2\kappa M}{c^2r}\right)+\left[\frac{A^2_2}{r^4}(r')^2\left(\frac{2\kappa M}{c^2r}\right)\right].[/math]

In this representation, the geodesic for a photon is immediately obtained by setting A

_{1}equal to zero (which corresponds to no initial motion in the tau direction: i.e., the photon has zero rest mass). Again, except for the term in square brackets, the result is exactly the same result Schwarzschild obtains by setting his invariant interval to zero and resolving the problem. Here also, the term in square brackets can be seen as an energy adjustment related to that part of the photons energy which can be seen as due to it's radial motion. Clearly the deflection of star light by the sun does not test the existence of my additional term as, in grazing the sun, the impact of this term is negligible as the radial motion of the photon is only comparable to the angular motion for large r.

With this final result, I have clearly demonstrated that Einstein's assertion of Minkowski space-time is in no way necessary to explain the validity of either the Lorentz transformations or the so called tests of general relativity. I think it is very possible that Einstein would have given me some serious thought had he lived to see what I have done.

I find it interesting that Einstein himself had doubts concerning field theory. In particular with regard to Qfwfq's insistence that the correct way of doing quantum mechanics was to quantize the fields: i.e., he considered field theory to be more fundamental than quantum mechanics whereas my position was quite the reverse. (Qfwfq never responded to that post.)after failing to unify the electro and gravity field, Einstein wrote ...

All these fifty years of conscious brooding have brought me no nearer to the answer to the question, 'What are light quanta?' Nowadays every Tom, Dick and Harry thinks he knows it, but he is mistaken. … I consider it quite possible that physics cannot be based on the field concept, i.e., on continuous structures. In that case, nothing remains of my entire castle in the air, gravitation theory included, [and of] the rest of modern physics. (Albert Einstein, 1954)

In addition to the above, there might be some evidence that Einstein's theory is actually wrong. Note that, if my work is correct (and I certainly admit I could be wrong) the difference between my solution and Einstein's solution would be that simple factor multiplying the [math](r')^2[/math] term

[math]\left(1-\frac{2\kappa M}{c^2r}\right)[/math].

If you solve the Schwarzschild solution to Einstein's field equations for “dr” and then divide by “dt” you will obtain

[math]\frac{dr}{dt}= \frac{r^2}{h}\left[c^2l^2-\left(1-\frac{2\kappa M}{c^2 r}\right)\left(1+\frac{h^2}{r^2}\right)\right]^{\frac{1}{2}}\frac {d\theta}{dt}[/math]

Add to this the fact that conservation of angular momentum requires [math]v_\theta = r\frac{d\theta}{dt}[/math] = a constant or [math]\frac{d\theta}{dt}=\frac{C}{r}[/math] and we have,

[math]\frac{dr}{dt}= \frac{r}{h}\left[c^2l^2-\left(1-\frac{2\kappa M}{c^2 r}\right)\left(1+\frac{h^2}{r^2}\right)\right]^{\frac{1}{2}}C=F(r).[/math]

If you use exactly the same procedure to solve my equation for [math]\frac{dr}{dt}\sqrt{1-\frac{2\kappa M}{c^2 r}}[/math] you will obtain exactly the same result: i.e.,

[math]\frac{dr}{dt}\sqrt{1-\frac{2\kappa M}{c^2 r}}=F(r),[/math]

where F( r ) is exactly the same function obtained from Schwarzschild's solution. This implies that the radial velocity obtained from my solution multiplied by [math]\sqrt{1-\frac{2\kappa M}{c^2 r}}[/math] must be exactly Schwarzschild's solution. That, in turn, implies that the radial velocity for a given radius will be greater in my case than it was in Schwarzchild's solution (or Newton's solution for that matter); however, if the object is far far away, the accurate measurement here is the radial velocity itself, not the exact value of r. The radial velocity is determined via the Doppler effect and, with an accurate clock on board the object, its velocity can be measured quite accurately.

The radius on the other hand would be estimated by solving for the orbital motion of the object (using either Einstein's field theory or Newtonian gravitational theory). Since the velocity is slowing as an object leaves the solar system, if my equation is the correct calculation, Schwarzschild would presume the object was somewhat further away than I would (I would have obtained that same value at a somewhat closer radius). If the object were actually slightly closer to the sun than they thought, its deceleration would be slightly greater. That is apparently what the measurements on that recent satellite leaving the solar system have yielded.

It appears (at least to me at the moment) that the effect of that extra term in my solution is to make the gravitational field appear to be slightly stronger than estimated via Einstein's field theory. If that conclusion is correct, then it could also explain the “dark matter” problem.

My final comment is, once again, that my presentation is

**not a theory**but a purely deductive conclusion from required symmetries and thus stands on much firmer ground than any theory including Einstein's. Of course, I could have made a mistake in my deductions but, if no one ever checks that deduction, how will anyone ever know? Am I really a crackpot? Or have I gone where no one else has tread?

Anssi, if you find any errors, I of course put them there to test you.

Have fun -- Dick