Okay, let's get to it...

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.e. even if the coordinate system b was moving inside the coordinate system a in an arbitrary manner.

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.

Since you mention momentum

*quantization* along the tau axis, I stopped to think about that a bit. I am not sure why was it quantized instead of being a continuous variable. I suppose it had got something to do with the uncertainty in the position of tau being infinite (since the momentum is constant/known?), but my understanding is quite shaky when it comes to the details of how the uncertaintly principle plays out here.

Also, I suspect this issue related to the use of dirac constant in the definition of mass operator [imath]-i\frac{\hbar}{c}\frac{\partial}{\partial \tau}[/imath]...?

Nevertheless, yes I remember how special relativistic transformation came into play.

.

.

.

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.

Yup, thank you about that whole explanation. It might be interesting to take a quick look at that Maupertuis' proof but I did not find it... Nevertheless, I have pretty good suspicion about how the inclusion of time axis and its relativistic transformation can get around proofs that rely on newtonian definition of time.

I don't have very good idea about the problems with the traditional attempts of gravity quantization. Wikipedia only vaguely mentions problems with re-normalization.

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.

i.e. whatever velocity they are "missing" in the [imath]x,y,z[/imath]-axes, is to be attributed to their velocity along the [imath]\tau[/imath]-axis...

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...

Hmmm, well not very obvious to me, as I don't know how his proof worked exactly...

I'm guessing what you are referring to is that, since Maupertuis' proof had something to do with the inability to "reduce two different objects' behaviour to geodesic motion in the same reference frame", this presentation form with a [imath]\tau[/imath]-axis (that gets "projected out"), will allow two different objects with different velocities in [imath]x,y,z[/imath] space to have the same total velocities when the velocity along [imath]\tau[/imath] is included.

Not really sure yet how that will circumvent the proof.

I understand though, that in the standard GR presentation form, the objects in gravitational free fall are following geodesic paths in the general relativistic coordinate system.

but, since energy is now not a function of velocity,

...i.e. not a function of the total velocity in the [imath]x,y,z,\tau[/imath]-space...? Are you saying that because, for instance, an object is considered to gain (kinetic) energy when it gains velocity in the [imath]x,y,z[/imath] directions, while its total velocity remains unchanged?

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);

...since velocities never change in the [imath]x,y,z,\tau[/imath]-space...?

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.

Yup.

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 [imath]\vec{F}=-m\vec{\nabla}\Phi [/imath] where [imath]\Phi(\vec{x})[/imath] is the gravitational potential.

Had a little adventure in Wikipedia about the definition of gravitational potential etc, and yes that seems to make perfect sense to me.

I gathered that the "gradient of a gravitational potential" is essentially the radial derivative of the function describing gravitational potential... I.e. the "slope" of a graph describing that function.

Also I gathered that the definition of gravitational potential is the "potential energy" PER unit mass, so that's why you are including the "m" there in your equation (I was getting a bit confused at first).

So in other words, [imath]\vec{F}=-m\vec{\nabla}\Phi [/imath] refers to the strength of the "fictional force" that affects an object with given mass in a given location in a gravitational field. (Just thought I'll say that out loud to benefit other people who are starting with almsot zero physics knowledge, like me

)

Any freshman physics text will provide an excellent derivation of centrifugal force. The result is the quite simple form, [imath]\;\vec{F}=m\omega^2\hat{r}\;[/imath] where omega is the angular velocity and [imath]\hat{r}[/imath] is a unit vector in the radial direction.

I'll take that on faith for now.

It follows that the analogous gravitational representation implies that the required [imath]\Phi(\vec{x})[/imath] (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].

I.e. the radial derivative, or the "slope" of the gravitational potential, is related to the angular velocity somehow?

Hmmm, you must be referring to a situation where some object is orbiting the center of the gravitational field, so then its angular velocity and the distance from the center are related to the strength of the associated "fictional force", which is keeping it in stable orbit.

Intuitively, that makes sense, but I am not entirely sure how you got that exact expression (specifically, the "r" on the right side).

I mean, I understand that [imath]\vec{\nabla}\Phi = \frac{\partial}{\partial r}\Phi(r)[/imath] since its a radial function, and I guess we are essentially associating gravitational force with a centrifugal force capable of negating it...?

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

...I guess...

?

And you have removed the "m" from both sides, but I am not sure how the "r" finds its way to the right side the way it did. I mean, I understand the associated fictional force is a function of the radius, but how do we know the [imath]\omega^2[/imath] is simply multiplied by the radius?

I'll pause here as I am getting the feeling I may well be interpreting you wrong already...

-Anssi