Introduction
In my previous writings (Hirano [1,2]), I have repeatedly stated my conjecture that the space-time assumed by the theory of relativity is a conical coordinate system:
\[ \small x^2+y^2+z^2+c^2T^2 = c^2t^2, \]
and that space-time is a space with time removed, i.e., space is a three-dimensional sphere:
\[ \small \hat{x}^2+\hat{y}^2+\hat{z}^2+\hat{s}^2 = 1. \]
However, although some speculations are described about what the equations of motion and conserved quantities in this space-time might be, I feel that the discussion is somewhat lacking in clarity. Therefore, I would like to summarize the discussion here once again and point out the issues.
Velocity and Momentum in the Theory of Relativity
To simplify the discussion, let us consider motion without potential (acceleration). The author’s argument was that in the theory of relativity, time is two-dimensional, and a different time \(\small T_{xyzt}\) exists at each coordinate, while in classical mechanics, time is recognized as local time \(\small T_{xyzt}\) defined for each coordinate, rather than a common time \(\small t\). Therefore, in order to deal with motion in the theory of relativity, the equations of motion, which were functions of \(\small T_{xyzt}\), need to be redefined as functions of \(\small t\).
To do this, consider a point mass (particle) moving in a conical coordinate system. Unlike classical mechanics, this point mass is subject to the constraint that it must remain within the conical coordinate system regardless of any movement it undergoes. To identify the equations of motion, we differentiate the coordinates in the conic coordinate system with respect to \(\small t\) to get
\[ \small x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}+c^2T\frac{dT}{dt} = c^2t. \]
Differentiating again gives us
\[ \small \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2+c^2\left(\frac{dT}{dt}\right)^2\qquad\qquad \\ \small \qquad\qquad+x\frac{d^2x}{dt^2}+y\frac{d^2y}{dt^2}+z\frac{d^2z}{dt^2}+c^2T\frac{d^2T}{dt^2} = c^2. \]
Since all the second derivatives are 0 by assumption,
\[ \small \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2+c^2\left(\frac{dT}{dt}\right)^2 = c^2 \]
can be obtained. These two equations are the equations that a particle moving in a conical coordinate system must satisfy.
Since the right-hand side is a constant, if there is a conserved quantity \(\small E\) for this motion, the equality will still hold even if we multiply both sides by that conserved quantity. At this time, the symbols will be replaced according to the following rules:
\[ \small p_\mu = \frac{E}{c^2}\frac{dq_\mu}{dt}, \quad \mu = x,y,z,T. \]
Here, \(\small q_\mu\) are the coordinate and \(\small (q_x,q_y,q_z,q_T) = (x,y,z,T)\). Multiplying both sides by \(\small E^2/c^2\) and substituting the equation gives us
\[ \small E^2 = (p_x^2+p_y^2+p_z^2+p_T^2c^2)c^2. \]
Inferring from the energy formula in the special theory of relativity, \(\small p_T=m\), we can see that the physical quantity known as mass is the momentum along the \(\small T\)-axis in a conical coordinate system. In addition, because the concept of time is two-dimensional, we can define two concepts for mass as well, and since the momentum of local time \(\small T\) is \(\small p_T\), we can also consider momentum \(\small p_t=M\) with respect to common time \(\small t\). By substituting these symbols into the above formula, it can be expressed as:
\[ \small E^2 = M^2c^4 = m^2c^4+\left(p_x^2+p_y^2+p_z^2 \right)c^2. \]
\(\small p_t=M\) corresponds to the physical quantity called invariant mass (relativistic mass).
I believe that when dealing with phenomena related to mass, such as gravity, the distinction between \(\small m\) and \(\small M\) must be strictly maintained. Depending on the book, there are different descriptions on whether the effect of gravity is greater or not depending on the speed (momentum)…If the assumption that the change in momentum over time (approximately the acceleration) is unrelated to the magnitude of one’s mass (equivalence principle) is correct, then gravitational mass is \(\small M\), and while the motion due to gravity itself is not affected by the magnitude of the speed (momentum), the magnitude of the energy changes depending on the magnitude of the speed (momentum). However, if we follow this line of thinking, even physical objects that have no mass \(\small m\), such as light, would have gravity (the textbook general theory of relativity seems to take this position).
If we consider that this does not match actual observations, then gravitational mass must be \(\small m\), but in this case the claim that the change in momentum over time from gravity is unrelated to the size of one’s own mass (equivalence principle) would be incorrect. Another hypothesis might be that the mass of the source of gravity that affects other matter is \(\small m\), but the mass of the matter affected by the gravity is \(\small M\). This justifies all three points: the equivalence principle is correct, light has no gravity, and light is bent by gravity. If we consider time to be two-dimensional, then mass is also two-dimensional in the general theory of relativity, and it may be necessary to use two masses appropriately. Well, to be honest, at this point I don’t have enough knowledge or conviction to determine which is the correct answer.
From the definition of momentum, the relationship between velocity and momentum is:
\[ \small E = Mc^2 = \sqrt{m^2c^4+p^2c^2}, \quad p^2 = p_x^2+p_y^2+p_z^2, \]
therefore,
\[ \small \begin{align*} &\frac{dx}{dt} = \frac{p_x}{M} = \frac{p_xc^2}{\sqrt{m^2c^4+p^2c^2}} \\ &\frac{dy}{dt} = \frac{p_y}{M} = \frac{p_yc^2}{\sqrt{m^2c^4+p^2c^2}} \\ &\frac{dz}{dt} = \frac{p_z}{M} = \frac{p_zc^2}{\sqrt{m^2c^4+p^2c^2}} \\ &\frac{dT}{dt} = \frac{m}{M} = \frac{mc^2}{\sqrt{m^2c^4+p^2c^2}} \end{align*} \]
can be obtained. This formula was used to calculate free fall in the theory of relativity, and was derived in this way. It is easy to see that velocity in classical mechanics can be calculated as:
\[ \small \frac{dx}{dT} = \frac{dx}{dt} \left(\frac{dT}{dt} \right)^{-1} = \frac{p_x}{m}. \]
Finally, we note that the properties of the conserved quantity \(\small E\) differ from those in classical mechanics. Assume that the momentum and energy of \(\small n\) particles are given by
\[ \small E_i^2 = M_i^2c^4 = m_i^2c^4+\left(p_{x_i}^2+p_{y_i}^2+p_{z_i}^2 \right)c^2, \quad i=1,\cdots,n \]
and that there is no interaction. In this case, how should we tally up the conserved quantity (energy) of this physical system? In classical mechanics it was
\[ \small E = \sum_{i=1}^n E_i = \sum_{i=1}^n\sqrt{m_i^2c^4+p_i^2c^2}-m_ic^2 \approx \sum_{i=1}^n\frac{p_i^2}{2m_i}, \]
but in the theory of relativity it would have to be:
\[ \small E^2 = \sum_{i=1}^n E_i^2 = \sum_{i=1}^n m_i^2c^4+p_i^2c^2. \]
If we consider the square root of \(\small E^2\) to be a conserved quantity, then it must be considered to be:
\[ \small E = \sqrt{\sum_{i=1}^nm_i^2c^4+p_i^2c^2}. \]
This is a value that cannot be approximated at all by the energy of classical mechanics. In the theory of relativity, these calculations are expressed as mass defects in relativistic mass, which means that mass cannot be calculated by adding it up. For example, when dealing with a system of two point masses with relativistic masses \(\small M_1,M_2\), it cannot be expressed as \(\small M = M_1+M_2\) and the mass is reduced in the form:
\[ \small M = \sqrt{M_1^2+M_2^2}\leq M_1+M_2. \]
In quantum mechanics, using the Dirac matrices are four-dimensional matrices:
\[ \small \gamma_{i,j}^2 = 1, \quad \gamma_{i,j}\gamma_{i,k} = 0,,\quad \gamma_{i,j}\gamma_{h,k} = 0, \quad i,h=1, \cdots,n,\;j,k=0,1,2,3 \]
that becomes 1 when squared, but becomes 0 when multiplied with different subscripts, the energy is represented as:
\[ \small \frac{E}{c} = \sum_{i=1}^n \gamma_{i,0}m_ic+\gamma_{i,1}p_{x_i}+\gamma_{i,2}p_{y_i}+\gamma_{i,3}p_{z_i}.\]
It can be understood that this is a formulation in which squaring both sides results in
\[ \small E^2 = \sum_{i=1}^n m_i^2c^4+p_i^2c^2. \]
In commonly accepted knowledge, classical mechanical energy and relativistic energy are explained as being approximately the same, but in reality they must be considered to be completely different things.
Time Derivatives of Acceleration and Momentum in the Theory of Relativity
From the discussion in the previous section, the relationship between velocity and momentum in the theory of relativity is given by
\[ \small \frac{dq_\mu}{dt} = \frac{p_\mu}{M} = \frac{p_\mu c^2}{\sqrt{m^2c^4+p^2c^2}}. \]
If we define the velocity in this way, then the velocity of a particle cannot exceed the speed of light, regardless of the magnitude of its momentum. Similarly, let us find the relationship between acceleration and the time derivative of momentum by differentiating the above equation with respect to time again. When actually calculated,
\[ \small \begin{align*} \frac{d^2q_\mu}{dt^2} &=\frac{c^2}{\sqrt{m^2c^4+p^2c^2}}\frac{dp_\mu}{dt} \\ & -\frac{p_\mu c^2}{(m^2c^4+p^2c^2)^{3/2}}\left[p_x\frac{dp_x}{dt}+p_y\frac{dp_y}{dt}+p_z\frac{dp_z}{dt}+mc^2\frac{dm}{dt} \right] \end{align*} \]
holds true. This equation is the relationship between acceleration and the time derivative of momentum in the theory of relativity.
An interesting example of this relationship is the Coulomb potential in electromagnetism, where the potential function is given by
\[ \small V = -\frac{e^2}{r}, \quad r = \sqrt{x^2+y^2+z^2}. \]
Assuming that the mass remains constant, let the time derivative of momentum be
\[ \small \frac{dp_\mu}{dt} = -\frac{\partial V}{\partial q_\mu} = -\frac{e^2}{r^3}q_\mu, \quad \frac{dm}{dt} = 0, \]
and by substituting this equation into the above equation, we obtain
\[ \small M\frac{d^2q_\mu}{dt^2}=-\frac{e^2}{r^3}q_\mu+\frac{p_\mu c^2}{m^2c^4+p^2c^2}\frac{e^2}{r^3}\left[p_xq_x+p_yq_y+p_zq_z \right]. \]
By rearranging the equation, we get
\[ \small \begin{align*} M\frac{d^2q_\mu}{dt^2}=&-\frac{e^2}{r^3}\frac{c^2}{m^2c^4+p^2c^2}\\ & \times \left[q_\mu m^2c^2+q_\mu (p_x^2+p_y^2+p_z^2)-p_\mu(p_xq_x+p_yq_y+p_zq_z) \right]. \end{align*} \]
Using the relationship between velocity and momentum, we convert momentum to velocity to get
\[ \small \begin{align*} M\frac{d^2q_\mu}{dt^2}=&-\frac{e^2}{r^3}\frac{q_\mu m^2c^4}{m^2c^4+p^2c^2} \\ &+\frac{e^2}{r^3c^2}\Biggl[q_\mu \left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right) \\ &\qquad\quad-\frac{dq_\mu}{dt}\left(q_x\frac{dq_x}{dt}+q_y\frac{dq_y}{dt}+q_z\frac{dq_z}{dt}\right) \Biggr]. \end{align*} \]
By approximating \(\small M\approx m\) and
\[ \small \frac{m^2c^4}{m^2c^4+p^2c^2} \approx 1, \]
we can obtained
\[ \small \begin{align*} m\frac{d^2q_\mu}{dt^2}=&-\frac{e^2}{r^3}q_\mu \\ &+\frac{e^2}{r^3c^2}\Biggl[q_\mu \left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right) \\ &\qquad \quad-\frac{dq_\mu}{dt}\left(q_x\frac{dq_x}{dt}+q_y\frac{dq_y}{dt}+q_z\frac{dq_z}{dt}\right) \Biggr]. \end{align*} \]
This equation corresponds to the Lorentz force equation and can be expressed as:
\[ \small \begin{align*} &m\frac{d^2q_\mu}{dt^2} = q(\mathcal{E}+v\times \mathcal{B}) \\ &\mathcal{E} = -\frac{e^2}{r^3} \\ &\mathcal{B} = \frac{1}{c^2} (v\times q\mathcal{E}). \end{align*} \]
Each term, \(\small \mathcal{E},\mathcal{B}\), corresponds to the electric field and magnetic field in electromagnetism. The conclusion that can be drawn from this is that the electric field and the magnetic field are both Coulomb potentials, and that the direct effect of this is recognized as the electric field and the relativistic effect as the magnetic field. This was Einstein’s idea of the electromagnetic field derived from his special theory of relativity. The formula derived above corresponds to such a calculation.
Next, let us find the equation for the constraint:
\[ \small \left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2+c^2\left(\frac{dT}{dt}\right)^2\qquad\qquad \\ \small \qquad\qquad+x\frac{d^2x}{dt^2}+y\frac{d^2y}{dt^2}+z\frac{d^2z}{dt^2}+c^2T\frac{d^2T}{dt^2} = c^2 \]
that the equation of motion must satisfy in the conical coordinate system by multiplying both sides by \(\small E^2/c^2\). Since the squared part of the velocity was as in the previous section, let us find the value of
\[ \small EV = \frac{E^2}{c^2}\left(x\frac{d^2x}{dt^2}+y\frac{d^2y}{dt^2}+z\frac{d^2z}{dt^2}+c^2T\frac{d^2T}{dt^2}\right). \]
In this case, \(\small E^2=M^2c^4 \neq m^2c^4+p^2c^2\), so the momentum can be expressed as:
\[ \small \begin{align*} &\frac{dq_\mu}{dt} = p_\mu\frac{ c^2}{E} \\ &\frac{d^2q_\mu}{dt^2}=\frac{c^2}{E}\frac{dp_\mu}{dt}-\frac{p_\mu c^2}{E^2}\frac{dE}{dt} \end{align*} \]
when simply expressed as velocity multiplied by \(\small E\). By calculating \(\small V\), we get
\[ \small \begin{align*} V &= \frac{E}{c^2}\left(x\frac{d^2x}{dt^2}+y\frac{d^2y}{dt^2}+z\frac{d^2z}{dt^2}+c^2T\frac{d^2T}{dt^2}\right) \\ &= x\frac{dp_x}{dt}+y\frac{dp_y}{dt}+z\frac{dp_z}{dt}+c^2T\frac{dm}{dt}-\frac{xp_x+yp_y+zp_z+c^2Tm}{E}\frac{dE}{dt} \\ &= x\frac{dp_x}{dt}+y\frac{dp_y}{dt}+z\frac{dp_z}{dt}+c^2T\frac{dm}{dt}-\frac{x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}+c^2T\frac{dT}{dt}}{c^2}\frac{dE}{dt}. \end{align*} \]
Since the numerator of the last term was
\[ \small x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}+c^2T\frac{dT}{dt} = c^2t, \]
we can obtain
\[ \small V = x\frac{dp_x}{dt}+y\frac{dp_y}{dt}+z\frac{dp_z}{dt}+c^2T\frac{dm}{dt}-t\frac{dE}{dt}. \]
By replacing \(\small E=Mc^2\), it can be also expressed as:
\[ \small V =\left(x\frac{dp_x}{dt}+y\frac{dp_y}{dt}+z\frac{dp_z}{dt}+c^2T\frac{dm}{dt}-c^2t\frac{dM}{dt} \right). \]
Of course, if we assume that \(\small E\) does not change over time, the last term will be zero.
If we consider this potential function (\small V) to be exogenously given, it follows that energy in the theory of relativity must satisfy
\[ \small E^2 = m^2c^4+p^2c^2+EV=K^2+EV. \]
The solution to this quadratic equation is:
\[ \small E = \sqrt{K^2+\frac{1}{4}V^2}+\frac{1}{2}V. \]
Relativistic mass also cannot be expressed by momentum alone
\[ \small M = \frac{E}{c^2}, \]
and since
\[ \small \frac{dq_\mu}{dt} = \frac{ p_\mu}{M}, \]
velocity is also strictly affected by the potential function. One might assume that these effects are treated in electromagnetism and quantum mechanics in the form of vector potentials.
Finally, as in the previous section, let us consider the aggregated values when the momentum and energy of \(\small n\) point particles are:
\[ \small E_i^2 = M_i^2c^4 = m_i^2c^4+\left(p_{x_i}^2+p_{y_i}^2+p_{z_i}^2 \right)c^2+E_iV_i, \quad i=1,\cdots,n. \]
If we aggregate them together, we get
\[ \small E^2 = \sum_{i=1}^n E_i^2 = \sum_{i=1}^n m_i^2c^4+p_i^2c^2+\sum_{i=1}^n E_iV_i. \]
With such an equation, it becomes difficult to consider the energy of each particle separately. Since the left side is a polynomial in \(\small E_i^2\) while the right side is a polynomial in \(\small E_i\), it seems difficult to express this equation as an aggregate value equation such as \(\small E=\sum E_i\) and \(\small V = \sum V_i\). In other words, it may not be possible in the theory of relativity to express the potential function as a function such as \(\small V\) that is shared by all physical objects. Approximating
\[ \small E^2 \approx \left(\sum_{i=1}^n \sqrt{m_i^2c^4+p_i^2c^2}+V_i\right)^2 \]
using some method (assumption) corresponds to classical mechanics.
Summary and Issues
As we have seen so far, contrary to commonly held belief, we can surmise that both energy and potential are physical quantities with completely different properties in classical mechanics and the theory of relativity. Therefore, if we want to treat a phenomenon like gravity with the theory of relativity, expressing it by adding a relativistic adjustment term to the Hamiltonian:
\[ \small H = \sum_{i=1}^n\frac{p_i^2}{2m_i}-\frac{1}{2}\sum_{i=1}^n\sum_{j=1,i\neq j}^n\frac{Gm_im_j}{r_{ij}} \]
does not seem to be an appropriate approach. Classical mechanics fundamentally misunderstands the physical phenomenon of gravity, and requires a completely different formulation. Of course, that would be the general theory of relativity, but the general theory of relativity is difficult to understand intuitively, and analytical solutions are rarely found, so calculations are often not easy either. I also get the impression that it’s difficult to understand what the essence of mass and gravity is (although it may just be that I don’t understand it). It is well known that in physics, the origin of mass and the mechanism by which gravity is generated have yet to be elucidated.
Probably the truth of what gravity and mass are is one of two things:
- There are elementary particles that give matter mass, and these elementary particles create gravity. Gravity occurs through elementary particles (gravitons) that transmit gravity, such as electromagnetic fields (photons).
- Concepts of time and space are different from what we perceive, and because we understand them as phenomena in Euclidean space, we are under the illusion that things like mass and gravity exist. In other words, the force of gravity does not exist in the first place, and the cause is that phenomena in curved space-time are treated as phenomena in a Cartesian coordinate system.
There are clear differences between these hypotheses, and if 1. is correct, gravity cannot travel faster than the speed of light. In addition, if we can control the elementary particles that transmit gravity, it may be possible to block it from transmitting gravity. On the other hand, in case 2, gravity would take no time to transmit and would be treated as if it acts uniformly on all matter in space. In this case, it would be impossible to interfere with the transmission of gravity by any means. It is likely that no definitive experimental evidence has yet been found to confirm these findings. For the time being, the hypothesis that I support is option 2, and specifically, that the cause is that a conical coordinate system (in spatial terms, a three-dimensional sphere) is transformed into Euclidean space and understood accordingly. However, to be honest, the author is currently not able to theorize this in a convincing manner.
Reference
[1] Hirano, Kaname (2022), Quantum Mechanics with Multi-Time Theory, Amazon Kindle Store.
[2] Hirano, Kaname (2024), Complex Functions and Schrödinger Equation, Amazon Kindle Store.
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