Uniform Circular Motion in Classical Mechanics
Let us consider uniform circular motion in three-dimensional space. In classical mechanics, the energy (Hamiltonian) of uniform circular motion is given by
\[ \small E = H(p,q) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2r^2. \]
For simplicity, we assume that the direction of rotation of the circular motion is in the \(\small xy\) plane and that \(\small z=0\). If we calculate the acceleration, it becomes
\[ \small \begin{align*} &-\frac{\partial H}{\partial x} = m\frac{d^2x}{dt^2} = -m\omega^2x \\ &-\frac{\partial H}{\partial y} = m\frac{d^2y}{dt^2} = -m\omega^2y \\ &-\frac{\partial H}{\partial z} = m\frac{d^2z}{dt^2} = -m\omega^2z=0. \end{align*} \]
There are an infinite number of general solutions that satisfy these differential equations, but we choose
\[ \small \begin{align*} &x(t) = r\cos(\omega t) \\ &y(t) = r\sin(\omega t) \\ &z(t) = 0 \end{align*} \]
as a particular solution. Let assume that \(\small r=\sqrt{x^2+y^2}\) is a constant.
Determine the velocity, momentum, and angular momentum of this motion. The velocity can be calculated as:
\[ \small \begin{align*} &v_x = \frac{dx(t)}{dt} = -r\omega\sin(\omega t) = -\omega y(t) \\ &v_y = \frac{dy(t)}{dt} = r\omega\cos(\omega t) = \omega x(t) \\ &v_z = \frac{dz(t)}{dt} = 0. \end{align*} \]
Therefore,
\[ \small v = \sqrt{v_x^2+v_y^2+v_z^2} = r\omega \]
can be obtained. Hence, the angular velocity \(\small \omega\) can be defined as:
\[ \small \omega = \frac{v}{r}. \]
The momentum can be calculated as:
\[ \small \begin{align*} &p_x = m v_x = -mr\omega \sin(\omega t) = -m\omega y(t) \\ &p_y = m v_y = mr\omega \cos(\omega t) = m\omega x(t) \\ &p_z = m v_z = 0 \\ &p=\sqrt{p_x^2+p_y^2+p_z^2} = mr\omega. \end{align*} \]
Calculating the angular momentum gives us
\[ \small \begin{align*} &l_x = yp_z – zp_y = 0 \\ &l_y = zp_x – xp_z = 0 \\ &l_z = xp_y – yp_x = m\omega r^2 \\ &L^2 = l_x^2+l_y^2+l_z^2 = m^2\omega^2r^4. \end{align*} \]
Substituting into the original energy equation, we get
\[ \small E = \frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2=\frac{L^2}{2mr^2}+\frac{1}{2}m\omega^2r^2=m\omega^2r^2. \]
In addition, since the circumference of a circle is \(\small 2\pi r \), the time (period) it takes to complete one revolution and return to the original coordinate is:
\[ \small vt = r\omega T = 2\pi r \Rightarrow T = \frac{2\pi}{\omega}. \]
Uniform Circular Motion in the Theory of Relativity
The theory of relativity asserts that the motion of any matter cannot exceed the speed of light \(\small c\), and in the special relativistic equations of motion, velocity can be expressed as:
\[ \small \frac{dq}{dt} = \frac{p_\mu c}{\sqrt{m^2c^2+p^2}}, \quad q=x,y,z \]
using momentum \(\small p\). As mentioned last time, this is an approximate calculation and strictly speaking it is necessary to solve a system of differential equations, but here we will find the equation of motion using an approximate calculation.
Since the momentum in uniform circular motion is:
\[ \small \begin{align*} &p_x = -mr\omega \sin(\omega t) \\ &p_y = mr\omega \cos(\omega t) \\ &p_z = 0, \end{align*} \]
we can substitute this into the formula for velocity in the theory of relativity to obtain the equation of motion. Substituting this, we get
\[ \small \begin{align*} &\frac{dx}{dt} = -\frac{mr\omega \sin(\omega t) c}{\sqrt{m^2c^2+m^2r^2\omega^2}} = -\frac{r\omega \sin(\omega t) c}{\sqrt{c^2+r^2\omega^2}} \\ &\frac{dy}{dt} = \frac{mr\omega \cos(\omega t) c}{\sqrt{m^2c^2+m^2r^2\omega^2}} = \frac{r\omega \cos(\omega t) c}{\sqrt{c^2+r^2\omega^2}} \\ &\frac{dz}{dt} = 0. \end{align*} \]
Since
\[ \small v = \sqrt{v_x^2+v_y^2+v_z^2} = r\omega\frac{c}{\sqrt{c^2+r^2\omega^2}} , \]
we can obtain
\[ \small \lim_{r,\omega\rightarrow \infty} v = c, \]
and confirm that the velocity does not exceed the speed of light. To find the equation of motion, we can calculate
\[ \small \begin{align*} &x(t) = \frac{rc}{\sqrt{c^2+r^2\omega^2}}\cos(\omega t) \\ &y(t) = \frac{rc}{\sqrt{c^2+r^2\omega^2}}\sin(\omega t) \\ &z(t) = 0 \end{align*} \]
by integrating this equation. This is the equation of motion for uniform circular motion in the theory of relativity. Note that in this case the radius of the circular motion is:
\[ \small R = \frac{rc}{\sqrt{c^2+r^2\omega^2}}, \]
and its maximum value given \(\small \omega\) is given by
\[ \small \lim_{r \rightarrow \infty} = R = \frac{c}{\omega}=\frac{\lambda}{2\pi}. \]
\(\small \lambda\) is a parameter that corresponds to the wavelength in wave terminology. If we think in this way, the parameters that describe the properties of light (photons) (wavelength, frequency, wave number, etc.) may also be interpreted as expressing the properties of motion such as uniform circular motion.
The hypothesis that interests the author is that relativistic space-time is represented by a conical coordinate system:
\[ \small x^2+y^2+z^2+c^2T_{xyzt}^2=c^2t^2, \]
and local time \(\small T_{xyzt}\) (note that the concept we recognize as time, \(\small t\) is not the time we recognize) is represented by a function. In this hypothesis, the local time is:
\[ \small T_{xyzt} = \sqrt{t^2-\frac{R^2}{c^2}}, \]
and the time flowing in the coordinates on the circular motion is a common value. It is clear from the equation that the further away from the origin you are, the slower time passes. Also, since the circumference of a circle is \(\small 2\pi R\), the time (period) it takes to complete one revolution and return to the original coordinates is:
\[ \small vt = R\omega T = 2\pi R \Rightarrow T = \frac{2\pi}{\omega} \]
when the speed is measured in terms of the time at the origin, \(\small t\). On the other hand, when measured in local time \(\small T_{xyzt}\), the period can be calculated as:
\[ \small vT_{xyzt} = r\omega T = 2\pi R \Rightarrow T = \frac{2\pi}{\omega}\frac{c}{\sqrt{c^2+r^2\omega^2}}, \]
and it can be thought that the period becomes shorter as the radius becomes smaller (as we approach the speed of light, time measured in local time progresses more slowly).
Finally, if we calculate the energy we get
\[ \small \begin{align*} E &= \sqrt{m^2c^4+p^2c^2}+\frac{1}{2}m\omega^2R^2 \\ &\approx mc^2+\frac{1}{2}m\omega^2r^2\left(1+\frac{c^2}{c^2+r^2\omega^2} \right), \end{align*} \]
and we can see that the result deviates (albeit slightly) from the energy of classical mechanics. This is thought to be due to the discrepancy between the momentum \(\small r\) and the orbital radius \(\small R\). This means that the calculations in this section are only approximations. In the previous article, I wrote that in order to perform an accurate calculation, the potential must be doubled and the energy must be defined as:
\[ \small E = \sqrt{m^2c^4+p^2c^2+\frac{1}{4}V^2}+\frac{1}{2}V. \]
Applying the above gives
\[ \small \begin{align*} &p = mr\omega \\ &V(q) = m\omega^2r^2, \end{align*} \]
and substituting it gives
\[ \small E = \sqrt{\left(mc^2+\frac{1}{2}m\omega^2r^2\right)^2}+\frac{1}{2}m\omega^2r^2=mc^2+m\omega^2r^2. \]
This energy represents uniform circular motion of radius \(\small r\). The equation of motion represented by this momentum and potential is:
\[ \small \begin{align*} &x(t) = r\cos(\omega t) \\ &y(t) = r\sin(\omega t) \\ &z(t) = 0, \end{align*} \]
but at present the author is unsure how to derive this from the energy equation. By the way, instead of doubling the potential and writing as:
\[ \small E = \sqrt{m^2c^4+p^2c^2+\frac{1}{4}V^2}+\frac{1}{2}V, \]
you might think that we should write it as:
\[ \small E = \sqrt{m^2c^4+p^2c^2+V^2}+V. \]
However, the above equation is the one that can be derived theoretically, and since the methods of deriving the equation of motion from energy differ between classical mechanics and the theory of relativity, it is not possible to derive a similar equation of motion unless it is written in this way.
Uniform Circular Motion Rotating in Any Direction
The purpose of this paper has been achieved by showing that uniform circular motion seems to be related to the motion of light (photons), but since this is a three-dimensional space, I cannot help but wonder if the assumption such as \(\small z=0\) is appropriate. Let us also show what happens when we extend the discussion of uniform circular motion to rotation in any direction.
I will omit the details, but it seems that all you need to do is multiply it by a rotational coordinate transformation matrix. It is as follows:
\[ \small \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \phi & \sin \phi \\ 0 & -\sin \phi & \cos \phi \end{array} \right] \left[ \begin{array}{ccc} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0 \\ -\sin \theta & 0 & \cos \theta \end{array} \right] \left[ \begin{array}{ccc} \cos \psi & \sin \psi & 0 \\ -\sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{array} \right] \]
These represent rotations around the \(\small x,y,z\) axes, respectively. Around the \(\small z\) axis, all that changes is the value of the phase (just replace \(\small \omega t\) with \(\small \omega t + \psi)\), i.e. the initial coordinates, so if we ignore this and calculate, the result can be transformed into:
\[ \small \begin{align*} &x(t) = r\cos\theta \cos(\omega t)\cos \phi + r\sin \phi \sin(\omega t) \\ &y(t) = r\cos\theta \cos(\omega t)(-\sin \phi) + r\cos \phi \sin(\omega t) \\ &z(t) = r\sin\theta \cos(\omega t). \end{align*} \]
The coordinate axes have been swapped so that the original equation results when \(\small \phi=\theta=0\). By changing the values of \(\small \theta,\phi\), it is possible to express uniform circular motion tilted in any direction.
Finally, find the velocity, momentum, and angular momentum. The velocity can be calculated as:
\[ \small \begin{align*} &v_x = \frac{dx(t)}{dt} = -r\omega\cos\theta \sin(\omega t)\cos \phi + r\omega\sin \phi \cos(\omega t) \\ &v_y = \frac{dy(t)}{dt} = -r\omega\cos\theta \sin(\omega t)(-\sin \phi) + r\omega\cos \phi \cos(\omega t) \\ &v_z = \frac{dz(t)}{dt} = -r\omega\sin\theta \sin(\omega t) \\ &v = \sqrt{v_x^2+v_y^2+v_z^2} = r \omega. \end{align*} \]
The momentum can be calculated using \(\small p=mv\). Angular momentum is calculated as:
\[ \small \begin{align*} &l_x = yp_z – zp_y = m\omega r^2\sin\theta\cos \phi \\ &l_y = zp_x – xp_z = m\omega r^2\sin\theta\sin \phi \\ &l_z = xp_y – yp_x = m\omega r^2\cos\theta \\ &L^2 = l_x^2+l_y^2+l_z^2 = m^2\omega^2r^4. \end{align*} \]
Thinking about it this way, it may be necessary to convert angular momentum into polar coordinates. The argument for the theory of relativity would similarly hold if we replaced \(\small r\) with:
\[ \small R = \frac{rc}{\sqrt{c^2+r^2\omega^2}}. \]
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