Free Fall in Classical Mechanics
The falling motion caused only by gravity, ignoring the effects of air resistance and friction, is called free fall. When gravity is acting, the potential energy can be expressed as:
\[ \small V=mr\frac{d^2r}{dt^2} = -\frac{GmM}{r}. \]
Here, \(\small G\) is a natural constant called the gravitational constant, \(\small m,M\) represent the masses of the two objects, and \(\small r\) represents the distance between the two objects. Therefore, the acceleration due to gravity is:
\[ \small \frac{d^2r}{dt^2} = -\frac{GM}{r^2}. \]
When gravity is acting on a very large object such as the Earth, (\small M) can be approximated by the mass of the Earth and (\small r=R) can be approximated by the radius of the Earth, so we assume that
\[ \small g = \frac{GM}{R^2} \]
is a constant. By approximating \(\small 1/r\approx r/R^2\), potential energy (The signs are confusing because the Earth’s coordinates are taken in the negative direction.) can be expressed as:
\[ \small V = \frac{GM}{R^2}mr = mgr, \]
which is the potential energy during free fall. To be precise, the distance from the Earth’s surface expressed as (\small r), it can be written as:
\[ \small V = \frac{GmMr}{R(R+r)}. \]
Let us find the equation of motion in this case.Since
\[ \small \frac{d^2r}{dt^2} = -g, \]
by calculating the integral,
\[ \small \begin{align*} &r(t) = -\frac{1}{2}gt^2 \\ &v=\frac{dr}{dt} = -gt \end{align*} \]
can be obtained. The integral constant is determined so that the initial coordinate is \(\small r(0)=0\). The energy is calculated as:
\[ \small E = \frac{1}{2}mv^2+mgr = \frac{1}{2}mg^2t^2-\frac{1}{2}mg^2t^2 = 0. \]
It was the equation of motion for free fall in classical mechanics.
Free Fall in the Theory of Relativity
Calculations of free fall in the theory of relativity are not widely available, since relativistic effects rarely have a meaningful effect on the motion of matter in free fall on Earth. Although it is often used to explain the equivalence principle in general relativity, it seems strange that the results of its calculations are rarely shown, but this is probably because it is thought to be not worth seriously calculating. If asked to show the equation of motion for free fall in the theory of relativity, I doubt there are many people who would be able to do so immediately. However, since this is thought to be a simple example that shows how classical mechanics misunderstands gravity, let us do a quick calculation here.
The theory of relativity asserts that the motion of any matter cannot exceed the speed of light \(\small c\), and in the relativistic equations of motion, velocity can be expressed as:
\[ \small \frac{dr}{dt} = \frac{pc}{\sqrt{m^2c^2+p^2}} \]
using momentum \(\small p\). If \(\small p\) is negligibly small compared to \(\small mc\), then
\[ \small \frac{dr}{dt} \approx \frac{p}{m}, \]
which was the equation of motion in classical mechanics.
Since the momentum of free fall is:
\[ \small p=m\frac{dr}{dt} = -mgt, \]
we can substitute this into the formula for velocity in the theory of relativity to obtain the equation of motion. By substituting, we get
\[ \small \frac{dr}{dt} = -\frac{mgct}{\sqrt{m^2c^2+m^2g^2t^2}} = -\frac{gct}{\sqrt{c^2+g^2t^2}}. \]
By approximating the denominator to \(\small c\), we can confirm that it matches the equation in classical mechanics. In addition, it can be confirmed that
\[ \small \lim_{t\rightarrow \infty} \frac{dr}{dt} = -c \]
is true and that the speed of light cannot be exceeded. To find the equation of motion, we simply integrate this equation and set the integral constant so that \(\small r(0)=0\). It can be calculated as:
\[ \small r(t) = \frac{c}{g}\left(c-\sqrt{c^2+g^2t^2}\right). \]
This is the equation of motion for free fall in the theory of relativity. If we approximate it to
\[ \small \sqrt{c^2+g^2t^2} \approx c\left(1+\frac{g^2t^2}{2c^2} \right), \]
we can confirm that it coincides with the equation of motion for free fall in classical mechanics.
By the way, the above solution is the solution to a quadratic equation:
\[ \small r^2-\frac{2c^2}{g}r-c^2t^2 = 0. \]
A different time flows at each coordinate \(\small r\), and if local time is defined as:
\[ \small T_{rt}^2 = -\frac{2r}{g} \rightarrow r = -\frac{1}{2}gT_{rt}^2, \]
it can be expressed as:
\[ \small r^2+c^2T^2_{rt}=c^2t^2. \]
If the concept of time can be defined as a function \(\small T_{rt}\), then we can infer that the space-time assumed by this coordinate system is a conical coordinate system. The classical mechanical approximation corresponds to \(\small T_{rt} \approx t\).
The Concept of Acceleration Is Inconsistent with the Theory of Relativity
In classical mechanics, gravity is considered to be the force that causes acceleration. In other words, when expressed in the coordinate equation of motion, it has the property of appearing as a term proportional to \(\small t^2\). However, in the theory of relativity, it is \(\small \sqrt{c^2+g^2t^2}\), which has the property that as it approaches the speed of light, it approaches a term proportional to \(\small t\). In classical mechanics, a value proportional to the square of a physical quantity approaches a value proportional to the first power when calculating motion close to the speed of light in a relativistic manner. This is probably an example of such a calculation.
The way classical mechanics misunderstands gravity is because the concept of acceleration is fundamentally incorrect and inconsistent with the theory of relativity. The concept of acceleration depends on the assumption that space can be represented in Euclidean space, but the actual properties of space may be different from those of Euclidean space. The reason classical mechanics has caused so much confusion is probably because we are trying to understand the phenomenon of gravity based on this incorrect concept. In fact, concepts such as force and work, which are derived from acceleration, almost never appear in the theory of relativity, and the difficulty of general relativity is that it must define phenomena such as gravity without using the concept of acceleration.
Hamiltonian in the Theory of Relativity
In Section 2, I wondered if there was a way to systematically derive the equation of motion for free fall in the theory of relativity, and the above calculation was obtained by setting the Hamiltonian in the theory of relativity to:
\[ \small H(p,q) = \sqrt{m^2c^4+p^2c^2} + V(q) \]
and calculating as follows：
\[ \small \begin{align*} &\frac{dq}{dt}=\frac{\partial H(p,q)}{\partial p} = \frac{pc^2}{\sqrt{m^2c^4+p^2c^2}} \\ &\frac{dp}{dt}= -\frac{\partial H(p,q)}{\partial q}. \end{align*} \]
It is important to note that the derivative of the Hamiltonian with respect to the coordinates does not result in acceleration, but only has the meaning of the time derivative of momentum. If you want to find the acceleration, you can simply differentiate \(\small dq/dt\) with respect to \(\small t\).
In fact, if we calculate the case of free fall, by calculating the momentum from
\[ \small H(p,q) = \sqrt{m^2c^4+p^2c^2} + mgr, \]
we get
\[ \small \frac{dp}{dt} = -\frac{\partial H(p,q)}{\partial q}=-mg \Rightarrow p(t) = -mgt. \]
The coordinate equation can be calculated as:
\[ \small \frac{dr}{dt}=\frac{\partial H(p,q)}{\partial p} = \frac{pc^2}{\sqrt{m^2c^4+p^2c^2}} = -\frac{gct}{\sqrt{c^2+g^2t^2}}. \]
Substituting the calculation result into the equation gives
\[ \small \begin{align*} E &= \sqrt{m^2c^4+p^2c^2} +mgr \\ &=\sqrt{m^2c^4+m^2c^2g^2t^2}+mg\frac{c}{g}\left(c-\sqrt{c^2+g^2t^2}\right)=mc^2, \end{align*} \]
which confirms that energy is a conserved quantity. In this case, the energy is expressed by the formula
\[ \small (E-V)^2=m^2c^4+p^2c^2. \]
It is unclear whether this calculation method is valid or whether the results are merely an approximation that happened to match.
In fact, the author’s opinion is negative. Although it is possible to easily derive a solution so that the speed of light is not exceeded, I speculate that this is merely an approximation and is not a conserved quantity in the original theory of relativity. If we assume that relativistic space-time is a conical coordinate system (space is a three-dimensional sphere), then the theoretical conserved quantity \(\small E\) is a value that satisfies
\[ \small E(E-V) = m^2c^4+p^2c^2, \]
and the solution is:
\[ \small E = \sqrt{m^2c^4+p^2c^2+\frac{1}{4}V^2}+\frac{1}{2}V. \]
The Hamiltonian would be:
\[ \small H(p,q) = \sqrt{m^2c^4+p^2c^2+\frac{1}{4}V^2(q)} + \frac{1}{2}V(q). \]
The potential contained in the square root is treated as a vector potential in electromagnetism and quantum mechanics and is thought to be expressed as:
\[ \small H(p,q) = \sqrt{m^2c^4+(p-eA)^2c^2} + \frac{1}{2}V(q). \]
Also, the potential in the theory of relativity is twice the potential in classical mechanics, so for example, in the case of free fall,
\[ \small V(q) = 2mgr \]
would be the potential in the theory of relativity. In this case, the canonical equation is:
\[ \small \begin{align*} &\frac{dp}{dt}= -\frac{\partial H(p,q)}{\partial q} =ー\frac{m^2g^2r}{\sqrt{m^2c^4+p^2c^2+m^2g^2r^2}}-mg \\ &\frac{dr}{dt}= \frac{\partial H(p,q)}{\partial p}= \frac{pc^2}{\sqrt{m^2c^4+p^2c^2+m^2g^2r^2}}, \end{align*} \]
and the exact solution requires solving this simultaneous differential equation. This may be difficult to solve analytically, and an exact solution may have to be obtained only through numerical calculations. Calculations that assume the energy to be
\[ \small (E-V)^2=m^2c^4+p^2c^2 \]
make an approximation of
\[ \small \begin{align*} &\frac{m^2g^2r}{\sqrt{m^2c^4+p^2c^2+m^2g^2r^2}} \approx \frac{mg^2r}{c^2} \approx 0 \\ &\frac{pc^2}{\sqrt{m^2c^4+p^2c^2+m^2g^2r^2}} \approx \frac{pc^2}{\sqrt{m^2c^4+p^2c^2}} ,\end{align*} \]
and ignore the term equivalent to the vector potential.
Comments