Many astronomy enthusiasts find cosmology “intimidating,” feeling that it is incomprehensible without a foundation in complex theories such as Einstein’s General Relativity. Indeed, this view is not entirely wrong; the current precise description of the universe in cosmology is indeed built upon theories like General Relativity and quantum mechanics. BoJone has only briefly browsed through books on the subject and cannot claim to have a deep understanding. However, for the average astronomy enthusiast, as long as they have a certain understanding of Newtonian mechanics and calculus, they can obtain a general description of our universe and reach many surprising conclusions. I believe that after performing this work, many enthusiasts will change their perspective: it turns out that cosmology is not that difficult... and they will be able to reach the following conclusion: although General Relativity fundamentally reformed Newton’s theory of gravity, from a mathematical standpoint, it merely corrected Newtonian mechanics.
The explanation of “Cosmology” on Wikipedia is:
Cosmology is the study of the universe as a whole, extending to the exploration of humanity’s place within it. Although the term cosmology is relatively recent, the study of the universe has a long history involving science, philosophy, esotericism, and religion.
Simply put, cosmology is a discipline dedicated to studying the physical origin and evolution of the universe, investigating the nature of the universe from the largest scales. Concepts we often hear about, such as “Hubble’s Law,” the “Big Bang model,” and “universal expansion,” all fall within the scope of cosmology. If we assume the universe is an isotropic sphere with radius R (R is a function of time) and total mass M (although mass can be converted into energy, for a mass as large as the universe, M can still be regarded as a constant over a long period), and only consider the gravitational interaction between matter, then for an object of mass m at the edge of the universe, we have: m\frac{d^2 R}{dt^2}=-\frac{GMm}{R^2} Thus, the basic equation of Newtonian cosmology is: \frac{d^2 R}{dt^2}=-\frac{GM}{R^2} \label{eq:1} We have discussed this type of equation before. By rewriting \frac{d^2 R}{dt^2} as \dot{R}\frac{d\dot{R}}{dR} and substituting it into [eq:1]: \dot{R}\frac{d\dot{R}}{dR}=-\frac{GM}{R^2} Integrating gives: \dot{R}^2=\left(\frac{dR}{dt}\right)^2=\frac{2GM}{R}+k \label{eq:2} where \dot{R}=V is the expansion velocity of the universe, and k is the integration constant whose value determines the direction of the universe’s evolution. In this equation, we ignore the influence of pressure; therefore, this model is also called the “zero-pressure universe.” It is divided into three cases:
When k = 0, V is always greater than 0, and the universe expands forever, but the velocity tends toward 0. This is what we call the flat universe case.
When k > 0, V is always greater than 0, and the universe expands forever, with the velocity tending toward k. Such a universe is an open three-dimensional hyperboloid, referred to as an open universe.
When k < 0, V is initially positive but gradually decreases until it becomes negative (negative represents contraction). Such a universe is a closed three-dimensional sphere, referred to as a closed universe.
Equation [eq:2] is also a manifestation of the universe’s energy. We can write [eq:2] as: \frac{1}{2}mV^2-\frac{GMm}{R}=\frac{k}{2} \label{eq:3} The left side of [eq:3] is exactly the sum of gravitational potential energy and kinetic energy. The sign of k reveals whether the universe is bound. Let V=HR and substitute it into [eq:3]. Rearranging gives: H^2=\frac{2GM}{R^3}+\frac{k}{R^2}=\frac{8}{3}\pi G \rho+\frac{k}{R^2} \label{eq:4} \rho is the average density of the universe, and H is what we call the Hubble constant. However, like \rho, it is a function of time t. Calling it a constant is only because it is a constant for the current universe; the current H is denoted as H_0. Space expands as a whole, and V=HR is the expansion velocity at the edge of the universe. It is not difficult to deduce that the expansion velocity (galactic recession velocity) at a distance D from the center of the universe is V=HD (you can imagine pulling a rubber band of length R fixed at one end at a speed HR; the speed at distance D from the fixed point is HD). This is “Hubble’s Law.” The isotropy of the universe tells us that every point in the universe is its center, so the recession velocity of a galaxy at distance D from Earth is also V=HD.
When k = 0, we obtain a universe model between the “open” and “closed” universes, also known as the “critical universe.” From [eq:4], we can easily find the current critical density of the universe: \rho_c=\frac{3H_0^2}{8\pi G} And we can solve this differential equation, which is the simplest form: \begin{aligned} \frac{dR}{dt}&=\sqrt{\frac{2GM}{R}}\\ dt&=\sqrt{\frac{1}{2GM}}R^{0.5}dR \end{aligned} Integrating gives: t=\frac{2}{3}\sqrt{\frac{1}{2GM}}R^{3/2}+C C is the integration constant. When t=0, R=0, which implies C=0. At this point, the above equation can be written as: \begin{aligned} R&=\left(\frac{9GM}{2}\right)^{3/4}t^{3/2}\\ \dot{R}&=\frac{3}{2}\left(\frac{9GM}{2}\right)^{3/4}t^{1/2} \end{aligned} Since HR=\dot{R}, then \frac{2}{3}H^{-1}=t. In other words, in this universe model, the age of the universe is only two-thirds of the reciprocal of the Hubble constant.
For the case where k \neq 0, we need to complete the integral t=\int (\frac{2GM}{R}+k)^{-0.5}dR. This is also a common integral (refer to integral tables), with different results for positive and negative k:
For k < 0: t=\frac{\sqrt{\frac{2GM}{R}+k}}{k}R+\frac{2GM\times \arctan\left(\frac{\sqrt{\frac{2GM}{R}+k}}{\sqrt{-k}}\right) }{k\sqrt{-k}}+C With initial conditions t=0, R=0, we find C=-\frac{GM\pi}{k\sqrt{-k}}.
For k > 0: t=\frac{\sqrt{\frac{2GM}{R}+k}}{k}R-\frac{GM\cdot \ln\left(\frac{\sqrt{\frac{2GM}{R}+k}+\sqrt{k}}{\sqrt{\frac{2GM}{R}+k}-\sqrt{k}}\right)}{k\sqrt{k}}+C Similarly, based on initial conditions t=0, R=0, we find C=0.
As we reach the end, we can compare our results from Newtonian mechanics with those from General Relativity. The result using General Relativity is: \frac{\ddot{R}}{R} =-\frac{4 \pi G}{3}\left(\rho+\frac{3p}{c^2}\right)+\frac{\Lambda c^2}{3}
where p is the pressure term and \Lambda is called the cosmological constant. It can be seen that General Relativity only adds two terms compared to Newtonian mechanics: pressure and the cosmological constant. The cosmological constant is an artificially added term, and Newtonian mechanics could do the same. Therefore, the real difference is the addition of the pressure term. This is a substantial improvement brought by General Relativity, which is crucial for the evolution of the universe: the appearance of pressure p allows us to have an equation of state, thereby describing the true form of matter in the universe.
The results obtained using Newtonian mechanics are basically the same as those from General Relativity, which is credited to the cosmological principle. The cosmological principle tells us that the state of motion in every local part of the universe is the same; therefore, we can study the expansion of the universe within a sufficiently small local range. Within a small range, the relative velocity of stars is less than the speed of light c, so Newtonian theory can be applied. However, for large-scale problems, such as high-redshift objects, distance, and luminosity, Newtonian mechanics is no longer applicable.
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