3. Elements of classical mechanics¶
3.1. Mechanics of a particle¶
Classical mechanics—as opposed to quantum mechanics—describes the motions of bodies under the influence of classical forces, such as gravitational or electromagnetic forces. The basic concepts are contained in Newton’s laws of motion. The time derivative of the location \(\vec{x}\) of a body with mass \(m\) is the velocity, \(\vec{v} = \dot{\vec{x}}\) (from now on we will often write a time derivative of a quantity as a dot over that quantity, and \(n\) time derivatives as \(n\) dots). An important, related concept is that of the momentum \(\vec{p}\) of the body: \begin{equation} \vec{p} = m\,\vec{v}\,. \end{equation} Newton’s laws of motion are then the following:
Newton’s first law of motion: In an inertial reference frame, unless acted upon by a force, a body remains at rest or moving at a constant velocity.
Newton’s second law of motion: In an inertial reference frame, the vector sum of all forces \(\vec{F}\) acting on a body is equal to the change in the body’s momentum in time: \begin{equation}\label{eq-classmech-newton2} \dot{\vec{p}} = \vec{F}\,. \end{equation}
Newton’s third law of motion: If a body \(i\) exerts a force \(\vec{F}_{ij}\) on a body \(j\), the second body simultaneously exerts a force \(\vec{F}_{ji} = -\vec{F}_{ij}\) equal in magnitude and opposite in sign on the first body.
For bodies with a constant mass, which is typically the case in galactic dynamics, Newton’s second law as stated above reduces to the more familiar \(\vec{F} = m\,\vec{a}\), with \(\vec{a} = \ddot{\vec{x}}\) the acceleration.
Newton’s laws of motion hold in an inertial reference frame. Newton considered these frames to derive from the absolute frame of the fixed stars, which he assumed to be stationary, homogeneous, isotropic, and unchanging in time. A frame moving at a constant velocity with respect to an inertial frame is an inertial frame itself; the transformation between frames is given by the Galilean transformation. To Newton, inertial frames were therefore those frames that move at constant velocity with respect to the fixed stars. We now know that Newton’s assumptions about the fixed stars were mistaken: the stars visible in the night sky are distributed in a quite non-homogeneous manner in the Milky Way and they all move within it at a small, but measurable rate. Furthermore, the general theory of relativity and cosmological world models based on it demonstrate that there is not a single, absolute reference frame. Thus, the notion of an inertial reference frame is problematic. However, much like Newtonian mechanics and gravitation is an approximation to the general theory of relativity, the notion of an inertial reference frame is still useful in an approximate manner.
To a physicist’s mind, clearly what matters is the magnitude of the acceleration of the reference frame itself versus that of bodies whose motions are described with respect to the frame. For example, much of classical mechanics pertains to the motion of objects on Earth (e.g., a ball rolling down an incline, or a pendulum swinging) and we use local patches of the Earth, such as a small laboratory, as the inertial reference frame in these cases. Why can we do this? Because the acceleration of the laboratory itself is small compared to the accelerations of the bodies under study within the frame. The relevant acceleration of the lab frame is the centripetal acceleration due to the rotation of the Earth, which we can compute as \(R_\mathrm{Earth}\,\left(2\pi/1\,\mathrm{day}\right)^2 \approx 0.034\,\mathrm{m\,s}^{-2}\):
[6]:
from astropy import constants
print(f'{(constants.R_earth*(2.*numpy.pi/(1.*u.day))**2.).to(u.m/u.s**2).value} m/s^2')
0.03373056189481955 m/s^2
This is much smaller than the acceleration due to gravity, which is \(9.80\,\mathrm{m\,s}^{-2}\) (so the centripetal acceleration is about 300 times smaller). Therefore, to a good approximation, the acceleration due to the rotation of the Earth does not matter for simple laboratory experiments and we may consider the frame as an inertial frame within which Newton’s laws of motion hold.
Moving to larger scales, there is a similar hierarchy at play. In the solar system, typical accelerations are like that at the Earth’s orbit, which we may estimate from the centripetal acceleration implied by the distance (1 AU) and period (1 year) of the Earth’s orbit around the Sun: \(1\,\mathrm{AU}\,\left(2\pi/1\,\mathrm{yr}\right) \approx 0.0059\,\mathrm{m\,s}^{-2}\):
[7]:
print(f'{(u.AU*(2.*numpy.pi/(1.*u.yr))**2.).to(u.m/u.s**2).value} m/s^2')
0.0059303075202325715 m/s^2
The entire solar system is accelerated by the mass distribution in the Milky Way, and at the location of the Sun—approximately 8 kpc from the center close to the mid-plane of the disk—we can compute this acceleration approximately from the MWPotential2014 Milky-Way model in galpy. Calculating the acceleration in the mid-plane, where its only non-zero component is the radial one, we find \(\approx 2\times 10^{-10}\,\mathrm{m\,s}^{-2}\):
[8]:
from galpy.potential import MWPotential2014, evaluateRforces
print(f'{numpy.fabs(evaluateRforces(MWPotential2014,8.*u.kpc,0.*u.pc,
ro=8.,vo=220.).to(u.m/u.s**2)).value} m/s^2')
1.960671470113841e-10 m/s^2
This acceleration is thirty million times smaller than that experienced by the Earth from the Sun. Therefore, we may typically consider the center of mass of the solar system (the solar system barycenter) as an inertial frame for the purposes of solar-system dynamics.
The acceleration of the Milky Way barycenter from the mass distribution of local galaxies surrounding the Milky Way is another three orders of magnitude smaller. This is much smaller than the acceleration within the Milky Way, even at a hundred kpc from the center. The situation is similar for other isolated galaxies and we can therefore typically consider the internal dynamics in galaxies as taking place in an inertial reference frame centered on their barycenters. But like any approximation, this may break down when the external acceleration of the frame becomes closer to the magnitude of internal accelerations and in those cases the acceleration of the frame should be taken into account. The motion of bodies in non-inertial reference frames can still be described in terms of forces and Newton’s laws of motion, but it is necessary to include fictitious forces to take the acceleration of the reference frame into account. Examples of these are the Coriolis force and the centrifugal force. We will come back to this when discussing the motion of bodies in a rotating reference frame in Chapter 19.4.2.1.
A very important quantity in the study of galactic orbits is the angular momentum, defined as \begin{equation} \vec{L} = \vec{x}\times\vec{p}\,, \end{equation} that is, the angular momentum is the cross product of the position and momentum vectors of a body. By taking the time derivative of the angular momentum, we can derive an equation for its evolution that is analogous to Newton’s second law: \begin{align} \dot{\vec{L}} & = \frac{\mathrm{d} (\vec{x}\times\vec{p})}{\mathrm{d} t} = \dot{\vec{x}}\times\vec{p} + \vec{x}\times\dot{\vec{p}} = \vec{x}\times\vec{F}\,, \end{align} where in the final step we have used that \(\dot{\vec{x}} = \vec{p}/m\), that the cross product of a vector with itself is zero, and that \(\dot{\vec{p}} = \vec{F}\) from Newton’s second law. The quantity \(\vec{x}\times\vec{F}\) is the torque \(\vec{N}\) and thus we have that \begin{equation}\label{eq-classmech-dldt-torque} \dot{\vec{L}} = \vec{N}\,. \end{equation} If the torque (or one of its components) is zero, the angular momentum (or one of its components) is conserved.
Another important mechanical concept is that of the work. The work \(W_{ij}\) done by a force \(\vec{F}\) in bringing a body from point \(\vec{x}_i\) to point \(\vec{x}_j\) is defined as \begin{equation} W_{ij} = \int_{\vec{x}_i}^{\vec{x}_j}\,\mathrm{d}\vec{x}\cdot\vec{F}\,. \end{equation} Using Newton’s second law, we can show that the work is equal to (assuming that the mass \(m\) of the body is constant) \begin{equation} W_{ij} = \int_{\vec{x}_i}^{\vec{x}_j}\,\mathrm{d}\vec{x}\cdot\vec{F} = m\int_{\vec{x}_i}^{\vec{x}_j}\,\mathrm{d}\vec{x}\cdot\dot{\vec{v}} = m\int_{\vec{x}_i}^{\vec{x}_j}\,\mathrm{d}t\,\vec{v}\cdot\dot{\vec{v}} = \frac{m|\vec{v}_j|^2}{2}-\frac{m|\vec{v}_i|^2}{2}\,. \end{equation} We define the kinetic energy \(T\) of a body as \begin{equation} T = \frac{m|\vec{v}|^2}{2}\,; \end{equation} the work is then equal to the difference in kinetic energy: \(W_{ij} = T_j-T_i\).
As we discussed in the previous chapter, the gravitational field is given by the gradient of a scalar function, the gravitational potential (see Equation 2.1). In this case, the work can also be written as \begin{equation} W_{ij} = \int_{\vec{x}_i}^{\vec{x}_j}\,\mathrm{d}\vec{x}\cdot\vec{F} = -m\int_{\vec{x}_i}^{\vec{x}_j}\,\mathrm{d}\vec{x}\cdot\nabla\Phi = m\Phi(\vec{x}_i)-m\Phi(\vec{x}_j)\,,\label{eq-work-potential-conservative} \end{equation} where the last equality follows from the gradient theorem from Equation (B.7) and assumes that the force is independent of time; the gradient theorem implies that the work is independent of the path taken from \(\vec{x}_i\) to \(\vec{x}_j\) and in this case the force is conservative. We then define the potential energy of a body in a gravitational field as \(V = m\Phi\). The specific potential energy—the potential energy per unit mass—is simply equal to the gravitational potential.
Equating the expressions for the work in terms of the kinetic and potential energies, we have that \begin{equation} T_i + V(\vec{x}_i) = T_j + V(\vec{x}_j)\,. \end{equation} The sum of the kinetic and potential energies of a body is the energy: \begin{equation} E = T + V\,. \end{equation} The energy is conserved when the force is conservative. When the force explicitly depends on time, Equation (3.9) only holds for instantaneous changes in position, that is, for constant time, and energy conservation therefore does not hold at different times. To see what happens to the energy, we can simply explicitly compute its time derivative \begin{align}\label{eq-energy-conserved} \dot{E} & = m\vec{v}\cdot\dot{\vec{v}} + m\nabla \Phi(\vec{x})\,\dot{\vec{x}}+m\frac{\partial \Phi(\vec{x})}{\partial t}= m\frac{\partial \Phi(\vec{x})}{\partial t}\,, \end{align} because \(m\nabla \Phi(\vec{x})\,\dot{\vec{x}} = -m\dot{\vec{v}}\cdot\vec{v}\). Thus, the total time derivative of the energy is mass times the partial time derivative of the gravitational potential. For a static potential, we recover the expected result that the energy is conserved.
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