Appendix 11 — Considerations on Rotation
Appendix 11.1 — Introduction
In this appendix, we provide an explanation of centrifugal and centripetal forces, first based on classical Newtonian mechanics, and later extend it to general relativity.
The centrifugal force is the apparent force acting outward from the center of rotation. The centripetal force is the real force directed toward the center, required to keep a particle in circular motion.
Appendix 11.2 — Momentum
According to Newton, a moving particle with mass \(m\) and velocity \(v\) has momentum:
\[ \vec{p} = m\vec{v}. \]
If no forces act on the particle, it moves uniformly in a straight line. Relative to a point at distance \(r\), the particle has angular momentum:
\[ \vec{L} = \vec{r} \times \vec{p}. \]
In the figure above, the angular momentum is:
\[ L = m v r \sin\varphi = m v b, \]
where \(b = r\sin\varphi\) is the perpendicular distance between the particle's path and the reference point.
Appendix 11.3 — Circular Motion
As mentioned earlier, a particle moves uniformly in a straight line if no force acts. Thus, if the particle's path is a circle, a force must be present: the centripetal force.
We start with a constant radius \(r\) and decompose the motion into x- and y-components. The position as a function of time is:
\[ x(t) = r\cos\varphi = r\cos(\omega t), \] \[ y(t) = r\sin\varphi = r\sin(\omega t), \]
where \(\omega\) is the angular velocity.
Velocities
\[ v_{x} = \frac{dx}{dt} = -\omega r \sin(\omega t), \qquad v_{y} = \frac{dy}{dt} = \omega r \cos(\omega t). \]Accelerations
\[ a_{x} = \frac{d^{2}x}{dt^{2}} = -\omega^{2} r \cos(\omega t), \qquad a_{y} = \frac{d^{2}y}{dt^{2}} = -\omega^{2} r \sin(\omega t). \]The total acceleration is:
\[ a = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{\omega^{4} r^{2}(\cos^{2}\omega t + \sin^{2}\omega t)} = \omega^{2} r. \]Since both components are negative (directed toward the center), the vector acceleration is:
\[ \vec{a} = -\omega^{2} r\, \hat{r}. \]Force
The total force on the particle is:
\[ \vec{F} = m\vec{a} = -m\omega^{2} r\, \hat{r}. \]The magnitude of the force is:
\[ F = m\omega^{2} r. \]Interpretation
The particle wants to move in a straight line according to the law of inertia. However, circular motion produces an outward-directed apparent force (centrifugal in the rotating frame):
\[ F_{\text{outward}} = m\omega^{2} r. \]To keep the particle on its circular path, this force must be exactly balanced by an inward-directed reaction force:
\[ F_{\text{centripetal}} = -m\omega^{2} r. \]This centripetal force keeps the particle on its circular orbit.
Appendix 11.4 — Rotation of a Sphere
Combined with the centripetal force, this results in the following force on a particle on a rotating sphere:
\[ F_{\text{radial}} = \frac{G m M}{r^{2}} - m \omega^{2} r \sin^{2}(\omega t) = m\left( \frac{G M}{r^{2}} - \omega^{2} r \sin^{2}(\omega t) \right). \]
Additionally, there is a tangential force toward the equator:
\[ F_{\text{tangential}} = m \omega^{2} r \cos(\omega t)\sin(\omega t). \]
Particles therefore experience a force component toward the equator, causing the sphere to deform into an ellipsoid. The distance from the center to the surface is smallest at the poles and largest at the equator, leading to variations in gravity.
Gravity also depends on the enclosed mass. Because the distance to the center at the poles is smaller, the enclosed mass is smaller there. Gravity increases due to the smaller distance but decreases due to the smaller enclosed mass.
The final shape is an ellipsoid where these effects are in balance. See also: Newtonian Gravity — Rotating Bodies .
Appendix 11.5 — Relation Between Angular Momentum and Energy
Difference in Kinetic Energy Between Two Circular Orbits
For a particle moving from a circle of radius \(r_{1}\) to a circle of radius \(r_{2}\), the difference in kinetic energy is:
\[ \Delta K = \frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{2}^{2}. \tag{1} \]
The angular momentum is constant:
\[ m v_{1} r_{1} = m v_{2} r_{2} \quad\Rightarrow\quad v_{2} = v_{1}\frac{r_{1}}{r_{2}}. \tag{2} \]Substitute (2) into (1):
\[ \Delta K = \frac{1}{2} m v_{1}^{2} - \frac{1}{2} m \left( v_{1}\frac{r_{1}}{r_{2}} \right)^{2} = \frac{1}{2} m v_{1}^{2} \left( 1 - \frac{r_{1}^{2}}{r_{2}^{2}} \right). \tag{3} \]Work Done by the Centripetal Force
The centripetal force is:
\[ F = -\, m \frac{v^{2}}{r}. \]The negative sign indicates that the force is directed inward. The work done by this force as the particle moves from \(r_{1}\) to \(r_{2}\) is:
\[ W = \int_{r_{1}}^{r_{2}} F\,dr = \int_{r_{1}}^{r_{2}} \left( -\, m \frac{v^{2}}{r} \right) dr. \]This work must equal the difference in kinetic energy:
\[ W = \Delta K. \]This establishes the connection between the change in kinetic energy and the work of the centripetal force.
The angular momentum is constant:
\[ m v r = \text{Const}. \]Hence: \[ v = \frac{\text{Const}}{m r}. \]
The work done by the centripetal force between \(r_{1}\) and \(r_{2}\) is:
\[ \int_{r_{1}}^{r_{2}} F\, dr = - \int_{r_{1}}^{r_{2}} m \frac{v^{2}}{r}\, dr. \]Substitute \(v = \frac{\text{Const}}{m r}\):
\[ \int_{r_{1}}^{r_{2}} F\, dr = - \int_{r_{1}}^{r_{2}} m \frac{1}{r} \left( \frac{\text{Const}^{2}}{m^{2} r^{2}} \right) dr = - \int_{r_{1}}^{r_{2}} \frac{\text{Const}^{2}}{m r^{3}}\, dr. \]This gives:
\[ \int_{r_{1}}^{r_{2}} F\, dr = \frac{\text{Const}^{2}}{2 m} \left( \frac{1}{r_{1}^{2}} - \frac{1}{r_{2}^{2}} \right). \]Since: \[ \text{Const} = m v_{1} r_{1}, \] we have:
\[ \frac{\text{Const}^{2}}{2 m} = \frac{m^{2} v_{1}^{2} r_{1}^{2}}{2 m} = \frac{1}{2} m v_{1}^{2} r_{1}^{2}. \]Thus:
\[ \int_{r_{1}}^{r_{2}} F\, dr = \frac{1}{2} m v_{1}^{2} \left( 1 - \frac{r_{1}^{2}}{r_{2}^{2}} \right). \tag{4} \]We see that formulas (3) and (4) are identical, so:
\[ \boxed{ \Delta K = \int_{r_{1}}^{r_{2}} F\, dr = \frac{1}{2} m v_{1}^{2} \left( 1 - \frac{r_{1}^{2}}{r_{2}^{2}} \right) } \]This confirms that the difference in kinetic energy exactly corresponds to the work of the centripetal force — a crucial step toward the effective potential in central force fields.