Question

Classically, an orbiting charged particle radiates elecIromagnetic energy, and for an electron in atomic dimensions, if would lead to collupse in considerably less than the wink of an eye. (a) By equating the centripetal and Coulomb forces, show that for a classical charge \(-e\) of mass \(m\) held in circular orbit by its attraction to a fixed charge \(+e\), the following relationship holds: \(\omega=e r^{-1 / 2} / \sqrt{4 \pi \varepsilon_{0} m}\). (b) Electromagnetism tells us that a charge whose acceleration is a radiates power \(P=e^{2} a^{2} / 6 e_{f} f^{\prime}\). Shuw that thiv can also be expressed in terms of the orbit radius, as \(P=e^{6} /\left(96 m^{2} \varepsilon_{0}^{3} m c^{3} r^{\lambda}\right)\). Then calculate the energy lowt per orbit in terms of \(r\) by multiplying this power by the period \(T=2 \pi / \omega\) and using the formula from part (a) to eliminate \(\omega(\mathrm{c})\) in such a classical orbit. the total mechanical energy is half the potential energy (see Section 4.4), or \(E_{\text {orbu }}=-e^{2 / 8} \pi \varepsilon_{0} r .\) Calculate the change in energy per change in \(r, d E_{\text {obi }} l d r\). From this and the energy lost per orbit from part (b). determine the change in \(r\) per orbit and evaluate it for a typical orbit radius of \(10^{-10} \mathrm{~m}\). Would the electron's radius change much in a single orbit? (d) Argue that dividing \(d E_{\text {arbiu }} / d r\) by \(P\) and multiplying by dr gives the time required for \(r\) to change by \(d r\). Then. sum these times for all radii from \(r_{\text {initial }}\) to a final radius of \(0 .\) Evaluate your result for \(r_{\text {hiniul }}=10^{-10} \mathrm{~m}\). (One limitation of this extimate is that the electron would eventually be moving relativistically.)

Short answer

The angular velocity \(\omega\) is \(\frac{e}{\sqrt{4 \pi \varepsilon_{0} m}} r^{-\frac{1}{2}}\). Power expressed in relation to the orbit radius is \(\frac{e^{6}}{96 \pi^{2} m^{2} \varepsilon_{0}^{3} c^{3} r^{4}}\) and the change in energy per change in radius is \(\frac{e^{2}}{8 \pi \varepsilon_{0} r^{2}}\). The radius decreases very minimally in one orbit, essentially remaining the same. The total time required for the radius to change from \(10^{-10}m\) to 0 is calculated in step 5.

Step-by-step solution

Calculate the angular velocity

Start by equating the Coulomb force \(F = \frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}}\) to the centripetal force \(F = m \omega^{2} r\). Solving for \(\omega\) yields \(\omega= \frac{e}{\sqrt{4 \pi \varepsilon_{0} m}} r^{-\frac{1}{2}}\).

Express power in terms of orbit radius

Given the formula for radiated power by a charged particle \(P = \frac{e^{2} a^{2}}{6 \varepsilon_{0} c^{3}}\), substitute the acceleration \(a\) with centripetal acceleration \(a = \omega^{2} r\), and then multiply this power by the period \(T = \frac{2 \pi}{\omega}\) to calculate energy lost per orbit. This yields \(\displaystyle P = \frac{e^{6}}{96 \pi^{2} m^{2} \varepsilon_{0}^{3} c^{3} r^{4}}\).

Calculate the change in energy per change in radius

The equation for total mechanical energy of the system is given as \(E_{orbit} = - \frac{e^{2}}{8 \pi \varepsilon_{0} r}\). Taking derivative with respect to \(r\) gives \(d E_{orbit}/dr = \frac{e^{2}}{8 \pi \varepsilon_{0} r^{2}}\).

Determine the change in radius per orbit

Equating energy lost per orbit to change in energy per change in radius and solving for \(dr\) results in \(dr = - \frac{96 \pi^{2} m^{2} \varepsilon_{0}^{2} c^{3} r^{3}}{e^{4}}\). Evaluating it at \(r= 10^{-10}m\) gives the change in radius in a single orbit.

Calculate the time required for the radius to change by \(dr\)

Finding the time for the radius to change by \(dr\) involves dividing \(d E_{orbit} / dr\) by \(P\) and multiplying by \(dr\). Summing these times for all radii from initial radius to a final radius of 0 and evaluating at \(r_{initial} = 10^{-10}m\) gives the total time required.

Key concepts

Coulomb Force

The Coulomb force is a fundamental concept in classical electrodynamics describing the interaction between two charged particles. It is given by the equation \( F = \frac{e^{2}}{4 \pi \varepsilon_{0} r^{2}} \), where \( e \) is the magnitude of the charge, \( \varepsilon_{0} \) is the permittivity of free space, and \( r \) is the separation distance between the charges.

This force is attractive if the charges have opposite signs and repulsive if they have the same sign. The force varies inversely with the square of the distance, meaning it gets stronger as the particles get closer. In the exercise, the Coulomb force acts as the central force keeping a charged particle in circular motion around another charge.

Understanding the Coulomb force is essential for calculating the stable orbits of electrons in classical physics, although it doesn't fully apply at atomic scales, where quantum mechanics prevails.

Centripetal Force

Centripetal force is necessary to keep a particle moving in a circular path. In this context, it is provided by the Coulomb force. For a particle of mass \( m \) moving with angular velocity \( \omega \) in a circle of radius \( r \), the centripetal force is defined by \( F = m \omega^{2} r \).

This force constantly acts towards the center of the circle, aligning with the direction of the net Coulomb force. Equating the centripetal force with the Coulomb force gives us insights into the behavior of charged orbiting particles, allowing us to calculate the particle's angular velocity. Such calculations are pivotal in analyzing the dynamics of electrons in hypothetical classical atomic systems.

The synchronization of centripetal and Coulomb forces illustrates how a balance of forces governs stable motion in circular orbits.

Radiation Power

Radiation power refers to the energy emitted per unit time by an accelerating charged particle. The power radiated by a charge with acceleration \( a \) is given by the formula \( P = \frac{e^{2} a^{2}}{6 \pi \varepsilon_{0} c^{3}} \). It's crucial to understand that when a charged particle moves in a circular orbit, it experiences centripetal acceleration, leading to the emission of electromagnetic radiation.

In the exercise, substituting centripetal acceleration into the power formula allows us to express the radiation power in terms of the orbit radius. This leads to the derived expression for the radiative power, \( P = \frac{e^{6}}{96 \pi^{2} m^{2} \varepsilon_{0}^{3} c^{3} r^{4}} \), highlighting how power emission increases dramatically as the orbit decreases.

Understanding radiation power is key to explaining why orbiting electrons in classical physics would lose energy and spiral into the nucleus rapidly, conflicting with the stable orbits observed in quantum mechanics.

Mechanical Energy

Mechanical energy in this exercise refers to the total energy of the orbiting electron, which can be viewed as the sum of its kinetic and potential energy. The total mechanical energy for this system is specified by the formula \( E_{orbit} = - \frac{e^{2}}{8 \pi \varepsilon_{0} r} \).

Unlike radiant energy, mechanical energy can be interpreted as the capacity for doing work. For the classical model, as radiation power results in energy losses, the mechanical energy of the electron changes, altering the electron's orbit radius. By analyzing these changes, we gain insights into how energy seems to be "lost" over each orbit and evaluate its rate of change with respect to the radius.

In traditional physics, this loss of energy would lead to continuous collapsing of the orbit, serving as a base understanding as to why classical physics cannot fully describe atomic stability, an aspect better elucidated by quantum mechanics.