# Kinetic and Radiative Energy of a Falling Electron

A charged particle that is accelerating emits radiation. Through the study of electrodynamics we often encounter moving charged particles without considering the issues of radiation. For example, the energy of a collection of charges may be calculated in part by solving for the energy involved in bringing those charges together from infinity. Radiation that they may emit along their journey is not part of the study. Not all moving charges are accelerating, however, as in the case of current flowing in a wire. Here, the motion of the electrons is modeled as a constant velocity, the drift velocity, through the wire. This is a zero acceleration case and there is no radiation from these electrons.

While the above is an example from each case, radiative and non-radiative, the vast majority of discussions in electrodynamics are actually about electrostatics since radiation is not considered. Whether this is a valid approximation will be discussed in this topic. We will examine the case of an electron undergoing a well known acceleration and then determine the energy involved in its radiation.

Consider an electron that is released from a zero velocity state (i.e. from rest) and allowed to fall to the Earth. Let this electron fall a distance of one meter under only the influence of gravity. Let us determine how much energy this electron will radiate and whether that should have any effect on the kinematics of its fall. The setup is shown very simply in figure 1.

Figure 1: Setup for this topic. The electron, a blue sphere, is held fixed at a position 1 m above the reference. When released, this electron will accelerate toward the ground and emit radiation.

This topic is essentially a one dimensional system. The kinematic equations for a one dimensional system (coordinates given by y, which is chosen because this is a vertical system) in which the acceleration (a) is constant are,

$v - v_0 = at \quad\text{Eq. 1}$

$y - y_0 = v_0t + \frac{1}{2}at^2 \quad\text{Eq. 2}$

where v is the velocity of the object, t is time, and a subscript 0 denotes the initial value of some variable.

One physical concept immediately presents itself. The expressions in Eqs. 1 and 2 assume that all of the kinetic potential energy, the gravitational potential energy in this case, is converted into kinetic energy. If the object converts its gravitational potential energy into other forms of energy, such as radiation, then the kinematic expressions will not accurately describe the object’s motion. From an experimentalist’s point of view we could set up a measurement of the electron’s motion during a one meter fall and then compare it to the motion predicted by the kinematic equations. The theoretical equivalent of this is to calculate the energy converted into radiation for an electron and then compare it to the kinetic energy.

If some of the gravitational potential energy of the electron is converted to radiative energy, then that means less is converted to kinetic energy. If less is converted to kinetic energy, then the particle will not be moving as fast as predicted by the kinetic equations. The kinetic energy, Wk, of the electron is given by,

$W_k = \frac{1}{2}m_ev^2 \quad\text{Eq. 3}$

where me is the electron mass. Radiative losses mean that Wk will be lower than expected, leading to the result that the electron’s velocity must also be lower†. The kinetic equations must therefore represent an upper bound to the velocity of the electron in which the actual velocity after accounting for radiative losses would be smaller. Our theoretical treatment may therefore consist of applying the kinetic equations to determine the maximum kinetic energy of the electron in the absence of radiation, and then compare that value to the radiation energy of the electron (which is determined solely by the acceleration due to the gravitational field of the Earth).

Put another way, the maximum possible kinetic energy for the electron is determined by converting all of its gravitational potential energy into kinetic energy. The change in potential energy, ΔUg over the distance of the fall is given by,

$\Delta U_g = m_egy_f - m_egy_0$

$\Delta U_g = m_eg\left(y_f - y_0 \right)$

$\Delta U_g = m_eg\Delta y$

where yf is the final location of the electron and g is the acceleration due to gravity. The total distance fallen, Δy, is the one meter value that we just chose for this topic. This leads to a kinetic energy of,

$W_k = \Delta U_g = m_eg\Delta y$

$W_k = \left(9.11\times 10^{-31} [\text{kg}] \right) \left(9.8 [\text{m/s}^2] \right) \left(1 [\text{m}] \right)$

$W_k = 8.928 \times 10^{-30} \;[\text{J}]$

where the units of each variable is given in square brackets. This is an incredibly small amount of energy. Electrons have little mass so this is reasonable.

The next issue is to calculate the amount of energy converted to radiation by the electron. While the energy of radiation is not a commonly referenced expression, the power, P, radiated by a single charged particle undergoing acceleration is often used. This is known as the Larmor formula and is given by,

$P = \frac{\mu_0q^2a^2}{6\pi c}$

where μ0 is the permeability of free space, q is the charge of the particle, and c is the speed of light. In this instance the acceleration is that of the Earth’s gravitational field, a = g. The Larmor formula is valid for velocities that do not approach the speed of light (v << c). We know that is the case here by appealing to our conceptual understanding of physics. Since all objects are affected by the gravitational field in the same manner, the final velocity of the electron is the same as our own velocity would be if we fell a distance of one meter from rest. Jumping off of a ledge of this height is possible, and while the landing may be uncomfortable we have certainly not reached a velocity comparable to that of light. The electron’s velocity is well below the speed of light, just as ours.

The power radiated by the falling electron is,

$P = \frac{\left(4\pi \times 10^{-7} \;\text{[H/m]} \right)\left(1.602\times 10^{-19} \;\text{[C]} \right)^2 \left(9.8 \;[\text{m/s}^2] \right)^2}{6\pi \left(2.998 \times 10^8 \;\text{[m/s]} \right)}$

$P = \frac{3.097 \times 10^{-42}}{5.651 \times 10^{9}}\; \text{[W]}$

$P = 5.480 \times 10^{-50}\; \text{[W]}$

where the power is given in units of Watts [W], which are equivalent to energy per unit time (W ≡ E / t). By determining the time it takes the electron to fall the prescribed distance of one meter, we will be able to determine the total amount of energy that it has radiated away. Conceptually, be mindful that the electron radiates based on its acceleration, which is constant as it falls in the gravitational field. The electron emits radiation at a constant power during its fall. Knowing the total time of the fall allows us to calculate the total energy it radiated because it did so at a constant rate.

To determine the amount of time for which the electron fell, we return to the kinetic equations. In this case our approximations work against us. Since the kinetic equations provide the maximum velocity of the electron it is reasonable to assume that they similarly provide the minimum time for this fall. That would decrease the theoretical radiative energy. Being aware of that we should remember that our theoretical experiment here is to determine whether the radiative energy is anywhere close to the magnitude of the kinetic energy. If they are close then we must revisit all of this physics. If they are not comparable, then these corrections must be minor and generally ignorable.

The time of fall for the electron can be found in the expressions at the beginning of this problem. The initial velocity is zero as given in the setup.

$y - y_0 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2 \quad\text{since} \;v_0=0$

$\Delta y = \frac{1}{2}gt^2$

where there should be a negative sign on the left side (y – yo = yf – yo = -Δy), but that cancels with the negative sign of the acceleration. For a coordinate system in which the y = 0 position is below the starting position these negative values are present and cancel. We could have defined the zero position to be the starting point of the electron’s fall with it falling to larger, positive values of y. In that case the gravitational acceleration would also have been positive. Either way the coordinates are defined, since time has to be positive as the final result we can be sure we have this issue correct.

The time of fall for the electron is,

$\Delta y = \frac{gt^2}{2}$

$1 \;\text{[m]} = \frac{1}{2}\left( 9.8 \;[\text{m/s}^2]\right) t^2$

$0.204 \;[\text{s}]^2 = t^2$

$t = 0.452 \;\text{[s]}$

and this is not large enough to make the total energy radiated comparable to the kinetic energy.

The energy radiated, Wr, by the electron during this fall is,

$W_r = Pt$

$W_r = \left(5.480 \times 10^{-50} \;\text{[W]} \right)\left(0.452\;\text{[s]} \right)$

$W_r = 2.477 \times 10^{-50} \;\text{[J]}$

and now we may compare the different energy values.