332. The Yukawa Particle and the Heisenberg Uncertainty Principle RevisitedLearning Objectives
Particle physics as we know it today began with the ideas of Hideki Yukawa in 1935. Physicists had long been concerned with how forces are transmitted, finding the concept of fields, such as electric and magnetic fields to be very useful. A field surrounds an object and carries the force exerted by the object through space. Yukawa was interested in the strong nuclear force in particular and found an ingenious way to explain its short range. His idea is a blend of particles, forces, relativity, and quantum mechanics that is applicable to all forces. Yukawa proposed that force is transmitted by the exchange of particles (called carrier particles). The field consists of these carrier particles. Specifically for the strong nuclear force, Yukawa proposed that a previously unknown particle, now called a $$\text{\Delta}E\text{\Delta}t\ge \frac{h}{4\pi}\text{,}$$ where $h$ is Planck’s constant. Therefore, conservation of massenergy can be violated by an amount $\text{\Delta}E$ for a time $\text{\Delta}t\approx \frac{h}{\mathrm{4\pi \Delta}E}$ in which time no process can detect the violation. This allows the temporary creation of a particle of mass $m$, where $\text{\Delta}E={\mathrm{mc}}^{2}$. The larger the mass and the greater the $\text{\Delta}E$, the shorter is the time it can exist. This means the range of the force is limited, because the particle can only travel a limited distance in a finite amount of time. In fact, the maximum distance is $d\approx c\text{\Delta}t$, where c is the speed of light. The pion must then be captured and, thus, cannot be directly observed because that would amount to a permanent violation of massenergy conservation. Such particles (like the pion above) are called Example 1: Calculating the Mass of a PionTaking the range of the strong nuclear force to be about 1 fermi (${\text{10}}^{\text{15}}\phantom{\rule{0.25em}{0ex}}\text{m}$), calculate the approximate mass of the pion carrying the force, assuming it moves at nearly the speed of light. Strategy The calculation is approximate because of the assumptions made about the range of the force and the speed of the pion, but also because a more accurate calculation would require the sophisticated mathematics of quantum mechanics. Here, we use the Heisenberg uncertainty principle in the simple form stated above, as developed in Probability: The Heisenberg Uncertainty Principle. First, we must calculate the time $\text{\Delta}t$ that the pion exists, given that the distance it travels at nearly the speed of light is about 1 fermi. Then, the Heisenberg uncertainty principle can be solved for the energy $\text{\Delta}E$, and from that the mass of the pion can be determined. We will use the units of $\text{MeV}/{c}^{2}$ for mass, which are convenient since we are often considering converting mass to energy and vice versa. Solution The distance the pion travels is $d\approx c\mathrm{\Delta}t$, and so the time during which it exists is approximately $$\begin{array}{lll}\text{\Delta}t& \approx & \frac{d}{c}=\frac{{\text{10}}^{\text{15}}\phantom{\rule{0.25em}{0ex}}\text{m}}{3\text{.}0\times {\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}\\ & \approx & 3.3\times {\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{s.}\end{array}$$ Now, solving the Heisenberg uncertainty principle for $\text{\Delta}E$ gives $$\text{\Delta}E\approx \frac{h}{4\pi \mathrm{\Delta}t}\approx \frac{6\text{.}\text{63}\times {\text{10}}^{\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot \text{s}}{\mathrm{4\pi}\left(3\text{.}3\times {\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{s}\right)}\text{.}$$ Solving this and converting the energy to MeV gives $$\text{\Delta}E\approx \left(1\text{.}6\times {\text{10}}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{J}\right)\frac{1\phantom{\rule{0.25em}{0ex}}\text{MeV}}{1\text{.}6\times {\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}}=\text{100}\phantom{\rule{0.25em}{0ex}}\text{MeV}\text{.}$$ Mass is related to energy by $\text{\Delta}E={\text{mc}}^{2}$, so that the mass of the pion is $m=\text{\Delta}E/{c}^{2}$, or $$m\approx \text{100}\phantom{\rule{0.25em}{0ex}}\text{MeV/}{c}^{2}\text{.}$$ Discussion This is about 200 times the mass of an electron and about onetenth the mass of a nucleon. No such particles were known at the time Yukawa made his bold proposal. Yukawa’s proposal of particle exchange as the method of force transfer is intriguing. But how can we verify his proposal if we cannot observe the virtual pion directly? If sufficient energy is in a nucleus, it would be possible to free the pion—that is, to create its mass from external energy input. This can be accomplished by collisions of energetic particles with nuclei, but energies greater than 100 MeV are required to conserve both energy and momentum. In 1947, pions were observed in cosmicray experiments, which were designed to supply a small flux of highenergy protons that may collide with nuclei. Soon afterward, accelerators of sufficient energy were creating pions in the laboratory under controlled conditions. Three pions were discovered, two with charge and one neutral, and given the symbols ${\pi}^{+}\text{,}\phantom{\rule{0.25em}{0ex}}{\pi}^{}\text{, and}\phantom{\rule{0.25em}{0ex}}{\pi}^{0}$, respectively. The masses of ${\pi}^{+}$ and ${\pi}^{}$ are identical at $\text{139}\text{.}6\phantom{\rule{0.25em}{0ex}}\text{MeV/}{c}^{2}$, whereas ${\pi}^{0}$ has a mass of $\text{135}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{MeV/}{c}^{2}$. These masses are close to the predicted value of $\text{100}\phantom{\rule{0.25em}{0ex}}\text{MeV/}{c}^{2}$ and, since they are intermediate between electron and nucleon masses, the particles are given the name The pions, or $\pi $mesons as they are also called, have masses close to those predicted and feel the strong nuclear force. Another previously unknown particle, now called the muon, was discovered during cosmicray experiments in 1936 (one of its discoverers, Seth Neddermeyer, also originated the idea of implosion for plutonium bombs). Since the mass of a muon is around $\text{106}\phantom{\rule{0.25em}{0ex}}\text{MeV/}{c}^{2}$, at first it was thought to be the particle predicted by Yukawa. But it was soon realized that muons do not feel the strong nuclear force and could not be Yukawa’s particle. Their role was unknown, causing the respected physicist I. I. Rabi to comment, “Who ordered that?” This remains a valid question today. We have discovered hundreds of subatomic particles; the roles of some are only partially understood. But there are various patterns and relations to forces that have led to profound insights into nature’s secrets. Summary
Problems & ExercisesExercise 1A virtual particle having an approximate mass of ${\text{10}}^{\text{14}}\phantom{\rule{0.25em}{0ex}}\text{GeV/}{c}^{2}$ may be associated with the unification of the strong and electroweak forces. For what length of time could this virtual particle exist (in temporary violation of the conservation of massenergy as allowed by the Heisenberg uncertainty principle)? Show/Hide Solution Solution$3\times {\text{10}}^{\text{39}}\phantom{\rule{0.25em}{0ex}}\text{s}$ Exercise 2Calculate the mass in $\text{GeV/}{c}^{2}$ of a virtual carrier particle that has a range limited to ${\text{10}}^{\text{30}}$ m by the Heisenberg uncertainty principle. Such a particle might be involved in the unification of the strong and electroweak forces. Exercise 3Another component of the strong nuclear force is transmitted by the exchange of virtual Kmesons. Taking Kmesons to have an average mass of $\text{495}\phantom{\rule{0.25em}{0ex}}\text{MeV/}{c}^{2}$, what is the approximate range of this component of the strong force? Show/Hide Solution Solution$1\text{.}\text{99}\times {\text{10}}^{\text{16}}\phantom{\rule{0.25em}{0ex}}\text{m}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}(0\text{.}2\phantom{\rule{0.25em}{0ex}}\text{fm})$
