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Ludwik Kowalski Montclair State University, Montclair, NJ, 07055 In December of 2007 this note was submitted to American Journal of Physics. But it was rejected, after an interesting exchange of messages between me and the editor of
the journal. The issue had nothing to do with Figier’h hypothesis; our discussion focused on Section II, more specifically, on the issue of stability of my macroscopic
three-body system. I plan to discuss this topic (Section II only) on a forum for physics teachers. The outcome of that discussion, if any, will be appended at the end of
this unit. Figure 1 Caption: A three-body system to be examined in this note. Emission of electromagnetic radiation by centripetally accelerating charges will be ignored. = = = = = = = = = = = = = = = = = = = = = = = = For the set of three electric charges shown in Figure 1, the potential energy is: U = k*Q ^{2} / (2*R) - k*Q^{2} / R - k*Q^{2} / R =
-(3/2)*k*Q^{2}/R (1)where Q is the magnitude of each charge, k is the Coulomb law constant and R is the orbit radius. Negative terms prevail over the positive term and U is always negative. The central particle is at rest and the combined kinetic energy is K=m*v ^{2}The net force on each positive particle, according to Coulomb’s law, is 0.75*k*Q ^{2} / R^{2}. The speed can thus be calculated by writing 0.75*k*Q ^{2} / R^{2} = m*v^{2}^{ }/ R (2)This also shows that K = 0.75*k*Q ^{2}/R (3)The total relevant energy is E = U + K = - (3/4)*k*Q ^{2}/R (4)The angular momentum of our simple system is L = 2*m*v*R (5) A numerical illustration: Suppose that the mass of each particle is 1 kg and the magnitude of each charge is 1 C. What should the speeds of positive particles be if the desired radius of the
circular orbit is 100 meters? The answer, according to Formula (2) is 8216 m/s. Our use of non-relativistic formulas is justified because v is much smaller than the speed
of light. Note that the answer does not depend on the mass of the central particle. The potential and kinetic energies of the system, according to Formulas (1) and (3),
are -13.5*10^{7} and 6.75*10^{7} joules, respectively. The total relevant energy, according to Formula (4) is
-6.75*10^{7} joules. The angular momentum of the system, according to Formula (5), is 1.64*10^{6} J*s; this is
about 39 orders of magnitude larger than Planck’s constant.The negative sign indicates that the system is bound in two dimensions. But is it also bound in three dimensions? The answer seems to be positive. To justify this answer, consider the configuration shown in Figure 2. The horizontal segment is the side view of the orbit shown in Figure 1. The cross represents the location of the central charge resulting from a virtual displacement. The distances between charges become d _{1}, d_{2} and 2*R. The potential energy of the new configuration isU' = k*Q ^{2} / (2*R) - kQ^{2}/d_{1} - k*Q^{2}/
d_{2}_{
}Figure 2 Caption: What happens to potential energy of the system when the central particle is virtually displaced in the the direction perpendicular to the orbit, either up or down? It becomes higher (less negative). = = = = = = = = = = = = = = = = = = = = = = = = The positive term does not change but the sum of negative terms becomes smaller. In other words, U' is higher (less negative) than our previous U. A vertical displacement of the central charge by 10 meters, either up or down, corresponds to the increase of potential energy by 9*10 ^{5} joules. A
diagram similar to that in Figure 2 can be sketched for a positive charge being displaced perpendicularly with respect to the horizontal orbit. The potential energy for
that configuration, U", also turns out to be less negative that what one finds when the displacement is zero. For a vertical displacement of 10 meters, for example,
the value of U" is by 4.5*10^{5} joules higher than U. It is a numerical problem and numerical illustrations are appropriate. III. ELEMENTARY QUANTUM MECHANICAL CONSIDERATIONSSome students might be familiar with elementary quantum mechanics (Bohr's angular momentum postulate and Heisenberg’s Uncertainty Principle). These conceptual
tools can be used to analyze a hypothetical submicroscopic system in which two protons orbit an electron. In a classroom debate, students can be guided to show that such
a system could not exist. The set of formulas to be used in such activity is show below.According to Bohr’s angular momentum postulate, L=n * h (6) where h is the Planck's constant divided by 2*π, and n is a quantum number for the orbit. Find the smallest R for the p-e-p system (two protons orbiting an electron). The method introduced by Bohr can be used to discuss this hypothetical system. Equating right sides of the Equations (5) and (6) one has v = h*n / (2*m*R) . . . (7) Substituting this into the Formula (2) gives R = h ^{2}*n^{2} / (3*k*e^{2}*m) . . . . . (8)The energy for that orbit can be calculated by substituting this radius into Formula (4). The result is E = - 2.25*m*k ^{2}*Q^{4} / (h^{2}*n^{2}) . . . . (9)For each proton m=1.6710 ^{-27} kg and Q=+1.6*10^{-19} C. The charge of the electron is
-1.6*10^{-19} C. Knowing that h=1.05*10^{-34} and k=9*10^{9}, both in SI units, one can calculate R, v and E. For the ground state orbit (n=1) the result are: R=9.55*10 ^{-15} m (9.55 fm), v=3.29*10^{6}
m/s (1.1% of the speed of light) and E=-1.81*10^{-14} J (113 keV). Note that for larger n the radius becomes n^{2}
times larger than for the ground state while the absolute value of E, not surprisingly, becomes n^{2} times smaller (E approaches zero when n
becomes very large). In other words, according to the semi-classical Bohr’s theory, the suggested p-e-p structure would be much smaller than the hydrogen atom. The
binding energy of that system (the absolute value of E at n=1) would be much larger than for the hydrogen atom (where it is only 13.6 eV).It is interesting that the distance between protons, on the ground state orbit, 19.1 fm, is still much larger than the range of the strong nuclear force (~2 fm). Only electrical forces are of any significance in this problem, as initially assumed. Also note that the energy at n=2 is -113/4=-28.2 keV. The difference, 84.7 keV, corresponds to the lowest energy photon in the Balmer series. For hydrogen, spectral lines of that series are in the ultraviolet region; for the p-e-p system they would be in the X-ray region. Speculations about significance of such findings can be educationally valuable. The above considerations could naturally lead to Heisenberg’s Uncertainty Principle, dp*dr ~ h. Is that principle (HUP) consistent with the existence of the conceived p-e-p structure? The answer turns out to be negative. To demonstrate this numerically one could begin with E=-1.81*10 ^{-14}
J (113 keV), as above. That represents the depth of the ground state potential energy well. According to initial consideration (see Figures 1 and 2) the electron was assumed
to be at rest in the center. That implies that both location and momentum are accurately specified (r=0 and p=0). But this is impossible, according to HUP. By decreasing the
uncertainty in r we are always increasing the uncertainty in p, and vice versa. The maximum possible uncertainty (fuzziness) in p can be estimated from the depth of the well. An electron, randomly oscillating about the center, would escape from the well if its kinetic energy, K, were larger than the depth. This gives us the maximum possible K = 1.81*10 ^{-14} J. The corresponding linear momentum, p,
for an electron, is 1.81*10^{-22} kg*m/s. Using that value in HUP one has dr=5.80*10^{-13} m. This is 61 times larger
than the above-calculated radius of the p-e-p structure. The initial assumption (electron being confined to a small sphere at the center of the orbit) is in conflict with HUP.
It is easy to show that, in order to be consistent with HUP, the mass of the electron would have to be at least as large as the mass of an alpha particle. Motivation for
this note, by the way, was provided by such discussion among several researchers. The idea of a tiny particle in which two protons orbit an electron was actually suggested
by Jean-Pierre Vigier (5), a French theoretical physicist who worked with Louis de Broglie. No one seriously claimed that such particles exist in nature. But theoretical
speculations about them can be pedagogically instructive. IV. CONCLUSION a) Linking of similar classical and non-classical problems has obvious pedagogical advantages in undergraduate physics courses.b) The idea of the possible existence of a stable p-e-p structure, conceived and published by Vigier (5), offers a rare case in which an interesting hypothesis can be numerically investigated by using elementary QM tools. Students familiar with such tools are likely to benefit from a properly structured discussion of Vigier's hypothesis. Note that the hypothetical p-e-p system is similar, but not identical, to the neutral p-e system, analyzed by Bohr. c) The mass of the central particle, totally irrelevant in the macroscopic problem, becomes highly relevant in the analogous submicroscopic one. The idea of hidden parameters is interesting. V. ACKNOWLEDGMENTContributions from Scott Little (USA) and Akito Takahashi (Japan) are highly appreciated. Information about Vigier was provided by Jean-Paul Biberian (France) and by Andrew Meulenberg (USA). Should I interest comments about Bohr’s model and HUP, found in (6) ? References:1) Paul A. Tipler and Gene Mosca, “Physics for scientists and engineers;” 5th edition, W.H. Freeman and Company, New York, 2004 (pages 1173-1178). 2) Hugh D. Young and Roger A. Freeman, “Sears and Zemansky’s university physics with modern physics,” tenth edition. Addison-Wesley, San Francisco, 2000 (1246-1251) 3) Arthur Beiser, “Concepts of modern physics,” fifth edition, McGraw-Hill, Inc., New York, 1995 (129-141) 4) T. R. Sandin, “Essentials of modern physics,” Addison-Wesley Publishing Company, Reading, Massachustess, 1989 (123-143) 5) Dragic A., Maric Z. and Vigier J.P,; “New quantum mechanical tight bond states and ‘cold fusion’ experiments;” Phys. lett. A 2000, 265, page 163. 6) Yalendra Pal Varshni, “The uncertainty principle and the Bohr theory of the hydrogen atom,” Am. J. Phys. (22), 1954,568-569. VI. APPENDIX I was not aware, when the above was composed, that the proposed classical system would be highly unstable. In real life, unlike in idealized world of mathematical ideas and simulations, inevitable small perturbations would quickly destroy the system. This would become visible in less than one period of one idealized (perturbation-free) revolution. The corresponding two-body system, like a star and a planet, is expected to be stable under such perturbations. But this is not true for the orbiting three-body system, like three stars, or three charged macroscopic spheres. This appendix was inspired by messages posted by several physics teachers during an interesting Internet debate about stability and chaos. = = = = = = = TO BE SHOWN LATER = = = = = = = Click to see the list of links |