Abstract Generated abstract
The paper studies a spherical wave associated with a scalar plankeon in the curved spacetime of the Einstein universe, using a scalar wave equation with curvature coupling. By separating variables in the metric, the author reduces the equation to a radial ordinary differential equation, gives its general solution in terms of angular momentum quantum numbers, and examines the spherically symmetric case. Relating the wave normalization and the trace of the energy momentum tensor to the curvature yields Planck scale values for the mass and radius, with the radius equal to the gravitational radius. The resulting de Broglie wave is argued to be localized within this radius, suggesting a stable, nonspreading wave packet maintained by its own gravitational field.
Full Text
UDC 530.12
PHYSICS
K. P. STANYUKOVICH
SPHERICAL WAVE OF A SCALAR PLANKEON
(Presented by Academician L. I. Sedov, 23 IV 1969)
The basic equation for studying a spherical wave corresponding to a “scalar” plankeon has the form (^1)
\[ \left(g^{ik}\hat p_i \hat p_k + m^2c^2 + \hbar^2 R/6\right)\psi = 0. \tag{1} \]
The interval corresponding to the “Einstein universe” is
\[ -ds^2 = -c^2 d\tau^2 + \frac{dr^2}{1-r^2/a^2} + r^2 d\Omega^2 . \tag{2} \]
In this case
\[ R = 6/a^2 . \tag{3} \]
Since for the scalar wave function
\[ g^{ik}\hat p_i\hat p_k = -\hbar^2 g^{ik}\nabla_i\nabla_k = -\hbar^2 g^{ik}\left( \frac{\partial^2}{\partial x^i \partial x^k} -\Gamma^m_{ik}\frac{\partial}{\partial x^m} \right), \]
equation (1) takes the form
\[ \left[ g^{ik}\left( \frac{\partial^2}{\partial x^i\partial x^k} -\Gamma^m_{ik}\frac{\partial}{\partial x^m} \right) -\left(\frac{m^2c^2}{\hbar^2}+\frac{1}{a^2}\right) \right]\psi = 0. \tag{4} \]
For the metric (2), the nonzero Christoffel symbols have the form
\[ \Gamma^1_{11}= \frac{r}{a^2(1-r^2/a^2)},\qquad \Gamma^1_{22}=r\left(1-\frac{r^2}{a^2}\right),\qquad \Gamma^1_{33}=r\sin^2\theta\left(1-\frac{r^2}{a^2}\right), \]
\[ \Gamma^2_{12}=\Gamma^3_{13}=1/r,\qquad \Gamma^2_{33}=-\sin\theta\cos\theta,\qquad \Gamma^3_{23}=\operatorname{ctg}\theta . \]
Now (4) can be written in explicit form
\[ \left[ -\frac{\partial^2}{c^2\partial \tau^2} +\left(1-\frac{r^2}{a^2}\right) \left(\frac{\partial^2}{\partial r^2} +\frac{2}{r}\frac{\partial}{\partial r}\right) -\frac{r}{a^2}\frac{\partial}{\partial r} +\frac{1}{r^2} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2} \right] -\left(\frac{m^2c^2}{\hbar^2}+\frac{1}{a^2}\right) \right]\psi = 0. \tag{5} \]
Since
\[ \frac{\partial^2\psi}{\partial r^2} +\frac{2}{r}\frac{\partial\psi}{\partial r} = \frac{1}{r}\frac{\partial^2(r\psi)}{\partial r^2}, \qquad r\frac{\partial\psi}{\partial r} = \frac{\partial(r\psi)}{\partial r}-\psi, \]
then (5) takes the form
\[ \left[ -\frac{\partial^2(r\psi)}{c^2\partial \tau^2} +\left(1-\frac{r^2}{a^2}\right) \frac{\partial^2(r\psi)}{\partial r^2} -\frac{r}{a^2}\frac{\partial(r\psi)}{\partial r} +\frac{r\psi}{a^2} +\frac{1}{r^2} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial(r\psi)}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2(r\psi)}{\partial\varphi^2} \right] -\left(\frac{m^2c^2}{\hbar^2}+\frac{1}{a^2}\right)r\psi \right]=0, \]
or
\[ \left[ -\frac{\partial^2}{c^2\partial \tau^2} +\left(1-\frac{r^2}{a^2}\right)\frac{\partial^2}{\partial r^2} -\frac{r}{a^2}\frac{\partial}{\partial r} +\frac{1}{r^2} \left( \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2} \right) -\frac{m^2c^2}{\hbar^2} \right]\varphi = 0, \tag{6} \]
where \(\varphi=\psi r\).
Set \(r=a\sin\chi\); then (6) takes the form
\[ \left[ -\frac{\partial^2}{c^2\partial \tau^2} +\frac{1}{a^2}\frac{\partial^2}{\partial \chi^2} +\frac{1}{a^2\sin^2\chi} \left( \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) +\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2} \right) -\frac{m^2c^2}{\hbar^2} \right]\varphi=0. \tag{7} \]
The operator
\[ \frac{1}{\sin\theta} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) \right] +\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2} =-\hat l^{\,2}=-l(l+1), \]
where \(l=0,1,2,\ldots\) is the angular momentum (orbital quantum number).
Therefore (7) takes the form:
\[ \left[ -\frac{\partial^2}{c^2\partial \tau^2} +\frac{1}{a^2}\frac{\partial^2}{\partial\chi^2} -\left( \frac{m^2c^2}{\hbar^2} +\frac{l(l+1)}{a^2\sin^2\chi} \right) \right]\varphi=0. \tag{8} \]
Let
\[ \varphi=A(\chi)e^{-\frac{i}{\hbar}E\tau}\mathcal{Y}_{lm}(\theta;\varphi), \tag{9} \]
where
\[ \mathcal{Y}_m=\theta_{lm}e^{im\varphi}; \tag{10} \]
\[ \theta_{lm}= \frac{(-1)^{l+m}}{2^l l!} \sqrt{\frac{(2l+1)(l-m)!}{2(l+m)!}}\, \sin^m\theta\, \frac{d^{\,l+m}\sin^{2l}\theta}{(d\cos\theta)^{l+m}}, \]
\(m=0,\pm1,\pm2,\ldots\) is the magnetic quantum number.
Substituting (9) into (8), we obtain
\[ \left[ \left(\frac{E^2}{c^2}-m^2c^2\right)\frac{a^2}{\hbar^2} -\frac{l(l+1)}{\sin^2\chi} \right]A+A_{\chi\chi}=0. \tag{11} \]
Denote
\[ (E^2/c^2-m^2c^2)a^2/\hbar^2=\xi^2 \tag{12} \]
and instead of \(l(l+1)\) introduce \(n(n-1)=l(l+1)\), whence \(n_1=l+1,\ n_2=-l\); then (11) assumes the standard form
\[ \sin^2\chi A_{\chi\chi}+\left[\xi^2\sin^2\chi-n(n-1)\right]A=0. \tag{13} \]
The general solution of equation (13), as is known, has the form \((^2)\)
\[ A=\sin^n\chi \left(\frac{1}{\sin\chi}D\right)^n \left(\bar A_1 e^{-i\xi\chi}+\bar A_2 e^{i\xi\chi}\right), \tag{14} \]
where
\[ \left(\frac{1}{\sin\chi}D\right)^n = \frac{1}{\sin\chi}\frac{d}{d\chi} \left[ \frac{1}{\sin\chi}\frac{d}{d\chi} \left( \frac{1}{\sin\chi}\frac{d}{d\chi}(\cdots) \right) \right]. \tag{15} \]
Thus we find that
\[ \psi= \frac{\bar A_1}{r}\theta_{lm}e^{-\frac{i}{\hbar}E\tau+im\varphi} \sin^{l+1}\chi \left(\frac{1}{\sin\chi}D\right)^{l+1} e^{-i\xi\chi} = \frac{\varphi}{r}, \]
\[ \psi^{*}= \frac{\bar A_2}{r}\theta_{lm}^{*}e^{\frac{i}{\hbar}E\tau-im\varphi} \sin^{-l}\chi \left(\frac{1}{\sin\chi}D\right)^{-l} e^{-i\xi\chi} = \frac{\varphi^{*}}{r}, \tag{16} \]
where \(\xi=pa/\hbar,\ p=\sqrt{E^2/c^2-m^2c^2}\) is the momentum.
For \(l=0,\ m=0\ (n=0)\) we shall have
\[ \psi=\frac{A_1}{r}e^{-\frac{i}{\hbar}(E\tau+pa\chi)},\qquad \psi^{*}=\frac{A_2}{r}e^{\frac{i}{\hbar}(E\tau+pa\chi)}. \tag{17} \]
The trace of the energy–momentum tensor corresponds to the given wave equation (1) and has the form \((^3)\)
\[ -T=\frac{m^2c^2}{\hbar^2}\langle \psi\psi^{*}\rangle. \tag{18} \]
If we choose \(A_1A_2=\beta\dfrac{m^2c^3a^2}{\hbar}\), where \(\beta=\mathrm{const}\) is determined below, then \(\nabla\psi\nabla\psi^*\) will have the dimension of energy density (as it should). In this case
\[ \langle\psi\psi^*\rangle =\frac{8\pi\beta m^2c^3a^2}{\pi^2a^3\hbar} \int_0^a \frac{dr}{\sqrt{1-r^2/a^2}} =\frac{8m^2c^3\beta}{\pi\hbar}\chi\Bigg|_0^{\pi/2} =\frac{4\beta m^2c^3}{\hbar}; \tag{19} \]
\[ -T=4\beta m^4c^5/\hbar^3; \tag{20} \]
\[ R=6/a^2=-\chi T=\frac{32\pi\beta m^4c}{\hbar^3}G. \tag{21} \]
If we define \(-T=2\varepsilon\) (with \(3p+\varepsilon=0\)), which holds for the Einstein space with metric (2), then we find
\[ -T=3mc^2/2\pi a^3. \tag{22} \]
From (20) and (22) we have \(mca=\hbar(3/8\pi\beta)^{1/3}\). Put \(\beta=3/8\pi\); then
\[ mca=\hbar. \tag{23} \]
From (21) and (23) we have
\[ m=\sqrt{\frac{c\hbar}{2G}},\qquad a=L=\sqrt{\frac{2G\hbar}{c^3}}=2Gm/c^2=r_g, \tag{24} \]
i.e., for \(m\) and \(a\) we obtain the Planck values (the values of the mass and size of the planckeon), with \(a=L=r_g\)—the gravitational radius, as must be the case for an Einstein universe. At the same time, the result obtained checks the calculations performed.
Denote
\[ \frac{1}{\hbar}[E\tau+pa\chi] =\frac{1}{\hbar}\left[E\tau+pa\arcsin\frac{r}{a}\right] =\alpha, \]
then (17), for \(m=0,\ l=0\ (n=0)\), takes the form
\[ \psi=\frac{A_1}{r}e^{-i\alpha}. \tag{25} \]
The de Broglie wave will propagate according to the law:
\[ E\tau+pa\arcsin\frac{r}{a}=\alpha\hbar, \]
or
\[ r=a\sin\frac{E\tau-\alpha\hbar}{pa} =a\sin\frac{E(\tau-\tau_0)}{pa} =a\sin\left[\frac{E}{pc}\frac{c(\tau-\tau_0)}{a}\right]. \tag{26} \]
Hence the phase velocity is
\[ \frac{V_{\mathrm{ph}}}{c} = \sqrt{1+\frac{m^2c^2}{p^2}}\, \cos\left[ \sqrt{1+\frac{m^2c^2}{p^2}}\, \frac{c(\tau-\tau_0)}{a} \right], \]
the group velocity is
\[ \frac{V_{\mathrm{gr}}}{c} = \frac{1}{\sqrt{1+m^2c^2/p^2}}. \]
Expression (26) shows that the de Broglie wave corresponding to the planckeon is localized in the region \(0\le r\le a=r_g=L\) and that the wave packet, held by its own gravitational field, is stable and does not spread.
For values \(c\tau/a\ll 1\),
\[ r=c(\tau-\tau_0)E/pc =E(\tau-\tau_0)/p =c(\tau-\tau_0)\sqrt{1+m^2c^2/p^2}, \tag{27} \]
i.e., for \(a\to\infty\), when gravity is switched off, the usual-
…situation of spreading of the wave packet, since as \(\tau \to \infty\), \(r \to \infty\). The conclusion regarding the stability of the wave packet in its own gravitational field of a definite energy is fundamental, although quite expected. This conclusion is significant in studying the structure of elementary particles in their own gravitational field.
Earlier M. A. Markov \({}^{6}\) proposed that the maximon (a particle equivalent to the planckeon) is described by a particle-like solution of the Einstein and Dirac equations in the form of a limiting state of a wave packet whose energy is gravitationally closed in a region of size \(L\).
Received
22 IV 1969
CITED LITERATURE
\({}^{1}\) E. A. Tagirov, P. A. Chernikov, Preprint R2-3777, Dubna, 1968. \({}^{2}\) E. Kamke, Handbook of Ordinary Differential Equations, Part 3, Vol. I, IL, 1950, p. 666 (2.424). \({}^{3}\) K. A. Bronnikov, V. N. Melnikov, K. P. Stanyukovich, Preprint ITF, 88–89, Kiev, 1968. \({}^{4}\) K. P. Stanyukovich, DAN, 168, No. 4 (1966). \({}^{5}\) K. P. Stanyukovich, V. G. Lapchinsky, Abstracts of the V International Conference on Gravitation, 1968. \({}^{6}\) M. A. Markov, ZhETF, 51, No. 9 (1966).