第十三章 黎曼球面泡的 Zitterbewegung 颤动
Chapter 13 Zitterbewegung of the Riemann-sphere-bubble
薛定谔指出,
| 凡是既有内部自由度又有外部自由度且两者间存在能量交换的物理系统都会具有颤动现象. (8.3.3-1) |
Schrӧdinger pointed out that
| Any physical system with both internal and external degrees of freedom and energy exchange between the two will have the phenomenon of trembling motion. (8.3.3-1) |
m惯性 = ± ( h' c2 / (b G v ) )(1/2). (11.2(5)-1)
minertial = ± ( h' c2 / (b G v ) )(1/2). (11.2(5)-1)
根据(10.2-2),黎曼球面泡的惯性质量是 m ( v2 / c2 ).
According to (10.2-2), the inertial mass of a Riemann-sphere-bubble is m ( v2 / c2 ).
根据量子力学,
| 自由粒子的 Zitterbewegung 颤动的平均速率是±c. (11.2(2)-3) |
According to quantum mechanics,
| The average speed of Zitterbewegung of the free particle is ±c . (11.2(2)-3) |
粒子的 Zitterbewegung 是由正负能态的干涉导致的.
The Zitterbewegung of a particle is caused by interference between the positive and negative energy states.
根据量子力学和相对论, ω ħ = E = m c2 .
According to quantum mechanics and relativity, ω ħ = E = m c2 .
对于 Zitterbewegung 而言,因为正负能态的能量差为 2 m c2 ,所以 Zitterbewegung 颤动的角频率是
ω ≈ 2 m c2 / ħ ; (13-1)
For Zitterbewegung, the energy difference between positive and negative states is 2 m c2, and hence the angular frequency of Zitterbewegung is
ω ≈ 2 m c2 / ħ ; (13-1)
请参看张永德.《量子菜根谭:量子理论专题分析》[1]第3版第61页.
Refer to ZHANG Yongde. Talks on the "Roots of Vegetables" in the Quantum Theory: Thematic Analyses on the Modern Quantum Theory[1]. Third edition. p. 61. (In Chinese.)
Paul Strange [2] 指出,Zitterbewegung 颤动的振幅是
A ≈ h / ( 4 m c ); (13-2)
Paul Strange [2] pointed out that the amplitude of Zitterbewegung is
A ≈ h / ( 4 m c ); (13-2)
在反引力场中,粒子的黎曼球面泡的 Zitterbewegung 颤动可以称为黎曼球面泡的 Zitterbewegung 颤动(Riemann-sphere-bubble Zitterbewegung, 缩写为 RSBZ).
The Zitterbewegung of the Riemann-sphere-bubble of a particle can be called Riemann-sphere-bubble Zitterbewegung, abbreviated as RSBZ.
设 RSBZ 的角频率是 ωRSBZ ,于是根据(13-1)、(12-5)和(10.2-2)有 ωRSBZ ≈ 2 m ( v粒子2 / c2 ) c2 / ħ' = 2 m v粒子2 ( 2 ) / h' .
ωRSBZ ≈ 4 m v粒子2 / h' . (13-3)
因为 νRSBZ = ωRSBZ / ( 2 ), 所以
νRSBZ ≈ 2 m v粒子2 / h' . (13-4)
式中 νRSBZ 是粒子的RSBZ的频率, m 是粒子(例如分子)的原质量, v 是粒子的速率.
Let the angular frequency of the RSBZ be ωRSBZ . Then from (13-1), (12-5) and (10.2-2), there is ωRSBZ ≈ 2 m ( vparticle2 / c2 ) c2 / ħ' = 2 m vparticle2 ( 2 ) / h' .
ωRSBZ ≈ 4 m vparticle2 / h' . (13-3)
Since there is νRSBZ = ωRSBZ / ( 2 ), there is
νRSBZ ≈ 2 m vparticle2 / h' , (13-4)
where νRSBZ is the frequency of RSBZ of the particle, m is the original mass of the particle (e.g., a molecule), and v is the speed of the particle.
因为 TRSBZ = 1 / νRSBZ ,所以
TRSBZ ≈ h' / ( 2 m v粒子2 ), (13-5)
式中 TRSBZ 是粒子的 RSBZ的周期.
Since there is TRSBZ = 1 / νRSBZ , there is
TRSBZ ≈ h' / ( 2 m vparticle2 ), (13-5)
where TRSBZ is the period of RSBZ of the particle.
由(13-5),
v粒子 ≈ ( h' / ( 2 m TRSBZ ) )1/2. (13-6)
From (13-5), there is
vparticle ≈ ( h' / ( 2 m TRSBZ ) )1/2. (13-6)
由(8.6.2-4)和(13-5),得到
TRSBZ ≈ t'lifespan ; (13-7)
t'lifespan ≈ h' / ( 2 m vparticle2 ). (13-8)
From (8.6.2-4) and (13-5), one can obtain
TRSBZ ≈ t'lifespan ; (13-7)
t'lifespan ≈ h' / ( 2 m v粒子2 ). (13-8)
另请参看(8.4.4.3-3).
See also (8.4.4.3-3).
因为 T = 1 / ν , 由(13-7)和(11.3-3)得到
νRSBZ = (1 / ) ( 1 / t' ) ≈ (1 / ) (2 v3 ) / (b G m ) = 2 v3 / ( b G m ).
νRSBZ ≈ 2 v3 / ( b G m ). (13-9)
Since T = 1 / ν , from (13-7)和(11.3-3), one finds
νRSBZ = (1 / ) ( 1 / t' ) ≈ (1 / ) (2 v3 ) / (b G m ) = 2 v3 / ( b G m ).
νRSBZ ≈ 2 v3 / ( b G m ). (13-9)
由(13-7)和(11.3-3)得到
TRSBZ ≈ b G m / ( 2 v3 ). (13-10)
From (13-7)和(11.3-3), one finds
TRSBZ ≈ b G m / ( 2 v3 ). (13-10)
由(11.2(2)-3),
| 对于在RSBZ中的粒子而言,瞬时速率是 ±c , 平均速率是 v . (13-11) |
According to (11.2(2)-3),
| In the case of the particle in the RSBZ, the instantaneous speed is ±c , and the average speed is v . (13-11) |
基于(13-2)和(12-5),由(10.2-2),有ARSBZ ≈ h' / ( 4 m v2 c / c2 ),即
ARSBZ ≈ h' c / ( 4 m v2 ). (13-12)
Based on (13-2) and (12-5), from (10.2-2), there is ARSBZ ≈ h' / ( 4 m v2 c / c2 ), i.e.
ARSBZ ≈ h' c / ( 4 m v2 ). (13-12)
比较(8.6.2-4)和(13-12),由(13-10)得到
ARSBZ ≈ c Δt' / 2 = b c G m / ( 4 v3), (13-13)
Comparing (8.6.2-4) and (13-12), from (13-10) one obtains
ARSBZ ≈ c Δt' / 2 = b c G m / ( 4 v3), (13-13)
| 1. 地球的RSBZ是导致大地震的一个因素. 2. 在地震前,反引力导致一些反引力现象,例如地震光.在地震发生时,| ∑R |非常大,于是根据(9.2(4)-1b),反引力现象消失. 3. 另请参看(8.4.4.5-8)、(12.1-3)、(15.1-5). (13-14) |
| 1. RSBZ of the Earth is a factor in causing large earthquakes. 2. Before the earthquake, antigravitation brings some antigravitational phenomena, e.g., earthquake lights. During the earthquake, | ∑R | is so large that, according to (9.2(4)-1b), antigravitational phenomena disappear. 3. See also (8.4.4.5-8), (12.1-3), and (15.1-5). (13-14) |
处于RSBZ中的粒子在实数时空与扭量空间之间振荡.
A particle in RSBZ oscillates between the real-number-spacetime and twistor space.
由于 ν' = 1 / t间隔 ,由(13-10),有
ν' = 2 v3 / ( b G m ), (在扭量空间里(就像在黎曼球面上那样)), (13-15)
式中 ν' 既是波的频率,又是类雾体由于圆运动而导致的的振荡频率.
请参看Paul Strange. Relativistic Quantum Mechanics[2]. (相对论量子力学.)第214页. Fig. 7.1.
Refer to Paul Strange. Relativistic Quantum Mechanics[2]. p. 214. Fig. 7.1.
Since ν' = 1 / tgap , from (13-10), there is
ν' = 2 v3 / ( b G m ), (in twistor space (like on the Riemann sphere)), (13-15)
where ν' is both the frequency of a wave, and the frequency of oscillation, of foggoid, caused by circular motion.
| 类雾体的振荡频率是 ν' = 2 v3 / ( b G m ). (13-16) |
| The oscillation frequency of foggoid is ν' = 2 v3 / ( b G m ). (13-16) |
| λ' = v / ν'
= b G m / (
2 v2 ), 式中 λ' 是类雾体粒子的波长. (13-17) |
|
λ' = v / ν' = b G m / ( 2 v2 ), where λ' is the wavelength of a foggoid particle. (13-17) |
| 当一个人处于类雾态时,他(或她)的可见光与可闻声的频率与强度范围与当他(或她)处于常态时不同. (13-18) |
| When a human being is in the foggoid state, his or her ranges of frequencies and intensities of visible light and audible sound are different from when he or she is in the normal state. (13-18) |
基于(8.10-4)和(11.3-3)可以知道,
| 为了持续关注, m 应该较大,而 v 应该较小. (13-19) |
Based on (8.10-4) and (11.3-3), one can know that
| For the purpose of paying sustained attention, m should be large, while v should be small. (13-19) |
参考文献
References
[1] 张永德.量子菜根谭:量子理论专题分析.北京:清华大学出版社,2016年11月第3版.
[1] ZHANG Yongde. Talks on the "Roots of Vegetables" in the Quantum Theory: Thematic Analyses on the Modern Quantum Theory. Third edition. Beijing: Tsinghua University Press, Nov., 2016. (In Chinese.)
[2] Paul Strange. Relativistic Quantum Mechanics. (相对论量子力学.) 英文. 北京：世界图书出版公司北京公司.2008年8月第1次印刷. 第210页.
[2] Paul Strange. Relativistic Quantum Mechanics. Cambridge University Press. 1998. Section 7.3 Zitterbewegung.
问题13(1) 地球的RSBZ和大地震
Problem 13(1) The Zitterbewegung of the Riemann-sphere-bubble of the Earth and large earthquakes
已知相对于邻近恒星,太阳的速度是 19.7 km/s.[1] 假设相对于邻近恒星,地球的速度是 v地球 = 19.7 km/s. 设 m地球 是地球的质量, T'地球 是在反引力场中地球RSBZ的周期.已知 m地球 = 5.972 19×1024 kg. 给出 v地球 = 19.7×103 m/s, 求 T'地球 .
It is known that the Sun has a velocity V, with respect to nearby stars, of 19.7 km s-1 .[1] Assume that vEarth, the Earth's speed, is 19.7 km/s with respect to nearby stars. Let mEarth be the mass of the Earth, and T'Earth be the period of the Earth's RSBZ in the antigravitational field. It is known that mEarth = 5.972 19×1024 kg. Given vEarth = 19.7×103 m/s, find T'Earth .
由(13-10), T'地球 ≈ b G m地球 / ( 2 v地球3 ),
T'地球 ≈ 6.112 s.
From (13-10), T'Earth ≈ b G mEarth / ( 2 vEarth3 ),
T'Earth ≈ 6.112 s.
郝晓光和胡小刚指出,在某些强震前两天左右,在南极中山站的 L & R ET-21重力仪侦测到持续60小时的不明稳定扰动,信号周期在 4~8 s 左右.[2]
Hao Xiaoguang and Hu Xiaogang pointed out that two days before some strong earthquakes, an unidentified 60-h stable disturbance was detected with the L & R ET-21 gravity meter at Zhongshan Station, Antarctica. The signal period is around 4~8 s.[2]
[1] Hughes, D.W. On hyperbolic comets. Journal of the British Astronomical Association, vol. 101, no. 2, p. 119-120.
http://adsabs.harvard.edu/full/1991JBAA..101..119H
[1] Hughes, D.W. On hyperbolic comets. Journal of the British Astronomical Association, vol. 101, no. 2, p. 119-120.
http://adsabs.harvard.edu/full/1991JBAA..101..119H
[2] 郝晓光,胡小刚.《强震短临前兆探索—“震前扰动”现象研究》.北京:测绘出版社.2012年12月.
[2] Hao Xiaoguang, Hu Xiaogang. Study of the short-term anomalous tremors before large earthquakes. Beijing: Surveying and Mapping Press. December, 2012.
问题13(2) 太阳的RSBZ和太阳五分钟振荡
Problem 13(2) The RSBZ of the Sun and the solar five-minute oscillations
已知太阳相对于宇宙微波背景辐射的运动速度是 370 km/s. 设 m太阳 是太阳的质量, v太阳 是太阳的运动速度, T'太阳 是太阳的 RSBZ 的周期.已知 m太阳 = 5.972 19 1024 kg, v太阳 = 370×103 m/s.根据(11.2(3)-7), b = 0.234 464. 求 T'太阳 .
It is known that the sun is moving at a speed around 370 km/s relative to the cosmic microwave background. Let mSun be the mass of the Sun, vSun the speed of the Sun, and T'Sun the period of the Sun's RSBZ. Given mSun = 5.972 19 1024 kg, vSun = 370×103 m/s. According to (11.2(3)-7), b = 0.234 464. Find T'Sun .
由(13-10), T'太阳 ≈ b G m太阳 / ( 2 v太阳3 ),
T'太阳 ≈ 5.12 min.
这与太阳的五分钟振荡是一致的.
From (13-10), T'Sun ≈ b G mSun / ( 2 vSun3 ),
T'Sun ≈ 5.12 min.
This is consistent with the solar five-minute oscillations.
问题13(3) 银河系银盘的厚度
Problem 13(3) The thickness of the stellar disk of the Milky Way Galaxy
银河系的银核的质量是m = 1.989 1×1030×7×109 kg.相对于宇宙微波背景辐射的静止参照系,银河系的速度是 v = 552×103 m/s[1].设银盘近核处的厚度是 d, 在该处 RSBZ 的振幅是 ARSBZ .求 d .
The mass of the Milky Way Galaxy's bar is m = 1.989 1×1030×7×109 kg. The speed of the Milky Way Galaxy relative to CMB rest frame is v = 552×103 m/s[1]. Let the thickness of the Milky Way Galaxy's stellar disk at the locations near the bar be d, and the amplitude of the RSB of the stellar disk at those locations be ARSBZ . Find d.
将以上数值代入(13-13), 得到
ARSBZ ≈
3.09×1019 m.Substituting the above numerical values into (13-13), one obtains
ARSBZ ≈ 3.09×1019 m.
根据(9.4-5),有 d = 2 ARSBZ , 即
d ≈
6.181×1019 m ≈ 2 003.13 秒差距.According to (9.4-5), there is d = 2 ARSBZ , i.e.
d ≈ 6.181×1019 m ≈ 2 003.13 pc.
观测值是,对于银河系而言,银盘近核球处的厚度约为2000 秒差距.[2]
The observed value is that in the case of the Milky Way Galaxy, the thickness of the stellar disk near the bar
is about 2 000 pc.[2][1] Wikipedia. Milky Way. http://en.wikipedia.org/wiki/Milky_Way
[1] Wikipedia. Milky Way. http://en.wikipedia.org/wiki/Milky_Way
[2] 苏宜. 天文学新概论 (第四版). 北京：科学出版社.2009年8月.第412页.
[2] Su Yi. A New Survey of Astronomy (Fourth Edition). Beijing: Science Press. August, 2009. p. 412. (In Chinese.)