# T_NAKÄ[uO

## ɁuȂFCURVED SPACE-TIMEvǂށBiQj

<<   쐬 F 2016/10/19 00:01   >>

 "Remarks on the Equivalence of Inertial and Gravitational Masses and on the Accuracy of Einstein's Theory of Gravity." CURVED SPACE-TIME 𑱂܂B [ȂFCURVED SPACE-TIMEiQ-Pj]======================================================== To find the solution of this system of equations is not difficult, but somewhat tedious. ̂̃VXẻ邱Ƃ́AA炩ދłB Certain precautions in the procedure of finding the solution must also be observed. 菇̊mȗpSAώ@Ȃ΂ȂȂB It proves beneficial to start with EL equation related to the time coordinate, followed by the coordinate constraint equation, and finally followed by solving the last EL equation. ܂ɎԍWɊ֘AELLvłƂƂؖAW񎮂ƁAŌELƂɂāAŌɊmFB This procedure thus first leads to: (29) ̎菇́AŏɎ𓱂F @$\frac{d}{d\tau }\left (2e^{A(x)}c^{2} \frac{dt}{d\tau } \right )= 0\; \; \; \; \; \; (29)$ Eq.29 can be easily integrated and using the condition that the space-time at infinity is flat, it is possible to write: (30) (29)͊ȒPɐϕłA̎󂪕łƂԂgpāÂ悤ɏƂłF @$\frac{dt}{d\tau }= e^{-A(x)}\; \; \; \; \; \; (30)$ This result can then be used with Eq.26 to obtain: (31) ̌ʂɁA(26) gƂɂĎF @$\left ( \frac{dx}{d\tau } \right )^{2}= e^{-B(x)}c^{2}\left ( e^{-A(x)}-1\right )\; \; \; \; \; \; (31)$ This procedure has now yielded all the first derivatives of coordinates and ensured that the relativistic constraint and thus LC is satisfied. ̎菇ɂAׂĂ̍ŴPK^AmɑΘ_Iƃ[csϐ邱ƂɂȂB Finally, the second EL equation leads to: (32)where the prime represents d/dx. ŌɁA2EL͎ƂȂF @$-\frac{d}{d\tau }\left ( 2e^{B(x)}\frac{dx}{d\tau } \right )=[{A}'(x)-{B}'(x)] e^{-A(x)}c^{2}+{B}'(x)c^{2}\; \; \; \; \; \; (32)$ ŃvC(') d/dx \킵ĂB =============================================================================================== @$L=e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-e^{B(x)}\left ( \frac{dx}{d\tau } \right )^{2}$ @$\frac{\partial L}{\partial \frac{\partial t}{\partial \tau }}=2c^{2}e^{A(x)}\left ( \frac{dt}{d\tau } \right )\; \; ,\; \; \frac{\partial L}{\partial \frac{\partial x}{\partial \tau }}= -2e^{B(x)}\left ( \frac{dx}{d\tau } \right )$ @$\frac{\partial L}{\partial t}= 0$ @$\frac{\partial L}{\partial x}={A}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-{B}'(x)e^{B(x)}\left ( \frac{dx}{d\tau } \right )^{2}$ @@$={A}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-{B}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}+{B}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-{B}'(x)e^{B(x)}\left ( \frac{dx}{d\tau } \right )^{2}$ @@$={A}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-{B}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}+{B}'(x)e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-{B}'(x)e^{B(x)}\left ( \frac{dx}{d\tau } \right )^{2}$ @@$=[{A}'(x)-{B}'(x)]e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}+{B}'(x)\left \{ e^{A(x)}\left ( \frac{cdt}{d\tau } \right )^{2}-e^{B(x)}\left ( \frac{dx}{d\tau } \right )^{2} \right \}$ @@$=[{A}'(x)-{B}'(x)]e^{A(x)}c^{2}\left ( \frac{dt}{d\tau } \right )^{2}+{B}'(x)c^{2}$ Ȃ̂ŁAEL @$\frac{d}{d\tau }\left ( \frac{\partial L}{\partial \frac{dt}{d\tau }} \right )= \frac{\partial L}{\partial t}$ A @$\frac{d}{d\tau }\left (2e^{A(x)}c^{2} \frac{dt}{d\tau } \right )= 0\; \; \to \; \; \frac{d}{d\tau }\left (e^{A(x)} \frac{dt}{d\tau } \right )= 0\; \; \to \; \; e^{A(x)} \frac{dt}{d\tau } = K$ K ͒萔ɂȂ܂Ał̓OWA~RtXL[ eA(x) 1 łAŗLтƍW t ݂͓̍ƂƁAK = 1 ł傤Bȉ͍Ă邩HMȂBBjƁA @$\frac{dt}{d\tau } = e^{-A(x)}$ ƂȂł傤B (26) @$\left ( \frac{dx}{d\tau } \right )^{2}= e^{-B(x)}c^{2}\left \{ e^{A(x)}\left ( \frac{dt}{d\tau } \right )^{2}-1 \right \}= e^{-B(x)}c^{2}(e^{-A(x)}-1)$ ܂A @$\frac{\partial&space;L}{\partial&space;x}=[{A}'(x)-{B}'(x)]e^{-A(x)}c^{2}+{B}'(x)c^{2}$ ܂B܂EL @$\frac{d}{d\tau }\left ( \frac{\partial L}{\partial \frac{dx}{d\tau }} \right )= \frac{\partial L}{\partial x}$ @$-\frac{d}{d\tau }\left ( 2e^{B(x)}\frac{dx}{d\tau } \right )=[{A}'(x)-{B}'(x)] e^{-A(x)}c^{2}+{B}'(x)c^{2}$@ ƂȂ܂B [ȂFCURVED SPACE-TIMEiQ-Qj]======================================================== By carrying out the differentiation and using Eq.31 and Eq.27, it is possible, after some algebra, to derive the following simple result: (33) ƁA(31)(27)gƂɂāA኱̑㐔vŽAȉ̒PȌʂoƂłF @$c^{2}{B}'(x)+2e^{B(x)}\frac{\kappa M_{s}}{x^{2}}= [{A}'(x)+{B}'(x)]e^{-A(x)}c^{2}\; \; \; \; \; \; \; (33)$ The use of Eq.27 for the second derivative of coordinate x in Eq.32 is the crucial step in obtaining the correct solution. (32) x ̂QK̂߂ (27) gƂ́A𓾂邽߂̏dvȃXebvłB If the second derivative were, for example, calculated from Eq.31 the result would be that any functions A(x) and B(x) satisfy the equations. Ƃ΁AQK (31) vZƂƁAʂ A(x) B(x) ǂȊ֐ł̕𖞂ł낤B This is understandable, since all the free falling coordinate systems, which transform out the force of gravity, are equivalent. d͂Ă邷ׂĎRn͓Ȃ̂ŁA͗łB The system of equations for the functions A(x) and B(x) without Eq.27 is thus not well defined. (27) Ȃł́A֐ A(x) B(x) ̃̕VXe͂͂Ƃ͒ȂB To specify the second derivative is,therefore, an interesting necessary and a crucial requirement for finding the functions A(x) and B(x). ĂQKw肷邱Ƃ́A֐ A(x) B(x) 肷邽߂̋[dvȏłB Appendix A explains that for a correct relativistic Lagrangian in the laboratory coordinate system it can be also specified that: (34) t^ł́An̐Θ_IOW֐̂߂ɁAȉ̎w肳邱ƂĂF @$A(x)=-B(x)\; \; \; \; \; \; \; (34)$ This condition actually guarantees that the Jacobian of the coordinate transformation from the free falling test body coordinate system to the laboratory coordinate system is equal to unity. R鎎̂̍WnWnւ̍Wϊ̃Rrs񎮂Pł邱ƂȀԂ͎ۂɕۏ؂B Using this condition in Eq.33 leads to a simple differential equation for B(x). (33) ł̏ԂgƁAB(x) ̂߂ɒPȔ𓱂B This is easily solved when the initial condition of flat space at large distances is again used: (35) Ȃ̋ɂĕȋԂ̏ĂюgƂA͊ȒPɉF $e^{B(x)}=\frac{1}{1-\frac{2\kappa M_{s}}{c^{2}x}}\; \; \; \; \; \; \; (35)$ =============================================================================================== ܂ (32) ̍ӂvZ܂B @$-\frac{d}{d\tau }\left ( 2e^{B(x)}\frac{dx}{d\tau } \right )=-2e^{B(x)}{B}'(x)\left ( \frac{dx}{d\tau } \right )^{2}-2e^{B(x)}\frac{d^{2}x}{d\tau ^{2}}$ ̉Eӂ (31) (27) Kp @$-\frac{d}{d\tau }\left ( 2e^{B(x)}\frac{dx}{d\tau } \right )=-2{B}'(x)(e^{-A(x)}-1)c^{2}-2e^{B(x)}\left ( -\frac{\kappa M_{s}}{x^{2}} \right )$ @@$=-2{B}'(x)e^{-A(x)}c^{2}+2{B}'(x)c^{2}+2e^{B(x)}\frac{\kappa M_{s}}{x^{2}}$ ƂȂA (32) @$-2{B}'(x)e^{-A(x)}c^{2}+2{B}'(x)c^{2}+2e^{B(x)}\frac{\kappa M_{s}}{x^{2}}= [{A}'(x)-{B}'(x)] e^{-A(x)}c^{2}$ @@$+{B}'(x)c^{2}$ ƂȂ܂Bڍ @$c^{2}{B}'(x)+2e^{B(x)}\frac{\kappa M_{s}}{x^{2}}= [{A}'(x)+{B}'(x)]e^{-A(x)}c^{2}$ (33) oĂ܂B ĕt^ŏoĂ֌W $A(x)=-B(x)$ i[ĂȂ̂łAAj (33) ɓKp @$c^{2}{B}'(x)+2e^{B(x)}\frac{\kappa M_{s}}{x^{2}}= 0\; \; \to \;\; e^{-B(x)}{B}'(x)= -2\frac{\kappa M_{s}}{c^{2}x^{2}}$ Ƃ悤ȕϐ̔ɂȂ܂Bϕ @$e^{-B(x)}= K-2\frac{\kappa M_{s}}{c^{2}x}$ ƂȂ܂Ał̓OWA~RtXL[ eB(x) 1 ł邱ƂlƁAK = 1 ł傤BāA @$e^{-B(x)}= 1-2\frac{\kappa M_{s}}{c^{2}x}\; \; \to \; \; e^{B(x)}= \frac{1}{1-\frac{2\kappa M_{s}}{c^{2}x}}$ ƂȂ܂B ͂̕ӂŁBB @

## uOC

NbNċC`悤I
OCăNbN΁ÃuOւ̃Nt܂B
OC

^Cg i{j uO^

 ^Cg {@

e jbNl[^

## Rg

 jbNl[ {@
ɁuȂFCURVED SPACE-TIMEvǂށBiQj T_NAKÄ[uO/BIGLOBEEFuuO
TCYF