王學仁

摘要：我們將介紹我們的工作：①把9個關聯(lián)系數分成兩組：涉及?（線加速度）的A組包括Γ?0?01?，Γ?1?00?， Γ?1?11?， Γ?1?22?， 和 Γ?1?33?;不涉及的B組包括Γ?2?12?， Γ?2?33?， Γ?3?13?， 和 Γ?3?23?。②回顧關系式（）?r=（）?Nγ?3，這里（）?r是相對論力學中的線加速度，（）?N是牛頓力學中的線加速度，γ是洛倫茲因數。③我們已經確定（Γ?0?01?）?N=-b?′b?-1?，（Γ?1?00?）?N=-c?2b?′b?-5?，（Γ?1?11?）?N=b?′b?-1?，（Γ?1?22?）?N=-rb?-2?，（Γ?1?33?）?N=-rb?-2?sin?2θ。④我們已經確定（Γ?0?01?）?r=（-b?′b?-1?）γ?3， （Γ?1?00?）?r=（-c?2b?′b?-5?）γ?3，（Γ?1?11?）?r=（b?′b?-1?）γ?3，（Γ?1?22?）?r=（-rb?-2?）γ?3，（Γ?1?33?）?r=（-rb?-2?sin?2θ）γ?3。⑤我們已經證明Γ?0?01?=A?′/（2A）=?-b?′b?-1?，?Γ?1?00?=A?′/（2B）=-c?2b?′b?-5?，Γ?1?11?=b?′/（2B）=b?′b?-1?，Γ?1?22?=-rb?-1?=-rb?-2?，Γ?1?33?=-rb?-1?sin?2θ=-rb?-2?sin?2θ。（6）我們已經確定（Γ?1?12?）?N?r=r?-1?，（Γ?2?33?）?N?r=?sin?θ?cos?θ，（Γ?3?13?）?N?r=r?-1?，（Γ?3?23?）?N?r=?cot?θ。⑥我們分析了?Schwarzschild?解并得出兩個結論：（?a?）B =（1-2GM/（c?2r））?-1?=（1-?r·?2/c?2）?-1?=γ?2，這表明它和牛頓守恒定律有關。（?b?）對于弱引力場GM/（c?2r）1， B=（1-2GM/（c?2r））?-1?≈?1+?2GM/（c?2r）≈1+2GM/（c?2r）+（GM/（c?2r））?2=（1+GM/（c?2r））?2=γ?2，因此，γ=1+GM/（c?2r）=b，這個關系式對強引力場也適用。把這些需要的表達方式帶入方程式，并應用關系式，我們能簡化方程式。我們已經得到了相對論解：-c?2?d?τ?2=c?2（1+GM/（c?2r））?-2?d?t?2-（1+GM/（c?2r））?2?d?r?2-r?2?d?θ?2-r?2?sin?2θ?d?φ?2?.

關鍵詞：廣義相對論;場方程;引力理論

DOI：10.15938/j.jhust.2019.01.024

中圖分類號： O412.1

文獻標志碼： A

文章編號： 1007-2683（2019）01-0145-04

A Study of Einstein Field Equation

WANG Xue?ren

（School of Applied Science， Harbin University of Science and Technology， Harbin 150080， China）

Abstract：

In this paper we present our research work： ①We divide 9 connection coefficients into two groups： Group A involving??（the linear acceleration） includes?Γ?0?01?，Γ?1?00?， Γ?1?11?， Γ?1?22?， and?Γ?1?33?; Group B not involving??includes?Γ?2?12?， Γ?2?33?， Γ?3?13?， and?Γ?3?23?. ②Recalling the relation formula：?（）?R=（）?Nγ?3?， where?（）?R?is the linear acceleration in relativistic mechanics，?（）?N?the linear acceleration in Newtoneon mechanics， and?γ?Lorrentz factor. ③We have determined that?（Γ?0?01?）?N= -b?′b?-1?， （Γ?1?00?）?N= -c?2b?′b?-5?， （Γ?1?11?）?N= b?′b?-1?， （Γ?1?22?）?N= -rb?-2?， （Γ?1?33?）?N= -rb?-2?sin?2θ?. ④We have determined that?（Γ?0?01?）?R = （-b?′b?-1?）γ?3， （Γ?1?00?）?R = （-c?2b?′b?-5?）γ?3， （Γ?1?11?）?R = （b?′b?-1?）γ?3， （Γ?1?22?）?R = （-rb?-2?）γ?3， （Γ?1?33?）?R = （-rb?-2?sin?2θ）γ?3?. ⑤We have proved that?Γ?0?01?=A?′/（2A）=-b?′b?-1?， Γ?1?00?= A?′/（2B） = -c?2b?′b?-5?， Γ?1?11?= B?′/（2B） = b?′b?-1?， Γ?1?22?= -rB?-1?= -rb?-2?，Γ?1?33?= -rB?-1?sin?2θ= -rb?-2?sin?2θ?. （6）We have determined that?（Γ?1?12?）?N?R = r?-1?， （Γ?2?33?）?N?R =?sin?θ?cos?θ， （Γ?3?13?）?N?R= r?-1?， （Γ?3?23?）?N?R =?cot?θ?. ⑥We have analyzed Schwarzschild solution， and drawn two conclusions： （a）?B=（1-2GM/（c?2r））?-1?= （1-?r·?2/c?2）?-1?= γ?2?， indicating that it involves Newtoneon formula of energy conservation. （b）In the case of the weak gravitational field，?GM/（c?2r）1， B =?（1-2GM/（c?2r））??-1?≈ 1 + 2GM/（c?2r） ≈ 1 + 2GM/（c?2r） +?（GM/（c?2r））?2 =?（1+GM/（c?2r））?2 = γ?2?， therefore，?γ= 1 + GM/（c?2r） = b?， it holds too， in the case of the strong gravitational field. （8）Substituting the needed expressions into equations， and applying the relation formulas， we can simplify the equations， Obtaining the relativistic solution：

-c?2?d?τ?2 = c?2（1+GM/（c?2r））?-2?d?t?2 - （1+GM/（c?2r））?2?d?r?2-r?2?d?θ?2-r?2?sin?2θ?d?φ?2

Keywords：general relativity; field equation; gravitational theory

0Introduction

Schwarzschild solution for Einstein field equation has made an important contribution to development of general relativity， but we should point out Schwarzschild solution belongs to Newtonian mechanics， and it is suitable only to the weak gravitational field， but not suitable to the strong gravitational field. In this paper we will give sufficient basis for this conclusion.

1Classification of connection coefficients

We divide 9 connection coefficients into two groups： group?A?involving??（the linear acceleration）， see Table1; group?B?not involving??， see Table 1 and 2.

Table1. Group?A?involving

Γ?0?01?（=A?′/（2A）=-B?′/（2B）） = -Γ?1?11?=?/（r·r·）， Γ?1?00?= -/（?t?·t·）， Γ?1?11?=-/（r·r·）， Γ?1?22?=-/（θ?·θ?·），Γ?1?33?=-/（φ·φ·）

Table2. Group?B?not involving

Γ?2?12?=-θ?¨/（r·θ?·）， Γ?2?33?=-θ?¨/（φ·φ·）， Γ?3?13?=-/（r·φ·）， Γ?3?23?=-/（θ?·φ·）

2?R=?Nγ?3

R?is linear acceleration in relativistic mechanics，?N?is linear acceleration in Newtoneon mechanics， the relation formula between??R?and??N?is as below （references[2]， P38）

R=?Nγ?3?（1）

where?γ?is Lorentz factor. So that the expressions of group?A?in two kinds of mechanics are different.

3The expressions of group A in Newtonian mechanics

Example 1： Determine the expression of?（Γ?1?11?）?N?.

Solution： From?-c?2?d?τ?2 = -b?2?d?r?2?（see line element Exp. （4））

r·=cb?-1?. From??N+?（Γ?1?11?）?N（r·r·）=0?，we can get

（Γ?1?11?）?N = -?d?r?d?τ?d?r·?d?r/（r·（cb?-1?））=-（r·（-cb?-2?b?′）/（r·（cb?-1?））=b?′b?-1?.

Following Example 1， we can obtain Table 3.

Table 3. The expressions of Group A in Newtonian mechanics

（Γ?0?01?）?N=-b?′b?-1?，?（Γ?1?00?）?N=-c?2b?′b?-5?，?（Γ?1?11?）?N=b?′b?-1?，?（Γ?1?22?）?N=-rb?-2?，?（Γ?1?33?）?N=-rb?-2??sin??2θ.

4The expressions of group A in relativistic mechanics

Example 2： Determine the expression of??（Γ?1?11?）?R?.

Solution： From??（）?R+?（Γ?1?11?）?R（r·r·）=0?we can get??（Γ?1?11?）?R=-?（）?R/（r·r·）=γ?3（-?N/（r·r·）=?（Γ?1?11?）?Nγ?3?.

Following Example 2 we can obtain Table 4.

Table4. The expressions of group A in relativistic mechanics

（Γ?0?01?）?R=（-b?′b?-1?）γ?3，?（Γ?1?00?）?R=（-c?2b?′b?-5?）γ?3，?（Γ?1?11?）?R=（b?′b?-1?）γ?3，

（Γ?1?22?）?R=（-rb?-2?）γ?3，?（Γ?1?33?）?R=（-rb?-2??sin??2θ）γ?3

5Group A used in Schwarzschild solution

Example 3. Prove?Γ?1?00?=A?′/（2B）= -c?2b?′b?-5

Proof.

Γ?1?00?= A?′/（2B）=?（c?2B?-1?）?′/（2B）=?（c?2b?-2?）?′/（2b?2）=-2c?2b?-3?b?′/（2b?2）=-c?2b?′b?-5

Following Example 3 we can obtain Table 5.

Table 5. Equivalence

Γ?0?01?=A?′/（2A）=-b?′b?-1?， Γ?1?00?=A?′/（2B）= -c?2b?′b?-5?，

Γ?1?11?=B?′/（2B）=b?′b?-1?， Γ?1?22?=-rB?-1?=-rb?-2?，

Γ?1?33?=-rB?-1??sin??2θ=-rb?-2??sin??2θ

6The expression of Group B

The expressions of group B in two kinds of mechanics have no difference. See Table 6.

Table 6. The expression of group B

（Γ?2?12?）?N?R=r?-1?，?（Γ?2?33?）?N?R=?sin?θ?cos?θ，?（Γ?3?13?）?N?R=r?-1?，?（Γ?3?23?）?N?R=?cot?θ

7Analyzing Schwarzschild Solution

Schwarzschild solution is a very important result in general relativity， practice has proved that it is suitable for the weak gravitational field very well.

As a part of Schwarzschild solution?B =?（1-2GM/（c?2r））??-1??is related closely with Lorentz factor，

γ?2=?（1-?r·?2/c?2）??-1?=?（1-2GM/（c?2r））??-1?=B=b?2

It is very clear that Schwarz Schild solution involves a Newtoneon formula of energy conservation.

r·?2/2=GM/r?（2）

In the case of the weak gravitational field，?GM/（c?2r）<<1，

b?2=B=?（1-2GM/（c?2r））??-1?≈1+2GM/（c?2r）≈

1+2GM/（c?2r））+?（GM/（c?2r））?2=?（1+GM/（c?2r））?2=γ?2

That is

γ=1+GM/（c?2r）=b?（3）

Form. （3） holds still in the case of the strong gravitational field.

8Solving Einstein field equation

Line element is expressed as below：

-c?2?d?τ?2=a?2（r）?d?t?2-b?2（r）?d?r?2-r?2?d?θ?2-r?2?sin??2θ?d?φ?2?（4）

In the following we will write a（r） and b（r） as a and b. The relation formulas between a and b are as following：

ab=c

a?′b+ab?′=0?（5）

Einstein field equation can be written as below：

R?μν?=?νΓ?σ?μσ?-?σ?Γ?σ?μν?+Γ?ρ?μσ?Γ?σ?ρν?-Γ?ρ?μν?Γ?σ?ρσ?=0?（6）

Eq. （6） can be rewritten as following：

R?00?=-?rΓ?1?00?+Γ?1?00?（Γ?0?01?-Γ?1?11?）-Γ?1?00?（Γ?2?12?+Γ?3?13?）=0

R?11?=?rΓ?0?01?+Γ?0?01?（Γ?0?01?-Γ?1?11?）-Γ?1?11?（Γ?2?12?+Γ?3?13?）=0

R?22?=-?rΓ?1?22?+?θ?Γ?3?23?+Γ?3?23?Γ?3?23?=0?（7）

Substituting the needed expressions of Tables 4 and 6 into Eq. （7）， we obtain：

R?00?=c?2b?″b?-5?γ?3-c?2b?′b?-5?γ?5?（5b?′b?-1?γ-3γ?′-2b?′b?-1?γ）+2c?2b?′b?-5?γ?3r?-1?=0

R?11?=-b?″b?-1?γ?3+b?′b?-1?γ?4?（b?′b?-1?γ?2-3γ?′γ+2b?′b?-1?γ?2）-2b?′b?-1?γ?3r?-1?=0

R?22?=?（rb?-2?γ?3）?′-1=0?（8）

Applying?γ= b?（Form. 3）， Eqs. （8） can be simplified as：

b?″+2b?′r?-1?=0

b?″+2b?′r?-1?=0

（rb）?′-1=0.?（9）

From Eq.（9） we can get

b=1+C/r?（10）

Comparing Exp.（10） with Form. （3） we can determine：

C=GM/c?2?，

so that

b（r）=1+GM/（c?2r）?（11）

using relation Form （5a） we can get

a（r）=c?（1+GM/（（c?2r））??-1?（12）

The relativistic solution can be expressed as

-c?2?d?τ?2=c?2?（1+GM/（c?2r））??-2?d?t?2-?（1+GM/（c?2r））?2?d?r?2-r?2?d?θ?2-r?2?sin??2θ?d?φ?2.?（13）

Appendix A.Schwarzschild Solution

Line element can be written as below：

-c?2?d?τ?2=A（r）?d?t?2-B（r）?d?r?2-r?2?d?θ?2-r?2?sin??2θ?d?φ?2?（A1）

In the following we will write?A（r）?and?B（r）?as?A?and?B?. Therelation formulas between A and B as following：

AB=c?2?（A2a）

A?′B+Ab?′=0?（A2b）

The expressions of connection coefficients are as below：

Γ?0?01?=A?′/（2A）=12A?′A?-1?， Γ?1?00?=A?′/（2B）=12c?2A?′A，

Γ?1?11?=B?′/（2B）=-12A?′A?-1?， Γ?1?22?=-rB?-1?=-c?-2?rA，

Γ?1?33?=-rB?-1??sin??2θ=-c?2rA?sin??2θ?（A3a）

Γ?2?12?=r?-1?， Γ?1?33?=?sin?θ?cos?θ， Γ?3?13?=r?-1?， Γ?3?23?=?cot?θ?（A3b）

Substituting the needed expressions in （A3a） and （A3b） into Eqs. （7）， ones can obtain：

R?00?=-?r（12c?-2?A?′A）+（12c?-2?A?′A）（12A?′A?-1?-（-12A?′A?-1?））-（12c?-2?A?′A）（r?-1?+r??-1?）=0?（A4a）

R?11?=?r（12A?′A?-1?）+（12A?′A?-1?）（12A?′A?-1?-（-12A?′A?-1?））-（-12A?′A?-1?）（r?-1?+r??-1?）=0?（A4b）

R?22?=-?r（-rA/c?2）+?θ?cot?θ+?cot?θ?cot?θ=0?（A4c）

Eqs.（A4a）～（A4c）can be simplified as

A?″+2A?′r?-1?=0?（A5a）

A?″+2A?′r?-1?=0?（A5b）

（rA）?′/c?2-1=0?（A5c）

From Eq. （A5c） ones can get

A=c?2（1+C/r）.

Analyzing the gravitational field， a conclusion is given （[1]， p.121）

h?00?=-2GM/（c?2r）=C/r，

So that

C =-2GM/c?2，

A（r）=c?2（1-2GM/（c?2r））?（A6a）

Applying relation Form. （A2a）， ones can get

B（r）=?（1-2GM/（c?2r））??-1?（A6b）

Therefore Schwarzschild solution can be written as

-c?2?d?τ?2=c?2（1-2GM/（c?2r））?d?t?2-?（1-2GM/（c?2r））??-1?d?r?2-r?2?d?θ?2-r?2?sin??2θ?d?φ?2?（A7）

References：

[1]FOSTER J.， Nightingale J.D.A short Course in General Relativity[M]. SecondEdition. 1995. New York： Springer Verlag， Chaps 2， 3.

[2]RINDLER W. Introduction to Special Relativity[M]. Oxford： Clarendon press，1982.