Exact Solution Of Einstein's Equation To The Gravitational Field Of ...

Ozean Journal of Applied Sciences 1(1), 200822 1 ⎛ 1−x ⎞2 ∂2∴ Γ33 = −2 2x22⎜−f⎟⎝ 2 ⎠ ∂x3 1 3k⎛ ∂g3k∂gk1∂g31⎞Γ31= g ⎜ + − ⎟21 3 k⎝ ∂x∂x∂x⎠3 1 1 ∂2Γ 31 =( f 2 ((1− x 2 ))2 2 1f2(1 − x2) ∂x3 1 1 ∂f2∴ Γ31= −2 f12 ∂x3 1 3k⎛ ∂g3k∂gk2∂g32⎞Γ32= g ⎜ + − ⎟2 2 3 k⎝ ∂x∂x∂x⎠3 1 1 ∂2Γ 32 = −( − f (((12 ))223− xf2(1− x2) ∂x23 x21∴ Γ32=21−x2Also, we get the non-zero components of Ricci tensor by applying Einstein law,0 1 2 3R = R + R + R +(13)11 110 111 112R1132( − f (1 − ))1 ⎛ 1 ∂f0⎞ ∂ ⎛ 1 ∂f1⎞ 1 ⎛ 1 ∂f1⎞ ⎛ 1 ∂f2⎞∴ R11=2⎜⎟ −⎜⎟ ++0 11 1 12⎜⎟⎜⎟(14)⎝ f ∂x⎠ ∂x⎝ f ∂x⎠ ⎝ f1∂x1⎠ ⎝ f2∂x1⎠0 1 2 3R 22 = R220+ R221+ R222+ R223(15)∴ ∂ ⎛ 1 1 ∂f2 ⎞ ⎛ 1 1 ∂f2 ⎞⎛1 1 ∂f2 ⎞22 =⎜−+ 21 2⎟ +⎜1 1 2⎟⎜1 1 2⎟∂x⎝ f ∂x⎠ ⎝ f ∂x⎠⎝f2∂x1⎠(16)R 33 = 0(17)0 1 2 300 = R000+ R001+ R002R003(18)R +∂ ⎛ 1 1 ∂f0⎞ 1 1 ∂f0 1 1 ∂f0∴ R00 = −x⎜12 f1x⎟ +(19)∂ ⎝ ∂1 ⎠ 2 f0∂x12 f1∂x1Due to rotational symmetry around the origin it is sufficient to write the field equations only for the equator( x 2 = 0); therefore, since they will be differentiated only once,Now we get the field equations from (14), (16), (17) and (18) by applying R = 0∂∂x1∂∂x∂∂x11⎛ 1 ∂f⎜⎝ f1∂x11⎛ 1 ∂f⎜⎝ f2∂x⎛ 1 ∂f⎜⎝ f1∂x01⎞ 1 ⎛ 1 ∂f⎟ =2⎜⎠ ⎝ f0∂x21⎞⎟ = 2 +⎠⎞ 1⎟ =⎠ f1f01f f1021⎛ ∂f⎜⎝ ∂x01⎞⎟⎠2⎛ ∂f⎜⎝ ∂x⎞⎟⎠21 ⎛ 1 ∂f1⎞+2⎜⎟⎝ f1∂x1⎠212⎞⎟⎠22⎛ 1 ∂f+⎜⎝ f2∂x212⎞⎟⎠αβ2(20a)(20b)(20c)34