Thus the energy-momentum does not conserve in General Relativity. People have de- rived quantities, which conserve in time (e.g., [1]) and have claimed, that ...
Does energy conserve in General Relativity, or the Day the Einstein turns wrong (Dated: March 1, 2018)
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I.
THE PROBLEM
In Special Relativity holds T,νµ ν = 0 , Thus, from Gauss theorem I
T
µν
dσν =
Z
T,νµ ν dt dV = 0 ,
where dσν is the surface element, we can have conservation of energy-momentum: Z
T µ t dV = const ,
µ = t, 1, 2, 3 .
However, in General Relativity holds T,νµ ν 6= 0 .
T;νµ ν = 0 ,
Thus the energy-momentum does not conserve in General Relativity. People have derived quantities, which conserve in time (e.g., [1]) and have claimed, that they are energymomentum. But it is just a wishfull thinking, because, e.g., gµµ = 4 = const, but it is not energy-momentum. Because there are singularities in General Relativity, latter can not be regarded as predictable and conservative theory. Indeed, the vanishing of a body sometimes (miraculously) occurs at finite curvature of spacetime: From the [2] are presented the velocity components in the Kerr spacetime √ dr/dτ = − B/(r2 + a2 cos2 θ) , dθ/dτ =
q
L − cos2 θ (a2 (1 − E 2 ) + L2z /sin2 θ)/(r2 + a2 cos2 θ) ,
(1) (2)
dφ/dτ = (−(a E − Lz /sin2 θ) + a P/∆)/(r2 + a2 cos2 θ) ,
(3)
dt/dτ = (−a (a E sin2 θ − Lz ) + (r2 + a2 ) P/∆)/(r2 + a2 cos2 θ) ,
(4)
where B := P 2 − ∆ (r2 + (Lz − a E)2 + L), P := E (r2 + a2 ) − Lz a − q Q r, ∆ := r2 − 2 M r + a2 + Q2 . Here the charge of BH is Q = 0. The e is the electric charge of the falling body. The system of three equations dφ/dτ = 0, dθ/dτ = 0, dr/dτ = 0 taken at r = 20 and θ = π/4 √ are giving us the three constants: L = 249620/155672961, Lz = −80/ 155672961, and the 2
√ E = (1/12801) 155672961. The radial coordinate velocity of a test-body (falling from a large distance r0 = 20 with zero initial velocity and the θ0 = π/4) in Kerr BH with mass M = 1/2 and rotation a = 1/4 is √ B dr =− 2 . u ≡ dτ r + (1/16) cos2 θ r
B=−
(5)
640 4 742460 2 12481 62405 r + r3 − r + r− . 12801 155672961 194576 622691844
(6)
Therefore, must be B ≥ 0, but in r < rm = 1/640 the B < 0, so there is no falling body in 0 ≤ r < rm .
A.
The Navier-Stokes equation is not compatible with curved spacetime laws
Indeed, the N–S, Quantum Mechanics, and the perfect fluid equations are derived in assumption of energy-momentum conservation. But the General Relativity vanishes energy. Thus, indeed, there is conflict: [3]
[1] Arthur Komar, Covariant Conservation Laws in General Relativity, Phys. Rev. 113, 934, Published 1 February 1959. [2] Lightman AP, Press WH, Price RH, Teukolsky SA. Problem Book in Relativity
and
Gravitation.
Princeton,
Princeton
University
Press,
1975.
Available
at
http://apps.nrbook.com/relativity/index.html [3] Navier Stockes Equation, Integrals of Motion and Generalization of the Equation of Continuity of the Flow of Matter to the Theory of Relativity, vixra.org/abs/1712.0566
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