Question

What is the physical significance of the fermi-field asymmetric stress-energy tensor?

Answer 1

Dark energy is generally thought to be nothing more than a nonzero cosmological constant. The signature of its stress-energy tensor should be +--- or -+++ depending on your favorite convention.


form

Tμνspin one=⎡⎣⎢⎢⎢⎢18π(E2+B2)Sx/cSy/cSz/cSx/c−σxx−σyx−σzxSy/c−σxy−σyy−σzySz/c−σxz−σyz−σzz⎤⎦⎥⎥⎥⎥Tspin oneμν=[18π(E2+B2)Sx/cSy/cSz/cSx/c−σxx−σxy−σxzSy/c−σyx−σyy−σyzSz/c−σzx−σzy−σzz]

created from a boson spin 1 electromagnetic field F=E+iBF=E+iB and changes the spin of FF from spin one to spin half, {F,A}→{ϕ+,ϕ−}{F,A}→{ϕ+,ϕ−} using here, then Tμνspin oneTspin .

Answer 2

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What is the physical significance of the fermi-field asymmetric stress-energy tensor?

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Here,

d = 4×10^(-3) m

t = 3×10^(-3) m

K = 3

C = ε0A/{d – t(1- 1/K)}

=> C = ε0A/{4×10^(-3) – 3×10^(-3)×(1- 1/3)}

Let x be the new distance between the plates

2/3C = ε0A/{x – 3×10^(-3)×(1- 1/3)}

=> 2/3 ε0A/{4×10^(-3) – 3×10^(-3)×(1- 1/3)} = ε0A/{x – 3×10^(-3)×(1- 1/3)}

=> (2/3)× 1/{4×10^(-3) – 3×10^(-3)×(1- 1/3)} = 1/{x – 3×10-3×(1- 1/3)}

Solving for x

=> x = 1/200 m

=> x = 5 mm