Exceeding speed of light using mass-energy continuum
If speed of light is c m/s, then E=mc^2
If speed of light is c^2 m/s, then E=m(c^2)^2=mc^4
If speed of light is c^3 m/s, then E=m(c^3)^2=mc^6
If speed of light is c^(n+1) m/s, then
E=m(c^(n+1))^2=mc^(2n+2)
If n=0, E=mc^(2x0+2)=mc^2
If n=1, E=mc^(2x1+2)=mc^4
If n=2, E=mc^(2x2+2)=mc^6
If n=n, E=mc^(2n+2)
If n=0, let c0 be the default field of nth field, then speed of light will be,
c^(0+1) m/s=c m/s. Therefore, speed of light in c0 field is c m/s .
If n=1, speed of light in c1 field is, c^(1+1) m/s=c^2 m/s
If n=2, speed of light in c2 field is, c^(2+1) m/s=c^3 m/s
Therefore, if n=n, speed of light in nth field will be, c^(n+1) m/s
These fields follow conservation law of energy.
If m0 be the mass in c0 field then in nth field mass will be m0c^(-2n) . This can be proved by following.
If E0 be the energy of m0 mass in c0 field and En be the energy of m mass in cn field then,
according to conservation law of energy, E0=En
Or,m0c^2 = mc^(2n+2)
Or,m0=mc^2n
Or,m=m0c^(-2n)
Therefore if m0 be the mass of c0 field, in cn field it will be m0c^(-2n) .
It also follows the conservation law of energy.
=m0c^(-2n) c^(2n+2) [Since, m=m0c^(-2n)
=m0c^2
Therefore, En=m0c^2
It proves, E=En
Therefore, these fields always follow conservation law of energy.
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