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x^2 + y^2 = c^2
x=a+b*i, y=u+v*i, c=n+m*i
(a+b*i)^2 + (u+v*i)^2 = (n+m*i)^2
(a^2 + abi + abi - b^2) + (u^2 + uvi + uvi -v^2) = n^2 + nmi + nmi - m^2
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2nmi - 2abi - 2uvi
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2i(nm - ab - uv)
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
___________________________________ = i
2(nm - ab - uv)
There are three cases of special relativity that remain untreated, first, when the
square root itself is negative, secondaly when the number beneath the root is imaginary.
Thirdly, when the root is negative and the lorentz factor imaginary.
First imaginary numbers result when (v^2/c^2) > 1
Using Relativistic_kinetic_energy_of_rigid_bodies from
en.wikipedia.org/wiki/Kinetic_…
M(1-v^2/c^2)^-1/2*v^2*i + Mc^2(1-v^2/c^2)^1/2*i ± E0 (the constant of integration can be negative)
Where M is mass, c is speed of light. It simplifies to
M(1-v^2/c^2)^-1/2*c^2*i ± E0 When (v^2/c^2) > 1
And when (1-v^2/c^2)^-1/2 is negative
Since "Every positive number a has two square roots: "(a)^1/2, which is positive, and -(a)^1/2, which is negative."
en.wikipedia.org/wiki/Square_r…
-M(1-v^2/c^2)^-1/2*c^2 ± E0 = kinetic energy
And the situation when the number is both imaginary and the root negative.
-M(1-v^2/c^2)^-1/2*c^2*i ± E0 = kinetic energy
And using the pythagore, vector combo theorem from above, we get real numbers again.
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
±M(1-v^2/c^2)^-1/2*c^2 *___________________________________ ± E0 = kinetic energy
2(nm - ab - uv)
x=a+b*i, y=u+v*i, c=n+m*i
(a+b*i)^2 + (u+v*i)^2 = (n+m*i)^2
(a^2 + abi + abi - b^2) + (u^2 + uvi + uvi -v^2) = n^2 + nmi + nmi - m^2
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2nmi - 2abi - 2uvi
a^2 - b^2 + u^2 - v^2 - n^2 + m^2 = 2i(nm - ab - uv)
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
___________________________________ = i
2(nm - ab - uv)
There are three cases of special relativity that remain untreated, first, when the
square root itself is negative, secondaly when the number beneath the root is imaginary.
Thirdly, when the root is negative and the lorentz factor imaginary.
First imaginary numbers result when (v^2/c^2) > 1
Using Relativistic_kinetic_energy_of_rigid_bodies from
en.wikipedia.org/wiki/Kinetic_…
M(1-v^2/c^2)^-1/2*v^2*i + Mc^2(1-v^2/c^2)^1/2*i ± E0 (the constant of integration can be negative)
Where M is mass, c is speed of light. It simplifies to
M(1-v^2/c^2)^-1/2*c^2*i ± E0 When (v^2/c^2) > 1
And when (1-v^2/c^2)^-1/2 is negative
Since "Every positive number a has two square roots: "(a)^1/2, which is positive, and -(a)^1/2, which is negative."
en.wikipedia.org/wiki/Square_r…
-M(1-v^2/c^2)^-1/2*c^2 ± E0 = kinetic energy
And the situation when the number is both imaginary and the root negative.
-M(1-v^2/c^2)^-1/2*c^2*i ± E0 = kinetic energy
And using the pythagore, vector combo theorem from above, we get real numbers again.
a^2 - b^2 + u^2 - v^2 - n^2 + m^2
±M(1-v^2/c^2)^-1/2*c^2 *___________________________________ ± E0 = kinetic energy
2(nm - ab - uv)
Suggested Collections
What I used to make this one. I was also thinking about how we catch baseballs.
en.wikipedia.org/wiki/Kinetic_…
en.wikipedia.org/wiki/Square_r…
en.wikipedia.org/wiki/Kinetic_…
en.wikipedia.org/wiki/Square_r…
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