Saturday, July 29, 2017

A Simple Derivation of General Relativity

According to Einstein's equivalence principle, a person accelerating upwards in an elevator (in outer space with no gravity) cannot distinguish it from gravity (downwards). Then acceleration and gravity are physically equivalent.

Assume a (laser) light send horizontally from one side (wall) of the elevator to other side (wall).

What is the Y coordinate of the beam for given X or T, if upwards constant speed of elevator is V?
x=c*t (assuming x is positive towards right)
y=v*t (assuming y is positive downwards)
m=y/x=(v*t)/(c*t)=v/c
Applying parametric to implicit conversion:
x=c*t => t=x/c => y=v*(x/c)=(v/c)*x=m*x => line with tangent m

What is the Y coordinate of the beam for given X or T, if upwards constant acceleration of elevator is A?
x=c*t (assuming x is positive towards right)
y=a*t^2 (assuming y is positive downwards)
Applying parametric to implicit conversion:
x=c*t => t=x/c => y=a*(x/c)^2=(a/c^2)*x^2 (parabola)
Geometry says:
if a parabola is y=x^2/(4*f) => f: focal length
The focal length of a parabola is half of its radius of curvature at its vertex => f=r/2
The radius of curvature is the reciprocal of the curvature (curvature of circle: 1/r)
Then:
y=(a/c^2)*x^2=x^2/(4*f) => a/c^2=1/(4*f) => 4*f*a/c^2=1 => f=c^2/(4*a)
r=2*f=c^2/(2*a) => curvature=1/r=1/(c^2/(2*a))=(1/1)/(c^2/(2*a))=(1/1)*((2*a)/c^2)=(2*a)/c^2
Newton's laws say: F=G*M*m/d^2 and F=m*a => Acceleration for unit mass in gravitational field of mass m:
a=F/m=F/1=G*M*1/d^2=G*M/d^2
Then:
curvature=(2*a)/c^2=(2*G*M/d^2)/c^2=2*G*M/c^2/d^2

Is this formula to calculate spacetime curvature correct (using mass of the object (star, planet etc) and distance from its gravitational center)? I have no idea. I searched online to find a similar formula to compare but could not found it.
If the formula is wrong I would like to know its correct expression (using same input variables M and d) of course. And also then, if it is possible to derive that formula from the same thought experiment.



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