One dimensional two-body function and equivalence principle
We will discuss the equality of two kinds of masses and the elevator trick.
１）A differential equation y"=-k^2/y^2
Put y'=v, then dv/dt=-k^2/y^2 →y^2dv=-k^2dt……①From dy/dt=v,dy=vdt……②.Divide ① by ②,we get y^2dv/dy=-k^2/v. Separate variables and integrate, then we get 1/2*v^2=-k/y+C（C is an integral constant）
The solution would be an implicit one, but it is interesting to chase the phase even at this stage. When y→∞,ｖ→constant. When y→constant,y→0. Change k^2 to GM and set the integral constant properly,and we could enjoy applying this equation to various situations. For example ,suppose C equals toc^2. When y→∞, v→ｃ.When y→2MG/c^2（Schwarzshild radius）,v→0..
２）The origin of this equation
y"=-k^2/y^2 is the equation of motion for the one dimensional two-body problem. The solution has been gained by use of Keplerian transformation
３）From ｍα=GMm/R^2……A to my"=GMm/y^2……B
The equation Ｂ is not the one which connects seemingly different kinds of terms like the equation A.
B is a differential equation concerning a function y=y(t).
Now let`s take a point of view that A is the result of B. B must have been fixed for a long time for Earth. The problem that why inertial mass equals to gravitational mass, ie why m=mi on the left side and m=mg on the right side are same does vanish automatically from our view point. m is only a multiplier to both sides. If mi does not equal to mg, the equality of the equation does not work. Whether A or Ｂ may be concerned,the right side stands for gravitational field and the left side stands for force.
But B indicates the following:
If a gravitational field exists,crossing the equal from the right to the left,we could naturally set a force. If a force exists,dividing it by m,we could gain α＝α（ｔ）.Let`s consider it as y" and integral it twice to get y. Using y, y" and G, we could make up M=M(t).
The right side shows the gravitational field.
note) About m.We should be careful whether the condition m<
4) The reason why my"=-GMm/y^2 is extended to my"=±GMm/y^2
It is not only a problem of attaching of the other signature.The gravitational field which inertial force brings up is located opposite to the "natural" one. So signature must be distinguished on the same co-ordinate system.Where there is a force,there is a gravitational field. Inertial force is also the case.Only the center of the field is located opposite.
５）How to use my"=±GMm/y^2
We cannot distinguish gravity（by my"=＋GMm/y^2）from inertial force（by my"=－GMm/y^2）because there are no other forces other than these two but microscopic,electromagnetic,or classically local.
For example, imagine we are standing in an elevator.
The center of the mass of the stable gravitational field exists under our feet. This is expressed by the right side of the plus equation. Our weight is ｍｇ.
The elevator begins to go down, and we feel a little bit uneasy. The inertial force has been born and is growing up. Just at the same time,the mass M~=M(t)~ has been born and is growing up,up over our heads. The stronger the inertial force becomes, the bigger M~ becomes and the nearer it comes down .At last mg＋my" equals to 0 .By the minus equation ,ｙ"=-g=-GM`m/y^2 is established .M~ and y are also decided so that M`/y^2=M/R^2 is held.
We cannot notice the invasion of the minus equation.
The ghost is over our heads, but we never know.
(no references but A Einstein`s The Meaning of Relativity, Princeton University Press,1922 )
We will discuss the equality of two kinds of masses and the elevator trick.
１）A differential equation y"=-k^2/y^2
Put y'=v, then dv/dt=-k^2/y^2 →y^2dv=-k^2dt……①From dy/dt=v,dy=vdt……②.Divide ① by ②,we get y^2dv/dy=-k^2/v. Separate variables and integrate, then we get 1/2*v^2=-k/y+C（C is an integral constant）
The solution would be an implicit one, but it is interesting to chase the phase even at this stage. When y→∞,ｖ→constant. When y→constant,y→0. Change k^2 to GM and set the integral constant properly,and we could enjoy applying this equation to various situations. For example ,suppose C equals toc^2. When y→∞, v→ｃ.When y→2MG/c^2（Schwarzshild radius）,v→0..
２）The origin of this equation
y"=-k^2/y^2 is the equation of motion for the one dimensional two-body problem. The solution has been gained by use of Keplerian transformation
３）From ｍα=GMm/R^2……A to my"=GMm/y^2……B
The equation Ｂ is not the one which connects seemingly different kinds of terms like the equation A.
B is a differential equation concerning a function y=y(t).
Now let`s take a point of view that A is the result of B. B must have been fixed for a long time for Earth. The problem that why inertial mass equals to gravitational mass, ie why m=mi on the left side and m=mg on the right side are same does vanish automatically from our view point. m is only a multiplier to both sides. If mi does not equal to mg, the equality of the equation does not work. Whether A or Ｂ may be concerned,the right side stands for gravitational field and the left side stands for force.
But B indicates the following:
If a gravitational field exists,crossing the equal from the right to the left,we could naturally set a force. If a force exists,dividing it by m,we could gain α＝α（ｔ）.Let`s consider it as y" and integral it twice to get y. Using y, y" and G, we could make up M=M(t).
The right side shows the gravitational field.
note) About m.We should be careful whether the condition m<
4) The reason why my"=-GMm/y^2 is extended to my"=±GMm/y^2
It is not only a problem of attaching of the other signature.The gravitational field which inertial force brings up is located opposite to the "natural" one. So signature must be distinguished on the same co-ordinate system.Where there is a force,there is a gravitational field. Inertial force is also the case.Only the center of the field is located opposite.
５）How to use my"=±GMm/y^2
We cannot distinguish gravity（by my"=＋GMm/y^2）from inertial force（by my"=－GMm/y^2）because there are no other forces other than these two but microscopic,electromagnetic,or classically local.
For example, imagine we are standing in an elevator.
The center of the mass of the stable gravitational field exists under our feet. This is expressed by the right side of the plus equation. Our weight is ｍｇ.
The elevator begins to go down, and we feel a little bit uneasy. The inertial force has been born and is growing up. Just at the same time,the mass M~=M(t)~ has been born and is growing up,up over our heads. The stronger the inertial force becomes, the bigger M~ becomes and the nearer it comes down .At last mg＋my" equals to 0 .By the minus equation ,ｙ"=-g=-GM`m/y^2 is established .M~ and y are also decided so that M`/y^2=M/R^2 is held.
We cannot notice the invasion of the minus equation.
The ghost is over our heads, but we never know.
(no references but A Einstein`s The Meaning of Relativity, Princeton University Press,1922 )