Appendix 2                                                                                     

An intuitive way to show that the dimension of mass can be   L³/T²

In a three-dimensional space, we will consider a volume V (a sphere of radius R, for example). This volume decreases continuously with the passage of time (the sphere shrinks) dV / dT. The volume decreases by a constant value every second and as a consequence the radius R decreases as well:

dV/dT = ( 4 p R² ) ( dR/dT ) (surface of the sphere . radius variation)
dR/dT = dV/dT / ( 4 p R² )

A point situated on the sphere will approach its centre at a velocity dR/dT, in other words:

  • According to the variation of volume in relation to time (dV / dT)
  • Inversely to the square of the distance from this point to the centre of the sphere (1 / 4 p R²).


So that it can remain at the same distance from the centre of the sphere, this point should distance itself from the centre at velocity dR/dT

  • Proportional to the variation of volume in relation to time (dV / dT)
  • Inversely proportional to the square of this distance (1 / 4 p R²)

Everything happens as if the area around P "swallowed" the space around it at a speed of dV / dT.



We will now consider the same sphere shrinking at an acceleration dV / dT² .We have :

dV/dT² = (4 p R² ) dR/dT² or :

dR/dT² = dV/dT² / (4 p R² )

To remain at the same distance from the centre of the sphere this point will distance itself from it with a constant acceleration of dR/dT²

  • Proportional to the variation of volume in relation to time (dV / dT²)
  • Inversely proportional to the square of the distance ( 1 / 4 p R² )

And exactly what a mass centred around P produces as a result: an acceleration

  • Proportional to the mass
  • Inversely proportional to the square of the distance 1 /4 p R²)

The value of m³ "swallowed" by (seconds)² by the mass of the Earth will be:

dV/dt² = 4 p Rt² g = (surface of the Earth) . g
Where g = ( K Mt ) / Rt ² (gravitational acceleration) thus

dV/dt² = ( 4 p Rt²) . ( ( K Mt ) / Rt ² ) = 4 p K Mt ( dV/dt² has the same dimensions as Mt )
By kg this results in (divided by the mass of the Earth Mt )
dV/dt² = 4 K = 8.3868 10^-10 m ³ / s ²
Where:

  • Rt = Earth's radius
  • 4p Rt² = surface of the Earth
  • Mt = Earth's mass
  • g = acceleration of gravity over the Earth = ( K Mt ) / Rt ² = 9.81 m / s ²
  • K = Newton constant = 6.674 10^-11

Mass can therefore be considered as the second derivative from the relationship between volume and time dV / dT² as dimension L³ / T²
( 1 kg = 8.3868 10^(-10 ) m³ / s ² or 1 m³ / s ² = 1.19235 10^9 kg )

The mass will be simply "a region of space that swallows the space around it faster and faster". This point of view corresponds exactly to what we observed in a mass at a certain distance: its power to accelerate bodies in its direction.

We often consider mass as a local deformation of "space-time". Why then wouldn't mass by itself have a dimension that depends exclusively on space and time?

We can therefore define mass as:

A region of space that swallows the space around it faster and faster

The natural unit of mass will be m³ / s² which will be worth 1 / (4 p K )  kg  =  1.19235 10^9  kg

With this unit of mass the Newton constant will be equivalent to 1 / 4 p

mon masse dimension mass dimension dimension mass dimension mass dimension mass dimension dimension mass dimension mass dimension mass dimension mass dimension mass dimensionension

Introduction
Space - Time
Electric Charge
Appendix 1