3.5.10 Energy and Force Density
Classical mechanics teaches that the energy expended in forcing an object to move a given distance is force times distance, and conversely, the force is proportional to the slope of the energy 'hill' it is 'climbing'. This is consistent with Newton's first law which states that, effectively, in flat space, an object will maintain a steady velocity. However, this is not the case in the Earth's gravity, where forces act in a direction towards the Earth's center. This implies very strongly that gravitation is associated with sloping space, as shown in the previous chapter.
Though it is clear that gravity must act, simultaneously, on all points within an object, gravitational force is assumed to act at a single point within a material object. Newton justified this assumption by showing that all the incremental forces acting within an object can be combined as a single force, \(F\), acting through a single point, which he called 'the center of gravity'. Newton calculated this force to be \(F= m \times GM_e/r^2 \), where \(m\) is the mass of the object, \(M_e\), is the mass of the Earth, \(G\), is the universal gravitational constant, and \(r\) is the distance from the object to the Earth's center.
It has been shown in the chapter on gravity that gravitational acceleration is related to the gradient of displacement, \(A/r^2\), where \(A\), is a constant related to the Earth's mass. Energy density within the gravitational structure is that of the to background energy density, which is \(\rho_o \).
The slope of energy density in the gravitational structure is the result of displacement as follows. \begin{align} force\ density\ f =&\ \nabla (\rho_o \times displacement) \ = \nabla( \rho_o\ \frac{A}{r}) = \rho_o \frac{A}{r^2}\nonumber \\ and\ since A=&\ \frac{M_e G}{H c}\ \ \ \ f= \frac{ \rho_o M_e G}{H c r^2} \nonumber \end{align} .
In order to calculate the gravitational force, \(F\), acting on an object via force density, \(f\), it is necessary to find an effective volume, \(V_{eff}\) of the object, so that \begin{align} Gravitational\ Force\ =&\ force\ density\ \times effective\ volume\ \nonumber \\ that\ is\ F= m \frac{GM_e}{r^2} =&\ f \times V_{eff} = \frac{ \rho_o M_e G}{H c r^2} \ V_{eff}\ \ \ \nonumber \\ consequently\ v_{eff}=&\ m \frac{GM_e}{r^2} \div \frac{ \rho_o M_e G}{H c r^2} = m \frac{H c}{\rho_o} \nonumber \end{align}
This result is incomplete since mass, \(m\), is in Kg, and \(\rho_o\) is in Joules per cubic meter. To bring these units into alignment, it is necessary to multiply \(\rho_o\) by the number of Joules in one Kg, that is \begin{align} \frac{energy}{mass} = c^2 \nonumber \end{align}. The final result is then\begin{align} V_{eff} = m \frac{Hc}{c^2 \rho_o} \nonumber \end{align}.
Since all material objects consist of atoms and molecules, the mass of an object can be represented by the number of atomic units, say \(N_a \), contained within a material object, multiplied by the mass of an atomic unit. The different densities of different materials is assumed to be due to the spacing of nucleons within the material. This implies that the effective volume of a material is the volume of its nucleons. To complete this analysis, it is necessary to find the volume of a nucleon.
The mass of an atomic unit, or nucleon, \(m_n\), is that of a neutron (or proton+electron pair), and is equal to \(1.675\ 10^{-27} \) Kg. So that the number of nucleons in a mass, \(m_o\), of any material is, \(N_a =m_o/ m_n\). Hence, the effective volume of the object, \(V_{eff} = N_a \times V_n\), where \(V_n\) is the volume of a single nucleon. For a single neutron, its effective volume, \(V_n\), is obtained by putting the mass of a nucleon, \(m_n\), into the expression, \begin{align} V_n =m_n \frac{H}{c \rho_o} \nonumber \end{align}
Using the following values, \(m_n = 1.675\ 10^{-27}\ \) Kg; \(H =2.64\ 10^{-18}\) per second; \(c=2.998\ 10^8 \) meters per second; \(rho_o = 7.49\ 10^{-10}\) Joules per cubic meter. \(V_n\) is calculated to be \( 19.69\ 10^{-45}\ cubic\ meters \).
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