13. Space and time energy

In the artlcle Lorentz effect in the gravitational field, it can be proven to some extent that elementary particles, especially such as quarks, have a large difference in mass between when they are gathered together and when they are separated. When these particles gather, their mass increases roughly exponentially rather than proportionally to their number. Here, we can see that there are two types of mass. One is the inertial mass ($m$), which refers to an algebraic sum of the elementary mass($m_i$), and the mass due to gravity or acceleration ($m_g$). Expressing this in distance is as follows:

 

$m=\Sigma{m_i}$

 

(1)  $m_g=me^\theta= m(1+\theta+\theta^2/2+\theta_d) = m/\beta, m \gamma$, ($\theta=Gm/rc^2$)

 

In subatomic particles, $m_g$ is $m/\beta$, but in a gravitational field it is expressed as $m \gamma$. The difference between $m$ and $m_i$ in the gravitational field is not clear. But $m_g = m(1+\theta+\theta^2/2+\theta_d)$ and it appears that the first term represents the intrinsic energy of $m$, the second term represents the effect of the atoms, and the third represents the effect of the subatomic particles. In other words, all of these particles react individually to gravity. When $r$ is large, $m_g$ becomes almost equal to $m$.

Figure 1. Density of mass as a function of distance radius

 

In Figure 1, the mass is clear up to a certain radius, but beyond that, it is often unclear, like the atmosphere on Earth. And this would be the same in space and the subatomic world. In other words, $m_g \approx m$ can be interpreted to mean that $\Delta m$ becomes $0$ compared to the change in diameter. To explain this, we have no choice but to think that there is some invisible mass, like the Higgs boson, that continuously interacts with the mass $m_i$. Therefore, considering that $m$ is not clearly distinguished, it can be expressed in the same form as $m_g$. Normally, movement due to attraction between two objects is expressed as one revolving around the other, but in reality, the two objects move relative to the center of mass, and gravity can be considered as differentiating the product of two gravitational masses ($m_{g1}, m_{g2}$) by distance. That is,

 

$m_{g12}=\frac{m_1m_2}{m_1+m_2}e^{\theta_1+\theta_2} \approx m_1e^{\theta_2}$ if $m_2 \geqslant m_1, e^{\theta_1} \approx 1 $

 

$F_g= c^2 dm_{g12}/dr= c^2m_1\frac{d(e^{\theta_2})}{dr}$

 

$\approx -G m_1m_2/r^2$ if $e^{\theta_2} \approx 1 $

 

Rest mass $m_0$ is different from intrinsic mass $m$, but they are almost same in the place where gravity is small enough. The potential between two objects is expressed as a scalar, but if there are three or more objects, it will have to be processed as a vector.

 

Meanwhile, I have not made an exact calculation, but as the diameter increases, the number of $m_i$ increases, so $m$ will increase, but it is expected that there will still be a difference from $m_g$, like the term $\theta_d$ in the equation (1). $\theta_d$ can be $0$ or not.

 

Case 1.

Assuming that the low density of energy does not mean that the density of neighboring Planck particles decreases, but that their motions become less active, it can be concluded that their energy can't be estimated by mass, or other energy form until the motions reach to the certain quantized energy level. 

 

Case 2.

Energy is emitted in forms of energy that are rarely measured, such as bosons or gravitational waves.

 

In both cases, even though it is small and does not exert sufficient attractive force to attract other particles, it appears to be involved in the bending or time delay of light and the distance contraction effect in space through which light passes. Therefore this can be seen as space and time energy in the place where mass can't be estimated. Kinetic energy is expressed as velocity. Energy due to velocity is $m/\beta$ and $\gamma m$, which is in contrast to the expression of potential as an exponential function for distance. The reason appears to be that the contraction or expansion factors of distance and time are the same. That is, $v=v', dr'/dt'=dr/dt$. If you want to know the reason more deeply, refer to the next article 14.

Space_and_time_energy.pdf

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Last Modified in 2023.12.05