Moving Rocks And Sun Worshippers

My take on moving rocks:  thermal gravity on the heated side of the rock in the Sun.

A layer of condensation under the rock frozen over in the early morning will aid its movement towards the Sun.  As the sun passes low over the horizon sometime blocked by the surrounding mountain might explain the sudden change in direction as evidenced by the trail behind the rock.   The Sun will be blocked over long period of time (weeks/months) for the Sun to change its rising position on the horizon noticeably.

All trails will have a slight curve as the Sun rises at a slightly different position each day and move slowly across the sky.  Movement stops when friction under the rock increases as the condensation melts and dries up.  Movement starts up again the next day on sunrise, towards a slightly different sunrise position, when the bottom of the rock is once again on a thin layer of frozen condensation.

And the psychological disorder to want to explain everything was once called religion.

Why Melt? Melting Without High Temperature

From the post "Young And On Heat"

$F=4\pi E_s\Delta L=F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

We see that by driving the system with  varying   $T$,  we can set the system into resonance at

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{4\pi E_s}{{4}\pi\rho}}$

where  $m={4}\pi\rho$,  a sphere of radius 1, mass per unit length,

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{E_s}{\rho}}$

$E_s=\sqrt{3}E$

where  $E$ is the linear Young's Modulus and  $\sqrt{3}$  the 3D constant for the case of a sphere.

($E_s=E$ for the case of a rod.)

It might be that the sphere will melt at this thermal resonance frequency irrespective of the magnitude of the temperature.

If we consider that on increasing temperature, atoms in a solid vibrates at higher and higher frequency about its mean position in the solid lattice, this would suggest that at a particular temperature when the atoms are vibrating at this resonance frequency,  $f_{res}$,  the atom will simply break away from the lattice.  That is to say, the solid melts.

So melting is a case of the atoms vibrating at resonance frequency in heat.  And that it is possible to melt without high temperature but by applying  $T$  at the correct resonance frequency.

Young And On Heat

Consider Young's Modulus in the case of a sphere,  $E_s$

$F_L=E_s\cfrac{A_o}{L_o}\Delta L$

when  $L_o=1$  and  $A_o=4\pi(1)^2$

$F_L=4\pi E_s\Delta L=F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

for the case of such a sphere under high temperature.  If we have,

$E_s=\sqrt{3}E$

where  $E$ is the linear Young's Modulus and  $\sqrt{3}$  the 3D constant, then

$4\pi E\Delta L=\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

We have,

$D.\cfrac{\partial (d_s)}{\partial T}=4\pi E\cfrac{\alpha}{\rho}\Delta L$  --- (*)

where  $\Delta L$  is the change in radius of the sphere,  $\rho$ is the density of the element,  $\alpha$
 is the linear thermal coefficient, and  $E$  the Young Modulus.

If both  $D$  and  $\cfrac{\partial (d_s)}{\partial T}$  are instead independent of the specific element, then the last formula provides a simple way of finding,

$D.\cfrac{\partial (d_s)}{\partial T}$    in general.

$\cfrac{\partial (d_s)}{\partial T}$  is a function in  $T$  and  $\Delta L$ changes with  $T$.  In which case  $D$  can be found by plotting  $f(T)$  vs  $\Delta L$.  Where  $f(T)$  is a function in  $T$  and  $f(T)$  vs  $\Delta L$  is a line through the origin with gradient,

$gradient=4\pi E\cfrac{\alpha}{\rho D}$

It should be noted that expression (*) is within the frame work of a solid lattice, an array of atoms in 3D.  The limiting case of an infinitely sparse lattice will then approximate the situation of free space.

Have a nice day.

Stress When The Heat Is On, Thermal Stress

From the post "Gravity Exponential Form Again",

$g_T=D.\cfrac{d(d_s(x))}{dx}$

And from the post "Possibilities, Possibilities, And God Created More Possibilities",

${d}_{s}(T)=\cfrac{({d}_{T_{max}}-{d}_{n})}{T_{max}}.T+d_n$

In general,

${d}_{s}=\cfrac{\partial (d_s)}{\partial T}.T+d_n$

which is basically the equation for a straight line when    $\cfrac{\partial (d_s)}{\partial T}$    is a constant.

$\cfrac{\partial ({d}_{s})}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}+\cfrac{\partial }{\partial x}\left\{\cfrac{\partial (d_s)}{\partial T}\right\}.T$

$\cfrac{\partial ({d}_{s})}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}+\cfrac{\partial }{\partial T}\left\{\cfrac{\partial (d_s)}{\partial x}\right\}.T$

If we assume that  $d_s$  itself does not varies in space,  but its distribution is wholly due to the variations in temperature in space, ie.  when

$\cfrac{d T(x)}{d x}=0$

$\cfrac{\partial ({d}_{s})}{\partial x}=0$

ie.  $d_s$  is uniform but redistribute solely due to  $T$,    so

$\cfrac{\partial }{\partial T}\left\{\cfrac{\partial (d_s)}{\partial x}\right\}.T=0$

(Note: There can be other treatment than setting this term to zero.)

That is to say,

$d_s(T)=d_s(T(x))=d_s(x)=d_s$

$\cfrac{\partial ({d}_{s}(T))}{\partial x}=\cfrac{\partial ({d}_{s}(x))}{\partial x}=\cfrac{\partial ({d}_{s})}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}$

As such,

$g_T=D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{\partial T(x)}{\partial x}$

If we assume a linear temperature profile along a elements/metal rod of original length  $L_o$,  then

$\Delta L=L_o.\alpha\Delta T$

where  $\alpha$  is the linear thermal expansion coefficient.

$\cfrac{\Delta T}{\Delta L}=\cfrac{1}{L_o.\alpha}$   ie

$\cfrac{d T(x)}{d x}=\begin{matrix} lim \\ \Delta L\rightarrow 0 \end{matrix}\cfrac{\Delta T}{\Delta L}=\cfrac{1}{L_o.\alpha}$

$g_T=D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{L_o.\alpha}$

A force per unit volume develops,

$\rho\sqrt{3}g_T=F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{L_o.\alpha}$

that is countered by the expanding structure (lattice) of the element,  where  $\sqrt{3}$ converts 1D to 3D

and  $\rho$  the density of the element.  Note that  $\alpha$  is the linear coefficient not the volumetric coefficient.  We are also assuming that the element expands uniformly in 3D, isotropic.

Thermal gravity then provides an explanation for the thermal expansion of elements, where for a sphere of  radius 1m,

$F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

This force pulls on the element causing it to expand.  Like a spring, a counter force,  $F_L$   acts against  $F_T$  due the increasing atomic distance in the lattice of the structure.  The result is a sphere of greater radius,  when both forces balances.   $F_T$  might be called thermal stress.  The resulting expansion is of course thermal expansion.

$\rho$  is specific to a particular element.  $D$  is the constant of proportionality in the gravity-space density formulation.  $\cfrac{\partial (d_s)}{\partial T}$  might be specific to a particular lattice structure and atom population.  And  $\alpha$  is the constant of proportionality in this formula; at this point it may not be element specific.  The point is to move all specificity to  $\cfrac{\partial (d_s)}{\partial T}$  or  $\rho$ instead, if possible.  Note:  $\alpha$  is element specific, more accurately specific to a particular lattice structure.   $\cfrac{\partial (d_s)}{\partial T}$   is approximately that of free space as the atoms themselves are mainly free space.

It's ALIVE! Gravity Lives, It Has A Pulse.

So, a LF oscillator at 7.5 Hz levitate because of the heat in the coil, not for the low frequency electromagnetic waves that it generates.  The lifting effect is due to the coil because we observed that the coil is upward always.  Lift does not occur immediately after the oscillator is turned on because heat has to buildup in the coil for a temperature gradient with the ambient.  This lift is on the hot body, in this case the oscillator coil and is proportional to the gradient of space density around the hot body.  Space density is perturbed because of the higher temperature.  But in order for observable lift,  this heat is to pulse at 7.489 Hz.  A sort of resonance occurs with Earth's gravitational wave also pulsing at 7.489 Hz.

But why?  Why would resonance generate greater lift?  This phenomenon suggests that earth is pulling at not a constant force but at a pulsated rate of 7.489 Hz.   Pushing against this force in phase with its pulse rate negate gravity.  But one cannot generate greater gravity by going anti-phase, with such explanation/understanding.

A simple electromagnet pulsing at 7.489 Hz acts to project a gravity beam.  It seems that instead of push it is possible to pull when a combination of three gravity beams are used.

How to pull?  Is magnet a natural space pump?  A natural gravity generator?   Why would two ends sucking at each other repel?  A strong enough magnet will stick to anything!?

Questions and more questions.

Seeding For Clouds

If hydrogen from the Sun is burning up where there's oxygen, then clouds formation is the result of H₂O condensing from this combustion process.  We have a new way of seeding for clouds and so rain, oxygen itself.  Oxygen introduced into the higher atmosphere will cause more hydrogen to burn and form clouds.

Not a difficult notion to test.

Flying Toaster

From the post "Gravity Wave and Schumann Resonance", Earth has a gravity wave at 7.489 Hz.  If there is a heat source powered by a 7.489 Hz pulsing source, so we have a  $g_T$  wave also at  7.489 Hz but attenuating, would that generate lift.  A flying toaster?    YES!

Erratum, 2D Flat and Flatulent

From the post "Wait A Magnetic Moment",

Since there are two degrees of freedom on a sphere,

The magnetic moment should be,

$S_e=\sqrt{2}q\omega r^2_{es}$

At light speed,

$S_e=\sqrt{2}qc r_{es}$

where  $\sqrt{2}$  is a constant in 2D, as a result of an added degree of freedom.

Out, Out And Away

When  $T=0$,  $g_T≠0$  so we have,  beyond the hot zone,

$g=g_T(T=0)e^{-\cfrac{x}{L}}$

where  $x=L$   is the extend of the hot zone.  If,

$g_T=\cfrac{G_{To}}{x}$  up to the hot zone,  then

$g_T(T=0)=\cfrac{G_{To}}{L}$

and

$g=\cfrac{G_{To}}{L}e^{-\cfrac{x}{L}}$

where  $x=0$  is on the boundary of  $T=0$.

Using the flux formulation, assuming no losses from  $x$  and beyond,

$g=\cfrac{G_{T}}{x^2}$

where  $G_T=L^2g_T(T=0)=LG_{To}$

$g=\cfrac{LG_{To}}{x^2}$

This gravity is positive outwards and drives all particles from within the hot zone into outer space.  Acting in the same region are, the gravity due to the mass of the hot body and gravity due to the mass of   $T$  (if this gravity exist).  Both of which are acting inwards.

Unfortunately,  $G_{To}=0$.

My Very Own Spin

This is how a electron might flip its spin, if spin to to be interpreted in such a naive way.

A photon at half the threshold frequency for photoelectric effect is directed at the atom,  their electrostatic interaction on just one hemisphere causes the atom as a whole to flip and thus the electron spin to be reversed.

In this case, the energy required for a spin reversal depended on the "lattice energy" that holds the atom in place as well as the moment of inertia of the atom itself.

Have a nice day.

Wait A Magnetic Moment

Consider an elemental ring of radius  $r_{es}$  on a sphere,  an electron is spinning around this ring with an angular velocity,  $\omega$,

The current due to this is,

$I=q\cfrac{\omega}{2\pi}$

The magnetic moment as a result of this current in a particular direction,

$M=I\pi r^2_{es}cos(\theta)=q\cfrac{\omega}{2\pi}\pi r^2_{es}cos(\theta)$

$M=\cfrac { 1 }{ 2 } q\omega { r }^{ 2 }_{es}cos(\theta)$

Now we consider the summation of all such moments confined to a sphere of radius  $r_{es}$,  up to  $\theta=\cfrac{\pi}{2}$

$s_e=\int^\frac{\pi}{2}_0{\frac { 1 }{ 2 } q\omega r^2_{es}cos(\theta)}dr$

$s_e=\frac { 1 }{ 2 } q\omega r^2_{es}\int^\frac{\pi}{2}_0{cos(\theta)}dr$

$s_e=\frac { 1 }{ 2 } q\omega r^2_{es}$

And so, the spin of an electron is,

$S_e=2s_e=q\omega r^2_{es}$

At light speed,

$S_e=qc r_{es}$

And given  $n$  valence electrons,

$S_e=2ns_e=nq\omega r^2_{es}$

This is not true, the effects of more than one orbiting electrons are not likely to sum simply.  In fact less than n times.

It does sum simply! The electrons are ATTRACTED to one another because of the B field they create, like two parallel current carrying wire.  The electron orbit in parallel and their magnetic moment sum numerically.

At light speed,

$S_e=nqc r_{es}$

Which is very interesting, provided  $e=e_s4\pi r^2_{es}$  when  $r\rightarrow0$.  So, why and how would a magnetic moment change?

Roasted In A Slow Burn

If the particles driven out of the Sun's hot zone is hydrogen then there 's a lot more burning out oxygen than fossil fuel, which make plant life even more important.

The spectrum of the Sun fits well with that of H₂O  but yet this combustion cannot occur beyond our atmosphere.  The spectrum of the sun in outer space must be different from within Earth's atmosphere.

If the heat in our ambient is due to hydrogen burning in the atmosphere that might explain why it get colder with higher altitude as oxygen is low.  And in areas where there's bright sunshine will have a low oxygen level, in spite of forestation and jungles.

One way to dispel these is to trap such particles in space, and see if they burn in oxygen rich environment.   Hydrogen or not.

Star Dust

Earth is not within the Sun's hot zone.  We are beyond the boudnary where  $T=0$.  We are not heated by  $T$  from the Sun, but we are bombard by the particles from the Sun that still experience finite  $g_T$ at the boundary  $T=0$.

Such particles enter into Earth's atmosphere and burnt up generating light and heat.  That is why you don't find photons in out space but a flare of them within the atmosphere.  It also explains why earth cools rapidly on the shadow side.   If we are within the hot zone,  $T$  should be in thermal equilibrium with Earth all around it,  no matter which direction you are facing.  $T$  is like a fluid that flows around  and engulf  all inside the hot zone.  Earth like a ball in water, there is no shadow side.  The dynamics of Earth's spin and rotation however might affect the flow and ebb of the star dust and so effect temperature immediately.

But what is the nature of this star dust?  If this is true.  Hydrogen  atoms?  And so the sky is blue?  Did Rayleigh scattering explained away everything?

Making it So, Squeeze!

$F_c=\cfrac{e^2}{\varepsilon_o}.\cfrac{1}{4\pi r^2}$

$a_c=\cfrac{F_c}{e}=\cfrac{e}{\varepsilon_o}.\cfrac{1}{4\pi r^2}$

From the post "General Field Equation (factor 2 correction)", where the maximum  $a_c$  is,

$\begin{matrix} \lim   \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e}{\varepsilon_o}.\cfrac{1}{4\pi r^2}\right\}\rightarrow c^\cfrac{3}{2}$

Consider,

$e=e_{s}4\pi r_{es}^2$

then, if we squeeze the electron to obtain the maximum acceleration as we would Earth to create a black hole,

$\begin{matrix} \lim   \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e}{\varepsilon_o}.\cfrac{1}{4\pi r^2}\right\}=\begin{matrix} \lim   \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e_{s}}{\varepsilon_o}\right\}=\cfrac{e_{s}}{\varepsilon_o}$

$\cfrac{e_s}{\varepsilon_o}\approx c^\cfrac{3}{2}$

$e_s\approx c^\cfrac{3}{2}\varepsilon_o$

This would suggest charge is a surface phenomenon.

Otherwise,

$e=e_v.\cfrac{4}{3}\pi r_{v}^3$

then

$\begin{matrix} \lim   \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{e}{\varepsilon_o}.\cfrac{1}{4\pi r^2}\right\}=\begin{matrix} \lim   \\ r\rightarrow 0 \end{matrix}\left\{\cfrac{1}{3}\cfrac{e_{v}}{\varepsilon_o}r_{v}\right\}\rightarrow 0$

However, by analogy this would also suggest that mass is a surface phenomenon too.

$m_{ss}\approx \cfrac{1}{4\pi G}c^{\cfrac{3}{2}}$

where  $m_{ss}$   is defined as

$m_{s}=m_{ss}4\pi r_{ss}^2$

That gravitational effects of a mass shields the effects of the mass below it, when $r\rightarrow 0$, at the atomic level.

And so,

${4\pi G}.m_{ss}\approx \cfrac{1}\varepsilon_o e_s$

Equivalence between mass and charge, at the atomic,  $r\rightarrow0$,  surface density level.

My guess is,

$m_{ss}\approx e_s$     and that,

${4\pi G}\approx \cfrac{1}\varepsilon_o$,    as    $r\rightarrow 0$

The question of how much for points is answered by volume density.  The same question for lines is answered by area density, for an area defined perpendicular to the lines.  Lines in a volume density, has redundant length information and is blind to changing area densities within the volume.

Have a nice day.

War Monger At Large

In a totally analogous way,

$F_f=4\pi G_{To}x.m_f\Phi$

where $4\pi G_{To}x$ is the thermal gravitation source and  $\Phi$ is the flux component,

$\Phi=\cfrac{1}{4\pi x^2}$

More importantly,  $F_f$  is the repulsive force between a thermal source and a hypothetical flame element  $m_f$,

This hopefully, will allow us to calculate the force down a barrel of a rifle, at the end of a rocket combustion chamber, inside a grenade or inside a bullet cartridge.  Together with polar plots of gravity like the post "Have A Heart, While It's Hot", we have a good view of the thermal gravitational force inside an exploding containment.

From previous posts, we see that if we cool the bullet or the solid projectile first before igniting the charge behind it,  the greater temperature gradient between it and the fire ball upon ignition will generate a greater thermal gravity that would drive the projectile to greater speed.

And from,

$G_{To}=\cfrac{G_T}{x_oln(T^2_{max})}$

$x_o$,  being the radius of the radiating body, suggests that the end seal of the explosion chamber should not be flat but instead of finite curvature, rounded and small.  $G_{To}$ is then higher to generate higher  $g_T$

Discrete From The Start

Why would you ever multiple two mass values or charge values?

Four force lines joins the right singular point mass or charge to the four on the left in the top picture.  When the two groups of points are far a part and very small, these force lines are parallel and are just summed to given the total force on the single point.  What if we have two other points on the right?  This group will experience a total 3 x 4 force lines.  If fact given two groups of mass/charges numbered at M and N, the total force lines between them is MxN.

Given a configuration of charges in two or more groups, MxN the total number of force lines between them are fixed.  Let's define a forces line per unit area term,  $\Phi$  over the area of a sphere around one of the group

${\Phi_1}.{4\pi r^2_1}={\Phi_2}.{4\pi r^2_2}=M.N$

and we have,

$\Phi_1=\cfrac{{\Phi_2}.{4\pi r^2_2}}{4\pi r^2_1}=\cfrac{M.N}{4\pi r^2_1}$

In general,

$\Phi=\cfrac{M.N}{4\pi r^2}$

${\Phi}$ =  number of force line per unit area = flux,

What is this force lines per unit area over a sphere?  Image a force being divided over small points covering a sphere.  Each of these points joins the center of the sphere along a radius.  The force at each of these points is a force along the radial line of the sphere.

Compare this with,

$F_c=\cfrac{q_M.q_N}{4\pi {\varepsilon_o} r^2}=\cfrac{Me.Ne}{{\varepsilon_o}}\cfrac{1}{4\pi  r^2}=\cfrac{e^2}{{\varepsilon_o}}\cfrac{M.N}{4\pi  r^2}=\cfrac{e^2}{{\varepsilon_o}}\Phi$

We found that the  $\cfrac{1}{r^2}$ dependence is the result of the total flux being conserved and that the source of this total flux is,

$F_{co}=\cfrac{e^2}{{\varepsilon_o}}$    and so,

$F_c=F_{co}\Phi=\cfrac{e^2}{{\varepsilon_o}}\Phi$

In a similar way,

$F_m=\cfrac{Gm_Mm_N}{r^2}=4\pi G\cfrac{Mm_s.Nm_s}{4\pi r^2}=4\pi Gm^2_s\cfrac{MN}{4\pi r^2}$

$F_m=4\pi Gm^2_s\Phi$

where $m_s$ is an elemental mass,  $G$ is the gravitational constant.  Let

$F_{mo}=4\pi Gm^2_s$    and so,

$F_m=F_{mo}\Phi=4\pi Gm^2_s\Phi$

The point is the basic number counting in such formula.  They assume that charges or mass are made up of countable elemental point quantities and there are distinct countable force lines.  In the first place, discrete.  And that the formulation can be separated into a source part and a flux part (flux are just lines).

If such formula are experimental and valid, then an elemental mass  $m_s$  exists already, thanks to Newton.

Flame Blue and Naked

This is a picture of a flame in zero gravity from the internet (howspectacular.com)

Flame In Zero Gravity

The original credits goes to NASA.

Flame In Zero Gravity Upside Down

Another flame upside down.

A temperature gradient is set up around the center of the flames,  temperature decreases by $\cfrac{1}{\sqrt{x}}$, spherically around the center.

One possible explanation is that  $g_{T}$ drives all the reactants beyond the hot zone before they have a chance to combust.  Under normal gravity reactants at the top are slowed by gravity and remains in the hot zone longer.  They combust, releasing even more energy and raise the temperate further.  $g_T$  increase further (before $\Delta d_s = d_n$), in this top region and the result is an elongated flame upwards.

Flame Under Normal Gravity

The bottom part of the flame where most of the reactants escape resembles the blue flame in zero gravity.

In the second picture, where the flame is upside down under zero gravity, some of the reactant bounced off the top hard surface and re-enter the hot zone.  Further combustion causes the color change inside.

Have a nice day.

Have A Heart, While It's Hot

The following is a plot of

$g^2_{net}=g^2_Tsin^2(\theta)+(g_Tcos(\theta)-g)^2$

as polar plot,  r = (A^2(sin(t))^2+(Acos(t)-10)^2)^(1/2),  where  g = 10,  and A = 0.5, 5, 9, 12, 20, 30

where for values of  A less than  g,  the curve is plotted in red.  Looks like the zones of a candle flame.

Hot air raises because of  $g_T$.  Hot air having less density so raises only if it is contained.  Otherwise fast molecules just go anywhere.

Have a nice day.

Where Most People Crack Too, Up Top.

$g_T$ could be the reason why surface crack on heating.

$g_{T}$  greatest where  $x_o$  is small resulting in high curvature.

For the case of a sphere,

$g^2_{net}=g_Tsin^2(\theta)+(g_Tcos(\theta)-g)^2$

If we let,

$2g_{net}\cfrac{\partial g_{net}}{\partial\theta}=2g_Tsin(\theta)cos(\theta)-2(g_Tcos(\theta)-g)sin(\theta)=0$

$2gsin(\theta)=0$

$\theta$ = 0^o   or   90^o

Now consider,

$2(\cfrac{\partial g_{net}}{\partial\theta})^2+2g_{net}\cfrac{\partial^2 g_{net}}{\partial\theta^2}=2gcos(\theta)$

when  $\theta$ = 0 ^o

$\cfrac{\partial^2 g_{net}}{\partial\theta^2}>0$

We know that at 0^o,  $g_{net}$ is a minimum  from $\cfrac{\partial^2 g_{net}}{\partial\theta^2}$

This means, forces at the side of the surface beyond 0^o to the central line are pulling the outwards with higher strength.  This force distribution tends to open the top of the heated body where  $g_{net}$  is minimum.

Levitating Hot Body

Is it possible that a hot body levitate,  since the gravity it produces is opposite to that of Earth's.

For this construct to lift,

$Mg_T-(M+m)g>0$

We assume a linear temperature profile,

$T=T_{max}(1-\cfrac{x}{d})$

and a log squared space density profile,

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(\cfrac { 1 }{ T_{ max }^{ 2 } } ) } ln(\cfrac { 1 }{ T^{ 2 }(x) } )+{ d }_{ n }$

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(\cfrac { 1 }{ T_{ max }^{ 2 } } ) } ln(\cfrac { 1 }{ T^2_{max}(1-\cfrac{x}{d})^2 } )+{ d }_{ n }$

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } ln(T^{ 2 }_{ max }(1-\cfrac { x }{ d } )^{ 2 })+{ d }_{ n }$

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (ln(T^{ 2 }_{ max })+ln((1-\cfrac { x }{ d } )^{ 2 }))+{ d }_{ n }\}$

$\cfrac { d(d_{ s }(T)) }{ dx } =\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (-\cfrac { 1 }{ (1-\cfrac { x }{ d } )^{ 2 } } \cfrac { 2 }{ d } (1-\cfrac { x }{ d } ))$

$\cfrac { d(d_{ s }(T)) }{ dx } =\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (-\cfrac { 2 }{ (d-x ) } )$

And we have,

$g_{ T }=\cfrac { { G }_{ T } }{ { x }_{ o }({ d }_{ n }-{ d }_{ T_{ max } }) } \cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (-\cfrac { 2 }{ (d-x) } )$

$g_{ T }=\cfrac { 2{ G }_{ T } }{ { x }_{ o }ln(T_{ max }^{ 2 }) }\cfrac { 1}{ (d-x) }$

Therefore,  at  $x=0$,

$M\cfrac { 2{ G }_{ T } }{ { x }_{ o }ln(T_{ max }^{ 2 }) } \cfrac { 1}{ d }-(M+m)g>0$

From this equation we see that small  $x_o$  (radius of body)  and small  $d$  provide for greater lift.  We shall now optimize  $T_{max}$  for greater lift.

It seem odd that  $g_T$  should be inversely proportional to  $T_{max}$.  In free space where  $T_{max}$  is allowed to fall over a  distance  $x=L$  to reach  $T=0$  increasing  $T_{max}$  increases  $L$  but  $d_s$  cannot fall below zero.  $\Delta d_s= d_n$  remains fixed,  as such,

$\cfrac{d(d_s(x))}{dx}=\cfrac{d_n}{L}$    when we use a linear approximation, decreases with increasing  $T_{max}$.  And so,  $g_T$  decreases.

We can fix  $L$  by placing a cold body at distance  $d$  but still  $\Delta d_s= d_n$  remains the same.

So in practice  $T_{max}$  should be increased until  $g_T$  and so lift is at a maximum and then no further.  At this point  $\Delta d_s=d_n$  and cannot be increased further.

Thermal Mass

If we start from,

$T(x)=\cfrac{A}{\sqrt{x}}$

A body experiences gravitational effect wholly due the mass behind it,  the net effect of all mass above the sphere containing the hot mass is zero.

If we introduce the concept of thermal mass, where

$M_{thermal}=T^4(x)=\left\{\cfrac{A}{\sqrt{x}}\right\}^4$,    numerically

We have the total  $M_{thermal}$  behind the body at point  $x$ from the center of  $T$.

$M_T=   \int _{ 0 }^{ x }{ 4\pi x^{ 2 }T^4(x) } dx=C\int _{ 0 }^{ x }{ 4\pi x^{ 2 }\cfrac { 1 }{ x^2  }  } dx=C.4\pi x$

where  $C=A^4$  is a constant.

Then we consider a hot body of total equivalent mass  $C.4\pi x$,  what is its thermal gravity outwards?   If we consider spheres of area

$4\pi x^2$

and the total thermal gravitational flux through them from a unit mass of  $T$,  assuming space is empty and nothing else affects the flux between  $x_o$  and  $x$,

$g_{To}.4\pi x^2_o=g_T.4\pi x^2$

and that at  $x=x_o$,  $G_{T}=g_{To}.4\pi x^2_o$

$g_T=\cfrac{G_{T}}{4\pi x^2}$    per unit mass

$G_{T}$  increases with equivalent mass by  $C.4\pi x$

$g_T=\cfrac{G_{T}}{4\pi x^2}C.4\pi x$

Let  $G_{To}=G_TC$

$g_T=\cfrac{G_{To}}{ x}$

This gravity is in the direction of  $x$,  outwards.  This is the result from the post "Thermal Gravity And Lost Innocence" previously,  that is to say the equivalent Thermal Mass concept

$M_{thermal}=T^4(x)$  

assumes that  $g_T=\cfrac{G_{To}}{ x}$.  This Thermal Mass generates a positive gravity outwards.

Space Density Under Temperature

From the post "Possibilities, Possibilities, And God Created More Possibilities",

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ -ln({ T_{ max }^{ 2 } } ) }( ln(x )-ln(A^2))+{ d }_{ n }$

${ d }_{ s }(T)=A.ln(x )+B$

A plot of log( x ) + 1 is shown below

When  $T=T_{max}$  space density is  $d_{Tmax}$  when  $T=0$,  $d_s=d_n$  at normal space density.  The derivative of  space density,

$\cfrac{d(d_s(x))}{dx}$  gives thermal gravity.

This is assuming that  $g_T=\cfrac{G_{To}}{x}$.

And Newton Rolls Over In His Grave

Just you might be happy with the previous derivations,

If we start from,

$T(x)=\cfrac{A}{\sqrt{x}}$

A body experiences gravitational effect wholly due the hot mass behind it,  the net effect of all hot mass above the sphere containing the hot mass is zero.

We have the total  $T(x)$  behind the body at point  $x$ from the center of  $T$.

Total  $T(x)$  in a sphere of   $T(x)$   of radius  $x$,

$M_T=   \int _{ 0 }^{ x }{ 4\pi x^{ 2 }T(x) } dx=\int _{ 0 }^{ x }{ 4\pi x^{ 2 }\cfrac { A }{ \sqrt { x }  }  } dx=\cfrac { 8A\pi  }{ 5 } { x }^{ \cfrac { 5 }{ 2 }  }$

Then we consider a hot body of total equivalent mass  $M_T$,  what is its gravity inward?   If we consider spheres of area

$4\pi x^2$

and the total gravitational flux through them from a unit mass of  $T$,  assuming space is empty,

$g_{o}.4\pi x^2_o=g.4\pi x^2$

and that at  $x=x_o$,  $G=g_{o}.4\pi x^2_o$

$g=\cfrac{G}{4\pi x^2}$    per unit mass

$G$  increases with equivalent mass by  $\cfrac { 8A\pi  }{ 5 } { x }^{ \cfrac { 5 }{ 2 }  }$  as  $x$  increases.

$g_T=\cfrac{G}{4\pi x^2}\cfrac { 8A\pi  }{ 5 } { x }^{ \cfrac { 5 }{ 2 }  }=G_{o}\sqrt{x}$

where  $G_{o}=\cfrac{2AG}{5}$

This is the gravity due to the mass of  $T$  which is here assumed to be proportional to  $T$.  It is assumed here that gravity is directly proportional to its mass.  $g_T$  increase in the square root of $x$ until  $T=0$  at some distance  $x=L$  from the hot body.

Tectonic Shells And Dancing Flames

The interaction of the two types of gravity,  $g_t$    and   $g$   that leads to the formation of rings and spheres around the planet might also results in spherical shell structure of Earth's interior when the planet cooled from a hot mass over time.

If this is true than the zero gravity cross over points that marks the gap boundaries might still mark the boundaries of the Earth's tectonic shells.  The variations in net gravity can also account for the relative rotation speeds of these shells.

Furthermore, this thermal gravity is readily observed in a naked candle flame as oblong shaped spheres standing upright inside the flame.  $g_T$  accelerates heavier particles less; light particles are driven further out the flame.  This is especially evident on the side and bottom of the flame where  $\mathbf g_{net}=\mathbf g_T+\mathbf g$ is high.  Thermal gravity is responsible for the flame layered combustion zones.

Gravity Exponential Form Again

Consider s space density function of the form,

${ d }_{ s }(x)=A{ e }^{ -bx }+B$  where A, B and b are constant to be determined.

At  ${ d }_{ s }(0)={ d }_{ e }$  space is compressed, its space density is  ${d}_{s}$,  on surface of earth. And  ${ d }_{ s }(x\rightarrow \infty )={ d }_{ n }$  where at long distance, space is relaxed and the corresponding space density is ${d}_{n}$.

We have,

${ d }_{ s }(x)={ e }^{ -bx }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }$ ----(1)

Then we let the inverse relationship between time speed squared  ${v}^{2}_{t}$  and space density,  ${ d }_{ s }(x)$   to be,

${v}_{t}^{2}=C-D{ d }_{ s }(x)$   where  $C, D$   are to be determined.

We know that as  $x\rightarrow \infty, { d }_{ s }(x)\rightarrow{ d }_{ n }$  where space is relaxed and time speed is $c$,

${v}_{t}^{2}={c}^{2}$,  ${ d }_{ s }(x)={ d }_{ n }$

So,

$C={c}^{2}+D{d}_{n}$    then,

${v}_{t}^{2}={c}^{2}-D({ d }_{ s }(x)-{d}_{n})$

Differentiating with respect to time,

$2{v}_{t}\cfrac{d{v}_{t}}{dt}=-D\cfrac{d({d}_{s}(x))}{dx}\cfrac{dx}{dt}=-D\cfrac{d({d}_{s}(x))}{dx}v_s$ ---(2)

But from the energy equation,

$2{v}^{2}_{t}+{v}^{2}_{s}={c}^{2}$ differentiating it

$4{v}_{t}\cfrac{d{v}_{t}}{dt} + 2{v}_{s}\cfrac{d{v}_{s}}{dt}=0$, since $\cfrac{d{v}_{s}}{dt}=g$

$2{v}_{t}\cfrac{d{v}_{t}}{dt} = -g.{v}_{s}$    substitute into (2)

and so,

$g=D.\cfrac{d({d}_{s}(x))}{dx}$

From (1), differentiating with respect to x,

$\cfrac{d({d}_{s}(x))}{dx}=-b{e}^{-bx}({d}_{e}-{d}_{n})$ subsitute into the above we have,

$g=-D({d}_{e}-{d}_{n}).b{e}^{-bx}$

when  $x=0$,    $g=-g_o$

$g_o=\cfrac{G_o}{r^2_e}=D({d}_{e}-{d}_{n}).b$

$g=-g_o{e}^{-bx}=-\cfrac{G_o}{r^2_e}{e}^{-bx}$

And if  we consider

$\int^{\infty}_0{g}dx=\cfrac{G_o}{r_e}=\int^{\infty}_0{\cfrac{G_o}{r^2_e}{e}^{-bx}}dx$

$\cfrac{G_o}{r_e}=\cfrac{G_o}{r^2_e}[^{\infty}_0-{e}^{-bx}\cfrac{1}{b}]$

$r_e=\cfrac{1}{b}$

$b=\cfrac{1}{r_e}$

So,

$g=-g_oe^{-\cfrac{x}{r_e}}=-\cfrac{G_o}{r^2_e}{e}^{-\cfrac{x}{r_e}}$

This expression is based on compression of free space, in a exponential manner.

Storing Heavy Elements, Where It Get Larger

From the post "Temperature, Space Density And Gravity",

If an electron collide into the nucleus,

$r_o=\cfrac{q^2}{4\pi \varepsilon_o{m_ec^2}}<r_{n}$

where $r_{n}$ is the radius of the positive nucleus.

We see that as  $c$  increases in less dense space  $r_o$,  the orbital radius of electron decreases and it might clash into the nucleus and annihilate with a positive charge there in a matter/antimatter fashion.

This suggests that heavier elements of larger  $r_n$, nucleus radius,  is less stable from such possible collisions.  It also suggest that if we are able to pump space into a containment and make space inside denser  ($c$  decreases),  heavy elements in it will be more stable.

And oddly, in denser space, atoms are larger because of greater electron orbits.

General Field Equation (factor 2 correction)

In the case for gravity, time slows down in denser space, but the total energy of the body is conserved, so a decrease in time speed leads to an increase in space speed.  The body accelerates, without the action of a immediate force.  No touch, touch.  Similarly, the present of a charge slows down time... an assumption of course.  In both repulsion and attraction time slows down, and the charge body experiences acceleration in the space dimension.  The question remains however, as to which direction is this acceleration.  ( the anser is in the post "i'm Imagining It...")  The initial condition for the specific case (same sign charges or different sign charges) will provide for the direction of the acceleration.  Combining both scenarios, we have the following postulation:

If the field effects such as gravity and charge, where, spreaded out in a region of space, a body experience acceleration without a contact force, is due only to time variation expressed through the conservation of energy equation across the time and space dimensions,

$qv^2_{t} + \frac{1}{2}qv^2_c=\frac{1}{2}qc^2=constant$

$2v^2_{t} + v^2_c=c^2$

The factor of two before time velocity,  $v_t$,  is due to the fact that the kinetic energy along the time dimension at speed is not

$E=\frac{1}{2}mv^2_t$  but

$E=mv^2_t$

Differentiating with respect to $x$,

$4v_{ t }\cfrac { dv_{ t } }{ dx } +2v_{ c }\cfrac { dv_{ c } }{ dx } =0$

$2 v_{ t }\cfrac { dv_{ t } }{ dx } +v_{ c }\cfrac { dv_{ c } }{ dt } \cfrac { dt }{ dx } =0$

$2 v_{ t }\cfrac { dv_{ t } }{ dx } +a_{ c }=0$  ----(1)

$2v_{ t }\cfrac { dv_{ t } }{ dt } \cfrac { dt }{ dx } +a_{ c }=0,$

$2\cfrac { v_{ t } }{ v_{ c } } \cfrac { dv_{ t } }{ dt } +a_{ c }=0$

Differentiating (1) again with respect to $x$,

$2(\cfrac { dv_{ t } }{ dx } )^{ 2 }+2v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } +\cfrac { da_{ c } }{ dx } =0$

$-\cfrac { da_{ c } }{ dx } =2(\cfrac { dv_{ t } }{ dx } )^{ 2 }+2v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } }$

From the post "Calculus We Have A Problem",

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$

So,

$-\cfrac { da_{ c } }{ dx } =2(\cfrac { dv_{ t } }{ dt } \cfrac { dt }{ dx } )^{ 2 }+2\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$

$-\cfrac { da_{ c } }{ dx } =2(\cfrac { 1 }{ v_{ c } } )^{ 2 }(\cfrac { dv_{ t } }{ dt } )^{ 2 }+2\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$ ----(*)

Differentiating the conservation of energy equation with respect to time $t$ twice,

$4v_{ t }\cfrac { dv_{ t } }{ dt } +2v_{ c }\cfrac { dv_{ c } }{ dt } =0,$

$2v_{ t }\cfrac { dv_{ t } }{ dt } +v_{ c }.a_{ c }=0$

$2(\cfrac { dv_{ t } }{ dt } )^{ 2 }+2v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +a^{ 2 }_{ c }+v_{ c }\cfrac { da_{ c } }{ dt } =0$

$2\cfrac { v_{ t } }{ v^2_{ c } } \cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +\cfrac { 2 }{ v^2_{ c } } (\cfrac { dv_{ t } }{ dt } )^{ 2 }=-\cfrac{a^2_{ c }}{v^2_c}-\cfrac { 1}{ v_{ c } } \cfrac { da_{ c } }{ dt }$

Substitute into (*)

$-\cfrac { da_{ c } }{ dx } =2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }-\cfrac{a_{ c }}{v^2_c}-\cfrac { 1}{ v_{ c } } \cfrac { da_{ c } }{ dt }$

$-\cfrac { da_{ c } }{ dx } =-\cfrac { da_{ c } }{dt}\cfrac{ dt }{dx}=-\cfrac{1}{v_c}\cfrac { da_{ c } }{dt}$

$2\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }=\cfrac{a^2_{ c }}{v^2_c}$

since, ${v}^2_{c}={c}^{2}-{v}^{2}_{t}$

$({c}^{2}-{v}^{2}_{t})2{ v_t }\cfrac { dv_{ t } }{ dt }=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})\cfrac { d(v^2_{ t }) }{ dt }=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})\cfrac { d(v^2_{ t }) }{ dx }\cfrac{dx}{dt}=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})v_c\cfrac { d(v^2_{ t }) }{ dx }=a^3_{ c }$

$({c}^{2}-{v}^{2}_{t})\sqrt{{c}^{2}-{v}^{2}_{t}}\cfrac { d(v^2_{ t }) }{ dx }=a^3_{ c }$

but, from (1)

$2v_t\cfrac{d{v}_{t}}{dx}=-{a}_{c}$

$\cfrac{d({v}^2_{t})}{dx}=-{a}_{c}$

$({c}^{2}-{v}^{2}_{t})^{\cfrac{3}{2}}(-a_c)=a^3_{ c }$

$a^2_c=-({c}^{2}-{v}^{2}_{t})^{\cfrac{3}{2}}$

$a^2_c=-c^3(1-{\gamma}^{2})^{\cfrac{3}{2}}$

$a_c=ic^\cfrac{3}{2}(1-{\gamma}^{2})^{\cfrac{3}{4}}$

which is true because  $a_c$  is perpendicular to  $v_t$.

$|a_{c max}|=c^\cfrac{3}{2}$    when   $\gamma=0$

Maximum  $a_c$ not  $c^2$  but  $c^\cfrac{3}{2}$.

Calculus, We Have A Problem

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } = v_{ t }\cfrac { d }{ dx } (\cfrac { dv_{ t } }{ dx } )$

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } = v_{ t }\cfrac { d }{ dx } (\cfrac { dv_{ t } }{ dt } \cfrac { dt }{ dx } )$

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =v_{ t }(\cfrac { d }{ dx } (\cfrac { dv_{ t } }{ dt } )\cfrac { dt }{ dx } +\cfrac { dv_{ t } }{ dt } \cfrac { d }{ dx } (\cfrac { dt }{ dx } ))$

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =v_{ t }(\cfrac { d }{ dt } (\cfrac { dv_{ t } }{ dt } )(\cfrac { dt }{ dx } )^{ 2 }+\cfrac { dv_{ t } }{ dt } \cfrac { d^{ 2 }t }{ dx^{ 2 } } )$

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =v_{ t }(\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } (\cfrac { dt }{ dx } )^{ 2 }+\cfrac { dv_{ t } }{ dt }\cfrac { 1 }{ a_c } )$

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =v_{ t }(\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } \cfrac { 1 }{ v^2_c }+\cfrac { dv_{ t } }{ dt } \cfrac { 1 }{ a_c } )$

$v_{ t }\cfrac { d^{ 2 }v_{ t } }{ dx^{ 2 } } =\cfrac { v_t }{ v^2_c }\cfrac { d^{ 2 }v_{ t } }{ dt^{ 2 } } +\cfrac { v_t }{ a_c }\cfrac { dv_{ t } }{ dt }$

In a Vaccuum

Find  $g_T$  vs  $T$.

Possibilities, Possibilities, And God Created More Possibilities

$T(x)=\cfrac{A}{\sqrt{x}}$ --- (1)

From the post "Gravity Exponential Form",

$g=\cfrac{{G}_{o}}{{r}_{e}({d}_{e}-{d}_{n})}.\cfrac{d({d}_{s}(x))}{dx}$,

this expression considered space density and conservation of energy,  there is a discrepancy where

${v}^{2}_{t}+\frac{1}{2}{v}^{2}_{s}={c}^{2}$  instead of  ${v}^{2}_{t}+ {v}^{2}_{s}={c}^{2}$

The extra factor of  $\frac{1}{2}$  in front of   $v^2_s$  is eventually absorbed into  $G_o$.

We will here consider

${v}^{2}_{t}+\frac{1}{2}{v}^{2}_{rg}={c}^{2}$

where  $v^2_{rg}$  is deemed to be thermal and is responsible for manifesting thermal gravity as time speed  $v_t$  slows in denser space.  ($v^2_{rc}$  was used in charges that annihilate along the charge time dimension.  The charges collided with  opposite velocities along the charge time dimension (matter/antimatter).  The loss of K.E was the gain in rotational K.E,  $v^2_{rc}$, which manifest itself as  $T$ temperature.)

We have a similar expression for thermal gravity,  $g_T$

$g_T=\cfrac{{G}_{T}}{{x}_{o}({d}_{n}-{d}_{T_{max}})}.\cfrac{d({d}_{s}(x))}{dx}$

where  $G_{T}$  is a constant,  $x_o$  is the radius of the hot spot.

$T=T_{max}$,  where $x=x_o$  on the boundary of the hot spot.

$d_{T_{max}}$  is the density of space at temperature  $T_{max}$  and  $d_n$  is the normal space density.  The term,

${d}_{n}-{d}_{T_{max}}>0$

because temperature thins out space and the resulting thermal gravity is positive in the direction of  $x$.  If we assume that  $d_s(x)$  is linear in  $T$.

Space Density, ds vs Temperature, T

${d}_{s}(T)=\cfrac{({d}_{T_{max}}-{d}_{n})}{T_{max}}.T+d_n$

Space density  varies with  $T$,  temperature,  given the temperature profile  (1).

${d}_{s}(x)=\cfrac{({d}_{T_{max}}-{d}_{n})}{T_{max}}.\cfrac{A}{\sqrt{x}}+d_n$

$\cfrac{d({d}_{s}(x))}{dx}=-A\cfrac{({d}_{T_{max}}-{d}_{n})}{2T_{max}}.\cfrac{1}{({x})^{{3}/{2}}}$

And we have,

$g_T=\cfrac{{G}_{T}}{{x}_{o}({d}_{n}-{d}_{T_{max}})}(-A\cfrac{({d}_{T_{max}}-{d}_{n})}{2T_{max}}.\cfrac{1}{({x})^{3/2}})$

$g_T={{G}_{To}}\cfrac{1}{{x}^{3/2}}$

where  $G_{To}$  replaces $\cfrac{A{G}_{T}}{2{x}_{o}{T_{max}}}$

In fact, for $n=1,2,3...$

${d}_{s}(T)=\cfrac{{d}_{T_{max}}-{d}_{n}}{T^n_{max}}.T^n+d_n$

$g_T=\cfrac{nA^n{G}_{T}}{2{x}_{o}{T^n_{max}}}\cfrac{1}{\sqrt[n+2]{x}}$

The following is a set of curves showing the dependence of  Space Density,  $d_s$  on  Temperature  $T^n$.

Space Density vs Temperature

It also show  $g_T$  for the case of linear dependence of  $d_s$  on  $T$.   The graph below shows  $g_T$  in the first few  $n$  and  $g=e^{-x}$.

gT(x) field equation

To obtain the answer calculated previously where  $g_T=\cfrac{A}{x}$.  We have to consider,

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(\cfrac { 1 }{ T_{ max }^{ 2 } } ) } ln(\cfrac { 1 }{ T^{ 2 }(x) } )+{ d }_{ n }$

then substitute (1) in,

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ -ln({ T_{ max }^{ 2 } } ) }( ln(x )-ln(A^2))+{ d }_{ n }$

$\cfrac{d({d}_{s}(x))}{dx}=-\cfrac{({d}_{T_{max}}-{d}_{n})}{   ln( { T_{ max }^{ 2 } } ) }.\cfrac{1}{x}$

And we have,

$g_{ T }=\cfrac { { G }_{ T } }{ { x }_{ o }({ d }_{ n }-{ d }_{ T_{ max } }) } (-\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln({ T_{ max }^{ 2 } }) } ).\cfrac { 1 }{ x }$

$g_T={{G}_{To}}\cfrac{1}{{x}}$

where  $G_{To}$  replaces $\cfrac{{G}_{T}}{{x}_{o}{ ln({ T_{ max }^{ 2 } }) }}$

So we have,  $g_T=\cfrac{G_{To}}{\sqrt[n+2]{x}}$,  for  $n=1,2,3..$  or  $g_T={{G}_{To}}\cfrac{1}{{x}}$

(Note:  The  $G_{To}$s  are to be determined respectively.)

Each of these suggests a unique variation of space density,  $d_s(x)$,  with temperature,  $T$.  Which one is the right one?  The only way to determine which is correct is to measure  $g_T$  or  $g_T-g$  and obtain its variation with  $T$  experimentally.

Thermal Gravity And Lost Innocence

From the web,  The surface temperature of planet as a function of   $x$  distance from the Sun is given by,

$T(x) = \cfrac{A}{\sqrt{x}}=A(x)^{-\cfrac{1}{2}}$

$A$  is a constant of proportionality.  Differentiating with respect to  $x$,

${ T }^{ ' }(x)\quad =-\cfrac { A }{ 2 } { (x) }^{ -\cfrac { 3 }{ 2 }  }$

${ T }^{ '' }(x)\quad =\cfrac { 3A }{ 4 } { (x) }^{ -\cfrac { 5 }{ 2 }  }$

and from the post "What Thermal Gravity?  Too Hot To Handle..."

$g_{ T } =v^2\cfrac{x}{T(x)}\cfrac { d^{ 2 }T(x) }{ d x^{ 2 } }$

$g_{ T } =v^2\cfrac{x(x)^\cfrac{1}{2}}{A}\cfrac { 3A }{ 4 } { (x) }^{ -\cfrac { 5 }{ 2 }  }$

$g_{ T }=\cfrac { 3 }{ 4x } v^{ 2 }=\cfrac{a}{x}=S_TT^2$

where $v$ is the speed of $T$ as it travels across the temperature profile,  $a$  a constant,  and  $S_T$  a constant of proportionality with the unit ms^-2J^-2.  The last equation relates  $g_T$ directly to  $T$  this will bring us to a relationship for space density and temperature in another post.

Unfortunately,  a plot of a family of curves of $\cfrac{a}{x}$  together with  $\cfrac{1}{x^2}$,

shows that   $g_T={a}/{x}$  does not intersect  $g={1}/{x^2}$  again after its goes above ${1}/{x^2}$   ie after $g_T>g$  along  $x$.  $g_T$  has to be greater than  $g$  at the start  $x=x_o$  for  $T$  to flow.  This is the case for nebula where the boundary,  $g_T-g=0$  does not exist, but not a general case of planetary spheres and rings where  $g_T>g$  for  $T$ to flow outwards and then  $g_T-g=0$ at some distance $x_{ring}$ beyond which  $g_T$  oscillate up and down  $g$.  It is not appropriate to repesent gravity of the hot spot here using  $1/x^2$.

However, if instead we use $e^{-x}$ to represent gravity,

Both Nebula and planetary spheres and rings scenarios are encountered as ${1}/{x}$ is scaled appropriately.   The curve marked as  ${a}/{x}$ is higher than the gravity curve  $g$  and does not intersect  $g$.  This is the Nebula scenario where  $T$ spreads over vast distances.

All other curves below  ${a}/{x}$ cross  $g$ twice.  The second intersection further out along  $x$ marks a gap where the net gravity due to  $g$  and  $g_T$  points away from it, before and after that point.

It should be noted that, beyond the first intersection,  $T$  redistributes and  $g_T$  is no longer of the form  ${a}/{x}$  but is expected to have a sinusoidal component and intersect  $g$  again multiple times.

Below is a plot of (1/x)*cos(2*x)^2/(cos(2)^2) to show how  $g_T$  might look like,

If  $T(x)$  can be determined from observations,  then logically  $g_T$  should come after  $T(x)$  has been formulated.  We shall do that in the next post.

$g_T$ was derived from an expression for  $T$  temperature, based on an moving mass analogy for $T$, together with the assumption that  $T$  propagates as a wave, at a constant velocity  $v$.  If  $T(x)$  can be obtained empirically,  then  $g_T$  can be derived directly with just one assumption of how  $T$  affects  $g_T$  through  interacting with space.

Slimming Sun, Solar Wind, Where All The Power Go?

But let's look at the boundary at  $x=L$.   $T$  is being accelerated towards this boundary, at the boundary itself,  $T=0$  but its velocity is not zero.  If this suggests that materials are lost from the hot zone, they are not taking heat from it, because  $T=0$  and  $T\ge0$.   So, at the boundary,

$\cfrac{dT_L}{dt}=0$

But when happens to all the power from the Sun  ($\cfrac{dT_S}{dt}$ at the Sun is finite)?

$g_T$  can apply in two possible ways.

Firstly,  $g_T$  applies on  $T$  only,  this is to suggest that  $T$  has an existence apart from  $m$,  and that  $m$,  the inertia along the gravity time dimension,  is affected by  $g_T$  only after acquiring  $T$.  In this scenario, a body dropped into a hot zone will not be accelerated outwards immediately.  The body will reach thermal equilibrium and then be accelerated outwards.  During the time the body is heating up it will drop further into the hot spot under  $g$.  As it heats up  $g_T$ increases and it falls with less acceleration  $g-g_T$.  There will be a point where  $g-g_T$ is zero.  At thermal equilibrium with its surroundings, it will experience a net gravity outwards.  So the body initially drops further towards the hot spot, stop and is then repelled away from the hot spot.  It initially heats up but on its path away from the hot spot it is cooling down as it gains velocity; the temperature profile decreases with increasing distance from  $x$.  Thermal energy is converted to kinetic energy in this thermal gravitational field.

Even after the temperature profile has been established  $g_T$  does not disappear.  Otherwise, planetary rings and spheres are just transients.   The temperature profile stop growing beyond  $x=L$ because

  $\cfrac{dT}{dt}$  is finite.   Power is finite.

Secondly,  it is possible that $g_T$  affects  $m$  directly as long as it is in the hot zone where  $g_T$  has already been established.  $m$  will  experience an net gravity due to  thermal $g_T$  and  normal gravity  $g$  irrespective of its temperature.  A body drop into a hot zone around a hot spot will be accelerated out immediately at the same time acquiring heat to establish thermal equilibrium with its changing ambient.  Eventually, the body will start losing temperature as it gains speed under the effect of  $g_T- g$, at distances further and further from the hot spot where the ambient temperature is lower.

The difference is,  in the first case each body has a personal space around it effected only by its  $T$,  in the second case a third medium exist between bodies shared by all bodies, changes in this medium by any body is felt by all sharing the medium.  It is more likely that space exist as a third medium effected by all and shared by all.

In both scenario, material are driven away from the hot zone.  The sun is losing mass.  $g_T$ is likely to be the origin of solar wind.

In both scenario, the hot body under $g_T-g$ gains speed but loses temperature (given time for the body to establish thermal equilibrium with its ambient) inside the thermal gravitational field.

$\cfrac{1}{2}mv^2\rightleftharpoons T$

Which is consistent with  $T$  being defined as an energy term,

$T_c=\cfrac{1}{2}qv^2_{rc}$    and    $T_g=\cfrac{1}{2}mv^2_{rg}$

in the post "Hot Topic, Mumble and Jumble..."

More importantly,

$\cfrac{dT_S}{dt}$  at the Sun = rate of loss of kinetic energy at the boundary  $x=L$.

That's why the temperature profile around the Sun does not grow further, and its power  ($\frac{dT}{dt}$)  is finite.  Power is loss from the hot zone in the form of kinetic energy.  The power from the Sun equals the loss of mass with K.E pass  $x=L$.  Something is wrong here.

Have a nice day.

Burning Sun

The post "Temperature", which served well and led to many possibilities, however is too assuming.  Specifically the expression that led to the sign before the gravity term to flip is,

$kg = \rho X_A x$

which suggests that, if we apply the analogy now,

$T(x) = \rho X_A x$

Strictly speaking this is not a valid use of analogy.  In normal application, all terms are reduced first before a replacement is made.   The flip in sign for the gravity term result as

$T(x) = T_{max}-\rho X_A x$

in the target analogous case.   Temperature decreases with distance  $x$  from the hot spot.  This is why $-g_T$  was used.  The point is,  $T$  is not linear in  $x$ throughout the distance  $L$.  However, for the first part of the  $g_T$  curve just before the intersection with  $g$,  the curve can be approximated as a straight line.  All expressions derived based on a linear dependence on  $x$  are still valid up to boundary  $g_T-g=0$.

Once the temperature profile is established,  between a hot spot at $T=T_{max}$  and some point  $x=L$ where  $T=0$,  it does not change with time.

There is however a $\cfrac{dT}{dt}$  term, that indicates the flow of  $T$  down the temperature profile.  Once the temperature profile is established, however,

 $\cfrac{dT}{dt}=constant$

If,

 $\cfrac{dT}{dt}\rightarrow\infty$

then the temperature profile will keep growing,  $L$ reaching further out into space.

But does  $T$  behave like a wave while its flows?  Post "Heat Wave" and "shhh... A Big Secret...No Wave Overhead, Just In Your Face"  try to find a wave travelling radially down to establish the temperature profile.  The former post approximated for  $T(x)$  and showed that  $g_T(x)$, the field equation for thermal gravity is needed.  A case of Chicken needing an Egg...  The latter post suggests that there is no radial wave assuming that  $g_T$ is high and that the hot spot gravity is ignored.  The post "Where the Sun Don't Shine And The Chill Cold Space" considered the hot spot gravity and finds that there may be a wave along the radial line near when $g_T-g=0$.  $T$ seem to perform SHM in and out of this boundary.  The gravity reversal at this zero points caused gravity bands to form around the hot spot.  The radial wave that exist over this short region will have the same form as the concentric wave from the post "Heat Wave Again", except  $x$  is now along the radius.  It was proposed in that post that once a temperature profile has been established, concentric waves travel around the hot spot and multiple concentric waves extends all the way till  $T=0$  when  $x=L$.

$\cfrac{dT}{dt}$  exist up till  $x$  reaches  $L$.  Once  $x$ reaches  $L$,  $T$ stops.  Like a hot body that melts because of its temperature, it collapses but eventually stops deforming.   Even if there is a standing wave between  $T=T_{max}$  and  $T=0$, it does not deliver a net flow of   $T$.

So the Sun is just a hot spot with an established  $T$  profile,  finite  $\cfrac{dT}{dt}$  is needed from it.  The Sun will last a long time,  it is not required to burn fuel at an infinite rate.  All that is required of the Sun is just to be hot.  Very HOT.

Hold Your Lembu...

Rings?  Shouldn't they be spheres?

This might explain why planets like Jupiter seemed over-sized.  It is very much less dense than Earth, given its observed size and its orbit (and period) around the Sun.  We are looking not at the planet's surface but at a gravitational phenomenon that traps debris around the planet.   These spheres are much bigger that the planet and shade the whole planet.

A planet in rotation will budge at the equatorial plane.  It is expected that the heat radiating from the planet is maximum at the equatorial plane.   As the interaction between gravity,  $g$  and thermal gravity  $g_T$ is strongest on this plane,  the formation of rings will reach furthest into space in the equatorial plane; and then spheres around the planet will form at distances closer to the planet's center.

Other factors may also be at play.  The planet magnetic field for example.

It would be interesting to find a planet that change size as it orbits around its sun.  Gravity around the planet might just be strong enough to effect the formation of such spheres.  If this happens, the planet will change size periodically as it revolve around its sun.

Hot Rings Come Get Some

Here's a representation of rings around a hot planet,

Have a nice day.

Gravity Due To the Mass of $T$

This is what gravity due to the mass of  $T$  would look like?  Possibly too small to be consider in the first place.  Thermal gravity must be greater than the hot body mass gravity  $g$  for $T$ to propagate outwards.  $T$  has low mass.   The gravity pull developed inside $T$  as  $T$  spread out, points in the opposite direction to  $g_T$.  This gravity is low.  Nonetheless, the following is a likely plot of gravity due to the mass of  $T$  inside $T$.

The initial increase in gravity is due to the increase in mass behind, as we approaches  $L$.  The gravity drops are due to the gaps that develops in  $T$,  over the region where there is no  $T$.  Gravity raises again at the end of the gap, as we approach the center of the next band and drops once we are at the next gap again.  Such gravity drop may not reach zero gravity.

Rings...Who's Calling? Saturn Who? Saturn Lee...

What if  $g_T$  and  $g$ crosses not just once but multiple times!  This is possible because  $g_T$  depends on  the distribution of  $T(x)$.  With reference to a high $T_o=T_{max}$,  $g_T$  increases with lower  $T(x)$.  We would expect  $T$  to bunch around the point  $g_t-g=0$ because of  the conditions for SHM.   We would also expect  $T$  to thin out at the farthest along its SHM path.  As such   $g_T$ varies with local conditions and its curve is not fixed by just the boundary conditions.

If  $g_t$  and $g$  crosses multiple times,  $T$  will then perform SHM about the multiple zeros.   And so bunches around the same zeros.  If we can see such heat as plasma than we will see them as rings around suns.

In general, if a planet radiates heat into outer space, at distances far beyond where  $g$  is low and  $g_T$  is non zero we might see $g_T$  and  $g$ crosses multiple times and results in gravity bands.  After all, both  $g_T$  and  $g$  are gravity terms.  Such gravity belts are illustrated below

This gravity profile looks like air pressure profile in a longitudinal sound wave.  Mass may be capture around the centers of the gravity bands and form rings around the heat radiating planet.  Where gravity points away in both directions, we will see the gaps that separate one ring from another.  Bear in mind that in outer space where no other forces are acting, small strength gravity bands do have a visible effect.

So, we might see suns and heat radiating planets with such gravity bands around it, from considering thermal gravity and mass gravity.

Suddenly, I'm in motion,  Oh ohoh oh oh...And there are rings... around Ju...pit...er...

Nebula When $g_T$ Is Too High

Without greater knowledge of thermal gravity $g_T$,  it is actually not wise to speculate that,  $g$  and $g_T$  have an intersection in general.  Both curves could be almost parallel but not cross as  $x\rightarrow\infty$.  In which case,  $T$ travels with decreasing acceleration that approaches zero as  $x\rightarrow\infty$.   $T$ is not decelerated.  Eventually, $T$ is zero (not zero velocity but zero value) when $x$ reaches the mathematical $L$.

The Sun will then display itself as a splat of heat.  When the heat is high we might see plasma.  Depending on the composition of materials inside this hot zone, we will see various display of colors and lights.

A Nebula!

The epicenter of such a Nebula will then be like the mass core of our Sun; we cannot see our Sun's core, only the boundary where $g_T-g=0$.  A Nebula is a sun where its thermal gravity  $g_T$  is always higher than its mass gravity, $g$ before $T$ reaches zero.  The curves of $g_T$  and  $g$ may not meet at all; ie no zero cross over point, or may meet at a point  $x>L$,  beyond  $T=0$.

Have a nice day.

Where the Sun Don't Shine And The Chill Cold Space

From the post "shhh... A Big Secret...No Wave Overhead, Just In Your Face",

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2v\cfrac { dv }{ dt } \cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (\cfrac { \partial T }{ \partial t } ) \right\}$

If

$\cfrac{dv}{dt}=g_T-g$

where $g$ is the normal gravity of a massive hot spot, instead of,

$2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } =0$

we have,

$2x(\cfrac{g_T-g}{g_T})vT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }=0$

Which means, if

$g_T-g=0$

just at the boundary where $T$ stop accelerating forward, but obviously has some velocity radially away from the hot spot, we can have radial wave behavior (a wave along the radial line) round this boundary.

Beyond this point,  $g$  is higher because  $g_T$ is higher before this zero-cross-over point, otherwise  $T$ would not be propagating outwards, $T$ emerge decelerating, and would eventually stop under the massive hot spot's gravity pull.

So, there is a distance beyond where the Sun's heat don't reach, but still visible.  Bright spot of light but not much heat!

What is even more interesting is that around the boundary where

$g_T-g=0$

the conditions are just right for Simple Harmonic Motion along a radial line in and out of this spherical boundary.  $T$  travels beyond this boundary but now decelerating.   At the farthest point beyond the boundary  $T$  has zero velocity, but still decelerating under the net effect of $g$  and  $g_T$ ($g$   and  $g_T$  both being monotonously decreasing quantities will not cross again   Note:  This is not strictly true.), races back into the boundary.  Inside the boundary it experiences again deceleration because it is now travelling inwards.  $g_T-g$ is positive outwards inside the boundary.  $T$ eventually stops and accelerates outwards, towards the boundary again.  SHM.

THIS BOUNDARY IS THE RADIUS OF THE SUN!

The actual mass radius of the Sun is smaller, below this boundary.

Some $T$ will escape.   $T$ with velocities greater than the Sun's escape velocity will escape from the Sun.  This means $T$ must have a minimum energy to reach us from the Sun.  If none of $T$ have enough escape velocities, the Sun will just be a bright spot in the sky with no heat!  The Sun don't have to turn dark when that happens.

This escape velocity is less than the escape velocity of the Sun considering Sun's mass alone.  This escape velocity is proportional to the area between the curves,  $g$ and  $g_T$ starting at the boundary and beyond.  Its value is expected to be much smaller than just the area under the  $g$  curve.

What if most of  $T$  can escape?  There would not be much SHM in and out of the boundary $g_T-g=0$.

shhh... A Big Secret...No Wave Overhead, Just In Your Face.

From previous episodes of Amnesia's Dream,

$\cfrac { \partial T }{ \partial t } =\cfrac { x }{ g_{ T } } \cfrac { \partial ^{ 3 }T }{ \partial ^{ 3 }t }$

and so,

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \cfrac { \partial  }{ \partial t } (\cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } } )$

From the wave equation,

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \cfrac { \partial  }{ \partial t } (v^{ 2 }\cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } )$

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2v\cfrac { dv }{ dt } \cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (\cfrac { \partial T }{ \partial t } ) \right\}$

If we assume that acceleration is wholly due to $g_T$,

$\cfrac{dv}{dt}=g_T$

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2vg_{ T }T(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (T(x)\cfrac { dT(t) }{ dt } ) \right\}$

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2vg_{ T }T(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +v^{ 2 }\cfrac { dT(t) }{ dt } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }  \right\}$

$\cfrac { dT(t) }{ dt } T(x)=2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +\cfrac { dT(t) }{ dt } v^{ 2 }\cfrac { x }{ g_{ T } } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }$

Compare this with,

$T(x)= v^{ 2 }\cfrac { x }{ g_{ T } } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }$

this suggests,

$2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } =2xv\cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } =\cfrac { 2x }{ v } \cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } } =0$

which might mean,

$\cfrac { \partial^{ 2 }T }{ \partial x^{ 2 } } =\cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } }=0$

then there is no wave in the radial direction, not of the form $T=T(t)T(x)$.  Just attenuation.

What Thermal Gravity? Too Hot To Handle...

From the post "Heat Wave"

$\cfrac { \partial T }{ \partial t } =\cfrac { x }{ g_{ T } } \cfrac { \partial^{ 3 }T }{ \partial^{ 3 }t }$

$\int  \partial T=\cfrac { x }{ g_{ T } } \int { \cfrac { \partial^{ 3 }T }{ \partial^{ 3 }t }  } \partial t$

$T(t)T(x)=\cfrac { x }{ g_{ T } } (\cfrac { \partial^{ 2 }T }{ \partial t^{ 2 } } +C(x) +C)$ ---(*)

where $C(x)$ is a function only in $x$ and $C$ is a numerical constant.

If we insist that $T$ propagate as a wave at velocity $v$,

$\cfrac { \partial^{ 2 }T }{ \partial t^{ 2 } }={ v }^{ 2 }\cfrac { \partial^{ 2 }T }{ \partial x^{ 2 } }$

Substitute the above into (*)

$T(t)T(x)=\cfrac { x }{ g_{ T } } ({ v }^{ 2 }\cfrac {\partial^{ 2 }T }{ \partial x^{ 2 } } +C(x) +C)$

when we let $T=T(t)T(x)$.

If  $T(t)$ has no constant term,

$C(x)=C=0$,

$T(t)T(x) ={ v }^{ 2 }\cfrac{ x } { g_{ T } }T(t)\cfrac { d^{ 2 }T(x) }{ d x^{ 2 } }$

$g_{ T } =v^2\cfrac{x}{T(x)}\cfrac { d^{ 2 }T(x) }{ d x^{ 2 } }$

The biggest assumption here is that $T$ propagate as a wave at velocity $v$.  We know that $g_T\rightarrow 0$   as  $x\rightarrow \infty$.    Thermal gravity is positive, and points away from higher temperature.  It is in the same direction as $x$.  $T(x)$ as a product with $T(t)$ satisfies the wave equation.  Just on the boundary of the hot spot, $g_T$ is maximum.   Thereafter,  $g_T$   decreases monotonously.

Still looking for $g_T(x)$...

Because the immediate answer is,

 $T(x)={ e }^{ -Ax }$

where

$A=\cfrac { 1 }{ v }\sqrt{\cfrac { { g }_{ T } }{ L}}$

which leads right back to

$g_{ T } =v^2x(\cfrac { 1 }{ v }\sqrt{\cfrac { { g }_{ T } }{ L}})^2$

$g_{ T } = x{\cfrac { { g }_{ T } }{ L}}$

$x=L$

which is undeniable.  Very Hot...

Temperature and Space Density.

This would suggest that if we measure gravity in a hot oven, it is going to be less than that obtained at room temperature.

From the post "Temperature",

$\cfrac{\partial T}{\partial t}=\cfrac{L}{g_T}\cfrac{\partial^3 T}{\partial t^3}$

But how does $g_T$ behaves?  Under normal circumstances, we would have to consider both gravity terms due to $v^2_s$ and $v^2_{rg}$ in the energy conservation equation,

$(mv^2_t+\cfrac{1}{2}mv^2_s+\cfrac{1}{2}mv^2_{rg})+(qv^2_{tc}+\cfrac{1}{2}qv^2_c+\cfrac{1}{2}qv^2_{rc})=c^2$

where normal gravity is attributed to   $v^2_s$,   and thermal gravity from the term  $v^2_{rg}$.

Intuitively, high temperature make space less dense and a gravity pointing towards denser/colder space pulls the heat away from the hot spot.

This gravity was approximated as $g_T=\cfrac{H_o}{r^2}$, which is inadequate.  Temperature spread out in space,  thermal gravity within it is like gravity inside a solid with a material density distribution that in general thins out into space.

Inside Earth, assuming uniform density,

$g=G.r=\cfrac{g_o}{r_e}r$

gravity increases linearly with the radius from Earth's center.

If temperature is uniform, then thermal gravity within it is zero.   We expect thermal gravity to be positive, pointing towards lower temperature, and decreases to zero at the far end.

The gravity profile we seek will be between zero and a positive monotonously decreasing bound, possibly $\cfrac{1}{r^2}$.  But, since  $T$  is propagating outward we expect the total net gravity to be outwards;  $g_T$  should be higher than the gravity of the hot body and gravity that develops within  $T$  as it spreads outwards.