What about this?
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
\(\cfrac{\partial\,x}{\partial\,t}=c\)
So,
\(\cfrac{\partial\,x}{\partial\,t_g}=\cfrac{\partial\,t}{\partial\,t_g}.\cfrac{\partial\,x}{\partial\,t}=1\)
Not quite! We have to consider the fact that \(x\) is orthogonal to \(t_g\) and in general,
\(\cfrac{\partial\,t_g}{\partial\,t}\neq c\)
In fact, the expression,
\(\cfrac{\partial\,x}{\partial\,t_g}\)
means change in \(x\) for every change in time \(t_g\) (\(t_g\) being the independent variable), just as
\(\cfrac{\partial\,x}{\partial\,t}\)
means change in \(x\) for every change in time \(t\) except we have to consider the fact that,
\(i\Delta t_g= \Delta x\)
\(c\) per change in \(t_g\) along \(x\) must be multiplied by \(i\) to reflect the fact that \(t_g\) is orthogonal to \(x\).
\(ic.\Delta t_g= \Delta x\)
and so,
\(ic= \lim\limits_{\Delta t_g\to0}\cfrac{\Delta x}{\Delta t_g}=\cfrac{\partial\,x}{\partial\,t_g}\)
We are now at time speed \(c\) does not mean we cannot achieve light speed in space. If we are constrained by energy conservation, as the case of a body at constant velocity, free motion, enters a conservative gravitational field, speeds up in space slows down time.
\(v_t^2+v_s^2=c^2\) --- (*)
where \(c\) is a constant, \(v_t\), time speed and \(v_s\), speed in space.
And at light speed \(c\), in space, time speed is zero. When we are not constrained by energy conservation and energy is pumped into the body, space speed increases without regard to time speed! Einstein was working in a conservative gravitational field, ie accelerating under gravity without external energy input, as such total K.E is conserved including K.E along the time dimension (as in expression *).
Tuesday, May 26, 2015
More Of The Same
Consider this,
a wave oscillating in two time dimensions with light speed in space. Obviously,
\(c^2\cfrac{\partial^2\psi}{\partial\,x^2}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
We replace one of the space dimension by a time dimension \(t_g\),
\(c^2\cfrac{\partial^2\psi}{\partial\,x\partial\,t_g}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
This relationship holds because,
\(c^2\cfrac{\partial}{\partial\,x}\left\{\cfrac{\partial\psi}{\partial\,t_g}\right\}=\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial\,x}{\partial\,t}\cfrac{\partial}{\partial\,x}\left\{\cfrac{\partial\psi}{\partial\,t_g}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,t}\right\}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
\(\cfrac{\partial\,x}{\partial\,t}=c\)
and assuming that,
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
Furthermore,
\(t_g=it_c\)
\(\cfrac{\partial\,t_g}{\partial\,t_c}=i\)
and
\(\cfrac{\partial\,x}{\partial\,t_c}=\cfrac{\partial\,x}{\partial\,t_g}=ic\)
where \(i\) is used to express orthogonality explicitly. We replaces \(t\),
\(\cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial \, t_{ g } }{ \partial \, t } .\cfrac { \partial \, x }{ \partial \, t } \cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, x } \right\} =\cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial }{ \partial \, t } \left\{ \cfrac { \partial \psi }{ \partial \, t } \right\} \)
\( \cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial \, x }{ \partial \, t } \cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, x } \right\} =\cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, t } \right\} \)
\( \cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, x } \right\} =\cfrac { \partial ^{ 2 }\psi }{ \partial \, t_{ g }^{ 2 } } \)
\( ic\cfrac { \partial ^{ 2 }\psi }{ \partial \, x\partial \, t_{ g } } =\cfrac { \partial ^{ 2 }\psi }{ \partial \, t_{ g }^{ 2 } } \)
which is the same wave equation as when \(\psi\) is travelling down \(t_g\) at light speed and oscillating on \(x\). This wave however, has light speed in space, the other can be stationary. In both cases, \(\psi\) wraps around the particle as a stationary wave, in a sphere; refer to post "Standing Waves, Particles, Time Invariant Fields" dated 20 Nov 14.
This wave although in space at light speed has energy oscillating in two time dimensions. These energies is coupled to the space dimension only via other particles that is oscillating in one of the respective time dimensions and the other in space, or by collisions. During collisions, energy oscillating in the two time dimensions is transferred/shared by the colliding particles.
This is highly speculative, but none the less, specific properties/behaviors of these waves, in general, depend on which dimensions the energy oscillates between and which dimension the wave is at light speed. These properties are not captured by the wave equations.
If we do not assume
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
but let
\(\cfrac{\partial\,t_g}{\partial\,t}=u\)
then the final expression for the wave is simply,
\( i\cfrac{c^2}{u}\cfrac { \partial ^{ 2 }\psi }{ \partial \, x\partial \, t_{ g } }=\cfrac { \partial ^{ 2 }\psi }{ \partial \, t_{ g }^{ 2 } } \)
where \(c\) is replaced by the factor \(\cfrac{c^2}{u}\).
It seems that swapping the oscillation dimension with the light speed dimension of the wave does not change the form of the wave equation. Both types of wave with its distinct characteristics share the same wave equation.
a wave oscillating in two time dimensions with light speed in space. Obviously,
\(c^2\cfrac{\partial^2\psi}{\partial\,x^2}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
We replace one of the space dimension by a time dimension \(t_g\),
\(c^2\cfrac{\partial^2\psi}{\partial\,x\partial\,t_g}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
This relationship holds because,
\(c^2\cfrac{\partial}{\partial\,x}\left\{\cfrac{\partial\psi}{\partial\,t_g}\right\}=\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial\,x}{\partial\,t}\cfrac{\partial}{\partial\,x}\left\{\cfrac{\partial\psi}{\partial\,t_g}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,t}\right\}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
\(\cfrac{\partial\,x}{\partial\,t}=c\)
and assuming that,
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
Furthermore,
\(t_g=it_c\)
\(\cfrac{\partial\,t_g}{\partial\,t_c}=i\)
and
\(\cfrac{\partial\,x}{\partial\,t_c}=\cfrac{\partial\,x}{\partial\,t_g}=ic\)
where \(i\) is used to express orthogonality explicitly. We replaces \(t\),
\(\cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial \, t_{ g } }{ \partial \, t } .\cfrac { \partial \, x }{ \partial \, t } \cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, x } \right\} =\cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial }{ \partial \, t } \left\{ \cfrac { \partial \psi }{ \partial \, t } \right\} \)
\( \cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial \, x }{ \partial \, t } \cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, x } \right\} =\cfrac { \partial \, t }{ \partial \, t_{ g } } .\cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, t } \right\} \)
\( \cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial }{ \partial \, t_{ g } } \left\{ \cfrac { \partial \psi }{ \partial \, x } \right\} =\cfrac { \partial ^{ 2 }\psi }{ \partial \, t_{ g }^{ 2 } } \)
\( ic\cfrac { \partial ^{ 2 }\psi }{ \partial \, x\partial \, t_{ g } } =\cfrac { \partial ^{ 2 }\psi }{ \partial \, t_{ g }^{ 2 } } \)
which is the same wave equation as when \(\psi\) is travelling down \(t_g\) at light speed and oscillating on \(x\). This wave however, has light speed in space, the other can be stationary. In both cases, \(\psi\) wraps around the particle as a stationary wave, in a sphere; refer to post "Standing Waves, Particles, Time Invariant Fields" dated 20 Nov 14.
This wave although in space at light speed has energy oscillating in two time dimensions. These energies is coupled to the space dimension only via other particles that is oscillating in one of the respective time dimensions and the other in space, or by collisions. During collisions, energy oscillating in the two time dimensions is transferred/shared by the colliding particles.
This is highly speculative, but none the less, specific properties/behaviors of these waves, in general, depend on which dimensions the energy oscillates between and which dimension the wave is at light speed. These properties are not captured by the wave equations.
If we do not assume
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
but let
\(\cfrac{\partial\,t_g}{\partial\,t}=u\)
then the final expression for the wave is simply,
\( i\cfrac{c^2}{u}\cfrac { \partial ^{ 2 }\psi }{ \partial \, x\partial \, t_{ g } }=\cfrac { \partial ^{ 2 }\psi }{ \partial \, t_{ g }^{ 2 } } \)
where \(c\) is replaced by the factor \(\cfrac{c^2}{u}\).
It seems that swapping the oscillation dimension with the light speed dimension of the wave does not change the form of the wave equation. Both types of wave with its distinct characteristics share the same wave equation.
More Corrections For Nothing
From the post "We Have A Problem, Coulomb's Law" dated 20 Nov 14,
\(F_v=2\cfrac { qc^{ 2 } }{ x } e^{i\pi/2}\)
and from "Wrong, Wrong, Wrong" dated 25 May 15,
\(F=\int{F_v}\,d\,x\)
The force in the field due to a particle is,
\(F=2\int{\cfrac { qc^{ 2 } }{ x } }\,d\,x.={2{ qc^{ 2 } }{ ln(x) } }+C\)
where we have dropped the phase information, \(e^{i\pi/2}\).
but \(F=-\psi\), so,
\(\psi=-2{ qc^{ 2 } }{ ln(x) }+A\)
However, \(\psi\) cannot be less than zero. So, both \(F\) and \(\psi\) is valid up till \(\psi=0\).
This is not the inverse square law! And what happens to \(F\) after \(\psi=0\)?
\(F_v=2\cfrac { qc^{ 2 } }{ x } e^{i\pi/2}\)
and from "Wrong, Wrong, Wrong" dated 25 May 15,
\(F=\int{F_v}\,d\,x\)
The force in the field due to a particle is,
\(F=2\int{\cfrac { qc^{ 2 } }{ x } }\,d\,x.={2{ qc^{ 2 } }{ ln(x) } }+C\)
where we have dropped the phase information, \(e^{i\pi/2}\).
but \(F=-\psi\), so,
\(\psi=-2{ qc^{ 2 } }{ ln(x) }+A\)
However, \(\psi\) cannot be less than zero. So, both \(F\) and \(\psi\) is valid up till \(\psi=0\).
This is not the inverse square law! And what happens to \(F\) after \(\psi=0\)?
Mistake In \(F\)
The respective posts where this mistake propagates has been changed.
From the post "Opps! Lucky Me", we suggested that,
\(F=-\cfrac{d\,E}{d\,r}=-\psi_A\)
where \(E\) is the energy in a infinitely thin spheric shell passing through a point \(r\), \(r\) distance from a point particle center. And
\(\cfrac{d\,E}{d\,r}\)
is the change in \(E\) along the radial line.
This is again wrong. This force must be divided by \(4\pi r^2\), as this total force is redistributed over the surface area of the sphere, to obtain \(F\), the force along a radial line.
\(F=-\cfrac{1}{4\pi r^2}\cfrac{d\,E}{d\,r}=-\cfrac{\psi_A}{4\pi r^2}\)
and we have,
\(F=-\psi\)
when
\(\psi_A=\psi.4\pi r^2\)
And for \(F\) to have a inverse square law dependence,
\(\psi\propto\cfrac{1}{r^2}\)
This hopefully is the last of the issue.
From the post "Opps! Lucky Me", we suggested that,
\(F=-\cfrac{d\,E}{d\,r}=-\psi_A\)
where \(E\) is the energy in a infinitely thin spheric shell passing through a point \(r\), \(r\) distance from a point particle center. And
\(\cfrac{d\,E}{d\,r}\)
is the change in \(E\) along the radial line.
This is again wrong. This force must be divided by \(4\pi r^2\), as this total force is redistributed over the surface area of the sphere, to obtain \(F\), the force along a radial line.
\(F=-\cfrac{1}{4\pi r^2}\cfrac{d\,E}{d\,r}=-\cfrac{\psi_A}{4\pi r^2}\)
and we have,
\(F=-\psi\)
when
\(\psi_A=\psi.4\pi r^2\)
And for \(F\) to have a inverse square law dependence,
\(\psi\propto\cfrac{1}{r^2}\)
This hopefully is the last of the issue.
Photon, More Lights, Big Mistake
Consider the wave equation,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\)
And to go where no man has gone before,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x_1}{\partial\,t}\cfrac{\partial\,x_2}{\partial\,t}\cfrac{\partial^2\psi}{\partial\,x_1\partial\,x_2}\)
I have made a mistake in previous posts!
In the case where the wave is travelling in the time dimension, \(t_g\) at speed \(c\), in the wave \(\psi\) varies with space \(x\),
\(\cfrac{\partial\psi}{\partial\,x}\)
and this variation changes with \(t_g\) as the wave travels down \(t_g\),
\(\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}\)
Obviously,
\(\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}\)
and
\(\cfrac{\partial\,x}{\partial\,t}.\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,t}\right\}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
since,
\(\cfrac{\partial\,x}{\partial\,t}=c\)
and the wave is at light speed down the time dimension, \(t_g\),
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
as such
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x}{\partial\,t}.\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=c.c\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=c^2\cfrac{\partial^2\psi}{\partial\,t_g\partial\,x}\)
which is the wave equation for a wave travelling down the time dimension \(t_g\). And when we substitute for specific time \(t\), for example \(t_g\), we obtain,
\(\cfrac{\partial^2\psi}{\partial\,t^2_g}=ic\cfrac{\partial^2\psi}{\partial\,x\partial\,t_g}\) --- (*)
using
\(t_c=\cfrac{1}{\sqrt{2}}t.e^{-i\pi/4}\)
\(t_g=\cfrac{1}{\sqrt{2}}t.e^{+i\pi/4}\),
\(t_c=t_ge^{-i\pi/2}\)
\(t_g=it_c\)
\(\cfrac{\partial\,t_g}{\partial\,t_c}=i\)
and
\(\cfrac{\partial\,x}{\partial\,t_c}=\cfrac{\partial\,x}{\partial\,t_g}=ic\)
where \(i\) is used to express orthogonality explicitly. The details are in the post "Not A Wave, But Work Done!" dated 18 Nov 14.
The problem is at \(\cfrac{\partial\,x}{\partial\,t}=c\). The particle is also at light speed in space, as would be the case of electrons in orbit around a nucleus, but not a stationary charge.
If \(\cfrac{\partial\,x}{\partial\,t}=0\) there is no wave. So what about the electrostatic field around a stationary charge then? Luckily, we are always in motion; not relative motion between two interacting particles but absolute motion of the particles involved.
Notice that in expression (*), only one of the two dimensions between which energy is oscillating is explicitly in the expression. This is because the oscillation is fully defined by what happens on just one of these two dimensions. So, in the case where another space dimension \(x\) has been replaced by a time dimension, expression (*) still applies. As long as one space dimension \(x\) remains as part of the oscillation pair, expression (*) is valid.
So the new particles in the post "Wrong Wrong Wrong", has the same wave equation as the above. It is just (*). And all derivations based on (*) is still valid.
This lead us to the wave equation for photons. Although one of the dimensions with oscillatory energy is replaced with a time dimension, the remaining space dimension \(x\) of the pair together with the fact that this wave is travelling in space at light speed \(c\), means that the wave equation is just,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x_1}{\partial\,t}\cfrac{\partial\,x_2}{\partial\,t}\cfrac{\partial^2\psi}{\partial\,x_1\partial\,x_2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\)
the same general wave equation as before. Swapping one space dimension in the wave for a time dimension has no effect on the wave equation, but we understand that photons are able to impart potential energies (gravitational, electric and heat) because of this.
\(\cfrac{\partial^2\psi}{\partial\,t^2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\)
And to go where no man has gone before,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x_1}{\partial\,t}\cfrac{\partial\,x_2}{\partial\,t}\cfrac{\partial^2\psi}{\partial\,x_1\partial\,x_2}\)
I have made a mistake in previous posts!
In the case where the wave is travelling in the time dimension, \(t_g\) at speed \(c\), in the wave \(\psi\) varies with space \(x\),
\(\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}\)
Obviously,
\(\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}\)
and
\(\cfrac{\partial\,x}{\partial\,t}.\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,t}\right\}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
since,
\(\cfrac{\partial\,x}{\partial\,t}=c\)
and the wave is at light speed down the time dimension, \(t_g\),
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
as such
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x}{\partial\,t}.\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=c.c\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=c^2\cfrac{\partial^2\psi}{\partial\,t_g\partial\,x}\)
which is the wave equation for a wave travelling down the time dimension \(t_g\). And when we substitute for specific time \(t\), for example \(t_g\), we obtain,
\(\cfrac{\partial^2\psi}{\partial\,t^2_g}=ic\cfrac{\partial^2\psi}{\partial\,x\partial\,t_g}\) --- (*)
using
\(t_c=\cfrac{1}{\sqrt{2}}t.e^{-i\pi/4}\)
\(t_g=\cfrac{1}{\sqrt{2}}t.e^{+i\pi/4}\),
\(t_c=t_ge^{-i\pi/2}\)
\(t_g=it_c\)
\(\cfrac{\partial\,t_g}{\partial\,t_c}=i\)
and
\(\cfrac{\partial\,x}{\partial\,t_c}=\cfrac{\partial\,x}{\partial\,t_g}=ic\)
where \(i\) is used to express orthogonality explicitly. The details are in the post "Not A Wave, But Work Done!" dated 18 Nov 14.
The problem is at \(\cfrac{\partial\,x}{\partial\,t}=c\). The particle is also at light speed in space, as would be the case of electrons in orbit around a nucleus, but not a stationary charge.
If \(\cfrac{\partial\,x}{\partial\,t}=0\) there is no wave. So what about the electrostatic field around a stationary charge then? Luckily, we are always in motion; not relative motion between two interacting particles but absolute motion of the particles involved.
Notice that in expression (*), only one of the two dimensions between which energy is oscillating is explicitly in the expression. This is because the oscillation is fully defined by what happens on just one of these two dimensions. So, in the case where another space dimension \(x\) has been replaced by a time dimension, expression (*) still applies. As long as one space dimension \(x\) remains as part of the oscillation pair, expression (*) is valid.
So the new particles in the post "Wrong Wrong Wrong", has the same wave equation as the above. It is just (*). And all derivations based on (*) is still valid.
This lead us to the wave equation for photons. Although one of the dimensions with oscillatory energy is replaced with a time dimension, the remaining space dimension \(x\) of the pair together with the fact that this wave is travelling in space at light speed \(c\), means that the wave equation is just,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x_1}{\partial\,t}\cfrac{\partial\,x_2}{\partial\,t}\cfrac{\partial^2\psi}{\partial\,x_1\partial\,x_2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\)
Monday, May 25, 2015
Wrong, Wrong, Wrong
The derivation of \(F_v\), force density is not wrong but the conversion of force density to force, \(F\) is wrong.
Since,
\(F_v=-\cfrac{d\,\psi}{d\,r}\)
\(\int{F_v}\,d\,r=-\psi\)
and since,
\(F=-\psi\) --- (*)
\(F=\int{F_v}\,d\,r\)
for a particle.
The formulation,
\(F=F_{v}\cfrac{1}{12\pi r}\)
is wrong. However, \(\psi\) does not go on forever, at some point along \(r\), \(\psi=0\) and remains zero thereafter. In addition,
\(\int^\infty_0{\psi}\,d\,r\nrightarrow\infty\)
The expression (*) is then valid up to \(r\) where \(\psi\ge 0\), thereafter, for a point particle, \(\psi\) remains zero, the change in \(\psi\) is zero and so \(F=0\). This is different from Newton's Law, Coulomb's Law and the constant flux derivation, all of which suggest that \(F\) is nonzero as \(r\to\infty\).
It is classic in physics to assert that there cannot be infinite amount of energy. It follows then, that a force field as a result of a spread of energy with an associated energy density distribution around a particle must not be infinite too. And the equivalent mass (\(E=mc^2\)) cannot be infinite.
What happened?
If \(\psi=\cfrac{D}{r^2}\)
at \(r=0\), \(\psi=C=\cfrac{D}{r_o^2}\), (ie the graph is shifted left)
then \(\int^\infty_0{\psi}\,d\,r\)
is actually finite but \(\psi\to0\) as \(r\to\infty\); \(\psi\) has a infinite tail and the force field extends to infinity.
There is still a problem with \(F\). \(F\) is derived from \(F_v\) which is obtained by differentiating \(\psi\) along \(r\).
\(F_v=-\cfrac{d\,\psi}{d\,r}\)
\(\psi\) is the energy density of a wave travelling along a time dimension at light speed \(c\). It is not a wave at light speed in space. In the latter case, \(\psi\) in along the wave. In the case of a wave with light speed in time, \(\psi\) was taken to be around a point particle. The two big assumptions here are that the time wave manifest itself in space as a point particle and that the energy density, \(\psi\) associated with such wave is distributed uniformly around the point mass. The post "Standing Waves, Particles, Time Invariant Fields" dated 20 Nov 14 shows that \(\psi\) is a standing wave around the point particle.
Is an explicit transformation needed between a time wave and a space wave? A transform that will morph \(\psi\) from a cylinder to a sphere? No, the post "Standing Waves, Particles, Time Invariant Fields" dated 20 Nov 14 shows that \(\psi\) is a standing wave around the point particle.
Furthermore, in the post "Difficult To Correct Oneself", it was proposed that instead of a wave oscillating in two space dimensions and travelling down a third time dimension at light speed, the wave in time that manifest itself as a particle is oscillating between one space and one time dimension. Fortunately, the wave equation is the same for both cases, from a later post "Photon, More Lights, Big Mistake".
These new particles are one dimensional entities; 1D particles not at light speed in space. These particles can provide potential energy (electrical, gravitational and heat) to other particles that they collide with. These particles could be responsible for induction/radiation. Are they also responsible for the force field around a mass or charge? It is possible that an oncoming mass be radiated with gravitational 1D particles and so gain gravitational potential energy and be bounced off.
Next stop, wave equations for photons.
Since,
\(F_v=-\cfrac{d\,\psi}{d\,r}\)
\(\int{F_v}\,d\,r=-\psi\)
and since,
\(F=-\psi\) --- (*)
\(F=\int{F_v}\,d\,r\)
for a particle.
The formulation,
\(F=F_{v}\cfrac{1}{12\pi r}\)
is wrong. However, \(\psi\) does not go on forever, at some point along \(r\), \(\psi=0\) and remains zero thereafter. In addition,
\(\int^\infty_0{\psi}\,d\,r\nrightarrow\infty\)
The expression (*) is then valid up to \(r\) where \(\psi\ge 0\), thereafter, for a point particle, \(\psi\) remains zero, the change in \(\psi\) is zero and so \(F=0\). This is different from Newton's Law, Coulomb's Law and the constant flux derivation, all of which suggest that \(F\) is nonzero as \(r\to\infty\).
It is classic in physics to assert that there cannot be infinite amount of energy. It follows then, that a force field as a result of a spread of energy with an associated energy density distribution around a particle must not be infinite too. And the equivalent mass (\(E=mc^2\)) cannot be infinite.
What happened?
If \(\psi=\cfrac{D}{r^2}\)
at \(r=0\), \(\psi=C=\cfrac{D}{r_o^2}\), (ie the graph is shifted left)
then \(\int^\infty_0{\psi}\,d\,r\)
is actually finite but \(\psi\to0\) as \(r\to\infty\); \(\psi\) has a infinite tail and the force field extends to infinity.
There is still a problem with \(F\). \(F\) is derived from \(F_v\) which is obtained by differentiating \(\psi\) along \(r\).
\(F_v=-\cfrac{d\,\psi}{d\,r}\)
\(\psi\) is the energy density of a wave travelling along a time dimension at light speed \(c\). It is not a wave at light speed in space. In the latter case, \(\psi\) in along the wave. In the case of a wave with light speed in time, \(\psi\) was taken to be around a point particle. The two big assumptions here are that the time wave manifest itself in space as a point particle and that the energy density, \(\psi\) associated with such wave is distributed uniformly around the point mass. The post "Standing Waves, Particles, Time Invariant Fields" dated 20 Nov 14 shows that \(\psi\) is a standing wave around the point particle.
Is an explicit transformation needed between a time wave and a space wave? A transform that will morph \(\psi\) from a cylinder to a sphere? No, the post "Standing Waves, Particles, Time Invariant Fields" dated 20 Nov 14 shows that \(\psi\) is a standing wave around the point particle.
Furthermore, in the post "Difficult To Correct Oneself", it was proposed that instead of a wave oscillating in two space dimensions and travelling down a third time dimension at light speed, the wave in time that manifest itself as a particle is oscillating between one space and one time dimension. Fortunately, the wave equation is the same for both cases, from a later post "Photon, More Lights, Big Mistake".
These new particles are one dimensional entities; 1D particles not at light speed in space. These particles can provide potential energy (electrical, gravitational and heat) to other particles that they collide with. These particles could be responsible for induction/radiation. Are they also responsible for the force field around a mass or charge? It is possible that an oncoming mass be radiated with gravitational 1D particles and so gain gravitational potential energy and be bounced off.
Next stop, wave equations for photons.
Flux It Too
From the post "Opps! Lucky Me",
\(F=-\psi\)
this expression gives the force along a radial line As we know \(\psi\to 0\) as \(r\to\infty\) as such \(F\to 0\) and we are safe from an ever expanding universe.
And \(\psi\) takes the form,
\(\psi=\cfrac{D}{r^2}\)
Consider again,
\(F_s=-\cfrac{d\,E}{d\,r}=-\psi_A\)
\(F_s\) is defined wholly by a thin spherical shell of \(\psi\) passing through point \(r\) (ie. of radius \(r\)) center at the point particle. This is different from the notion that all \(\psi\) within the sphere contributes to \(F_s\). If we have,
\(\psi_A=\psi.4\pi r^2\)
And so,
\(F_s=\psi.4\pi r^2\) and
\(F=\cfrac{F_s}{4\pi r^2}=\psi\)
(We have dropped the negative sign here; \(\psi_A\) and \(\psi\) have no direction and cannot be negative. We have lost the direction information that was in the gradient of \(E\)).
Comparing the above with \(F=\cfrac{m}{4\pi \varepsilon_o r^2}\) where \(m\) establishes the field, just as charge \(q\) establishes an electric field, is a constant inside the bounding sphere of radius \(r\),
\(F_s=\psi_A=\cfrac{m}{\varepsilon_o}\)
that,
\(\psi_A=constant\)
That the surface energy density on any bounding sphere around the particle is a constant.
Furthermore, if we change the notion that all \(\psi\) within the sphere contributes to \(F_s\), but instead only \(\psi_A\), the energy density on the surface of the sphere contributes to \(F_s\) then,
\(F=\cfrac{F_s}{4\pi r^2}=\cfrac{\psi_A}{4\pi r^2}=\cfrac{m_sc^2}{4\pi r^2}\) and
\(F=\cfrac{m_s}{4\pi \varepsilon_o r^2}\)
where \(m_s\) is the equivalent mass density of the energy density on the surface of the sphere. We have
\(\varepsilon_o=\cfrac{1}{c^2}\)
This is consistent with the post "Maxwell, Planck And Particles" dated 27 Dec 14, where \(\mu_o=4\pi\times10^{-7}\) was found to be unnecessary and that we set \(\mu_o=1\), here.
And when we compare with
\(F=G\cfrac{m_s}{r^2}\)
\(G=\cfrac{c^2}{4\pi}\)
But it is true that only the surface energy density on the thin spherical shell, just next to \(r\), effects \(F\)?
\(F=-\psi\)
this expression gives the force along a radial line As we know \(\psi\to 0\) as \(r\to\infty\) as such \(F\to 0\) and we are safe from an ever expanding universe.
And \(\psi\) takes the form,
\(\psi=\cfrac{D}{r^2}\)
Consider again,
\(F_s=-\cfrac{d\,E}{d\,r}=-\psi_A\)
\(F_s\) is defined wholly by a thin spherical shell of \(\psi\) passing through point \(r\) (ie. of radius \(r\)) center at the point particle. This is different from the notion that all \(\psi\) within the sphere contributes to \(F_s\). If we have,
\(\psi_A=\psi.4\pi r^2\)
And so,
\(F_s=\psi.4\pi r^2\) and
\(F=\cfrac{F_s}{4\pi r^2}=\psi\)
(We have dropped the negative sign here; \(\psi_A\) and \(\psi\) have no direction and cannot be negative. We have lost the direction information that was in the gradient of \(E\)).
Comparing the above with \(F=\cfrac{m}{4\pi \varepsilon_o r^2}\) where \(m\) establishes the field, just as charge \(q\) establishes an electric field, is a constant inside the bounding sphere of radius \(r\),
\(F_s=\psi_A=\cfrac{m}{\varepsilon_o}\)
that,
\(\psi_A=constant\)
That the surface energy density on any bounding sphere around the particle is a constant.
Furthermore, if we change the notion that all \(\psi\) within the sphere contributes to \(F_s\), but instead only \(\psi_A\), the energy density on the surface of the sphere contributes to \(F_s\) then,
\(F=\cfrac{F_s}{4\pi r^2}=\cfrac{\psi_A}{4\pi r^2}=\cfrac{m_sc^2}{4\pi r^2}\) and
\(F=\cfrac{m_s}{4\pi \varepsilon_o r^2}\)
where \(m_s\) is the equivalent mass density of the energy density on the surface of the sphere. We have
\(\varepsilon_o=\cfrac{1}{c^2}\)
This is consistent with the post "Maxwell, Planck And Particles" dated 27 Dec 14, where \(\mu_o=4\pi\times10^{-7}\) was found to be unnecessary and that we set \(\mu_o=1\), here.
And when we compare with
\(F=G\cfrac{m_s}{r^2}\)
\(G=\cfrac{c^2}{4\pi}\)
But it is true that only the surface energy density on the thin spherical shell, just next to \(r\), effects \(F\)?
Opps! Lucky Me
And we have a problem,
\(F=\cfrac{1}{12\pi r}\cfrac{d\,\psi}{d\,r}\)
\(RHS=m^{-1}.Jm^{-3}m^{-1}\)
\(J=kgm^2s^{-2}\)
\(RHS=kgm^{-3}s^{-2}\)
\(LHS=kgms^{-2}\)
\(m^{-4}\) missing?
WTF! This also means that the post "Flux It" dated 21 Nov 14 is also wrong.
But \(\varepsilon_o\) to the rescue! It is possible to introduce a constant \(\varepsilon_o\) to remove the inconsistency in unit dimensions and at the same time introduce the unit for coulomb charge \(C\). But it is first, more important to realize that \(\psi\) due to a point source is distributed uniformly in a sphere with the point source as center. The force due a change in \(\psi\) along a radial line, at \(r\) distance from the center, is also divided evenly over the surface area of the sphere \(4\pi r^2\). As such the force at \(r\), along a radial line, is the force due to \(\psi\) per unit surface area of the sphere of radius \(r\). In this way, strictly speaking,
\(LHS.m^{-2}=RHS\)
\(LHS.m^{-2}=kgms^{-2}m^{-2}=kgm^{-1}s^{-2}\)
still \(m^{-2}\) missing!
What happened? Then comes \(\varepsilon_o\) with units of \(C^2N^{-1}m^{-2}\). Since \(\varepsilon_o\) would be in the denominator on the RHS, we have \(N.C^{-2}m^{2}\). This is not satisfactory.
Maybe the second division by \(4\pi r^2\) in the derivation for \(F\) along a radial line is not required?
We could have, energy density over a surface area of \(4\pi r^2\), \(\psi_A\) as
\(\psi_A=\psi.4\pi r^2\)
If \(\Delta E\) is the energy in the thickness of the thin spherical shell around the center particle of radius \(r\),
\(\Delta E=\psi_A.\Delta r\)
When this shell is infintely thin, \(\Delta r\to 0\),
\(\cfrac{d\,E}{d\,r}=\lim\limits_{\Delta r\to0}{\cfrac{\Delta E}{\Delta r}}=\psi_A\)
From which we obtain,
\(F_s=-\cfrac{d\,E}{d\,r}=-\psi_A\)
Strictly speaking we have lost the direction of \(F_s\) when we substituted in \(-\psi_A\) because
\(\psi_A\) cannot be negative,
\(|F_s|=\psi.4\pi r^2\)
But this force is distributed over the surface of the sphere of radius \(r\), as such the force along a radial line is,
\(|F|=\cfrac{|F_s|}{Area}=\cfrac{1}{4\pi r^2}\psi.4\pi r^2\)
\(|F|=\psi\)
If \(F\) is to have a inverse square law dependence then,
\(\psi=\cfrac{D}{r^2}\)
where \(D\) is a constant. Using \(E=mc^2\) we have
\(m=\cfrac{D}{c^2r^2}\)
but \(m\), mass density is in the time dimension so we have instead,
\(m=\cfrac{D}{c^2t^2}\)
which then suggests that \(m\) stretches out in time in a corresponding way as \(\psi\) stretches out in space. And instead of just a equivalence relationship we have a transform,
\(E(r)=m(t)c^2\)
from space (\(r\)) to time (\(t\)).
Simplicity rides again.
\(F=\cfrac{1}{12\pi r}\cfrac{d\,\psi}{d\,r}\)
\(RHS=m^{-1}.Jm^{-3}m^{-1}\)
\(J=kgm^2s^{-2}\)
\(RHS=kgm^{-3}s^{-2}\)
\(LHS=kgms^{-2}\)
\(m^{-4}\) missing?
WTF! This also means that the post "Flux It" dated 21 Nov 14 is also wrong.
But \(\varepsilon_o\) to the rescue! It is possible to introduce a constant \(\varepsilon_o\) to remove the inconsistency in unit dimensions and at the same time introduce the unit for coulomb charge \(C\). But it is first, more important to realize that \(\psi\) due to a point source is distributed uniformly in a sphere with the point source as center. The force due a change in \(\psi\) along a radial line, at \(r\) distance from the center, is also divided evenly over the surface area of the sphere \(4\pi r^2\). As such the force at \(r\), along a radial line, is the force due to \(\psi\) per unit surface area of the sphere of radius \(r\). In this way, strictly speaking,
\(LHS.m^{-2}=RHS\)
\(LHS.m^{-2}=kgms^{-2}m^{-2}=kgm^{-1}s^{-2}\)
still \(m^{-2}\) missing!
What happened? Then comes \(\varepsilon_o\) with units of \(C^2N^{-1}m^{-2}\). Since \(\varepsilon_o\) would be in the denominator on the RHS, we have \(N.C^{-2}m^{2}\). This is not satisfactory.
Maybe the second division by \(4\pi r^2\) in the derivation for \(F\) along a radial line is not required?
We could have, energy density over a surface area of \(4\pi r^2\), \(\psi_A\) as
\(\psi_A=\psi.4\pi r^2\)
If \(\Delta E\) is the energy in the thickness of the thin spherical shell around the center particle of radius \(r\),
\(\Delta E=\psi_A.\Delta r\)
When this shell is infintely thin, \(\Delta r\to 0\),
\(\cfrac{d\,E}{d\,r}=\lim\limits_{\Delta r\to0}{\cfrac{\Delta E}{\Delta r}}=\psi_A\)
From which we obtain,
\(F_s=-\cfrac{d\,E}{d\,r}=-\psi_A\)
Strictly speaking we have lost the direction of \(F_s\) when we substituted in \(-\psi_A\) because
\(\psi_A\) cannot be negative,
\(|F_s|=\psi.4\pi r^2\)
But this force is distributed over the surface of the sphere of radius \(r\), as such the force along a radial line is,
\(|F|=\cfrac{|F_s|}{Area}=\cfrac{1}{4\pi r^2}\psi.4\pi r^2\)
\(|F|=\psi\)
If \(F\) is to have a inverse square law dependence then,
\(\psi=\cfrac{D}{r^2}\)
where \(D\) is a constant. Using \(E=mc^2\) we have
\(m=\cfrac{D}{c^2r^2}\)
but \(m\), mass density is in the time dimension so we have instead,
\(m=\cfrac{D}{c^2t^2}\)
which then suggests that \(m\) stretches out in time in a corresponding way as \(\psi\) stretches out in space. And instead of just a equivalence relationship we have a transform,
\(E(r)=m(t)c^2\)
from space (\(r\)) to time (\(t\)).
Simplicity rides again.
Stretching Out In Time
This post is wrong.
From the post "The Distribution Of \(\psi\) Again",
\(\psi=D.ln(r)+C\) since \(E=mc^2\),
we have equivalently,
\(m=A.ln(r)+B\)
where \(m\) is now mass density,
\(A=\cfrac{D}{c^2}\) and \(B=\cfrac{C}{c^2}\)
This mass is the neutral mass along a time dimension that serve to define energy via Einstein's \(E=mc^2\) in that time dimension.
The problem is, \(\psi\) is in space but \(m\) is in the time dimension. Could it be that,
\(m=A.ln(t)+B\)
instead. That \(m\) stretches out in time in a corresponding way as \(\psi\) stretches out in space? That instead of a simple equivalence relationship we have a transform,
\(E(r)=m(t)c^2\)
from the space domain (\(r\)) to the time domain (\(t\))?
What would be the significance of such a notion?
From the post "The Distribution Of \(\psi\) Again",
\(\psi=D.ln(r)+C\) since \(E=mc^2\),
we have equivalently,
\(m=A.ln(r)+B\)
where \(m\) is now mass density,
\(A=\cfrac{D}{c^2}\) and \(B=\cfrac{C}{c^2}\)
This mass is the neutral mass along a time dimension that serve to define energy via Einstein's \(E=mc^2\) in that time dimension.
The problem is, \(\psi\) is in space but \(m\) is in the time dimension. Could it be that,
\(m=A.ln(t)+B\)
instead. That \(m\) stretches out in time in a corresponding way as \(\psi\) stretches out in space? That instead of a simple equivalence relationship we have a transform,
\(E(r)=m(t)c^2\)
from the space domain (\(r\)) to the time domain (\(t\))?
What would be the significance of such a notion?
The Distribution Of \(\psi\) Again
This post is wrong. The derivation for force density is wrong. Please refer to following post "Opps! Lucky Me".
Consider this,
\(\psi =\cfrac { E }{ Vol } =\cfrac { 3E }{ 4\pi { r }^{ 3 } } \)
where \(E\) is the energy at a point \(r\) from a particle, ie \(E(r)\). Energy density at \(r\) is thus consider to be \(E\) redistributed in a sphere of radius \(r\) centered at the particle.
The force density as a result of changing \(\psi\) along \(r\), the radial line is,
\( F_{ v }=\cfrac { d\, \psi }{ d\, r } =-3\cfrac { 3E }{ 4\pi { r }^{ 4 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } =-\cfrac { 9E }{ 4\pi { r }^{ 4 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } \)
The total force in such a sphere is,
\(F=F_v.Vol\)
and so the total flux emanating through the surface area of this sphere is,
\( \phi _{ total }=\cfrac { F }{ Area } =F_{ v }.Vol.\cfrac { 1 }{ Area }\)
\( \phi _{ total }=-\cfrac { 9E }{ 4\pi { r }^{ 4 } } .\cfrac { 4 }{ 3 } \pi { r }^{ 3 }.\cfrac { 1 }{ 4\pi { r }^{ 2 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } .\cfrac { 4 }{ 3 } \pi { r }^{ 3 }.\cfrac { 1 }{ 4\pi { r }^{ 2 } } \)
And so the Newtonian force, total flux per unit area is,
\( F=\phi _{ A }=\phi _{ total }\cfrac { 1 }{ Area } =\left( -\cfrac { 3E }{ { 4\pi r }^{ 3 } } +\cfrac { d\, E }{ d\, r } .\cfrac { 1 }{ 4\pi { r }^{ 2 } } \right) \cfrac { 1 }{ 4\pi { r }^{ 2 } }\)
\(F=\left( -\cfrac { 3E }{ { 4\pi r }^{ 4 } } +\cfrac { d\, E }{ d\, r } .\cfrac { 1 }{ 4\pi { r^{ 3 } } } \right) \cfrac { 1 }{ 4\pi { r } } \)
which is actually,
\(F=\cfrac { 1 }{ 4\pi { r } } .\cfrac { d\, }{ d\, r } \left\{ \cfrac { E }{ 4\pi { r }^{ 3 } } \right\} =\cfrac { 1 }{ 12\pi { r } } .\cfrac { d\, }{ d\, r } \left\{ \cfrac {3 E }{ 4\pi { r }^{ 3 } } \right\}\)
\(F=\cfrac { 1 }{ 12\pi { r } } . \cfrac { d\,\psi }{ d\, r }\)
So, the Newtonian force at point \(r\) is the change of energy density along the radius at \(r\), per perimeter of an enclosing circle passing through \(r\), centered at the particle, with a factor of \(\cfrac{1}{6}\).
But we know that \(F\) follows Coulomb's inverse square law, ie
\(F\propto\cfrac{1}{r^2}\)
so
\(\cfrac { d\,\psi }{ d\, r }=\cfrac{D}{r}\)
where \(D\) is an arbitrary constant,
Integration both sides,
\(\psi=D.ln(r)+C\)
It is likely that,
\(\psi=C-D.ln(r+r_o)\)
for positive \(\psi\), such that
\(\psi_o=C-Dln(r_o)\) when \(r=0\)
so that \(\psi\) is not infinite at \(r=0\).
And at \(r=r_e\),
\(\psi=0=C-D.ln(r_e+r_o)\)
so that \(\psi\) is bounded between \(r=0\) and \(r=r_e\). An illustrative plot is made below,
The gradient of \(\psi\) is such that \(F\) obeys the inverse square law.
The likelihood that \(\psi\) does not extend to infinity but instead ends abruptly at \(r=r_e\), suggests that \(F\) might not obey the inverse square law when \(r\lt r_e\). For \(r\gt r_e\), we know that the flux emanating from a sphere with radius \(r_e\), bounding \(\psi\), is a constant, as such \(F\) beyond \(r_e\) obeys the inverse square law. These two points taken together, we might have a near field and far field phenomenon.
Note: A density value at a point of distance \(r\) from a center \(O\), is dealt with, first by letting the value fills a sphere of radius \(r\) centered at \(O\) (ie. multiplied by the total volume of the sphere). Then the total value bounded by the volume of this sphere is redistributed onto the surface area of the sphere on a infinitely thin shell (the first division by area \(4\pi r^2\)). At this point the analysis has moved from a point at \(r\) to values on the surface of a sphere of radius \(r\), centered at \(O\). We then calculate the value per unit area on the surface (the second division by area \(4\pi r^2\).
Consider this,
\(\psi =\cfrac { E }{ Vol } =\cfrac { 3E }{ 4\pi { r }^{ 3 } } \)
where \(E\) is the energy at a point \(r\) from a particle, ie \(E(r)\). Energy density at \(r\) is thus consider to be \(E\) redistributed in a sphere of radius \(r\) centered at the particle.
The force density as a result of changing \(\psi\) along \(r\), the radial line is,
\( F_{ v }=\cfrac { d\, \psi }{ d\, r } =-3\cfrac { 3E }{ 4\pi { r }^{ 4 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } =-\cfrac { 9E }{ 4\pi { r }^{ 4 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } \)
The total force in such a sphere is,
\(F=F_v.Vol\)
and so the total flux emanating through the surface area of this sphere is,
\( \phi _{ total }=\cfrac { F }{ Area } =F_{ v }.Vol.\cfrac { 1 }{ Area }\)
\( \phi _{ total }=-\cfrac { 9E }{ 4\pi { r }^{ 4 } } .\cfrac { 4 }{ 3 } \pi { r }^{ 3 }.\cfrac { 1 }{ 4\pi { r }^{ 2 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } .\cfrac { 4 }{ 3 } \pi { r }^{ 3 }.\cfrac { 1 }{ 4\pi { r }^{ 2 } } \)
And so the Newtonian force, total flux per unit area is,
\( F=\phi _{ A }=\phi _{ total }\cfrac { 1 }{ Area } =\left( -\cfrac { 3E }{ { 4\pi r }^{ 3 } } +\cfrac { d\, E }{ d\, r } .\cfrac { 1 }{ 4\pi { r }^{ 2 } } \right) \cfrac { 1 }{ 4\pi { r }^{ 2 } }\)
\(F=\left( -\cfrac { 3E }{ { 4\pi r }^{ 4 } } +\cfrac { d\, E }{ d\, r } .\cfrac { 1 }{ 4\pi { r^{ 3 } } } \right) \cfrac { 1 }{ 4\pi { r } } \)
which is actually,
\(F=\cfrac { 1 }{ 4\pi { r } } .\cfrac { d\, }{ d\, r } \left\{ \cfrac { E }{ 4\pi { r }^{ 3 } } \right\} =\cfrac { 1 }{ 12\pi { r } } .\cfrac { d\, }{ d\, r } \left\{ \cfrac {3 E }{ 4\pi { r }^{ 3 } } \right\}\)
\(F=\cfrac { 1 }{ 12\pi { r } } . \cfrac { d\,\psi }{ d\, r }\)
So, the Newtonian force at point \(r\) is the change of energy density along the radius at \(r\), per perimeter of an enclosing circle passing through \(r\), centered at the particle, with a factor of \(\cfrac{1}{6}\).
But we know that \(F\) follows Coulomb's inverse square law, ie
\(F\propto\cfrac{1}{r^2}\)
so
\(\cfrac { d\,\psi }{ d\, r }=\cfrac{D}{r}\)
where \(D\) is an arbitrary constant,
Integration both sides,
\(\psi=D.ln(r)+C\)
It is likely that,
\(\psi=C-D.ln(r+r_o)\)
for positive \(\psi\), such that
\(\psi_o=C-Dln(r_o)\) when \(r=0\)
so that \(\psi\) is not infinite at \(r=0\).
And at \(r=r_e\),
\(\psi=0=C-D.ln(r_e+r_o)\)
so that \(\psi\) is bounded between \(r=0\) and \(r=r_e\). An illustrative plot is made below,
The gradient of \(\psi\) is such that \(F\) obeys the inverse square law.
The likelihood that \(\psi\) does not extend to infinity but instead ends abruptly at \(r=r_e\), suggests that \(F\) might not obey the inverse square law when \(r\lt r_e\). For \(r\gt r_e\), we know that the flux emanating from a sphere with radius \(r_e\), bounding \(\psi\), is a constant, as such \(F\) beyond \(r_e\) obeys the inverse square law. These two points taken together, we might have a near field and far field phenomenon.
Note: A density value at a point of distance \(r\) from a center \(O\), is dealt with, first by letting the value fills a sphere of radius \(r\) centered at \(O\) (ie. multiplied by the total volume of the sphere). Then the total value bounded by the volume of this sphere is redistributed onto the surface area of the sphere on a infinitely thin shell (the first division by area \(4\pi r^2\)). At this point the analysis has moved from a point at \(r\) to values on the surface of a sphere of radius \(r\), centered at \(O\). We then calculate the value per unit area on the surface (the second division by area \(4\pi r^2\).
Sunday, May 24, 2015
Photon Bodies and Material Things
The following diagram shows a basic particle and a photon; the basic particle is a wave in time, at time speed \(c\); in space it is stationary; the photon is a wave in space at speed \(c\).
It would seem that it is possible to send a basic particle to light speed in space and so convert it into a corresponding photon. There is no ambiguity when the photon is converted back to a basic particle at the end of the trip. There is enough information in all basic particles to make such conversions back and forth consistently.
In this case, the space dimension around \(t_T\) is uncoiled and set off at light speed; the particle now is a photon. The reversion uncoils the time dimension around \(x\) and the photon is again a basic particle. Since K.E along the time dimension is twice that along the space dimension at the same speed \(c\), energy is removed from the particle on conversion to a photon.
\(E_{to\,\,ph}=-\cfrac{1}{2}mc^2\)
And positive energy input is required to convert the photon back into the corresponding basic particle.
How then to uncoil a space dimension warp around time, and vice versa? Massive collisions will work as there are no ambiguities. But with what particles/photons?
When energy of this particle oscillates from \(t_T\) to \(x_2\) at a high rate, \(\cfrac{d\,E}{d\,t_g}\gt\gt 0\), it is possible then that \(x_2\) attains light speed \(c\). At the reversion, the same particle is used but the oscillation is when K.E along \(x_2\) reverts to energy along \(t_T\) also at a high rate.
This would require technology to manipulate the energy oscillations of the particles.
This photon will also work in a similar way.
It is important that as this photon passes, energy is transferred to the space dimension on forward conversion from the basic particle to the corresponding photon; and energy is oscillating from the space dimension to \(t_T\) on reversion back from the photon to a basic particle.
Both such particles that convert a basic particle to a photon and back will be oscillating slowly and will not interact with the fast oscillations of the basic particles, hopefully.
It is likely to use a particle beam on conversion to photon and on reversion, a photon beam to avoid photon contamination and particle contamination of the respective outputs.
Beam me up again , Scotty!
It would seem that it is possible to send a basic particle to light speed in space and so convert it into a corresponding photon. There is no ambiguity when the photon is converted back to a basic particle at the end of the trip. There is enough information in all basic particles to make such conversions back and forth consistently.
In this case, the space dimension around \(t_T\) is uncoiled and set off at light speed; the particle now is a photon. The reversion uncoils the time dimension around \(x\) and the photon is again a basic particle. Since K.E along the time dimension is twice that along the space dimension at the same speed \(c\), energy is removed from the particle on conversion to a photon.
\(E_{to\,\,ph}=-\cfrac{1}{2}mc^2\)
And positive energy input is required to convert the photon back into the corresponding basic particle.
How then to uncoil a space dimension warp around time, and vice versa? Massive collisions will work as there are no ambiguities. But with what particles/photons?
When energy of this particle oscillates from \(t_T\) to \(x_2\) at a high rate, \(\cfrac{d\,E}{d\,t_g}\gt\gt 0\), it is possible then that \(x_2\) attains light speed \(c\). At the reversion, the same particle is used but the oscillation is when K.E along \(x_2\) reverts to energy along \(t_T\) also at a high rate.
This would require technology to manipulate the energy oscillations of the particles.
This photon will also work in a similar way.
It is important that as this photon passes, energy is transferred to the space dimension on forward conversion from the basic particle to the corresponding photon; and energy is oscillating from the space dimension to \(t_T\) on reversion back from the photon to a basic particle.
Both such particles that convert a basic particle to a photon and back will be oscillating slowly and will not interact with the fast oscillations of the basic particles, hopefully.
It is likely to use a particle beam on conversion to photon and on reversion, a photon beam to avoid photon contamination and particle contamination of the respective outputs.
Beam me up again , Scotty!
Saturday, May 23, 2015
Sunshine and Proton Beam
This post is confused. Charges exist in the \(t_c\) time dimension and have oscillatory energy in an orthogonal time dimension. Electrons have oscillatory energy in \(t_T\) and have light speed along \(t_g\) and protons have oscillatory energy in \(t_g\) and have light speed along \(t_T\). On collisions energy along the various dimensions are released with characteristic of that dimension. In particular, \(t_c\) electric potential, \(t_g\) gravitational potential and \(t_T\) heat.
Both these particles impart electric potential energy,
A bold guess is that electron that also carries heat and has low mass is the left particle and proton is the right particle, where \(t_g\) manifest itself perpendicular to two parallel, current carrying wires. They are not equivalent.
So, what's is the difference between existing in the \(t_g\) time dimension, as in the case of the electron, and being a wave along \(t_g\), as like a proton?
Energy in the wave can be dissipated and reduced, the particle still exist. Energy that marks the particle existence (\(E=mc^2\)) cannot be dissipated, the particle disappears from existence if that happens.
The space dimensions are curled along the time dimensions \(t_g\) and \(t_T\) respectively. The \(x_1\)s in the diagram are 90o apart. When we align the two waves such that the time dimensions, \(t_c\) and \(t_g\) are parallel, we find that \(x_1\) for the case of proton is going down \(-t_T\) with respect to the electron's \(+t_T\) time axis. (After note: Why is \(x_1\) not along \(t_g\) instead?? It does not matter here, as long as \(x_1\) is in negative time, the force will have an opposite sign. The time dimensions, \(t_c\), \(t_g\) and \(t_T\) are not differentiated as far as our senses are concerned. However, this could just be a graphical trick as a result of the right hand screw rule. More convincing proof is needed.) This is consistent with the fact that the charges have opposite force fields.
If this model is true, then electron give heat but not gravitational potential energy and proton gives gravitation potential energy but not heat; in addition to the electrical fields around them. Energy oscillating between \(t_c\) and \(x_1\) forms the electrical field. Energy along \(t_g\) or \(t_T\), that is the kinetic energy of the wave, is extracted by collisions.
So a proton beam will give you a positive light feeling. Have a nice day and plenty of eternal sunshine!
Both these particles impart electric potential energy,
A bold guess is that electron that also carries heat and has low mass is the left particle and proton is the right particle, where \(t_g\) manifest itself perpendicular to two parallel, current carrying wires. They are not equivalent.
So, what's is the difference between existing in the \(t_g\) time dimension, as in the case of the electron, and being a wave along \(t_g\), as like a proton?
Energy in the wave can be dissipated and reduced, the particle still exist. Energy that marks the particle existence (\(E=mc^2\)) cannot be dissipated, the particle disappears from existence if that happens.
The space dimensions are curled along the time dimensions \(t_g\) and \(t_T\) respectively. The \(x_1\)s in the diagram are 90o apart. When we align the two waves such that the time dimensions, \(t_c\) and \(t_g\) are parallel, we find that \(x_1\) for the case of proton is going down \(-t_T\) with respect to the electron's \(+t_T\) time axis. (After note: Why is \(x_1\) not along \(t_g\) instead?? It does not matter here, as long as \(x_1\) is in negative time, the force will have an opposite sign. The time dimensions, \(t_c\), \(t_g\) and \(t_T\) are not differentiated as far as our senses are concerned. However, this could just be a graphical trick as a result of the right hand screw rule. More convincing proof is needed.) This is consistent with the fact that the charges have opposite force fields.
If this model is true, then electron give heat but not gravitational potential energy and proton gives gravitation potential energy but not heat; in addition to the electrical fields around them. Energy oscillating between \(t_c\) and \(x_1\) forms the electrical field. Energy along \(t_g\) or \(t_T\), that is the kinetic energy of the wave, is extracted by collisions.
So a proton beam will give you a positive light feeling. Have a nice day and plenty of eternal sunshine!
Thursday, May 21, 2015
Ripples On Golden Pond
How to find and prove the "ripples" mentioned in the post "Difficult To Correct Oneself"?
Energy in the space dimension is just K.E that manifest as changes in momentum. Such energy oscillating in 2 space dimensions will result in the particle changing speed along these dimensions.
Brownian motions at the particle level! More correctly, particles that give a body Brownian motion property, just as a charge particle gives a body charge property.
Which opens up the possibility of damping Brownian motion by manipulating such ripple particles.
震子！哈哈。
Temperature \(T\) is not directly the issue involved, although it seems, decreasing \(T\) reduces vibrations. Maybe it is possible to achieve an Einstein condensate at room temperature! Does particles in closer proximity share a common quantum state (not necessarily the lowest) and so coalesce?
Energy in the space dimension is just K.E that manifest as changes in momentum. Such energy oscillating in 2 space dimensions will result in the particle changing speed along these dimensions.
Brownian motions at the particle level! More correctly, particles that give a body Brownian motion property, just as a charge particle gives a body charge property.
Which opens up the possibility of damping Brownian motion by manipulating such ripple particles.
震子！哈哈。
Temperature \(T\) is not directly the issue involved, although it seems, decreasing \(T\) reduces vibrations. Maybe it is possible to achieve an Einstein condensate at room temperature! Does particles in closer proximity share a common quantum state (not necessarily the lowest) and so coalesce?
Holographic World, Hide and Seek In Time, Rigidity
The unit vectors that represent \(t_c\), \(t_g\) and \(t_T\) forms a time cube. This cube moves forward un-deformed from moment to moment. Carried in this time cube is our 3D world, as space and time dimensions warp around each other. In this way we live in a holographic world where if we travel forward or backward in time, we find a complete world just as the present "now", to interact with.
Just like a hologram where each constituent parts carries information of the complete whole, we live in a holographic world in time. When we slice time into small pieces we find a complete 3D world in the small pieces.
Is there a limit to the size of such slices? Is time quantized, that there is a minimum time slice length?
Imagine hiding between time slices.
The key point here is that \(t_c\), \(t_g\) and \(t_T\) forms a immutable time cube. Any effort to time travel by manipulating \(t_c\), \(t_g\) or \(t_T\) must return the time cube to its original form, else the world will be distorted. Maybe it is because of this rigidity that we experience time as a singular whole undifferentiated into \(t_c\), \(t_g\) and \(t_T\).
Have a nice day.
Just like a hologram where each constituent parts carries information of the complete whole, we live in a holographic world in time. When we slice time into small pieces we find a complete 3D world in the small pieces.
Is there a limit to the size of such slices? Is time quantized, that there is a minimum time slice length?
Imagine hiding between time slices.
The key point here is that \(t_c\), \(t_g\) and \(t_T\) forms a immutable time cube. Any effort to time travel by manipulating \(t_c\), \(t_g\) or \(t_T\) must return the time cube to its original form, else the world will be distorted. Maybe it is because of this rigidity that we experience time as a singular whole undifferentiated into \(t_c\), \(t_g\) and \(t_T\).
Have a nice day.
Wednesday, May 20, 2015
The Answer To The Question: What is energy?
Energy is differentiated as along \(t_c\), \(t_g\) and \(t_T\). They are coupled onto a space dimension via oscillations between the respective time dimension and a space dimension in a wave that is a photon (where the third dimension of the wave is at light speed in space) or a basic particle (where the third dimension of the wave is at light space in time, as such a stationary charge, mass or heat in space).
The space dimension is visible/measurable to us; it is in 3D space that we commonly conduct physics experiments. Time dimension experiments not common, in fact \(t_c\), \(t_g\) and \(t_T\) are not differentiated directly as 3D space.
Energy along the time dimensions can be materialized as mass \(m_E\), from Einstein,
\(E=m_Ec^2\)
so energy along the time dimension is just mass accelerated to light speed \(c\) along the time dimension. Such mass \(m_E\), are property neutral mass without associated property of gravity, charge nor temperature.
And so we go full circle, ENERGY is just a lump of inert mass \(m_E\), except now they are at light speed \(c\) along a time dimension, \(t_c\), \(t_g\) or \(t_T\) not space.
From which we resolve the ambiguity; energy is \(m_E\) at light speed \(c\) along a time dimension. In the space dimension where a particle is in a helical path its terminal speed is \(\sqrt{2}.c\). A particle in space is also travelling along the time dimension at \(\sqrt{2}.c\). This is to be consistent, since kinetic energy along any dimension is,
\(K.E=\cfrac{1}{2}mv^2=\cfrac{1}{2}m(\sqrt{2}.c)^2=mc^2\)
when time and space dimensions are considered equivalent. This is not the case.
The problem was that time appears in the denominator in the definition of velocity.
As we speed up along a time dimension, the measure of time, unit time = standard unit time * time velocity lengthens and so velocity decreases (velocity=distance per unit time). It would seems that we are going no where, but as proven previously, it requires twice the amount of energy to accelerated to time speed \(c\) along the time dimension (post "No Poetry for Einstein") and K.E along any time dimension is
\(K.E=E=mc^2\)
and in space
\(K.E=\cfrac{1}{2}mc^2\)
In which case, the terminal speed is \(c\) in both space and time dimensions.
This definition of energy, for the time being, avoids the circular, self referencing discourse of energy potential changes due to energy imparted, but energy potential is in fact due to oscillating energy between dimensions in the wave model of particles.
Energy exists along the time dimensions \(t_c\), \(t_g\) and \(t_T\), when coupled to the space dimension it imparts its property of charge, gravity or temperature upon a particle as changes in the corresponding particle's property energy potential (ie. electric potential energy, gravitational potential energy or a temperature gradient) .
So we sweep energy into the carpet, as mere existence (\(m\)), along the time dimensions, \(t_c\), \(t_g\) and \(t_T\). This energy presents itself when a photon/basic particle comes along and couples it onto a space dimension. It is then observable in our 3D space reality.
This definition of energy points to the fact that we do not interact with the space dimension and the time dimension in the same way. \(t_c\), \(t_g\) and \(t_T\) is undifferentiated and time appears to us as a singular 1D stream, whereas we move and interact in 3D space.
The space dimension is visible/measurable to us; it is in 3D space that we commonly conduct physics experiments. Time dimension experiments not common, in fact \(t_c\), \(t_g\) and \(t_T\) are not differentiated directly as 3D space.
Energy along the time dimensions can be materialized as mass \(m_E\), from Einstein,
\(E=m_Ec^2\)
so energy along the time dimension is just mass accelerated to light speed \(c\) along the time dimension. Such mass \(m_E\), are property neutral mass without associated property of gravity, charge nor temperature.
And so we go full circle, ENERGY is just a lump of inert mass \(m_E\), except now they are at light speed \(c\) along a time dimension, \(t_c\), \(t_g\) or \(t_T\) not space.
From which we resolve the ambiguity; energy is \(m_E\) at light speed \(c\) along a time dimension. In the space dimension where a particle is in a helical path its terminal speed is \(\sqrt{2}.c\). A particle in space is also travelling along the time dimension at \(\sqrt{2}.c\). This is to be consistent, since kinetic energy along any dimension is,
\(K.E=\cfrac{1}{2}mv^2=\cfrac{1}{2}m(\sqrt{2}.c)^2=mc^2\)
when time and space dimensions are considered equivalent. This is not the case.
The problem was that time appears in the denominator in the definition of velocity.
As we speed up along a time dimension, the measure of time, unit time = standard unit time * time velocity lengthens and so velocity decreases (velocity=distance per unit time). It would seems that we are going no where, but as proven previously, it requires twice the amount of energy to accelerated to time speed \(c\) along the time dimension (post "No Poetry for Einstein") and K.E along any time dimension is
\(K.E=E=mc^2\)
and in space
\(K.E=\cfrac{1}{2}mc^2\)
In which case, the terminal speed is \(c\) in both space and time dimensions.
This definition of energy, for the time being, avoids the circular, self referencing discourse of energy potential changes due to energy imparted, but energy potential is in fact due to oscillating energy between dimensions in the wave model of particles.
Energy exists along the time dimensions \(t_c\), \(t_g\) and \(t_T\), when coupled to the space dimension it imparts its property of charge, gravity or temperature upon a particle as changes in the corresponding particle's property energy potential (ie. electric potential energy, gravitational potential energy or a temperature gradient) .
So we sweep energy into the carpet, as mere existence (\(m\)), along the time dimensions, \(t_c\), \(t_g\) and \(t_T\). This energy presents itself when a photon/basic particle comes along and couples it onto a space dimension. It is then observable in our 3D space reality.
This definition of energy points to the fact that we do not interact with the space dimension and the time dimension in the same way. \(t_c\), \(t_g\) and \(t_T\) is undifferentiated and time appears to us as a singular 1D stream, whereas we move and interact in 3D space.
Brownian Motion, Short Time Travel, O particles
Then we come to the mistake I have made,
What are these?
We experience \(t_c\), \(t_g\) and \(t_T\) as a whole, just as in any direction space is just \(x\); then \(t_c\), \(t_g\) and \(t_T\) are undifferentiated and so are these particles. However, if our sight is a charge phenomenon (along \(t_c\) only) then \(o_{t_T}\)(\(x\),\(t_c\),\(t_g\)) and \(o_{t_g}\)(\(x\),\(t_c\),\(t_T\)) can impart energy along \(t_c\) and maybe slow/speedup \(p_{t_c}\) and induce invisibility. The notation \(o_{t_x}\) is used here.
These O particles can still affect the space dimensions via the space-time energy oscillations in photons. And can couple energy directly onto the time dimension. Indirectly, they require a photon or a basic particle to be detected in space. In the case of the left most O particle in the diagram, \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)), \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) or \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)); photons with energy oscillating in either \(t_g\) or \(t_c\) time dimension and a space dimension, or a stationary basic particle where the first entry inside the parenthesis is replaced with a time dimension and has energy oscillation between \(t_c\) or \(t_g\) and a space dimension. That is, \(q\)(\(t_T\),\(x\),\(t_g\)), \(T\)(\(t_c\),\(x\),\(t_g\)), \(T\)(\(t_g\),\(x\),\(t_c\)) and \(g\)(\(t_T\),\(x\),\(t_c\)).
OK?!
What are these?
We experience \(t_c\), \(t_g\) and \(t_T\) as a whole, just as in any direction space is just \(x\); then \(t_c\), \(t_g\) and \(t_T\) are undifferentiated and so are these particles. However, if our sight is a charge phenomenon (along \(t_c\) only) then \(o_{t_T}\)(\(x\),\(t_c\),\(t_g\)) and \(o_{t_g}\)(\(x\),\(t_c\),\(t_T\)) can impart energy along \(t_c\) and maybe slow/speedup \(p_{t_c}\) and induce invisibility. The notation \(o_{t_x}\) is used here.
These O particles can still affect the space dimensions via the space-time energy oscillations in photons. And can couple energy directly onto the time dimension. Indirectly, they require a photon or a basic particle to be detected in space. In the case of the left most O particle in the diagram, \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)), \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) or \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)); photons with energy oscillating in either \(t_g\) or \(t_c\) time dimension and a space dimension, or a stationary basic particle where the first entry inside the parenthesis is replaced with a time dimension and has energy oscillation between \(t_c\) or \(t_g\) and a space dimension. That is, \(q\)(\(t_T\),\(x\),\(t_g\)), \(T\)(\(t_c\),\(x\),\(t_g\)), \(T\)(\(t_g\),\(x\),\(t_c\)) and \(g\)(\(t_T\),\(x\),\(t_c\)).
OK?!
More Science Fictions
From the previous post "Why \(\cfrac{1}{3}\) Einstein?",
\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)) could be the charge-pair which existed all along but detectable separately only when high energy beam is scattered.
An orbiting electron in the vicinity of \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)) or \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)) will receive electric potential energy and be ejected as in photoelectric effects.
Compton shift is then a detectable shift in the oscillating frequency of \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_c}\)(\(x_1\),\(x_2\), \(t_T\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) or \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_T\)) that results from mechanical collisions of orbiting electrons and these particles. Since \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)) will eject the electrons and not readily collide with it, Compton scattering is mainly upon the rest of the \(\cfrac{2}{3}\) photons.
If we can detect the partial charge on the photons and can accurately count such particles, these issues can be cleared experimentally.
They are fictions as they are from my mind. Whether it is science or delusion depends on whether you have stood on a platform with a speaker mounted on it, pulsating at 7.489 Hz in "zero gravity". You have to remember to switch off the speaker when you return, otherwise you wouldn't land readily.
Why are you in my dream anyway?
\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)) could be the charge-pair which existed all along but detectable separately only when high energy beam is scattered.
An orbiting electron in the vicinity of \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)) or \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)) will receive electric potential energy and be ejected as in photoelectric effects.
Compton shift is then a detectable shift in the oscillating frequency of \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_c}\)(\(x_1\),\(x_2\), \(t_T\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) or \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_T\)) that results from mechanical collisions of orbiting electrons and these particles. Since \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)) will eject the electrons and not readily collide with it, Compton scattering is mainly upon the rest of the \(\cfrac{2}{3}\) photons.
If we can detect the partial charge on the photons and can accurately count such particles, these issues can be cleared experimentally.
They are fictions as they are from my mind. Whether it is science or delusion depends on whether you have stood on a platform with a speaker mounted on it, pulsating at 7.489 Hz in "zero gravity". You have to remember to switch off the speaker when you return, otherwise you wouldn't land readily.
Why are you in my dream anyway?
Difficult To Correct Oneself
Then we have,
where the particles are at speed \(c\) in one time dimension and have energy oscillating between one time and one space dimension. Which means the particles are stationary and energy is constantly being coupled to a linear dimension (1D) around it. If the particles are modeled as spheres or points, these lines would be the radial lines emanating from it. Particles along such lines it will be oscillating with electric, gravitational or heat (temperature) potential.
These are the particles responsible for electric induction, heat radiation and gravitational pull. In the last aspect, a body receiving gravitational potential energy from any of the two particles on the top left of the diagram will fall towards the particles. In effect, the small body is attracted to the particles imparting gravitational potential energy upon it. If there is an accumulation of energy on the body then it will gain distance from the particles. This gravitational potential energy is oscillating along a radial line between space and time. Given a large particle, as earth was modeled as a particle with gravitational wave at 7.489 Hz previously, such oscillations along the radial line might be detectable, as in the case of the planet Mercury and the Sun.
Furthermore, in the case where the particles exist in charge time \(t_c\) (middle diagrams), we know that energy along \(t_g\) is orthogonal to energy along \(t_T\) just as \(E\) field is perpendicular to \(B\) field.
These could be the respective models for charge, gravitational mass and heat, instead. But what then are these?
These particles are stationary in space, and have oscillatory energy between two space dimensions.
Ripples!!
What were proposed as basic particles are instead ripples! We have seen the like of these in the post "Quantum and A Packet Of Space" and we have mistakenly called them "Photon Flakes" in the post "Photon Flakes".
These particles being mechanical in nature opens up the possibility of mechanic waves affecting electric/magnetic, heat/orthogonal heat and gravity/orthogonal gravity.
And so a speaker pulsating at the right frequency can provide gravity/anti-gravity. A piece of quartz vibrated generates a electric potential. Sound waves along a piece of metal generate heat. And vibrations on a sheet of material generate heat and so material fatigue.
Have a nice day.
where the particles are at speed \(c\) in one time dimension and have energy oscillating between one time and one space dimension. Which means the particles are stationary and energy is constantly being coupled to a linear dimension (1D) around it. If the particles are modeled as spheres or points, these lines would be the radial lines emanating from it. Particles along such lines it will be oscillating with electric, gravitational or heat (temperature) potential.
These are the particles responsible for electric induction, heat radiation and gravitational pull. In the last aspect, a body receiving gravitational potential energy from any of the two particles on the top left of the diagram will fall towards the particles. In effect, the small body is attracted to the particles imparting gravitational potential energy upon it. If there is an accumulation of energy on the body then it will gain distance from the particles. This gravitational potential energy is oscillating along a radial line between space and time. Given a large particle, as earth was modeled as a particle with gravitational wave at 7.489 Hz previously, such oscillations along the radial line might be detectable, as in the case of the planet Mercury and the Sun.
Furthermore, in the case where the particles exist in charge time \(t_c\) (middle diagrams), we know that energy along \(t_g\) is orthogonal to energy along \(t_T\) just as \(E\) field is perpendicular to \(B\) field.
These could be the respective models for charge, gravitational mass and heat, instead. But what then are these?
These particles are stationary in space, and have oscillatory energy between two space dimensions.
Ripples!!
What were proposed as basic particles are instead ripples! We have seen the like of these in the post "Quantum and A Packet Of Space" and we have mistakenly called them "Photon Flakes" in the post "Photon Flakes".
These particles being mechanical in nature opens up the possibility of mechanic waves affecting electric/magnetic, heat/orthogonal heat and gravity/orthogonal gravity.
And so a speaker pulsating at the right frequency can provide gravity/anti-gravity. A piece of quartz vibrated generates a electric potential. Sound waves along a piece of metal generate heat. And vibrations on a sheet of material generate heat and so material fatigue.
Have a nice day.
Friday, May 8, 2015
What is energy?
The point of a wave is that it naturally exert a force field around it. In the case of EMW, a \(E\) field and a \(B\) field. To explain why a stationary electron or mass also has a force field around it, we swapped one of the 3 dimensions that constitute a wave (2 dimensions between which energy oscillates and a third at light speed) for a time dimension. The wave/particle is now stationary but with the respective electric or gravitational field around it. The premise of this model as such, is that energy is already oscillating between two dimensions, and so this model cannot explain how energy might be oscillating between any dimensions.
In order to obtain a fuller picture, one need to step back and re-look energy and its existence along a dimension.
So then...
What is energy?
In order to obtain a fuller picture, one need to step back and re-look energy and its existence along a dimension.
So then...
What is energy?
Tuesday, May 5, 2015
Why \(\cfrac{1}{3}\) Einstein?
In a similar way, in photoelectric effect, only \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)) will eject electrons from the metal by imparting electric potential energy upon them.
The remaining \(\cfrac{2}{3}\) (\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_T\)) ) will not eject electrons. \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\))/\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)) is the partial charge measured/calculated previously.
In this way we account for the missing \(\cfrac{2}{3}\) of the total photon count illuminating the metal in the photoelectric experiment.
Hello Albert again.
The remaining \(\cfrac{2}{3}\) (\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_T\)) ) will not eject electrons. \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\))/\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)) is the partial charge measured/calculated previously.
In this way we account for the missing \(\cfrac{2}{3}\) of the total photon count illuminating the metal in the photoelectric experiment.
Hello Albert again.
Friday, May 1, 2015
Not All Equal
If these photons are created in equal numbers then only \(\cfrac{1}{3}\) of the total, \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) can create anti gravitational effects. The remaining \(\cfrac{2}{3}\) photons do not carry oscillating gravitational energy oscillating between the one space and one time dimensions, as part of the 3D wave at light speed. Collisions with these photons do not increase gravitational potential energy of the body/particles as a whole. They do not have an anti-gravity effects when pulsed at earth's gravitational resonance at 7.489 Hz.
This is the answer to the \(\cfrac{2}{3}\) verses \(\cfrac{1}{3}\) question.
This is the answer to the \(\cfrac{2}{3}\) verses \(\cfrac{1}{3}\) question.
Stealing Cows and Alien Abductions
The photons are repeated below,
1D particles travelling at light speed and have energy oscillating between one time and one space dimension. Energy can be transferred between a space dimension and a time dimension, then photons \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) both carry gravitational energy that can be coupled to a space dimension by collisions. In the notation used, \(p_{t_x}\) denotes photons that exist along \(t_x\) timeline. The triplet in the parenthesis denotes the wave in which the first element is at light speed and the following two dimensions carry the oscillating energy.
When \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) are pulsed at 7.489 Hz and directed at an object, they will create an anti-gravitational effect. The object will be lifted.
An extra womb, anyone?
1D particles travelling at light speed and have energy oscillating between one time and one space dimension. Energy can be transferred between a space dimension and a time dimension, then photons \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) both carry gravitational energy that can be coupled to a space dimension by collisions. In the notation used, \(p_{t_x}\) denotes photons that exist along \(t_x\) timeline. The triplet in the parenthesis denotes the wave in which the first element is at light speed and the following two dimensions carry the oscillating energy.
When \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)) are pulsed at 7.489 Hz and directed at an object, they will create an anti-gravitational effect. The object will be lifted.
An extra womb, anyone?