If,
"re" is " 鲁"，
"pu" is "国"，
and
"lic" is "历", in Cantonese.
But I have also come to know that,
"weapon" is "尾巴" in Mandarin.
Given the first two above,
"weapon" is "武功" in Cantonese.
this means it is possible to convert Cantonese to Mandarin by notching out certain frequencies or shifting frequencies in Cantonese.
Cantonese and Mandarin is the same language. Physiological and possibly anatomical differences between the ancient species that spoke them, heard the languages differently and so sound them off differently.
Have a nice day.
Note: Plato's Republic and 孔子 - 鲁国历 penned by 乐欬 (ngok6 koi3), if "Plato" is Cantonese in ancient phonetics with "pla" pronounced as "glo" and "to" pronounced as "ko". 卜商 is also possible but unlikely because the "n" sound is missing.
高柴! gou1 caai2 where "p" is a "g", "a" is a "o" as above and "t" is a "j".
Tuesday, August 23, 2016
Wednesday, August 10, 2016
Other Dimensions?
There is no need for a separate space time particle, all particles have a space and time component. These particles, free to move in a 3D volume, generate three space dimensions. Around each space dimension, a time dimension wraps tightly in a close spiral.
Does the volume contains the three space dimensions, or do the three space dimensions extend the volume?
The volume and the three space dimensions both describe the same entity.
If the time dimension, as one of the two orthogonal dimensions about which energy oscillates in a particle, is available us, then the other space dimension in the same oscillation should also be accessible. And as we measure \(E\), \(g\) and \(T\) field around the particle, we access this space dimension. Conceptually, the energy in these fields push a marker along a linear 1D scale de-marking a value indicative of the energy level in that field. When such a scale is located physically around a particle, we obtain an indication of the field around that particle, along the direction of the scale.
Is it possible that other dimensions exist, yet undetected? This is a different question from whether
other energies (energy pair) exist. Each such energy has an associated particle, a force field and an orthogonal oscillating energy that manifest itself when the particle goes into spin. This new particle is still in the old space time dimension framework. Whereas a new dimension would imply a new space time framework, not necessarily with new energy pair. (Such energy pairs are needed only for wave/particles to exist. Is free energy without a particle embodiment possible?)
Have a nice day.
Does the volume contains the three space dimensions, or do the three space dimensions extend the volume?
The volume and the three space dimensions both describe the same entity.
If the time dimension, as one of the two orthogonal dimensions about which energy oscillates in a particle, is available us, then the other space dimension in the same oscillation should also be accessible. And as we measure \(E\), \(g\) and \(T\) field around the particle, we access this space dimension. Conceptually, the energy in these fields push a marker along a linear 1D scale de-marking a value indicative of the energy level in that field. When such a scale is located physically around a particle, we obtain an indication of the field around that particle, along the direction of the scale.
Is it possible that other dimensions exist, yet undetected? This is a different question from whether
other energies (energy pair) exist. Each such energy has an associated particle, a force field and an orthogonal oscillating energy that manifest itself when the particle goes into spin. This new particle is still in the old space time dimension framework. Whereas a new dimension would imply a new space time framework, not necessarily with new energy pair. (Such energy pairs are needed only for wave/particles to exist. Is free energy without a particle embodiment possible?)
Have a nice day.
Thursday, August 4, 2016
ABC Con Gatcha
For all triples \((a,\, b,\, c)\) of coprime positive integers, with \(a + b = c\),
we see that for \(\varepsilon\gt 0\) where
\(c\gt rad(abc)^{1+\varepsilon}\)
with the radical of a positive integer \(n\), the radical of \(n\), denoted \(rad(n)\), is the product of the distinct prime factors of \(n\);
is the small blue circle smaller than the red circle of radius \(a+b=c\).
The diagram does not count the valid number of triples for which the blue circle is small, but if we consider all valid triples of co-prime and scale each corresponding green circle \(\small{abc}\) to a uniform size, and allow all with smaller blue circle to sieve through the red circle, then obviously, by the pigeonhole principle, since the green circles are bigger and the blue circle can be as big as the green circle, only some blue circles will fall through the red circle at the center.
Is the number of blue circles that fell through finite?
We note that \(abc\gt \sqrt{ab^2+a^2b}\gt c \gt rad(abc)^{1+\varepsilon}\), in other words, the blue circle is the smallest and has radius less than \(\sqrt{abc}\).
Using Euler's totient function,
\(\varphi(ab)=\varphi(a)\varphi(b)\)
where \(a\) and \(b\) are co-prime. We may,
\(\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)\)
where \(a\), \(b\) and \(c\) are co-prime. As, \(n\gt\varphi(n)\)
\(abc\gt\varphi(abc)\)
and
\(abc\gt c\)
So, we have either,
\(abc\gt\varphi(abc)\gt c\gt\varphi(c)\)
or
\(abc\gt c\gt \varphi(abc)\gt\varphi(c)\)
Since, \(abc\) is not prime, \(\varphi(abc)\) is finite; that the number of positive integer not greater than \(abc\), and relatively prime with \(abc\) is countable, the number of blue circles that fell through is also countable. If \(abc\) is prime (which is impossible as it has three factors) than \(\varphi(abc)=abc-1\), which is as large as \(abc\).
When
\(abc\gt c\gt \varphi(abc)\gt rad(abc)^{1+\varepsilon}\)
which is the same as the first case, the position of \(c\) is irrelevant. In both cases, however, the count of terms \(\varphi(abc)\) is greater than the product of terms \(rad(abc)^{1+\varepsilon}\), which can be true (in the limiting case of equality \(\varphi(abc)=rad(abc)^{1+\varepsilon}\) ) only for the first two prime numbers \(1\) and \(2\).
When
\(abc\gt c\gt rad(abc)^{1+\varepsilon}\gt \varphi(abc)\gt\varphi(c)\)
we see that for \(\varepsilon\gt 0\) where
\(c\gt rad(abc)^{1+\varepsilon}\)
with the radical of a positive integer \(n\), the radical of \(n\), denoted \(rad(n)\), is the product of the distinct prime factors of \(n\);
is the small blue circle smaller than the red circle of radius \(a+b=c\).
The diagram does not count the valid number of triples for which the blue circle is small, but if we consider all valid triples of co-prime and scale each corresponding green circle \(\small{abc}\) to a uniform size, and allow all with smaller blue circle to sieve through the red circle, then obviously, by the pigeonhole principle, since the green circles are bigger and the blue circle can be as big as the green circle, only some blue circles will fall through the red circle at the center.
Is the number of blue circles that fell through finite?
We note that \(abc\gt \sqrt{ab^2+a^2b}\gt c \gt rad(abc)^{1+\varepsilon}\), in other words, the blue circle is the smallest and has radius less than \(\sqrt{abc}\).
Using Euler's totient function,
\(\varphi(ab)=\varphi(a)\varphi(b)\)
where \(a\) and \(b\) are co-prime. We may,
\(\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)\)
where \(a\), \(b\) and \(c\) are co-prime. As, \(n\gt\varphi(n)\)
\(abc\gt\varphi(abc)\)
and
\(abc\gt c\)
So, we have either,
\(abc\gt\varphi(abc)\gt c\gt\varphi(c)\)
or
\(abc\gt c\gt \varphi(abc)\gt\varphi(c)\)
because,
\(\varphi(abc)=\varphi(a)\varphi(b)\varphi(c)\gt\varphi(c)\)
When,
\(abc\gt\varphi(abc)\gt c\gt rad(abc)^{1+\varepsilon}\)
When
\(abc\gt c\gt \varphi(abc)\gt rad(abc)^{1+\varepsilon}\)
which is the same as the first case, the position of \(c\) is irrelevant. In both cases, however, the count of terms \(\varphi(abc)\) is greater than the product of terms \(rad(abc)^{1+\varepsilon}\), which can be true (in the limiting case of equality \(\varphi(abc)=rad(abc)^{1+\varepsilon}\) ) only for the first two prime numbers \(1\) and \(2\).
When
\(abc\gt c\gt rad(abc)^{1+\varepsilon}\gt \varphi(abc)\gt\varphi(c)\)
we have,
\(c\gt rad(abc)^{1+\varepsilon}\gt \varphi(c)\) --- (*)
where \(rad(abc)^{1+\varepsilon}\) is squeeze between \(c\) (\(c\gt a\) and \(c\gt b\)) and \(\varphi(c)\) which makes \(rad(abc)^{1+\varepsilon}\) countable.
For expression (*) to be possible, when \(c\) is large but not divisible by large powers of prime then, as \(rad(abc)^{1+\varepsilon}\) exists in the range from \(c\) to \(\varphi(c)\), this implies that \(a\) and \(b\) must be of high powers of primes, because \(a\), \(b\) and \(c\) are co-prime and factor \(a\) and \(b\) must contribute less to \(rad(abc)\) than \(c\), on condition that \(rad(abc)\lt c\) is true for the triple \((a,\, b,\, c)\).
This is not to be taken too seriously...
Wednesday, August 3, 2016
When Time Passes in \(c\)?
Just as observing \(B\) spinning around a dipole, at the dipole, yields
\(B=-i\cfrac{\partial E}{\partial x}\)
and observing \(B\) away from the dipole, at a location fixed in space as the dipole passes,
\(\cfrac{\partial B}{\partial t}=-i\cfrac{\partial E}{\partial x}\)
as in the post "Magnetic Field In General, HuYaa" dated 13 Oct 2014.
For time, \(\tau\) at the spacetime particle,
\(\tau=f(x)\)
at a point away from the spacetime particle,
\(\cfrac{\partial \tau}{\partial t}=f(x)\)
as the particle passes. This provides intuitively, an explanation for the time derivative needed, after the expression for time has been derived at the spacetime particle and we needed an expression observing the spacetime particle from afar away from the particle, in a second perspective.
Time at light speed is particles carrying the time field passing by us at light speed.
Which lead us back to the notion of time travel by isolation; a solid block of lead with 42 cm thick wall, where inside the walled cell, time stood still.
And the oldest cat is 41 yrs...poor thing.
\(B=-i\cfrac{\partial E}{\partial x}\)
and observing \(B\) away from the dipole, at a location fixed in space as the dipole passes,
\(\cfrac{\partial B}{\partial t}=-i\cfrac{\partial E}{\partial x}\)
as in the post "Magnetic Field In General, HuYaa" dated 13 Oct 2014.
For time, \(\tau\) at the spacetime particle,
\(\tau=f(x)\)
at a point away from the spacetime particle,
\(\cfrac{\partial \tau}{\partial t}=f(x)\)
as the particle passes. This provides intuitively, an explanation for the time derivative needed, after the expression for time has been derived at the spacetime particle and we needed an expression observing the spacetime particle from afar away from the particle, in a second perspective.
Time at light speed is particles carrying the time field passing by us at light speed.
Which lead us back to the notion of time travel by isolation; a solid block of lead with 42 cm thick wall, where inside the walled cell, time stood still.
And the oldest cat is 41 yrs...poor thing.