When we shrink Earth radius, from 6371000 m to a value of 0.00886 m, increasing its material density, we created a black hole with a gravity value of 5.07071e18 ms-2.
A decrease in radius of,
\(-\Delta x\) = 6371000 m (ignoring 0.00886)
results in an increase in force per unit mass of,
\(\Delta \cfrac{F}{m}\) = 5.07071e18 ms-2 (ignoring 9.80665)
Consider the region just above and below the surface of such a planet, where space density is equal. The space density there decreases in both direction above and below the surface. Gravity also decreases in both direction above and below the surface. Gravity is maximum in the region. It is reasonable to assume that increasing space density requires a force. This is the same force that hold the material of the planet together; it is gravity. An opposite and equal force pushes space around the planet surface back. If space behave like a spring, we might have,
\(\cfrac{-k}{m}= \cfrac{\Delta F}{m \Delta x}\)
\(\cfrac{k}{m}\) = 5.07071e18/6371000 = 7.9590e11 Nm-1kg-1
And therefore resonance,
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}\)
\(f_{res}\) = sqrt(5.07071e18/6371000)/2/pi
\(f_{res}\) = 1.4199e5 Hz = 141.99 kHz
If we compare this with the frequency of the gravitational wave, \(f_{BH}\), from a black hole,
\(f_{BH}=\cfrac{1}{2\pi}\cfrac{c}{r_{eo}}\) = 1/(2pi)*299792458/0.00886 = 5.3853e9 Hz
They are not the same. Waves at resonance frequency travels further with the least attenuation. The wave from the black hole decrease exponentially from the source. However if we have a planet of the size,
\(\cfrac{1}{2\pi}\cfrac{c}{r_{res}}=f_{res}\)
\(r_{res}=\cfrac{1}{2\pi}\cfrac{c}{f_{res}}\)
\(r_{res}\) = 1/(2pi)*299792458/1.4199e5 = 336.033 m
At this radius, it will generate resonance gravity waves that travels over long distances. A gravity wave antenna will have such a dimension; radius of 336.033 m
Thursday, July 31, 2014
Wednesday, July 30, 2014
Who's Ringing? Wobbly Hydrogen?
From the post "Beijing Mask Opera and Wobbling Hydrogen",
\(F_{ t }=m_{ e }a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 } } cos(\theta )\) --- (1)
When \(v=c\),
\(r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } sin^{ 2 }(\theta )=\cfrac { x }{ cos(\theta ) } \) --- (2)
\(x=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } sin^{ 2 }(\theta )cos(\theta )\)
\(\cfrac { dx }{ d\theta } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2{ \cos ^{ 2 } \left( \theta \right) \sin \left( \theta \right) -\sin ^{ 3 } \left( \theta \right) })\)
The negative sign is the result of \(x\) decreasing as \(\theta\) increases.
\(\cfrac { d^2x }{ d\theta^2 } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } }( 2cos^3(\theta)-7cos(\theta)sin^2(\theta))\)
\(\cfrac { d^{ 2 }x }{ d\theta ^{ 2 } } \cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } =\cfrac { dx^{ 2 } }{ dt^{ 2 } } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2cos^{ 3 }(\theta )-7cos(\theta)sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } \) --- (*)
Substitute (2) into (1)
\(m_{ e }a=\cfrac { 4\pi \varepsilon _{ o }m^{ 2 }_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } \)
\(a=\cfrac { 4\pi \varepsilon _{ o }m_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } =\cfrac { dx^{ 2 } }{ dt^{ 2 } } \)
From (*),
\(\cfrac { 4\pi \varepsilon _{ o }m_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2cos^{ 3 }(\theta )-7cos(x)sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } \)
\( \cfrac { 16\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } =-sin^{ 4 }(\theta )(2cos^{ 2 }(\theta )-7sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } \)
\(\cfrac { 16\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } \iint _{ 0 }^{ T }{ d^{ 2 }t } =-\iint _{ 0.9553 }^{ \frac { \pi }{ 2 } }{ sin^{ 4 }(\theta )(2cos^{ 2 }(\theta )-7sin^{ 2 }(\theta)) } d\theta ^{ 2 }\)
This was solved on the web,
\(-\int _{ 0.955 }^{ \frac { \pi }{ 2 } } \int \sin ^{ 4 } \left( \theta \right) \left( 2\cos ^{ 2 } \left( \theta \right) -7\sin ^{ 2 } \left( \theta \right) \right) d\theta d\theta\: =\: \cfrac { 2358870+3 }{ 1280000 } -\cfrac { 33\pi ^{ 2 } }{ 128 } +\cfrac { 103\cos \left( \cfrac { 191 }{ 100 } \right) }{ 128 } -\cfrac { 23\cos \left( \cfrac { 191 }{ 50 } \right) }{ 256 } +\cfrac { \cos \left( \cfrac { 573 }{ 100 } \right) }{ 128 } \)
\(\cfrac { 8\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } { T }^{ 2 }\)=0.8927
And the period of this oscillation is \(4T\),
\(T=\sqrt{0.8927\cfrac { q^{ 4 } }{ 8\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }} \)
\(T=\sqrt{\cfrac{0.8927}{8}}\cfrac { q^{ 2 } }{ \pi \varepsilon _{ o }m_{ e }c^{ 3 } } \)
\(T_p = 4*T\) = 4*( 0.8927/8)^(1/2)*(1.602176565e-19)^2/(pi*(8.854187817e-12)*(9.10938291e-31)*(299792458)^3)
\(T_p\) = 5.02388e-23 s
And the resonance frequency is,
\(f\) = 1.9905e22 Hz
Comparing this to the previous estimate, this value is fair. It is expected that the oscillation frequency be higher because the system is not linear (\(\frac{1}{r^2}\)) as was the first estimate. The correction factor is pi/4*(8/0.8927)^(1/2) = 2.3511.
When \(v=c\),
\(r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } sin^{ 2 }(\theta )=\cfrac { x }{ cos(\theta ) } \) --- (2)
\(x=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } sin^{ 2 }(\theta )cos(\theta )\)
\(\cfrac { dx }{ d\theta } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2{ \cos ^{ 2 } \left( \theta \right) \sin \left( \theta \right) -\sin ^{ 3 } \left( \theta \right) })\)
The negative sign is the result of \(x\) decreasing as \(\theta\) increases.
\(\cfrac { d^2x }{ d\theta^2 } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } }( 2cos^3(\theta)-7cos(\theta)sin^2(\theta))\)
\(\cfrac { d^{ 2 }x }{ d\theta ^{ 2 } } \cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } =\cfrac { dx^{ 2 } }{ dt^{ 2 } } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2cos^{ 3 }(\theta )-7cos(\theta)sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } \) --- (*)
Substitute (2) into (1)
\(m_{ e }a=\cfrac { 4\pi \varepsilon _{ o }m^{ 2 }_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } \)
\(a=\cfrac { 4\pi \varepsilon _{ o }m_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } =\cfrac { dx^{ 2 } }{ dt^{ 2 } } \)
From (*),
\(\cfrac { 4\pi \varepsilon _{ o }m_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2cos^{ 3 }(\theta )-7cos(x)sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } \)
\( \cfrac { 16\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } =-sin^{ 4 }(\theta )(2cos^{ 2 }(\theta )-7sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } \)
\(\cfrac { 16\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } \iint _{ 0 }^{ T }{ d^{ 2 }t } =-\iint _{ 0.9553 }^{ \frac { \pi }{ 2 } }{ sin^{ 4 }(\theta )(2cos^{ 2 }(\theta )-7sin^{ 2 }(\theta)) } d\theta ^{ 2 }\)
This was solved on the web,
\(-\int _{ 0.955 }^{ \frac { \pi }{ 2 } } \int \sin ^{ 4 } \left( \theta \right) \left( 2\cos ^{ 2 } \left( \theta \right) -7\sin ^{ 2 } \left( \theta \right) \right) d\theta d\theta\: =\: \cfrac { 2358870+3 }{ 1280000 } -\cfrac { 33\pi ^{ 2 } }{ 128 } +\cfrac { 103\cos \left( \cfrac { 191 }{ 100 } \right) }{ 128 } -\cfrac { 23\cos \left( \cfrac { 191 }{ 50 } \right) }{ 256 } +\cfrac { \cos \left( \cfrac { 573 }{ 100 } \right) }{ 128 } \)
\(\cfrac { 8\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } { T }^{ 2 }\)=0.8927
And the period of this oscillation is \(4T\),
\(T=\sqrt{0.8927\cfrac { q^{ 4 } }{ 8\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }} \)
\(T=\sqrt{\cfrac{0.8927}{8}}\cfrac { q^{ 2 } }{ \pi \varepsilon _{ o }m_{ e }c^{ 3 } } \)
\(T_p = 4*T\) = 4*( 0.8927/8)^(1/2)*(1.602176565e-19)^2/(pi*(8.854187817e-12)*(9.10938291e-31)*(299792458)^3)
\(T_p\) = 5.02388e-23 s
And the resonance frequency is,
\(f\) = 1.9905e22 Hz
Comparing this to the previous estimate, this value is fair. It is expected that the oscillation frequency be higher because the system is not linear (\(\frac{1}{r^2}\)) as was the first estimate. The correction factor is pi/4*(8/0.8927)^(1/2) = 2.3511.
To Root Two Or Not To Root Two
An electron helical path has two velocity components, perpendicular to each other. One responsible for circular motion the other translation along the direction of the spiral. Since there is no reason that they experience different drag in uniform space, they are equal, \(c\). And so their resultant is \(\sqrt{2}c\).
When an electron is at simple circular motion around a nucleus, its terminal velocity is \(\sqrt{2}c\).
A photon is in a helical path, light speed so measured is the translation velocity along the direction of the helix. Light speed is \(c\), not the resultant \(\sqrt{2}c\). Light speed is not the true terminal velocity given that space is a thin fluid. The true terminal velocity of space is \(\sqrt{2}c\)...here's the bomber...at a given gravity, ie. a given space density!
Have a nice day...
When an electron is at simple circular motion around a nucleus, its terminal velocity is \(\sqrt{2}c\).
A photon is in a helical path, light speed so measured is the translation velocity along the direction of the helix. Light speed is \(c\), not the resultant \(\sqrt{2}c\). Light speed is not the true terminal velocity given that space is a thin fluid. The true terminal velocity of space is \(\sqrt{2}c\)...here's the bomber...at a given gravity, ie. a given space density!
Have a nice day...
Beijing Mask Opera and Wobbling Hydrogen
From the post "Temperature, Space Density And Gravity",
The electron fly past the positive nucleus and is retarded by the attractive force. At some point \(r_o\) the electron has speed \(v_i=0\) and begins its return to the nucleus.
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }+\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_a}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_o }\) --- (1)
where the second term on the LHS is the potential energy of the electron at a distance \(r_a\) right above the nucleus.
The centripetal force, \(F_{c}\) is given by,
\(F_c=\cfrac{q^2}{4\pi\varepsilon_o r^2}sin(\theta)=\cfrac{m_e v^2}{r^{'}}\)
where \(v\) is the circular velocity of the electron.
\(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } sin(\theta )\frac { { r }^{ ' } }{ r } =m_{ e }v^{ 2 }r\)
\( r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } sin^{ 2 }(\theta )\) --- (*)
At \(\theta=\frac{\pi}{2}\), ie. right above the nucleus,
\( r_a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } }\)
Substitute into (1)
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }+m_{ e }v^{ 2 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_o }\)
\(r_o=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_e(\frac{c^2}{2}+v^2)}\)
At \(r = r_o\),
\(\cfrac { v^{ 2 } }{ sin^{ 2 }(\theta ) } =\frac { 1 }{ 2 } c^{ 2 }+v^{ 2 }\)
\(v^{ 2 }=\frac { 1 }{ 2 } c^{ 2 }\cfrac { sin^{ 2 }(\theta ) }{ 1-sin^{ 2 }(\theta ) } \)
\(v^{ 2 }=\frac { 1 }{ 2 } c^{ 2 }tan^{ 2 }(\theta )\)
\(v=\cfrac { c }{ \sqrt{2} }tan(\theta)\)
Since \(v≤c\),
\(tan(\theta)≤\sqrt{2}\)
When \(v=c\), ie the electron is in circular motion with circular velocity \(c\),
\( tan(\theta_o)=\sqrt{2}\)
\(\theta_o\) = 0.9553 r
Since the electron takes very little distance to accelerate up to \(c\) (from the post "To Light Speed and Beyond"), the case of \(v<c\) does not happen.
Consider expression (*) again. A polar plot of r=0.28(sin(theta))^2 for the 0.9553^<theta<2.187
The electron will resonate about the positive charge in a spiral with the above curve as envelope. The circular speed of the electron remains at \(c\), the velocity component that is always perpendicular to the circular path oscillate between zero,\(c\), zero and \(-c\).
A plot of this oscillation is attempted below,
The electron is in a circular spiral, the smallest circle at the extreme ends is given by,
\(r_{small}=r_osin(\theta)=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_e(\frac{c^2}{2}+c^2)}sin(\theta_o)\)
\(r_{small}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_ec^2}.\frac{2}{3}sin(0.9553)\)
\(r_{small}\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*(299792458^2))*(2/3)*sin(0.9553)
\(r_{small}\) = 1.5339e-15 m
And the largest at \(r_a\),
\( r_a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } }\)
\(r_a\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*(299792458^2))
\(r_a\) = 2.818e-15 m
The ratio of these two radii is about 0.5443.
Under such oscillations, the hydrogen would seems to be wobbling. We have considered the drag force on the electron when we set the speed limit to be \(c\). Until next time...
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }+\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_a}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_o }\) --- (1)
where the second term on the LHS is the potential energy of the electron at a distance \(r_a\) right above the nucleus.
The centripetal force, \(F_{c}\) is given by,
\(F_c=\cfrac{q^2}{4\pi\varepsilon_o r^2}sin(\theta)=\cfrac{m_e v^2}{r^{'}}\)
where \(v\) is the circular velocity of the electron.
\(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } sin(\theta )\frac { { r }^{ ' } }{ r } =m_{ e }v^{ 2 }r\)
\( r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } sin^{ 2 }(\theta )\) --- (*)
At \(\theta=\frac{\pi}{2}\), ie. right above the nucleus,
\( r_a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } }\)
Substitute into (1)
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }+m_{ e }v^{ 2 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r_o }\)
\(r_o=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_e(\frac{c^2}{2}+v^2)}\)
At \(r = r_o\),
\(\cfrac { v^{ 2 } }{ sin^{ 2 }(\theta ) } =\frac { 1 }{ 2 } c^{ 2 }+v^{ 2 }\)
\(v^{ 2 }=\frac { 1 }{ 2 } c^{ 2 }\cfrac { sin^{ 2 }(\theta ) }{ 1-sin^{ 2 }(\theta ) } \)
\(v^{ 2 }=\frac { 1 }{ 2 } c^{ 2 }tan^{ 2 }(\theta )\)
\(v=\cfrac { c }{ \sqrt{2} }tan(\theta)\)
Since \(v≤c\),
\(tan(\theta)≤\sqrt{2}\)
When \(v=c\), ie the electron is in circular motion with circular velocity \(c\),
\( tan(\theta_o)=\sqrt{2}\)
\(\theta_o\) = 0.9553 r
Since the electron takes very little distance to accelerate up to \(c\) (from the post "To Light Speed and Beyond"), the case of \(v<c\) does not happen.
Consider expression (*) again. A polar plot of r=0.28(sin(theta))^2 for the 0.9553^<theta<2.187
The electron will resonate about the positive charge in a spiral with the above curve as envelope. The circular speed of the electron remains at \(c\), the velocity component that is always perpendicular to the circular path oscillate between zero,\(c\), zero and \(-c\).
\(r_{small}=r_osin(\theta)=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_e(\frac{c^2}{2}+c^2)}sin(\theta_o)\)
\(r_{small}=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_ec^2}.\frac{2}{3}sin(0.9553)\)
\(r_{small}\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*(299792458^2))*(2/3)*sin(0.9553)
\(r_{small}\) = 1.5339e-15 m
And the largest at \(r_a\),
\( r_a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } }\)
\(r_a\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*(299792458^2))
\(r_a\) = 2.818e-15 m
The ratio of these two radii is about 0.5443.
Under such oscillations, the hydrogen would seems to be wobbling. We have considered the drag force on the electron when we set the speed limit to be \(c\). Until next time...
Too Hot, Too Cold, Auntie's Gravy
Space around the Sun is less dense, that is the reason why the gravity around it is less. High temperature thin out space. At the other end of the spectrum, extreme cold puncture a hole in space, everything in the vicinity get drawn in and there is fusion. So a hot and cold spectrum will also set up a gravity gradient.
The cold body has denser space around it, The solar sail absorbs solar radiation and heats up, space thins out around it. There is a net gravitation force along the trust holding the cold body towards the cold body. Both the sail and the cold body are attached to the same structure.
Another variation shown above is in a dense bell structure. The hot body might be heated by plasma and the cold body, part of heat distribution/cooling system. The dense bell "contains" denser space between it and the cold body. A high gravity point forms in this space and should be above the C.G. of the superstructure of which the bell is part. It may be made of Lead, Pb alloy or other dense materials.
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| Solar Sail |
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| Plasma Bell |
Monday, July 28, 2014
When Drag Is Good For You, Collapsing Universe Otherwise
From the previous post "Drag and A Sense of Lightness"
\(F_D=Av^2\)
If the helical path of the electron at terminal velocity is due only to the drag force.
\(\cfrac{m_ev^2}{r}=Av^2=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}v^2\)
\(\cfrac { 1 }{ r } =(1-\cfrac { r_{ ec } }{ r_{ e } } )\cfrac { 1 }{ r_{ e } } \)
\( r=r_{ e }\cfrac{1}{(1-\cfrac { r_{ ec } }{ r_{ e } } )}\)
\(r\) = 2.5e-11/(1-5.636e-5) = 2.50014e-11 m
Since,
\((1-\cfrac { r_{ ec } }{ r_{ e } } )<1\)
\(r>r_e\)
That means, under normal circumstances, the electron will never be in a helical radius smaller than the atomic radius. It will not crash into the positive nucleus. The presence of a positively charged nucleus adds to the centripetal force and the radius of circular motion is smaller.
Have a nice day.
\(F_D=Av^2\)
If the helical path of the electron at terminal velocity is due only to the drag force.
\(\cfrac{m_ev^2}{r}=Av^2=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}v^2\)
\(\cfrac { 1 }{ r } =(1-\cfrac { r_{ ec } }{ r_{ e } } )\cfrac { 1 }{ r_{ e } } \)
\( r=r_{ e }\cfrac{1}{(1-\cfrac { r_{ ec } }{ r_{ e } } )}\)
\(r\) = 2.5e-11/(1-5.636e-5) = 2.50014e-11 m
Since,
\((1-\cfrac { r_{ ec } }{ r_{ e } } )<1\)
\(r>r_e\)
That means, under normal circumstances, the electron will never be in a helical radius smaller than the atomic radius. It will not crash into the positive nucleus. The presence of a positively charged nucleus adds to the centripetal force and the radius of circular motion is smaller.
Have a nice day.
Drag and A Sense of Lightness
As the electron accelerates towards a positive charge, a perpendicular velocity component develops as a result of instability. Due to the need to satisfy both conservation of momentum and conservation of total energy, as the electron speed increases continuously, there are unstable speed points along the way that the electron experience a oscillatory force perpendicular to its direction of travel. This force redistribute the total kinetic energy and split the velocity into two perpendicular components in order that both conservation laws are satisfied.
Because of this perpendicular velocity component, the electron move in a helical path centered at the positive nucleus too.
We know that the electron experiences a drag force that eventually hold its velocity at terminal velocity,
drag force, \(F_D=Av^2\)
where \(A\) is a constant and \(v\), the electron velocity.
If we factor this force into the electron orbit around a positive charge.
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } } +Av^{ 2 }\)
\( (m_{ e }-Ar_{ e })r_{ e }v^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } \)
\( r_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }(m_{ e }-Ar_{ e })v^{ 2 } } \)
Compare this with,
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ ec } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ ec } } \)
\( r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } \)
Is seem that the electron has lost mass,
\(m_e>m_{ e }-Ar_{ e }\)
If we take the ratio of the two values,
\(\cfrac{r_{ec}}{r_e}=\cfrac{m_{ e }-Ar_{ e }}{m_{ e }}\)
\(\cfrac{r_{ec}}{r_e}=1-\cfrac{A}{m_e}r_{e}\)
where \(r_e\) is the measured atomic radius of hydrogen and \(r_{ec}\) is the calculated value.
\( r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } \)
\(r_{ ec }\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*2*(299792458^2))=1.4090e-15m
Notice that the new terminal velocity, \(2c^2\) has been used.
The measured \(r_{e}\) = 5.3e-11 m
\(\cfrac{r_{ec}}{r_e}\) = 1.4090e-15/5.3e-11=2.658e-5 m
\(A=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}\)
\(A\) = (1-2.658e-5)*9.10938291e-31/5.3e-11=1.719e-20 Nm-2s2
This drag coefficient seems low but for the mass of an electron (9.10938291e-31 kg) it is significant.
Because of this perpendicular velocity component, the electron move in a helical path centered at the positive nucleus too.
We know that the electron experiences a drag force that eventually hold its velocity at terminal velocity,
drag force, \(F_D=Av^2\)
where \(A\) is a constant and \(v\), the electron velocity.
If we factor this force into the electron orbit around a positive charge.
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } } +Av^{ 2 }\)
\( (m_{ e }-Ar_{ e })r_{ e }v^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } \)
\( r_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }(m_{ e }-Ar_{ e })v^{ 2 } } \)
Compare this with,
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ ec } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ ec } } \)
\( r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } \)
Is seem that the electron has lost mass,
\(m_e>m_{ e }-Ar_{ e }\)
If we take the ratio of the two values,
\(\cfrac{r_{ec}}{r_e}=\cfrac{m_{ e }-Ar_{ e }}{m_{ e }}\)
\(\cfrac{r_{ec}}{r_e}=1-\cfrac{A}{m_e}r_{e}\)
where \(r_e\) is the measured atomic radius of hydrogen and \(r_{ec}\) is the calculated value.
\( r_{ ec }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }v^{ 2 } } \)
\(r_{ ec }\) = (1.602176565e-19)^2/(4pi*8.854187817e-12*9.10938291e-31*2*(299792458^2))=1.4090e-15m
Notice that the new terminal velocity, \(2c^2\) has been used.
The measured \(r_{e}\) = 5.3e-11 m
\(\cfrac{r_{ec}}{r_e}\) = 1.4090e-15/5.3e-11=2.658e-5 m
\(A=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}\)
\(A\) = (1-2.658e-5)*9.10938291e-31/5.3e-11=1.719e-20 Nm-2s2
This drag coefficient seems low but for the mass of an electron (9.10938291e-31 kg) it is significant.
Sunday, July 27, 2014
To Light Speed and Beyond
However, an electron under the attraction of a single positive charge experiences acceleration given by,
\(m_ea_e=\cfrac{q^2}{4\pi\varepsilon_o r^2}\)
\(a_e=\cfrac{q^2}{4\pi\varepsilon_o m_e r^2}\)
\(\cfrac{d^2r}{dt^2}r^2=\cfrac{q^2}{4\pi\varepsilon_o m_e}\)
\(\iint _{ 0 }^{ r }{ r^{ 2 } } d^{ 2 }r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \iint _{ 0 }^{ t }{ 1. } (dt)^{ 2 }\)
\( \int _{ 0 }^{ r }{ \cfrac { r^{ 3 } }{ 3 } } dr=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \int _{ 0 }^{ t }{ t+C } dt\)
\( \cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } +Ct+B \right\} \) --- (1)
When \(t=0\), \(r=r_o\)
\( B=\cfrac { { r }_{ o }^{ 4 }\pi \varepsilon _{ o }m_{ e } }{ 3q^{ 2 } } \)
Differentiating (1) wrt time \(t\),
\( \cfrac { { r }^{ 3 } }{ 3 } \cfrac { dr }{ dt } =\cfrac { { r }^{ 3 } }{ 3 } v=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ t+C \right\} \)
Suppose \(t=0\), \(v=0\) therefore \( C=0\) and we have
\( \cfrac { { r }^{ 3 } }{ 3 } v=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } }t \)
\( \cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } \right\} +\cfrac { { r }_{ o }^{ 4 } }{ 12 } \)
And if the electron is accelerated to velocity \(c\),
\( t=t_{ f }\) \(v=c\) \( r=r_{ f }\)
\( \cfrac { r^3_{ f } }{ 3 } c=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } t_{ f }\)
\({ t }_{ f }=\cfrac { 4\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } }r^3_{ f } c\) --- (*)
and
\( \cfrac { r^4_{ f } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t_{ f } }^{ 2 } }{ 2 } \right\} +\cfrac { { r }_{ o }^{ 4 } }{ 12 } \)
\(\cfrac { t_{ f }^2 }{ 2 } =\cfrac { \pi \varepsilon _{ o }m_{ e } }{ { 3q }^{ 2 } } \{ { r_{ f } }^{ 4 }-{ r_{ o } }^{ 4 }\} \)
\(\cfrac { { t_{ f } }^{ 2 } }{ 2 } =\cfrac { \pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } } ({ r_{ f } }-{ r_{ o } })({ r }_{ f }+{ r }_{ o })({ r_{ f } }^{ 2 }+{ r_{ o } }^{ 2 })\)
\(r_f<(r_f+r_o)\) and \(r_f^2<({ r_{ f } }^{ 2 }+{ r_{ o } }^{ 2 })\)
\( { (r_{ f } }-{ r_{ o } }).r_f^3=\Delta r.r_f^3<\cfrac { 3{ q }^{ 2 } }{ 2\pi \varepsilon _{ o }m_{ e } } { t_{ f } }^{ 2 }\)
Substitute (*) into the above,
\( \Delta r<\cfrac { 8\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } } { r_{ f } }^{ 3 }c^{ 2 }\)
\(\cfrac { 8\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } }c^{ 2 }\)=8pi*8.854187817e-12*9.10938291e-31*(299792458)^2/(3*(1.602176565e-19)^2)=2.3658e14 m-5
If \(r_f\) is a thousand times the atomic radius of hydrogen then
\(r_f\) = 1000*5.29e-11
\( \Delta r\)<(5.29e-8)^3*2.3658e14
\( \Delta r\)<3.502233061962e-8 m
If \(r_f\) is a hundred times the atomic radius of hydrogen then
\(r_f\) = 100*5.29e-11
\( \Delta r\)<(5.29e-9)^3*2.3658e14
\( \Delta r\)<5.185e-11 m which means, an electron has to accelerated over a distance equal to the atomic radius of hydrogen in order to gain light speed at a distance of 100 times hydrogen radius from the hydrogen.
If \(r_f\) is a ten times the atomic radius of hydrogen then
\(r_f\) = 10*5.29e-11
\( \Delta r\)<(5.29e-10)^3*2.3658e14
\(m_ea_e=\cfrac{q^2}{4\pi\varepsilon_o r^2}\)
\(a_e=\cfrac{q^2}{4\pi\varepsilon_o m_e r^2}\)
\(\cfrac{d^2r}{dt^2}r^2=\cfrac{q^2}{4\pi\varepsilon_o m_e}\)
\(\iint _{ 0 }^{ r }{ r^{ 2 } } d^{ 2 }r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \iint _{ 0 }^{ t }{ 1. } (dt)^{ 2 }\)
\( \int _{ 0 }^{ r }{ \cfrac { r^{ 3 } }{ 3 } } dr=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \int _{ 0 }^{ t }{ t+C } dt\)
\( \cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } +Ct+B \right\} \) --- (1)
When \(t=0\), \(r=r_o\)
\( B=\cfrac { { r }_{ o }^{ 4 }\pi \varepsilon _{ o }m_{ e } }{ 3q^{ 2 } } \)
Differentiating (1) wrt time \(t\),
\( \cfrac { { r }^{ 3 } }{ 3 } \cfrac { dr }{ dt } =\cfrac { { r }^{ 3 } }{ 3 } v=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ t+C \right\} \)
Suppose \(t=0\), \(v=0\) therefore \( C=0\) and we have
\( \cfrac { { r }^{ 3 } }{ 3 } v=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } }t \)
\( \cfrac { { r }^{ 4 } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t }^{ 2 } }{ 2 } \right\} +\cfrac { { r }_{ o }^{ 4 } }{ 12 } \)
And if the electron is accelerated to velocity \(c\),
\( t=t_{ f }\) \(v=c\) \( r=r_{ f }\)
\( \cfrac { r^3_{ f } }{ 3 } c=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } t_{ f }\)
\({ t }_{ f }=\cfrac { 4\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } }r^3_{ f } c\) --- (*)
and
\( \cfrac { r^4_{ f } }{ 12 } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e } } \left\{ \cfrac { { t_{ f } }^{ 2 } }{ 2 } \right\} +\cfrac { { r }_{ o }^{ 4 } }{ 12 } \)
\(\cfrac { t_{ f }^2 }{ 2 } =\cfrac { \pi \varepsilon _{ o }m_{ e } }{ { 3q }^{ 2 } } \{ { r_{ f } }^{ 4 }-{ r_{ o } }^{ 4 }\} \)
\(\cfrac { { t_{ f } }^{ 2 } }{ 2 } =\cfrac { \pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } } ({ r_{ f } }-{ r_{ o } })({ r }_{ f }+{ r }_{ o })({ r_{ f } }^{ 2 }+{ r_{ o } }^{ 2 })\)
\(r_f<(r_f+r_o)\) and \(r_f^2<({ r_{ f } }^{ 2 }+{ r_{ o } }^{ 2 })\)
\( { (r_{ f } }-{ r_{ o } }).r_f^3=\Delta r.r_f^3<\cfrac { 3{ q }^{ 2 } }{ 2\pi \varepsilon _{ o }m_{ e } } { t_{ f } }^{ 2 }\)
Substitute (*) into the above,
\( \Delta r<\cfrac { 8\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } } { r_{ f } }^{ 3 }c^{ 2 }\)
\(\cfrac { 8\pi \varepsilon _{ o }m_{ e } }{ 3{ q }^{ 2 } }c^{ 2 }\)=8pi*8.854187817e-12*9.10938291e-31*(299792458)^2/(3*(1.602176565e-19)^2)=2.3658e14 m-5
If \(r_f\) is a thousand times the atomic radius of hydrogen then
\(r_f\) = 1000*5.29e-11
\( \Delta r\)<(5.29e-8)^3*2.3658e14
\( \Delta r\)<3.502233061962e-8 m
If \(r_f\) is a hundred times the atomic radius of hydrogen then
\(r_f\) = 100*5.29e-11
\( \Delta r\)<(5.29e-9)^3*2.3658e14
\( \Delta r\)<5.185e-11 m which means, an electron has to accelerated over a distance equal to the atomic radius of hydrogen in order to gain light speed at a distance of 100 times hydrogen radius from the hydrogen.
If \(r_f\) is a ten times the atomic radius of hydrogen then
\(r_f\) = 10*5.29e-11
\( \Delta r\)<(5.29e-10)^3*2.3658e14
\( \Delta r\)<3.502e-14 m which is a thousand times less that the atomic radius of hydrogen.
If \(r_f\) is a 2 times the atomic radius of hydrogen then
\(r_f\) = 2*5.29e-11
\( \Delta r\)<(1.058e-10)^3*2.3658e14
This shows that an electron takes very little distance to accelerate to light speed, \(c\). It is reasonable to assume that an electron in motion under the attraction of a positive charge is already at light speed.If \(r_f\) is a 2 times the atomic radius of hydrogen then
\(r_f\) = 2*5.29e-11
\( \Delta r\)<(1.058e-10)^3*2.3658e14
\( \Delta r\)<2.802e-16 m which is a hundred thousand times less that the atomic radius of hydrogen.
Terminal Velocity Revisited
Terminal velocity in a helical path,
Terminal velocity as defined is translation along \(c_1\),
Terminal Velocity = \(c_1\)
Not \(c_2\) which is circular speed, changing direction all the time. Nor it is \(c\) the instantaneous resultant of \(c_1\) and \(c_2\)
\(c^2=c^2_1+ c^2_2\)
Light speed will be \(c_1\). But terminal velocity as a limiting velocity, as a consequent of space density/viscosity then it should be \(c\). That is,
\(c^2\propto drag force\)
Terminal Velocity = \(c\) and
\(c>c_1\)
But \(c_1\) is the measured value, as such light speed is less than \(c\). By symmetry,
\(c_1=c_2\)
because moving in one direction is the same as moving in another in uniform space.
\(c^2=2c^2_1\)
\(c=\sqrt{2}c_1=\sqrt{2}.light speed\)
The actual terminal velocity of space is \(\sqrt{2}.light speed\). This is the velocity by which electrons orbit around the nucleus. This is also the velocity that prevents the atom from collapsing.
\(\cfrac{m_e.2c^2}{r_o}=\cfrac{q^2}{4\pi \varepsilon_o r^2_o}\) and
\(r_o=\cfrac{q^2}{8\pi \varepsilon_o{m_ec^2}}<r_{n}\)
where \(r_{n}\) is the radius of the positive nucleus.
In these expressions, \(c^2\) has been replaced with \(2c^2\) and in the case for \(c\), to be replaced by \(\sqrt{2}c\). The speed limit for charge particles in general is \(\sqrt{2}c\).
Terminal velocity as defined is translation along \(c_1\),Terminal Velocity = \(c_1\)
Not \(c_2\) which is circular speed, changing direction all the time. Nor it is \(c\) the instantaneous resultant of \(c_1\) and \(c_2\)
\(c^2=c^2_1+ c^2_2\)
Light speed will be \(c_1\). But terminal velocity as a limiting velocity, as a consequent of space density/viscosity then it should be \(c\). That is,
\(c^2\propto drag force\)
Terminal Velocity = \(c\) and
\(c>c_1\)
But \(c_1\) is the measured value, as such light speed is less than \(c\). By symmetry,
\(c_1=c_2\)
because moving in one direction is the same as moving in another in uniform space.
\(c^2=2c^2_1\)
\(c=\sqrt{2}c_1=\sqrt{2}.light speed\)
The actual terminal velocity of space is \(\sqrt{2}.light speed\). This is the velocity by which electrons orbit around the nucleus. This is also the velocity that prevents the atom from collapsing.
\(\cfrac{m_e.2c^2}{r_o}=\cfrac{q^2}{4\pi \varepsilon_o r^2_o}\) and
\(r_o=\cfrac{q^2}{8\pi \varepsilon_o{m_ec^2}}<r_{n}\)
where \(r_{n}\) is the radius of the positive nucleus.
In these expressions, \(c^2\) has been replaced with \(2c^2\) and in the case for \(c\), to be replaced by \(\sqrt{2}c\). The speed limit for charge particles in general is \(\sqrt{2}c\).
Saturday, July 26, 2014
More Fairly Tales, Plasma and Neutrons
Positive and negative charges interact in our reality. Their interactions can be measured and manipulated. Positive and negative charge with opposing time speed along the charge time dimension, annihilate each other in a matter/anti-matter manner at the component time dimension level not at the total resultant time speed level. The result of matter/anti-matter interaction is pure energy, more precisely the kinetic energy along the charge time component is released. Plasma is not the result of matter/anti-matter interaction, the heat in it is.
If this is true, then matter/anti-matter interaction can result in charge neutral particles. Such neutral particles have no time speed component along the charge time dimension and so display no charge properties. Such particle still have time speed component along the gravitational time dimension and so have mass. The following is an updated time diagram for an electron and a proton.
Consider the momentum along the charge time dimension, upon collision,
\(qc + q(-c) = 0\)
This means after the collision, the particle loses its charge property. The energy released is \(E=2qc^2\), seriously.
And along the gravitational time dimension, upon collision,
\(m_ec + m_pc=(m_e+m_p)c\)
This means after the collision the particle has a mass slightly higher that proton. More accurately, the mass of an electron plus a proton. This particle might just be the fabled neutron.
This is not the final picture of multiple time dimensions. Momentum used to be added like vectors. \(qc+mc\) however does not make sense. Until next time.., try to have a nice day.
If this is true, then matter/anti-matter interaction can result in charge neutral particles. Such neutral particles have no time speed component along the charge time dimension and so display no charge properties. Such particle still have time speed component along the gravitational time dimension and so have mass. The following is an updated time diagram for an electron and a proton.
Consider the momentum along the charge time dimension, upon collision,
\(qc + q(-c) = 0\)
This means after the collision, the particle loses its charge property. The energy released is \(E=2qc^2\), seriously.
And along the gravitational time dimension, upon collision,
This means after the collision the particle has a mass slightly higher that proton. More accurately, the mass of an electron plus a proton. This particle might just be the fabled neutron.
This is not the final picture of multiple time dimensions. Momentum used to be added like vectors. \(qc+mc\) however does not make sense. Until next time.., try to have a nice day.
Rest Mass and Sleeping Beauty
Along the charge time dimension, the inertia is charge \(q\) not mass \(m\). So we have for kinetic energy along charge time as,
\(K.E_c=qc^2\)
As along gravitational time, the inertia is mass, \(m\), as such,
\(K.E_g=mc^2\)
To consider both at the same time is like putting a charge body on a weighting machine under gravity and at the same time place an opposite charge underneath. The body will now experience acceleration due to both gravity and electrostatic attraction. And then to consider the resulting force, \(F_{humbug}\) on the weighting machine to be,
\(F_{humbug}=m_{humbug}(a_c+a_g)\)
\(m_{humbug} = \cfrac{F_{humbug}}{a_c+a_g}\)
Then we define rest mass, \(m_{rest}\) to be when,
\(a_c=0\)
\(m_{rest}=\cfrac{F_{humbug}}{a_g}\)
But if there are no other forces but gravity,
\(F_{humbug}=mg\) and \(a_g=g\)
and so,
\(m_{rest}=\cfrac{F_{humbug}}{a_g}=\cfrac{mg}{g}=m\)
And that is the story of humbug mass. Hopefully charge as inertia along the charge time dimension has woken. She is a beauty. Unfortunately, sleeping beauty has a serious units/dimension problem and is still comatose. Another kiss!
\(K.E_c=qc^2\)
As along gravitational time, the inertia is mass, \(m\), as such,
\(K.E_g=mc^2\)
To consider both at the same time is like putting a charge body on a weighting machine under gravity and at the same time place an opposite charge underneath. The body will now experience acceleration due to both gravity and electrostatic attraction. And then to consider the resulting force, \(F_{humbug}\) on the weighting machine to be,
\(F_{humbug}=m_{humbug}(a_c+a_g)\)
\(m_{humbug} = \cfrac{F_{humbug}}{a_c+a_g}\)
Then we define rest mass, \(m_{rest}\) to be when,
\(a_c=0\)
\(m_{rest}=\cfrac{F_{humbug}}{a_g}\)
But if there are no other forces but gravity,
\(F_{humbug}=mg\) and \(a_g=g\)
and so,
\(m_{rest}=\cfrac{F_{humbug}}{a_g}=\cfrac{mg}{g}=m\)
And that is the story of humbug mass. Hopefully charge as inertia along the charge time dimension has woken. She is a beauty. Unfortunately, sleeping beauty has a serious units/dimension problem and is still comatose. Another kiss!
Oops! May be not SHM
There is no SHM about the nucleus. The electron will simply be in circular motion around the nucleus at a radius \(r_o\) and a velocity of \(c\), as given by.
\(r_o=\cfrac{q^2}{4\pi \varepsilon_o{m_ec^2}}\)
However if the electron does not approach close enough ie \(> r_o\) then it may be possible that the electron over shoot the nucleus only to reverse direction at some distance \(x\) then,
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }x } =\cfrac { 1 }{ 2 } kx^{ 2 }\)
As a first approximation, the resonance is liken to be from Hooke's spring formulation. Speed in the perpendicular direction does not change and so does not concern us here.
\( \cfrac { k }{ { m }_{ e } } =\cfrac { c^{ 2 } }{ x^{ 2 } } =c^{ 2 }\left\{ \cfrac { 2\pi \varepsilon _{ o } }{ q^{ 2 } } m_{ e }c^{ 2 } \right\} ^{ 2 }\)
\(f=\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { k }{ { m }_{ e } } }=\cfrac { 1 }{ 2\pi } c\left\{ \cfrac { 2\pi \varepsilon _{ o } }{ q^{ 2 } } m_{ e }c^{ 2 } \right\} \)
\(f=\cfrac{m_{ e }c^{ 3 }\varepsilon _{ o }}{q^2}\)
\(f\)=9.10938291e-31*(299792458)^3*8.854187817e-12/(1.602176565e-19)^2=8.4660e21 Hz
Was this resonance from hydrogen ever detected? The sun might have this resonance.
\(r_o=\cfrac{q^2}{4\pi \varepsilon_o{m_ec^2}}\)
However if the electron does not approach close enough ie \(> r_o\) then it may be possible that the electron over shoot the nucleus only to reverse direction at some distance \(x\) then,
\(\cfrac { 1 }{ 2 } m_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }x } =\cfrac { 1 }{ 2 } kx^{ 2 }\)
As a first approximation, the resonance is liken to be from Hooke's spring formulation. Speed in the perpendicular direction does not change and so does not concern us here.
\( \cfrac { k }{ { m }_{ e } } =\cfrac { c^{ 2 } }{ x^{ 2 } } =c^{ 2 }\left\{ \cfrac { 2\pi \varepsilon _{ o } }{ q^{ 2 } } m_{ e }c^{ 2 } \right\} ^{ 2 }\)
\(f=\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { k }{ { m }_{ e } } }=\cfrac { 1 }{ 2\pi } c\left\{ \cfrac { 2\pi \varepsilon _{ o } }{ q^{ 2 } } m_{ e }c^{ 2 } \right\} \)
\(f=\cfrac{m_{ e }c^{ 3 }\varepsilon _{ o }}{q^2}\)
\(f\)=9.10938291e-31*(299792458)^3*8.854187817e-12/(1.602176565e-19)^2=8.4660e21 Hz
Was this resonance from hydrogen ever detected? The sun might have this resonance.
Temperature, Space Density And Gravity
Low temperature, collapsed space, black hole, fusion.
High temperature, gravity lower than expected, space portal, Sun.
Hot ball, cold ball, denser space, gravity toward cold ball, solar sail.
There is a relationship between temperature, space density and gravity...
But what???
We know that temperature below absolute zero, space collapses. A black hole forms and we eventually have fusion. This is the point where temperature rolls back, where extreme low temperature meets with extreme high temperature. This circular connectivity where the start meets with the end could be the characteristic of some dimensions. Other dimensions are rolled out with infinite length in both directions.
When space is pump out of a containment, as space thins out matter losses their magnetic and electric properties. Atom begins to uncoil sending the orbital electrons wayward but in distinct geometrical shape/pattern, and the nucleus disassociate into free protons. Thin space is similar to high temperature space associated with plasma. It could be that plasma is a thin space phenomenon not a high temperature phenomenon. In which case we will observe cold plasma in a thin space containment, where space has been partially pumped out. We have measured the temperature of free space to be about 3 K, at the low end of the Kelvin Scale, so normal space is cold temperature space.
Under normal space, gravitational forces are weak, beyond a threshold, at temperature beyond absolute zero (T < 0 K), gravitational forces are stronger than electromagnetic forces and everything collapses. Fusion that results from such a collapse suggests that positive and negative charges are matter-antimatter pair, travelling in opposite directions along a time dimension (charge time, \(tc\)) perpendicular to gravitational time,\(tg\). (Post "Charge Kinetic Energy")
At this point things are really awkward, matter and anti matter have just been defined as masses travelling along some time dimension in opposing directions \(E=mc^2\) being kinetic energy along the time dimension (Post "No Poetry for Einstein") and two opposing masses colliding, stopping both in time will release both their kinetic energy and will disappear from our time (both totally annihilated).
The introduction of the concept of an orthogonal time dimension allows for both negative and positive time speed to be perceived at the same time (along \(tg\)), and also allows for a consistent treatment of gravitation field and electrostatic field. Both field forces are the results of time speed slowing down in the respective time dimensions, and the conservation of energy across time and space. More importantly, since time speed for opposing charges are opposite, we perceive their force actions to be opposite. In other words, opposing time speed results directly in their opposing charge behavior. (Post "General Field Equation" and "General Field Equation Duos")
If positive and negative charges are matter and anti-matter pair, why don't they annihilate immediately?
The answer is they cannot get close enough. Because of the light speed limit, the negative charge experiences a centripetal force that sets it into circular motion about the positive charge as the pair approach each other. They cannot collide.
This is the reason why for thin space, when space density is reduced, either by high temperature or simple pumping space out of a containment, plasma occurs. At that instant, terminal velocity of space is raised and light speed is higher. Negative and positive charges can collide and annihilate each other. The heat in plasma is then the result of matter and antimatter annihilation.
If all this is true, then immediately we have an expression for which the collapse is imminent; Consider a negative charge approaching a positive nucleus. As the negative charge approaches it is accelerated to light speed, the terminal velocity of space of a given space density. At this point, a perpendicular velocity component develops; the charge move away from the line joining the centers of the two charges, this is the start of its helical path at terminal velocity. The force components on the negative charge is shown below,
The drag force will always balance the forward component of the electrostatic force at terminal velocity, light speed. The perpendicular component of the electrostatic force provides for the centripetal force and set the negative charge into circular motion. As the negative charge approaches, the centripetal force increases, the circular velocity increases until that too reaches terminal velocity light speed. (We have seen that acceleration can reach up to \(c^2\) from the post "General Field Equation") And we have,
\(\cfrac{m_ec^2}{r_o}=\cfrac{q^2}{4\pi \varepsilon_o r^2_o}\)
when the negative charge is right over the positive nucleus.
It is likely that the negative charge will over shoot the positive charge and perform simple harmonic motion about the positive charge.
The negative charge does not change rotation sense but the horizontal velocity component reverses direction. Its helical path is bounded by the curves above and below. This envelop is wrong. The concave envelop is derived later in the post "Bejing Mask" The negative weave back and forth about the positive charge in a helical path never changing its sense of rotation. We see that the charges don't collide given the light speed limit under normal space density. Collision occurs when,
\(r_o=\cfrac{q^2}{4\pi \varepsilon_o{m_ec^2}}<r_{n}\)
where \(r_{n}\) is the radius of the positive nucleus.
This can happen when \(c\), the terminal velocity of space is increased by reducing space density with high temperature or simply draining out space in a suitable containment (42 cm of Lead, Pb, remember the cat in a corner stone experiment?). We can also see that if \(r_n\) is higher as in the case of heavy elements, collisions are more likely.
This SHM about the nucleus is by itself very interesting. And relativistic concerns are not necessary so far... The issues with electron and proton mass at light speed, may be the result of their time speed component along the orthogonal charge time frame that is once removed from gravitational time, from which we derive mass. Until next time, have a nice day...
High temperature, gravity lower than expected, space portal, Sun.
Hot ball, cold ball, denser space, gravity toward cold ball, solar sail.
There is a relationship between temperature, space density and gravity...
But what???
We know that temperature below absolute zero, space collapses. A black hole forms and we eventually have fusion. This is the point where temperature rolls back, where extreme low temperature meets with extreme high temperature. This circular connectivity where the start meets with the end could be the characteristic of some dimensions. Other dimensions are rolled out with infinite length in both directions.
When space is pump out of a containment, as space thins out matter losses their magnetic and electric properties. Atom begins to uncoil sending the orbital electrons wayward but in distinct geometrical shape/pattern, and the nucleus disassociate into free protons. Thin space is similar to high temperature space associated with plasma. It could be that plasma is a thin space phenomenon not a high temperature phenomenon. In which case we will observe cold plasma in a thin space containment, where space has been partially pumped out. We have measured the temperature of free space to be about 3 K, at the low end of the Kelvin Scale, so normal space is cold temperature space.
Under normal space, gravitational forces are weak, beyond a threshold, at temperature beyond absolute zero (T < 0 K), gravitational forces are stronger than electromagnetic forces and everything collapses. Fusion that results from such a collapse suggests that positive and negative charges are matter-antimatter pair, travelling in opposite directions along a time dimension (charge time, \(tc\)) perpendicular to gravitational time,\(tg\). (Post "Charge Kinetic Energy")
At this point things are really awkward, matter and anti matter have just been defined as masses travelling along some time dimension in opposing directions \(E=mc^2\) being kinetic energy along the time dimension (Post "No Poetry for Einstein") and two opposing masses colliding, stopping both in time will release both their kinetic energy and will disappear from our time (both totally annihilated).
The introduction of the concept of an orthogonal time dimension allows for both negative and positive time speed to be perceived at the same time (along \(tg\)), and also allows for a consistent treatment of gravitation field and electrostatic field. Both field forces are the results of time speed slowing down in the respective time dimensions, and the conservation of energy across time and space. More importantly, since time speed for opposing charges are opposite, we perceive their force actions to be opposite. In other words, opposing time speed results directly in their opposing charge behavior. (Post "General Field Equation" and "General Field Equation Duos")
If positive and negative charges are matter and anti-matter pair, why don't they annihilate immediately?
The answer is they cannot get close enough. Because of the light speed limit, the negative charge experiences a centripetal force that sets it into circular motion about the positive charge as the pair approach each other. They cannot collide.
This is the reason why for thin space, when space density is reduced, either by high temperature or simple pumping space out of a containment, plasma occurs. At that instant, terminal velocity of space is raised and light speed is higher. Negative and positive charges can collide and annihilate each other. The heat in plasma is then the result of matter and antimatter annihilation.
If all this is true, then immediately we have an expression for which the collapse is imminent; Consider a negative charge approaching a positive nucleus. As the negative charge approaches it is accelerated to light speed, the terminal velocity of space of a given space density. At this point, a perpendicular velocity component develops; the charge move away from the line joining the centers of the two charges, this is the start of its helical path at terminal velocity. The force components on the negative charge is shown below,
The drag force will always balance the forward component of the electrostatic force at terminal velocity, light speed. The perpendicular component of the electrostatic force provides for the centripetal force and set the negative charge into circular motion. As the negative charge approaches, the centripetal force increases, the circular velocity increases until that too reaches terminal velocity light speed. (We have seen that acceleration can reach up to \(c^2\) from the post "General Field Equation") And we have,
\(\cfrac{m_ec^2}{r_o}=\cfrac{q^2}{4\pi \varepsilon_o r^2_o}\)
when the negative charge is right over the positive nucleus.
It is likely that the negative charge will over shoot the positive charge and perform simple harmonic motion about the positive charge.
\(r_o=\cfrac{q^2}{4\pi \varepsilon_o{m_ec^2}}<r_{n}\)
where \(r_{n}\) is the radius of the positive nucleus.
This can happen when \(c\), the terminal velocity of space is increased by reducing space density with high temperature or simply draining out space in a suitable containment (42 cm of Lead, Pb, remember the cat in a corner stone experiment?). We can also see that if \(r_n\) is higher as in the case of heavy elements, collisions are more likely.
This SHM about the nucleus is by itself very interesting. And relativistic concerns are not necessary so far... The issues with electron and proton mass at light speed, may be the result of their time speed component along the orthogonal charge time frame that is once removed from gravitational time, from which we derive mass. Until next time, have a nice day...
Friday, July 25, 2014
Twinkle, Twinkle Little Stars.
Gravity around a black hole
\(g=-g_{eo}e^{-\cfrac{1}{r_{eo}}x}\)
where,
\(g_{eo}\) = 5.07071e18 ms-2 and \(r_{eo}\) = 0.00886 m
If we approximate this gravity curve at \(x=0\) with a straight line, we have for small \(x\), \(\Delta x\),
\(\Delta g=\cfrac{g_{eo}}{r_{eo}}e^{-\cfrac{1}{r_{eo}}(x=0)}\Delta x=\cfrac {g_{eo}}{r_{eo}}\Delta x\)
From the posts "In, Up, Out and Away..." and "A Right Turn, A Wrong Turn" we know that photons are repelled away from strong gravitational region, as such
\(\Delta g=-\cfrac {g_{eo}}{r_{eo}}\Delta x\)
If we compare this to Hooke's Law,
\(F=ma=-kx\) or \(a=-\cfrac{k}{m} x\)
We find that, for small \(x\), \(\Delta x\)
\(\cfrac{k}{m}=\cfrac{g_{eo}}{r_{eo}}\)
That is to say, photons around a black hole or massive planet with high gravitational field have a resonance frequency of,
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{g_{eo}}{r_{eo}}}\)
At some nominal surface gravity \(g\), as \(\Delta x\) increases \(\Delta g\) increases in the opposite direction, some photons fall back with less acceleration, as \(\Delta x\) decreases \(\Delta g\) decreases in the opposite direction and the same photons accelerate forward again. The result is photons cluster into groups or packets.
For the case of a black hole,
\(f\)=1/(2pi)*(5.07071e18/0.00886)^(1/2) = 3.807e9 Hz = 3.807 GHz
For any other planet,
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{g_{p}}{r_{p}}}\)
where \(g_p\) is the planet surface gravity and \(r_p\) the planet radius.
In the case of Earth,
\(f\)=1/(2pi)*(9.80665/6371000)^(1/2) = 0.000197 Hz or 1 pulse every 5064.3 s
This frequency is not to be confused with light frequency but pulses of photons that escapes the pull of a strong gravitational field. It is possible that such pulses be detectable with our naked eye given the right values for high \(g_p\) and small \(r_p\).
Twinkle, Twinkle Little Stars.
\(g=-g_{eo}e^{-\cfrac{1}{r_{eo}}x}\)
where,
\(g_{eo}\) = 5.07071e18 ms-2 and \(r_{eo}\) = 0.00886 m
If we approximate this gravity curve at \(x=0\) with a straight line, we have for small \(x\), \(\Delta x\),
\(\Delta g=\cfrac{g_{eo}}{r_{eo}}e^{-\cfrac{1}{r_{eo}}(x=0)}\Delta x=\cfrac {g_{eo}}{r_{eo}}\Delta x\)
From the posts "In, Up, Out and Away..." and "A Right Turn, A Wrong Turn" we know that photons are repelled away from strong gravitational region, as such
\(\Delta g=-\cfrac {g_{eo}}{r_{eo}}\Delta x\)
If we compare this to Hooke's Law,
\(F=ma=-kx\) or \(a=-\cfrac{k}{m} x\)
We find that, for small \(x\), \(\Delta x\)
\(\cfrac{k}{m}=\cfrac{g_{eo}}{r_{eo}}\)
That is to say, photons around a black hole or massive planet with high gravitational field have a resonance frequency of,
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{g_{eo}}{r_{eo}}}\)
At some nominal surface gravity \(g\), as \(\Delta x\) increases \(\Delta g\) increases in the opposite direction, some photons fall back with less acceleration, as \(\Delta x\) decreases \(\Delta g\) decreases in the opposite direction and the same photons accelerate forward again. The result is photons cluster into groups or packets.
For the case of a black hole,
\(f\)=1/(2pi)*(5.07071e18/0.00886)^(1/2) = 3.807e9 Hz = 3.807 GHz
For any other planet,
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{g_{p}}{r_{p}}}\)
where \(g_p\) is the planet surface gravity and \(r_p\) the planet radius.
In the case of Earth,
\(f\)=1/(2pi)*(9.80665/6371000)^(1/2) = 0.000197 Hz or 1 pulse every 5064.3 s
This frequency is not to be confused with light frequency but pulses of photons that escapes the pull of a strong gravitational field. It is possible that such pulses be detectable with our naked eye given the right values for high \(g_p\) and small \(r_p\).
Twinkle, Twinkle Little Stars.
Die Glocke, And Long Live the Bell
Notice that a strong magnet going down a copper or aluminium tube slows down. A force is acting against gravity on the magnet that is proportional to the rate of change of magnetic flux between the magnet and the pipe wall. If the rate of change of this flux is increased then the force will increase proportionally and generate greater lift.
This bell has pairs of magnets arranged in a spiral with similar poles facing outwards (This may not be necessary, a normal NS pole arrangement might do just as well). The spiral is rotated at high speed. It would seem that the magnets are falling downwards perpetually. A total lift force acts along the axial of the screw, hopefully lifting the whole bell. For stability, with an upward acceleration, the C.G of this device will have to be high. Lift can be adjusted either through the strength of the magnets (electromagnets) or the spin velocity of the spiral. Two of such bells with opposing spin may be used in a superstructure without the need for a counter spinning weight within each bell.
This is the Spiral Wind Charm Bell. Have a nice day.
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| Oriental Bell |
This bell has pairs of magnets arranged in a spiral with similar poles facing outwards (This may not be necessary, a normal NS pole arrangement might do just as well). The spiral is rotated at high speed. It would seem that the magnets are falling downwards perpetually. A total lift force acts along the axial of the screw, hopefully lifting the whole bell. For stability, with an upward acceleration, the C.G of this device will have to be high. Lift can be adjusted either through the strength of the magnets (electromagnets) or the spin velocity of the spiral. Two of such bells with opposing spin may be used in a superstructure without the need for a counter spinning weight within each bell.
This is the Spiral Wind Charm Bell. Have a nice day.
Wind Under Your Wings Always
Here is an aerofoil with an air slit in the front. Compressed air is released through this slit uniformly down its width.
Air flows at high speed uniformly across the width of the wing, generating a region of low pressure. Lift develops as a result of the pressure differential above and under the wing. The good thing about this set up is that the speed of the air flow across the width of the wing is independent of its forward speed. Lift can be directly controlled by changing the speed of the compressed air. Although lift will drop as air speed under the wing increases with forward speed and the pressure underneath the wing drops. Otherwise, this wing will always have lift.
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| Active Wing |
Wednesday, July 23, 2014
Fe, Ni and Co... the Three Musketeers and Al, Who's Gd?
Magnetism is attributed to the 3d shell electrons. When these electrons are at the outer most and are all aligned into aggregated domains which in turns are aligned, strong magnetism develops.
Iron, Fe
[Ar] 3d6 4s2
Cobalt, Co
[Ar] 4s2 3d7
Nickel, Ni
[Ar] 4s2 3d8
Gadolinium, Gd
[Xe] 4f7 5d1 6s2
Aluminium, Al [Ne] 3s2 3p1, does not have 3d shells electrons but if thin sheets of it were to be friction welded or, with smoothed surfaces, pressed hard against a strong magnet; Aluminium is able to generate stronger magnetism especially in the presence of free electrons from an electric current through it. Stack of alternating magnetic discs and aluminium discs, with Aluminium press against the upper OR lower magnets will have an magnetic amplifying effect. It was proposed that a free electron (possibly a pair) is now in orbit in a 3d shell around the Aluminium nucleus and it is this loosely held electron(s) that amplify(ies) the magnetic field.
It may be possible to friction weld sheets of Zinc, Zn [Ar] 3d10 4s2 and Al [Ne] 3s2 3p1 together and make the resulting sandwich magnetic. Both Zinc and Aluminium by themselves are not ferromagnetic. An electric current might help an electron from the 3d10 shell of Zinc jump to Al 3d orbits. If the 4s2 electrons in Zinc now sinks below the 3d electrons then Zinc may now be magnetic. Aluminium is now magnetic having received a free electron in its 3d orbit as in the previous case.
Copper Cu [Ar] 3d10 4s1 is also a candidate for such a sandwich with Aluminium. In this case, if the layers become magnetic then Cu electronic configuration is likely to have changed to [Ar] 4s2 3d8+ like Nickel after an electron jumped to the Aluminium side.
As a whole the layers are still charge neutral.
It is also interesting that Chromium,Cr [Ar] 3d5 4s1, is an anti-ferromagnetic material, although Chromium(IV) oxide (CrO2) is highly magnetic.
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| This particular d shell |
Iron, Fe
[Ar] 3d6 4s2
Cobalt, Co
[Ar] 4s2 3d7
Nickel, Ni
[Ar] 4s2 3d8
Gadolinium, Gd
[Xe] 4f7 5d1 6s2
Aluminium, Al [Ne] 3s2 3p1, does not have 3d shells electrons but if thin sheets of it were to be friction welded or, with smoothed surfaces, pressed hard against a strong magnet; Aluminium is able to generate stronger magnetism especially in the presence of free electrons from an electric current through it. Stack of alternating magnetic discs and aluminium discs, with Aluminium press against the upper OR lower magnets will have an magnetic amplifying effect. It was proposed that a free electron (possibly a pair) is now in orbit in a 3d shell around the Aluminium nucleus and it is this loosely held electron(s) that amplify(ies) the magnetic field.
It may be possible to friction weld sheets of Zinc, Zn [Ar] 3d10 4s2 and Al [Ne] 3s2 3p1 together and make the resulting sandwich magnetic. Both Zinc and Aluminium by themselves are not ferromagnetic. An electric current might help an electron from the 3d10 shell of Zinc jump to Al 3d orbits. If the 4s2 electrons in Zinc now sinks below the 3d electrons then Zinc may now be magnetic. Aluminium is now magnetic having received a free electron in its 3d orbit as in the previous case.
Copper Cu [Ar] 3d10 4s1 is also a candidate for such a sandwich with Aluminium. In this case, if the layers become magnetic then Cu electronic configuration is likely to have changed to [Ar] 4s2 3d8+ like Nickel after an electron jumped to the Aluminium side.
As a whole the layers are still charge neutral.
It is also interesting that Chromium,Cr [Ar] 3d5 4s1, is an anti-ferromagnetic material, although Chromium(IV) oxide (CrO2) is highly magnetic.
Gravity Wave
How to produce Gravity Wave?...
The problem is gravity is weak. Gravity of earth itself can be countered by your pair of legs jumping up and about.
It is important to realize that gravity wave as it manifest itself as Schumann Resonance is detected as electromagnetic waves. This would mean that GW will response to both a E field and a B field.
Take a electromagnet and energize it at a frequency of 7.489 Hz. Set up the circuit such that the phase of the excitation can be reversed and adjusted. The point is to have the B field pushing against space downwards interact with earth GW at 7.489 Hz. Schumann Resonance sits on a DC value of \(g_o\) and is detected as EMW at 7.83 Hz. With the electromagnet at the correct phase this EMW will increase in amplitude, still at 7.83 Hz
Space is magnetic, a strong magnet will push out space and so pushes out gravity waves.
All hypothetical of course.
The problem is gravity is weak. Gravity of earth itself can be countered by your pair of legs jumping up and about.
It is important to realize that gravity wave as it manifest itself as Schumann Resonance is detected as electromagnetic waves. This would mean that GW will response to both a E field and a B field.
Take a electromagnet and energize it at a frequency of 7.489 Hz. Set up the circuit such that the phase of the excitation can be reversed and adjusted. The point is to have the B field pushing against space downwards interact with earth GW at 7.489 Hz. Schumann Resonance sits on a DC value of \(g_o\) and is detected as EMW at 7.83 Hz. With the electromagnet at the correct phase this EMW will increase in amplitude, still at 7.83 Hz
Space is magnetic, a strong magnet will push out space and so pushes out gravity waves.
All hypothetical of course.
Gravity Wave and Schumann Resonance
From the post "Gravity Exponential Form",
\(g=-g_{ o }e^{ -\cfrac { x }{ r_{ e } } }\)
Consider,
\(g=-g_{ o }e^{ -i\cfrac { x }{ r_{ e } } }\)
where \(ix\) is perpendicular to \(x\) and if \(x\) is a set of radial lines from a origin then \(ix\) is the circular front with radius \(x\). On this front, the value of \(g\) is a constant, given by the expression above.
If we vary gravity by a driving force such that gravity varies in time,
\( g_{ w }=ge^{ iwt }=-g_{ o }e^{i (-\cfrac { x }{ r_{ e } } +wt) }\)
Consider,
\( \cfrac { \partial ^{ 2 }g_{ w } }{ \partial t^{ 2 } } =w^{ 2 }g_{ o }e^{i (-\cfrac { x }{ r_{ e } } +wt) }\)
and
\( \cfrac { \partial ^{ 2 }g_{ w } }{ \partial x^{ 2 } } =\cfrac { g_{ o } }{ r^{ 2 }_{ e } } e^{i (-\cfrac { x }{ r_{ e } } +wt) }\)
If
\( w^{ 2 }g_o=c^{ 2 }\cfrac { g_{ o } }{ r^{ 2 }_{ e } } \)
then,
\( \cfrac { \partial ^{ 2 }g_{ w } }{ \partial t^{ 2 } } =c^{ 2 }\cfrac { \partial ^{ 2 }g_{ w } }{ \partial x^{ 2 } } \)
That is to say, we have a gravity wave that satisfy the above wave equation. It is assumed here that gravity wave travels at light speed \(c\).
\(w^2=(2\pi f)^2=\cfrac { c^{ 2 } }{ r^{ 2 }_{ e } } \)
then
\(f=\cfrac{1}{2\pi}\cfrac{c}{r_e}\)=1/(2pi)*299792458/6371000=7.489 Hz
This value is very close to the fundamental Schumann resonances at 7.83 Hz and could the underlying cause of it. For the case of Mars of radius, \(r_m\) = 3390000 m, the fundamental Schumann Resonance will then be,
\(f_{mars}=\cfrac{1}{2\pi}\cfrac{c}{r_m}\)=1/(2pi)*299792458/3390000=14.075 Hz
If Gravitational Wave (GW) can be detected as electromagnetic wave (EMW), that means space is a carrier of both EMW and GW. Could it be that EMW and GW is one and the same thing. From previously, we also have
\(f=\cfrac{1}{2\pi}\cfrac{c}{r}\)
where \(r\) is the radius of the helical path of the dipole not the wavelength of the EMW.
\(g=-g_{ o }e^{ -\cfrac { x }{ r_{ e } } }\)
Consider,
\(g=-g_{ o }e^{ -i\cfrac { x }{ r_{ e } } }\)
where \(ix\) is perpendicular to \(x\) and if \(x\) is a set of radial lines from a origin then \(ix\) is the circular front with radius \(x\). On this front, the value of \(g\) is a constant, given by the expression above.
If we vary gravity by a driving force such that gravity varies in time,
\( g_{ w }=ge^{ iwt }=-g_{ o }e^{i (-\cfrac { x }{ r_{ e } } +wt) }\)
Consider,
\( \cfrac { \partial ^{ 2 }g_{ w } }{ \partial t^{ 2 } } =w^{ 2 }g_{ o }e^{i (-\cfrac { x }{ r_{ e } } +wt) }\)
and
\( \cfrac { \partial ^{ 2 }g_{ w } }{ \partial x^{ 2 } } =\cfrac { g_{ o } }{ r^{ 2 }_{ e } } e^{i (-\cfrac { x }{ r_{ e } } +wt) }\)
If
\( w^{ 2 }g_o=c^{ 2 }\cfrac { g_{ o } }{ r^{ 2 }_{ e } } \)
then,
\( \cfrac { \partial ^{ 2 }g_{ w } }{ \partial t^{ 2 } } =c^{ 2 }\cfrac { \partial ^{ 2 }g_{ w } }{ \partial x^{ 2 } } \)
That is to say, we have a gravity wave that satisfy the above wave equation. It is assumed here that gravity wave travels at light speed \(c\).
\(w^2=(2\pi f)^2=\cfrac { c^{ 2 } }{ r^{ 2 }_{ e } } \)
then
\(f=\cfrac{1}{2\pi}\cfrac{c}{r_e}\)=1/(2pi)*299792458/6371000=7.489 Hz
This value is very close to the fundamental Schumann resonances at 7.83 Hz and could the underlying cause of it. For the case of Mars of radius, \(r_m\) = 3390000 m, the fundamental Schumann Resonance will then be,
\(f_{mars}=\cfrac{1}{2\pi}\cfrac{c}{r_m}\)=1/(2pi)*299792458/3390000=14.075 Hz
If Gravitational Wave (GW) can be detected as electromagnetic wave (EMW), that means space is a carrier of both EMW and GW. Could it be that EMW and GW is one and the same thing. From previously, we also have
\(f=\cfrac{1}{2\pi}\cfrac{c}{r}\)
where \(r\) is the radius of the helical path of the dipole not the wavelength of the EMW.
Monday, July 21, 2014
One Machi Going Up, Upside Down
Wait a minute... if,
\(r=a+bcos(\theta)\) and \(a<0\), \(|a|>b>0\)
then the curve should be upside down,
The tube should lift this way up where the upside \(\frac{v^2}{r}\) has a higher value as \(r\) is smaller, \(v\) being constant throughout the tube.
\(r=a+bcos(\theta)\) and \(a<0\), \(|a|>b>0\)
then the curve should be upside down,
The tube should lift this way up where the upside \(\frac{v^2}{r}\) has a higher value as \(r\) is smaller, \(v\) being constant throughout the tube.
Machi and Garden Hose
From the post "And Some More Auntie Gravy", we have
\(a<0\), \(|a|≥\cfrac{L}{\pi}\) and
\(|a|>b>0\)
for
\(r=a+bcos(\theta)\)
If we plot r=-5+2.5cos(x) on a polar plot we have,
The high curvature up top compared to the low flat line below suggest that this shape might lift. Ever seen a garden hose rise when turned on? Below shows the same curve with a circle superimposed.
Very nice machi shaped curve, indeed.
\(a<0\), \(|a|≥\cfrac{L}{\pi}\) and
\(|a|>b>0\)
for
\(r=a+bcos(\theta)\)
If we plot r=-5+2.5cos(x) on a polar plot we have,
The high curvature up top compared to the low flat line below suggest that this shape might lift. Ever seen a garden hose rise when turned on? Below shows the same curve with a circle superimposed.
Very nice machi shaped curve, indeed.
Sunday, July 20, 2014
Anti Gravity, Mechanically Yours
From the Post "Anti Gravity", the expression for lift is given by,
\(F_l=\rho\pi a^2_o(2v^2-\frac{1}{2}g\pi r_o)\)
Smaller \(r_o\) provides greater lift, as such it may be better to break one big curve into smaller curves as shown.
Although we see that the main contribution to lift is still \(v\) the velocity with which the fluid is moving through the tube.
In this case, the total lift is \(16*F_l\).
\(F_l=\rho\pi a^2_o(2v^2-\frac{1}{2}g\pi r_o)\)
Smaller \(r_o\) provides greater lift, as such it may be better to break one big curve into smaller curves as shown.
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| Anti Gravity II |
Although we see that the main contribution to lift is still \(v\) the velocity with which the fluid is moving through the tube.
In this case, the total lift is \(16*F_l\).
Saturday, July 19, 2014
And Some More Auntie Gravy
From the post "More Anti Gravity, Have a Heart Too",
\(a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }\)
If much of the weight of the device is in the fluid then
\(m_t=\rho L.\pi a^2_o\)
where \(L\) is the total length of the tube. Then,
\(\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }=\cfrac { \rho L.\pi a^2_o }{ \rho \pi^2 a^{ 2 }_o }=\cfrac{L}{\pi}\)
and so,
\(|a|≥\cfrac{L}{\pi}\)
But the total length of a polar curve is,
\(L=\int{\sqrt{r^2+(\cfrac{dr}{d\theta})^2}}d\theta\)
In this case,
\(r=a+bcos(\theta)\)
\(\cfrac{dr}{d\theta}=-bsin(\theta)\)
So,
\(L=2\int^{\pi}_{0}{\sqrt{(a+bcos(\theta))^2+b^2sin^2(\theta)}}d\theta\)
\(L=2\int^{\pi}_{0}{\sqrt{a^2+b^2+2abcos(\theta)}}d\theta\)
For positive lift,
\(\pi|a|≥2\int^{\pi}_{0}{\sqrt{a^2+b^2+2abcos(\theta)}}d\theta\)
The expression on the Right Hand Side cannot be easily evaluated, but we know that \(a\) and \(b\) are of opposite sign,
\(\pi|a|≥2\int^{\pi}_{0}{\sqrt{a^2+b^2-2|a||b|cos(\theta)}}d\theta\)
When \(b=0\),
\(RHS=2|^{\pi}_0a=2\pi a\) which the perimeter of a circle but is greater than LHS.
So, \(b≠0\); the tube in the shape of a circle does not provide a net lift.
If we consider, \(\cfrac{dL}{db}=0\) for a minimum \(L\) given \(a\)
\(\cfrac{dL}{db}=\int^{\pi}_{0}{(a^2+b^2-2|a||b|cos(\theta))^{-\cfrac{1}{2}}(2b-2{|a|cos(\theta))}}d\theta\)
\(\cfrac{dL}{db}=2\int^{\pi}_{0}{(a^2+b^2-2|a||b|cos(\theta))^{-\cfrac{1}{2}}(b-{|a|cos(\theta))}}d\theta\)
For any possibility of \(\cfrac{dL}{db}=0\), \(b<|a|\) such that the expression,
\((b-|a|cos(\theta))\) can change sign.
So, \(|a|>b>0\)
For the case of \(|a|=b\),
\( \cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi }{ (b^{ 2 }-b^{ 2 }cos(\theta ))^{ -\cfrac { 1 }{ 2 } }(b-{ bcos(\theta )) } } d\theta \)
\( \cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi }{ (1-cos(\theta ))^{ -\cfrac { 1 }{ 2 } }(1-{ cos(\theta )) } } d\theta \)
\( \cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi }{ (1-cos(\theta ))^{ \cfrac { 1 }{ 2 } } } d\theta \)
\(\cfrac { dL }{ db }=\sqrt{2}|^{\pi}_{0}\left(-\sqrt{2}{\cdot}\cos\left(x\right)-\sqrt{2}\right){\cdot}\sin\left(\dfrac{\mathrm{arctan2}\left(\sin\left(x\right), \cos\left(x\right)\right)+{\pi}}{2}\right)+\sqrt{2}{\cdot}\sin\left(x\right){\cdot}\cos\left(\dfrac{\mathrm{arctan2}\left(\sin\left(x\right), \cos\left(x\right)\right)+{\pi}}{2}\right)\)
\(\cfrac { dL }{ db }=4≠0\) but \(L=(2\sqrt{2})^2a=8a\).
For the case of \(b=0\),
\( \cfrac { dL }{ db } =2\int^{\pi}_{0}{-cos(\theta)}d\theta=0\)
This means for a given \(a\), \(b=0\) is a turning point for \(L\) which is \(2\pi a\), the perimeter of a circle of radius \(a\). This is a local maximum value for \(L\) because,
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\int _{ 0 }^{ \pi }{ \cfrac { 2a^{ 2 }sin^{ 2 }(\theta ) }{ (a^{ 2 }+b^{ 2 }-2abcos(\theta ))^{ \cfrac { 3 }{ 2 } } } } d\theta \)
When \(b=0\),
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{2}{a}\int _{ 0 }^{ \pi }{sin^{ 2 }(\theta ) }d\theta\)
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{1}{a}|^{\pi}_{0}\{\theta-\cfrac{1}{2}sin(2\theta)\}\)
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{\pi}{a}\)
Since, \(a<0\), \(\cfrac { d^{ 2 }L }{ db^{ 2 } }<0\) for \(b=0\), \(L\) is a maxima.
When \(b=|a|\), \(L=8a\), when \(b=0\), \(L=2\pi a<8a\) and at \(b=|a|\),\(\cfrac{dL}{db}>0\). This means there is at least one minimum before \(b=|a|\) so that the curve turns upward again after the maximum at \(b=0\). So, there is still the possibility of satisfying the positive lift condition. Hold the gravy.
\(a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }\)
If much of the weight of the device is in the fluid then
\(m_t=\rho L.\pi a^2_o\)
where \(L\) is the total length of the tube. Then,
\(\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }=\cfrac { \rho L.\pi a^2_o }{ \rho \pi^2 a^{ 2 }_o }=\cfrac{L}{\pi}\)
and so,
\(|a|≥\cfrac{L}{\pi}\)
But the total length of a polar curve is,
\(L=\int{\sqrt{r^2+(\cfrac{dr}{d\theta})^2}}d\theta\)
In this case,
\(r=a+bcos(\theta)\)
\(\cfrac{dr}{d\theta}=-bsin(\theta)\)
So,
\(L=2\int^{\pi}_{0}{\sqrt{(a+bcos(\theta))^2+b^2sin^2(\theta)}}d\theta\)
\(L=2\int^{\pi}_{0}{\sqrt{a^2+b^2+2abcos(\theta)}}d\theta\)
For positive lift,
\(\pi|a|≥2\int^{\pi}_{0}{\sqrt{a^2+b^2+2abcos(\theta)}}d\theta\)
The expression on the Right Hand Side cannot be easily evaluated, but we know that \(a\) and \(b\) are of opposite sign,
\(\pi|a|≥2\int^{\pi}_{0}{\sqrt{a^2+b^2-2|a||b|cos(\theta)}}d\theta\)
When \(b=0\),
\(RHS=2|^{\pi}_0a=2\pi a\) which the perimeter of a circle but is greater than LHS.
So, \(b≠0\); the tube in the shape of a circle does not provide a net lift.
If we consider, \(\cfrac{dL}{db}=0\) for a minimum \(L\) given \(a\)
\(\cfrac{dL}{db}=\int^{\pi}_{0}{(a^2+b^2-2|a||b|cos(\theta))^{-\cfrac{1}{2}}(2b-2{|a|cos(\theta))}}d\theta\)
\(\cfrac{dL}{db}=2\int^{\pi}_{0}{(a^2+b^2-2|a||b|cos(\theta))^{-\cfrac{1}{2}}(b-{|a|cos(\theta))}}d\theta\)
For any possibility of \(\cfrac{dL}{db}=0\), \(b<|a|\) such that the expression,
\((b-|a|cos(\theta))\) can change sign.
So, \(|a|>b>0\)
For the case of \(|a|=b\),
\( \cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi }{ (b^{ 2 }-b^{ 2 }cos(\theta ))^{ -\cfrac { 1 }{ 2 } }(b-{ bcos(\theta )) } } d\theta \)
\( \cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi }{ (1-cos(\theta ))^{ -\cfrac { 1 }{ 2 } }(1-{ cos(\theta )) } } d\theta \)
\( \cfrac { dL }{ db } =\sqrt{2}\int _{ 0 }^{ \pi }{ (1-cos(\theta ))^{ \cfrac { 1 }{ 2 } } } d\theta \)
\(\cfrac { dL }{ db }=\sqrt{2}|^{\pi}_{0}\left(-\sqrt{2}{\cdot}\cos\left(x\right)-\sqrt{2}\right){\cdot}\sin\left(\dfrac{\mathrm{arctan2}\left(\sin\left(x\right), \cos\left(x\right)\right)+{\pi}}{2}\right)+\sqrt{2}{\cdot}\sin\left(x\right){\cdot}\cos\left(\dfrac{\mathrm{arctan2}\left(\sin\left(x\right), \cos\left(x\right)\right)+{\pi}}{2}\right)\)
\(\cfrac { dL }{ db }=4≠0\) but \(L=(2\sqrt{2})^2a=8a\).
For the case of \(b=0\),
\( \cfrac { dL }{ db } =2\int^{\pi}_{0}{-cos(\theta)}d\theta=0\)
This means for a given \(a\), \(b=0\) is a turning point for \(L\) which is \(2\pi a\), the perimeter of a circle of radius \(a\). This is a local maximum value for \(L\) because,
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\int _{ 0 }^{ \pi }{ \cfrac { 2a^{ 2 }sin^{ 2 }(\theta ) }{ (a^{ 2 }+b^{ 2 }-2abcos(\theta ))^{ \cfrac { 3 }{ 2 } } } } d\theta \)
When \(b=0\),
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{2}{a}\int _{ 0 }^{ \pi }{sin^{ 2 }(\theta ) }d\theta\)
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{1}{a}|^{\pi}_{0}\{\theta-\cfrac{1}{2}sin(2\theta)\}\)
\(\cfrac { d^{ 2 }L }{ db^{ 2 } } =\cfrac{\pi}{a}\)
Since, \(a<0\), \(\cfrac { d^{ 2 }L }{ db^{ 2 } }<0\) for \(b=0\), \(L\) is a maxima.
When \(b=|a|\), \(L=8a\), when \(b=0\), \(L=2\pi a<8a\) and at \(b=|a|\),\(\cfrac{dL}{db}>0\). This means there is at least one minimum before \(b=|a|\) so that the curve turns upward again after the maximum at \(b=0\). So, there is still the possibility of satisfying the positive lift condition. Hold the gravy.
Friday, July 18, 2014
More Anti Gravity, Have a Heart Too
Consider a elemental fluid mass, \(\Delta m\) in circular motion inside a tube, where the centripetal force is provided by the reaction force from the surface of the tube on the rotating mass,
\(\Delta m = \rho\pi a^2_o r\Delta\theta\)
\(F_c=\Delta m\cfrac{v^2}{r}-\Delta mgcos(\theta)\)
The vertical lift on the wall of the tube is given by,
\(F_wcos(\theta)=F_ccos(\theta)=\Delta m\{\cfrac{v^2}{r}-gcos(\theta)\}cos(\theta)\)
Over the full length of the tube, the total upward force is given by
\(F_{ l }=2\int _{ 0 }^{ \pi }{ \rho \pi a^2_o r\{ \cfrac { v^{ 2 } }{ r } -gcos(\theta )\} cos(\theta ) } d\theta -m_{ t }g\)
where \(m_t\) is the total mass of the contraption.
\( F_{ l }=2\rho \pi a^2_o\int _{ 0 }^{ \pi }{ r\{ \cfrac { v^{ 2 } }{ r } -gcos(\theta )\} cos(\theta ) } d\theta -m_{ t }g\)
\( F_{ l }=2\rho \pi a^2_o\int _{ 0 }^{ \pi }{ v^{ 2 }cos(\theta )-grcos^{ 2 }(\theta ) } d\theta -m_{ t }g\)
\( F_{ l }=2\rho \pi a^2_o\{ v^{ 2 }sin(\theta )|^{ \pi }_{ 0 }-\int _{ 0 }^{ \pi }{ grcos^{ 2 }(\theta ) } d\theta \} -m_{ t }g\)
\( F_{ l }=-2\rho g\pi a^2_o\int _{ 0 }^{ \pi }{ rcos^{ 2 }(\theta ) } d\theta -m_{ t }g\)
For positive overall lift,
\(F_l≥0\)
\( \int _{ 0 }^{ \pi }{ rcos^{ 2 }(\theta ) } d\theta \le \cfrac { -m_{ t } }{ 2\rho \pi a^2_o } \)
\( \int _{ 0 }^{ \pi }{ rcos^{ 2 }(\theta ) } d\theta \le -A\)
\( A=\cfrac { m_{ t } }{ 2\rho \pi a^2_o } \)
Let,
\( \int { rcos^{ 2 }(\theta ) } d\theta =B(\theta )\)
where \( B(\pi )-B(0)\le -A\) and is symmetrical about \(\theta=0\),
\( rcos^{ 2 }(\theta )=\cfrac { dB(\theta ) }{ d\theta } \)
Therefore,
\( r=\cfrac { 1 }{ cos^{ 2 }(\theta ) } \cfrac { dB(\theta ) }{ d\theta }\)
Consider,
\(B(\theta )=\cfrac { a }{ 2 }\theta +\cfrac { 3b }{ 4 }sin(\theta) +\cfrac { a }{ 4 }sin(2\theta) +\cfrac { b }{ 12 }sin(3\theta)\)
\( B(\pi )-B(0)=\cfrac { a\pi }{ 2 }≤-A\)
\(a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^2_o }\)
\(\cfrac{dB(\theta)}{d(\theta)}=acos^{ 2 }(\theta )+bcos^{ 3 }(\theta ) \)
So,
\( r=\cfrac { 1 }{ cos^{ 2 }(\theta ) } \cfrac { dB(\theta ) }{ d\theta }=a+bcos(\theta) \)
where \(a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }\) and \(b\) is arbitrary. \(b\) is in the expression for total length of the tube and so effects the total mass \(m_t\) of the device. In this way \(b\) effects lift.
A example polar plot of such a Limaçon, r=-5+7cos(t) is shown below,
This shape is symmetric about the center line and so we know that the resultant will be acting through the C.G. of the tube along the center line and that there is only one upward resultant force. All horizontal force components cancel. To counter the tendency to rotate, two such tubes with counter flowing fluid are used.
Have a nice day.
\(F_c=\Delta m\cfrac{v^2}{r}-\Delta mgcos(\theta)\)
The vertical lift on the wall of the tube is given by,
\(F_wcos(\theta)=F_ccos(\theta)=\Delta m\{\cfrac{v^2}{r}-gcos(\theta)\}cos(\theta)\)
Over the full length of the tube, the total upward force is given by
\(F_{ l }=2\int _{ 0 }^{ \pi }{ \rho \pi a^2_o r\{ \cfrac { v^{ 2 } }{ r } -gcos(\theta )\} cos(\theta ) } d\theta -m_{ t }g\)
where \(m_t\) is the total mass of the contraption.
\( F_{ l }=2\rho \pi a^2_o\int _{ 0 }^{ \pi }{ r\{ \cfrac { v^{ 2 } }{ r } -gcos(\theta )\} cos(\theta ) } d\theta -m_{ t }g\)
\( F_{ l }=2\rho \pi a^2_o\int _{ 0 }^{ \pi }{ v^{ 2 }cos(\theta )-grcos^{ 2 }(\theta ) } d\theta -m_{ t }g\)
\( F_{ l }=2\rho \pi a^2_o\{ v^{ 2 }sin(\theta )|^{ \pi }_{ 0 }-\int _{ 0 }^{ \pi }{ grcos^{ 2 }(\theta ) } d\theta \} -m_{ t }g\)
\( F_{ l }=-2\rho g\pi a^2_o\int _{ 0 }^{ \pi }{ rcos^{ 2 }(\theta ) } d\theta -m_{ t }g\)
For positive overall lift,
\(F_l≥0\)
\( \int _{ 0 }^{ \pi }{ rcos^{ 2 }(\theta ) } d\theta \le \cfrac { -m_{ t } }{ 2\rho \pi a^2_o } \)
\( \int _{ 0 }^{ \pi }{ rcos^{ 2 }(\theta ) } d\theta \le -A\)
\( A=\cfrac { m_{ t } }{ 2\rho \pi a^2_o } \)
Let,
\( \int { rcos^{ 2 }(\theta ) } d\theta =B(\theta )\)
where \( B(\pi )-B(0)\le -A\) and is symmetrical about \(\theta=0\),
\( rcos^{ 2 }(\theta )=\cfrac { dB(\theta ) }{ d\theta } \)
Therefore,
\( r=\cfrac { 1 }{ cos^{ 2 }(\theta ) } \cfrac { dB(\theta ) }{ d\theta }\)
Consider,
\(B(\theta )=\cfrac { a }{ 2 }\theta +\cfrac { 3b }{ 4 }sin(\theta) +\cfrac { a }{ 4 }sin(2\theta) +\cfrac { b }{ 12 }sin(3\theta)\)
\( B(\pi )-B(0)=\cfrac { a\pi }{ 2 }≤-A\)
\(a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^2_o }\)
\(\cfrac{dB(\theta)}{d(\theta)}=acos^{ 2 }(\theta )+bcos^{ 3 }(\theta ) \)
So,
\( r=\cfrac { 1 }{ cos^{ 2 }(\theta ) } \cfrac { dB(\theta ) }{ d\theta }=a+bcos(\theta) \)
where \(a≤-\cfrac { m_{ t } }{ \rho \pi^2 a^{ 2 }_o }\) and \(b\) is arbitrary. \(b\) is in the expression for total length of the tube and so effects the total mass \(m_t\) of the device. In this way \(b\) effects lift.
A example polar plot of such a Limaçon, r=-5+7cos(t) is shown below,
This shape is symmetric about the center line and so we know that the resultant will be acting through the C.G. of the tube along the center line and that there is only one upward resultant force. All horizontal force components cancel. To counter the tendency to rotate, two such tubes with counter flowing fluid are used.
Have a nice day.
Anti-gravity
Consider a body in circular motion in a circular tube at high velocity, \(v\). Where
\(mg=m\cfrac{v_g^2}{r}\), \(v_g=\sqrt{g.r}\)
\(v>>v_g\)
The body pushes against the outer wall of the tube and a reaction force from the wall pushes it towards the center of the circular path. It is this force on the body together with gravity that provides for the centripetal force.
\(m\cfrac{v^2}{r}=F_b+mgcos(\theta)\)
Numerically,
\(F_w=F_b=m\cfrac{v^2}{r}-mgcos(\theta)\)
where \(F_b\) and \(F_w\) are action and reaction pair. The vertical lift created by the reaction force on the tube wall by \(F_w\) is,
\(F_l=F_wcos(\theta)=m\left\{\cfrac{v^2}{r}-gcos(\theta)\right\}cos(\theta)\)
The horizontal component that comes with this lift can be counter-balanced by another similar tube with the body rotating in the opposite sense. At high velocity \(v>>v_g\) such that we can ignore the term \(mgcos^2(\theta)\), then
\(F_l=m\cfrac{v^2}{r}cos(\theta)\)
If the body is a ferromagnetic fluid or mercury driven by a electromagnetic force, then it is possible to change its velocity \(v\) by changing the hollow radius of the tube or \(r\) the path of the tube, since velocity is inversely proportional to the radius squared,
\(\cfrac{v_o}{v}=\cfrac{a^2}{{a_o}^2}\)
If
\(\cfrac{a^2}{{a_o}^2}=\sqrt{cos(\theta)}\), then
\(v^2={v_o}^2\left\{\cfrac{{a_o}^2}{a^2}\right\}^2={v_o}^2\cfrac{1}{cos(\theta)}\)
and so,
\(F_l=m\cfrac{v^2}{r}cos(\theta)=m\cfrac{{v_o}^2}{r}\)
which is a constant provided \(r\) is a constant, ie. a circular path. \(F_l\) can then be varied by changing \(v_o\). A pair of such a device that compensate each other laterally, will then generate a total lift of \(2*F_l\). It is not possible to make the tube with a hollow radius of zero at \(\theta=\cfrac{\pi}{2}\), but it can be at a practical minimum, \(a_f\) at some \(\alpha_f\).
On the lower half of the tube,
\(m\cfrac{v^2}{r}=F_b-mgsin(\alpha)\)
\(F_w=F_b=m\cfrac{v^2}{r}+mgsin(\alpha)\)
When \(\theta\) is measured from the vertical axis, then
\(\theta=\alpha+\cfrac{\pi}{2}\)
So,
\(sin(\alpha)=sin(\theta-\cfrac{\pi}{2})=sin(-(\cfrac{\pi}{2}-\theta))=-sin(\cfrac{\pi}{2}-\theta)=-cos(\theta)\)
Still numerically,
\(F_w=m\cfrac{v^2}{r}-mgcos(\theta)\) for \(\cfrac{\pi}{2}<\theta<\pi\)
This force is downwards, an expression for vertical lift is given by,
\(F_l=-F_wcos(\theta)=mcos(\theta)\left\{-\cfrac{v^2}{r}+gcos(\theta)\right\}\)
which is always negative given the value of \(r\) and \(\theta\). It is possible to change \(r\) with \(\theta\) to obtain a path over which the body in motion inside is always experiencing a positive lift. In this case however we will bent the tube such that there is no lower half and at the same time provide for a tube that runs in the opposite sense to compensate for the horizontal component of \(F_w\).
And we consider the whole tube filled with fluid in circular motion, then
\(\Delta F_l=\Delta m\cfrac{v^2}{r}cos(\theta)\)
where \(\Delta m=\rho.\pi {a^2}r\Delta \theta\) and \(\rho\), the density of the fluid.
\(F_l=2\int^{\pi/2}_{\alpha_f}{\rho.\pi a^2r\cfrac{v^2}{r}cos(\theta)}d\theta\)
\(F_l=2\rho.\pi a^2{v^2}\int^{\pi/2}_{\alpha_f}{cos(\theta)}d\theta\)
\(F_l=2\rho.\pi a^2{v^2}\{1-sin(\alpha_f)\}\)
\(F_l=2\rho.\pi a^2{v^2}\) when \(\alpha_f=0\)
In the picture above, the device can then provide a constant lift of \(6*F_l\) at all time. Unfortunately, nobody can patent this but me.
In the case where \(v\) is varied, for a tube filled with fluid in circular motion, we let,
\(\cfrac{{a_o}^2}{a^2}={cos^{-1}(\theta)}\), then since,
\(\cfrac{v_o}{v}=\cfrac{a^2}{{a_o}^2}\)
\(v^2={v_o}^2\left\{\cfrac{{a_o}^2}{a^2}\right\}^2={v_o}^2cos^{-2}(\theta)\)
and so,
\(\Delta F_l=\Delta m\cfrac{v^2}{r}cos(\theta)=\Delta m\cfrac{v^2_o}{r}cos^{-1}(\theta)\)
where \(\Delta m=\rho.\pi a^2 r\Delta \theta\) and \(\rho\), the density of the fluid.
\(F_l=2\int^{\pi/2}_{\alpha_f}{\rho.\pi a^2 r\cfrac{v^2_o}{r}cos^{-1}(\theta)}d\theta\)
Since \(a^2=a_o^2{cos(\theta)}\),
\(F_l=2\rho.\pi a^2_o{v^2_o}\int^{\pi/2}_{\alpha_f}{}d\theta\)
\(F_l=2\rho.\pi a^2_o{v^2_o}\{\cfrac{\pi}{2}-\alpha_f\}\)
\(F_l=\rho.\pi^2 a^2_o{v^2_o}\) when \(\alpha_f=0\)
It is not necessary to vary the hollow radius of the tube. However \(F_l\) independent of \(\theta\) would mean that lift is uniform throughout the length of the tube.
If we do not ignore the effect of gravity, there is a negative term given by,
\( F_{ g }=-2\rho g\pi a^2_o\int _{ 0 }^{ \cfrac{\pi}{2} }{ rcos^{ 2 }(\theta ) } d\theta\)
When \(r\) is a constant \(r_o\)
\( F_{ g }=-2\rho g\pi a^2_o r_o\int _{ 0 }^{ \cfrac{\pi}{2} }{ cos^{ 2 }(\theta ) } d\theta\)
\( F_{ g }=-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o\)
And the total lifts for the two cases are given by
\(F_l=2\rho\pi a^2_o{v^2}-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o=\rho\pi a^2_o(2v^2-\cfrac{1}{2}g\pi r_o)\)
\(F_l=\rho\pi^2 a^2_o{v^2_o}-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o=\rho\pi^2a^2_o(v^2_o-\cfrac{1}{2}g r_o)\)
respectively.
\(mg=m\cfrac{v_g^2}{r}\), \(v_g=\sqrt{g.r}\)
\(v>>v_g\)
The body pushes against the outer wall of the tube and a reaction force from the wall pushes it towards the center of the circular path. It is this force on the body together with gravity that provides for the centripetal force.
Numerically,
\(F_w=F_b=m\cfrac{v^2}{r}-mgcos(\theta)\)
where \(F_b\) and \(F_w\) are action and reaction pair. The vertical lift created by the reaction force on the tube wall by \(F_w\) is,
\(F_l=F_wcos(\theta)=m\left\{\cfrac{v^2}{r}-gcos(\theta)\right\}cos(\theta)\)
The horizontal component that comes with this lift can be counter-balanced by another similar tube with the body rotating in the opposite sense. At high velocity \(v>>v_g\) such that we can ignore the term \(mgcos^2(\theta)\), then
\(F_l=m\cfrac{v^2}{r}cos(\theta)\)
If the body is a ferromagnetic fluid or mercury driven by a electromagnetic force, then it is possible to change its velocity \(v\) by changing the hollow radius of the tube or \(r\) the path of the tube, since velocity is inversely proportional to the radius squared,
\(\cfrac{v_o}{v}=\cfrac{a^2}{{a_o}^2}\)
If
\(\cfrac{a^2}{{a_o}^2}=\sqrt{cos(\theta)}\), then
\(v^2={v_o}^2\left\{\cfrac{{a_o}^2}{a^2}\right\}^2={v_o}^2\cfrac{1}{cos(\theta)}\)
and so,
\(F_l=m\cfrac{v^2}{r}cos(\theta)=m\cfrac{{v_o}^2}{r}\)
which is a constant provided \(r\) is a constant, ie. a circular path. \(F_l\) can then be varied by changing \(v_o\). A pair of such a device that compensate each other laterally, will then generate a total lift of \(2*F_l\). It is not possible to make the tube with a hollow radius of zero at \(\theta=\cfrac{\pi}{2}\), but it can be at a practical minimum, \(a_f\) at some \(\alpha_f\).
On the lower half of the tube,
\(m\cfrac{v^2}{r}=F_b-mgsin(\alpha)\)
\(F_w=F_b=m\cfrac{v^2}{r}+mgsin(\alpha)\)
When \(\theta\) is measured from the vertical axis, then
\(\theta=\alpha+\cfrac{\pi}{2}\)
\(sin(\alpha)=sin(\theta-\cfrac{\pi}{2})=sin(-(\cfrac{\pi}{2}-\theta))=-sin(\cfrac{\pi}{2}-\theta)=-cos(\theta)\)
Still numerically,
\(F_w=m\cfrac{v^2}{r}-mgcos(\theta)\) for \(\cfrac{\pi}{2}<\theta<\pi\)
This force is downwards, an expression for vertical lift is given by,
\(F_l=-F_wcos(\theta)=mcos(\theta)\left\{-\cfrac{v^2}{r}+gcos(\theta)\right\}\)
which is always negative given the value of \(r\) and \(\theta\). It is possible to change \(r\) with \(\theta\) to obtain a path over which the body in motion inside is always experiencing a positive lift. In this case however we will bent the tube such that there is no lower half and at the same time provide for a tube that runs in the opposite sense to compensate for the horizontal component of \(F_w\).
 |
| Anti Gravity Device I |
And we consider the whole tube filled with fluid in circular motion, then
\(\Delta F_l=\Delta m\cfrac{v^2}{r}cos(\theta)\)
where \(\Delta m=\rho.\pi {a^2}r\Delta \theta\) and \(\rho\), the density of the fluid.
\(F_l=2\int^{\pi/2}_{\alpha_f}{\rho.\pi a^2r\cfrac{v^2}{r}cos(\theta)}d\theta\)
\(F_l=2\rho.\pi a^2{v^2}\int^{\pi/2}_{\alpha_f}{cos(\theta)}d\theta\)
\(F_l=2\rho.\pi a^2{v^2}\{1-sin(\alpha_f)\}\)
\(F_l=2\rho.\pi a^2{v^2}\) when \(\alpha_f=0\)
In the picture above, the device can then provide a constant lift of \(6*F_l\) at all time. Unfortunately, nobody can patent this but me.
In the case where \(v\) is varied, for a tube filled with fluid in circular motion, we let,
\(\cfrac{{a_o}^2}{a^2}={cos^{-1}(\theta)}\), then since,
\(\cfrac{v_o}{v}=\cfrac{a^2}{{a_o}^2}\)
\(v^2={v_o}^2\left\{\cfrac{{a_o}^2}{a^2}\right\}^2={v_o}^2cos^{-2}(\theta)\)
and so,
\(\Delta F_l=\Delta m\cfrac{v^2}{r}cos(\theta)=\Delta m\cfrac{v^2_o}{r}cos^{-1}(\theta)\)
where \(\Delta m=\rho.\pi a^2 r\Delta \theta\) and \(\rho\), the density of the fluid.
\(F_l=2\int^{\pi/2}_{\alpha_f}{\rho.\pi a^2 r\cfrac{v^2_o}{r}cos^{-1}(\theta)}d\theta\)
Since \(a^2=a_o^2{cos(\theta)}\),
\(F_l=2\rho.\pi a^2_o{v^2_o}\int^{\pi/2}_{\alpha_f}{}d\theta\)
\(F_l=2\rho.\pi a^2_o{v^2_o}\{\cfrac{\pi}{2}-\alpha_f\}\)
\(F_l=\rho.\pi^2 a^2_o{v^2_o}\) when \(\alpha_f=0\)
It is not necessary to vary the hollow radius of the tube. However \(F_l\) independent of \(\theta\) would mean that lift is uniform throughout the length of the tube.
If we do not ignore the effect of gravity, there is a negative term given by,
\( F_{ g }=-2\rho g\pi a^2_o\int _{ 0 }^{ \cfrac{\pi}{2} }{ rcos^{ 2 }(\theta ) } d\theta\)
When \(r\) is a constant \(r_o\)
\( F_{ g }=-2\rho g\pi a^2_o r_o\int _{ 0 }^{ \cfrac{\pi}{2} }{ cos^{ 2 }(\theta ) } d\theta\)
\( F_{ g }=-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o\)
And the total lifts for the two cases are given by
\(F_l=2\rho\pi a^2_o{v^2}-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o=\rho\pi a^2_o(2v^2-\cfrac{1}{2}g\pi r_o)\)
\(F_l=\rho\pi^2 a^2_o{v^2_o}-\cfrac{1}{2}\rho g\pi^2 a^2_o r_o=\rho\pi^2a^2_o(v^2_o-\cfrac{1}{2}g r_o)\)
respectively.
Saturday, July 12, 2014
Not So Pointy
This is a picture of the wire antenna (a solid) generated by rotating the area below (10/x)^(1/2) about the x-axis.
Such a antenna so shaped, will hopefully have a \(E=\cfrac{V}{d^2}.x\) profile along its length and so all energy fed into it is radiated off. Close to 100% efficiency.
But under normal circumstances what happened to the rest of the energy? A good guess is that the signal in the antenna not radiated, bounces up and down the wire and generates echos and noise. If it is possible to attain higher efficiency, there should also be a corresponding reduction in noise.
But under normal circumstances what happened to the rest of the energy? A good guess is that the signal in the antenna not radiated, bounces up and down the wire and generates echos and noise. If it is possible to attain higher efficiency, there should also be a corresponding reduction in noise.
Friday, July 11, 2014
A Pointy Wire...
Since, resistance \(R\) is inversely proportional to cross section area \(A\),
\(R\propto \cfrac{1}{A}\)
We let,
\(R=\cfrac{k}{\pi r^2}\) where \(k\) is a constant of proportionality.
If \(r=\sqrt{\cfrac{d}{x}}\) then
\(R=\cfrac{k}{\pi d}.x\)
then
\(V=IR=I\cfrac{k}{\pi d}.x\)
In which case,
\(E=\cfrac{V}{d}=I\cfrac{k}{\pi d^2}.x\)
Given a constant \(I\),
\(E=D\cfrac{V}{d^2}.x\)
where \(D=\cfrac{k}{R\pi }\) is a constant and \(R\) is the total resistance of the wire.
A plot of \(r=\sqrt{\cfrac{d}{x}}\) is given below, the actual function plotted is (10/x)^(1/2), with d =10,
This graph is rotated about the x-axis to generate a solid that is the antenna. Since the driving signal \(V\) is sinusoidal, it can be applied to the boarder end all the same.
\(R\propto \cfrac{1}{A}\)
We let,
\(R=\cfrac{k}{\pi r^2}\) where \(k\) is a constant of proportionality.
If \(r=\sqrt{\cfrac{d}{x}}\) then
\(R=\cfrac{k}{\pi d}.x\)
then
\(V=IR=I\cfrac{k}{\pi d}.x\)
In which case,
\(E=\cfrac{V}{d}=I\cfrac{k}{\pi d^2}.x\)
Given a constant \(I\),
\(E=D\cfrac{V}{d^2}.x\)
where \(D=\cfrac{k}{R\pi }\) is a constant and \(R\) is the total resistance of the wire.
A plot of \(r=\sqrt{\cfrac{d}{x}}\) is given below, the actual function plotted is (10/x)^(1/2), with d =10,
This graph is rotated about the x-axis to generate a solid that is the antenna. Since the driving signal \(V\) is sinusoidal, it can be applied to the boarder end all the same.
Wait Another Minute
From the post "Wait a Minute",
\(E=\cfrac{V}{d^2}.x\) ---(*)
comparing this expression with the normal expression,
\(E=\cfrac{V}{d}\)
we see that \(d\) is replaced with
\(\cfrac{d^2}{x}\)
A comparative graph of the two expressions,
shows that expression (*) has only half the energy. This means only half the energy delivered to the wire is radiated under normal circumstances (superposition). This could mean that, if a material follows instead a material density profile of
such that \(E\) is related to the applied \(V\) given by (*) then we may have close to 100% efficiency for radiated energy.
\(E=\cfrac{V}{d^2}.x\) ---(*)
comparing this expression with the normal expression,
\(E=\cfrac{V}{d}\)
we see that \(d\) is replaced with
\(\cfrac{d^2}{x}\)
A comparative graph of the two expressions,
shows that expression (*) has only half the energy. This means only half the energy delivered to the wire is radiated under normal circumstances (superposition). This could mean that, if a material follows instead a material density profile of
 |
| Graph of d2/x, d=10 |
such that \(E\) is related to the applied \(V\) given by (*) then we may have close to 100% efficiency for radiated energy.
Wait A Minute...
From previously,
\( B=\cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E\).
since \(x^{'}=i.2\pi r\), \(x^{'}\) is tangential to \(r\) everywhere
\( B=-i\cfrac { d }{ Rv } .\cfrac { 1 }{ \varepsilon _{ o }x^{'} } E\)
where \(i\) was mulitplied to both the denominator and numerator.
Comparing this expression with,
\(B=-i\cfrac{\partial E}{\partial x^{'}}\)
which leads to the wave equation
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v^{ 2 }\),
with a typical solution of the form,
\(E(x,t)=\cfrac{A}{x}.e^{i(\omega t\pm kx)}\)
We find that, since \(E\) can be replaced with \(D.E\) where \(D\) is a constant, and still satisfy the wave equation,
\(-i\cfrac{\partial E}{\partial x^{'}}=-i\cfrac{E}{x^{'}}\)
and
\(D=\cfrac { d }{ Rv \varepsilon _{ o }}\)
where \(-i\) rotates \(x^{'}\) to the \(x\) direction,
\(\cfrac{\partial E}{\partial x}=\cfrac{E}{x}\)
\(\int{\cfrac{1}{E}}\partial E=\int{\cfrac{1}{x}}\partial x\)
If we consider only the spatial component of \(E\),
\(E=Cx\) over the length of the wire. Since \(V=d.E\),
\(C=\cfrac{V}{d^2}\)
ie.,
\(E=\cfrac{V}{d^2}.x\) where \(d\) is the length of the wire.
\(D.E=\cfrac { V }{ dRv \varepsilon _{ o }}.x\\=\cfrac{J}{\varepsilon _{ o }}.\cfrac{x}{d}\)
Since, \(V=d.D.E\)
\(V=\cfrac{J}{\varepsilon _{ o }}.x\)
where \(V\) is the voltage along the length of the wire.
This is a very interesting result. It is tempting to suggest that this \(E\) field/\(V\) voltage profile up the wire antenna is most efficient for radiation at the end of the wire. Although it is true that this expression for \(E\) is derived from \(B=-i\cfrac{dE}{dx^{'}}\), that leads to the wave equation directly. This seem like impedance matching, but
\(V=IR=\cfrac{I}{\varepsilon _{ o }v}.d\)
\(R=\cfrac{d}{\varepsilon _{ o }v}\) where \(v\) is the charge average velocity along the wire that changes with the driving voltage at the end of the wire.
Have a nice day.
\( B=\cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E\).
since \(x^{'}=i.2\pi r\), \(x^{'}\) is tangential to \(r\) everywhere
\( B=-i\cfrac { d }{ Rv } .\cfrac { 1 }{ \varepsilon _{ o }x^{'} } E\)
where \(i\) was mulitplied to both the denominator and numerator.
Comparing this expression with,
\(B=-i\cfrac{\partial E}{\partial x^{'}}\)
which leads to the wave equation
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v^{ 2 }\),
with a typical solution of the form,
\(E(x,t)=\cfrac{A}{x}.e^{i(\omega t\pm kx)}\)
We find that, since \(E\) can be replaced with \(D.E\) where \(D\) is a constant, and still satisfy the wave equation,
\(-i\cfrac{\partial E}{\partial x^{'}}=-i\cfrac{E}{x^{'}}\)
and
\(D=\cfrac { d }{ Rv \varepsilon _{ o }}\)
where \(-i\) rotates \(x^{'}\) to the \(x\) direction,
\(\cfrac{\partial E}{\partial x}=\cfrac{E}{x}\)
\(\int{\cfrac{1}{E}}\partial E=\int{\cfrac{1}{x}}\partial x\)
If we consider only the spatial component of \(E\),
\(E=Cx\) over the length of the wire. Since \(V=d.E\),
\(C=\cfrac{V}{d^2}\)
ie.,
\(E=\cfrac{V}{d^2}.x\) where \(d\) is the length of the wire.
\(D.E=\cfrac { V }{ dRv \varepsilon _{ o }}.x\\=\cfrac{J}{\varepsilon _{ o }}.\cfrac{x}{d}\)
Since, \(V=d.D.E\)
\(V=\cfrac{J}{\varepsilon _{ o }}.x\)
where \(V\) is the voltage along the length of the wire.
This is a very interesting result. It is tempting to suggest that this \(E\) field/\(V\) voltage profile up the wire antenna is most efficient for radiation at the end of the wire. Although it is true that this expression for \(E\) is derived from \(B=-i\cfrac{dE}{dx^{'}}\), that leads to the wave equation directly. This seem like impedance matching, but
\(V=IR=\cfrac{I}{\varepsilon _{ o }v}.d\)
\(R=\cfrac{d}{\varepsilon _{ o }v}\) where \(v\) is the charge average velocity along the wire that changes with the driving voltage at the end of the wire.
Have a nice day.
Thursday, July 10, 2014
Pointing Where? No Side Issue, Please
\(x^{'}\quad =\quad 2\pi r\)
\( dx^{'}\quad =\quad 2\pi dr\)
\( B=-\cfrac { dE }{ dx^{'} } =-\cfrac { 1 }{ 2\pi } \cfrac { dE }{ dr } \) ----(*)
Consider a wire carrying current, \(I\).
\(B=\cfrac { I }{ 2\pi \varepsilon _{ o }r } \cfrac { 1 }{ v } \)
where \(r\) is the distance perpendicular to the wire.
\(V=IR\), a wire of length \(d\), \(I=\cfrac{E.d}{R}\)
\( B=\cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E\)
Equating to (*)
\( \cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E =-\cfrac { 1 }{ 2\pi } \cfrac { dE }{ dr } \)
\( -\cfrac { d }{ R\varepsilon _{ o }v } \int {\cfrac { 1 }{ r }} dr=\int{\cfrac { 1 }{ E }} dE\)
\( -\cfrac { d }{ R\varepsilon _{ o }v } ln(r)=ln(E)\)
\( E=C{ r }^{ -\cfrac { d }{ R\varepsilon _{ o }v } }\)
Since \(E=0\) as \(r\rightarrow\infty\) the only reasonable solution is \(C=0\). Which means \(E\) is zero for all \(r\) through the length of the conductor. This is consistent with the fact that the E field outside of a current carrying wire is zero.
But from, \( B=-\cfrac { dE }{ dx^{'} }\) we are able to derive,
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v^{ 2 }\)
That is to say, when a potential function \(V=d.E\) is applied to the wire, that \(E\) satisfies the wave equation above then we have a wave propagating in the \(x\) direction parallel to the length of the wire. Combining this with the fact that E field outside of the conductor along its length is zero,
one possibility is that the wave is emanating from the end of the wire in semi-spherical wave front and is zero below that point. The intensity of the wave decreases by a factor of \(\cfrac{1}{2}*4\pi r^2=2\pi.r^2\) where \(r\) is the radial distance from the end of the wire. Another wire slightly lower than the transmitting wire antenna will not be affect by it, reflected signals from the ionosphere aside. The receiving wire must be higher that the transmitting wire to receive from it, especially if the signal is too weak to bounce off the ionoshpere. A reflecting back plane at the transmitting wire is useless because the wave does not reach behind the end point. The wire is not behaving like a point source. The intensity of the wave from the end of the wire is twice that from a point source.
So, if you want to receive from a straight antenna directly you have to be higher than the antenna, otherwise point your receiving antenna upwards to receive reflected signals from the ionosphere or at some tall building higher than the antenna, reflecting signals from the antenna.
\( dx^{'}\quad =\quad 2\pi dr\)
\( B=-\cfrac { dE }{ dx^{'} } =-\cfrac { 1 }{ 2\pi } \cfrac { dE }{ dr } \) ----(*)
Consider a wire carrying current, \(I\).
\(B=\cfrac { I }{ 2\pi \varepsilon _{ o }r } \cfrac { 1 }{ v } \)
where \(r\) is the distance perpendicular to the wire.
\(V=IR\), a wire of length \(d\), \(I=\cfrac{E.d}{R}\)
\( B=\cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E\)
Equating to (*)
\( \cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E =-\cfrac { 1 }{ 2\pi } \cfrac { dE }{ dr } \)
\( -\cfrac { d }{ R\varepsilon _{ o }v } \int {\cfrac { 1 }{ r }} dr=\int{\cfrac { 1 }{ E }} dE\)
\( -\cfrac { d }{ R\varepsilon _{ o }v } ln(r)=ln(E)\)
\( E=C{ r }^{ -\cfrac { d }{ R\varepsilon _{ o }v } }\)
Since \(E=0\) as \(r\rightarrow\infty\) the only reasonable solution is \(C=0\). Which means \(E\) is zero for all \(r\) through the length of the conductor. This is consistent with the fact that the E field outside of a current carrying wire is zero.
But from, \( B=-\cfrac { dE }{ dx^{'} }\) we are able to derive,
\( \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v^{ 2 }\)
That is to say, when a potential function \(V=d.E\) is applied to the wire, that \(E\) satisfies the wave equation above then we have a wave propagating in the \(x\) direction parallel to the length of the wire. Combining this with the fact that E field outside of the conductor along its length is zero,
 |
| Radiation Pattern of a Wire Antenna |
one possibility is that the wave is emanating from the end of the wire in semi-spherical wave front and is zero below that point. The intensity of the wave decreases by a factor of \(\cfrac{1}{2}*4\pi r^2=2\pi.r^2\) where \(r\) is the radial distance from the end of the wire. Another wire slightly lower than the transmitting wire antenna will not be affect by it, reflected signals from the ionosphere aside. The receiving wire must be higher that the transmitting wire to receive from it, especially if the signal is too weak to bounce off the ionoshpere. A reflecting back plane at the transmitting wire is useless because the wave does not reach behind the end point. The wire is not behaving like a point source. The intensity of the wave from the end of the wire is twice that from a point source.
So, if you want to receive from a straight antenna directly you have to be higher than the antenna, otherwise point your receiving antenna upwards to receive reflected signals from the ionosphere or at some tall building higher than the antenna, reflecting signals from the antenna.