风带云烟不着地
仲夏宁夜热依然
孤寂本不泛思潮
心自多情谁念谁 《想？雪》
Monday, June 16, 2014
Wanting More, All Over And Twisted
In the last post the expression,
\(\boxed{-i\cfrac{\partial E_i}{\partial x^{'}} =} \cfrac{\partial( i.-E_i )}{\partial x^{'}}=-i\cfrac{\partial E_i}{\partial x}=-i\cfrac{1}{2\pi\varepsilon_o r}.\cfrac{-\partial\lambda}{\partial x}=-i\cfrac{J}{2\pi\varepsilon_o r}=B\)
\(-i\cfrac{\partial E_i}{\partial x^{'}}\) is wrong.
is really clumsy because we know \(E\) along \(r\) to be orthogonal to \(x\) and so orthogonal to \(J\), at the same time \(B\) is orthogonal to both \(E\) and \(J\). All conforming to the right hand rule. But these relationships are expressed in part as the partial derivatives and the term \(-i\) which rotates everything clockwise 90o on the side of the equation it is applied to. The negative sign in \(\cfrac{-\partial\lambda}{\partial x}\) is because of \(E\) being negative in the original formulation. And the whole term is rotated by \(-i\) to be along \(B\) (ie. \(\partial x\rightarrow\partial x^{'}\)), eventually. So it might be clearer, to express
\(-i\cfrac{\partial E_i}{\partial x^{'}}\rightarrow \cfrac{\partial ( i. -E_i )}{\partial x^{'}}\) This is why.
instead where \(i\) is applied to \(-E_i\) to be in the direction of \(B\) along \(x^{'}\). And so
\( \cfrac{\partial ( i. -E_i )}{\partial x^{'}} = -i.\left\{\cfrac{\partial E_i}{\partial x}\right\}\)
where \(x^{'}\) has been rotated by \(i\) to \(\partial x\) in reverse as \(-i\) is brought out to apply to both denominator and numerator. (\(i .-i = 1\)). If, however \(i\) or \(-i\) were to apply to the numerator only by convention, then an expression like \(i\cfrac { \partial ^{ 2 }E }{ \partial x^{'} \partial x^{'}}\cfrac{\partial x^{'}}{\partial t }\) will be confusing because \(x^{'}\) is in both the numerator and denominator. All such \(i\) rotations, jump from axis to axis on the right hand frame.
So we have, hopefully in its final form,
\(B=\cfrac{\partial ( i.E )}{\partial x^{'}}=i\cfrac{\partial E}{\partial (-x)}\neq -i\cfrac{\partial E}{\partial x } \)
but
\(B= -i\cfrac{\partial (-E)}{\partial x } \)
Both \(E\) and \(-x\) are rotated by \(i\) to the \(x^{'}\) direction along \(B\). In the second instance, \(-E\) and \(x\) are rotated by \(-i\) to the \(x^{'}\) direction. Still \(i*i=-1\), which should be anticlockwise 90o rotation twice and not reverse, in the negative direction. This ambiguity means \(i\) should not be used as a notation for rotation. So,
\(B=-\cfrac{\partial E}{\partial x'}\)
and all directions are discussed separately. \(B\) is perpendicular to \(E\), \(x'\) is along \(B\) and \(B \times E\) gives the \(x\) direction that is perpendicular to \(x'\)
The cross product, does not lend itself to manipulation. The two terms involved are stuck on each other. The \(i\) term at least allows a partial derivative in between. But the cross product allows for non-orthogonal vectors with a \(cos(\theta)\) term, where \(\theta\) is the angle between the two vectors.
There has to be a better way that all this is represented, specially when they are written out as a computer program.
\(\boxed{-i\cfrac{\partial E_i}{\partial x^{'}} =} \cfrac{\partial( i.-E_i )}{\partial x^{'}}=-i\cfrac{\partial E_i}{\partial x}=-i\cfrac{1}{2\pi\varepsilon_o r}.\cfrac{-\partial\lambda}{\partial x}=-i\cfrac{J}{2\pi\varepsilon_o r}=B\)
\(-i\cfrac{\partial E_i}{\partial x^{'}}\) is wrong.
is really clumsy because we know \(E\) along \(r\) to be orthogonal to \(x\) and so orthogonal to \(J\), at the same time \(B\) is orthogonal to both \(E\) and \(J\). All conforming to the right hand rule. But these relationships are expressed in part as the partial derivatives and the term \(-i\) which rotates everything clockwise 90o on the side of the equation it is applied to. The negative sign in \(\cfrac{-\partial\lambda}{\partial x}\) is because of \(E\) being negative in the original formulation. And the whole term is rotated by \(-i\) to be along \(B\) (ie. \(\partial x\rightarrow\partial x^{'}\)), eventually. So it might be clearer, to express
\(-i\cfrac{\partial E_i}{\partial x^{'}}\rightarrow \cfrac{\partial ( i. -E_i )}{\partial x^{'}}\) This is why.
instead where \(i\) is applied to \(-E_i\) to be in the direction of \(B\) along \(x^{'}\). And so
\( \cfrac{\partial ( i. -E_i )}{\partial x^{'}} = -i.\left\{\cfrac{\partial E_i}{\partial x}\right\}\)
where \(x^{'}\) has been rotated by \(i\) to \(\partial x\) in reverse as \(-i\) is brought out to apply to both denominator and numerator. (\(i .-i = 1\)). If, however \(i\) or \(-i\) were to apply to the numerator only by convention, then an expression like \(i\cfrac { \partial ^{ 2 }E }{ \partial x^{'} \partial x^{'}}\cfrac{\partial x^{'}}{\partial t }\) will be confusing because \(x^{'}\) is in both the numerator and denominator. All such \(i\) rotations, jump from axis to axis on the right hand frame.
So we have, hopefully in its final form,
\(B=\cfrac{\partial ( i.E )}{\partial x^{'}}=i\cfrac{\partial E}{\partial (-x)}\neq -i\cfrac{\partial E}{\partial x } \)
but
\(B= -i\cfrac{\partial (-E)}{\partial x } \)
Both \(E\) and \(-x\) are rotated by \(i\) to the \(x^{'}\) direction along \(B\). In the second instance, \(-E\) and \(x\) are rotated by \(-i\) to the \(x^{'}\) direction. Still \(i*i=-1\), which should be anticlockwise 90o rotation twice and not reverse, in the negative direction. This ambiguity means \(i\) should not be used as a notation for rotation. So,
\(B=-\cfrac{\partial E}{\partial x'}\)
and all directions are discussed separately. \(B\) is perpendicular to \(E\), \(x'\) is along \(B\) and \(B \times E\) gives the \(x\) direction that is perpendicular to \(x'\)
The cross product, does not lend itself to manipulation. The two terms involved are stuck on each other. The \(i\) term at least allows a partial derivative in between. But the cross product allows for non-orthogonal vectors with a \(cos(\theta)\) term, where \(\theta\) is the angle between the two vectors.
There has to be a better way that all this is represented, specially when they are written out as a computer program.
Running Along a Line of Charge
The expresison for \(B\) is consistent because, if
\(\lambda=\cfrac{Q}{x}\) then,
\(\cfrac{\partial\lambda}{\partial x}=\cfrac{\partial}{\partial x}\left\{\cfrac{Q}{x}\right\}=-\cfrac{Q}{x^2}=-J\)
\(x\) is in a direction perpendicular to \(r\). One way to interpret the negative sign is to see that \(x\) is rotated twice anticlockwise 90o about the right hand axis. This may be the reason why \(i\) is abandoned, because \(i*iJ=-J\) is ambiguous. \(-J\) should be in the negative \(J\) direction not \(x^{'}\).
Since the E field due to a line of charges of line charge density \(\lambda\) is,
\(E=\cfrac{\lambda}{2\pi\varepsilon_o r}\) outwards,
\(E_i=-\cfrac{\lambda}{2\pi\varepsilon_o r}\) inwards,
Because the expression for \(B\) was written for \(E\) inwards with a negative sign to conform with the right hand rule.
\(\boxed{-i\cfrac{\partial E_i}{\partial x^{'}} =} \cfrac{\partial( i.-E_i )}{\partial x^{'}}=-i\cfrac{\partial E_i}{\partial x}=-i\cfrac{1}{2\pi\varepsilon_o r}.\cfrac{-\partial\lambda}{\partial x}=-i\cfrac{J}{2\pi\varepsilon_o r}=B\)
\(-i\cfrac{\partial E_i}{\partial x^{'}}\) is wrong.
as before with the understanding that \(B\) is orthogonal to \(J\), by the right hand rule. The field \(E\), is established by the moving charges \(q\) in \(J\), it is not the emf pushing the charges along the line. \(B\) is orthogonal to both \(E\) and \(J\) by the right hand rule.
Maybe it is best to calculate for absolute values and then apply the right hand rule to find direction of the vectors involved separately. So,
\(B=-\cfrac{dE}{dx}\)
where E is perpendicular to B and conforms to the right hand rule. And \(dx\) is along B.
\(\lambda=\cfrac{Q}{x}\) then,
\(\cfrac{\partial\lambda}{\partial x}=\cfrac{\partial}{\partial x}\left\{\cfrac{Q}{x}\right\}=-\cfrac{Q}{x^2}=-J\)
\(x\) is in a direction perpendicular to \(r\). One way to interpret the negative sign is to see that \(x\) is rotated twice anticlockwise 90o about the right hand axis. This may be the reason why \(i\) is abandoned, because \(i*iJ=-J\) is ambiguous. \(-J\) should be in the negative \(J\) direction not \(x^{'}\).
Since the E field due to a line of charges of line charge density \(\lambda\) is,
\(E=\cfrac{\lambda}{2\pi\varepsilon_o r}\) outwards,
\(E_i=-\cfrac{\lambda}{2\pi\varepsilon_o r}\) inwards,
Because the expression for \(B\) was written for \(E\) inwards with a negative sign to conform with the right hand rule.\(\boxed{-i\cfrac{\partial E_i}{\partial x^{'}} =} \cfrac{\partial( i.-E_i )}{\partial x^{'}}=-i\cfrac{\partial E_i}{\partial x}=-i\cfrac{1}{2\pi\varepsilon_o r}.\cfrac{-\partial\lambda}{\partial x}=-i\cfrac{J}{2\pi\varepsilon_o r}=B\)
\(-i\cfrac{\partial E_i}{\partial x^{'}}\) is wrong.
as before with the understanding that \(B\) is orthogonal to \(J\), by the right hand rule. The field \(E\), is established by the moving charges \(q\) in \(J\), it is not the emf pushing the charges along the line. \(B\) is orthogonal to both \(E\) and \(J\) by the right hand rule.
Maybe it is best to calculate for absolute values and then apply the right hand rule to find direction of the vectors involved separately. So,
\(B=-\cfrac{dE}{dx}\)
where E is perpendicular to B and conforms to the right hand rule. And \(dx\) is along B.
Sunday, June 15, 2014
Use J Not I
From,
\(B=-i\cfrac{dE}{dx}\)
\(B=-i\cfrac{dE}{dt}\cfrac{dt}{dx}\)
\(B=-i\cfrac{dE}{dt}\cfrac{1}{v}\)
If we are dealing with a wire carrying moving charges, we know that the effects of the moving charges cancels before and after the point where the normal drop from the point \(B\) is to be calculated. The only effect on \(B\) is confined to a disc directly under, at the base of the normal. By conservation of flux from a fixed source, in this case a fixed current density,
\(E=\cfrac{q}{2\pi\varepsilon_o r}\) as oppose to \(E=\cfrac{q}{4\pi\varepsilon_o r^2}\)
a disc of perimeter \(2\pi r\) not a sphere of area \(4\pi r^2\).
So,
\(B=-i\cfrac{d}{dt}\left\{{\cfrac{q}{2\pi\varepsilon_o r}}\right\}\cfrac{1}{v}\)
where \(v\) is the average speed of the moving charges and yet \(\varepsilon_o\) is unchanged.
\(B=-i\cfrac{1}{2\pi\varepsilon_o r}\cfrac{dq}{dt}\cfrac{1}{v}\)
\(B=-i\cfrac{1}{2\pi\varepsilon_o r}\cfrac{I}{v}\)
\(B=-i\cfrac{J}{2\pi\varepsilon_o r}\)
\(|B|=\cfrac{J}{2\pi\varepsilon_o r}\)
Which is not the often published formula of \(B=\cfrac{\mu_o I}{2\pi r}\). \(-i\) signifies that \(B\) is perpendicular to \(J\) in the right hand sense. The scaling factor of \(\cfrac{1}{\varepsilon_o}\) as opposed to \(\mu_o\) is easily forgiven except for energy formulae, in which case, another constant is applied to get rid of this artificial \(\mu_o\). For example, dividing by the wave impedence defined as \(Z=\sqrt{\cfrac{\mu_o}{\varepsilon_o}}\).
The real concern is \(J\), current density instead of \(I\), current. If we compare this with the E field due to a line charge,
\(E=\cfrac{\lambda}{2\pi\varepsilon_o r}\)
where \(\lambda\) is the line charge density. We see a good analogy between E and B field if we drop \(\mu_o\) and its unit all together.
\(B=-i\cfrac{dE}{dx}\)
\(B=-i\cfrac{dE}{dt}\cfrac{dt}{dx}\)
\(B=-i\cfrac{dE}{dt}\cfrac{1}{v}\)
If we are dealing with a wire carrying moving charges, we know that the effects of the moving charges cancels before and after the point where the normal drop from the point \(B\) is to be calculated. The only effect on \(B\) is confined to a disc directly under, at the base of the normal. By conservation of flux from a fixed source, in this case a fixed current density,
\(E=\cfrac{q}{2\pi\varepsilon_o r}\) as oppose to \(E=\cfrac{q}{4\pi\varepsilon_o r^2}\)
a disc of perimeter \(2\pi r\) not a sphere of area \(4\pi r^2\).
So,
\(B=-i\cfrac{d}{dt}\left\{{\cfrac{q}{2\pi\varepsilon_o r}}\right\}\cfrac{1}{v}\)
where \(v\) is the average speed of the moving charges and yet \(\varepsilon_o\) is unchanged.
\(B=-i\cfrac{1}{2\pi\varepsilon_o r}\cfrac{dq}{dt}\cfrac{1}{v}\)
\(B=-i\cfrac{1}{2\pi\varepsilon_o r}\cfrac{I}{v}\)
\(B=-i\cfrac{J}{2\pi\varepsilon_o r}\)
\(|B|=\cfrac{J}{2\pi\varepsilon_o r}\)
The real concern is \(J\), current density instead of \(I\), current. If we compare this with the E field due to a line charge,
\(E=\cfrac{\lambda}{2\pi\varepsilon_o r}\)
where \(\lambda\) is the line charge density. We see a good analogy between E and B field if we drop \(\mu_o\) and its unit all together.
Saturday, June 14, 2014
Something Familiar, Everything Else Not the Same
The significance of \(p\) being inversely proportional to \(T\) is, in the expression for
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
If we let \(p=\cfrac{A}{T}\), where \(A\) is any constant.
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega \cfrac{A}{T})^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 4 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega A)^{ 2 }+T^2(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
At near resonance \(\omega_o\simeq \omega\), the term \(T^2(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 }\) is small, and
\(PE_{ e }\simeq \cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 4 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega A)^{ 2 }} \)
\(PE_{ e }\propto T^4\)
And at resonance,
\(\omega=\sqrt{\omega^2_o-2p^2}\), \(\omega^2_o-\omega^2=2p^2\)
\((2wA)^2=4A^2(\omega_o^2-2p^2)\)
\((2wA)^{ 2 }+T^{ 2 }(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 }=4A^{ 2 }(\omega ^{ 2 }_{ o }-2\cfrac { A^{ 2 } }{ T^{ 2 } } )+4\cfrac { A^{ 4 } }{ T^{ 2 } } =4A^{ 2 }(\omega ^{ 2 }_{ o }-\cfrac { A^{ 2 } }{ T^{ 2 } } )\)
\(PE_{ e }= \cfrac { 1 }{ 8 }\cfrac{1}{A^{ 2 }} \cfrac { m_{ e }c^{ 2 } }{{ r }^{ 2 }_{ e }}\cfrac{1}{(\omega ^{ 2 }_{ o }-\cfrac { A^{ 2 } }{ T^{ 2 } } ) }.T^4 \)
When \(\omega_o=\omega\),
\(PE_{ e }= \cfrac { 1 }{ 8 }\cfrac{1}{A^{ 2 }} \cfrac { m_{ e }c^{ 2 } }{{ r }^{ 2 }_{ e }}\cfrac{1}{\omega ^{ 2 }_{ o }}.T^4 \)
This relation is at near resonance and at resonance, but not for all frequencies of the driving force. And definitely not for all radiated "frequencies", the two frequency ranges are not the same thing.
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
If we let \(p=\cfrac{A}{T}\), where \(A\) is any constant.
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega \cfrac{A}{T})^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 4 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega A)^{ 2 }+T^2(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
At near resonance \(\omega_o\simeq \omega\), the term \(T^2(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 }\) is small, and
\(PE_{ e }\simeq \cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 4 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega A)^{ 2 }} \)
\(PE_{ e }\propto T^4\)
And at resonance,
\(\omega=\sqrt{\omega^2_o-2p^2}\), \(\omega^2_o-\omega^2=2p^2\)
\((2wA)^2=4A^2(\omega_o^2-2p^2)\)
\((2wA)^{ 2 }+T^{ 2 }(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 }=4A^{ 2 }(\omega ^{ 2 }_{ o }-2\cfrac { A^{ 2 } }{ T^{ 2 } } )+4\cfrac { A^{ 4 } }{ T^{ 2 } } =4A^{ 2 }(\omega ^{ 2 }_{ o }-\cfrac { A^{ 2 } }{ T^{ 2 } } )\)
\(PE_{ e }= \cfrac { 1 }{ 8 }\cfrac{1}{A^{ 2 }} \cfrac { m_{ e }c^{ 2 } }{{ r }^{ 2 }_{ e }}\cfrac{1}{(\omega ^{ 2 }_{ o }-\cfrac { A^{ 2 } }{ T^{ 2 } } ) }.T^4 \)
When \(\omega_o=\omega\),
\(PE_{ e }= \cfrac { 1 }{ 8 }\cfrac{1}{A^{ 2 }} \cfrac { m_{ e }c^{ 2 } }{{ r }^{ 2 }_{ e }}\cfrac{1}{\omega ^{ 2 }_{ o }}.T^4 \)
This relation is at near resonance and at resonance, but not for all frequencies of the driving force. And definitely not for all radiated "frequencies", the two frequency ranges are not the same thing.
Thursday, June 12, 2014
A Whacko Jump
If electrons is at light speed, how easy it is to gain a bit more speed and slow time. Since everything else is in motion, when the electron returns from such short trip in time, it well find itself free from the nucleus as everything have literally moved on. This is a Whacko Jump. A jump in time, resulting in ionization.
Why not then all other particles, protons do the same? Too Much Mass??
Hydrogen has an exposed nucleus, maybe its proton can be made to jump by shooting a high speed particle at it.
Mayby then we can begin to quantify such jumps and have a better understanding of how mass is involved in the process. Obviously, momentum and kinetic energy, where else does the mass term appears.
Why not then all other particles, protons do the same? Too Much Mass??
Hydrogen has an exposed nucleus, maybe its proton can be made to jump by shooting a high speed particle at it.
Mayby then we can begin to quantify such jumps and have a better understanding of how mass is involved in the process. Obviously, momentum and kinetic energy, where else does the mass term appears.
The Other Force
From the post "And There Was Light, Upon the Words", we have,
\(k=\cfrac{{m}_{e}c^2}{{r}^2_{e}}\)
It is natural then to consider damped forced motion and the associated resonance. But what would be the driving function? Photons on passing, blob the electrons up and down, but there is another driving force, an increase in temperature, \(T\). Increasing temperature increases random motions across a wide range of frequency. As a driving force, it acts over a wide range of frequency at the same time. The resulting response under such a driving force is also simultaneously over the same range of frequency. So, we have, based on classical mechanics,
\(x^{''}+2px^{'}+\omega^2_ox={T}cos(\omega t)\)
where \(\omega_o =\sqrt{\cfrac{k}{m_e}}\) and \(p=\sqrt{\cfrac{k_1}{2m_e}}\), \(k_1\) the damping factor is to be to be determined and given physical interpretation later. We will used a solution of the form,
\(x=\cfrac{T}{\sqrt{(2\omega p)^2+(\omega^2_o-\omega^2)^2}}cos(\omega t-\gamma)\)
where \(tan(\gamma)=\cfrac{2\omega p}{(\omega^2_o-\omega^2)}\)
The electrons at increase temperature are driven to an amplitude above it normal orbital position. It has a maximum potential energy of (\(\cfrac{1}{2}kx^2\)),
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { { m }_{ e }c^{ 2 } }{ { r }^{ 2 }_{ e } } .\left\{ \cfrac { T }{ \sqrt { (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } } \right\} ^{ 2 }\)
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
They fall from such amplitude and emits heavy light of energy (\({m}_{e}{v}^2\)) equals to this amount. Therefore,
\({m}_{e}v^2=PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e } } .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
\({ v }^{ 2 }=\cfrac { 1 }{ 2 } \cfrac { T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e } } .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
A plot \(\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } }\) is shown below on the left. The actual parameter used are \({r}_{e}\) = 53e-12 m for hydrogen, \(c\) = 299792458 ms-1,
The actual graph plotted is, 10^7*1/((2*(x*100)*a)^2+(299.792458^2/53-(x*100)^2)^2) on the left, where \(a\) range from 0.5 to 20.5 in increment of 4. The y-axis is proportional to energy \(\) and the x-axis, \(\omega\). \(p\) is the damping factor per unit mass, which is related to the \(T\), equivalently the velocity of the electrons, and the damping force.
By comparing with graphs of black-body radiation with the x-axis plotted in frequency, it seems that this damping factor decreases with increasing temperature, \(T\).
We can form the refractive index,
\(\cfrac { c }{ v } ={ n }_{ T }=\sqrt { 2 }\cfrac{1 }{ T } r_{ e } .\sqrt { \{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 }\} } \)
A plot of \(n_T\) vs \(\omega\), ((2*(x*1000)*20)^2+(299.792458^2/53-(x*1000)^2)^2)^(1/2)*10^(-15) is shown below. The value of \(n_T\) increases with \(\omega\). This is the reverse of the case for spectra line where \(n_n\) decreases with increasing \(n\).
The amount of deflection, \(\theta\) at one medium interface \(n_T | n_p\) is
\(n_T sin(\theta_i)=n_p sin(\theta_r)\)
and at the exit interface \(n_p | n_T\),
\(n_p sin(\Phi-\theta_r)=n_T sin(\theta_e)\),
where \(\Phi\) is the angle between the two interfaces. As \(n_T\) increases \(\theta_e\) decreases and the spectrum is sweep from right to left towards the normal. This is the reverse of spectra lines.
And at resonance, we have,
\(\omega=\sqrt{\omega^2_o-2p^2}\)
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{{c^2}}{{r}^2_{e}}-2p^2}\)
This frequency is however, not the frequency of the emitted light spectrum. The spectrum spread after the prism is the direct result of increasing speed of the electrons with increasing \(\omega\). The relative speed of the electrons just before and after the prism determines the amount of deflection and the direction of deflection relative to the normal at the point of incidence. High temperature drive the system over a wide spread of frequencies, one of which is this resonance value that corresponds to a high energy output (max. amplitude). The spread of velocities after the prism is related to the driving frequency due to the raised temperature, but they are not equivalent. One driving frequency gives raise to a particular amplitude of oscillation and so a value of velocity and a particular deflection point on the screen. All such driving frequencies are present at the same time with raised temperature. This driving frequency, \(\omega\) is not unlimited and neither is the \(x\), the amplitude of the oscillation. Both are valid up to point when the electron is pull from the nucleus, that the electron has gain enough energy to escape the attraction of the nucleus.
However, the resonance frequency, undamped, provide information to obtain \({r}_{e}\), the atomic radius of the element concerned, because,
\(\omega_o =\sqrt{\cfrac{k}{m_e}}\)
\(\omega^2_o =\cfrac{c^2}{r_e}\)
\({r}_{e}=\left (\cfrac{c}{\omega_o}\right)^2\)
In conclusion, the x-axis of all black body plot is wrong. The observed spectrum spread and the driving force frequencies and observed color of the spectrum are not the same thing.
The damping factor in relation to temperature needs to be further investigated. It seems \(p\) is inversely proportional to \(T\).
It should be noted that \(n_p\), the refractive index of the prism used should be high. When \(n_T\) is bigger than \(n_p\) the direction of deflection reverses, and the spectrum is refracted away from the normal at the point of incidence and towards the normal at the point of exit.
Have a nice day.
\(k=\cfrac{{m}_{e}c^2}{{r}^2_{e}}\)
It is natural then to consider damped forced motion and the associated resonance. But what would be the driving function? Photons on passing, blob the electrons up and down, but there is another driving force, an increase in temperature, \(T\). Increasing temperature increases random motions across a wide range of frequency. As a driving force, it acts over a wide range of frequency at the same time. The resulting response under such a driving force is also simultaneously over the same range of frequency. So, we have, based on classical mechanics,
\(x^{''}+2px^{'}+\omega^2_ox={T}cos(\omega t)\)
where \(\omega_o =\sqrt{\cfrac{k}{m_e}}\) and \(p=\sqrt{\cfrac{k_1}{2m_e}}\), \(k_1\) the damping factor is to be to be determined and given physical interpretation later. We will used a solution of the form,
\(x=\cfrac{T}{\sqrt{(2\omega p)^2+(\omega^2_o-\omega^2)^2}}cos(\omega t-\gamma)\)
where \(tan(\gamma)=\cfrac{2\omega p}{(\omega^2_o-\omega^2)}\)
The electrons at increase temperature are driven to an amplitude above it normal orbital position. It has a maximum potential energy of (\(\cfrac{1}{2}kx^2\)),
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { { m }_{ e }c^{ 2 } }{ { r }^{ 2 }_{ e } } .\left\{ \cfrac { T }{ \sqrt { (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } } \right\} ^{ 2 }\)
\(PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e }} .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
They fall from such amplitude and emits heavy light of energy (\({m}_{e}{v}^2\)) equals to this amount. Therefore,
\({m}_{e}v^2=PE_{ e }=\cfrac { 1 }{ 2 } \cfrac { m_{ e }T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e } } .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
\({ v }^{ 2 }=\cfrac { 1 }{ 2 } \cfrac { T^{ 2 }c^{ 2 } }{{ r }^{ 2 }_{ e } } .\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } } \)
A plot \(\cfrac { 1 }{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 } }\) is shown below on the left. The actual parameter used are \({r}_{e}\) = 53e-12 m for hydrogen, \(c\) = 299792458 ms-1,
The actual graph plotted is, 10^7*1/((2*(x*100)*a)^2+(299.792458^2/53-(x*100)^2)^2) on the left, where \(a\) range from 0.5 to 20.5 in increment of 4. The y-axis is proportional to energy \(\) and the x-axis, \(\omega\). \(p\) is the damping factor per unit mass, which is related to the \(T\), equivalently the velocity of the electrons, and the damping force.
We can form the refractive index,
\(\cfrac { c }{ v } ={ n }_{ T }=\sqrt { 2 }\cfrac{1 }{ T } r_{ e } .\sqrt { \{ (2\omega p)^{ 2 }+(\omega ^{ 2 }_{ o }-\omega ^{ 2 })^{ 2 }\} } \)
A plot of \(n_T\) vs \(\omega\), ((2*(x*1000)*20)^2+(299.792458^2/53-(x*1000)^2)^2)^(1/2)*10^(-15) is shown below. The value of \(n_T\) increases with \(\omega\). This is the reverse of the case for spectra line where \(n_n\) decreases with increasing \(n\).
The amount of deflection, \(\theta\) at one medium interface \(n_T | n_p\) is
\(n_T sin(\theta_i)=n_p sin(\theta_r)\)and at the exit interface \(n_p | n_T\),
\(n_p sin(\Phi-\theta_r)=n_T sin(\theta_e)\),
where \(\Phi\) is the angle between the two interfaces. As \(n_T\) increases \(\theta_e\) decreases and the spectrum is sweep from right to left towards the normal. This is the reverse of spectra lines.
And at resonance, we have,
\(\omega=\sqrt{\omega^2_o-2p^2}\)
\(f=\cfrac{1}{2\pi}\sqrt{\cfrac{{c^2}}{{r}^2_{e}}-2p^2}\)
This frequency is however, not the frequency of the emitted light spectrum. The spectrum spread after the prism is the direct result of increasing speed of the electrons with increasing \(\omega\). The relative speed of the electrons just before and after the prism determines the amount of deflection and the direction of deflection relative to the normal at the point of incidence. High temperature drive the system over a wide spread of frequencies, one of which is this resonance value that corresponds to a high energy output (max. amplitude). The spread of velocities after the prism is related to the driving frequency due to the raised temperature, but they are not equivalent. One driving frequency gives raise to a particular amplitude of oscillation and so a value of velocity and a particular deflection point on the screen. All such driving frequencies are present at the same time with raised temperature. This driving frequency, \(\omega\) is not unlimited and neither is the \(x\), the amplitude of the oscillation. Both are valid up to point when the electron is pull from the nucleus, that the electron has gain enough energy to escape the attraction of the nucleus.
However, the resonance frequency, undamped, provide information to obtain \({r}_{e}\), the atomic radius of the element concerned, because,
\(\omega_o =\sqrt{\cfrac{k}{m_e}}\)
\(\omega^2_o =\cfrac{c^2}{r_e}\)
\({r}_{e}=\left (\cfrac{c}{\omega_o}\right)^2\)
In conclusion, the x-axis of all black body plot is wrong. The observed spectrum spread and the driving force frequencies and observed color of the spectrum are not the same thing.
The damping factor in relation to temperature needs to be further investigated. It seems \(p\) is inversely proportional to \(T\).
It should be noted that \(n_p\), the refractive index of the prism used should be high. When \(n_T\) is bigger than \(n_p\) the direction of deflection reverses, and the spectrum is refracted away from the normal at the point of incidence and towards the normal at the point of exit.
Have a nice day.
Tuesday, June 10, 2014
Frequently Troubled By Frequencies, Heavy Light
How then is \(\Delta PE\) related to photon frequencies? Are we back to square one?
\(\Delta PE_{n}= { m }_{ e }c^{ 2 }\left\{1 -\cfrac { 1 }{ { {n}^{ 2 } } } \right\}\)
It is not related to photon frequency, but it behaves like light in that the electron move in a helical path characterized by frequency, \(f\), a wavelength \(\lambda\), and a radius of circular motion, \(r\) and a speed lower than light speed, \(v\)
The total energy of such a particle is,
\(v=\lambda.f\), if the electron circular velocity is also \(v\) then \(v=2\pi r.f\)
\(E_e=\cfrac{1}{2}{m}_{e}{v_e}^2+\cfrac{1}{2}{m}_{e}{v_e}^2\)
\(E_e={m}_{e}{v_e}^2\)
So,
\( { m }_{ e }{v_e}^2={ m }_{ e }c^{ 2 }\left\{ 1-\cfrac { 1 }{ { { n }^{ 2 } } } \right\}\)
\( {v_e}^2=c^{ 2 }\left\{ 1-\cfrac { 1 }{ { { n }^{ 2 } } } \right\} \)
If we formulate,
\(\cfrac{c}{{v_e}} = \begin{matrix} 1 \\ \overline { \sqrt{ 1-\cfrac { 1 }{ { n }^{ 2 } }}} \end{matrix}=n_n\)
then we see that "this pretending to be light ray" is travelling at a low speed as if light traveling in a medium with reflective index given by the expression above. We know that refraction depends on the relative speed of the light before entering and after entering the medium boundary. The greater the difference in speed the greater the deflection. Equivalently, the greater the difference in refractive index between the two mediums the greater the deflection on screen. The sign of the differences determine whether the refraction is away or towards the normal at the point of incident. A plot of this \(n_n\) is given below, n is discrete, \(n=2,3,4,5...\)
So, given a prism of refractive index, \(n_p\), all spectra lines with \(n_n\) less than \(n_p\) will refract towards the normal and all spectra lines with \(n_n\) greater than \(n_p\) will reflect away from the normal. Since \(n\) is discrete, we then see discrete lines spaced apart, all deflected to different angles depending on \(n_n\).
It also means, according to this model, that sometimes, part of spectra lines (corresponding to higher values of \(n\)) are positioned before other lines with lower \(n\), because \(n_n\) is now smaller than the prism refractive index \(n_p\). All such lines with smaller \(n_n\) will be deflected backwards and overlap the lines that comes before them. The direction of deflection on screen reverses as \(n_n\) decreases with increasing \(n\) below \(n_p\), the refractive index of the prism used.
The question is; do we sense such electrons as light with all its colors? This Heavy Light are electrons not photons. It is too heavy to be diffracted on a diffraction grating. The channels formed in space by the grating will not channel the electrons at high speed to form patterns on screen.
Heavy light and we are not on square one.
\(\Delta PE_{n}= { m }_{ e }c^{ 2 }\left\{1 -\cfrac { 1 }{ { {n}^{ 2 } } } \right\}\)
It is not related to photon frequency, but it behaves like light in that the electron move in a helical path characterized by frequency, \(f\), a wavelength \(\lambda\), and a radius of circular motion, \(r\) and a speed lower than light speed, \(v\)
The total energy of such a particle is,
\(v=\lambda.f\), if the electron circular velocity is also \(v\) then \(v=2\pi r.f\)
\(E_e=\cfrac{1}{2}{m}_{e}{v_e}^2+\cfrac{1}{2}{m}_{e}{v_e}^2\)
\(E_e={m}_{e}{v_e}^2\)
So,
\( { m }_{ e }{v_e}^2={ m }_{ e }c^{ 2 }\left\{ 1-\cfrac { 1 }{ { { n }^{ 2 } } } \right\}\)
\( {v_e}^2=c^{ 2 }\left\{ 1-\cfrac { 1 }{ { { n }^{ 2 } } } \right\} \)
If we formulate,
\(\cfrac{c}{{v_e}} = \begin{matrix} 1 \\ \overline { \sqrt{ 1-\cfrac { 1 }{ { n }^{ 2 } }}} \end{matrix}=n_n\)
then we see that "this pretending to be light ray" is travelling at a low speed as if light traveling in a medium with reflective index given by the expression above. We know that refraction depends on the relative speed of the light before entering and after entering the medium boundary. The greater the difference in speed the greater the deflection. Equivalently, the greater the difference in refractive index between the two mediums the greater the deflection on screen. The sign of the differences determine whether the refraction is away or towards the normal at the point of incident. A plot of this \(n_n\) is given below, n is discrete, \(n=2,3,4,5...\)
So, given a prism of refractive index, \(n_p\), all spectra lines with \(n_n\) less than \(n_p\) will refract towards the normal and all spectra lines with \(n_n\) greater than \(n_p\) will reflect away from the normal. Since \(n\) is discrete, we then see discrete lines spaced apart, all deflected to different angles depending on \(n_n\).
It also means, according to this model, that sometimes, part of spectra lines (corresponding to higher values of \(n\)) are positioned before other lines with lower \(n\), because \(n_n\) is now smaller than the prism refractive index \(n_p\). All such lines with smaller \(n_n\) will be deflected backwards and overlap the lines that comes before them. The direction of deflection on screen reverses as \(n_n\) decreases with increasing \(n\) below \(n_p\), the refractive index of the prism used.
The question is; do we sense such electrons as light with all its colors? This Heavy Light are electrons not photons. It is too heavy to be diffracted on a diffraction grating. The channels formed in space by the grating will not channel the electrons at high speed to form patterns on screen.
Heavy light and we are not on square one.
Seriously Series, Spectra
Consider the case of a single positive charge center and a single negative charge in circular motion at speed \(c\).
\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}\)
In general for a electrically neutral system of \(2n\) charges, \(n\) positive and \(n\) negative,
\(\cfrac{{m}_{e}c^2}{{r}_{n}}=\cfrac{n^2q^2}{4\pi\varepsilon_o{r}^2_{n}}\)
\({r}_{n}=\cfrac{n^2q^2}{4\pi\varepsilon_o{m}_{e}c^2}\)
\({ r }_{ n+1 }=\cfrac { (n+1)^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \)
obviously \({ r }_{ n+1 } \gt { r }_{ n}\), but the corresponding potential energy term,
\(PE_{n}= \cfrac{n^2q^2}{4\pi\varepsilon_o{r}_{n}}={m}_{e}c^2\) which is a constant
Suppose a nucleus of \(n+1\) positive charge, emits a positive-negative charge pair, both charges leave the atom, then \(n\) negative charges will be caught at a higher orbit. This can be the case of disassociated hydrogen gas where a number of the ionized atoms come together to for a loose nucleus, under high temperature.
The potential energy difference is,
\(\Delta PE=\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ n } }-\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ n+1 } } \)
\(=\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o } } \left\{ \cfrac { 1 }{ { r }_{ n } } -\cfrac { 1 }{ { r }_{ n+1 } } \right\} \)
\(=n^{ 2 }{ m }_{ e }c^{ 2 }\left\{ \cfrac { 1 }{ n^{ 2 } } -\cfrac { 1 }{ { (n+1)^{ 2 } } } \right\} \)
In general, for \(n_1\) remaining positive charges from a nucleus originally containing \(n_2\) positive charges,
\(\Delta PE_{n_1,n_2}= {n_1}^{ 2 }{ m }_{ e }c^{ 2 }\left\{ \cfrac { 1 }{ {n_1}^{ 2 } } -\cfrac { 1 }{ { {n_2}^{ 2 } } } \right\} \)
obviously \(n_1 \lt n_2 \). The final position of the negative charge has a lower potential than before the charge pair leaves the atom. The speed of the electrons at both positions is \(c\), unchanged. So, the energy difference \(\Delta PE\) is totally emitted, as with the charge pair. Where did this energy go? How did the energy went?
If similar process happens for other elements of higher atomic numbers, where bare nuclei come together to form loose positive charge centers, than it is expected that \({n}_{2}\) be a multiple of \({n}_{1}\). ie.
\(\Delta PE_{n}= { m }_{ e }c^{ 2 }\left\{1 -\cfrac { 1 }{ { {n}^{ 2 } } } \right\} \)
where \(n_2=n.n_1\) and \(n=2,3,4...\)
In this case, \(\Delta PE_n\) seems to have lost its atomic number signature. The last expression assume that the nucleus of the elements are not restricted by their geometry and come together freely. In actual cases, some \(n\) are not possible for some elements which gives that element its characteristic spectra lines.
\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}\)
In general for a electrically neutral system of \(2n\) charges, \(n\) positive and \(n\) negative,
\(\cfrac{{m}_{e}c^2}{{r}_{n}}=\cfrac{n^2q^2}{4\pi\varepsilon_o{r}^2_{n}}\)
\({r}_{n}=\cfrac{n^2q^2}{4\pi\varepsilon_o{m}_{e}c^2}\)
\({ r }_{ n+1 }=\cfrac { (n+1)^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \)
obviously \({ r }_{ n+1 } \gt { r }_{ n}\), but the corresponding potential energy term,
\(PE_{n}= \cfrac{n^2q^2}{4\pi\varepsilon_o{r}_{n}}={m}_{e}c^2\) which is a constant
Suppose a nucleus of \(n+1\) positive charge, emits a positive-negative charge pair, both charges leave the atom, then \(n\) negative charges will be caught at a higher orbit. This can be the case of disassociated hydrogen gas where a number of the ionized atoms come together to for a loose nucleus, under high temperature.
The potential energy difference is,
\(\Delta PE=\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ n } }-\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ n+1 } } \)
\(=\cfrac { n^{ 2 }q^{ 2 } }{ 4\pi \varepsilon _{ o } } \left\{ \cfrac { 1 }{ { r }_{ n } } -\cfrac { 1 }{ { r }_{ n+1 } } \right\} \)
\(=n^{ 2 }{ m }_{ e }c^{ 2 }\left\{ \cfrac { 1 }{ n^{ 2 } } -\cfrac { 1 }{ { (n+1)^{ 2 } } } \right\} \)
In general, for \(n_1\) remaining positive charges from a nucleus originally containing \(n_2\) positive charges,
\(\Delta PE_{n_1,n_2}= {n_1}^{ 2 }{ m }_{ e }c^{ 2 }\left\{ \cfrac { 1 }{ {n_1}^{ 2 } } -\cfrac { 1 }{ { {n_2}^{ 2 } } } \right\} \)
obviously \(n_1 \lt n_2 \). The final position of the negative charge has a lower potential than before the charge pair leaves the atom. The speed of the electrons at both positions is \(c\), unchanged. So, the energy difference \(\Delta PE\) is totally emitted, as with the charge pair. Where did this energy go? How did the energy went?
If similar process happens for other elements of higher atomic numbers, where bare nuclei come together to form loose positive charge centers, than it is expected that \({n}_{2}\) be a multiple of \({n}_{1}\). ie.
\(\Delta PE_{n}= { m }_{ e }c^{ 2 }\left\{1 -\cfrac { 1 }{ { {n}^{ 2 } } } \right\} \)
where \(n_2=n.n_1\) and \(n=2,3,4...\)
In this case, \(\Delta PE_n\) seems to have lost its atomic number signature. The last expression assume that the nucleus of the elements are not restricted by their geometry and come together freely. In actual cases, some \(n\) are not possible for some elements which gives that element its characteristic spectra lines.
Hydrogen, And Everyone Has His Spins
For hydrogen,
Consider the forces on the hydrogen nulceus,
\(\cfrac{q.q}{4\pi\varepsilon_o(2{r}_{pe}sin(\theta))^2}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}sin(\theta)\)
Both electrons are in circular motion about the axis joining the two positive nucleus at radius \(\cfrac{1}{2}{r}_{ee}\),
\(\cfrac{2{m}_{e}c^2}{{r}_{ee}}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}.cos(\theta)-\cfrac{q.q}{4\pi\varepsilon_o{r}^{2}_{ee}}\)
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ pe }cos(\theta ) } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }(2{ r }_{ pe }cos(\theta ))^{ 2 } } \)
\( { r }_{ pe }=\cfrac { q^{ 2 }cos(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos(\theta )-\cfrac { 1 }{ (2cos(\theta ))^{ 2 } } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { \sqrt { 3 } }{ 2 } \{ 2\cfrac { \sqrt { 3 } }{ 2 } -\cfrac { 1 }{ 3 } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 6 } \} \)
\( { r }_{ pe }\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3/2-(3)^(0.5)/6)
\( { r }_{ pe }\) = 3.41e-15 m, \( { r }_{ ee }\) = 5.91e-15 m, \(\theta\) = 30o
The problem with this calculation is, why should not the hydrogen be in circular motion also? If the hydrogen nucleus is in circular motion about the axis joining the two electrons, then instead of the first equation,
\(\cfrac{{m}_{pr}c^2}{{r}_{pe}sin(\theta)}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}sin(\theta)-\cfrac{q.q}{4\pi\varepsilon_o(2{r}_{pe}sin(\theta))^2}\)
\({ r }_{ pe }=\cfrac { q^{ 2 }sin(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ pr }c^{ 2 } } \{ 2sin(\theta )-\cfrac { 1 }{ (2sin(\theta ))^{ 2 } } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ pr }c^{ 2 } } \{ 2sin^{ 2 }(\theta )-\cfrac { 1 }{ 4sin(\theta ) } \} \) ----(*)
Together with,
\({r}_{ee}=2{r}_{pe}cos(\theta)\) and,
\(\cfrac { 2{ m }_{ e }c^{ 2 } }{ { r }_{ ee } } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ ee } } \)
or,
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ pe }cos(\theta ) } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }(2{ r }_{ pe }cos(\theta ))^{ 2 } } \)
\( { r }_{ pe }=\cfrac { q^{ 2 }cos(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos(\theta )-\cfrac { 1 }{ (2cos(\theta ))^{ 2 } } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos^2(\theta )-\cfrac { 1 }{ 4cos(\theta ) } \} \) ----(**)
Dividing (*) b (**)
\(1=\cfrac{{m}_{e}}{{m}_{pr}}\begin{matrix}{2sin^{ 2 }(\theta )-\cfrac { 1 }{ 4sin(\theta ) }} \\ \overline { {2cos^2(\theta )-\frac { 1 }{ 4cos(\theta ) }}} \end{matrix}\)
\(\cfrac{{m}_{pr}}{{m}_{e}}\) = 1.67262178e-27/(9.10938291e-31) = 1836.152
If we let \(x=sin(\theta)\),
\(1836.152=\begin{matrix}{2x^{ 2 }-\cfrac { 1 }{ 4x}} \\ \overline { {2(1-x^2)-\cfrac { 1 }{ 4\sqrt{1-x^2} }}} \end{matrix} \)
\(x\) has 3 solutions,
\(x\) = -0.0000778025 \(x\) = -0.865838,
\(\theta\) = -0.00445775489o and -60o
and
\(x\) = sin(\theta) = 0.865898, \(\theta\) = 59.985o
This is twice the angle when the positive charges were considered not spinning.
\({r}_{pe}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*1.67262178e-27*299792458^2)*(2*(0.865898)^2-1/(4*0.865898))
\({r}_{pe}\) = 1.85e-18 m, \({r}_{ee}\) = 2*1.85e-18*(1-(0.865898)^2)^(0.5) = 1.85e-18 m
This distances are both shorter than the electron orbit a hydrogen atom by itself. The configuration is a two equilateral triangles with a common side. Both electrons and proton are in circular motion at light speed in planes that are perpendicular to each other, about a common center. Would they collide?
Distance between positive charges \({r}_{pp}\) is,
\({r}_{pp}=2*sin(\theta)*{r}_{pe}\) = 2*0.865898*1.85e-18 = 3.20e-18 m
The charges will always keep their distances and spin about the common center that is also their C.G.
Consider the forces on the hydrogen nulceus,
\(\cfrac{q.q}{4\pi\varepsilon_o(2{r}_{pe}sin(\theta))^2}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}sin(\theta)\)
\(\cfrac { 1 }{ (2sin(\theta ))^{ 2 } } =2sin(\theta )\)
\( sin(\theta )=\cfrac { 1 }{ 2 } ,\quad \quad \quad cos(\theta )=\cfrac { \sqrt { 3 } }{ 2 } , \theta=30^o\)
\({r}_{ee}=2{r}_{pe}cos(\theta)=\sqrt{3}{r}_{pe}\)
Both electrons are in circular motion about the axis joining the two positive nucleus at radius \(\cfrac{1}{2}{r}_{ee}\),
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ pe }cos(\theta ) } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }(2{ r }_{ pe }cos(\theta ))^{ 2 } } \)
\( { r }_{ pe }=\cfrac { q^{ 2 }cos(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos(\theta )-\cfrac { 1 }{ (2cos(\theta ))^{ 2 } } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { \sqrt { 3 } }{ 2 } \{ 2\cfrac { \sqrt { 3 } }{ 2 } -\cfrac { 1 }{ 3 } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 6 } \} \)
\( { r }_{ pe }\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3/2-(3)^(0.5)/6)
\( { r }_{ pe }\) = 3.41e-15 m, \( { r }_{ ee }\) = 5.91e-15 m, \(\theta\) = 30o
The problem with this calculation is, why should not the hydrogen be in circular motion also? If the hydrogen nucleus is in circular motion about the axis joining the two electrons, then instead of the first equation,
\(\cfrac{{m}_{pr}c^2}{{r}_{pe}sin(\theta)}=2\cfrac{q.q}{4\pi\varepsilon_o{r}^2_{pe}}sin(\theta)-\cfrac{q.q}{4\pi\varepsilon_o(2{r}_{pe}sin(\theta))^2}\)
\({ r }_{ pe }=\cfrac { q^{ 2 }sin(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ pr }c^{ 2 } } \{ 2sin(\theta )-\cfrac { 1 }{ (2sin(\theta ))^{ 2 } } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ pr }c^{ 2 } } \{ 2sin^{ 2 }(\theta )-\cfrac { 1 }{ 4sin(\theta ) } \} \) ----(*)
Together with,
\({r}_{ee}=2{r}_{pe}cos(\theta)\) and,
\(\cfrac { 2{ m }_{ e }c^{ 2 } }{ { r }_{ ee } } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ ee } } \)
or,
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ pe }cos(\theta ) } =2\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ pe } } .cos(\theta )-\cfrac { q.q }{ 4\pi \varepsilon _{ o }(2{ r }_{ pe }cos(\theta ))^{ 2 } } \)
\( { r }_{ pe }=\cfrac { q^{ 2 }cos(\theta ) }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos(\theta )-\cfrac { 1 }{ (2cos(\theta ))^{ 2 } } \} \)
\( { r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \{ 2cos^2(\theta )-\cfrac { 1 }{ 4cos(\theta ) } \} \) ----(**)
Dividing (*) b (**)
\(1=\cfrac{{m}_{e}}{{m}_{pr}}\begin{matrix}{2sin^{ 2 }(\theta )-\cfrac { 1 }{ 4sin(\theta ) }} \\ \overline { {2cos^2(\theta )-\frac { 1 }{ 4cos(\theta ) }}} \end{matrix}\)
\(\cfrac{{m}_{pr}}{{m}_{e}}\) = 1.67262178e-27/(9.10938291e-31) = 1836.152
If we let \(x=sin(\theta)\),
\(1836.152=\begin{matrix}{2x^{ 2 }-\cfrac { 1 }{ 4x}} \\ \overline { {2(1-x^2)-\cfrac { 1 }{ 4\sqrt{1-x^2} }}} \end{matrix} \)
\(x\) has 3 solutions,
\(x\) = -0.0000778025 \(x\) = -0.865838,
\(\theta\) = -0.00445775489o and -60o
and
\(x\) = sin(\theta) = 0.865898, \(\theta\) = 59.985o
This is twice the angle when the positive charges were considered not spinning.
\({r}_{pe}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*1.67262178e-27*299792458^2)*(2*(0.865898)^2-1/(4*0.865898))
\({r}_{pe}\) = 1.85e-18 m, \({r}_{ee}\) = 2*1.85e-18*(1-(0.865898)^2)^(0.5) = 1.85e-18 m
This distances are both shorter than the electron orbit a hydrogen atom by itself. The configuration is a two equilateral triangles with a common side. Both electrons and proton are in circular motion at light speed in planes that are perpendicular to each other, about a common center. Would they collide?
Distance between positive charges \({r}_{pp}\) is,
\({r}_{pp}=2*sin(\theta)*{r}_{pe}\) = 2*0.865898*1.85e-18 = 3.20e-18 m
The charges will always keep their distances and spin about the common center that is also their C.G.
Monday, June 9, 2014
And There Was Light, Upon the Words
In summary, a photon travels in a helical path, the radius of which is related to frequency by
\(2\pi.r=f\)
A frequency increases \(r\) decreases. When a photon approaches an orbiting electron, their interaction is electrostatic. The electron is pushed further into the nucleus and like a spring gains energy. When the photon passes, the electron bounces backup. If the stored energy is greater than ionization energy, (ie. the electron's PE at normal orbital radius) the electron is ejected from the atom.
The electron is at light speed around the nucleus. When pushed further into the nucleus, its speed tends to propels it outwards, back to normal orbit. A force exist that oppose the motion of the electron downwards. Such a force is represented by,
\({F}_{s}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\Delta {r}_{e}\)
where \(\Delta {r}_{e}\) is the displacement downward along the orbital radius from the normal radius \({r}_{e}\). (Normally, \(\Delta {r}_{e}\) is define positve upwards. In which case a negative sign is required on the RHS.) The work done against this force is the energy stored. This stored energy is responsible for ejecting the electron, after the photon has passed.
Let's look at this force further. Like a spring,
\(k=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\)
and resonance is at
\({f}_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{c}{2\pi{r}_{e}}\), \(m={m}_{e}\)
At this excitation frequency electrons are removed from the atom readily. When electrons are excited by a driving force at this frequency,\({f}_{res}\) with light at the threshold photon frequency, it will resonate. Short pulses at \({f}_{res}\) of light at the metal photoelectric threshold frequency.
A typical value of \({f}_{res}\) is,
For Iron, Fe,
\({f}_{res}\) = 299792458/(2*pi*140e-12) =3.4081e17 Hz, over 340 PHz (peta Hz).
For Gold, Au,
\({f}_{res}\) = 299792458/(2*pi*135e-12) =3.5343e17 Hz, over 353 PHz.
For Copper, Cu,
\({f}_{res}\) = 299792458/(2*pi*135e-12) =3.5343e17 Hz, over 353 PHz.
If this range of oscillation is possible, metal will melt in light.
\(2\pi.r=f\)
A frequency increases \(r\) decreases. When a photon approaches an orbiting electron, their interaction is electrostatic. The electron is pushed further into the nucleus and like a spring gains energy. When the photon passes, the electron bounces backup. If the stored energy is greater than ionization energy, (ie. the electron's PE at normal orbital radius) the electron is ejected from the atom.
The electron is at light speed around the nucleus. When pushed further into the nucleus, its speed tends to propels it outwards, back to normal orbit. A force exist that oppose the motion of the electron downwards. Such a force is represented by,
\({F}_{s}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\Delta {r}_{e}\)
Let's look at this force further. Like a spring,
\(k=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\)
and resonance is at
\({f}_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}=\cfrac{c}{2\pi{r}_{e}}\), \(m={m}_{e}\)
At this excitation frequency electrons are removed from the atom readily. When electrons are excited by a driving force at this frequency,\({f}_{res}\) with light at the threshold photon frequency, it will resonate. Short pulses at \({f}_{res}\) of light at the metal photoelectric threshold frequency.
A typical value of \({f}_{res}\) is,
For Iron, Fe,
\({f}_{res}\) = 299792458/(2*pi*140e-12) =3.4081e17 Hz, over 340 PHz (peta Hz).
For Gold, Au,
\({f}_{res}\) = 299792458/(2*pi*135e-12) =3.5343e17 Hz, over 353 PHz.
For Copper, Cu,
\({f}_{res}\) = 299792458/(2*pi*135e-12) =3.5343e17 Hz, over 353 PHz.
If this range of oscillation is possible, metal will melt in light.
Funny, Very Funny...
Strangely, ionization energy can be further simplified. Since,
\({ r }_{ ef }={ r }_{ eo }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
and
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}\}\)
We have,
\({ E }_{ s }={ m }_{ e }c^{ 2 }\{ ln{ (e^{ \cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }) }\} \)
\( { E }_{ s }={ m }_{ e }c^{ 2 }\{ \cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } \} \)
\( { E }_{ s }=\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o } } \cfrac { 1 }{ { r }_{ eo } } \)
which is very funny indeed.
\({ r }_{ ef }={ r }_{ eo }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
and
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}\}\)
We have,
\({ E }_{ s }={ m }_{ e }c^{ 2 }\{ ln{ (e^{ \cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }) }\} \)
\( { E }_{ s }={ m }_{ e }c^{ 2 }\{ \cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } \} \)
\( { E }_{ s }=\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o } } \cfrac { 1 }{ { r }_{ eo } } \)
which is very funny indeed.
Foresee, For C, For E
For the factor \(C\), we consider the potential energy in bringing a charge from infinity to its location \({r}_{eo}\) in the configuration around the positive charge along the line joining it and the positive charge.
In the case of electrostatic consideration,
\(PE=\lim_{r\rightarrow\infty}{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )}\)
For the case developed in the post "Like Wave, Like Particle, Not Attracted to Electrons"
\(PE={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r } }{ { r_{ eo } } } ) }+C({ r }_{ eo }-{ r })\} } \)
Equating the two,
\({ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{ { r } } { { r_{ eo } } } ) }+C({ r }_{ eo }-{ r })\} } =\lim_{r\rightarrow\infty}{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )}\)
\(\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r } } { r_{ eo } } ) }+C({ r }_{ eo }-{ r })\} } =\lim_{r\rightarrow\infty}{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )}\)
\(C=\lim _{ r\rightarrow \infty }{ \{ } \cfrac { 1 }{ ({ r }_{ eo }-{ r }) } \cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )-\cfrac { 1 }{ ({ r }_{ eo }-{ r }) } ln{ (\cfrac{ { r } } { { r_{ eo } } } ) }\} \)
\( C=\lim _{ r\rightarrow \infty }{ \{ } \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo }r } -\cfrac { 1 }{ ({ r }_{ eo }-{ r }) } ln{ (\cfrac { { r } }{ { r_{ eo } } } ) }\} \)
\( C=\lim _{ r\rightarrow \infty }{ \{ } -\cfrac { 1 }{ ({ r }_{ eo }-{ r }) } ln{ (\cfrac{ { r } } { { r_{ eo } } } ) }\}=0 \)
For the case developed in the post "Like Wave, Like Particle, Not Attracted to Electrons II"
\({ r }_{ ef }e^{ -C.{ r }_{ ef } }={ r }_{ eo }e^{ -C.{ r }_{ eo } }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
When \(C\) = 0,
\({ r }_{ ef }={ r }_{ eo }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
Which means \({r}_{ef}\) can be determined theoretically once the electron configuration is known (for \(A\)), because \({r}_{eo}\) can also be calculated given the electron configuration. And the ionization energy that was developed in the post "Like Wave, Like Particle, Not Attracted to Electrons", can also be obtained,
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}\}\)
since, \({r}_{ef}\lt{r}_{eo}\), \({E}_{s}\) is positive as expected.
In the case of electrostatic consideration,
\(PE=\lim_{r\rightarrow\infty}{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )}\)
For the case developed in the post "Like Wave, Like Particle, Not Attracted to Electrons"
\(PE={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r } }{ { r_{ eo } } } ) }+C({ r }_{ eo }-{ r })\} } \)
Equating the two,
\({ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{ { r } } { { r_{ eo } } } ) }+C({ r }_{ eo }-{ r })\} } =\lim_{r\rightarrow\infty}{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )}\)
\(\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r } } { r_{ eo } } ) }+C({ r }_{ eo }-{ r })\} } =\lim_{r\rightarrow\infty}{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )}\)
\(C=\lim _{ r\rightarrow \infty }{ \{ } \cfrac { 1 }{ ({ r }_{ eo }-{ r }) } \cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } (\cfrac { 1 }{ { r }_{ eo } } -\cfrac { 1 }{ r } )-\cfrac { 1 }{ ({ r }_{ eo }-{ r }) } ln{ (\cfrac{ { r } } { { r_{ eo } } } ) }\} \)
\( C=\lim _{ r\rightarrow \infty }{ \{ } \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo }r } -\cfrac { 1 }{ ({ r }_{ eo }-{ r }) } ln{ (\cfrac { { r } }{ { r_{ eo } } } ) }\} \)
\( C=\lim _{ r\rightarrow \infty }{ \{ } -\cfrac { 1 }{ ({ r }_{ eo }-{ r }) } ln{ (\cfrac{ { r } } { { r_{ eo } } } ) }\}=0 \)
For the case developed in the post "Like Wave, Like Particle, Not Attracted to Electrons II"
\({ r }_{ ef }e^{ -C.{ r }_{ ef } }={ r }_{ eo }e^{ -C.{ r }_{ eo } }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
When \(C\) = 0,
\({ r }_{ ef }={ r }_{ eo }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
Which means \({r}_{ef}\) can be determined theoretically once the electron configuration is known (for \(A\)), because \({r}_{eo}\) can also be calculated given the electron configuration. And the ionization energy that was developed in the post "Like Wave, Like Particle, Not Attracted to Electrons", can also be obtained,
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}\}\)
since, \({r}_{ef}\lt{r}_{eo}\), \({E}_{s}\) is positive as expected.
Sunday, June 8, 2014
Like Wave, Like Particle, Not Attracted to Electrons II
From the previous post "Like Wave, Like Particle, Not Attracted to Electrons", the energy in required in moving from \(r\) at \(\infty\) to \({r}_{eo}\) is
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r}}{{r_{eo} }})}+C({ r }_{ eo}-{ r })\}}\)
Energy required in moving from \(\infty\) to \({r}_{ef}\) is
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r}}{{r_{ef} }})}+C({ r }_{ ef}-{ r })\}}\)
So, the energy required in moving the electron from from \({r}_{eo}\) to \({r}_{ef}\) is
\(E_s =PE_e({r}_{ef})-PE_e({r}_{eo})\)
\(E_s={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r }}{{r}_{ef}})}-ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ ef}-{ r })-C({ r }_{ eo}-{ r })\}}\)
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r }_{ e o}}{{r}_{ef}})}+C({ r }_{ ef }-{ r }_{ e o})\}\)
which is the same expression as \({E}_{s}\) as before.
From previous calculation of atomic radius we find that, the centripetal force is of the form,
\(\cfrac{{m}_{e}c^2}{{r}_{eo}}=\cfrac{q^2}{4\pi\varepsilon_o {r}^2_{eo}}.(4-\cfrac{3}{\frac{8}{3}}).\cfrac{\sqrt{3}}{2\sqrt{2}}\cfrac{3}{2}\)
and
\(\cfrac{m_ec^2}{{r}_{eo}}= \cfrac{q^2}{4\pi\varepsilon_o{r}^2_{eo}}(3-\cfrac{\sqrt{3}}{3}) \)
In general,
\(\cfrac{m_ec^2}{{r}_{eo}}= \cfrac{q^2}{4\pi\varepsilon_o{r}^2_{eo}}.A\)
where \(A\) is a numerical constant dependent on the configuration of the electrons around the positive charge. Using this expression to bring a charge from infinity to its final configuration position along the line joining it and the positive center.
\(PE=-A\int^{{r}_{eo}}_{r\rightarrow\infty}{ \cfrac{-q^2}{4\pi\varepsilon_o{r}^2}d r}\)
\(PE= -\cfrac{q^2A}{4\pi\varepsilon_o}\{ \cfrac{1}{r}\}|^{{r}_{eo}}_{r\rightarrow\infty}\)
\(PE= -\cfrac{q^2A}{4\pi\varepsilon_o} \cfrac{1}{{r}_{eo}}\)
The negative sign indicates that this potential is lower than that at \(\infty\) which is zero. Numerically, the energy stored when the electron is pushed to a lower orbit, (photon and electron repel) is equal to this PE when the electron is ejected (ie the atom is ionized) after the photon has passed.
\(E_s ={ m }_{ e }c^{ 2 }\{ ln{( \cfrac{{r }_{ e o}}{{r}_{ef}})}+C({ r }_{ ef}-{ r }_{ eo})\}=\cfrac{q^2A}{4\pi\varepsilon_o} \frac{1}{{r}_{eo}}\)
\(ln{ ( \cfrac{{r }_{ e o}} {{r}_{ef}}e^{ C({ r }_{ ef }-{ r }_{ eo }) } })=\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } \)
\( \cfrac{{r }_{ e o}}{{r}_{ef}} e^{ C({ r }_{ ef }-{ r }_{ eo }) } =e^{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } }}\)
\({ r }_{ ef }e^{ -C.{ r }_{ ef } }={ r }_{ eo }e^{ -C.{ r }_{ eo } }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
because of the \(e^{ -\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\lt 1\) factor, and \({x}{e^{-Cx}}\) is monotonous increasing for \(x \lt 1\) for all values of \(C\). \({r}_{ef}\lt {r}_{eo}\) for \({r}\lt 1\). Which is an acceptable result.
Given an element the constant \(A\) is determined from distribution of electrons around the positive center. The factor \(C\)....
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r}}{{r_{eo} }})}+C({ r }_{ eo}-{ r })\}}\)
Energy required in moving from \(\infty\) to \({r}_{ef}\) is
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r}}{{r_{ef} }})}+C({ r }_{ ef}-{ r })\}}\)
So, the energy required in moving the electron from from \({r}_{eo}\) to \({r}_{ef}\) is
\(E_s =PE_e({r}_{ef})-PE_e({r}_{eo})\)
\(E_s={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r }}{{r}_{ef}})}-ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ ef}-{ r })-C({ r }_{ eo}-{ r })\}}\)
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r }_{ e o}}{{r}_{ef}})}+C({ r }_{ ef }-{ r }_{ e o})\}\)
which is the same expression as \({E}_{s}\) as before.
From previous calculation of atomic radius we find that, the centripetal force is of the form,
\(\cfrac{{m}_{e}c^2}{{r}_{eo}}=\cfrac{q^2}{4\pi\varepsilon_o {r}^2_{eo}}.(4-\cfrac{3}{\frac{8}{3}}).\cfrac{\sqrt{3}}{2\sqrt{2}}\cfrac{3}{2}\)
and
\(\cfrac{m_ec^2}{{r}_{eo}}= \cfrac{q^2}{4\pi\varepsilon_o{r}^2_{eo}}(3-\cfrac{\sqrt{3}}{3}) \)
In general,
\(\cfrac{m_ec^2}{{r}_{eo}}= \cfrac{q^2}{4\pi\varepsilon_o{r}^2_{eo}}.A\)
where \(A\) is a numerical constant dependent on the configuration of the electrons around the positive charge. Using this expression to bring a charge from infinity to its final configuration position along the line joining it and the positive center.
\(PE=-A\int^{{r}_{eo}}_{r\rightarrow\infty}{ \cfrac{-q^2}{4\pi\varepsilon_o{r}^2}d r}\)
\(PE= -\cfrac{q^2A}{4\pi\varepsilon_o}\{ \cfrac{1}{r}\}|^{{r}_{eo}}_{r\rightarrow\infty}\)
\(PE= -\cfrac{q^2A}{4\pi\varepsilon_o} \cfrac{1}{{r}_{eo}}\)
The negative sign indicates that this potential is lower than that at \(\infty\) which is zero. Numerically, the energy stored when the electron is pushed to a lower orbit, (photon and electron repel) is equal to this PE when the electron is ejected (ie the atom is ionized) after the photon has passed.
\(E_s ={ m }_{ e }c^{ 2 }\{ ln{( \cfrac{{r }_{ e o}}{{r}_{ef}})}+C({ r }_{ ef}-{ r }_{ eo})\}=\cfrac{q^2A}{4\pi\varepsilon_o} \frac{1}{{r}_{eo}}\)
\(ln{ ( \cfrac{{r }_{ e o}} {{r}_{ef}}e^{ C({ r }_{ ef }-{ r }_{ eo }) } })=\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } \)
\( \cfrac{{r }_{ e o}}{{r}_{ef}} e^{ C({ r }_{ ef }-{ r }_{ eo }) } =e^{\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } }}\)
\({ r }_{ ef }e^{ -C.{ r }_{ ef } }={ r }_{ eo }e^{ -C.{ r }_{ eo } }e^{ \cfrac { -q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\)
because of the \(e^{ -\cfrac { q^{ 2 }A }{ 4\pi \varepsilon _{ o }{ m }_{ e }c^{ 2 } } \cfrac { 1 }{ { r }_{ eo } } }\lt 1\) factor, and \({x}{e^{-Cx}}\) is monotonous increasing for \(x \lt 1\) for all values of \(C\). \({r}_{ef}\lt {r}_{eo}\) for \({r}\lt 1\). Which is an acceptable result.
Given an element the constant \(A\) is determined from distribution of electrons around the positive center. The factor \(C\)....
Like Wave, Like Particle, Not Attracted to Electrons
So far photon was treated as a particle, not wave. Photons interaction with electrons however is electrostatic, at a distance. This might seems that photon acts like wave. What is more important is that photons are accelerated in the direction of a reverse E-field; attracted to a positive charge. It should be repelled by electrons.
Does a electron really gets knocked out? If it does what actually happens?
Consider a electron in circular motion
\({F}_{c}=-{m}_{e}\cfrac{c^2}{{r}_{e}}\)
Light speed is the maximum, if this is an phenomenon of nature then, if \({r}_{e}\) is made smaller without changing \(c\), a force should develop to resist such a change in \({r}_{e}\). That is
\(\cfrac{\partial {F}_{c}}{\partial {r}_{e}}=-\cfrac{\partial}{\partial {r}_{e}}\left({m}_{e}\cfrac{c^2}{{r}_{e}}\right)\)
\(\cfrac{\partial {F}_{c}}{\partial {r}_{e}}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}} \) this force is in the same direction as \({r}_{e}\), when \({r}_{e}\) is compressed a force develops in the opposite direction, so
\({F}_{s}=\cfrac{\partial {F}_{c}}{\partial {r}_{e}}.\Delta {r}_{e}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\Delta {r}_{e}\)
where \(\Delta {r}_{e}\) is the small change in \({r}_{e}\) at \(r={r}_{e}\), in the negative \({r}_{e}\).
As the photon approaches, the electron from a position \(r={r}_{eo}\), is compressed to, \(r\rightarrow {r}_{ef}\), the electron behave just like a spring. After the photon passes, the stored energy is released and the electron is propelled upwards. If the store energy is greater than its potential energy, the electron leaves the pull of the nucleus. The atom is ionized.
Energy stored = \(\int^{{r}_{ef}}_{{r}_{eo}}{{F}_{s}}\partial {r}_{e}=\int^{{r}_{ef}}_{{r}_{eo}}{\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}(\partial {r}_{e})^2}\)
Energy stored, \(E_s ={ m }_{ e }{ c^{ 2 } }\int _{ {r}_{eo} }^{ { r }_{ ef } }{ (-\cfrac { 1 }{ { r }_{ e } } +C) \partial { r }_{ e } } \)
\(E_s ={ m }_{ e }{ c^{ 2 } }|\begin{matrix} { r }_{ ef } \\ {r}_{eo} \end{matrix}-ln({ r }_{ e })+C.{ r }_{ e }\)
\(E_s ={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\)
And ionization occurs when,
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\ge PE_e\) ----(*)
But what is potential energy of the electron orbiting at \({r}_{eo}\)? Potential Energy of the system is the energy stored when the electron is first placed at \(\infty\) and then moved to \({r}_{eo}\). ie. \({r}_{eo}\rightarrow \infty\) and \({r}_{ef}\rightarrow {r}_{eo}\)
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ eo}-{ r })\}}\) ----(**)
The idea is simply this, compress the spring to give it enough energy such that when it is release the electron will orbit at at a new orbit \(r \rightarrow\infty\). Consider (**) minus (*),
\({ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{r.{ r }_{ ef }}{r^2_{eo}}) }+C(-{ r }_{ ef }+2{ r }_{ eo }-r)\} }\)
\(={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{r.{ r }_{ ef }}{r^2_{eo}}e^{C(-{ r }_{ ef }+2{ r }_{ eo }-r)}) }\} }\)
\(\because \lim _{ r\rightarrow \infty }{r{ e }^{ -Cr } }\rightarrow 0\)
\(={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r }_{ ef } }{ r^{ 2 }_{ eo } } e^{ C(2{ r }_{ eo }-{ r }_{ ef }) }.r{ e }^{ -Cr } }\} } \rightarrow -\infty\)
ie. \(PE_s-E_s \lt 0\)
That is to say it is possible to ionize this way regardless of \({r}_{eo}\). That it is mathematically plausible. Given \({r}_{eo}\), there is an \({r}_{ef}\) to cause ionization.
Does a electron really gets knocked out? If it does what actually happens?
Consider a electron in circular motion
\({F}_{c}=-{m}_{e}\cfrac{c^2}{{r}_{e}}\)
Light speed is the maximum, if this is an phenomenon of nature then, if \({r}_{e}\) is made smaller without changing \(c\), a force should develop to resist such a change in \({r}_{e}\). That is
\(\cfrac{\partial {F}_{c}}{\partial {r}_{e}}=-\cfrac{\partial}{\partial {r}_{e}}\left({m}_{e}\cfrac{c^2}{{r}_{e}}\right)\)
\(\cfrac{\partial {F}_{c}}{\partial {r}_{e}}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}} \) this force is in the same direction as \({r}_{e}\), when \({r}_{e}\) is compressed a force develops in the opposite direction, so
\({F}_{s}=\cfrac{\partial {F}_{c}}{\partial {r}_{e}}.\Delta {r}_{e}=\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}\Delta {r}_{e}\)
where \(\Delta {r}_{e}\) is the small change in \({r}_{e}\) at \(r={r}_{e}\), in the negative \({r}_{e}\).
As the photon approaches, the electron from a position \(r={r}_{eo}\), is compressed to, \(r\rightarrow {r}_{ef}\), the electron behave just like a spring. After the photon passes, the stored energy is released and the electron is propelled upwards. If the store energy is greater than its potential energy, the electron leaves the pull of the nucleus. The atom is ionized.
Energy stored = \(\int^{{r}_{ef}}_{{r}_{eo}}{{F}_{s}}\partial {r}_{e}=\int^{{r}_{ef}}_{{r}_{eo}}{\cfrac{{m}_{e}{c^2}}{{r}^2_{e}}(\partial {r}_{e})^2}\)
Energy stored, \(E_s ={ m }_{ e }{ c^{ 2 } }\int _{ {r}_{eo} }^{ { r }_{ ef } }{ (-\cfrac { 1 }{ { r }_{ e } } +C) \partial { r }_{ e } } \)
\(E_s ={ m }_{ e }{ c^{ 2 } }|\begin{matrix} { r }_{ ef } \\ {r}_{eo} \end{matrix}-ln({ r }_{ e })+C.{ r }_{ e }\)
\(E_s ={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\)
And ionization occurs when,
\({E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\ge PE_e\) ----(*)
But what is potential energy of the electron orbiting at \({r}_{eo}\)? Potential Energy of the system is the energy stored when the electron is first placed at \(\infty\) and then moved to \({r}_{eo}\). ie. \({r}_{eo}\rightarrow \infty\) and \({r}_{ef}\rightarrow {r}_{eo}\)
\(PE_e={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ eo}-{ r })\}}\) ----(**)
The idea is simply this, compress the spring to give it enough energy such that when it is release the electron will orbit at at a new orbit \(r \rightarrow\infty\). Consider (**) minus (*),
\({ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{r.{ r }_{ ef }}{r^2_{eo}}) }+C(-{ r }_{ ef }+2{ r }_{ eo }-r)\} }\)
\(={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac{r.{ r }_{ ef }}{r^2_{eo}}e^{C(-{ r }_{ ef }+2{ r }_{ eo }-r)}) }\} }\)
\(\because \lim _{ r\rightarrow \infty }{r{ e }^{ -Cr } }\rightarrow 0\)
\(={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty }{ \{ ln{ (\cfrac { { r }_{ ef } }{ r^{ 2 }_{ eo } } e^{ C(2{ r }_{ eo }-{ r }_{ ef }) }.r{ e }^{ -Cr } }\} } \rightarrow -\infty\)
ie. \(PE_s-E_s \lt 0\)
That is to say it is possible to ionize this way regardless of \({r}_{eo}\). That it is mathematically plausible. Given \({r}_{eo}\), there is an \({r}_{ef}\) to cause ionization.
A Party of Four
Obviously,
That is why in a true vacuum, devoid of space itself, where charges loses their electrostatic property and uncoils from the nucleus of the atom, we see geometrical shape such as this one. So if Beryllium is placed inside a true vacuum, take a picture with the cap still on the lens, the film will capture a tetrahedron such as the one shown atop. Basically, the electrons that hits the metal containment generates X-rays. Does it work with a digital camera? May be not.
Since this is a tetrahedron,
\(\cfrac{{r}_{ee}}{{r}_{pe}}=\cfrac{2\sqrt{2}}{\sqrt{3}}\)
\({r}^2_{ee}=\cfrac{8}{3}{r}^2_{pe}\)
\(sin(\theta)=\cfrac{{r}_{pe}}{{r}_{ee}}.\cfrac{3}{2}=\cfrac{3\sqrt{3}}{4\sqrt{2}}\)
Consider centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q.4q}{4\pi\varepsilon_o{r}^2_{pe}}-\cfrac{3q.q}{4\pi\varepsilon_o{r}^2_{ee}}.sin(\theta)\)
\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q^2}{4\pi\varepsilon_o {r}^2_{pe}}.(4-\cfrac{3}{\frac{8}{3}}).\cfrac{\sqrt{3}}{2\sqrt{2}}\cfrac{3}{2}\)
\({{r}_{pe}}= \cfrac{q^2}{4\pi\varepsilon_o m_ec^2}(4-\cfrac{ 9 }{8}).\cfrac{3\sqrt{3}}{4\sqrt{2}} \)
\({{r}_{pe}}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(4-9/8)*(3)^(0.5)/(2)^(0.5)*(3/4))
\({r}_{pe}\) = 7.4418e-15 m
And the published atomic radius of Be, Beryllium is 112e-12 m.
That is why in a true vacuum, devoid of space itself, where charges loses their electrostatic property and uncoils from the nucleus of the atom, we see geometrical shape such as this one. So if Beryllium is placed inside a true vacuum, take a picture with the cap still on the lens, the film will capture a tetrahedron such as the one shown atop. Basically, the electrons that hits the metal containment generates X-rays. Does it work with a digital camera? May be not.
Since this is a tetrahedron,
\(\cfrac{{r}_{ee}}{{r}_{pe}}=\cfrac{2\sqrt{2}}{\sqrt{3}}\)
\({r}^2_{ee}=\cfrac{8}{3}{r}^2_{pe}\)
\(sin(\theta)=\cfrac{{r}_{pe}}{{r}_{ee}}.\cfrac{3}{2}=\cfrac{3\sqrt{3}}{4\sqrt{2}}\)
Consider centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q.4q}{4\pi\varepsilon_o{r}^2_{pe}}-\cfrac{3q.q}{4\pi\varepsilon_o{r}^2_{ee}}.sin(\theta)\)
\(\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q^2}{4\pi\varepsilon_o {r}^2_{pe}}.(4-\cfrac{3}{\frac{8}{3}}).\cfrac{\sqrt{3}}{2\sqrt{2}}\cfrac{3}{2}\)
\({{r}_{pe}}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(4-9/8)*(3)^(0.5)/(2)^(0.5)*(3/4))
And the published atomic radius of Be, Beryllium is 112e-12 m.
A Party of Three
From the case of three electrons,
\(2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) = \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}\)
\(\cfrac{2}{{r}^{2}_{ee}}\cfrac{\sqrt{3}}{2} = \cfrac{3}{{r}^2_{pe}}\)
\(\sqrt{3}{r}^2_{ee}={r}^2_{pe}\)
\(\sqrt[4]{3}{r}_{ee}={r}_{pe}\)
but from geometry,
\(2.{r}_{pe}cos(30^o)={r}_{ee}\)
\(\sqrt{3}{r}_{pe}={r}_{ee}\)
This means the configuration is not static, the electrons are in circular motion, instantaneously at least, in order not to fall into the positive charge. Let assume orbit speed is \(c\)
\(\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) \)
\(\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{3q^2}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{q^2}{4\pi\varepsilon_o 3{r}^{2}_{pe}} .\cfrac{\sqrt{3}}{2} \)
\({{r}_{pe}}= \cfrac{q^2}{4\pi\varepsilon_o m_ec^2}(3-\cfrac{\sqrt{3}}{3}) \)
\({{r}_{pe}}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3-(3)^(0.5)/3)
\({{r}_{pe}}\) = 6.8269e-15 m
But the atomic radius of Lithium. Li is 1.52e-10 m. There is one possibility, that the electrons are not orbiting a light speed \(c\). What speed are the electron orbiting at then? Then again electron mass has a relative uncertainty of 4.2e-10, so what of e-15. This value can not be more accurate than electron mass.
\(2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) = \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}\)
\(\cfrac{2}{{r}^{2}_{ee}}\cfrac{\sqrt{3}}{2} = \cfrac{3}{{r}^2_{pe}}\)
\(\sqrt{3}{r}^2_{ee}={r}^2_{pe}\)
\(\sqrt[4]{3}{r}_{ee}={r}_{pe}\)
but from geometry,
\(2.{r}_{pe}cos(30^o)={r}_{ee}\)
\(\sqrt{3}{r}_{pe}={r}_{ee}\)
This means the configuration is not static, the electrons are in circular motion, instantaneously at least, in order not to fall into the positive charge. Let assume orbit speed is \(c\)
\(\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) \)
\(\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{3q^2}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{q^2}{4\pi\varepsilon_o 3{r}^{2}_{pe}} .\cfrac{\sqrt{3}}{2} \)
\({{r}_{pe}}= \cfrac{q^2}{4\pi\varepsilon_o m_ec^2}(3-\cfrac{\sqrt{3}}{3}) \)
\({{r}_{pe}}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3-(3)^(0.5)/3)
But the atomic radius of Lithium. Li is 1.52e-10 m. There is one possibility, that the electrons are not orbiting a light speed \(c\). What speed are the electron orbiting at then? Then again electron mass has a relative uncertainty of 4.2e-10, so what of e-15. This value can not be more accurate than electron mass.
Saturday, June 7, 2014
Again About Electron Shell Again
There is one other case to consider,
where one of the electron is slightly displaced as shown. We see that resultant forces act on the electrons to keep them furthest apart at a distance of 2\({r}_{e}\).
So, for the case where the electrons are on opposite sides of the positive charge,
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ e } } =\cfrac { q.2q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ e } } -\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ (2{ r }_{ e })^{ 2 } } } \)
\({ { r }_{ e } } =\cfrac { 7q^2 }{ 16\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } } \)
\({ { r }_{ e } }\) = 7*(1.60217657e-19)^2/(16*pi*8.8541878176e-12*9.10938291e-31*299792458^2)
\({ { r }_{ e } }\) = 4.9314e-15 m
If we check this against the atomic radius of He, 31e-12 m. This value is much lower. Mass of the electrons are too small to consider gravitational pull. If we consider the fact that the electron charge may be higher because of the dialectic effects of oil in the oil drop experiment, \(e \rightarrow 3e\).
\({ { r }_{ e } }\) = 4.4382e-14 m
is still lower, than the published He atomic radius.
where one of the electron is slightly displaced as shown. We see that resultant forces act on the electrons to keep them furthest apart at a distance of 2\({r}_{e}\).
So, for the case where the electrons are on opposite sides of the positive charge,
\(\cfrac { { m }_{ e }c^{ 2 } }{ { r }_{ e } } =\cfrac { q.2q }{ 4\pi \varepsilon _{ o }{ r }^{ 2 }_{ e } } -\cfrac { q.q }{ 4\pi \varepsilon _{ o }{ (2{ r }_{ e })^{ 2 } } } \)
\({ { r }_{ e } } =\cfrac { 7q^2 }{ 16\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } } \)
\({ { r }_{ e } }\) = 7*(1.60217657e-19)^2/(16*pi*8.8541878176e-12*9.10938291e-31*299792458^2)
\({ { r }_{ e } }\) = 4.9314e-15 m
If we check this against the atomic radius of He, 31e-12 m. This value is much lower. Mass of the electrons are too small to consider gravitational pull. If we consider the fact that the electron charge may be higher because of the dialectic effects of oil in the oil drop experiment, \(e \rightarrow 3e\).
\({ { r }_{ e } }\) = 4.4382e-14 m
is still lower, than the published He atomic radius.
Again About Electron Shell
Consider two electrons in circular motion about a positive charge \(2q\), at distance \({r}_{e1}\) and \({r}_{e2}\). Both charges are at light speed \(c\), and can attain has no higher speed. At the extreme distances,
for \({r}_{e1}\), we consider the centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}+{r}_{e2})}^2}\)
for \({r}_{e2}\),
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}+{r}_{e2})}^2}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac {2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } } }{ \cfrac {2 }{ { r }^{ 2 }_{ e2 } } -\cfrac { 1 }{ { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } } } \)
\( \cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { { 2{ ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\(\cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { { 2{ ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\( x=\cfrac { 2(x+1)^{ 2 }-x^2 }{ 2(x+1)^{ 2 }-1 } \)
\(x\) has three solutions. The only positive solution is,
\(x\) = 1, \({r}_{e1}={r}_{e2}\)
this suggest that this configuration is possible.
And for the next extreme case, without lost of generality, \({r}_{e1}\gt{r}_{e2}\), the diagram on the right
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}-{r}_{e2})}^2}\)
and
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}+\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}-{r}_{e2})}^2}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac { 2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } } }{ \cfrac { 2 }{ { r }^{ 2 }_{ e2 } } +\cfrac { 1 }{ { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } } } \)
\( \cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }+{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\(\cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }+{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\( x=\cfrac {2 (x-1)^{ 2 }-x^2 }{2 (x-1)^{ 2 }+1 } \)
\(x\) also has three solution, a complex pair, and a positive solution,
\(x = 0.3684\), \({r}_{e1}=0.3684*{r}_{e2}\)
Since we have assumed that \({r}_{e1}\gt{r}_{e2}\), this solution is not admissible. This configuration is not possible. This might suggest that the electrons will always remain on the opposite side of the positive charge.
We will consider the limiting case of,
For \({r}_{e1}\), again we look at the centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}-\cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac { 2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ \sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) } }{ \cfrac { 2 }{ { r }^{ 2 }_{ e2 } } -\cfrac { 1 }{ { \sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) } } } \)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e1 } } }{ { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\( \cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e1 } } }{ { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\(x=\cfrac { { 2\sqrt { 2 } ({ x }^{ 2 }+1^{ 2 }) }-{ { x }^{ 2 } } }{ { 2\sqrt { 2 } ({ x }^{ 2 }+1^{ 2 }) }-1 }\)
\(x\) also has three solution, a complex pair, and a positive solution,
\(x =1\), \({r}_{e1} = {r}_{e2}\)
This is possible. In this case there is a tangential force component equal to
\({F}_{t}= \cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
but the speed of the electron will not increase further. Speed of the electron is at light speed \(c\). As the electron move relative to one another, the centripetal force changes and the orbital radius changes. If we now consider a small displacement \(\Delta a\) towards the other electron,
We see that the electrons will experience a force that drive them further apart. And the electrons will remain on opposite sides of the positive charge.
for \({r}_{e1}\), we consider the centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}+{r}_{e2})}^2}\)
for \({r}_{e2}\),
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}+{r}_{e2})}^2}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac {2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } } }{ \cfrac {2 }{ { r }^{ 2 }_{ e2 } } -\cfrac { 1 }{ { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } } } \)
\( \cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { { 2{ ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\(\cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { { 2{ ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }+{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\( x=\cfrac { 2(x+1)^{ 2 }-x^2 }{ 2(x+1)^{ 2 }-1 } \)
\(x\) has three solutions. The only positive solution is,
\(x\) = 1, \({r}_{e1}={r}_{e2}\)
this suggest that this configuration is possible.
And for the next extreme case, without lost of generality, \({r}_{e1}\gt{r}_{e2}\), the diagram on the right
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}-{r}_{e2})}^2}\)
and
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}+\cfrac{q.q}{4\pi\varepsilon_o {({r}_{e1}-{r}_{e2})}^2}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac { 2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } } }{ \cfrac { 2 }{ { r }^{ 2 }_{ e2 } } +\cfrac { 1 }{ { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } } } \)
\( \cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }+{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\(\cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }-{ { r }^{ 2 }_{ e1 } } }{ {2 { ({ r }_{ e1 }-{ r }_{ e2 }) }^{ 2 } }+{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\( x=\cfrac {2 (x-1)^{ 2 }-x^2 }{2 (x-1)^{ 2 }+1 } \)
\(x\) also has three solution, a complex pair, and a positive solution,
\(x = 0.3684\), \({r}_{e1}=0.3684*{r}_{e2}\)
Since we have assumed that \({r}_{e1}\gt{r}_{e2}\), this solution is not admissible. This configuration is not possible. This might suggest that the electrons will always remain on the opposite side of the positive charge.
We will consider the limiting case of,
For \({r}_{e1}\), again we look at the centripetal force,
\(\cfrac{{m}_{e}c^2}{{r}_{e1}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e1}}-\cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
\(\cfrac{{m}_{e}c^2}{{r}_{e2}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e2}}-\cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { \cfrac { 2 }{ { r }^{ 2 }_{ e1 } } -\cfrac { 1 }{ \sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) } }{ \cfrac { 2 }{ { r }^{ 2 }_{ e2 } } -\cfrac { 1 }{ { \sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) } } } \)
\(\cfrac { { r }_{ e2 } }{ { r }_{ e1 } } =\cfrac { { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e1 } } }{ { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e2 } } } .\cfrac { { r }^{ 2 }_{ e2 } }{ { r }^{ 2 }_{ e1 } } \)
\( \cfrac { { r }_{ e1 } }{ { r }_{ e2 } } =\cfrac { { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e1 } } }{ { 2\sqrt { 2 } ({ r }^{ 2 }_{ e1 }+{ r }^{ 2 }_{ e2 }) }-{ { r }^{ 2 }_{ e2 } } } \)
Let \(x=\cfrac{{r}_{e1}}{{r}_{e2}}\)
\(x=\cfrac { { 2\sqrt { 2 } ({ x }^{ 2 }+1^{ 2 }) }-{ { x }^{ 2 } } }{ { 2\sqrt { 2 } ({ x }^{ 2 }+1^{ 2 }) }-1 }\)
\(x\) also has three solution, a complex pair, and a positive solution,
\(x =1\), \({r}_{e1} = {r}_{e2}\)
This is possible. In this case there is a tangential force component equal to
\({F}_{t}= \cfrac{1}{\sqrt{2}}.\cfrac{q.q}{4\pi\varepsilon_o ({r}^2_{e1}+{r}^2_{e2})}\)
but the speed of the electron will not increase further. Speed of the electron is at light speed \(c\). As the electron move relative to one another, the centripetal force changes and the orbital radius changes. If we now consider a small displacement \(\Delta a\) towards the other electron,
We see that the electrons will experience a force that drive them further apart. And the electrons will remain on opposite sides of the positive charge.
Friday, June 6, 2014
About Electronic Shell
Most present electrons around a proton nucleus as shown where the electrons are furthest apart from each other. But if we write down the force equation based on simple electrostatic then.\(\cfrac{{q}_{e}{q}_{e}}{4\pi \varepsilon_o {r}^2_{ee}}=\cfrac{2{q}_{p}{q}_{e}}{4\pi \varepsilon_o {r}^2_{pe}}\)
if the basic positive and negative charge magnitudes are the same \({q}_{e}={q}_{p}\) then,
\(\cfrac{1}{{r}^2_{ee}}=\cfrac{2}{{r}^2_{pe}}\)and
\(\sqrt{2}{r}_{ee}={r}_{pe}\)
This is possible only if the electrons are all on the same side.
But, we see that the center negative charge, experiences a net force towards the positive end.
The configuration is impossible unless, the center charge can fall no further towards the positive charge.
Consider a negative charge accelerated by the pull of a positive charge, moves towards the charge at greater and greater speed. What if the negative charge reaches light speed before colliding with the positive charge? The negative charge will go into circular motion around the positive charge where the attractive force between the charges provides for the centripetal force. And the negative charge is at light speed. In this case,
\(-\cfrac{{m}_{e}c^2}{{r}_{e}}=-\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e}}-\cfrac{q.q}{4\pi\varepsilon_o {r}^2_{ee}}\)
\(\because {r}_{pe}={r}_{ee}+{r}_{e}\)
We have then,
\(\cfrac{{r}_{ee}+{r}_{e}}{{r}_{ee}}=\sqrt{2}\)
\({r}_{ee}=\cfrac{1}{\sqrt{2}-1}.{r}_{e}\)
\(\cfrac{{m}_{e}c^2}{{r}_{e}}=\cfrac{q.2q}{4\pi\varepsilon_o {r}^2_{e}}+\cfrac{q.q}{4\pi\varepsilon_o {r}^2_{e}}.({\sqrt{2}-1})^2\)
\( { m }_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ e } } (2+({ \sqrt { 2 } -1 })^{ 2 })\)
\({ m }_{ e }c^{ 2 }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ r }_{ e } } (5-2\sqrt { 2 } )\)
\({ r }_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } } (5-2\sqrt { 2 } )\)
\({ r }_{ ee }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } }\cfrac{ (5-2\sqrt { 2 } )}{\sqrt{2}-1}\)
\({ r }_{ pe }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }{ { m }_{ e }c^{ 2 } } }\cfrac{ (5-2\sqrt { 2 } )}{\sqrt{2}-1}.\sqrt{2}\)
We find that the resultant force on the furthest negative charge has a acceleration component towards the positive charge. This charge is still falling towards the positive charge. This configuration is stable only if this charge is also performing circular motion.
This invalidates the the first equation where the net force on this charge was considered zero. We will start again.
\({m}_{p}\), Just Madness Pal
There is a clear dependence on the value of \({r}_{e}\) in the table from the post "More Photon Mass, More Power!", for \({m}_{p}\). The value of \({m}_{p}\) decreases with increasing \({r}_{e}\).
Since Planck's Constant is not a constant in this model, all calculations based on a constant \(h\) will not be valid.
The most logically valid answer is,
\({m}_{p}\) = 6.2948e-34 kg
from the post "Light, indeed Light".
Since Planck's Constant is not a constant in this model, all calculations based on a constant \(h\) will not be valid.
The most logically valid answer is,
\({m}_{p}\) = 6.2948e-34 kg
from the post "Light, indeed Light".
More Photon Mass, More Power!
Below is a table of elements with known Work Function \(\Phi \) and Atomic radii \({r}_{e}\). Values of photon mass \({m}_{p}\) are calculated and averaged, using
\({m}_{p}= \cfrac{1}{c^2}.(\cfrac{h.c}{4{r}_{e}}+\Phi)\),
where Planck's Constant \(h\) = 4.135667516e-15 eV, light speed \(c\) = 299792458 m s-1
The average \({m}_{p}\) value is 3.74315E-033 kg. It is noted that C, Carbon, B, Boron, Si, Silicon and Be, Beryllium provided exceptionally high values of \({m}_{p}\) because of their low \({r}_{e}\) values. And Cs, Caesium provided the lowest value because of its high atomic radius \({r}_{e}\).
\({m}_{p}= \cfrac{1}{c^2}.(\cfrac{h.c}{4{r}_{e}}+\Phi)\),
where Planck's Constant \(h\) = 4.135667516e-15 eV, light speed \(c\) = 299792458 m s-1
| Atomic Symbol | Work Function \(\Phi \) | Atomic Radius \(r_e\) | \({m}_{p}\) |
| Ag | 4.260 | 160 | 3.46106E-033 |
| Al | 4.060 | 125 | 4.42768E-033 |
| As | 3.750 | 115 | 4.81151E-033 |
| Au | 5.100 | 135 | 4.10209E-033 |
| B | 4.450 | 85 | 6.50858E-033 |
| Ba | 2.520 | 215 | 2.57451E-033 |
| Be | 4.980 | 105 | 5.27130E-033 |
| Bi | 4.310 | 160 | 3.46115E-033 |
| C | 5.000 | 70 | 7.90255E-033 |
| Ca | 2.870 | 180 | 3.07486E-033 |
| Cd | 4.080 | 155 | 3.57214E-033 |
| Ce | 2.900 | 185 | 2.99195E-033 |
| Co | 5.000 | 135 | 4.10191E-033 |
| Cr | 4.500 | 140 | 3.95484E-033 |
| Cs | 2.140 | 260 | 2.12903E-033 |
| Cu | 4.530 | 135 | 4.10107E-033 |
| Eu | 2.500 | 185 | 2.99124E-033 |
| Fe | 4.670 | 140 | 3.95514E-033 |
| Ga | 4.320 | 130 | 4.25812E-033 |
| Gd | 2.900 | 180 | 3.07492E-033 |
| Hf | 3.900 | 155 | 3.57182E-033 |
| Hg | 4.475 | 150 | 3.69168E-033 |
| In | 4.090 | 155 | 3.57216E-033 |
| Ir | 5.000 | 135 | 4.10191E-033 |
| K | 2.290 | 220 | 2.51569E-033 |
| La | 3.500 | 195 | 2.83985E-033 |
| Li | 2.900 | 145 | 3.81589E-033 |
| Lu | 3.300 | 175 | 3.16334E-033 |
| Mg | 3.660 | 150 | 3.69022E-033 |
| Mn | 4.100 | 140 | 3.95413E-033 |
| Mo | 4.360 | 145 | 3.81849E-033 |
| Na | 2.360 | 180 | 3.07396E-033 |
| Nb | 3.950 | 145 | 3.81776E-033 |
| Nd | 3.200 | 185 | 2.99249E-033 |
| Ni | 5.040 | 135 | 4.10198E-033 |
| Os | 5.930 | 130 | 4.26099E-033 |
| Pb | 4.250 | 180 | 3.07732E-033 |
| Pd | 5.220 | 140 | 3.95612E-033 |
| Pt | 5.120 | 135 | 4.10213E-033 |
| Rb | 2.261 | 235 | 2.35533E-033 |
| Re | 4.720 | 135 | 4.10141E-033 |
| Rh | 4.980 | 135 | 4.10188E-033 |
| Ru | 4.710 | 130 | 4.25882E-033 |
| Sb | 4.550 | 145 | 3.81883E-033 |
| Sc | 3.500 | 160 | 3.45971E-033 |
| Se | 5.900 | 115 | 4.81534E-033 |
| Si | 4.600 | 110 | 5.03143E-033 |
| Sm | 2.700 | 185 | 2.99160E-033 |
| Sn | 4.420 | 145 | 3.81860E-033 |
| Sr | 2.590 | 200 | 2.76739E-033 |
| Ta | 4.000 | 145 | 3.81785E-033 |
| Tb | 3.000 | 225 | 2.46115E-033 |
| Te | 4.950 | 140 | 3.95564E-033 |
| Th | 3.400 | 180 | 3.07581E-033 |
| Ti | 4.330 | 140 | 3.95454E-033 |
| Tl | 3.840 | 190 | 2.91503E-033 |
| U | 3.630 | 175 | 3.16393E-033 |
| V | 4.300 | 135 | 4.10066E-033 |
| W | 4.320 | 135 | 4.10070E-033 |
| Y | 3.100 | 180 | 3.07527E-033 |
| Yb | 2.600 | 175 | 3.16209E-033 |
| Zn | 3.630 | 135 | 4.09947E-033 |
| Zr | 4.050 | 155 | 3.57209E-033 |
| Average | 3.74315E-033 | ||
| Std Deviation | 9.16011E-034 | ||
| Max | 7.90255E-033 | ||
| Min | 2.12903E-033 |
The average \({m}_{p}\) value is 3.74315E-033 kg. It is noted that C, Carbon, B, Boron, Si, Silicon and Be, Beryllium provided exceptionally high values of \({m}_{p}\) because of their low \({r}_{e}\) values. And Cs, Caesium provided the lowest value because of its high atomic radius \({r}_{e}\).
Three Times And I am Light
Since most have found \(h\) the Planck's Constant to be accurate at,
\(h\) = 4.135667516e-15 eV
If we assume that photon-electron interaction is the same in metals with a threshold frequency, as of photoelectric effect. That the mechanism of photon and electron collision is the same across such metals and so share the same probability function. Then,
\(E_{max}(f)=E[KE(f)]=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }.f=h.f\)
\(h=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }\)
\(KE_{max}=\cfrac{h.c}{4{r}_{e}}\)
and from
\(KE_{max}={m}_{p}c^2-\Phi\)
\({m}_{p}= \cfrac{{KE}_{max}+\Phi}{c^2}\)
\({m}_{p}= \cfrac{1}{c^2}.(\cfrac{h.c}{4{r}_{e}}+\Phi)\)
and so, using potassium as an example, \({r}_{e}\) = 235 pm and \(\Phi\) = 2.29 V,
\({m}_{p}\) = (4.135667516e-15*299792458/(4*235e-12)+2.29)/(299792458)^2*(1.60217657e-19)
\({m}_{p}\) = 2.3554e-33 kg
This is my third attempt at \({m}_{p}\). From this value we obtain,
Energy of a photon is
\({E}_{p}={m}_{p}c^2\) = (2.3554e-33)*299792458^2 = 2.1169e-16 J
Momentum of a photon is
\({P}_{p}={m}_{p}c\) = (2.3554e-33)*299792458 = 7.0613e-25 kgm s-1
Try once, try twice and maybe third time lucky.
\(h\) = 4.135667516e-15 eV
If we assume that photon-electron interaction is the same in metals with a threshold frequency, as of photoelectric effect. That the mechanism of photon and electron collision is the same across such metals and so share the same probability function. Then,
\(E_{max}(f)=E[KE(f)]=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }.f=h.f\)
\(h=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }\)
\(KE_{max}=\cfrac{h.c}{4{r}_{e}}\)
and from
\(KE_{max}={m}_{p}c^2-\Phi\)
\({m}_{p}= \cfrac{{KE}_{max}+\Phi}{c^2}\)
\({m}_{p}= \cfrac{1}{c^2}.(\cfrac{h.c}{4{r}_{e}}+\Phi)\)
and so, using potassium as an example, \({r}_{e}\) = 235 pm and \(\Phi\) = 2.29 V,
\({m}_{p}\) = (4.135667516e-15*299792458/(4*235e-12)+2.29)/(299792458)^2*(1.60217657e-19)
\({m}_{p}\) = 2.3554e-33 kg
This is my third attempt at \({m}_{p}\). From this value we obtain,
Energy of a photon is
\({E}_{p}={m}_{p}c^2\) = (2.3554e-33)*299792458^2 = 2.1169e-16 J
Momentum of a photon is
\({P}_{p}={m}_{p}c\) = (2.3554e-33)*299792458 = 7.0613e-25 kgm s-1
Try once, try twice and maybe third time lucky.
Not Oh P lausible
Energy of a photon is
\({E}_{p}={m}_{p}c^2\) = (6.2948e-34)*299792458^2 = 5.6575e-17 J
Momentum of a photon is
\({P}_{p}={m}_{p}c\) = (6.2948e-34)*299792458 = 1.8871e-25 kgm s-1
Mass of a photon is
\({m}_{p}\) = 6.2948e-34 kg
Charge on a photon dipole, which was derived using Planck's relation, is
\(q\) = 1.4745*10^(-43) C this may not be true.
Planck's relation is still true beyond the plateau region of the Electron E_max vs Photon Frequency graph in the post "Fodo Electric Effect"graph reproduced below,
However, \({E}_{max}\) dependence on \(f\) incident photon frequency, is probabilistic and statistical. Changing \(f\) changes the radius of the photon helix path and so changes the probability of colliding with an orbital electron. That in turns changes the mean value of the measured \({E}_{max}\).
Planck's relation is not a intrinsic property of photons. Photons have constant energy, \({m}_{p}c^2\).
We can check
\(h=4{ r }_{ e } {m}_{p}c-\cfrac { 4 }{ c }{r}_{e}.\Phi\)
derived in the post "Light Indeed Light" for the value of Planck's constant,
\(h\) = 4*235e-12*(6.2948e-34*299792458)/(1.60217657e-19)-4*235e-12*2.29/299792458
\(h\) = 1.1000e-15 eV s-1
Based on this model Planck's Constant is no longer a constant, and its calculation and the calculation for \({m}_{p}\) relies a lot on the collision model between photons and orbital electrons. The paper from which the data was taken estimates \(h\) via two methods producing values of (0.94+.48)e-15 and (2.92+0.77)e-15.
\({E}_{p}={m}_{p}c^2\) = (6.2948e-34)*299792458^2 = 5.6575e-17 J
Momentum of a photon is
\({P}_{p}={m}_{p}c\) = (6.2948e-34)*299792458 = 1.8871e-25 kgm s-1
Mass of a photon is
\({m}_{p}\) = 6.2948e-34 kg
Charge on a photon dipole, which was derived using Planck's relation, is
\(q\) = 1.4745*10^(-43) C this may not be true.
Planck's relation is still true beyond the plateau region of the Electron E_max vs Photon Frequency graph in the post "Fodo Electric Effect"graph reproduced below,
However, \({E}_{max}\) dependence on \(f\) incident photon frequency, is probabilistic and statistical. Changing \(f\) changes the radius of the photon helix path and so changes the probability of colliding with an orbital electron. That in turns changes the mean value of the measured \({E}_{max}\).
Planck's relation is not a intrinsic property of photons. Photons have constant energy, \({m}_{p}c^2\).
We can check
\(h=4{ r }_{ e } {m}_{p}c-\cfrac { 4 }{ c }{r}_{e}.\Phi\)
derived in the post "Light Indeed Light" for the value of Planck's constant,
\(h\) = 4*235e-12*(6.2948e-34*299792458)/(1.60217657e-19)-4*235e-12*2.29/299792458
\(h\) = 1.1000e-15 eV s-1
Based on this model Planck's Constant is no longer a constant, and its calculation and the calculation for \({m}_{p}\) relies a lot on the collision model between photons and orbital electrons. The paper from which the data was taken estimates \(h\) via two methods producing values of (0.94+.48)e-15 and (2.92+0.77)e-15.
Thursday, June 5, 2014
Light, Indeed Light.
From the post "Planck Constant not Constant"
\(E_{max}(f)=E[KE(f)]=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }.f\)
From the web, a set of photoelectric data by Lulu Liu (Partner: Pablo Solis)∗
MIT Undergraduate of MIT 2007 "Determination of Planck’s Constant Using the Photoelectric Effect". The target metal was potassium, K e(2,8,8,1) of work function 2.29 eV, and atomic radius 235 pm.
wavelength frequency Hz Stoppping Volatge
365.0 nm 8.2135e14 0.76 V
404.7 nm 7.4078e14 0.60 V
546.1 nm 5.4897e14 0.44 V
577.0 nm 5.1957e14 0.44 V
The last entry in the table seem to suggest saturation where \(\cfrac{r_e}{r}>1\). We will not use this point. Further more the data seem to be for the region beyond the first plateau for the Electron E_max vs Photon Frequency graph in the post "Fodo Electric Effect".
A plot of stopping voltage (V) vs frequency (*e14 Hz) is reproduced below,
The regression line is y=0.11x-0.18 error in slop = +0.0371, error in y intercept = +0.2647
Therefore, from the gradient of the regression line,
\(\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }=0.11e-14\)
\(KE_{max}=\cfrac {(0.11e-14)*c}{ 4{ r }_{ e } }\)
where \({r}_{e}\) is the atomic radius of K.
\({KE}_{max}\)=(0.11e-14)*299792458/(4*(235e-12))
Since,
\(KE_{max}={m}_{p}c^2-\Phi\)
\({m}_{p}=\cfrac{KE_{max}+\Phi}{c^2}\)
So,
\({m}_{p}\) = ((0.11e-14)*299792458/(4*(235e-12))+2.29)/299792458^2 *(1.60217657e-19)
the last factor converts eV to Joules (J). And so my second attempt at mass of a photon is,
\({m}_{p}\) = 6.2948e-34 kg
compared to that of an electron, 9.10938291e-31 kg and the mass of a proton 1.67262178e-27 kg.
Light, indeed light.
\(E_{max}(f)=E[KE(f)]=\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }.f\)
From the web, a set of photoelectric data by Lulu Liu (Partner: Pablo Solis)∗
MIT Undergraduate of MIT 2007 "Determination of Planck’s Constant Using the Photoelectric Effect". The target metal was potassium, K e(2,8,8,1) of work function 2.29 eV, and atomic radius 235 pm.
wavelength frequency Hz Stoppping Volatge
365.0 nm 8.2135e14 0.76 V
404.7 nm 7.4078e14 0.60 V
546.1 nm 5.4897e14 0.44 V
577.0 nm 5.1957e14 0.44 V
The last entry in the table seem to suggest saturation where \(\cfrac{r_e}{r}>1\). We will not use this point. Further more the data seem to be for the region beyond the first plateau for the Electron E_max vs Photon Frequency graph in the post "Fodo Electric Effect".
A plot of stopping voltage (V) vs frequency (*e14 Hz) is reproduced below,
The regression line is y=0.11x-0.18 error in slop = +0.0371, error in y intercept = +0.2647
Therefore, from the gradient of the regression line,
\(\cfrac { 4{ r }_{ e } }{ c } .KE_{ max }=0.11e-14\)
\(KE_{max}=\cfrac {(0.11e-14)*c}{ 4{ r }_{ e } }\)
where \({r}_{e}\) is the atomic radius of K.
Since,
\(KE_{max}={m}_{p}c^2-\Phi\)
\({m}_{p}=\cfrac{KE_{max}+\Phi}{c^2}\)
So,
\({m}_{p}\) = ((0.11e-14)*299792458/(4*(235e-12))+2.29)/299792458^2 *(1.60217657e-19)
the last factor converts eV to Joules (J). And so my second attempt at mass of a photon is,
\({m}_{p}\) = 6.2948e-34 kg
compared to that of an electron, 9.10938291e-31 kg and the mass of a proton 1.67262178e-27 kg.
Light, indeed light.