In the case of a circular path, \(n=1\)
\(h=2\pi a_\psi.mv\)
\(\hbar=\cfrac{h}{2\pi}= a_\psi.mv\)
in this case \(a_\psi\) is parallel to \(x\), but perpendicular to \(\lambda\),
When \(n=2\),
\(\Delta s=\sqrt { a^{ 2 }_{ \psi }sin^{ 2 }(2\theta )(\Delta \theta )^{ 2 }+a^{ 2 }_{ \psi }cos^{ 2 }(2\theta )(\Delta \theta )^{ 2 } } \)
\( \Delta s=a_{ \psi }\Delta \theta \sqrt { sin^{ 2 }(2\theta )+cos^{ 2 }(2\theta ) } \)
\( \Delta s=a_{ \psi }\Delta \theta \)
\( s=2\int _{ 0 }^{ \pi }{ a_{ \psi } } d\theta =2\pi a_{ \psi }\)
Strangely both paths have the same length. In fact,
\(\Delta s=\sqrt { a^{ 2 }_{ \psi }sin^{ 2 }(n\theta )(\Delta \theta )^{ 2 }+a^{ 2 }_{ \psi }cos^{ 2 }(n\theta )(\Delta \theta )^{ 2 } } \)
\(\Delta s=a_{ \psi }\Delta \theta \sqrt { sin^{ 2 }(n\theta )+cos^{ 2 }(n\theta ) }\)
\(\Delta s=a_{ \psi }\Delta \theta \)
Which means,
\(\hbar=a_\psi.mv\)
is valid for all \(n\). The path length being the same for all \(n\) also means that for all \(n\), given a constant velocity at light speed, the period, \(T\) to traverse all paths is the same, hence, \(\psi\) has only one fundamental frequency, \(f_{n=1}\).
\(f_{n=1}=\cfrac{1}{T}\)
This is consistent with the fact that,
\(p=mc\) and \(E=mc^2\)
which are both constants, independent of \(n\). The amplitude of the wave changes such that the energy of the wave is fully defined by,
\(E=hf_{n=1}\) --- (*)
For all cases of \(n\),
\(c=nf_{n=1}.\cfrac{\lambda_{n=1}}{n}=f_{n=1}.\lambda_{n=1}\)
Both \(p\) and \(E\) remain constant. In this scenario where \(n\) wavelengths are packed into a perimeter of \(2\pi a_\psi\) all harmonics generated has the same energy given by (*).
The paths for \(n=1\) and \(n=2\) and the reference circle is generated using the following code in Scilab.
t = 0:0.01:(2.0*%pi);
x = 3*cos(t);
y = 3*sin(t);
z = 10*0.2*sin(t.*2);
param3d(x,y,z,45,60,"",[2,0]);
p=get("hdl"); //get handle on current entity (here the polyline entity)
p.foreground=2;
t = 0:0.01:(2.0*%pi);
x = 3*cos(t);
y = 3*sin(t);
z = 10*0.2*sin(t.*1);
param3d(x,y,z,45,60,"",[2,0]);
p=get("hdl"); //get handle on current entity (here the polyline entity)
p.foreground=5;
t = 0:0.01:(2.0*%pi);
x = 3*cos(t);
y = 3*sin(t);
z = 0.000002*sin(t.*1);
param3d(x,y,z,45,60,"",[2,0]);
p=get("hdl"); //get handle on current entity (here the polyline entity)
p.foreground=1;
Have a nice day.
