momentum kills, well momentum has not killed a single cat as it is a mathematical construct only. But knowing about momentum p=mv can kill people and cats, some think twice the speed, twice the momentum, twice the impact... (not that any physicist think so, but still). nothing measurable correspond directly to the standard momentum (de Broglie related)...it is a derivative, that can be replaced by better stuff.

on the other hand kinetic energy, Compton momentum and also kinetic mass (yes a new useful concept) are all linked to v^2 (when v<<c), so yes if aware of any of them one will not double the speed. And yes we can measure such effects directly: Crash!

momentum came out from the time it was big confusion around if kinetic energy and moving masses where related to v^2 or v...it is a math artifact, that survived. If you take a cow and multiply it with the speed of light and divide it by 2/3 of a banana, then this construct is not a cow, even if it embedded has information about the cow. Sure it is useful to model the cow, if u don't know better. Better is to observe a cow and model it directly!

Old Bernoulli had a farm
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]
And on his farm he had a cow
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]
With a [$]\partial/\partial x[$] [$]\partial/\partial x[$] here
And a [$]\partial/\partial t[$] [$]\partial/\partial t[$] there
Everywhere a [$]\partial \partial \partial[$]
Old Bernoulli had a farm
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]

Old Schrödinger had a farm
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]
And on his farm he had a pig
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]
With [$]\partial/\partial x[$] and [$]\partial/\partial x[$] here
And only a [$]\partial/\partial t[$] there
Everywhere a [$]\partial \partial \partial[$]
Old Schrödinger had a farm
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]

Old Klein and Gordon had a farm
[$]c\hbar[$]-[$]c\hbar[$]-[$]m_{OO}[$]
And on their farm he had a horse
[$]c\hbar[$]-[$]c\hbar[$]-[$]m_{OO}[$]
With [$]\partial/\partial x[$] [$]\partial/\partial x[$] here
And [$]\partial/\partial t[$] [$]\partial/\partial t[$] there
Everywhere a [$]\partial \nabla \partial\nabla [$]
Klein and Gordon had a farm (in Brokeback Mountain)
[$]c\hbar[$]-[$]c\hbar[$]-[$]m_{OO}[$]

Old Dirac had a farm
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]
And on his farm he had no soul
[$]i\hbar[$]-[$]i\hbar[$]-[$]\hat O[$]
With nothing here
And nothing there
Nothing nowhere (I hope you note the Floyds inspiration)
So he d-slashed others' animals out of spite!
[$]U^\dagger[$]-[$]U^\dagger[$]-[$]U^\dagger[$]

(Thanks to Collector for spotting factual errors in the lyrics.)

Conclusion: seems like everyone with a fam get's old, and most partial derived!

Old Bill Euler had a farm
[$]e^i e^ i e^i O[$]
And on his farm he had a cow
[$]e^i e^ i e^i O[$]
With a [$]\pi \pi[$] here
And a [$]\pi \pi[$] there
Everywhere a [$]\pi \pi[$]
Old Bill Euler had a farm
[$]e^i e^ i e^i \Omega[$]

Allegro ma non troppo; con brio.

"Is this 'new' equation rigorous? "  as far as a farm boy can see yes!

Here in West Friesland we have a saying "the dumbest farmers have the biggest potatoes"
And farmers have that native cunning, especially in Hooterville.

My favourite show

www.youtube.com/watch?v=umS3XM3xAPk

"Is this 'new' equation rigorous? "  as far as a farm boy can see yes!

Here in West Friesland we have a saying "the dumbest farmers have the biggest potatoes"
And farmers have that native cunning, especially in Hooterville.

My favourite show

www.youtube.com/watch?v=umS3XM3xAPk

I was planning to experiment with potato towers this year, my plan was to build a very tall potato tower...the custom materials never arrived (due to corona?) and all sette potatoes was for first time sold out (and potatoes from store not allowed due to not controlled for diseases etc.)...next year...or may be my luck I did not build it, very tall potato towers involve risks, they can collapse, if not properly constructed, and I am not exactly an engineer.

I'm only going to say one word, are you listening Ben? Excise.

Johnny tells it as it is

www.youtube.com/watch?v=Un1XyR7MDk4

wondered if my PDE says anything useful about hydrogen like atoms. I re-wrote it on polar coordinates, did some separation of variables, get 3 ODE's, that I all let Mathematica solve Schrodinger ODE's versus our ODE's 

Azimuthal equation

\( \frac{1}{\Phi}\frac{d^2 \Phi }{d \varphi^2}+B = 0, \qquad   \frac{1}{\Phi}\frac{d \Phi }{d \varphi}+B = 0 \)

Polar

\( (1-x^2) \frac{d^2 P }{d x^2}-2x\frac{d P}{d x}+\left(A+\frac{B}{1-x^2}\right)P =0, \qquad   \sqrt{1-x^2}\frac{d P }{d x}-\left(A+\frac{B}{\sqrt{1-x^2}}\right)P = 0  \)

Radial

\(\frac{d    }{d r}\left(r^2\frac{d R}{d r} \right)-\frac{2\mu r^2}{\hbar }\left(E-k_e\frac{Ze^2}{r}\right)-AR=0,\qquad \frac{r}{R}\frac{d R   }{d r}-\frac{ir}{\hbar c}\left(E-k_e\frac{Ze^2}{r}\right)-AR=0\)

\(k_e\frac{Ze^2}{r}\) is  just coulomb force, A and B separation constants.


tell me if u see my equations tell how to turn hydrogen into hydrogen like gold??? if so I should possibly keep it low key!

another side note, when working with e (2.71...) and in additional elementary charge: e, what is then best symbol to use for Exp?

Also in Schrodinger solution for Azimuthal  \( \Phi(\phi)=c_1e^{i\sqrt{B}\phi}+c_2e^{-i\sqrt{B}\phi} \) I see they used a bit "strange" argument to set \(c_2=0\), and by this basically getting 'same' solution as my Azimuthal equation gives. Anyone know foundation/argument for kicking out last term of their solution?

I need to triple check for calculation errors/typos, but many disturbing elements in the summer!

early draft

The Schrodinger Azimuthal ODE solution is  \(\Phi(\phi)=c_1\cos[\sqrt{B}\phi]+c_2\sin[\sqrt{B}\phi]\), by setting \(c_1=1\) and \(c_2=i\) we get (Eulers formula)

\(e^{i\sqrt{B}\phi}\)

Then I have checked with several sources, now they replace out of thin air the constant B with m^2 so they get 

\(e^{im\phi}\)

How can one just replace a constant B with such a trick? My point is it dose not come out of the equations, so it is guessing or fudging to make the equation to make sense compared to observations where one use m as Azimuthal quantum number. ?

However in my solution there is no such guessing, as my ODE solution is

\(c_1e^{-B\phi}\)

in other words after setting up the "boundary condition" that it should hold for any angle \(phi\) I get that B must be a positive integer (quantum number).  I do not need to replace B with a guessed function (B replaced with m^2 out from thin air).

not worked much with such solutions before, but is it normal to replace constants with guessed functions (or out of convenience as it seems some call it) ?

minor typo in my radial equation, that lead to the ugly complex solution...here I think fixed and A and B must be quantum numbers (positive integers), A from the radial solution, as \(\hbar\) must come in quanta, and B due to the angle boundary condition. The Schrodinger radial ODE is a beast to solve compared.

my Azimuth solution seems to make sense, need to investigate radial and polar, one can get bi-polar of less.

Also in Schrodinger solution for Azimuthal  \( \Phi(\phi)=c_1e^{i\sqrt{B}\phi}+c_2e^{-i\sqrt{B}\phi} \) I see they used a bit "strange" argument to set \(c_2=0\), and by this basically getting 'same' solution as my Azimuthal equation gives. Anyone know foundation/argument for kicking out last term of their solution?

This is the general solution of any 2nd order differential equation. In your problem, an electron orbits around the nucleus, so putting one of the $c$ constants 0 means choosing the direction of the electron's motion. You can see it if you calculate the orbital angular momentum of the wave function [$]L_z\Phi = -i\hbar \frac{d\Phi}{d\phi}[$]. The first term gives [$]\hbar m c_1e^{im\phi}[$] and the second gives [$]-\hbar mc_2e^{im\phi}[$]. Assuming that the electron orbit is perpendicular to the z axis, [$]\vec L_z = \vec p \times \vec r[$], so the two terms give the opposite direction of the electron motion.

Also in Schrodinger solution for Azimuthal  \( \Phi(\phi)=c_1e^{i\sqrt{B}\phi}+c_2e^{-i\sqrt{B}\phi} \) I see they used a bit "strange" argument to set \(c_2=0\), and by this basically getting 'same' solution as my Azimuthal equation gives. Anyone know foundation/argument for kicking out last term of their solution?

This is the general solution of any 2nd order differential equation. In your problem, an electron orbits around the nucleus, so putting one of the $c$ constants 0 means choosing the direction of the electron's motion. You can see it if you calculate the orbital angular momentum of the wave function [$]L_z\Phi = -i\hbar \frac{d\Phi}{d\phi}[$]. The first term gives [$]\hbar m c_1e^{im\phi}[$] and the second gives [$]-\hbar mc_2e^{im\phi}[$]. Assuming that the electron orbit is perpendicular to the z axis, [$]\vec L_z = \vec p \times \vec r[$], so the two terms give the opposite direction of the electron motion.

Thanks!! However still wondering about the mathematical "foundation" for going from \(\sqrt{B} \) to m (for me it almost seems like a smart trick to get away with switching from  \(\sqrt{B} \)  to B ?) . I read

"Again using an enlightened choice, we pick m^2 " and also

"Normally, an arbitrary separation constant, like K, is selected and then you solve for K later. In this example, we are instead going to stand on the shoulders of some of the physicists and mathematicians of the previous 300 years, and make the enlightened choice of l(l + 1) as the separation constant."  https://faculty.washington.edu/seattle/physics227/reading/reading-26-27.pdf

Sure smart choices, but it seems a bit vague for me that one not are solving the constants based on mathematical conditions given from for example boundary conditions, but perhaps that is what they do indirectly also here.  



 my azimuthal solution is \(c_1e^{-B\phi}\),  a wild guess, can I perhaps decide on electron motion direction with sign of \(c_1\)? (keep in mind my angular momentum operator will be a angular Compton momentum operator, and not a angular momentum operator (de Broglie momentum from quantum perspective), \(L_c=rmc\) rather than L=rmv when v<<c .

Also in Schrodinger solution for Azimuthal  \( \Phi(\phi)=c_1e^{i\sqrt{B}\phi}+c_2e^{-i\sqrt{B}\phi} \) I see they used a bit "strange" argument to set \(c_2=0\), and by this basically getting 'same' solution as my Azimuthal equation gives. Anyone know foundation/argument for kicking out last term of their solution?

This is the general solution of any 2nd order differential equation. In your problem, an electron orbits around the nucleus, so putting one of the $c$ constants 0 means choosing the direction of the electron's motion. You can see it if you calculate the orbital angular momentum of the wave function [$]L_z\Phi = -i\hbar \frac{d\Phi}{d\phi}[$]. The first term gives [$]\hbar m c_1e^{im\phi}[$] and the second gives [$]-\hbar mc_2e^{im\phi}[$]. Assuming that the electron orbit is perpendicular to the z axis, [$]\vec L_z = \vec p \times \vec r[$], so the two terms give the opposite direction of the electron motion.

Thanks!! However still wondering about the mathematical "foundation" for going from \(\sqrt{B} \) to m (for me it almost seems like a smart trick to get away with switching from  \(\sqrt{B} \)  to B ?) . I read

¡De nada! See yourself why B i¡ m^2. Simply substitute [$]\Phi[$] to the azimuth equation.

This is the general solution of any 2nd order differential equation. In your problem, an electron orbits around the nucleus, so putting one of the $c$ constants 0 means choosing the direction of the electron's motion. You can see it if you calculate the orbital angular momentum of the wave function [$]L_z\Phi = -i\hbar \frac{d\Phi}{d\phi}[$]. The first term gives [$]\hbar m c_1e^{im\phi}[$] and the second gives [$]-\hbar mc_2e^{im\phi}[$]. Assuming that the electron orbit is perpendicular to the z axis, [$]\vec L_z = \vec p \times \vec r[$], so the two terms give the opposite direction of the electron motion.

Thanks!! However still wondering about the mathematical "foundation" for going from \(\sqrt{B} \) to m (for me it almost seems like a smart trick to get away with switching from  \(\sqrt{B} \)  to B ?) . I read

¡De nada! See yourself why B i¡ m^2. Simply substitute [$]\Phi[$] to the azimuth equation.

by not substituting \(\sqrt{B}\) with m (in Schrodinger), we could hold on to that \(\sqrt{B}=\)Integer, because we know \(\sqrt{B}\) must come in quanta to satisfy full angle (360), B in Schrodinger naturally have to be 1, 4, 9, 16.... but likely harder to interpret then what B is, so for me it seems like B=m^2 only can come out of qualified guess and is done to simplify interpretation, one know from the quantum number (linked to the 360 angle, and also the quanta of h) is 1,2,3,4,5...so one re-write B=m^2 and say m is the quantum number. Since Schrodinger is rooted in standard momentum, that I am convinced is a derivative of the much more fundamental Compton momentum it seems plausible that B in Schrodinger also is a function of the real quantum number, my B, why I get B=m in my equation (why I dont need a m, as B is m), in my solution B=1,2,3,4,5..

OK, but insert the solution [$]\Phi(\varphi) = \frac{1}{\sqrt{2\pi}} e^{im\varphi}[$] into [$]\frac{1}{\Phi}\frac{d^2 \Phi }{d \varphi^2}+B = 0[$], do the differentiation and see what happens.