880 reputation
730
bio website ziga-lausegger.t15.org/strani/…
location Slovenia
age 28
visits member for 3 years
seen Nov 27 at 17:43

I love to program and crosscompile baremetal C programs for ARM based microcontrollers, I love physics and I love writing science documents/books in LaTeX. It amazes me how physics is connecting all science and is helping mathematics to evolve. In order for science profession to comunicate on a high level I advise everyone to use Linux, LaTeX and a good vector imaging program like Inkscape.


Sep
8
asked LS Coupling - weird image in the book
Sep
8
accepted Total angular momentum - single electron
Sep
8
comment Total angular momentum - single electron
This means that the $z$ axis must be defined in the same direction as the total angular momentum $\mathbf{J}$.
Sep
8
comment Total angular momentum - single electron
Spin magnitude of an electron is allways the same $S=\sqrt{s(s+1)}\hbar=\frac{\sqrt{3}}{2}\hbar$ where $s$ is the spin quantum number which for electron equals $1/2$. By $s$ you probably meant the $z$ component of the spin $S_z=m_s\hbar$ where $m_s= \pm s = \pm\frac{1}{2}$? So in one case $S_z = \frac{1}{2}\hbar$ while in the other case $S_z = -\frac{1}{2}\hbar$?
Sep
8
asked Total angular momentum - single electron
Sep
2
comment positron / electron annihilation - where is the invariant?
Thank you. So as long as I stay in the same coordinate system there is no need to use the equation $E^2={E_0}^2+(pc)^2$. But I have allso been dealing with systems where we used this equation within the same coordinate system to connect the momentum $p$ and energies - in this case it comes quite handy. Is my thinking ok now?
Sep
2
revised positron / electron annihilation - where is the invariant?
added 2 characters in body
Sep
2
comment positron / electron annihilation - where is the invariant?
I am sorry but I don't understand. Is it different fi I say the Lorentz invariance is: $E^2={E_0}^2+(pc)^2$ or if I say: ${E_0}^2 =E^2-(pc)^2$? It is the same equation. Is above example solved correctly?
Sep
1
asked positron / electron annihilation - where is the invariant?
Sep
1
revised Kaon spontaneously splits into two pions (just need an confirmation)
deleted 2 characters in body
Sep
1
comment Kaon spontaneously splits into two pions (just need an confirmation)
Sorry - it was a mistake which has now been fixed :)
Sep
1
revised Kaon spontaneously splits into two pions (just need an confirmation)
added 275 characters in body
Sep
1
comment Kaon spontaneously splits into two pions (just need an confirmation)
Thank gosh! I am so happy now, because i think i understand this :). What about if the kaon would spolit into two different particles with different mass and speeds. Look at the EDIT.
Sep
1
asked Kaon spontaneously splits into two pions (just need an confirmation)
Aug
23
accepted Transition integral from 1-D cartesian into 3-D polar coordinate system
Aug
16
revised Transition integral from 1-D cartesian into 3-D polar coordinate system
edited title
Aug
16
revised Transition integral from 1-D cartesian into 3-D polar coordinate system
added 147 characters in body
Aug
16
asked Transition integral from 1-D cartesian into 3-D polar coordinate system
Aug
15
comment The probability of finding the electron in the H-atom
So this is the full probability... What if i was looking for $P d\theta$? Would i just have to remove the $\int\limits_{0}^{\pi}$?
Aug
15
comment The probability of finding the electron in the H-atom
I know if i want to get the full probability i need to calculate: $$P=\int\limits_{V}|R(r)|^2|\Phi(\phi)|^2|\Theta(\theta)|^2dV = \int\limits_{V}|R(r)|^2|\Phi(\phi)|^2|\Theta(\theta)|^2\,\,r^2dr\, d\theta \sin\theta d\phi=\\ =\int\limits_{0}^{\infty}r^2|R(r)|^2dr \int\limits_{0}^{\pi}|\Theta(\theta)|^2 \sin\theta d\theta \int\limits_{0}^{2\pi}d\phi$$ I hope i wrote that right and i think that this full integral should equal 1. But somehow it is still not clear to me how to get the $P(r)dr$... How can i interpret the $dr$ after the $P(r)$. What do you think of when you see something like it?