848 reputation
428
bio website ziga-lausegger.netau.net/…
location Slovenia
age 28
visits member for 2 years, 10 months
seen Feb 24 at 9:43

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.


Oct
6
comment How do I find the electric field above the center of a square plate (rather than circular)?
Could you be a bit more speciffic?
Oct
6
comment Transmission Lines + DC Current
I edited your question using LaTeX formating. About your question... Do you refer to electric potential or voltage? Remember voltage is difference of electric potentials.
Sep
13
comment Drawing the circuit from a differential equation
Do you know any good book which teach to think this way?
Sep
13
comment Drawing the circuit from a differential equation
Why don't you try and ask here: electronics.stackexchange.com
Sep
8
comment Term symbol - how do we know the number of electrons $e^-$?
I copy/pasted the statement in the book. Now tell me if this book is a garbage?
Sep
8
comment Term symbol - how do we know the number of electrons $e^-$?
But in the book A. Beiser, Concepts of modern physics it says that multiplicity is $2S+1$ only in cases when $L>S$ while in cases when $L<S$ the multiplicity is $2L+1$. This is why we can't just state $4=2S+1$ as it could be $4=2L+1$... so can this even be solved?
Sep
8
comment LS Coupling - weird image in the book
So i did this like i should but forgot that there is no such thing as negative magnitude... so from all the solutions for $L=3\hbar, 2\hbar, 1\hbar, 0, -1\hbar, -2\hbar, -3\hbar$ i have to simply remove negative ones and 0.
Sep
8
comment LS Coupling - weird image in the book
Why do you all use $l$ (orbital quantum number) in place of $L$ (orbital angular momentum)? In the book it only says that $L$ can be 3 2 or 1 and not -3, -2, -1...
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
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
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
comment Kaon spontaneously splits into two pions (just need an confirmation)
Sorry - it was a mistake which has now been fixed :)
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.
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?
Aug
15
comment The probability of finding the electron in the H-atom
How do i do this?
Aug
15
comment The probability of finding the electron in the H-atom
Never mind i think i understand :)
Aug
15
comment The probability of finding the electron in the H-atom
I have one more question about this. Why did we use the diferential of volume when we are searching for $P(r)dr$
Aug
15
comment The probability of finding the electron in the H-atom
Thank you very much. So if we are satisfied with a fraction of a probability there is no need to integrate :) What confused me was that feeling that if we want to integrate the equation we have to do it on both sides and not only one side... It is weird to me that we integrate just part of the equation.