« first day (5089 days earlier)      last day (137 days later) » 

04:12
@SirCumference Oh, miao miao didn't think that was useful to students. Can add something like that to myow own book. But maybe you would want to read Feynman-Hibbs, because even though it doesnt even begin to talk about Hilbert spaces, the connection to Analytical mechanics is very obvious.
@SirCumference id wonder why someone would think otherwise. Even from the beginning, from Bohr's complementarity principle, the connection to classical mechanics is emphasised. But I can see that the problem was that both von Neumann and Dirac didnt see that it was important to present it that way, and so everybody else just accepted their treatment as gospel and left it at that.
@qwerty You have a full programming language in your hands right there; there ought to be nothing that you cannot handle. It is more a matter of convenience: is it easier to write some code every time? Or is it easier to open Excel and perform some quick calculation.
@qwerty Those hurdles existed because VBA is almost powerful enough to be a proper programming language, and that meant there were plenty of viruses. Business needs the viruses gone, so those protective hurdles got installed and never removed.
04:31
@Allie I think so too, see gpaw.readthedocs.io/documentation/ofdft/ofdft.html but you can read how to use atomic orbitals in the arXiv version of nature.com/articles/s43588-024-00605-8 ; a review is in pubs.acs.org/doi/10.1021/acs.chemrev.2c00758
05:02
Hi everyone, I have a basic question about Dirichlet boundary value problem in electromagnetism: I know how to derive potential outside a sphere given the potential on the boundary using Green's functions.
But if I am given a surface charge density on the sphere, instead of the potential on the it and I have derived the Dirichlet Green's function for a sphere, how do I find the potential outside the sphere?
05:25
@naturallyInconsistent what defines a "proper programming language" in your opinion?
06:19
Hello
I was just curious that we defined work because we found out that F.ds is very useful for solving our problems or is there something i am missing?
06:34
hi
@NOTEBook u r not missing anything
oh ok
@Ryder Rude
can you help on one more thing
Since work done = integration of F.ds... under what conditions can we directly say that work = F.S
one condition is angle between force and displacement is constant and force is constant
@NOTEBook When $\vec{F}$ is independent of $\vec{s}$ it comes out of the integral sign, so we can write $\vec{F} \cdot \vec{s}$
So if force is constant?
06:52
yes
who on PSE is an expert on measurement problem
i wanna discuss some stuff
do u know anyone
i also want book references
i need someone who follows the progress in quantum measurement and information stuff, but doesn't subscribe to a particular QM interpretation
123
123
07:16
Hello Everyone...
Hi @RyderRude , I have resolved the mystery of constants in term of one mass automatically become magnitude as 1 any one object. Also how momentum always conserved. : )
@123 great :)
@bolbteppa do u follow research progress on the measurement problem
 
3 hours later…
10:40
@qwerty Turing completeness. And not being so horrible as to be hilariously unusable. The latter condition rules out things like intercal, brainf*ck, and so forth.
11:39
have you read TAOCP by Knuth?
i think his concrete math is a prerequisite
123
123
12:05
Hello @RyderRude
I have few concerns and questions about above topic from taylors mechanics by imposing different situations.
1. In the above two particles interacting system, but $F_1^{ext}$ and $F_2^{ext}$ are two different particles having different positions and both external forces on both particles having different magnitude and direction at different location. How can we add both external forces to find resultant external force adding like a concurrent forces? $F^{ext} = F_1^{ext} + F_2^{ext}$
2. What if these two particles are noninteracting system. Then there is no $F_{12}$ and $F_{21}$ then how can we add external forces on both particles to find resultant external force?
@123 u r allowed to add forces at different locations, because this is a useful quantity. The total force on a system is the sum of forces, and it doesn't matter if the forces are at different locations
123
123
I think answer for question-1 lies in the center-of-mass COM . The external forces on both interacting particles considered as acting on a COM. By this way we can add both external forces as concurrent forces to find resultant. Am i correct?
@123 there is still F12 and F21. Both are equal to zero
@123 u can think of it in terms of the forces concurrently acting on CoM, yes
The CoM behaves as if all the forces were at that point
123
123
@RyderRude Ookay.
@RyderRude But in my second question both are noninteracting particles. So we don't have any F12 and F21.
12:21
@123 like I said, both are there but just equal to 0. The book's analysis applies all the same
book doesn't assume that F12 and F21 are non zero
123
123
@RyderRude So the behavior of motion is same in both situations?
@123 yes
@123 the analysis applies to both interacting and non interacting bodies in general
123
123
How can non interacting system and interacting have same motion?
if they assumed F12 not equal to 0, then it would only apply to interacting bodies
@123 it doesn't mean same motion
123
123
interacting system are locked each other by equal and opposite forces. Like thin massless rod joined both particles.
12:24
look at the conclusion. $\dot {P}= F1+ F2$. I'm just saying that this conclusion applies to all systems @123
the conclusion isn't that all systems have the same motion
it's just that all system's motion obey $\dot {P}= F1+F2$ @123
Doesn't mean all systems have the same motion
123
123
How can we compute different types of motion using same strategy and formula?
it is the power of mathematics.....
mathematics work with variables, not constants
Like F=Gm1m2/r2 applies to a variety of motions
like a stone as well as planets
123
123
How? we know F1ext and F2ext both are same in interacting and noninteracting situations . We will give same result.
if F1 and F2 are the same in both situations, u will get the same result of $\dot {P}$ for both situations. It doesn't mean u will get the same motion for both situations
123
123
Because we are only taking external forces to find the resultant
12:29
just because two systems have the same $\dot {P}$, do they have the same motion? @123
123
123
@RyderRude I think yes. because it depend on external forces.
no, that is wrong...
123
123
Pls explain
what does $\dot P$ mean?
123
123
Net rate of change in momentum.
12:32
right. So just because net rate of change of momentum is the same, does it mean individual rate of change of momentum is the same?
123
123
@RyderRude No.
right. So it's not the same motion. For it to be the same motion, everything wud need to be the same
123
123
But why could individual rate of change of momentum will be different , this is my question.
Because individual $\dot {p1}= F1+ F21$. Even if F1 is the same, F21 is different for an interacting system
So individual $\dot {p_1}$ is different for an interacting system
123
123
Aaaaaah i see.... here is the key
12:35
great
123
123
It means individual rates are different because of presence and absence of F12 and F21
Can someone recommend me a thermodynamics / statistical mechanics lecture series course? Something which is more advanced than the usual?
123
123
But net rate of change of momentum will be same.
@MoreAnonymous hi
123
123
It means N2L works for all interacting and noninteracting and partial-interacting systems.
12:38
yes
123
123
Finally how can we get different motions from same net external force in both situations?
@RyderRude hey!
123
123
Because net result does not contain interacting and noninteracting behavior. Both are zero
@MoreAnonymous physics.stackexchange.com/a/830294/156987 it is related to relativistic quantum measurements. U might be interested too ..
@MoreAnonymous i will star it so it doesn't get drowned
@RyderRude interesting... I wish I had done more math modules
@RyderRude thanks
12:44
@MoreAnonymous the paper doesn't have advanced math
it is for physicists
I was recently telling someone how I felt doing physics has become like pokemon knowledge... it's useless in my day to day life.. but alas my love
there r also some other papers on that thread
@RyderRude I'll have a look again
@MoreAnonymous lol
@MoreAnonymous yeah... I don't feel like I understand everyday world better because of physics
but physics still gives goosebumps sometimes
physics is great on its own, even if it doesn't apply to everyday life
123
123
I think to find the motion in both situations, we need to work on forces/rate of momentum for individual particles. We can not use net external force or net momentum change. Am i correct? @RyderRude
12:47
that paper gave me physics joy after a long time
@123 yes
@123 using net stuff, u can calculate motion of CoM
123
123
Thanks a lot. 🫡
123
123
@RyderRude and COM only compatible for fully interacting systems. Am i correct
@123 no... Again it applies to every system
@123 to see what things a mathematical analysis applies to, look at the assumptions listed
123
123
Oooh i see. It means in both situations COM movement is different?
12:50
if $F1+F2$ is different, CoM moves differently
123
123
But in my situation F1 + F2 is same in both situations.
@123 i mean that CoM of non interacting and interacting can move the same cuz CoM motion doesn't depend on internal forces like F12 and F21
@123 then CoM moves different
123
123
Ookay. means COM doesn't depend on interacting and noninteracting systems.
yes. Its motion only depends on the total external force and the total momentum and total mass
123
123
It depend on how masses change its positions.
$\Large{R = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}}$
@RyderRude How can the total external force and total momentum can have different values? They are equal in both situations.
Okay you mean to say total force and total momentum (not rate of change of momentum).
@RyderRude : ) you gave me new ideas to think and learn this.
13:01
could you two please take it to your personal room again? it's the same topic
123
123
It almost ends
@qwerty it is not...
@123 great :)
123
123
You are saying motion of COM depend on net external force , total momentum and total mass. Can you pls share the equation? Pls we discuss on personal room. It is almost end
13:22
@123 u can find it in the CoM discussion of the book...
123
123
Okay... You raised the answer about COM. but problem is that in free space two particles system will have motion both rotation and translation in both situations differently about COM.
Net external act at COM? if this is the case then only translation happen.
Okay here's something I'm confused about .. when I use the geodesic deviation vector for a family of particles ... Do these particles have to be of the same mass?
@123 i don't mean that the net external force acts on CoM. I mean that the CoM moves as if the net external force was acting on it.
@123 we should end this discussion rn...
123
123
Ookay NP
I'm really worried about the considerations of continuity in GR
13:32
iirc those are test particles right?
@qwerty ah I see ...
14:08
@MoreAnonymous Have you checked out mehren kardar's lectures?
I'm not sure if they are at the level you seek for though
14:26
@Arjun I haven't done statistical field theory
Actually ... not what I was thinking of but thanks
@MoreAnonymous Oh,he's got two courses,the first one is 'statistical physics of particles' .The second one being the more advanced statistical field theory.
@Arjun I see ... I'll have a look ... thanks
15:21
what is meant by parity/even-odd solutions for $\psi$ for bound states in finite well
nvm ill finish the section
Why would the author choose the imaginary component then want to disregard it later on
They could have easily just chose $\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$ no?
nvm
didn't read it correctly, when plugging in the boundary conditions it's e^ix-e^{-ix}
 
2 hours later…
17:51
@naturallyInconsistent yes that second article was authored by the professor im doing research with :)
thanks!
lol
@qwerty no, but it is always looming in the corner from some cosmic realm
18:28
what is the last thing u learned that gave u joy to learn
 
1 hour later…
19:52
How does one show the turning point condition in a 2 particle collision is Lorentz invariant? Like how do I know total energy = kinetic energy wont change if I do a boost?
Wait this is trivally true ... ignore me
lol
@naturallyInconsistent where does using only MOV fall? ;)
 
3 hours later…
23:11
is there a name for a distribution of probability distributions?
I am interested in, say, a random variable that follows a Bernoulli distribution characterized by probability $p(t)$ where $p$ itself is a time-dependent random variable

« first day (5089 days earlier)      last day (137 days later) »