1:32 AM
LOL.. I spent hours trying to do a java assignment only to find out and remember there's a thing called multidimensional arrays
successfully did it the wrong way ... yay

1:59 AM
@Obliv operation failed successfully?

1 hour later…
3:04 AM
Apparently not, the auto grader software says it expects a different amount of lines even though I've accomplished what it was asking
this class is an utter waste of my time and I wish I didn't have to take it
god that's so funny. In this show I'm watching there's a brief moment on screen where it shows some writing on a scroll that's supposed to look mysterious and complicated and I paused it and identified parts of maxwell's equations
I'm dying this is hilariously relevant for me since I got an EM exam on tuesday wow
@Slereah can you decipher these schematics? it seems we have a John Titor in the animation studio
am I imagining things or is that saying Div D is non negative and $\frac{\partial D}{\partial t} = \vec{j}$ the current density
it can't be, that'd make no sense. amp/m^2 from a time derivative da heck
wait im dumb $D$ is supposed to be an E field and $H$ is B field

3:53 AM
@Obliv No, it says $\text{div}\,D=\rho$ and $\text{rot}\,\vec H-\frac{\partial\vec D}{\partial t}=\vec j$

is it ok if I as an aspiring mathematical physicist get a D in a spectroscopy course?

@nickbros123 ;)
@nickbros123 totally fine. Maths phys does not really care about experimental stuff.

heisenberg nearly failed cuz he sucked at lab

Wherever there is a Yang, there is a bang

@Obliv the message im getting here is you either fix up in labs or intuit matrix mechanics

4:01 AM
yeah except its already been done so u cant really do that

@nickbros123 I wonder what is the proportion of physics professors who understand matrix mechanics. I've read what people have been explaining about it and none made the computation make sense.

u could save ur grade by developing new theory though and revolutionizing physics as we know it

@naturallyInconsistent im stealing this quote

can u still bring the D up to a C? surely theres a curve

@Obliv rofl I cant go thru existing theories without copious amounts of acid reflux

4:03 AM
@nickbros123 It isn't mine. Even Yang repeats it.

@nickbros123 sometimes it's hard to go thru material we aren't immediately interested in. It's not that it's hard to understand, but that you aren't interested in understanding it at that time or maybe it isn't time for it
just be like galois, ez
wait a minute, it says on ur profile ur self studying

@naturallyInconsistent acid base titration apparently releases 50 kJ, and water's specific heat capacity is 4kJ, which means, approx my body temp goes up by 12 deg (given i absorb al the heat), that doesnt sound so good....
@Obliv yyes

@nickbros123 drink water to cool down like a power plant

@nickbros123 nevermind, this calculation is just stupid

@nickbros123 Find something that bubbles. carbonate will do. burp the heat out.

4:10 AM
burp the heat out? what is he, a dragon?
but bubbly water does help, or tums (but lots of tums can cause kidney stones eventually)

@Obliv why not?

the interesting question is how something completely external and artificial, like the microscopic to macroscopic averaging for example, could cause such reactions in my body
also watching chemists mutilate physics causes this reaction as well

@nickbros123 mutilate maths to relieve stress

3 hours later…
6:58 AM
Apparently if you look at the internal logic of a geometric category, the weird adjunctions that nlab loves to talk about become modal operators
And you get to have things like "It is locally true that..." expressed in that logic

7:21 AM
@Obliv "auto grader software"... Man, this version of the future sucks

It seems to be about a logic related to the notion of points being connected by certain curves
And the modality here is expressed by a lift of the curve to the tangent bundle, where the necessity of a curve to go to another point depends on its lift to a certain distribution

7:48 AM
what does the curve represent?

From what I gather basically propositions of the kind "Point $p$ can be connected to point $q$"
By a curve of some specific property, presumably

1 hour later…
9:19 AM
Is there a quick way to compute $\langle n l m\lvert\hat{r}\rvert n l m \rangle$?
For a hydrogen atom of course

@SirCrackpot i think i saw a similar question on the forum here yesterday

9:51 AM
@lucabtz Not me :P
This is actually an XY problem. I'm wondering whether the dipole approximation for hydrogen holds true for big $n$

@SirCrackpot where have you been in the past few days?

I've been sick for a few days, then I had an exam and after exams I go non-interacting

@SirCrackpot ooo, easily computable!

10:08 AM
@naturallyInconsistent the average value? I don't doubt it is

@SirCrackpot omg double negative; feeling spicy, eh?

No, just using non geodesic language :P

exotic paths?

Cat shaped
So, I don't feel like dealing with Laguerre polynomials. Is that average value somewhere?

@SirCrackpot im sure you find it somewhere
it seems a basic useful fact about the hydrogen atom
but i dont know where

10:26 AM
Ok, I got it
Given that $\langle r \rangle_{\ell=0}\sim n^2 a_0$, this means that for high $n$ the dipole approx should break

@SirCrackpot why do you say so?

@naturallyInconsistent I'm kind of handwaving. I say so because this means that the electron is not mostly within $a_0$ as it happens for the gs
I'm not convinced at all smh

that's true, but the dominant behaviour is still dipole, since the monopole term is zero. You at least need to show that the quadrupole term grows important or something.
(and it does, since it grows as $n^4$)

Ok, well here it is. I was reading the historical article by Bethe in which he used OFPT to extimate Lamb shift. The idea was to perform a subtraction between the correction to the hydrogen energy levels and the free electron energy level. The formula Bethe uses contains the dipole approximation
Using the dipole approximation for a free electron is weird, so I was hoping somewhere else things would break
The first eq. is very clearly in the dipole approximation

10:50 AM
@SirCrackpot the subtraction is used to get a finite result?

11:18 AM
@lucabtz how dare you
@lucabtz In non relativistic QED you still get a logarhtimic divergence :P

im going crazy
whats wrong with you mathematica
@SirCrackpot yeah i think i recall something
but the way saw computation of the lamb shift was more modern i would say
they just applied time independent perturbation theory to the Uehling potential

Oh sure, this is "old" stuff for me too but I was asked this question and I didn't know what to say
That dipole approximation in the free particle is nasty

@SirCrackpot i mean though does it really matter?
as long as what you subtract gives you a finite result
the difference between all the finite results you get will be scheme dependent but should cancel in observables

11:47 AM
Hello Everyone...

@lucabtz a posteriori knowing renormalization, not really. I'm contesting the physical meaning of what Bethe subtracted

Hello @JohnRennie Sir...

12:29 PM
is Bohm's Implicate order just an extension of the hidden variable interpretation
it seems like he is not giving up the hidden variable interpretation

12:40 PM
How N2L and NGL (Newton's gravitation law) different. Why can we take both equal?
Hello @RyderRude

1:03 PM
hello

@lucabtz 0 v.s. O

@123 the three laws are valid for all types of forces. the gravity is just a specific type of force

@RyderRude can you pls explain more. N2L involves one one mass, but NGL involves two masses phenomenon. How can we take both equal?

1:22 PM
@naturallyInconsistent omg
I often type o instead of zero for some reason

1:36 PM
@123 consider only two objects in the universe and under gravity. we hav F12=ma2, F21=ma1 and F12 = - F21. substiute $F_{12}=\frac{Gm_1m_2}{r^2} \hat{n}_{21}$ to get the equations of motion
a simple case is acceleration due to gravity. the force on a falling object is $F=\frac{Gm_Em}{h^2}$ downward. Equate this to $ma$, and get the acceleration due to gravity @123
sorry i meant to write F12=m2a2 and F21=m1a1

1:58 PM
@123 u just write N2L seperately for both masses under gravity. F12=m2a2 and F21=m1a1. the gravity law just gives the formula of F12 or F21

@SirCrackpot i guess he just used something that would give a finite result, there is some sense to it even though maybe its not as meaningful as you would want it to be

@SirCrackpot the greatest of theoretical physicists know that whatever they are trying to do is imprecise and ad hoc, but capture enough of the most relevant and important physics as to obtain qualitatively correct results. Because then can other people try to make rigorous their findings.

@naturallyInconsistent yeah this is the point i was trying to make

1 hour later…
3:12 PM
$F_{12} = mg$ and $F_{21} = M_E a_E$ , N3L : $F_{12} = F_{21}$ then $a_E = \Big(\frac{m}{M_E}\Big)g$
If we take ball mass $m=1 Kg$ and $g = 10 m/s^2$ then we can find $a_E = 1.6\times 10^{-24} m/s^2$ . The above relation don't care about size of the Earth.
In other way. $F_{12} = NGL$ then $M_E a_E = G \frac{mM_E}{R_E^2}$ ball is near the Earth surface so $R_E + h = R_E$, then $a_E = G \frac{m}{R_E^2}$
If we consider size of the earth then $a_E = 1.6\times 10^{-24} m/s^2$
Both are approximately equal in that case. But
If we consider size of the as point particle then all mass of Earth concentrated at point and i am considering distance between Earth and ball $R_E + h = 1m$ then i found $a_E = G = 6.67\times 10^{-11} m/s^2$.
Why both results are so different?

the R^2 in the force formula is the distance between the objects. if R_E becomes zero then h becomes 1+R_E

3:27 PM
@RyderRude How $h = 1+R_E$?? If $R_E$ becomes zero

visualise it. u hav a ball 1m above Earth's surface. now shrink earth to a point. did the distance between Earth's center and the ball change?
@123 h is defined to be the distance from.the surface. so h increases when u shrink the surface to a point

Okay... I found the solution. Problem and solution lies in taking $g$, $a_E = \Big(\frac{m}{M_E}\Big)g$
if Earth were point particle then at $h = 1m$ then $g = 4\times 10^{14} m/s^2$ , and at surface $g = 10m/s^2$,

noo

@RyderRude I have calculated the result. Both become same by taking value of $g$ correctly.

i think u r considering the situation where u also moved the ball to 1m above earth after u shrink the earth?

3:35 PM
@RyderRude Yes you are right. Sorry about my explaination.

then h doesnt change, yeah. and u hav to change g now
then it's correct

I was wondered basically in how $a_E = \Big(\frac{m}{M_E}\Big)g$ this equation we can find exact answer of acceleration of the ball or Earth, because there is no information about distance between them. So i found the solution lies in taking the value of $g$.

yeah. g implicitly depends on R
u can do $\frac{m}{M_E}g=\frac{m}{M_E} \frac{GM_E}{R^2}= \frac{Gm}{R^2}$. so the two formulas are always the same

I found one answer here how $g$ is encounter $R$. but i still don't understand N2L has one mass, NGL involves two masses. How they are related.

does diff. form of maxwells equations specify every point in the fields or just the total
i.e does $\nabla \times E = 0$ mean this for every point or for the total region of the vector field

3:46 PM
@123 consider the meaning of N2L for a single particle. it says $m\vec{a}=\vec{F_{tot}}$. the rhs is the total force on mass $m$ due to all the other masses in the universe. so the rhs can depend on the masses and charges of the other particles
in electrostatics, the $\vec{F}_{tot}$ depends on the charges of all the particles

Why does extremization of integral of the Lagrangian density integrated over angular variables w.r.t. some constants that appear in the metric give the near horizon geometry?

@RyderRude Aaaah... Nice explaination.

btw i know div. / curl can be specified by a point but I mean could it be the case that the net div/curl satisfy the eq. but the individual points might not @naturallyInconsistent

But gravity also involve $m$ as well. $F_{tot}$ should be forces due to all other masses. Which should not include the object at which forces are applying.

@123 no, strength of the force is directly proportional to the mass of both particles.
note that the third law wud otherwise b violated. F12=Gm1/R^2 and F21=Gm2/R^2 @123

3:58 PM
@RyderRude Aah Ookay, it means we can say gravity is the type force which depend on both object masses.

yes. in electrostatics, Ftot is directly proportional to the charges of both the particles
masses showing in the rhs is a feature of gravity @123

@RyderRude Pls explain this gravity as per N3L

whats there to explain. |F12|=|F21|=Gm1m2/R2 and direction is attractive (hence opposite)
so gravity's formula obeys the third law

@RyderRude Okay this i can understand. What is the problem when one mass is taken.

@123 if the formula only had the mass of the other particle then, F21=Gm2/R2 and F12=Gm1/R2. this wouldve violated the third law
because the magnitudes are different now

4:03 PM
@RyderRude Aaah i see.. Okay

4:25 PM
Thanks

4:40 PM
I guess it's $\vec{\nabla}\times \vec{E}$ not $\overline{\vec{\nabla}\times \vec{E}}$

@Obliv what are you on about?

@Obliv yeah the equations are satisfied at each point individually, but note that $\vec{\nabla} \times \vec{E}=0$ isnt a Maxwell equation. the general eqn is $\vec{\nabla}\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$

5:26 PM
huh.. i must be misremembering the formulas in class
@NaturallyInconsistent I was wondering if $\vec{\nabla}\cdot \vec{B}$ could be $0$ overall but nonzero at individual points but ryder made it clear that it has to satisfy each point
but then why is it said that magnetic monopoles 'may' exist? if the eq. clearly says they cant

@SirCrackpot In L&L 4 section 123, there is a chapter on the Lamb shift that may be helpful. The section is titled "Radiative shift of atomic levels"

6:12 PM
@Obliv Maxwell's equations alone asserts that they do not exist. But people have studied that there is a hidden rotational-like symmetry of Maxwell's equations that interchange E and B fields, i.e. also electric and magnetic charges, in which case the extended Maxwell's equations can have magnetic monopoles.

6:32 PM
how seriously should one take statements like this? "A sound with an intensity of 1*10-12 W/m^2 corresponds to a sound that will displace particles of air by a mere one-billionth of a centimeter." The calculation works out, but talking about air displacements of 10 pm seems...dubious.

1 hour later…
7:40 PM
those are averages. They can be much smaller than any minimum length unit