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5:00 PM
@Obliv set theory is extremely complicated
 
user116211
@Obliv Hell no!!
 
user116211
Ask Asaf in Maths.SE.
 
user116211
@Obliv Are you talking about Naive Set Theory? I must add this is set theory for the dummies. There is much much more in Set Theory that needs to be explored.
 
@mafia can you give some examples
 
user116211
@Obliv Have you read ZF axioms?
 
5:03 PM
zfc?
 
user116211
@Obliv ZF and ZFC is different.
 
oh is that inner model set theory
 
user116211
The latter contains the Axiom of Choice.
 
I am TIRED of my analysis prof giving WRONG proof sketches
wtf is up with this
 
@0celo7 how do you suppose they're wrong'
 
user116211
5:05 PM
@0celo7 Why would he do that?
 
@JohnRennie there was a discussion about that here.
 
@Obliv because there are false statements in them
@MAFIA36790 he lectures from memory and gives hints on the homework which are just wrong
He knows they're wrong too, which is annoying
 
user116211
@0celo7 Then report it to the Dept!
 
user116211
He is potentially hazardous.
 
user116211
BTW, did he realize that he is providing wrong hints?
 
5:07 PM
He's my advisor
 
@MAFIA36790 were you talking about these axioms en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory
 
I'm not reporting him
 
user116211
._.
 
@MAFIA36790 He realizes they "take some work to be correct"
 
how old is he
 
user116211
5:08 PM
weird. very very weird.
 
maybe he just doesnt give a fuck anymore
 
user116211
@Obliv yes.
 
flagging
 
if u do that i'm not going to talk to you anymore
 
one less annoyance in my life...
oh crap I have a lab report to write
 
5:10 PM
what is it on @0celo7
 
proving that amps are related to Joules or some shit
it was calorimetry with a ceramic resistor
 
user116211
Hate these stuffs ;/
 
@0celo7 aren't the amps related to resistance which is related to the temperature of the object?
 
user116211
My lab works are due for a month or longer; hmm. No interest at all; but have to pass on those; damn ;/
 
@Obliv no clue
I need to read the lab again
we did it a week ago
 
5:15 PM
@mafia how many labs ?
 
user116211
@Obliv Don't know; have to check; all with pendulum and sort of that ;/
 
@mafia really? are you taking a mechanics class ?
 
user116211
@Obliv We have Newtonian Mechanics in 1st sem syllabus; so yes.
 
he's a freshman
 
o.e' you all are so held back by the education system lol..
you @0celo7 and @iceL that is
 
5:18 PM
no?
 
except 0celo7's taking hard classes actually
so nvm
 
my EM class is too hard
idk how the other engineers do it
 
user116211
@0celo7 I love it; waiting for the 2nd year.
 
at least in high school you were all held back.
 
aha
I figured out the problem
 
user116211
5:19 PM
I'm waiting when to completely read Jackson.
 
jesis
I couldn't read Jackson
way too hard
 
user116211
Do you guys have to do the exercises of Jackson?
 
@MAFIA36790 Jackson is PhD level, my class is much easier
who the hell does Jackson as an undergrad
 
user116211
@0celo7 But it is recommended for our under grad classes along with Purcell :(
 
India is messed up
 
user116211
5:22 PM
It's not the textbook; Purcell is. Jackson is just recommended; it is upto the prof who teaches whether he uses Jackson or not.
 
what's so hard about jackson @0celo7
 
user116211
@Obliv Did you check the problems? They are insane ;/
 
dling a pdf now
classical electrodynamics?
 
user116211
@Obliv yes.
 
people drop out of grad school because of Jackson
Let $(X,d)$ be a metric space. If a sequence $(f_n)\subset C_E(X)$ converges to $f\in C_E(X)$ uniformly on $X$, then the family $\mathscr F=\{f_1,\dotsc,f_n,\dotsc,f\}$ is equicontinuous at each $x\in X$.

*Proof.*
Let $\epsilon>0$ and fix $x\in X$. There is a $\delta_x(\epsilon)>0$ such that $||f(y)-f(x)||<\epsilon/3$ whenever $d(x,y)<\delta_x$ since $f$ is continuous at $x$. Also, since $(f_n)$ converges uniformly, there is an $N(\epsilon)\in\Bbb N$ such that $||f_n(y)-f(y)||<\epsilon/3$ whenever $n\ge N$, for any $y\in X$. Under these conditions,
fun times with analysis
 
5:29 PM
uh oh
 
what
 
I'm reading into jackson a bit..
it says its self contained but i don't even think it's deriving the equations it's giving
 
proof?
 
user116211
@Obliv 'Method of Image section is quite good.
 
first chapter i don't think he derives maxwell's equations but I didn't really read it much maybe he explains the motivation & background @0celo7
 
5:35 PM
you can't derive them
they are Empirical
deriving them from Yang-Mills doesn't count
 
@0celo7 i wouldn't know because my E&M class over the summer didn't have time to go over them lol
maybe I should learn them since my next lecture will be on light where he assumes we know the equations
 
user116211
@Obliv Use Purcell to learn.
 
@mafia is purcell good?
 
I've heard bad things about Purcell.
 
user116211
There is a better but older Dover text by Panofsky and Philips.
 
user116211
5:41 PM
@Obliv Really intuitive.
 
do i need to use it just to learn maxwell's equations?
 
user116211
@0celo7 No idea why the negative view.
 
user116211
@Obliv You can.
 
user116211
It'an introductory undergrad text.
 
user116211
The exercise is good; though I have done very few in my high school.
 
user116211
5:43 PM
There is a solution book written by Purcell himself.
 
jeeeez what was originally a 9mb djvu file turned into a 265mb pdf after conversion. I guess I can always refer to purcell whenever I want to know something not covered in my other textbook
which is many things
 
user116211
@Obliv WTH ;/
 
@mafia djvu is veerrrry compact for some reason. maybe because purcell has a lot of graphics?
 
user116211
@Obliv Lots of pics.
 
sign of a bad book
the best books have zero pictures
Li, Chow-Lu-Ni, Kobayashi-Nomizu, Helgason, Wolf.
 
5:49 PM
@0celo7 aren't there many pics in geo/top.
 
Not in good books, no.
 
wtf
 
Authors only need pictures when they're bad at explaining.
 
user116211
@0celo7 Bourbaki.
 
Jost is light on pictures
 
5:50 PM
I guess it does help the person get better at visualizing things themselves.
 
you should not have to visualize
let the axioms guide you
 
user218912
^
 
user116211
@0celo7 Purcell is really good book, no offense.
 
@0celo7 I'm not a computer A.I. like you so it helps to visualize ideas
 
user218912
@Obliv read tong's notes.
 
user116211
Purcell is an introductory undergrad book ---- for beginners. Pics are must for the discussion.
 
user116211
Even Jackson has pics.
 
Kolar et all have some pictures
 
@iceL I remember saving this damtp.cam.ac.uk/user/tong/dynamics/clas.pdf for when I finish calc 3
I think 0celo7 recommended it to me
 
user218912
I also recommended it to you.
 
user218912
5:54 PM
@Obliv didn't you finish calc 3?
 
user116211
@Obliv My pdf was 28006kb; I checked now.
 
Nope. Should have back in the spring though
 
user218912
@Obliv it's okay you don't need calc 3.
 
user218912
just learn it while you read the notes.
 
5:55 PM
@MAFIA36790 uhh for purcell? why is mine 265mb
 
user116211
@Obliv No for Jackson.
 
oh yeah jackson was 87mb
 
I'm stuck on my homework :/
 
user218912
@0celo7 that's me 24/7
 
Are you still taking analysis?
 
user218912
6:01 PM
no I'm taking advanced stat mech instead.
 
user116211
1
Q: Why is de Rham cohomology important in fundamental physics?

SchlomoI'm currently working on some mathematical aspects of higher-spin gravity theories and de Rham cohomology pops up quite often. I understand its meaning as the group of closed forms on some space, modulo the exact forms. Before diving into concrete mathematical details, I was wondering if anyone...

 
6:35 PM
Could anyone answer this?
0
Q: How could lithium burning take place in a quasi-star?

Sir CumferenceAccording to Begelman et al. (2008), one of the most distinguishing features of the hypothetical quasi-star is that it's supported by radiation pressure from the accretion disk of the black hole in its core. The paper suggests that these stars form from Population III stars. What's more, while a...

 
nope
 
user116211
@SirCumference No idea.
 
@sirC what I want to know is how a star forms from the collapse of a star into a blackhole
like what, is there a blackhole inside of it?
 
@Obliv Yep
 
what would stop it from collapsing into it?
 
6:43 PM
I'm lazy so I'll copy-paste from the paper
 
oh wait i think it says in your question. It's supported by radiation pressure from the accretion disk of the black hole
 
user116211
 
that's so .. strange o.e
 
Will continue on this tomorrow
 
> In outline, the BVR model envisages a three-stage process for black hole formation. First, gas in metal-free haloes with a virial temperature above $T = 10^4 K$ flows towards the centre of the potential as a result of gravitational instabilities, forming a massive, pressure-supported central object.
> Nuclear reactions may start, but the very high infall rate continues to compress and heat the core, precluding formation of an ordinary star. Eventually, when the core temperature attains $T ∼ 5 × 10^8 K$, neutrino losses result in a catastrophic collapse of the core to a black hole.
> We dub the resulting structure – comprising an initially low-mass black hole embedded within a massive, radiation-pressure-supported envelope – a quasi-star. Initially, the black hole is much less massive than the envelope. Over time, the black hole grows at the expense of the envelope, until finally the growing luminosity succeeds in unbinding the envelope and the seed black hole is unveiled.
 
user116211
6:44 PM
Why does the author say there is no algebraic formulation between $\phi$ and $\phi^\prime$?
 
Their relation is analytic, i.e. via the derivative
 
user116211
yes, so?
 
That's not algebraic
 
@secret is that a giant list of definitions of algebraic structures?
 
In general there is no algebraic function $f$ s.t. $\phi'=f(\phi)$.
algebraic being polynomials, rational functions, etc.
 
user116211
6:46 PM
@0celo7 somewhat got that.
 
user116211
@Secret You are crazy ;)
 
@Obliv Ya wanna know what's cooler?
 
@sirC why would blackholes emit radiation
 
Look how big quasi-stars would be
 
@MAFIA36790 is this the book that made that crazy claim about GR
 
6:49 PM
@Obliv The heat from the accretion disk emits light.
That's how quasars form.
The accretion disk gets insanely, unimaginably hot.
 
user116211
@0celo7 yes; Cornelius Lanczos' The Variational Principles of Mechanics.
 
probably not a great book
 
@sirC where would the lithium be burning, in your question?
 
user116211
@Obliv Hawking Radiation? I've no idea though.
 
@Obliv In the core, surrounding the black hole.
 
user116211
6:50 PM
@0celo7 Better than Goldstein.
 
I want to understand electrolysis of aquous copper II sulphate
Ca sumone explain it to me >
 
user116211
@Abcd Ask at Chem.
 
@Obliv Also, technically black holes emit Hawking radiation, but that's not what I'm talking about in the question.
 
thats a dormant site
 
@sirC my guess would be that the accretion disk gets hot enough from the in-fall of solar material that it fuses the hydrogen/helium into lithium then that burns :D?
 
user116211
6:52 PM
Although Lanczos does not contain Perturbation Theory.
 
@Obliv That actually is a very good suggestion, but the problem is that most core temperatures for quasi-stars lie between $10^5–10^6 \mathrm{K}$, and pp II fusion (which produces lithium) only happens at temperatures of at least $1.4 \times 10^7 \mathrm{K}$.
 
user116211
@SirCumference make the units upright.
 
@MAFIA36790 Fine
Good?
 
user116211
@SirCumference :)
 
7:15 PM
@sirC why would the temperature be less than pp II fusion
 
@Obliv What is this paper?
Er, it's actually a presentation...
 
it's on the efficiency of accretion
 
Even if they were high enough for lithium burning, they might not be hot enough for $\beta^-$ decay, which is necessary in this case before pp II fusion
(it usually comes after lithium burning in modern stars though)
 
rob
@SirCumference In what context does temperature affect beta-decay rates? Or do you mean electron capture?
 
@rob No. The reason why lithium burning takes place in pre-main sequence stars, before hydrogen fusion takes place, is because it does not involve weak interactions.
Therefore the cross-section at a given temperature is much higher than the first step in the pp chain, regardless of the stronger coulomb repulsion between the reactants.
Also just to mention, deuterium fusion takes place even before lithium burning.
The problem I find regarding quasi-stars is that the temperatures may not be hot enough to initiate hydrogen fusion, and there is not much inherit lithium or deuterium in Population III stars to start their respective types of fusion.
 
rob
7:33 PM
@SirCumference So, the competition between (free, in a stellar core) electron capture and positron decay of Be-7? I can see that being strongly dependent on temperature.
 
7:46 PM
"Lanczos" means "Chained" on Hungarian. And, "Erdos" is "Forester"
 
he was probably a prisoner then
 
@0celo7 is it bad to think of EM waves as propagating magnetic & electric fields instead of photons?
 
I'm not a physicist.
 
I think the latter description is more physical even tho they are massless.
dude @0celo7 help me think about this.. what the hell are EM waves if they don't displace any medium but supply energy to matter?
 
I'm busy.
 
8:00 PM
Hi, everybody.
 
good afternoon @daniel
 
rob
@Obliv The electromagnetic fields (and, while we're at it, the matter fields) are properties of the vacuum that arise when you demand that the vacuum be unchange under translations, rotations, and boosts.
 
I'm retracting my vote for @ACuriousMind because of his recent absence.
 
@rob unchanged? But, if a charged particle is accelerated it releases an EM wave that displaces other particles, no? Doesn't that count as being changed
 
I want my representative to actually use the site.
 
8:05 PM
wait translations,rotations,boosts of what @rob I should have asked first
 
rob
@Obliv It's relativistic invariance: all inertial observers have to see the same vacuum.
 
so EM radiation is a property by default of any vacuum that you consider to be relativistically invariant? @rob Hmmmmmmmmmmmm
 
Yeah I doubt that too.
There's no reason you need to have an EM field to begin with.
You can have a vacuum with no fields just fine.
 
rob
@Obliv The ability to support EM radiation: an EM field (which may be locally zero). Actual presence of EM radiation depends on nearby sources.
 
oh okay.. So I see where you're coming from but like I said, what do you mean by unchanged then? Even if every observer observes the same vacuum, do we say that an EM wave caused by accelerating some charged particle which then causes something else to accelerate leaves the vacuum unchanged?
 
rob
8:11 PM
Accelerated motion isn't a symmetry of the vacuum, so accelerated motion changes the EM field (for all observers).
 
@rob can the EM field be considered a medium through which EM waves travel? well this all makes sense but then why are we able to see the propagation through the EM field?
we can't see the field itself but only the propagation?
 
rob
"A medium" suggests that the medium could be removed from some underlying "true vacuum," which isn't the case.
The same symmetries that tell us we have to have the EM field tells us that the EM field must support oscillations.
 
@rob well if I wave my hand in an accelerating path, does that release EM waves?
 
rob
8:27 PM
It does if you're holding a charge. But you also used EM waves, traveling down your nerve fibers and activating your muscle chemistry, to move your hand in the first place. The theory is time-symmetric.
 
@rob two questions: 1. do neutrally charged particles, then, get excluded from this symmetry of the vacuum (i.e they can accelerate without any consequent release of energy)
2. What is time-symmetric? Time is relatively decided so I wonder if there is a geometric explanation of this symmetry instead
 
rob
All of the particles that we know about interact with some field, which is how we have found them.
Neutrinos do not directly couple to the EM field, which makes them hard to detect.
Neutrons do couple to thanks to magnetic moment.
 
If they do not directly couple to the EM field, how are they detected (what causes them to react?) Neutrons are not fundamental so they possess some hidden charge, right? Is that what you mean?
 
rob
Time symmetry means that the laws of QED don't discriminate between time running forward and time running backward.
However for large complex systems, entropy emerges and futures are overwhelmingly likely to be different from pasts.
Neutrinos exchange energy with more ordinary matter via the weak fields. The ordinary matter is charged.
The ordinary matter is electrically charged.
 
Are you saying complex systems aren't time symmetric @rob ? sorry I got confused. So then would charged particles interact with other charged particles within this weak field too?
 
rob
8:39 PM
We've started having two simultaneous discussions. Let 's shelve entropy and time symmetry for later. (There's a nice recent book by Sean Carroll.)
The weak fields are like the EM field: another property of the vacuum that arises from symmetry arguments.
 
Sure. Then the EM field isn't the only field in this proposed vacuum. In every case, though, there is a set of symmetry laws that cause each interaction? Would it be fair to say the neutral particles don't exist in the EM field?
 
rob
Any completely neutral particle would be invisible.
That's a proposed model for dark matter: no charge relative to the fields of the Standard Model, but a nonzero gravitational mass.
 
@rob I'm going to grossly simplify this then. Suppose you have an object $A$ that has properties in the set $P_A = \{a_1,a_2,a_3,...\}$ and suppose $P$ is finite. These properties in $P$ are only observed by observing the interaction between $A$ and another object $B$ with properties $P_B = \{b_1,b_2,b_3,...\}$. Through each observation, though, the cause of interaction is linked to the apparent symmetry in motion of the vacuum?
Suppose the properties in each set only interacted if they were both non-zero. I.e pretend $a_1$ and $b_1$ were charge. Then $A$ accelerating causes $B$ to accelerate iff both $a_1,b_1 \ne 0$?
 
rob
That's a very mathematical way to put it, but I think I agree with you
 
hold on
I'm not sure what I said was correct. I have no idea how our eyes & brain work actually.
Okay, yes @rob so our retinas detect EM waves which send signals to our brain (I think). What is required of an electrically-neutral particle to be able to weakly interact with another particle?
or rather what would the property be called?
 
rob
8:52 PM
The EM field and the weak fields are quite closely related.
There are two electrically charged weak fields, whose quantum excitations are called the W^\pm.
There's an electrically neutral weak field, whose quantum excitation is called the Z.
And the photon is the quantum of excitation of the electromagnetic field, which doesn't have any weak charge.
The difference is that the photon is massless, while the W and Z are quite heavy.
 
There are 3 weak fields in total, then? How would the neutral one be detected, if it doesn't interact with normal matter?
 
rob
Futhermore the matter fields have a different arrangement of weak charges than the weak field carriers themselves do. The mixing was first described by Weinberg, and the relevant parameter is called the Weinberg angle.
Coupling strength to the Z boson is sometimes called the "weak charge," see physics.stackexchange.com/a/158838/44126
You've probably heard of the selection rule in quantum mechanics that says you can't emit or absorb a photon without a change in angular momentum.
 
@rob Ok, but an indirect coupling, for example by virtual W bosons, isn't possible? It would mean some EM interaction of the neutrinos. It would mean that the neutrinos, although very weakly, but take part in EM interactions.
 
No, i wonder why that is though @rob
Matter fields..what are those?
 
rob
It's because the photon carries its own spin, which has to come from somewhere.
In a way, the photon is like a raising and lowering operator for spin-half particles
... and it's by considering electromagnetism and invariance under boosts that the Dirac equation predicts spin-half particle with associated antiparticles.
Similarly, the W particles act like a raising and lowering operator for another quantity called "weak isotopic spin"
 
9:02 PM
Why not have 0 spin?
 
rob
We usually think of it as particle flavor, but the particles come in related pairs.
Interacting with a W turns an electron into a neutrino, and vice-versa (so that the electric charges are conserved).
In fact it's only because the weak force is feeble that we can think of particle flavor as a relatively stable property of matter.
If the W were also massless, electrons and neutrinos could freely convert back and forth, and the universe would be quite different.
@Obliv The electromagnetic field transforms under rotations like a vector, and has to have spin 1.
@Obliv The theory of the zero-spin field, which transforms under rotations like a scalar, and its connection to the rest of the electroweak sector was worked out by, among others, Peter Higgs. You might have heard of it :-)
 
Sorry I don't understand.. @rob What do you mean by EM field transforms under rotations like a vector? I'm imagining a charged particle linearly accelerating, creating EM waves or photons if you will (is this correct so far). Then, the photons must have 1/2 spin? maybe I should search up what spin is
I'm just assuming it's related to angular momentum which, if it is, I don't see why a linear acceleration would create a disturbance that has angular momentum
 
rob
The electric field is a vector, yes?
 
Yes
 
rob
If you build a capacity and make it so that the electric field points up, and I look your capacitor while I'm standing on my head, we have to disagree about whether the direction of your field is "headward" of "footward."
That's what it means to transform under rotations like a vector.
The mass of your capacitor is a scalar: we agree on that even though I'm standing on my head.
There is a theorem in quantum field theory that says that any field that transforms under rotations like a vector must have unit spin.
 
9:16 PM
@rob What is the cause of this, though? Does this mean the emitting particle gains/loses angular momentum because of the photon's intrinsic spin?
I mean, can nothing, at least electrically charged, then accelerate without changing its angular momentum?
 
rob
With each photon, yes, there's a change in angular momentum. However if you emit many photons while accelerating (e.g. in brehmsstralung) there's no reason for all the photons to be polarized the same way.
 
user218912
I feel like I don't want to do condensed matter anymore ._.
 
rob
If all the photons are polarized the same way, you add up to what in classical physics is called circularly polarized light.
Linearly polarized light, or for that matter unpolarized light, is just a mix of the two circular polarizations.
 
Ah I see. @iceL why is that?
@rob I have more questions about field interactions (esp. matter & weak fields) but I'll leave those for later since I'm actually starving lol
Thank you very much for everything that you've shared with me thus far, though :)
 
rob
@IceLord The condensed matter folks get to look at these same issues with phonons and spin waves; that's part of what's neat about this year's Nobel
@IceLord You should read Robert Laughlin's "A different universe," which does a great job of making the connection.
@Obliv no problem. These discussions are useful for me to hash out how I think about these things too.
 
user218912
9:26 PM
@Obliv idk just ignore me I'm still thinking.
 
@IceLord There was a typo on the homework
4 hours GONE
 
user218912
@0celo7 lol
 
rob
@0cel07 No worries ... good preparation for research where your own typos will have you wasting your own time
 
the problem statement seemed unreasonably nice anyway
 

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