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12:36 AM
> The UT Police Department received a Campus Security Authority (CSA) report today of a fondling that occurred earlier this morning. A female reported that a male stranger grabbed her breasts without her consent while stating derogatory remarks.
 
user54412
@Danu It's less surprising when you think about a lack of chemical treatment we sometimes take for granted. Water served boiling hot is at least free of any living germs, whereas you can't be sure with cold water. I suspect that over the generations the old wives' tales about stomach aches and such just developed to justify these habits.
 
user54412
Habits die hard. I know plenty of first- and second- generation Chinese-Americans who can't imagine drinking unboiled tap water because they were trained to not trust it.
 
@ChrisWhite Do you know a lot of Asians?
 
user54412
If you rank-ordered people by how well I know them, I bet Asians dominate the top-N list for 10 <= N <= 100.
 
1:46 AM
@0celo7 @ChrisWhite I have two foreign exchange students in my AP Physics C class. They both will not drink tap water unless it has been boiled.
One is from Vietnam, the other is from China.
 
People from the most densely populated regions of the world have these habits for a reason.
 
I would expect an individual from New York City to hold to similar behaviors.
 
Indeed nyc is densely populated as well :-)
 
Okay, I'm lurking now...
 
 
1 hour later…
3:03 AM
Anyone know the answer to this?
0
Q: If a white dwarf collides with a giant star, could it create a TZO?

Sir CumferenceThorne–Żytkow objects (TZOs) form from collisions between neutron stars and main sequence or giant stars. Ultimately, the neutron star becomes the "core" of the giant star. However, could this also happen with white dwarfs, if the combined mass is below the TOV limit? White dwarfs are degenerate...

 
 
1 hour later…
user116211
4:26 AM
> There is no randomness in quantum mechanics, there is only uncertainty.
 
user116211
Hmmm.... is there no randomness in QM? What about the wavefunction-collapse? How is it related to uncertainty?
 
@MAFIA36790 Who said that?
 
Did you post your question from French's book? @MAFIA36790
 
user116211
@skillpatrol Ha! Nope!
 
user116211
2
Q: "Randomness" versus "uncertainty"

Mitchell PorterHighly rated PhysicsSE contributor @CuriousOne regularly makes the following claim about quantum mechanics (e.g. here): There is no randomness in quantum mechanics, there is only uncertainty. I want to know what this is supposed to mean.

 
user116211
4:39 AM
@SirCumference It's by CuriousOne.
 
4:58 AM
@dmckee @DavidZ the comments here are a disgrace. The question ought to be closed too but I'd like to ask that at least the speculative comments be removed.
 
Hi @DanielSank how are you feeling?
Back to 100%
:-)
 
@skillpatrol thanks for asking :-)
Yes, I'm mostly recovered.
 
nice to hear
 
Still a very intermittent cough but I'd say I'm 98% back to normal.
 
Ah, so it was a flu?
perhaps a super flu
those are nasty
 
5:02 AM
Definitely not flu.
I did not have body aches or any of the characteristic flu symptoms.
It was just a really, really bad cold.
 
What do they say? Feed a cold, starve a flu.
Never worked for me :P
flu fever
Who knows with all these mutated viruses floating around...
 
5:20 AM
@skillpatrol Yeah whoever had that never had flu.
 
user116211
@DanielSank: !!coffee
 
@MAFIA36790 wat?
 
user116211
It would make you feel better.
 
user116211
hmmm....
 
@MAFIA36790 Nah.
Coffee messes me up real bad.
 
5:20 AM
@barrycarter I start work at 6 a.m. BST so yes, I'm afraid I go to bed about 9 p.m.
 
user116211
@skillpatrol: You could understand that, don't you?
 
user116211
Okay, what's the difference between randomness and uncertainty?
 
user116211
The later is a consequence of the former, isn't it?
 
Uncertain is not always random, is it?
 
user116211
@DanielSank You don't like coffee ;/
 
user116211
5:24 AM
@TheGhostOfPerdition: o/
 
user116211
After a long time!
 
5:58 AM
@MAFIA36790 Yeah, done with your exams?
 
@ChrisWhite Nice characterization ;)
 
Do you? @danu :P
 
user116211
6:19 AM
@TheGhostOfPerdition kinda.
 
@
http://physics.stackexchange.com/questions/115151/is-quantity-a-dimension

Actually, how do we know there are only two types of physical dimensions, namely space and time, whereas anything else (e.g. color, temperature, energy, momentum, position, rotation, angle etc.) are all just parameters and unknowns (dimensions in some abstract space)?
 
6:37 AM
There are 7 internationally recognized units of measurement.
SI units
The rest are not just "parameters and unknowns" but derived units from those 7.
 
6:52 AM
@MAFIA36790 good :) all the best
 
user116211
@TheGhostOfPerdition :D
 
7:06 AM
@MAFIA36790 JEE is not over right? Are you planning to do an integrated MSc in physics?
 
user116211
@TheGhostOfPerdition MAINS is over.... Yes.
 
@MAFIA36790 ohh okay, I'm preparing for JEST, planning to take an unpaid long leave from my company for 2 years :D
 
user116211
@TheGhostOfPerdition WoW!
 
user116211
That's really awesome....
 
user116211
I salute to your attitude and decision....
 
user116211
7:09 AM
kudos.....
 
@MAFIA36790 yeah.. That is a safe option I have, I can always ge back to my company if I dont get a job in physics
 
user116211
I mean who has the guts to take a leave for 2yrs and pursue his education?
 
user116211
@TheGhostOfPerdition Be positive.... first target.... a decent degree.... then would come later....
 
@skillpatrol
No, I am talking about dimensions not in the sense of metrology, like how when non physicists talked about dimensions (or even for physicists when they do talk about dimensions such as compactified dimensions in some unified theory) there is always seemed to be only two types ever mentioned, i.e. space and time.

What prevent a 3rd type of continuum to exist?
 
@MAFIA36790 Naah I salute you, you know so much physics at such an young age, much more than me, you are a prodigy
_/_
 
user116211
7:11 AM
that is exaggeration....
 
user116211
@TheGhostOfPerdition stop it.
 
@MAFIA36790 no, I'm serious,
 
user116211
__
 
user116211
::runs::
 
@MAFIA36790 haha
 
7:30 AM
What is your definition of a dimension? @Secret
 
Hello
 
@skillpatrol
I only know of 4 distinct types:
1. Units, which determines how physical quantities affected by scaling. This type is most relevant in dimensional analysis
2. Parameters, which can be thought of as degrees of freedom in some abstract or mathematical space
3. (I don't know the umbrella term for this) the physical ones such as space (xyz, compactified spatial dimensions in string theory and other unified theories) and time, which together formed the continuum called spacetime)
4. Dimensions in the context of linear algebra, which is the size of a basis
 
3 is the same as 4
Space has 3 dimensions because that's the number of basis vectors to span the space
 
How do you define measurement? @Secret
Hello @Slereah
 
@Slereah
but why there are only two types: Space and Time. What prevent a 3rd type of dimension to exist without end up becoming something that live in an abstract space?

@skillpatrol
My memory is very hazy on that one, but from what I learnt in my analytical chemistry and a bit of first year physics. Measurement (not talking about the quantum mech one) is basically any quantity obtained that has a value and an uncertainty associated with it

For measurement in quantum systems, it is what happens when the wavefunction of the system get evolved into the eigenstate of some operator due to th
 
7:45 AM
@yuggib o/
 
@slereah More specifically (I am not sure whether the following makes sense in the physic's context, but the following is commonly heard in pop science and public silence talks) space and time differ from other dimensions is that it seemed to have a notion that it can contain or hold something, as if they are actual places that can be visited
 
@Secret metric signature - there's only positive and negative, corresponding to space and time respectively. (Or the other way around, depending on your conventions.)
 
You can add dimensions for other things if you want m8
you can add the dimensions of the various gauge bundles and whatnot
It's not particularly useful but knock yourself out
 
Is it not acceptable @DavidZ to define the real numbers, initially as, any number that is either positive, negative, or zero?
 
@David Z. Hmm, makes sense (and using conclusions from a few months ago with Danu and Slereah, the value of q in a metric signature for a general metric (p,q,r), which count the number of 0 eigenvalues, does not have physical consequences in those directions because they are degenerate (not exact phrase, will dig that phrase up in the h bar later))

@Slereah I am not sure how to respond to that answer, as I know nothing about gauge bundles, thus I don't know if that sense of "can contain something" is present in the concept "gauge bundles"
 
8:00 AM
Would zero then have a "metric signature"?
 
https://en.wikipedia.org/wiki/Metric_signature

If zero is the zero tensor, then it probably have metric signature of (0,q,0) where q= dimension of the tensor

According to Acuriousmind and Danu, the metric tensors we use in GR are always symmetric and real and nondegenerate.

There are some guys who do work with degenerate metric tensors
http://arxiv.org/pdf/1105.0201.pdf, http://publica-sbi.if.usp.br/PDFs/pd1193.pdf

and some ones who worked with complex valued metrics, as I just dug up from google
correction: the second wikipedia link should be "google complex valued metric tensor", a few things poped up, including arxiv.org/pdf/gr-qc/9911121.pdf";
more on complex valued metric tensor:books.google.com.au/…
 
8:16 AM
no you fool
Nooo
 
The thing I noticed: no physics
 
What the hell are they doing in studio art!
 
Not a lot of studying apparently
 
Good observation @Secret
 
8:38 AM
@skillpatrol probably
But you can't have a coordinate with zero metric signature because it wouldn't be part of the metric.
 
8:49 AM
@DavidZ

(I know I have not read those papers I linked above in full due to lack of time). But what about those notions of degenerate metric tensors that those guys are working on in semi riemannian manifold models, how does "can't have a coordinate with zero metric signature because it wouldn't be part of the metric." fit into that picture?
 
I don't know about that
 
@DavidZ
btw, for particle physics, when particles interact or decay to form new particles, is a notion of transition state exists in the step between the starting point and end point of the reaction?
that is, are there examples (theory or experiment) where there are observations of hadrons and elementary particles going over some energy barrier before they can react, or organised into some kind of transient state before resulting in the product particles we detect?
 
Yes, that's how collisions normally work. At least the interesting ones. All unstable particles are detected only as transient states.
 
https://en.wikipedia.org/wiki/Transition_state

I see

So I am guessing something like this happened in a decay, similar to how in chemistry when molecules fragment?
$$\pi^+\rightarrow \text{Some high energy intermediate state involving the superposition of states of $\pi^+$, $e^-$ and $\nu_e$} \rightarrow e^+ + \nu_e$$?
 
Is Introduction to Algorithms by Thomas H. Cormen worth it?
 
9:05 AM
@Secret The intermediate state need not have anything to do with the individual states for $\pi^+$, $e^+$, and $\nu_e$ - except that it obviously needs to satisfy conservation laws.
 
I see
 
In this case the intermediate state would be a $W^+$ (or at least that is the most likely option)
 
I wonder if we can ever unravel the underlying mechanism of these intermediate states

It seems particle physics is kinda like photochemistry (both have notions of branching ratios, decay channels, reaction barriers etc.) , except that it is relativistic, and that intermediate states does not need to be a combination of some constituents, unlike in chemistry where the transition states and intermediates always involve the atoms in the products and reactants

Perhaps, the analogous quantity in particle physics to atoms in photochemistry will be those conservation laws (e.g. color charge, iso
---
is there anything else I miss (or oversimplified) in the above compare and contrast between the two fields?
 
Well, there's quantum field theory. It doesn't necessarily give fully satisfying explanations at the lowest level, but it does explain to some extent why the processes work as they do.
 
Describing interacting QFT with particles is not gonna be particularly true, anyway
there are no particles involved in the "real" process
 
9:20 AM
I see

I am comfortable with the concept of nonlocality and fields, thus I am ok with "there are no particles involved in the "real" process"

But I agree that is a big contrast to photochemistry, because even for the highly quantum photochemical systems, you can still sort of predict things with the trajectories of the atoms in the molecule that is moving on some potential energy surface
 
(Beginning unrelated comment)
It always amaze me how invariant to scale some concepts and patterns really are. For example, the behaviour of electrons circulating in some solid is similar to what bacteria are dong when confined into some cavity.

It is possible that we are in a simulation, but I am not sure what we will do if this is found to be true
(End unrelated comment)
 
Btw when digging around some quantum stuff, I stumble upon this
https://en.wikipedia.org/wiki/Quantum_potential

It sounds very similar to that "trajectory like" interpretation I have a few months ago that discussed with acuriousmind and slereah

except that this is from pilot wave theory thus deterministic, unlike the copenhagen and other interpretations that there is genuine randomness involved
 
9:47 AM
http://arxiv.org/pdf/1601.02897v1.pdf

(Old news)
Need to revise about heat, cause still not able to shake away the feeling that if the heat bath is the surrounding universe, then the heat transferred between the two baths will be more efficient than the case where heat flow from a localised heat bath to the colder suurroundings
perhaps, the only thing that matters is the area of contact between the two baths
 
10:19 AM
56
Q: Why is quantum entanglement considered to be an active link between particles?

Andrey TatarinovFrom everything I've read about quantum mechanics and quantum entanglement phenomena, it's not obvious to me why quantum entanglement is considered to be an active link. That is, it's stated every time that measurement of one particle affects the other. In my head, there is a less magic explanat...

 
10:50 AM
Using the answers there (and the discussion with h bar throughout these months) I wonder, if entanglement can be interpreted as follows (combining nonlocality and nonrealism):

Suppose we have two photons which can have two different polarisations: Clockwise(+) and anticlockwise(-)

We first prepared two photons of opposite polarisations
$$\lvert + \rangle and \lvert - \rangle$$
Next we brought them closer together and then interact them using some machine (which its interaction is included in the hamiltonian as usual). This prepared the following entangled state
correction: $$A(\frac{1}{\sqrt{2}}(\lvert ++\rangle + \rvert -- \rangle)) $$
 
I have no idea what you're talking about. The entangled state is $\frac{1}{\sqrt{2}}\left(\lvert ++\rangle + \lvert --\rangle\right)$. If you measure some operator $A$ on it, the state after measurement will be an eigenstate of $A$. Why do you have so much more words in what you wrote there?
Also, when measuring $\lvert \psi \rangle$, note that $A\lvert \psi \rangle$ has nothing to do with the measurement process. You never write down the operator applied to the state when you're just interested in measuring the operator.
Also, all these newlines and uses of display instead of inline math make your posts harder, not easier to read, at least to my eyes.
 
I don't get the $A\lvert \Psi \rangle$ part, isn't that if you measure the entangled state (and the effect of the measurement is represented by the operator A) you decohere the system in the process thus changing $\lvert \Psi \rangle$ into some new state?
 
Yes. But that resultant state has nothing to do with $A\lvert \psi\rangle$:
Unless you do something weird, it's just one of the eigenstates of $A$.
 
I would say probably don't worry too much about quantum interpretation until you understand QM better
As the saying goes
Shut up and calculate
 
11:05 AM
@ACuriousMind cc @Secret I was just thinking the same thing. Long multi-line messages with line breaks are difficult to parse in chat.
 
QM interpretations get a bit easier to understand if you know the basic mathematics of QM
 
Let me think, is the basis that describes the eigenstates of A are ${\lvert ++ \rangle, \lvert +- \rangle, \lvert -+ \rangle,\lvert -- \rangle}$ instead of just $\lvert + \rangle$ and $\lvert - \rangle$?
I felt like that might be why I get confused by the $A|\Psi \rangle$ statement
 
No idea, you never defined the operator $A$.
 
I mean in general, e.g. suppose A is a general hermitian 2x2 matrix
 
If $A$ is a 2x2 matrix, it doesn't act on the 4-dim space that e.g. $\lvert ++\rangle$ lives in.
 
11:11 AM
I am super bad at doing maths whenever this symbol $\otimes$ is involved, always forgetting where a matrix is acting when it operates on a sum of some a $\otimes b$
Guess I will cleared up this bit before going back to this paragraph of stuff (which I had something I want to express, but I must ensure I got all the maths right before i get there)
It's so annoying to go home from uni today because the internet at home broke down
 
11:32 AM
Trying to redo all of Peskin step by step
Gonna take a while
I'm up to Noether's theorem :V
 
11:59 AM
Good morning miscreants
@ACuriousMind QM prof said he saw no reason to know functional analysis for QM
Also I think he thought I was a grad student for a few minutes
 
Schrodinger QFT is functional analysis
 
How's Peskin coming along
 
What I said
Up to Noether
Proving about conserved current for translation now
 
@dmckee The QM prof wanted to know if the sophomore E&M class I'm taking next semester is taught out of Jackson
Apparently he learned PDEs and stuff in high school...
 
Woo
Onto ACTUAL QUANTUM THINGS
The part I don't look forward to is chapter 6
that is when shit gets complicated
I should fix my Peskin
The cover is hanging a bit loose
 
user54412
12:52 PM
@FenderLesPaul How's UCSB? I see now why Gary couldn't be there, since he's here right now for a mini GR conference.
 
Is mini GR the study of tiny spacetimes
 
@ChrisWhite What are they talking about?
 
user54412
@Slereah You would come up with something weird like that, wouldn't you? ;)
 
weird imaginative
:P
 
user54412
@0celo7 Everything from waves to duality to astrophysics to computation. pdf of program
 
12:56 PM
will you be attending?
 
user54412
As a graduate student, I'm obliged to find the promised free food now.
 
@ChrisWhite Meh.
How the heck did they come up with that Lemma oO
 
It is Trivial
 
@Slereah Well if the derivatives of $f$ are bounded and $\dot x$ is just a derivative of $f$, it is certainly bounded as well?
 
Sounds alright
 
1:06 PM
Trivial for Leonard "J." Crabs :P
 
Wait, what the heck is $||\cdot||_{C^1(\mathbb{R},TX)}$
 
Hm
Trying to do the SHO
But for the Hamiltonian in ladder operators
I get $H = a^2 + {a^\dagger} ^2$
 
sounds about right
 
Since $H = x^2 + p^2 = (a + a^\dagger)^2 + (a - a^\dagger)^2$
 
what should it be
 
1:18 PM
$H = a^\dagger a + \frac{1}{2}$
 
oh right
let's get out THE BOOK
Ok, the trick is this:
calculate $a^\dagger a$
 
Hm
so you can't know it in advance?
 
you will get $$a^\dagger a=\frac{m\omega}{2\hbar}X^2+\frac{1}{2m\omega\hbar}P^2+\frac{\mathrm{i}}{2\hbar}‌​[X,P]=\frac{H}{\hbar\omega}-\frac{1}{2}$$
@skillpatrol define number?
 
@0celo7 real number?
 
@Slereah Try again, then. You can get $a^\dagger a + \frac{1}{2}$ by factorizing in the right way.
but it's easier to just compute $a^\dagger a$, as @0celo7 says
 
1:24 PM
Are the two equivalent?
 
@ACuriousMind would you happen to know about the morse thing above
 
is $a^2 + {a^\dagger}^2 = a^\dagger a$
 
@0celo7 What Morse thing?
@Slereah No
 
24 mins ago, by 0celo7
user image
 
Hm
Then what did I fuck up
 
1:25 PM
You can see that because the l.h.s. goes from the n particles space to the n+2 and the n-2 particle space, while the r.h.s. remains in the n particle space.
 
@skillpatrol you can't define real numbers as all the "numbers" when you haven't defined "numbers"
 
@0celo7 and what do you hope that I happen to know about it?
 
you need to construct the integers via sets or something
then the rationals as the fraction field
 
$H = x^2 + p^2 = (a + a^\dagger)^2 + (a - a^\dagger)^2 = a^2 + {a^\dagger}^2 + aa^\dagger + a^\dagger a + a^2 + {a^\dagger}^2 - a a^\dagger - a^\dagger = 2a^2 + 2 {a^\dagger}^2$
What part of it fucks up
 
and then dedekind cuts or limits to get the reals
@ACuriousMind I don't understand anything after the $\ddot x$ equation
 
1:28 PM
@0celo7 did you even read what I asked David?
 
@skillpatrol no
link?
 
@skillpatrol I can't find the question
 
i said nvm
 
I don't speak in acronyms
 
1:30 PM
@Slereah Hm
 
@0celo7 so you can dish them out but you can't take them?
 
@ACuriousMind I do not understand why $\dot x$ is bounded! I know that the derivatives of $f$ are bounded, but $\nabla f(x(t))$ depends on the flow lines
Or does it? I'm a little confused.
 
Lol. I just realized I am in the all time list of top optics answerers on this site:
 
The gradient is just a sum of derivative
 
1:31 PM
@skillpatrol When have I ever used acronyms?
 
The sum of bounded bits is bounded?
 
@0celo7 It's value depends on the flow lines, but the bound on the derivative of $f$ bounds it, too.
 
@ACuriousMind So $\nabla f(Y)$ is bounded for any $Y$ I want?
 
The worst part of the story, I looked at that list because I posted an optics question and was trying to figure out who is likely to come to my rescue!
 
@skillpatrol I would say I am rather anti-acronym
 
1:34 PM
(And on the all time list with 3 answers having a total score of 4 !!!)
 
Jul 15 '15 at 15:29, by 0celo7
Atlanta-based chapter of NAACP has proved once again that the acronym was well chosen. If you don't know, NAACP stands for "Blacks Are Always Creating Problems" and it is an organization of blacks who always want to create problems. (No, the actual expansion of the acronym is "National Association Against Caucasian People".)
 
@skillpatrol I was quoting Lubos there.
I don't see what your point is.
 
now you sound like Trump :P
4 mins ago, by 0celo7
@skillpatrol When have I ever used acronyms?
 
@skillpatrol Can I be rich and handsome like him too?
 
3 mins ago, by 0celo7
@skillpatrol I would say I am rather anti-acronym
 
1:37 PM
I would buy @Slereah a PhD if I had Trump money
 
@0celo7 Yes, that's what $\nabla f$ being bounded means!
 
@ACuriousMind I'm not convinced!
 
If $\nabla f (x) \leq c\forall x$ then of course $\nabla f(x(t))\leq c \forall t$. You just restricted the values of $x$ to $\{x(t)\mid t\in\mathbb{R}\}$, how should the bound disappear when making the set that we quantify over smaller?
 
would you make america great again, tho
 
Mar 26 at 14:06, by ACuriousMind
@Danu He has a strange aversion to being convinced by proofs :P
oh wait
oops
@ACuriousMind Yes, it's trivial, I agree.
@Slereah Do you own Courant and Hilbert?
 
1:45 PM
No
 
Oh, one of my profs owns all of the Spivak books.
They really do have those strange covers.
 
they worship his books in the math room
 
@skillpatrol I would get them but I have enough books for now
And I'm trying to get the department to get me some more books
 
ted helped him write the exercises
 
:3
Ted's an ass
 
1:48 PM
don't say that
 
why?
 
he's a nice guy
 
@0celo7 Maybe don't say that because it's not nice to speak badly about people where they are unaware of it?
Or maybe don't say that simply because it is an insult.
 
He got really upset at me for calling him "dude"
 
Who's Ted
 
1:54 PM
That's necessary and sufficient in my book.
 
@Slereah here
 
@0celo7 You feel uncomfortable with calling your profs by their first names when they ask you to, but you're fine with calling a professor emeritus you've never met "dude"?
 
@ACuriousMind Of course.
 
A professor emeritus on the internet is still just a random dude
 
^
 
1:56 PM
If the pope came here I would call him Duderino
 
@0celo7 well, you are OUT of my book ::on ignore::
 
Especially because I don't know him.
 
I don't get why you would think disrespect is acceptable just because you're on the internet.
 
Disrespect?
 
"dude" is not disrespectful for us younger folks
I assume it is among the aging old men
 
1:58 PM
@0celo7 You'd consider it disrecpectful to walk up to him and call him "dude" in real life, wouldn't you?
Or what is the reason you don't want to call profs by their first names?
 
whatever
 

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