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12:00 AM
@JaimeGallego The answer is simple: physics.
Why? Because if you pick physics you get to do engineering and math anyway.
@DanielSank Yes, the "first principles" physics approach to engineering is quite attractive.
@JaimeGallego Well also note that once you get to real life physics, at least as an experimentalist, you do plenty engineering anyway.
For the last couple weeks I'm working on a circuit board that processes RF signals.
It's all circuit design.
Plenty engineering.
The cool thing is that those RF signals carry the information about the quantum state of a superconducting qubit. So there you go: physics lets you do quantum mechanics and engineering at the same time!
'sup @rob?
@DanielSank Howdy
> Cosmic Bell Test: Measurement Settings from Milky Way Stars. Johannes Handsteiner et al., Phys. Rev. Lett. 118, 060401 (2017)
> Bell’s theorem states that some predictions of quantum mechanics cannot be reproduced by a local-realist theory. That conflict is expressed by Bell’s inequality, which is usually derived under the assumption that there are no statistical correlations between the choices of measurement settings and anything else that can causally affect the measurement outcomes.
> In previous experiments, this “freedom of choice” was addressed by ensuring that selection of measurement settings via conventional “quantum random number generators” was spacelike separated from the entangled particle creation. This, however, left open the possibility that an unknown cause affected both the setting choices and measurement outcomes as recently as mere microseconds before each experimental trial.
Oh goodness, this again.
12:14 AM
> Here we report on a new experimental test of Bell’s inequality that, for the first time, uses distant astronomical sources as “cosmic setting generators.” In our tests with polarization-entangled photons, measurement settings were chosen using real-time observations of Milky Way stars while simultaneously ensuring locality.
> Assuming fair sampling for all detected photons, and that each stellar photon’s color was set at emission, we observe statistically significant $\gtrsim7.31\sigma$ and $\gtrsim11.93\sigma$ violations of Bell’s inequality with estimated $p$ values of $\lesssim 1.8\times10^{-13}$ and $\lesssim4.0\times10^{-33}$, respectively, thereby pushing back by ∼600 years the most recent time by which any local-realist influences could have engineered the observed Bell violation.
↑ now that is one cheeky abstract =P
@DanielSank yeah, I imagined you'd have that kinda reaction to this one
I found the ~600 years thing a nice touch though
@EmilioPisanty I may just not appreciate the importance.
@DanielSank no, I kinda side with you tbh
I think Bell violation experiments are important! I just don't necessarily find it interesting to think up loopholes and then debunk them.
I dunno, maybe I'm missing something.
I mean, it's cool and all, but to a large extent this is Zeilinger bragging about the fancy stuff his group can do
but on the other hand, superdeterminism is way weirder than QM, I think
It just so happens that Marissa Giustina is in my office right now.
12:19 AM
I think this sort of work just tries to make that sword sharper
She was Zeilinger's student. She did their recent loophole-free Bell violation.
I will ask her to educate me about these matters.
@DanielSank any wisdom you glean, pass it on ;-)
I think that's her big paper.
Oh wait no that's not it.
heck, ask her if she wands to do an AMA here =P
I think that's it.
@EmilioPisanty I will.
Why are eigenvalue equations so counter-intuitive? At least for me, a beginner. $ab=cb$, but $a \neq c$. This absolutely blows my mind!
@loltospoon Huh?
I promise you they're not counterintuitive ;-)
@loltospoon It's just saying that two completely different geometrical transformations can still agree on certain directions.
$a$ is the operator, $b$ is the eigenfunction, $c$ is the eigenvalue.
12:27 AM
@loltospoon yes. they're wildly different types of object, and your notation obscures those distinctions.
When I was told to think of $a$ as the operator, it began to make a little more sense.
Oh you mean like $\hat{T}|v\rangle = t|v \rangle$?
you can rephrase it as $f(x)=g(x)$ if you want
@DanielSank yep. In the QM context.
@loltospoon You don't "think of $a$ as an operator"... $a$ is an operator!
@loltospoon There is absolutely nothing quantum about that equation.
12:28 AM
$f:\mathbb C^n\to\mathbb C^n$ is an arbitrary linear function, $g(x)=\lambda x$.
Bra-ket notation is for vectors and linear transformations. You don't need to be talking about quantum mechanics to talk about vectors and linear transformations.
But let's look at $5x$ where $5$ is the operator acting on some variable $x$. How can you tell me that the result is $cx$ where $c \neq5$???
@loltospoon ...so?
@loltospoon woah woah woah
matrices and numbers are different
12:29 AM
Think of this differently.
Define a function $\text{TIMES_FIVE}$ which is defined by the equation $\text{TIMES_FIVE}(x)=5x$.
Now it's obvious that $\text{TIMES_FIVE}$ and 5 are not the same thing.
@DanielSank but then do I still think of $TIMES_ FIVE(x)$ as an operator or no?
Can functions be operators?
Functions are operators, yes. They're not necessarily linear operators though.
Use \text{...}.
So the for eigenvalue equation using your function it would be $\text{TIMES_FIVE(x)}x = 5x$ ?
@loltospoon Why are there two x's on the left?
@DanielSank Functions are operators? Wu-what?
12:38 AM
@Danu I probably should have said operators are functions.
Because I realize now that I have no idea what mathematicians think "operator" means.
Functions on what? :P
I'm not having this discussion.
For me, a function means a map $X\to \Bbb R$ or $X\to \Bbb C$
Operator typically means "thing that maps from one space into itself"
Yea I'm only a beginner physicist, so deeper mathematical definitions will go over my head anyway
For me, a function means a map $\text{WHATEVER} \rightarrow \text{WHATEVER ELSE}$.
@Danu o
12:39 AM
@DanielSank Ah. That's what I call a "map".
^ that's the physicists way
I also used to do physics
to @DanielSank 's comment
But okay
Nice work in there Danu :-)
12:40 AM
Glad you approve, skill :P
@DanielSank Slightly more precisely, an operator is typically an endomorphism too. So that's a map $X\to X$ that preserves the relevant structure on $X$
Whatever. The point is that you can always think of an operator as a function/map/I don't care what word you use.
@Danu Ok.
For instance, an operator on a vector space is typically taken to be a linear map $V\to V$.
Nobody ever said than in my however many years in physics education ;)
12:41 AM
@DanielSank That's right :)
I know; I like establishing good terminology for mathematical concepts ;)
I think I had a discussion with ACM about what the fuck "functional" is supposed to mean, once upon a time
Now that's a lame word :P
It's a great word!
Functor and functional are awesome.
I think my conclusion was that it's just a function, but on a space of functions.
Functor is great.
Functional, not so much :P
obligatory, haha
That line is a little... sexually aggressive :P
It is?
12:44 AM
Not really
But when taken out of context it could be misconstrued.
"without loss of generality, I will assume you feel the same way"
@DanielSank Now, that's an AMA that I'd be interested in.
Actual physicist AMA is the best AMA
@rob Ok, I've asked her if she's interested. We'll see. She gets a lot of requests for speaking etc.
...and a lot of crank email, so I'm given to understand.
Is there a reason of why world lines are inverse to classic mechanics when talking about graphs? I learned in high school that the x-axis is time and the y-axis is position.
@DanielSank I can't even begin to imagine.
12:49 AM
God, it must be terrible, publishing on QM.
One more advantage of doing really mathy stuff... No one will bother you :p
Bell's inequality papers probably maximize your crank attraction.
@Danu I dunno, travelling for free sounds fun.
Right, the official speaking invitations are nice, I guess.
@PichiWuana that's a convenient way to display what is going on.
@PichiWuana I'm familiar with both conventions in different contexts.
In other words, there is not a real reason?
12:52 AM
Context is king!
It's a bit confusing because now lower slope is higher speed
You'll get used to it :-)
Clear thinking is the important thing.
I hope. I'm learning world lines / theory of relativity alone through Coursera, while at the same time my works are the inverse way
Make things easier for yourself and follow their convention.
It might be a good exercise to transform the different types of pictures into each other a few times.
12:59 AM
Which convention are you talking about?
The one in Coursera.
@Danu Good idea
@skillpatrol Ooh yeah
1:17 AM
@PichiWuana I wonder if this tradition dates back to early thought experiments in relativity about simultaneous things happening in different parts of trains?
You would draw a train on your paper, labeled "now"
another train above or below it, labeled "later" or "earlier" depending on what you're doing
then when you start measuring coordinates for events, you find that you've got a vertical time axis just because it was perpendicular to your horizontal train.
Never heard about that
Yup, that's how Einstein originally did it.
@PichiWuana A great book to read, if you're teaching yourself relativity.
1:22 AM
I thought that I would read it after the course so everything is clear when I read it
Gotta go, buh bye
@PichiWuana It's an explaining book. The author was something of an authority on the original development of relativity.
It was originally written in German and loses a lot in the English translation.
@skullpetrol I'm prepared to believe that ... but I got a lot out of it as an undergrad.
Well I saw a lot of math when I saw it in iBooks
Seemed to be things that would be hard to understand
1:27 AM
He presents special relativity and general relativity in such a concise manner in that book.
@skullpetrol Are you a German speaker?
Nope @rob
@skullpetrol Just curious whether your opinion of Einstein-as-translated was based on personal experience. I think languages are neat, but I don't plan on getting to German for a while yet.
1:40 AM
Yes, on personal experience.
@skullpetrol You thought the original in German was better than the English translation, but you don't know German? That's confusing to me.
Could someone help me understand what this phrase means in this context?
it's from Griffiths, Introduction to Quantum Mechanics, 2nd ed., pg. 307
@rob I meant the English sentences are missing something throughout.
He writes the the final value for the Hamiltonian is in units of $-E_1$, but I don't get what that means
@rob and I attribute that to translation.
Sorry for the confusion.
1:51 AM
@skullpetrol I suppose that's possible. I often find translated text to be more spare, in a sense, than text written originally in English, but a lot of the time the effect is charming. I don't remember the language in that book being especially remarkable either way.
But I think I read it in 1998, and perhaps again in 2006, so it's been a while.
1 hour later…
3:17 AM
@PichiWuana if it's lots of equations, you're probably fine
the math you have to be afraid of consists of words
@ACuriousMind My actual algebra question had to do with the group $\langle a,b|aba^{-1}b\rangle$. I think I'll need to think about some more first
That's not a bad group.
My topology prof (who does algebraic knot theory) has led me to believe that no free groups are good
2 hours later…
5:12 AM
Testing to see if I can talk here
Hmm it works
@user400188 Hi
Hey; I'm planning on asking my first physics question on SE; but I am having trouble wording it
mainly becuase I am unfamiliar with how things are done on in this community
Say it here.
People in the chat room will help if possible.
I'm typing it right now.
Okay. Go on.
5:18 AM
There is a minute physics video:
That shows that it is not possible to clone things using a proof by contradiction.

In the video; they talk about Superposition, Composite systems and the property of Transformation Distribution.

Eventualy, they arrive at the statement: $A^2+B^2=A^2+B^2+2AB$. At this point it is established that because $A^2+B^2\nequl A^2+B^2+2AB$ there is a contradiction.

However one may note that the above has a solution (namely A=B=0). I am aware that an event with 0 probability wont exist; but what I am confused about is why
Note: When I type this on SE I will use proper formating and I will re-write the proof for reference. It just wont fit here in chat.
That should be a $\neq$ sign there (in place of what shows up in red). I ran out of time to edit it correctly.
@user400188 For not equal to symbol use $\neq$. You can edit your message by clicking edit on the drop down menu on the left of each message. Also, your question seems more suitable for Math SE than Physics SE. BTW when you add links to any website add them using the hyperlink option which is given while asking questions.
Yeah I was planning on asking it in Math SE but the potential for an event with 0 probability to be cosidered meaningless made it seem unsuitable. Mathematicians don't just through away something becuase it is un-physical.
@user400188 Hmm, try asking it on Math SE. If you don't get good answers then ask a moderator to migrate your question to Physics SE. I am quite sure that this better suited for Math SE.
Would the question be suitable for the Mathematical physics tag on this site? I know they don't like people slapping it on everything here but I feel that those kind of people might be able to answer this question.
@user400188 You could try. I am not sure you will get a good response here.
Mathematical Physics is quite different from your type of question anyway.
5:27 AM
I thought so. To tell the truth I am not even sure what mathematical physics is...
Maybe you could tell me?
"DO NOT USE THIS TAG just because your question involves math! If your question is on simplification of a mathematical expression, please ask it at math.stackexchange.com. Mathematical physics is the mathematically rigorous study of the foundations of physics, and the application of advanced mathematical methods to problems in physics. Examples include partial differential equations (PDEs), functional analysis, variational calculus, and potential theory."
^ Mathematical Physics tag wiki
It talks about topics that most questions are on but doesn't seem to explain what Mathematical physics is.
Unless Mathematical Physics is in fact the rigorous study of the foundations and nothing else.
Mathematical physics refers to development of mathematical methods for application to problems in physics. The Journal of Mathematical Physics defines the field as "the application of mathematics to problems in physics and the development of mathematical methods suitable for such applications and for the formulation of physical theories". It is a branch of applied mathematics, but deals with physical problems. == Scope == There are several distinct branches of mathematical physics, and these roughly correspond to particular historical periods. === Classical mechanics === The rigorous, ...
@user400188 I need to go. Choose which site is better for your question and go with it. I gave you my inputs and advice already. Bye. :)
@anonymous Thank you for the help :)
5:58 AM
@anonymous got that question
@Koolman You solved it ?
I just woke up
@anonymous Ahh sorry :-(
@YashasSamaga Quite late :P
Hope you had a good sleep :)
5:59 AM
I actually need more
3 days of sleep
Why don't you post it on main site
@Koolman I know the solution. That question was for you guys :)
Oh nice
Is band gap of superconductors lesser than metals?
I mean are the valence and conduction bands more overlapping in semiconductors than metals?
Hello @JohnRennie
6:07 AM
Good morning. Where are you from? @JohnRennie
I live in a city called Chester in the UK.
@Ramanujan See his profile :P
@JohnRennie, oo
@JohnRennie, what are the differences between the image formed by objective lens and eye piece lens of astronomical telescope?
@Ramanujan I can't remember I'm afraid. We did study the optics of telescopes during my physics degree, but that was 40 years ago and I haven't studied optics since. I'm sure a quick Google would find you the answer.
6:12 AM
Morning @JohnRennie . Too early
Objective lens is just a convex lens. As light is coming from practically infinite distance so image is formed at focal plane (approx).
@JohnRennie, oo
@Mostafa 6 a.m. here :-)
@Kaumudi.H I was sleeping off a huge lunch, sorry :-) Anyway, congratulations :-)
@JohnRennie Ah, had that big lunch, eh? :-)
It was indeed, very exciting!!
^ :-)
Lucky day ^ :)
Never heard of that channel though
6:22 AM
Yes, lucky! :-)
@anonymous Description:
> Raising nerdy to the power of awesome.
Do we have Seebeck Effect in Jee @Kaumudi.H It came in 2002 paper but it is not in NCERT :/
@YashasSamaga ^
@Kaumudi.H I saw the channel just now :)
Good one
@anonymous I'm not sure, I'm yet to look through my handbook for that chapter.
The thermoelectric effect is the direct conversion of temperature differences to electric voltage and vice versa. A thermoelectric device creates voltage when there is a different temperature on each side. Conversely, when a voltage is applied to it, it creates a temperature difference. At the atomic scale, an applied temperature gradient causes charge carriers in the material to diffuse from the hot side to the cold side. This effect can be used to generate electricity, measure temperature or change the temperature of objects. Because the direction of heating and cooling is determined by t...
I'm having a hard time making sense out it
Why should current increase and then again decrease on increasing temperature of hot terminal ?
Its going over my head :P
@Kaumudi.H Okay
@JohnRennie Do you know about Seebeck Effect ?
@anonymous Yes, what about it?
6:28 AM
@JohnRennie Could you tell me why current decreases after increasing the temperature of the hot terminal to a certain point ? I mean why should the direction of current reverse?
Also, why does it reach a maximum and then again drop?
A: Thermo-Emf variation with temperature

John RennieThe calibration curve you show isn't typical. In fact there is no typical calibration curve because different types of thermocouples can have very different voltage:temperature dependences. There are lots and lots of articles giving different calibration curves - I selected this PDF as being fair...

@anonymous you have cbse board
@Koolman yes
Have you started studying for that
@Koolman yeah, i don't care about cbse
6:32 AM
Searched "are Americans" on Google and on YouTube:
"When you heat a metal you produce a population of electrons with energies above the Fermi energy, and that population is temperature dependant. The Seebeck effect arises because if two ends of a metal wire are at different temperatures the electrons at hot end will have higher energies than the ones at the bottom end and they can lower their energy by moving from the hot end to the cold end. This movement of charge creates a potential difference between the ends of the wire."
@JohnRennie I can't find the reason for decrease of emf after neutral point given in your answer....
And this pdf is full of so much maths (cdn.intechopen.com/pdfs/24115.pdf)
Can't make much sense out of it
I guess I'll have to leave the topic for the time being
@anonymous can you post the exact question
6:36 AM
Sorry ... distracted for a few moment
@Koolman where?
H bar
I asked already...
@Mostafa Try searching Europeans and Asians also :D
When a ball is pressed in water, why does it try to come up?
Archimedean force
6:46 AM
@anonymous haha xD
Look at this
Because it doesn't want to go down.
@anonymous: in a thermocouple you are measuring the difference in the Seebeck effect between the two metals of the thermocouple.
@Mostafa I asked you to search because of the first result only :D Hehe
6:47 AM
@anonymous, why it doesn't want to go down?
@Ramanujan Because it wants to go up. :P
I'm trolling now
Trumps Facts

Proposed Q&A site for given how many questions abe being asked about Trump on skeptics and so on, why not have a community to check Trump facts? In a Q&A style, not extended discussions

Closed before being launched.

@JohnRennie Okay. Makes sense. So, the Seeback effect one metal is initially greater than the other but on increasing temperature the situation reverses? Why is it so?
Suppose we take Cu and Fe
@Mostafa My goodness . Heights of insanity.
@anonymous The Seebeck voltage-temperature dependence is a complicated function of the band structure of the metal. So two different metals will in general have different temperature dependences.
As a result the difference in the Seebeck effect that a thermocouple measures will be temperature dependent.
@JohnRennie Alright. Seems to make some sense now. Thanks for the help :). I will get back in case of further confusion.
You must have heard of this guy in your early classes ^ :D
6:59 AM
Did they have rubber ducks in ancient Greece

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