« first day (2152 days earlier)   

4:17 AM
@BernardMeurer I wouldn't bother with the nvidia drivers. The laptop has dual GPUs - the nvidia and the built in HD4000, and the nvidia gpu isn't that great anyway. I would just use te HD4000 for graphics. It's plenty fast enough unless you want to use the laptop for gaming.
 
@JohnRennie are you familiar with the book on solid state physics by ashcroft and mermin?
 
@IceLord No, sorry :-(
 
because I need some help for condensed matter and idk if anyone here knows much about it.
I should learn to be more independent.
 
4:42 AM
What is this talking about:
0
Q: Motion in relation to gravity

Abdalla EidIs it possible that gravity could be a byproduct of mass moving through space and if we could stop a masses movement would it still have gravitational influence on other masses within its vicinity. Could it even be possible to slow an object speed to absolute zero everything else is moving we wou...

?
 
Absolute motion wrt. aether, it seems. Into the trash.
 
5:00 AM
Morning Kaumudi :-)
 
Hello all :-)
@JohnRennie Can u help me with something..?
 
@KaumudiHarikumar Yes, of course.
 
@JohnRennie I'm trying to solve problems involving forces that depend on both the x and y coordinates of the particle; to find the work done by forces of this kind.
 
OK, what's the problem?
 
The path is described in sentences such as "The particle is taken from point 1 to point 2" where the coordinates of the two points are given.
 
5:06 AM
@KaumudiHarikumar You need a line integral along the curve.
 
Presumably there is extra info e.g. whether it's a conservative force, and if not then what path is taken between the points ...
 
If it is conservative, then work with the gradient of the scalar potential.
 
@JohnRennie No information about the nature of the force has been provided and regarding the path, well, yes, it has been described in sentences.
 
and ...
 
@MAFIA36790 Riight. I watched a video by KhanAcademy about this concept but in his video, the path was actually described in terms of a position vector function with time as the independent variable...
 
5:09 AM
okay...
Can I see the function of the curve?
 
Can you post a picture of the problem?
 
@JohnRennie Sample sentence: "The particle is moved from the origin to a point (a,0) along the x-axis, after which it is moved to the point (a,a), parallel to the y-axis"
 
Well that's easy, because the path is just two straight lines.
Is the force constant, or does it depend on $x$ and $y$?
 
@JohnRennie It depends on $x$ and $y$.
 
5:13 AM
Take the first bit, along the $x$ axis. This means $y$ is just a constant - in fact it's zero - so the force is just a function of $x$.
 
@JohnRennie No no, see, I did that, and I did arrive at the answer...
 
We know $$W = \int_{x=0}^{x=a} F(x) dx $$ so it's a straightforward integral
 
I did that for the path along the x-axis first and then the one parallel to y-axis after that, taking y and x as constants for those two paths respectively and then I arrived at the correct answer...
 
OK, good, so you've done the problem!
 
@JohnRennie Well, yeah :P But I looked at the solutions and instead of seperately integrating the function w.r.t $dx$ and $dy$, they integrated wrt $xi+yj$...
And then they split the integral.
OK, for u to make better sense of that ^, I must mention that the function is given as $F=-K(yi+xj)$
So they did the dot product and split the integral.
How did they do that..? I mean, how can they simply integrate wrt $xi+yj$..?
 
5:21 AM
When you are moving a particle along a path the work is given by: $$W = \int_a^b \mathbf F(\mathbf r) \cdot d\mathbf r$$ where $\mathbf F$ an $\mathbf r$ are vectors and the $\cdot$ is the dot product.
So $\mathbf F = (F_x, F_y)$ and $\mathbf r = (x, y)$
And the dot product is: $$ \mathbf F(\mathbf r) \cdot d\mathbf r = F_xdx + F_ydy $$
When we're actually doing the integral you usually looks for tricks and shortcuts to help you. For example in this case the trick is to split it into two separate easier integrals, which is exactly what you did and it's what I would have done.
The actual trick used depends on the details of the problem. It's the sort of thing you pick up by doing lots of this type of problem.
Unless you have a specific problem in mind it's hard to say much more.
 
@JohnRennie I see. Well, I didn't do the dot product thing. I split it up by assuming one variable to be constant over a given path but OK, I see.
 
@KaumudiHarikumar actually you did use the dot product but without realising
 
@JohnRennie In hindsight, yes.
 
Along the first part $dy = 0$ so the dot product is just $F_xdx$. Likewise for the second part it's $F_ydy$.
 
In the video by Khan Academy, he differentiated the position function wrt $t$ to get the value of $dx$ and $dy$ and then he did the dot product.
 
5:32 AM
That's a trick called parameterisation.
 
@JohnRennie $t$, time, being the parameter in question, yeah..?
 
Suppose we can write the positions as a function of time i.e. $x(t)$ and $y(t)$
Then we can calculate $dx/dt = f_x(t)$ and $dy/dt = f_y(t)$ for some funcctions $f_x$ and $f_y$. OK so far?
 
Then we use the physicsts trick of treating the derivatives as fractions to get $dx = f_x(t)dt$ and $dy=f_y(t)dt$
And the dot product becomes $F_x(t)f_x(t)dt + F_y(t)f_ydt$ so it's only a function of $t$. Then the integral is just $\int dt$
 
@JohnRennie To do this, $F$ must also be a function of time, no..?
 
5:37 AM
Remember that $x$ and $y$ are functions of time, so $F(x,y)$ can be written as $F(x(t), y(t))$ where $x(t)$ and $y(t)$ are functions of time.
So we can write $F$ as a function just of time.
 
@JohnRennie Ohh, yeah. Stupid me. OK. I understand. Thanks so much! :-)
 
No, not stupid you, these are tricks we all had to learn along the way :-)
 
Who is stupid?
 
@JohnRennie I guess :-) BTW, in ur head, how do u pronounce my name?
 
You wouldn't normally learn this stuff until first year at university, so you're well ahead of the curve.
 
5:40 AM
@JohnRennie second
 
@KaumudiHarikumar I've never thought about it. "Cow" "Moody" I suppose ...
 
@JohnRennie do you know about correlation functions?
 
That's a slightly bad choice of words :-)
 
specifically wrt. the SHO
 
@0celo7 To be honest I've never really understood correlation functions, though if we're thinking about the same thing they're simple enough aren't they?
 
5:42 AM
Yes. But my answer disagrees with the solution
 
@JohnRennie Ah, that sucks :/ You're right about the "cow" part but not "moody" :P It's "Cow"-"moo" (lol)-"thee".
 
Aha, the "d" is a "th" sound.
 
It's a Sanskrit word that means moonlight so at least it's not completely random :P
@JohnRennie Yes!
 
Interestingly that was the case in Old English. The letter "d" was called eth and pronounced as a "th".
 
@JohnRennie Oh, that's very interesting!
 
5:45 AM
The Indo-European language theory strikes again! :-)
 
@JohnRennie I'm gonna have to look that up :P
 
"John" derives from an ancient Hebrew word meaning "praise be to Jehovah" or something like that.
 
@JohnRennie That last thing should be $0$
but it's not...
I don't see what I could have done wrong.
the $a^\dagger $ takes us to $|1\rangle$, and $a$ back down to $|0\rangle$.
So that's just $i\hbar/2$
 
@KaumudiHarikumar कौमुदी - the power of Google!
Though I'm guessing you speak Tamil (as well as Hindi) and that's a Dravidian language that isn't related to the Indo-European lanuages.
 
5:57 AM
@JohnRennie can I ask one personal question?
 
@JohnRennie enough with languages -- I need QM help
I'm not a physicist, this is confusing me
 
Where did you studied?from which university?
 
Yes of course, though whether i answer depends on how personal it is :-)
@Ramanujan Cambridge University in the UK
 
@JohnRennie OMG, are you doing research?
 
@Ramanujan no, he's too old for that
 
5:59 AM
@Ramanujan Not now. I did a PhD at Cambridge but then went to work as an industrial scientist studying colloid science. Now I'm part retired and work part time as a computer nerd.
 
OK sorry
 
Aha. The solution is wrong, as google shows.
 
These days I do physics for fun :-)
@0celo7 Link? Your working seemed OK to me ...
 
@JohnRennie The book solution is wrong.
I'm correct.
 
Aha! :-)
In colloid science we'd call C(t) an autocorrelation function. We use them in light scattering experiments where you're studying particle motion.
 
6:01 AM
the answer is $\frac{\hbar}{2m\omega}e^{-i\omega}$
the book omitted the complex part for some reason
@dmckee you're up...late
 
Aah and the second part of your answer provides the imaginary component.
 
exactly
but I have the solution manual (thanks Russian servers), and Sakurai says explicitly that $\langle px\rangle=0$.
tfw earned $0.18 interest.
pretty amazing tbh
 
@JohnRennie Ah, I see! I didn't know about this.
@JohnRennie Yes, yes! That's it :-)
@JohnRennie No, no, I speak Malayalam, Tamil and Hindi :-)
 
6:25 AM
I only speak English and Physics - two entirely different languages and not to be confused with each other :-)
 
@JohnRennie Haha :-) Didn't u ever learn any other languages?
 
@KaumudiHarikumar There's not much use in English-speaking countries.
 
It makes a big difference living on an island nation. Until the last decade or so there was very little mixing with mainland Europe to we rarely heard other languages.
While most people on the European mainland speak at least two languages in the UK most people only speak English.
The only exception is in Wales where a significant number of people speak Welsh as well as English.
 
Does "sheep" rhyme with "lady" in Welsh?
 
I had lessons in French and German at school, but that was forty years ago and I've forgottent what little I ever learned.
@0celo7 :-)
I can see the reputation the Welsh have for unsavoury activities is known even in the US :-)
Wow, while Googling the Welsh word for sheep I found ...
 
6:33 AM
@JohnRennie The sheep probably ate her boyfriend's joint.
 
I bet it makes the meat taste good :-)
 
I should go to sleep.
 
@0celo7 What time is it there?
 
2:37am
 
@JohnRennie I see :-) I learned German for a year but like you, have forgotten.
@0celo7 Holy crap.
 
6:37 AM
What?
I'm not sleepy at all, which is the problem.
I had a coffee at 6
 
@JohnRennie Whoa.
@0celo7 Read the newspaper, maybe?
 
@KaumudiHarikumar I was born in Sudan (my father was working there) and I spent my first nine years there. According to my mum I could speak Arabic quite well as a child, but I've since forgotten all but a few words of it.
 
@JohnRennie Oh! Yes, it's interesting how quickly we pick up languages as children but also, how quickly we can forget them. When I was 5/6, I learned the Hindi alphabet at school and was even able to read small books, but after vacationing with my grandparents for two months, I'd forgotten everything!
 
I used to know chemistry
I can't balance an equation any more.
 
@0celo7 Lol. Are ur classes usually not in the morning time? Seeing as u're usually here all night :P
 
6:45 AM
classes?
tomorrow is Sunday
 
I mean on the weekdays.
 
usually I only get up on weekends up because Rebecca tells me to, but she's not here
@KaumudiHarikumar I'm here until 1 at the latest, I have classes at 9
 
@0celo7 What do u mean "I only get on weekends"?
 
get up
 
Right.
@0celo7 Oh, OK.
Are u in ur 1st/2nd year of college?
I'm off to have lunch. Bye!
 
6:49 AM
@KaumudiHarikumar 2nd
@JohnRennie Do I need more books?
 
@0celo7 silly question, the answer is always "yes"
 
yes.
agreed.
but I will wait until my next paycheck
 
I have about 20,000 books and that's still too few
 
that way I do not need to move money around
you do not have 20,000 books
 
e-books
 
6:51 AM
oh, well, sure
I could download a few million
I'm talking about book books
 
I ony have books that I will in principle read one day, though I'll have to live several thousand years to do it. I don't keep books that are of no interest just so I can say I've got them.
 
I agree.
I have an immediately available shelf of 34 books that I use fairly regularly
 
@JohnRennie Man, you can't get satisfied without turning a page by your hand and enjoy the smell of the old pages of the book (most Dover books do have their signature oldish smell); that's why I prefer actual books to e-books unless the price of the book is beyond my reach.
 
Dover books do have a distinct smell.
 
Yes! Yes!!
 
6:54 AM
But these books are AMS Chelsea, which have amazing covers
they're textured, dark red, with gold letters
 
@MAFIA36790 I completely disagree. The only time I buy paper books is for science textbooks because ebooks are no good for flipping forwards and backwards quickly. For everything else I love reading on my tablet.
 
LOOK AT THAT
 
@JohnRennie I'm talking about science textbooks too.
@0celo7 WoW!!
@JohnRennie Most of my storybooks and sci-fi novels are in my Kindle reader.
 
Even for science textbooks I don't care what they look like. As long as they are printed clearly that's fine by me.
 
@JohnRennie oh come on
 
6:57 AM
ohh.
 
what's a crappy looking book...
 
Who flagged 0celo7's comment?
Good grief, have some sense of proportion!
 
I swear if I get banned
 
Flag things that are offensive not words you hear every day.
 
^terrible cover
 
6:59 AM
Hi Kaz. Responding to the flag?
 
wheeeeee! *crash*
yep
just wanted to pop in and make sure everything was okay
 
Silly flag IMHO though your mileage may vary
 
yeah, it got dismissed pretty fast, so it seems like everything's okay here.
 
@0celo7 Sadly, I still have not encountered any literature of AMS Chelsea; most of the books (e-books) I possess are of Springer, Wiley, Pearson, Addison-Wesley and Dover.
 
@KazWolfe Thanks for keeping an eye on us naughty physicists :-)
 
7:00 AM
@0celo7 Kiddish dull cover.
 
@JohnRennie It's my job! :D (I honestly don't know why people flag soft-curses, but meh)
have fun, and don't go crazy chasing the higgs boson agian.
 
@JohnRennie who is your favorite physicist?
 
These are the only books I have in paper form (apart from several large boxes of SF paperbacks stored in the loft):
user image
2
 
@JohnRennie Looking nice!
 
Bottom shelf science and technical, top shelf cultural.
Everything else is in e-book form
 
7:07 AM
@Ramanujan I think such questions are silly; you can't narrow down to a single physicist; every physicist has his own contribution in different aspects of the field. Choosing one favourite, seems to be comparing them with each other; which I really feel odd.
 
@MAFIA36790 hmm
 
@Ramanujan Although I love Lumo ;P
 
My favourite physicists are the ones who try to reach out. Feynmann was probably best at this, though obviously he's no longer around.
Susskind is probably the closest these days. He puts a lot of effort into writing books and giving lectures for young physicists.
 
@JohnRennie BTW, which optics book is that in the bottom shelf?
 
Though I have to say I find Sussking a difficult lecturer to follow. He seems to spend a long time on simple things then skip quickly through steps in proofs I find hard.
 
7:11 AM
Hecht?
 
@MAFIA36790 Hecht and Zajac - an absolute classic! Brilliant book.
 
@JohnRennie Ah!
But I'm really not in agreement with optics ;/
I find this a very dull topic ;/
 
I got temporarily interested in radio astronomy, and a lot of the stuff on Fourier optics discussed in Hecht and Zajac is relevant to that.
 
By far, the most interesting physics book for me is Lanczos' The Variational Principles of Mechanics.
@JohnRennie ohh.
 
But, like you, I find there are more exciting things in physics than classical optics.
 
7:13 AM
@JohnRennie yeh, yeh.
@JohnRennie, You should read Lanczos, I'm sure you'd find it interesting and worthy to read it.
 
If/when I find the time to get stuck into a physics book it will be a QFT book. QFT is the last bit of physics I really want to understand before I die.
 
@JohnRennie I did notice a book of QFT in your shelf, isn't it?
 
Tom Lancaster
Oxford University Press

Title: Quantum Field Theory for the Gifted Amateur Binding: Paperback Author: TomLancaster Publisher: OxfordUniversityPress,USA
 
@JohnRennie ah!
For the Gifted Amateur...
What does that mean?
 
Allegedly it's a good book for people who are just interested in how QFT works as opposed to wanting to learn it well enough to to do research in it.
 
7:18 AM
ohh.
Freeman Dyson also wrote a book on Advanced Quantum Mechanics which is practically a treatise on early QFT.
Freeman Dyson
World Scientific Publishing Company

Renowned physicist and mathematician Freeman Dyson is famous for his work in quantum mechanics, nuclear weapons policy and bold visions for the future of humanity. In the 1940s, he was responsible for demonstrating the equivalence of the two formulations of quantum electrodynamics - Richard Feynman's diagrammatic path integral formulation and the variational methods developed by Julian Schwinger and Sin-Itiro Tomonoga -showing the mathematical consistency of QED. This invaluable volume comprises the legendary lectures on quantum electrodynamics first given by Dyson at Cornell University in 1951. The late theorist Edwin Thompson Jaynes once remarked, "For a generation of physicists they were the happy medium: clearer and better motivated than Feynman, and getting to the point faster than Schwinger". This edition has been printed on the 60th anniversary of the Cornell lectures, and includes a foreword by science historian David Kaiser, as well as notes from Dyson's lectures at the Les Houches Summer School of Theoretical Physics in 1954. The Les Houches lectures, described as a supplement to the original Cornell notes, provide a more detailed look at field theory, a careful and rigorous derivation of Fermi's Golden Rule, and a masterful treatment of renormalization and Ward's Identity. Future generations of physicists are bound to read these lectures with pleasure, benefiting from the lucid style that is so characteristic of Dyson's exposition.
 
7:49 AM
0
Q: publish annual q/a journal

LelouchI was thinking of asking the moderators to publish a journal annually, containing the best 100 questions of the year , and their detailed discussion sessions, so that if someone searches for a particular topic 6 years later, it shows up in the journal , along with its detailed discussion.

 
 
2 hours later…
9:20 AM
Doesn't apply to us physicists of course :-)
 
:)))
 
10:19 AM
0
Q: How do you solve for the minimum kinetic energy given that you have an area of movement of the particle?

TESLAGENHow do you solve for the minimum kinetic energy given that you have an area of movement of the particle? It seems kind of random to me. I thought of using the momentum given that you have the position uncertainty and then calculating the lowest possible momentum from the deviation by subtracting ...

I think he is on the right track, but I should not just wrote an answer and said that "you have done this correctly exactly as you did"?
 
10:42 AM
@JohnRennie Pff, gaming. I want to use CUDA to speed up my image analysis project!
 
@BernardMeurer Ah OK it's access to the GPUs number crunching abilities you need.
But can't you install the libre drivers but still use the HD4000 to run the display?
 
@JohnRennie Yes. I need to run face-recognition on a lot of people from Facebook to find a suitable human to ask on a date
 
@BernardMeurer if you need a face recognition algorithm to find a girl suitable for you the simple answer is that there are no girls suitable for you :-)
 
@JohnRennie That's what NVidia Optimus does, but they never released good drivers for Linux. The Libre alternative was breaking my backlight control and there's an OSS third path called Bumblebee which seems to be abandoned
 
@BernardMeurer to be honest I've never understood how the E6530 manages the dual GPUs. I think under Windows it always uses the HD4000 but you can specify that running certain processes causes a switch to the nvidvia GPU.
 
10:47 AM
@JohnRennie Well, I'm just trying to use my time more effectively :p If I could make an algo that would detect facial traits which I like and add that to all the shit they make available on their facebooks to create a smaller pool of people to try and talk to it means I have more free time to work on kernels
@JohnRennie So, it has this Nvidia Optimus thing, which is what cause the famous "Fuck you Nvidia" from Torvalds. What it does is that both GPUs share the same framebuffer and, IIRC, Optimus has a specialized chip that will detect load intensity and automatically do load balancing between GPUs or make them shift seamlessly. That however requires drives. The original Nvidia Optimus drivers were shit, and Nvidia's response was basically WONTFIX which pissed everyone off.
At the time the nouveau drivers didn't have a suitable switching mechanics, so Bumblebee came up and implemented that, although from what I've heard it never worked really well. Nowadays nouveau fully supports Optimus technology, but it has a performance drop when compared to the binary-blob based Nvidia drivers
 
The bumblebee implementation looks ideal if you just want access to the GPU for number crunching. According to Google it doesn't support dynamic switching to the nvidia GPU, but that's fine for your application.
So long as CUDA works.
 
:: John checks the news websites for reports of exploding laptops in Lisbon ::
 
11:13 AM
John, how should i wrote an answer to a question where the OP basically have done all the steps correctly thus in effect answered his/her question?
 
@Secret Better not answer that.
 
ok
 
Because such questions are asking basically then to check where they mistook and Phys.SE is not meant for that.
 
12:00 PM
NB: This is a snipplet of a game, hence no point in giving a link
Caption: Sometimes incremental games like these reminds of power series and rates of change of a quantity
Consider a function $f(x)$ and its rate of change $\dot{f}(x)$. Now consider that x is actually a function $x(a(t),b(t),c(t),d(t),e(t),\dots)$. Letting $S=\{a(t),b(t),c(t),d(t),e(t),/dots\}$ and expanding $\dot{f}(x)$ you will obtain the following:
$$\dot{f}(x)=\frac{df}{dx}\sum_{i\in S}\partial_i x \dot{i}$$
 
@Secret \ldots
Hey @loong.
 
@MAFIA36790 hi
 
Now consider the individual $\dot{i}=\frac{di}{dt}$ terms. Let the values be expressed in terms of scientific notation $k_i\times 10^{n_i}$. Now observe that if one of the $\dot{i}$ terms is larger than another $\dot{j}$ term for some orders of magnitude (where $i,j\in S$) then no matter how much faster the smaller one is changing, the overall change will be dominated by the one with the larger change in the orders of magnitude
 
@Secret Use \times ;_;
;))
 
It might seemed straightforward to illustrate this dominance of the orders of magnitude by taking log both sides and make use of the rule $\ln(xy)=\ln x+\ln y$ and then apply a $\mathcal{O}(n_i)$ like argument to disregard lower or higher order terms. however one cannot simply take log on $\dot{f}(x)$ as $\ln(\text{a bunch of sums})$ have no easy formula.
 
12:16 PM
@Secret whispers to secret to use \ln ;P
;))
 
I might have to tidy this up a bit and ask the maths chat on how to compare between the relative rates of quantities and the conditions such that no matter how finitely fast the smaller one is rising, its change will be swamped by the change with a large order of magnitude
 
@Secret Currently the chat is empty...
 
It is, thus I am just kinda chilling out for now
In case the above maths makes no sense to some later readers: I am trying to investigate and quantify the observation in incremental games that if something gives you points of e.g. 1000/s vs one that gives you points of 10/s, then the former one will drown out the contribution from the latter one
 

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