9:00 AM
But suppose you take $dt$ and you multiply it by a constant with the dimensions of velocity.

Ah, correct.
but what is that constant?
How do we get its value?

Ah, this is where it get's interesting :-)
because the $-c^2dt^2$ term is negative it's possible for the value of $ds^2$ to become negative if $dt$ is big enough. Yes?

....
But.. it's squared. How can it be negative?
It is a magnitude
I see what you are saying but that shouldn't happen.

Correct. $ds$ is a real physically measurable property and it cannot make sense for it to be imaginary. The smallest value $ds^2$ can have is zero. This is important, so I need to make sure you're with me on this.

I am.

9:05 AM
OK so let's see what happens if we make $ds^2 = 0$. We get: $$0 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$

@JohnRennie I still don't understand this, but that's probably a conversation for another time

Or: $$c^2 = \frac{dx^2 + dy^2 + dz^2}{dt^2}$$

(As in, why can't a measurable value be complex?)

And $dx^2 + dy^2 + dz^2$ is just the distance moved in space so let's call it $dr^2$. Then the equation is: $$c^2 = \frac{dr^2}{dt^2}$$

Clearly dt can't be 0 then..

9:07 AM
So $c$ is obviously just a velocity because it's a distance divided by a time.

It is.

@Sid And what's more $c$ is the maximum velocity anything can have. Can you see why or do I need to make this explicit?

@Mithrandir24601 Technically the question does not make sense; $ds^2$ is not really length squared in any sense, it's a 2-tensor.

@BalarkaSen do you mind, we're physicists here! :-)

lol
Don't get me wrong, I do think the physicist point of view is interesting. I've been following your conversation with Sid incognito.

9:10 AM
Make it explicit. I don't quite see why.

OK. Let me put $ds^2$ back into the equation, which proceeding as before gives me: $$c^2 = \frac{dr^2 - ds^2}{dt^2}$$
That's just our original equation for $ds^2$ rearranged to give an expression for $c^2$. Is that OK?

since ds^2=0, c^2 has to be the max value, I understand?

Yes!

Hi, everyone! :-)

@Sid So by assuming the equation for the norm we find we automatically get a maximum speed that nothing can go faster than.
@Kaumudi.H Hi :-) Back in Chennai?

9:15 AM
(I did think of that before. I was then, like, "It can't be that simple".)

@JohnRennie For the weekend, yep! :-)

@Sid Well arguably it's not that simple. After all, this is a pretty weird way to calculate the length of a vector.

How far is your uni from Chennai
it's on Kerala right

Well, yeah. I mean, that "max part.:
@BalarkaSen *in

@BalarkaSen Yep, yep. A nine hour journey.

9:17 AM
@Sid But, this is the key point about special relativity, and actually general relativity too. It's that the norm of a vector is calculated using that weird equation.

@Sid ondeed. i am ondebted to you for catching that.

@Kaumudi.H You are going home for only 2 days and are willing to travel 18 hours for that?

@Kaumudi Ah, I see

@Sid Clearly :-P Not every weekend though.

@JohnRennie Okay.

9:18 AM
@Sid obviously that velocity $c$ is the speed of light - if you hadn't already guessed :-)

If it's max, it is c. :-)

@Sid emergency bicycle rescue mission!

@JohnRennie Hehe :-)

The equation for $ds^2$ is called the metric. Specifically in special relativity it is the Minkowski metric.
The difference between SR and GR is that in GR the equation for $ds^2$ becomes more complicated.
@sid if you're interested have a look at:
13

Suppose we have a rod of length $l$ at rest in the unprimed frame and we watch an observer in the primed frame speeding past: We'll take the origins in both frames to coincide when the observer in the primed frame passes the first end of the rod, so Event A is $(0, 0)$ in both frames. In the ...

This shows how you can use the expression for $ds^2$ to calculate the weird relativistic effects like time dilation and length contraction.

@BalarkaSen huh? The $ds^2$ that I was told about is a scalar... In any case I'm not specifically talking about that, but any measurements of an arbitrary physical thing

9:23 AM
@Sid is this still making sense, or have I just convinced you that Einstein (and the rest of us) are completely mad! :-)

@Mithrandir24601 It's not a scalar. Metrics are 2-tensors.

@Mithrandir24601 it's an assumption that all physically observable quantities are real.

Once you feed it two vectors, it becomes a scalar, because that's how tensors act on vectors.

That pervades all of physics - it's the reason observables have to be represnted by Hermitian operators in QM.

@JohnRennie See, all of us know that Einstein was mad. I don't know about the rest of you. :-)

9:24 AM
@Sid :-)

@Bernardo: I've got Python this sem. and yesterday, I was introduced to Linux.

@Sid But anyway, this all started when you asked about the Lorentz transformations ...

@BalarkaSen yeah, sorry - I just realised that the thing I was thinking of wasn't $ds^2$ - my bad!

@Kaumudi.H oh ... my ... god ... you're officially a real programmer :-)

No worries. Actually writing (semi)Riemannian metrics as ds^2 = stuff stuff confused me for some time quite a while back

9:26 AM
Although, wait... it's still a scalar...

@JohnRennie Hahaha :-) Do I need to actually have a Linux O.S to program in Python?

No, Python works on Windows and Mac as well.

Python is really easy to debug though..

I mean, it's not a metric - $g_{\mu\nu}$ is the metric...

@Mithrandir24601 Eg writing the Euclidean metric as $ds^2 = dx^2 + dy^2$ is just an infinitisimal representation of the quadratic form the identity matrix corresponds to: it's $ds \otimes ds = ds \otimes dx + dy \otimes dy$.

9:28 AM
@Kaumudi.H CS courses tend to like Linux because in some ways it's a lot simpler than Windows and OSX. It doesn't have all the consumer baggage like pretty colours etc.

No, I mean, $g_{\mu \nu}$ is the matrix corresponding to the metric.
It gives a quadratic form on each tangent space, which we represent in local coordinates by stuff squared stuff stuff

@JohnRennie Ah, I see.

@Kaumudi.H and of course it's free so the university don't have to pay :-)

I have a pressing question: I'm not too familiar with most things computer-related and this fact is scary.

Yes?

9:30 AM
@JohnRennie cc @Kaumudi.H Well it doesn't have to, but it certainly can.

@JohnRennie Exactly how can I expand my vocabulary in this regard?

@Kaumudi.H just have fun playing with computers and you'll pick up all the nerd talk without even trying.

> playing with computers

That's how Bernardo and I, and millions of other nerds enthusiasts did it :-)

Hmm...

9:32 AM
You need a second laptop so you can install Linux

@JohnRennie COmputer programmers i know seem to love Linux and hate Windows for some reason

@JohnRennie There was talk about something called Git...

Github?

@Sid there are a lot of sociological reasons for that. The truth is that both Linux and Windows have their strengths and weaknesses.

@BalarkaSen I get you now...

9:34 AM
@Kaumudi.H Git is a way for programmers to share working on project.

Github is a really useful site if you are working on a bot, though.
We did that back on Puzzling.

Anonymous
@ThomasKlimpel I'm using the CodeChef IDE

This is different from this, no?

@JohnRennie It's a version control system first, though
@Kaumudi.H correct, git and github are different

@Blue You are in codechef? i just joined today.

Anonymous
9:36 AM
@Kaumudi.H What? The first one is a tutorial for git

@Mithrandir24601 The $g_{\mu \nu}$ thing and the $ds^2$ thingymagicky is the manifold analogue of switching between a quadratic form and a symmetric matrix alternately.

Anonymous
Git is a version control system

But yeah, it was all a bit confusing at first.

148

I have recently added a new project to Git using Eclipse, but do not see the project appear in my GitHub account. Why do they have the same account information and different repositories? Isn't Git and GitHub the same thing?

Anonymous
@Sid Yeah. But presently doing the hacker rank exercises

Anonymous
9:38 AM
Hacker Rank 30 days of code is pretty nice

Oh, duh. I am simply doing beginner problems to refresh my memory of C++.

@Kaumudi.H a Git server is a big complicated scary system to implement. Github is a site that does all this scary stuff for you.
On your PC you can use the Git client tools and the Github server does all the hard work of handling the server side stuff.

@JohnRennie I...see(?)

Anonymous
There are some youtube tutorials on Git

Anonymous
Learn it as soon as possible. You'll need it

Anonymous
9:39 AM
You'll have to make contributions in GitHub to get selected for GSOC

@Kaumudi.H when you use the web, the client software is IE or Chrome, and the server software is some big complicated piece of software running on a big server at some company.

@Blue ^ This is really what I mean; my vocabulary and knowledge in this area is so shockingly poor that one would wonder why I took C.S in the first place :-/
@JohnRennie Ah.

@Kaumudi.H When you read web pages or fill in web forms you are sending data to and from the big server - that data being the form data or the web page contents.

Anonymous
@Kaumudi.H Don't worry. Even I didn't know much about Git till @JaimeGallego explained it to me :)

Anonymous
Just get started

9:41 AM
Wokay.
@JohnRennie Right, right.

@Kaumudi.H With git the data is your source code. The git client tools on your PC send/receive your code to/from a git server running on the GitHub servers.

learn homotopy type theory
ba dum tss

@JohnRennie Ah, right.

The point is that the GitHub servers provide a repository so:
(a) if your PC dies your code is safe
(b) other people can share you code by downloading it from the GitHub servers
But honestly don't start getting paranoid about all this stuff. It's easy to pick it up as you go along.
You only don't know about git and GitHub because no normal sane person ever needs to use it :-)

@JohnRennie I am starting to, for a few students in my class are already quite passionate about computers alone and they know this stuff.
@JohnRennie I see, I see.

9:45 AM
@Kaumudi.H that's because they are NERDS :-)
(just like me)

Anonymous
@Kaumudi.H You could also join the coding club in your university. The seniors will help you in learning these things (like Competitive Coding) which won't be taught in your normal CS curriculum. Open source contribution and competitive programming helps to get a good job. But if you are interested in higher studies then it might not be very useful for you. Rather spend that time doing research projects. (BTW I started competitive programming just for fun and it looks pretty exciting!)

@Blue competitive programming as in?

@JohnRennie Oh, well, sure, but I remain a massive nerd in other areas (unrelated to C.S, heavily related to C.N) so I'm not too sure just how into it I must get and how to do that too, even.

Anonymous
@Sid ACM-ICPC

Anonymous
Topcoder competitions

Anonymous
9:48 AM
etc

@Kaumudi.H What is C.N?

Anonymous
@Sid Cognitive Neuroscience

@Kaumudi.H obviously the minimum you need to do is what your course work demands, so do that. If you find you enjoy it then follow your interests (though note it can be a twisty and dangerous road). If you don't find it interesting then just do what the course demands.

@Blue Ah, that's cool! :-) Unfortunately, us freshers aren't allowed to join any of the clubs in the first sem. lest we get ragged (Sigh. Oh, in other news, I did get ragged yesterday) :-(

Wasn't there some form you had to sign to ban ragging or something in your college?

9:50 AM
@Kaumudi.H there will be people (generally male people :-) who have been hacking computers since birth and whose knowledge of computers will make you feel totally inadequate. It isn't that you're inadequate, it's that they are obsessed!

@JohnRennie Hmm, right, right.
@Sid Cognitive Neuroscience :-)

Anonymous
@Kaumudi.H You can contribute to open source projects of organizations which work in the area of medical research (maybe neuroscience too). Just go to the GSOC website and type "medical" or "neuroscience" in the organizations tab and check them out

@Kaumudi.H Also, what did they tell you to do?

How did you get ragged?

Anonymous
That way you will be working at the junction of C.N and C.S :)

9:51 AM
@Blue Ah, again, that's a great suggestion, thanks! I will be sure to check it out.

@JohnRennie There are way too many cyberpunks in the world, true.

@Sid Well, they did it anyway.

Anonymous
@BalarkaSen Says the mathpunk XD

@BalarkaSen nonsense, you can't have too many nerds. It is a noble and honourable calling :-)
5

On Fridays, we have a 2-hour break between 12 PM and 2 PM. Since I had been quite stressed all through the week, I decided to take this time to relax a bit so on my way back to the college from my hostel, I started reading Age of Reason.

9:53 AM
@Blue Well, CodeChef IDE is not a bad choice. After all, it is completely appropriate for your current goals. And it allows you to quickly try different languages, which is also useful for the goal of getting into programming.

Anonymous
@ThomasKlimpel Yup! :)

@Blue I'm a (pretentious) hippie, but I don't think mathematics has anything to do with that. :p

Another important fact to my ragging story is that I sit in the first bench in the middle column. I'd been sitting there, reading, when three seniors yelled for me to come outside.

Anonymous
HackerRank, CodeChef etc all of these are pretty exciting :D I'm bored of doing school level string, array programs

@BalarkaSen yeah - I'd just forgotten stuff I did last year in differential geometry about scalar/vector/one-form/tensor stuff :P

9:55 AM
differential geometry is a confusing puddle of things

@Blue and I am just refreshing my memory of them now. Thankfully, the rustiness seems to be wearing off faster than I had thought..

They "reprimanded" me by saying "Have you really come to college to read?" Then, they went on to ask me my name, the name of my staff advisor, etc.

I struggled with understanding what an object even is a few days ago. I knew it was a vector field, but I couldn't figure out what valued (it's a bundle-valued vector field).

Anonymous
@Kaumudi.H Hey, hey ! There's something for you :D

...that is ragging? They simply asked you your name..

9:56 AM
And then they asked me to sing a song for them so I went ahead and did just that. Unfortunately, I'd forgotten most of the lyrics to that particular song so they waved me off, saying "The next time we come around, you better have learned the lyrics".

Yep... Thankfully, the head of department created what felt like an incredibly good set of notes, then the lecturer was really good at teaching it in a clear way :)

@Blue Oh, wow, thanks!!

Anonymous
lol...few of our seniors came to take our intro...most of us just strolled out of the class XD

They asked you to sing a song? Weren't you getting a bit suspicious by that stage? :-)

Oh, then, that is ragging.

9:58 AM
@JohnRennie Oh, of course I was. I knew it the moment they called for me that I was going to get ragged :-P

Heh, when we had orientation, we got chocolates from our seniors. :P

I'd sing a death metal if I had the chance.

Anonymous
@Sid We got laddos :P

We got crap on a stick and a bucketful of pretentiousness.

Anonymous
Anyway, they'd have to get an ear drum operation if they asked me to sing

9:59 AM
@Kaumudi.H :-) It makes sense to just go along with it. Let them have their fun and hope they grow up one day. If you make a fuss it just draws attention to you.

^ My thoughts exactly.

@Kaumudi.H if there was ever an appropriate time to burst into full-blown opera, that was it :P

@Kaumudi.H it wasn't the wittiest of rags :-)

..Also, why on earth would you want to sit on the first bench where you are visible to everyone?

10:00 AM
@JohnRennie Thankfully not :-)

i mean it was a fantastic chance with a lot of potential
you could have rickrolled them
come on

^^My thoughts

@BalarkaSen that would have been fantastic

Anonymous
Mine too ^^

@Sid On the contrary, nobody can see your nervous face if you sit right at the front, especially if you're asked a question or something.

10:01 AM
Though, I doubt they would have understood what hit them

@BalarkaSen Oh, yeah, like they'd know that song.
In any case, it would've needed to be a Malayalam song.

Anonymous
You should have just sung "meine selfie leli aaj"...you'd become an instant celebrity

There you go

Oh damn. Talk about a missed opportunity. Gimme a break, guys, I was quite nervous.

@Blue Please... no. No No.

10:04 AM
johnwbarrett.wordpress.com/… @BalarkaSen (although you'll already know much more than in this)

Anonymous
On the contrary my school music teacher would get nervous while asking me to sing. My singing is so bad that he would usually stop me midway and tell me to sit down and say "Thank You" XD (I think he used to give me grades for not singing :P)

Hahaha x'D I sing OK, actually.
@JohnR: Are you particularly busy atm?

@Blue That is also why I skipped music classes after 4th grade. :P

@Mithrandir24601 Interesting, I'll have a look, thanks!

@Kaumudi.H no, I'm killing time before lunch. What's up?

10:06 AM
Gchat?

Anonymous
@Kaumudi.H Oh, and one more thing. Coursera has CS videos on almost all topics. If you are preparing for GSOC or Competitive Programming then it's gold. Also, you can audit all courses for free. Check it out sometime.

@Blue coursera asks to give your debit card/bank account details..

Anonymous
@Sid Nope.

Anonymous
You can audit all courses for free with any cards

Anonymous
I'm currently doing a Java course for free without having to give any card details

10:13 AM
@Blue Ah, I know about Coursera. Thanks! :-) Please do let me know about other related websites/programs about which you become aware.

@BalarkaSen haha I saw that just yesterday

Anonymous
@Kaumudi.H Another free website is edx. But I find Coursera better. Ravindrababu Ravula's video lectures are also quite nice but they are unfortunately paid

@Avantgarde i found it on my dank meme/shit videos/instant regret list on youtube
do you like Trump?
musically, I mean?

Anonymous

Anonymous
@Sid

Anonymous
10:18 AM
Click on "Audit" instead of "Start Free Trial"

I just tried CodeChef IDE with a small little fun C program. I was surprised how often I got a "Status Submissing limit Reached". But after a few times pressing "Run", it got scheduled. It computed 4 s. Then I changed it to take about twice as long. Now I got "Runtime Error": SIGTSTP! I tried it several times, always the same. Looks like the program should take less than 8 s to run...

@Avantgarde In any case, it doesn't matter. Have a look:

OK, the status actually says it quite explicit: "Status Time limit exceeded", "Time 5 s". Which is fine, many reasonable things can be done in less than 5 s.

"for" works in C++, I think?

Anonymous
10:22 AM
It should

@Blue Oh, cool, thanks :-) If you find anything else in the future, do tell.

@0celóñe7 u there?

@AccidentalFourierTransform :)

if $f:M\to N$ with $M,N$ some manifolds, what exactly do we mean by $f\in C^k(M,N)$? does it mean that $f\circ \phi^{-1}\in C^k(\mathbb R^m,\mathbb R^n)$ or something like that?

Almost, should be $\varphi \circ f \circ \phi^{-1}$, I think.
But yeah, it means in local coordinates it's a $C^k$ function.

10:25 AM
that makes sense
what if $N$ is some fibre over $M$? same thing I guess

Not sure what that means. You mean $N$ is a fiber bundle over $M$?

If you have a map $f$ going between two things, then it being C^k means locally it's a k times continuously differentiable function R^m --> R^n. Whether those two objects are related or not, that does not affect the meaning of "$f$ is $C^k$"
I suppose in particular you are looking at sections of fiber bundles.

yes, for some reason I expected the projection $\pi$ to play some role here, but I guess not (or does it?)
BTW

@AccidentalFourierTransform It does play a role, but on being a section, not on the section being $C^k$. If $\pi : E \to B$ is a bundle, a section is a map $f : B \to E$ such that $\pi \circ f$ is the identity map $B \to B$, after all.

Anonymous
10:32 AM

Anonymous
Heh

@BalarkaSen hmm that makes sense, thx

@Blue Low budget but fantastic -+++ meme. bangalir ccheler motoi dhadhabaji buddhi

Anonymous
:P

Anonymous
I didn't mean it to be a meme though. But, oh well

10:35 AM
all great things are discovered by accident
the important thing is to sell it afterwards
and sell it reaaal good

- my parents, 23 years ago

@AccidentalFourierTransform sorry for my next two sentences lol

@Balarka haha yeah, Trump songs are great

Anonymous
@AccidentalFourierTransform bwahahahahahahahahahahaha

@BalarkaSen Let me try to decode that. Buddhi=Clever/intelligence. Dhadhabaji=Grandpa? Bangalir=Some guy named that? Or simply Bengali?
@blue recommend some courses on coursera on programming. I am getting bored alone here..

Anonymous
10:40 AM
@Sid Which language do you know?

I know Java and a bit of C++.
a bit= can do easy to moderate programming problems
Right now, I would prefer C++ courses though.
Java takes way too long to debug for me.

@0celóñe7 ^

@Sid A non-literal translation would mean "ideas worthy of a bengali kid"

Anonymous
Then start with UC San Diego's "Object Oriented Programming in Java:Data Structures and Beyond Specialization"

Anonymous
@Sid

10:42 AM
Read above statement. Would prefer C++ as of now..

Anonymous
Try "C++ for C programmers" and then Princeton's or Stanford's Data Structures/Algorithms

Anonymous
Data Structures and Algo is a the backbone of CS...those courses are mostly language independent

0

It would be nice to have mhchem enabled, by adding TeX: { extensions: ["mhchem.js"] } to the mathjax configuration. This would be useful to write nuclear reactions.

Anonymous
11:03 AM

Anonymous
At 0K would the Pauli Exclusion principle be violated ?

Anonymous
Since, at $E_f$ there is a vertical line

Anonymous
I'm finding this part a bit confusing regarding Fermi-Dirac distribution

Anonymous
@JohnRennie Could you help?

Anonymous
I can't understand how to interpret the graph

11:06 AM
0K is impossible

Anonymous
Theoritically

Anonymous
I'm just trying to interpret the graph

0K is just as meaningless as ∞K
you see, the definition of temperature is actually wrong
the meaningful parameter is $\log T$, and not $T$ itself

Anonymous
Now I'm getting even more confused =P

not really. Once you accept that 0K is a logical impossibility, it all becomes clear
you wouldnt ask what happens at $T=\infty\,\mathrm{K}$, would you?
you can ask about $T\to\infty$, but that is a radically different question
the second one is meaningful and interesting, the first one makes no sense
now replace $\infty$ by $0$
the same thing happens
temperature is not like the rest of physical parameters. you can very well have zero mass, zero lenght (at least in principle)
temperature is logarithmic, zero is not in the domain

Anonymous
11:11 AM
The problem is $0K$ doesn't seem to be as much of a logical impossibility to as does $\infty K$. I am looking for some facts which will convince me. BTW, that was not even my question. I was just trying to understand the meaning of the vertical line in the graph.

Anonymous
@AccidentalFourierTransform Ok, I didn't know that. Any reference?

anyone tell me why this is wrong: codechef.com/viewsolution/14808264

"doesn't seem to be as much of a logical impossibility" it does not because you never thought about that. I'll take some time. It will eventually become obvious to you.

Anonymous
@AccidentalFourierTransform Maybe...do you suggest me to read something?

yes, the stranger, such a wonderful book

Anonymous
11:15 AM
........_...........

well it is

The question was to find the factorial of an entered number...

@Sid why do you think its wrong=
?

Anonymous
What is the meaning of "cin>>m;" ? Sorry, I don't know much C++

@AccidentalFourierTransform Because that's what codechef says to me..
@Blue Input a variable m.

11:18 AM
well it compiles, and it does take an integer from cin, and it does return its factorial
so it works :-)

Exactly. I don't understand why it is a wrong answer, though
Here's the question. codechef.com/problems/FCTRL2

¯\_(ツ)_/¯

@Sid you got an integer overflow... The problem seems to expect that you use a bigint representation to work around overflow issues.
Actually it makes sense. The problem says that n<=100. So you need a bigint representation which can still handle 100!

So... you are saying that I need to give an upper limit to the input value?

No, the problem statement gave the upper limit of n<=100. Your task is to output n! even for n=100. A normal int or long will not do the trick.

11:32 AM
@BalarkaSen haha

@ThomasKlimpel that is ridiculous. But, I can't complain since that's what the question says

Anonymous
What does one mean by "the wave number decides the electron state" ? @AccidentalFourierTransform Do you know?

Anonymous
What does electron state refer to?

context?
wave number= k?
free electrons, right?

Anonymous
@AccidentalFourierTransform Yes. I'm not being able to understand how the professor is deriving the concentration of free electrons

Anonymous
11:43 AM

a free electron is typically described by means of a plane wave, right?

Anonymous
@AccidentalFourierTransform Yup?

(you can also consider wave packets, but that's another story)
well, a plane wave is fully determined by the three numbers $k^1,k^2,k^3$, that is, by the vector $\boldsymbol k$, right?
a plane wave is, by definition, of the form $\mathrm e^{-i\omega_{\boldsymbol k}t-\boldsymbol k\cdot\boldsymbol x}$
where $\omega_{\boldsymbol k}\equiv\frac{\boldsymbol k^2}{2m}$
so if you know $\boldsymbol k$, you know the plane wave, and you know the state of the electron

Anonymous
@AccidentalFourierTransform Sorry. I don't know what $k^1,k^2$ and $k^3$ are. I just write the equation of plane wave as $A\sin(wt-kx)$ or $A\cos(wt-kx)$

sorry, I was talking about 3D electrons
you seem to be dealing with 1D electrons
so $k$ is just a number instead of a vector
same thing anyway
the plane wave is a sin or a cos of $\omega t-kx$, where $\omega=\frac{k^2}{2m}$
so given $k$, you know the plane wave, and you know the state of the electron
right?

Anonymous
11:47 AM
@AccidentalFourierTransform Oh. How to write the equation of an electron in 3D ?

Anonymous
@AccidentalFourierTransform yup!

Anonymous
So that is what he means by state of electron

@Blue essentially, the same thing, but with three numbers instead of just one: $\sin(\omega t-k^1x^1-k^2x^2-k^3x^3)$, where $\boldsymbol x=(x^1,x^2,x^3)$ are the coordinates
@Blue the correct statement is "the wave number determines the electron state"
that is, given the wave number you know what the electron is doing

Anonymous
@AccidentalFourierTransform Interesting. Thanks!

(assuming, once again, that we are dealing with plane waves instead of, say, wave packets)
a (1D) classical electron is determined by two numbers: its position and its velocity
if you know those two numbers, you know the sate of the electron
a quantum-mechanical electron is determined by a wave function
if you are dealing with free electrons, then its wave function is of the form $\sin(\omega t-kx)$, where $\omega=\frac{k^2}{2m}$
so in this case, it suffices to fix one number, $k$, and the state of the electron is determined
$k$ is similar to the classical velocity
in this case, you dont need to specify the position, nor anything similar, because you assumed the electron to be a plane wave, which means that its position is uncertain

Anonymous
11:51 AM
What about $t$ ?

Anonymous
Doesn't that also determine the position ?

no, in the quantum-mechanical case, you cannot determine the position
all you know, all you can attempt to know, is the wave function
this function gives you a complex number for each pair $(t,x)$
the wave function determines the state of the electron
in the plane wave case, the number $k$ determines the wave function
and therefore, this number determines the state of the electron

Anonymous
Right. Makes sense now :)

a bit confusing, I know
it takes time :-P