The h Bar

3:17 AM

@heather reopen my other question

@DanielSank : da

for me i like to pretend that i understand german even though im about to hit the 8 year mark since i last took classes

also, i cant believe i got through almost all of highschool and college until i learned about libgen

5:35 AM

Hi Guys, I need some help with 'things' most people find noobish. Anybody got free 10 minutes?

Alright.

Hi Aakash, what's up?

user228700

@JohnRennie: Morning :-)

@Kaumudi.H Morning. And the news is ...

user228700

Hm?

The laptop?

5:45 AM

Howdy John

hows your morning

Morning

user228700

Oh, right. They haven't contacted us since and the website still says "Held awaiting customs clearance"

@Skyler dark, cold and wet! But I've had breakfast and I'n enjoying my coffee, so that's all right :-)

@Kaumudi.H :-(

@AakashTomar what did you want to ask?

user228700

The cyclone was more devastating than it seemed during. The city has changed dramatically. Steel rods by the sides of flyovers were bent completely. More than 3000 trees fell and they have blocked many many many roads.

@Kaumudi.H Wow!

5:47 AM

@Kaumudi.H dang

not nearly as devastating a story, but once we had really bad winds that knocked out power in town, and brought down a good few trees, so me and my highschool buddies played a game called "fugitive"

Can anybody tell how they have written that step

Fugitive (game)

The game of Fugitive combines elements of a number of outdoor games such as capture the flag, cops and robbers, and Sharks and Minnows. In Fugitive, players divide into two teams, with each team playing the part either of the "fugitives" or of the "police". The cops are further divided into "runners", "drivers", and "riders." == Game Basics == The fugitives' objective is to run from a starting point to a finishing point without being tagged by any of the riders. The fugitive must remain on foot, no vehicles or alternate modes of transportation. There are no boundaries as to where the fugitive may...

user228700

We didn't have power all day yesterday, hence the absence. Power was restored only at around 11 PM last night after being gone for more than 36 hours.

user228700

5:50 AM

Today my mother had an item from Amazon delivered home so the roads have been cleared so as to allow motorcycles to pass through, I guess.

user228700

So I hope that once the laptop clears customs, it will be able to reach my home on the same day. But goddammit, how much longer before it effing clears customs?!

@Kaumudi.H incidentally, not even the British would use a pressure cooker to cook vegetables! :-)

user228700

Really? We use a pressure cooker to cook lentils and vegetables (potatoes, carrots, beans etc.) all the time. Every day, actually.

user228700

@Skyler Yeah :-/

user228700

Although I do understand why you didn't include them poor veggies.

user228700

5:55 AM

(:-P)

user228700

Say, @JohnRennie: Dyou think I ought to call and ask them about it?

user228700

And what's his name, THE LONE WOLF (OP) is from India, so :-)

@Kaumudi.H no, it will only annoy them.

I'd say they've been pretty good so far, so I'd let them get on with it.

user228700

:-| Okay.

oh hey @koolman give me a sec

i forgot to answer that

oh wait, its a little different then what I was thinking

6:01 AM

@Skyler Check if you can answer it

user228700

Can anybody recommend a great book for calculus? My textbook sucks so bad in this particular topic that I absolutely must refer to another book.

user228700

@koolman: Which book dyou use?

@koolman for the step you are confused on

what is p dot x

@JohnRennie btw do you remember how the heck you put vector arrows on your tex

\vec{x}

@Kaumudi.H my institutes

Allen

user228700

6:07 AM

@JohnRennie: Dyou know of any good book for calculus?

$\vec{test}$

user228700

@koolman Oh, okay.

@Kaumudi.H first off

user228700

Yeah..?

since the three vectors are the same magnitude lets make them all unit length

so we dont need to think too hard

6:07 AM

Ohk

we can always rescale at the end

Yes

they are orthogonal, so any dot product between them is 0

user228700

@koolman: I take it u weren't able to prove what I told u before, related to this problem?

you see that right @koolman ?

6:09 AM

@Kaumudi.H nope

@Skyler yes

@koolman so now what happens if you take $\vec{x} \cdot \vec{p}$

@JohnRennie what about dot product

gosh my vector tex sucks

\cdot

@Skyler don't know

@Skyler this is a useful link that I keep close to hand:

@JohnRennie WHY DO THEY PUT A C IN FRONT, gosh thats random

Q: MathJax basic tutorial and quick reference

6:11 AM

1632

To see how any formula was written in any question or answer, including this one, right-click on the expression it and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next point.) For inline formulas, enclose the formula in $......

@Skyler "centre dot" ?

@JohnRennie and thanks

@JohnRennie good point

@koolman think about it like you have a unit vector

what is a unit vector dot some random vector

x $\cos\theta$

@koolman think a bit more abstractly, what are you doing when you take the dot product?

say like you take the unit vector in the $\hat{x}$ axis off some vector $\vec{v}$

Take its projection

@koolman so you are taking the part of it in the direction of that vector right?

and assigning a scalar

6:17 AM

Yes

so now what do you have if you take that scalar and multiply by $\hat{x}$

Vector in that direction

@koolman now what if you did the exact same thing for $\hat{y}$ and $\hat{z}$

Vectors in all three directions

do you know latex?

6:19 AM

Not much

user228700

May I interrupt, just for a bit?

@koolman enough to write out these 3 vectors

@Kaumudi.H whats up

@koolman just the answers

user228700

I was able to get to a point where if $\hat{p} \cos \alpha + \hat{q} \cos \beta + \hat{r} \cos \gamma$ is $\hat{x}$ ($\alpha$, $\beta$ and $\gamma$ are the angles b/w $x$ and $p$, $x$ and $q$ and $x$ and $r$ respectively) is $\hat{x}$, we prove it.

@Kaumudi.H doing it with trig will waste ALOT of your time here

user228700

(Here I didn't start out with $p$, $q$ and $r$ being unit vectors, but that doesn't matter)

6:21 AM

@koolman and just for x y z, once you write that out i think things will become clear for you

user228700

@Skyler Yeah, Ik. So I was thinking and I'm unable to figure out how the sum of the projections of the unit vectors $\hat{p}$, $\hat{q}$ and $\hat{r}$ along $x$ gives $\hat{x}$

user228700

That's what I gotta prove and I haven't been able to convince myself that that (^) is true.

@Kaumudi.H you wont be able to prove that

user228700

Dang it, why not?

user228700

It looked like u were travelling along the same lines...

6:24 AM

because it gives you $\vec{x}$ not $\hat{x}$
=P

user228700

Oh, hang on, that works too. Maybe I've made a mistake somewhere, but can u tell me how it gives me $\vec{x}$?

how about you try it too, do $(\vec{v} \cdot \hat{x}) \hat{x}$

I'm gonna end up with carpal tunnel syndrome if I go on like that... Working 10+ hours on a laptop is not what I would call ergonomic :p

for x y and z

for notational convenience you can use $v_x$ to represent the projection of v and x

type out the answer here

@G.Bergeron what type of work

$ \widetilde{\mathcal{Z}}_S^N(z) = \frac{A_S}{\sqrt{2}} \Bigg\{ z^{1+t} (-1)^p B_S E_S^{-1} \xi_S^+ P_{n-s-p}^{(2s+2p+\mu_2+\mu_3,\mu_1-1/2+t)}(2z^2-1) P_{s+p-1}^{(\mu_3+1/2,\mu_2+1/2)}\left(\frac{z^2+1}{z^2-1}\right)(1-z^2)^{s+p-1}
+ z^{t} B_S E_S \xi_S^+ P_{n-s-p}^{(2s+2p+\mu_2+\mu_3,\mu_1-1/2+t)}(2z^2-1) P_{s+p}^{(\mu_3-1/2,\mu_2-1/2)}\left(\frac{z^2+1}{z^2-1}\right)(1-z^2)^{s+p}
- z^{1-t} B_S^{-1} F_S \xi_S^- P_{n+t-s-1}^{(2s+1+\mu_2+\mu_3,\mu_1+1/2-t)}(2z^2-1) P_{s}^{(\mu_3+1/2,\mu_2-1/2)}\left(\frac{z^2+1}{z^2-1}\right)(1-z^2)^s

6:29 AM

@Skyler I got it

Thanks

@koolman if @Kaumudi.H starts getting confused have her ask you for hints, if you can teach someone else it you learn it 10x better

in my opinion

@Skyler But yeah TeXing a paper and using mathematica on the side

@G.Bergeron what in gods name is that monster

user228700

@koolman How?

user228700

@Skyler *her.

6:31 AM

@Kaumudi.H noted =P

@Kaumudi.H I have done it in another way . Not like you have done

I don't know how programmers do it with all the weird characters... Maybe not having a French keybord helps!

user228700

@koolman Still, will u please share how u did it?

We know p.p = k

user228700

Right.

6:32 AM

Hence magnitude of p is root of k

Sorry for latex

user228700

Yep. NP.

@Skyler This : arxiv.org/abs/1605.03407 It's a reworking of this paper with new results

@G.Bergeron programming doesnt use way too many weird characters tbh, latex probably gets more messy than most programming syntax

Now we can write p.x as (root k)(|x|)$\cos\theta$

@AakashTomar I prefer to discuss things here where other people may be able to help.

user228700

6:34 AM

@koolman Yeah.

And when we multiply it with vector p

I didn't want to disturb the ongoing conversation.

Anyway, first, please guide me on how to delete a chatroom you create.

user228700

Yes..?

@AakashTomar It's OK, we commonly have many overlapping conversations :-)

We get a component of x in direction of p vector

6:35 AM

@AakashTomar you can't delete a chat room, but the system automatically deletes unused chat rooms after some set time.

@Skyler I hope for them! But sometimes I question myself why did I got into a field with so many nasty equations?

Similarly for other two we can say like this

user228700

:-| Alright, maybe I should've just brainstormed a little more.

Hmm, okay.

@G.Bergeron and gibberish sounding terms, Banana-Ito whatever you call it polynomials

6:36 AM

LOL

user228700

This is almost exactly what I've done, except for the last step, which I haven't been able to understand.

@Kaumudi.H have you understood till here

user228700

Thanks :-)

Welcome

@G.Bergeron im guessing its just some weird set of orthogonal polynomials that form some group and algebra right?

6:37 AM

@AakashTomar anyhow, what did you want to ask?

@Skyler They are the solution of the "double" Clebsch-Gordan problem for particles obeying weird statistics, basically

Or 3nj symbols, I don't know what makes more sense to you

user228700

@koolman: BTW, what u just did was prove what I had asked u help with, before :-)

@G.Bergeron this probably

Next, I was reading about a cyclotron. Actually, you can say I love building stuff, but till now, all I've made is some stupid DIY's. So, to summarize, firstly, is building a cyclotron possible, if I know the underlying principles? Or is it possible only for big science labs and scientists?

never heard of the double Clebsch-Gordan (though I know the regular case CG problem)

6:39 AM

Also, are there any particle accelerators which are simpler in design, as compared to cyclotrons?

@Skyler It's actually nice the thing about statistics: They go beyond the boson/fermion classification and can exist in 1D/2D confined systems

@Kaumudi.H oh yeah , I couldn't recognize it :)

For instance the excitation in the fractional quantum Hall effect are anyons, not the same statistics, but some other exotic statistics

@G.Bergeron whats so weird about the statistics in the double clebsch gordon

ah

@AakashTomar I don't really know to be honest. Cyclotrons are pretty simple so you probably could build one yourself. I'd have a Google to see if anyone has done it.

6:41 AM

@JohnRennie Yes

@AakashTomar make sure you have a grounding wristband

#1 on the list

Amateur scientist column of scientific american!

If you like tinkering buy the CD or whatever it is now of the whole archive

2MeV is a lot of arc length

@AakashTomar the simplest accelerators are probably linear accelerators using a pair of charged plates or a Van der Graaff generator.

I bought when in high school, really nice!

@JohnRennie There is also a linac

@Skyler The question is how do you relate the different possible basis when decomposing more than two representations

Ans I mean Clebsch-Gordan problem in the general sense, not just for spins.

@Skyler Interestingly, it has to do with the statistics the particle obeys.

6:44 AM

Van der Graff is huge power. 10^6 - 10^7 isn't small stuff. :-P

@AakashTomar Not much power necessarily

@AakashTomar super low current though iirc

microamps i think?

@AakashTomar VdG generators can build up high potentials, but the power they generate is only small.

@Skyler some giant ones get milliamps

Yes, sorry, I was talking about the voltage. ( That's 10^6-10^7).

6:45 AM

Or advanced designs like pelletron

@G.Bergeron yes, but you're talking massive generators for that.

@JohnRennie Yes :P

@AakashTomar it's that high voltage that accelerates the charged particles.

@AakashTomar I would go with a flyback transformer without the diode and then use a multiplier

@JohnRennie because F = qE

6:47 AM

If you go above around 100kV in hard vacuum, you need to worry about xray and dosimetry

@G.Bergeron xrays, really? and wth is dosimetry

like exposure to radiation

@Skyler To track your radiation dosage so you don't have too much

@G.Bergeron not at the hobbyist scale.

@JohnRennie i assumed hard vacuum covered that John =P

You're making this out to be more of a big deal than it is :-)

6:48 AM

can you even get to a hard vacuum with hobbyist things

@JohnRennie But if he gets a diffusion or turbo pump running on that you can accidentally build a pretty efficient xray generator

@Skyler Yes zeolith

So, you connect the terminals of the Vdg with two big plates, and create a vacuum, and that is the particle accelerator ( a very rough description)?

@AakashTomar yes

That you bake and then immerse in a liquid nitrogen bath

But you do need a forepump, like rotary vane

Wait, who'll give me liquid nitrogen? :-P

6:50 AM

But you can buy cheap AC units and modify them to have bigger throttle

@Skyler My first big purchase in life was a vacuum pump, :p

And does a vacuum cleaner create enough vacuum for 'hobbyist' experiments?

@AakashTomar nope

@AakashTomar No :-)

Lol, kidding.

I know :P

@G.Bergeron how "big" was the purchase?

6:51 AM

@JohnRennie But the thing is that you kind off need to get a mildly hard vacuum so that the mean free path is not too small

@G.Bergeron if you want to start having to deal with the headache of ionizing radiation that is harmful to your health

Roughly, I mean very roughly, what can be the order of dimensions of the chamber?

somehow it doesnt sound like the cost and effort is worth making your project more hazardous

@AakashTomar you could build a fusor

@Skyler I was beginning high school, so not too much but obviously a lot for me back then

like 350$

6:52 AM

symmetrymagazine.org/article/august-2010/…

That accelerates ions.

@Skyler I liked the hazard, my parents... a bit less

But still with fusor you need below 10-4 microns

hence the radiation risk... But with a fusor, you don't need a big acceleration potential

Wow, the fusor is a nice idea. @G.Bergeron can you suggest something which'll work according to you?

So not much problem

@AakashTomar Mechanical rotary vane vacuum pump, buy the one for the AC industry

Then modify the inlet to be able to put a BIG tube

ALready with this setup, you can do a lot of nice plasma experiement

Hey @JohnRennie how familiar are you with phase diagrams?

6:55 AM

and accelerate electrons and ions on short distance without any complication of deep vacuum

with a crookes tube like device

@Skyler phase diagrams of what? The phrase phase diagram covers a multitude of sins!

Shriek!!!

@JohnRennie like that

What I used younger for the chamber where soda-lime glass tubes and welding torches so that I can fuse wires inpute

You can use the dumet seal in incandescent light bulbs

6:56 AM

@Skyler OK, diagrams like that are pretty simple. What did you want to know?

Those seals have about the same dilatation coefficients as the soda-lime glass and are "wetted" by molten glass

@JohnRennie I'm going over my notes and trying to redo this problem, however I've completely forgotten what that dashed curve on the left means

was it metastability?

@AakashTomar I'd suggest you start playing around plasma discharge before going on with that

like whats the usual convention for dashed lines like that

@Skyler I don't think there is a convention for it. At first glance it looks a bit like a spinodal or binodal curve.

6:59 AM

i feel like there was SOMETHING tricky about this problem, and its not just usual eutectoid phase transition

Maybe have a look at a really good resource : belljar.net You can buy the collection of all the past published material

I'd REALLY recommend it!

@Skyler Are you studying for the phD exam or something?

@JohnRennie are there any things it could be where it would effect the type of transformation the arrow trajectory has

@G.Bergeron meeting with professor

Haha!

@Skyler I don't think so, or at least I can't see how. The arrow is in a region of the diagram where it behaves as a simple eutectic.

Ok, long enough break... see you later

7:03 AM

hmm

@JohnRennie well thats reassuring then

means i can just do the simple problem

Yes. As the temperature lowers you'll hit the coexistence curve and get two phases as usual.

@G.Bergeron i think ive found the big problem with trying to review kinetics problems, they can always be made a zillion times harder if you start to second guess yourself

@JohnRennie wait

actually nvm

@JohnRennie btw, when you worked with colloids how much did intrfacial stat mech and/or phase diagrams come up

I spent more hours than I care to remember calculating binodals and spinodals!

But it was generally for simple two component systems.

@JohnRennie alas, another thing im reviewing atm

Sorry guys, had to go for a while.

John, how old are you?

7:11 AM

@JohnRennie btw, that actually brings up another quick question i had

@Skyler Yes?

@JohnRennie (getting picture ready)

@AakashTomar 55 - an old man compared to the rest of you :-)

Wow, you must feel so awesome.

Being young isn't very nice.

Mostly I feel old :-)

@AakashTomar Booooooooooooooooooooooooooooooooooooooollocks! :-)

7:13 AM

No, seriously. Your parents are so worried all the time.

@AakashTomar true, my mom asked me not to come home tonight (couldnt anyways) because it was the 13th

@JohnRennie here we go

@AakashTomar so being old means you're worried all the time. That doesn't sound so great to me :-)

Like, they're worried you'll get injured, you'll fail, you won't get a job.

Is the free energy curve for gamma a simple parabola here

Oh, so you're a parent. Actually, I should call you sir or something, you're older than my dad.

7:15 AM

@Skyler You mean is $G(x)$ a parabola? As I recall the free energy curves generally look vaguely parabolic, but they aren't parabolas.

@JohnRennie parabolic but yea

monotonic curvy things

@AakashTomar no, I don't have any children. Just a niece. That's why I'm not worried :-)

@Skyler monotonic curvy things :-))))))

But yes, that's a good description.

@JohnRennie but theres no way the monoticity of gamma can change in this problem since there is nothing to indicate a spinodal decomposition right?

the only times you can have valleys and peaks is if you have two phases with the same structure or spinodal decomposition right

To be honest I can't really remember as it's 25 years since I did this stuff and my phase diagrams were generally simpler than yours.

I would risk misleading you if I answered without going away and thinking about it.

Which reminds me, sir, that I need your help again. My course book says that V=IR can be applied even on diodes. They're saying that V=IR is not a statement of the ohm's law, (necessarily) and it can be applied to all conducting devices. Is this correct?

7:20 AM

@AakashTomar Well you can measure the voltage V that you apply, and you can measure the current I that you get. Then you can define a resistance R using V=IR. The trouble is that the R defined this way is not necessarily constant but is a function of V and/or I.

@JohnRennie well at least its refreshing to have someone who instantly knows what spinodal decomposition is if i say it =)

@Skyler I'm not sure that's a great claim to fame :-)

Okay, so all ohm's law says is that the plot of V and I is linear, right?

@AakashTomar Yes, linear and goes through the origin.

@JohnRennie there are a good number of times where I just feel like if I ask a question people here will be like "whatcha talkin about"

7:22 AM

Yeah, there's no intercept. Also, what is the reason semiconductors follow ohm's law for only a limited range of electric fields?

the cross talk between materials science and condensed matter physicists is a bit less than i thought

@Skyler I think the condensed matter field tends to be highly theoretical these days while the materials guys are more experimental.

@JohnRennie yea, MS works its way up into CMT often, but the other way isnt that true

@AakashTomar I think pure semiconductors follow Ohms law pretty well don't they? Semiconductor junctions generally don't.

@AakashTomar basically they are generally very nonlinear, but a region is engineered where they can get a linear response in a region so they can make engineers lives way easier

lots of fancy statistical mechanics probably

for nonlinear region

7:26 AM

Yes, but what is the reason they deviate from that linear region for high electric fields?

@AakashTomar probably semiconductor electrons start getting stripped off

and flowing

Uh, but isn't that the very reason current flows? ( Sorry if I sound like I'm high )

@AakashTomar true in a sense, in semiconductors the flow of electrons is mediated by the holes moving around most of the time

so you move the holes to swap electrons

but once you break down the semiconductors (which may also alter the crystal structure, isotropy, and a zillion material properties), the simple hole mediated (and likely linear) response disappears

Okay, so you're saying, high electric fields change the material properties of the thing. ( Am I right)

the hole stuff isnt even generally linear, they just create a set of conditions (say on a transistor with some required range of gate voltages), where the answer is very close to linear, and work from there

@AakashTomar they very likely do

and can in ALOT of ways

7:30 AM

@AakashTomar Yes, I'd say that's a good summary

Okay, thanks a loooooot. I mean a lot. These things have been haunting my sleep.

There's one last thing.

@JohnRennie i havent paid attention to aakash level, do you think I should have just said it changes material properties or was the rest of the details worthwhile

@AakashTomar fire away

@Skyler it's alwas good to discuss things at lots of levels. You never know exactly what level is best and also someone else might be interested. I quite often find myself in discussions where two of us are explaining things at different levels.

These days, I've developed a lot of interest for particle physics. So, I've done electromagnetism, classical mechanics, thermodynamics, optics and a bit of modern physics. First, is this sufficient to start learning particle physics?

@AakashTomar how old are you? Are you in school, or college, or what?

7:34 AM

@AakashTomar lower division of each? like have you worked in dirac notation

or done lagrangian/hamiltonian mechanics at all

I'm 17, high school. No, :D no no. That's quantum mechanics. I've done things like Bohr's model, photoelectric effect. Basic, pretty basic stuff.

Technically, homeschool.

If you're really interested you need to learn the basics of quantum mechanics.

I mean, there's no teacher I can ever consult.

have you done linear algebra?

@JohnRennie i was going to suggest the lin alg of hilbert spaces in particular, wouldnt you agree

Does linear algebra include complex numbers, quadratic equations, that stuff?

7:36 AM

When people say particle physics it really means quantum field theory but that's a bit scary. Basic QM should be accessible at your level though.

Exactly, but I'm clueless about the next step.

@Skyler there are lots of different views about best how to approach QM and I'm honetsly not sure what is best.

@AakashTomar linear algebra basically generalizes the relationships you've observed in the mathematical objects you've worked with so far, and you'll find some really useful things are common across a lot of objects while if you make certain assumptions (axioms) for a type of object other useful properties pop up

I learned it starting with the Schrodinger equation, which is the traditional route that Griffiths etc tak. But for example @DanielSank thinks it's much better to learn it in a different way that concentrates on the mathematical structure.

in a fundamental sense particles in p.physics are mathematical objects so understanding linear algebra (and ideally some group theory) is really important for it

7:39 AM

I have to get back to work for about half an hour. See you later ..

@JohnRennie i go to a school where they do a special lin alg of hilbert spaces (optional) that just makes griffiths a walk in the park after

Okay, that's a pile of work. Yeah, I've to go too. Bye @Skyler @JohnRennie

@AakashTomar later, just make sure you look into linear algebra (its the most useful math youll learn imo)

Sure thing.