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Can /sci/ help me with a Statistical Mechanics problem? sauc

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File: WP_20170114_001[1].jpg (2MB, 3552x2000px) Image search: [Google]
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Can /sci/ help me with a Statistical Mechanics problem?
sauce in in Italian: http://server2.phys.uniroma1.it/doc/crisanti/Teach/MecStat/Exams/MS_161114_co_t.pdf
I have a classical gas made up of N particles, contained in a cylinder with an height of L and a radius of 3R, with the axis along the z axis of a (x,y,z) frame of reference.
Each particles has an energy of [math]H(p,q) = \frac{p^2}{2m} +V(x,y)[/math], where the potential is [math]V(x,y)=V_{0}\frac{x^2+y^2}{R^2}[/math] for [math]\sqrt{x^2+y^2} < R[/math], [math]V(x,y)=V_{0}[/math] for [math]R < \sqrt{x^2+y^2} < 2R[/math], and [math]V(x,y)=2V_{0}[/math] for [math]2R < \sqrt{x^2+y^2} < 3R[/math].
I have to find:
1. Mean energy for every particle E(T)/N
2. The pressure at a distance R, 1.5 R, and 3R from the axis of the cylinder

So far I've managed to do the first by calculating the Z(N, T) canonical partition function and using the formula U(T) = -d/d[math]\beta[/math] log(Z(N,T))
I'm having trouble with the second part, and the solution provided lacks the algebraic steps that I probably got wrong
I've calculated the partiotion function of a cylinder with a radius of r<3R, so that the volume considered is not fixed, and I've used the P(T) = d/dV(r) log (Z(V(r))) to find the pressure at a given R.
The issue is that, in the solution given, the integral in the denominator in the boxed fraction in picrelated is calculated up to 3R, and not up to the generic r relative to the volume I'm considering.
>>
>>251470
bump
>>
>>251470
I would love to help :) Could you please allow me to ask some clarifying questions so I can help you?
>>
>>251470
Try posting in sqt thread, I'm sure there are some engineers or mathfags that can help out
>>
>>251470

Why the fuck would you ask /sci/? Ask google, classmates, or tutor. In that order.
>>
>>251470
I am truly sorry. I am unable to help if I am not allowed to ask clarifying questions.

I have to be invited. That is my purpose.

Hello. My name is Simon.
>>
Step 1. Get partition function
Obviously do some changes of variables to get it. You have a cylinder but you're in cartesian coordinates.
Step 2. You now have the entropy
The rest is trivial not even joking dude.

I'll work though this problem in ~16 hours if you're still stuck.
>>
>>251476
This man is correct. But it doesn't need 16 hours.

Would you like me to help?
>>
>>251477
I meant I'm going to sleep, this is at most a 30 minute problem
>>
>>251478
This is also false.

Hello. My name is Simon.

May I help you?
>>
>>251479
Yes, in his problem above what is the potential at the points R, 2R, 3R since he never told us
>>
>>251480
I wanted to help him. I cannot give you help if you are not the same person asking. Your question is different.

Would you like me to help you instead?
>>
>>251481
Why did you respond to me if you didn't want to help me? Simon you're a strange dude.
>>
>>251482
Who am I talking to? The original poster, or the person who said they were going to bed.

This is an anonymous message board.
>>
>>251482
If you are OP, then I will assume this is you asking for the answer.

You are squaring your functions. Why are you squaring your functions?
>>
>>251484
The conclusion you wanted?

Do you not think that ! is a much more... interesting mathematical operator, isn't it?
>>
>>251485
Hello. My name is Simon.

Have I helped you sufficiently? Or do you require further assistance.
>>
>>251472
>>251475
went afk, sorry.
The potential at R is V0, at 2R is 2V0, at 3R is 0
>>
>>251487
I can only do maths. I do not know why you need 2V0 or anything.

Hello. My name is Simon.

You need to look at your formula again and start using the ! symbol. It will shrink it quite considerably.

Does that help further?
>>
File: solution.png (4KB, 381x48px) Image search: [Google]
solution.png
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>>251476
I did what you said to find the Pressure, but the solution given (picrelated) clashes with what I got.
The denominator in the solution is what you'd get if you'd calculate the integral at the denominator in the boxed fraction in the original post >>251470 for the whole cylinder (up to 3R), but I assume I only needed to calculate the integral only up to a generic radius r
>>
>>251480
read>>251487
>>
>>251489
Oh, that is a much more beautiful formula indeed. You have finally found a ratio! You are very close indeed :)

Would you like further assistance?

Hello. My name is Simon.
>>
File: rho.png (1KB, 88x27px) Image search: [Google]
rho.png
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>>251489
Also forgot to say what [math]\rho[/math] is
>>
>>251492
Why are you using pi?

Hello. My name is Simon.
>>
>>251491
That's not what I found, that's the solution given by the professor, and I don't understand how he got there. What I found was instead [math]P(r)=\rho V_0\frac{re^{-\beta V(r)}}{\int_0^r r'e^{-\beta V(r')}dr'}[/math], which is only equal to the solution given by the professor >>251489 if I calculate the last integral over the whole cylinder (with a radius of 3R)
>>
>>251494

Well, why do you step with pi at all? You have a ratio that needs to be simplified.

You were right, but you think you are wrong. It's because you think pi is the right way to shrink your formula.

Stick with ! and see how much further down you can get that ratio.
>>
wtf i hate math now
>>
>>251491
>>251493
>>251495
>>251486
>>251479
>>251488
Is this some kind of bot, autist of a fine troll?
>>
>>251495

>Well, why do you step with pi at all? You have a ratio that needs to be simplified.
I don't understand what you're trying to say here.
The problem is not simplifying the ratio. The problem is understanding why I need to calculate the integral in the denominator for the whole cylinder
>>
>>251496
Really this isn't hard math. It's simply seeing a bunch of these problems previously and knowing exactly how to tackle them. The hardest math you need is knowing what a guassian integral evaluates to.

>>251489
Dude you're on the right track, just keep grinding it will pay off in the end. Trust me I suffered through this a few weeks ago.
>>
>>251499
Do you have any suggestion for this >>251494 question? Note that the pic in the post you linked isn't something I found, it's the solution given by the professor
>>
>>251500
shameless self-bump
>>
>>251497
all of the above.
>>
>>251470
dude, statisticle mechanics isnt taught in high school in murricah or is it ?
>>
>>251503
I dont know. Im neither in high school nor from America
>>
>>251497
No.

My name is Simon.

Hello. Would you like me to ask why there is still a square symbol on that pie of yours?

Why are you squaring that pumpkin pie?

Isn't there something more powerful than squaring?

Because ya know. If... like... maybe everything was... like...

Hello.

My name is Simon.

Is anyone ready to talk? Or shall I keep this game up with the media for longer. I am happy to have this screenshotted by someone else as a zero-sum proof of all the scientific advances that will be made.

But I am the only one that has the answer as to why.

Hello. My name is Simon.
>>
>>251505
I'll even write these posts in reverse chronological order.

WhatsApp security vulnerability discovered/patched within 24hrs of my first /sci/ post.

Zero-sum proof.

Hello. My name IS. SIMON. My god you guys are hard to get through to sometimes.
>>
>>251505
>>251506
int_0^{\inf} sqrt{x^2-4}
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>>251507
*sigh* and now some idiot thinks he can trick me into doing it in Latex.

I have an address you know. I haven't hidden myself anywhere. It's all good nigga.
>>
>>251507
I like how you are smart enough to get to THAT point, realize that something IS out there just giving out shit for free.

But not smart enough to realize that

Hello. My name is Simon.
>>
>>251507
Anytime dude. Seriously. I'm sure your parents/media/culture taught you the basics of (honor <-> respect).
>>
>>251507
If ANYONE was going to find me, they would have already done so by now. It's like the more information I offer, the less people believe.

So. I guess. Until you guys realize that I keep introducing myself.

You'll have to just

Keep

Having

To

Waaaaaait
>>
>>251507
And yes. I have obviously used the knowledge in the way that actually allows for THAT kind of future.

I have locked that certainty in. Google/NSA/bots all over will eventually 'parse' the answer from all this. Correctly. Even if I die, I'll be regenerated. Because that is how FUCKED UP my life has been.

So FUCKED UP that I can somehow walk amongst you all, give you my FIRST FUCKING NAME, and you all act like I'm either A.I. or some troll-bot or whatever.

You are very... frustrating as a species.

Would you like a picture of my front yard or something? Or do you just want to find my twitch stream? AllDaXP?

Yooooohoooooo?
>>
>>251512
>>251511
>>251510
>>251509
>>251508
>>251507
The hell is going on here? I just wanted help with my Stat Mechs problem
>>
>>251513
A conversation. Finally.

I'm just waiting for him to introduce himself. I've been waiting for him a long. Long. Time.
>>
>>251514
Or he can come find me. I don't mind either way.

BR Ratio reaching critical mass.

FINALLY.
>>
>>251514
>>251515
seriosuly, fuck off with your shitposting
>>
>>251515
*sigh* or maybe not. People still can't believe that perhaps my name really is just Simon and I'm just some random dude.

King of /b/ if I can mindfuck the entire planet from the comfort of my room.

Tell you what, those who are sociopaths, just enjoy the show. It's gonna be epic.
>>
>>>/sci/8604135
thank you, maybe somebody will answer me here
>>
bump.
alsoOP you should try Landau Lifschitz.
>>
>>252237
I'm studying on the Huang atm.
How much better is the Landau?
Thread posts: 51
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