The h Bar

12:16 AM

Whoever designed the dialog interface in Slyrim should be shot.

3:50 AM

Mini-poll to satisfy my curiosity. Would any of the assembled wise describe spin (that QMical, intrinsic angular momentum thingy) as being "a one dimensional point particle"?

dmckee

I'll say why I asked sometime tomorrow if anyone is actually interested.

0celo7

4:37 AM

@dmckee I honestly have no idea what you're asking :/

dmckee

So, I can take that as a no, then?

0celo7

Could you maybe rephrase the question?

dmckee

I agree, BTW, it doesn't make any sense. But I want to know that this is not just me being the provincial lout that I am.

0celo7

English-wise I have no clue what you are asking.

dmckee

It's not my phrase, and I think the person to whom it belongs is far, far off the beaten track.

0celo7

4:41 AM

Is spin a particle?

Is that the gist?

Would any of the assembled wise confuses me. Do you mean "would any of the assembled wise men"?

dmckee

@0celo7 Well, if you would answer that as "no." then we can safely put down a "no" for the whole thing.

@0celo7 Well, the women are invited too.

0celo7

Is a goat a symphony?

dmckee

I thought I was channeling Tolkein on that.

0celo7

@dmckee I got that impression, actually.

Does spin have some sort of representation in phase space? No? It's not a particle.

Immersive weapons is a weird mod. Some of the weapons are really cool, like elven steel swords, but the katanas are plain strange...why are there katanas in Skyrim?

@ACuriousMind One of the many joys of Immersive Creatures I assume. Scared the crap out of me!

Recognize Evil as Waste

6:01 AM

What is the \dot m \Delta v in en.wikipedia.org/wiki/…? How is it related to m dv/dt?

skill patrol

6:39 AM

here

Recognize Evil as Waste

7:01 AM

Mass flow rate just says that \dot m = dm/dt. I am asking to clarify \dot m \Delta v = m dv/dt. I see nothing similar in the mass flow article.

David Z

8:14 AM

@RecognizeEvilasWaste I can't tell offhand how they get from $m\,\mathrm{d}v/\mathrm{d}t$ to $\dot{m}\Delta v$.

gonenc

11:17 AM

@KyleKanos you should know that i-people tend to do that, eg they say they want an ipod, what they really want is an mp3-player-like device.

i-people do it so that the customer would feel unique. no no no that is not a touchpad that is a trackpad cause it is attached to a mac pc

ACuriousMind

@0celo7 That looks indeed scary

@0celo7 Morrowind natively had tantos and dai-katanas, so they're only true to former lore, I guess

@dmckee To answer the question as I parse it: No, spin is not a point particle, it's a property of quantum objects that often can be imagined as particles.

ACuriousMind

11:50 AM

^the "that" there pertains to "quantum objects", not "property"

skill patrol

12:03 PM

What's the favorite food of a nuclear physicist?

gonenc

@skillpatrol fission chips?

skill patrol

:D

gonenc

@skillpatrol get out → :D

0celo7

12:49 PM

@ACuriousMind I was just walking along, with my posse (follower mod), and this thing popped out of the ground. Then, when I killed it (after dying a few times, SkyRe is a bitch), it resurrected itself.

ACuriousMind

Nice

0celo7

Multiple followers is nice, but they are a bit OP.

0celo7

1:08 PM

@ACuriousMind Found a problem - I can't get Elven smithing. I seem to have to buy the "basic smithing" perk from the other side of the tree first.

0celo7

1:21 PM

Holy crap! I just walked into a bandit camp only to find a group of guard had already dealt with them!!

Must be the AI overhaul, or maybe the guard dialogue overhaul.

0celo7

1:56 PM

@ACuriousMind Lore unfriendly?

ACuriousMind

@0celo7 As I said, they were there in Morrowind. Since some Dunmer now live in Skyrim, I see no lore-conflict.

0celo7

Cool, it's actually by far the best weapon I can make right now.

Windows making random noises again...is there a way to tell what it thinks happened?

ACuriousMind

What kind of...noises? oO

skill patrol

moaning -_-

0celo7

^^

Wait...how do you know?

Oh

Wat

That totally said mounting

ACuriousMind

2:02 PM

What kind of noise is "mounting"?

0celo7

@ACuriousMind Stick in a USB stick and yank it out. That noise.

ACuriousMind

Ah

0celo7

I want to know what it thinks was disconnected.

Mouse and keyboard work fine.

ACuriousMind

That's not the "mounting" sound, strictly speaking, but the detection of any device

0celo7

Regardless, how do I know what it thinks happened?

It's done it a few times.

ACuriousMind

2:05 PM

No idea, if no pop-up appears telling you "XY detected/disconnected", you're out of luck, I think

0celo7

Skyrim keeps crashing.

I've only had one in-game crash.

It refuses to let me get to the main screen.

Again...

ACuriousMind

yesterday, by Kyle Kanos

You keep breaking your games. You should stop doing that.

0celo7

Worked.

gonenc

3:12 PM

why are we allowing people to have the same user name?

ACuriousMind

@gonenc They are uniquely identified by their user ID

gonenc

3:40 PM

@ACuriousMind though this is correct I don't even know my user ID

so if I change my username to ACM and use exactly the same avatar I bet a lot of people think that I'm you :D

that would be a cheap imitation but I could, in theory, do that

0celo7

Eek, not one of these again! runs to the side By the Nine, another one. dies

This is awesome!

ACuriousMind

@gonenc You have too little rep :P

@gonenc It's the number in the url when you visit your profile

gonenc

@ACuriousMind yes but you got the point :D

@ACuriousMind I know it is that number, what I meant was, I don't know it by heart :) I don't think you don't either

ACuriousMind

50583

I've used it often enough in data.SE queries

gonenc

@ACuriousMind :D anyhow I don't know anyone else knows your ID by heart to distinguish you and me if we have the same name :D

0celo7

4:04 PM

@gonenc ACM is the only one here who has heard my voice.

Likewise, I have heard his voice.

A voice lineup could distinguish the two of you.

gonenc

@0celo7 be my guest, but are you gonna do that every time you talk with me (or him) on the chat?

though I think you get the point

0celo7

@gonenc I'd just use Steam chat.

gonenc

@0celo7 and you still don't know if they is a gal or a guy? :P :D :D

@0celo7 btw did you have time to look for carroll's exercise?

The Dark Side

physics.stackexchange.com/review/suggested-edits/89514

^ Somebody tell them this is not the way about it.

I'm so becoming an edit reject machine

gonenc

4:21 PM

@TheDarkSide to whom are you referring?

The Dark Side

@gonenc All. The proposer, and the approvers!

The original intent has to be preserved. If the original intent needs improvement, one should comment below the post, and OP will look into it.

But OP gets to decide in the end, not some reviewers.

HDE 226868

@TheDarkSide Confusing Neil Armstrong with Buzz Aldrin isn't a minor error, especially when the photograph take up a good 90% of the answer.

The Dark Side

@HDE226868 But the proposer has commenting privileges. This sort of editing can (at least in principle) lead to edit-wars.

HDE 226868

@TheDarkSide How so? Why would the answerer want to change it?

0celo7

@gonenc Forgot all about it!

Lemme look.

gonenc

4:30 PM

@0celo7 do you have the book or do you want a ss

0celo7

@gonenc I have it.

The Dark Side

@HDE226868 I said, in principle. Here, of course, it is possible to objectively judge it is Buzz not Neil. But there could be possibilities where it is not so objective. Then, with roll-backs and re-edits (happens when egos get pinched), you get an edit war.

0celo7

4.6?

Or 6.4?

HDE 226868

@TheDarkSide I just re-read your message and saw that.

The Dark Side

I think that has happened in Physics.SE, before.

0celo7

4:31 PM

@gonenc which exercise?

gonenc

@0celo7 give me a sec

The Dark Side

@HDE226868 - It could be that I'm evil, but I'll always reject this type of edit, if editor has more than 50 rep.

That's a personal thing.

gonenc

@0celo7 neither 3.6b

0celo7

I did that one!

I have the solution around here somewhere...

@gonenc $$\frac{\mathrm{d}\phi}{\mathrm{d}t}=\sqrt{\frac{GM}{r^3}}$$

gonenc

@0celo7 is $t$ your parameter or the time as in space-time

0celo7

4:41 PM

@gonenc I chose coordinate time to be the parameter, I think.

Take a look at the geodesic equation for the time component.

gonenc

Its solution is $t=\alpha s + \beta$

s being the parameter

0celo7

I just chose the solution $t=s$.

gonenc

@0celo7 on which basis?

0celo7

@gonenc Freedom of parametrization.

gonenc

cause if I choose the proper time to be the parameter then $\alpha$ is fixed

0celo7

4:43 PM

Every affine parametrization is equivalent.

gonenc

I think...

0celo7

If you don't do that, then you should get $$\frac{\mathrm{d}\phi}{\mathrm{d}s}=\sqrt{\frac{GM}{r^3}}\frac{\mathrm{d}t}{ \mathrm{d} s}$$

gonenc

What I did was, I have the metric and from it, I thought I can deduce what $dt/d\tau$ should be.

was that wrong?

0celo7

Did you get my answer?

@gonenc The parameter in the geodesic equation needn't be proper time.

Any parameter related to the proper time by an affine trafo works.

I just adjusted the parameter I used so $t=s$.

gonenc

@0celo7 I see...

0celo7

4:47 PM

Note that this is only possible because of the $t$ EoM, which relates it linearly to the parameter.

Since linear trafos are transitive, $t$ is then linearly related to the proper time, as well.

gonenc

I had $dt/d\tau$ to be $\sqrt{\frac 1 {1-3\phi}}$ if I am looking at the correct paper

which is then depends on $r$

0celo7

How could that be, if its second derivative wrt. $\tau$ vanishes?

gonenc

@0celo7 sorry that should have been $\Phi$ as in $\Phi= -GM/r$

0celo7

My approach used the following: the radius of the orbit is constant, so the $r$ equation is just $$\Gamma^r{}_{\mu\nu}u^\mu u^\nu=0$$

gonenc

@0celo7 yup I did the same

0celo7

4:53 PM

@gonenc Mine simplified down to the $\mathrm{d}\phi/\mathrm{d}s$ equation given above.

Do you want my Christoffel symbols?

gonenc

I also considered the $\phi$ equation since and it all boiled down to $$\frac{d^2\phi}{ds^2}= \implies \phi = C_1 s + C_2$$

for some constant $C_i$

0celo7

Oh really...

gonenc

@0celo7 I computed them correctly checked with my computer.

0celo7

Well that makes no sense.

gonenc

@0celo7 surprise surprise!

:22723532 nice move to remove that :D

0celo7

4:56 PM

@gonenc I was thinking a polar spiral for some reason :D

What are the $\Gamma^\phi{}_{\mu\nu}$?

All zero?

Must be for the geodesic equation to be flat.

gonenc

@0celo7 you gotta wait cause I've thrown them away :D

0celo7

@gonenc Unless they're all zero, your second order equation for $\phi$ is wrong!

gonenc

@0celo7 not all zero of course but has terms like $\cos \theta$ which vanish for $\theta = \pi/2$

0celo7

$\Gamma^\phi{}_{\mu\nu}|_{\theta=\pi/2}$ is what I meant.

@gonenc Yes.

This is correct.

gonenc

^the geodesic equation for $\phi$ written like $\varphi$

where $r_s$ denotes derivative wrt $s$

0celo7

5:02 PM

Note that since $r$ is constant, $d\phi/ds$ is constant too!

$C_1=\sqrt{GM/r^3}$

$C_2=0$

gonenc

@0celo7 not quite because in the equation for $r$ I have

alot of them vanish but notice the term $t_s ^2=( \partial t / \partial s )^2$

$p$ is btw $\Phi$

0celo7

I don't see the issue.

gonenc

well if I choose $s\neq t$ but $t=\tau$ I have an additional term

which depends on $\Phi$

0celo7

I guarantee the geodesic equation does not change if you use $s$ over $\tau$.

gonenc

@0celo7 what I wanna ask is this

0celo7

5:07 PM

Doesn't Carroll cover affine trafos?

gonenc

I used $\tau$

@0celo7 it covers affine parameters, I don't have a clue what a trafo is

0celo7

@gonenc Transformation

gonenc

I used $\tau$ and to get $dt/d\tau$ I used the metric. Is that allowed?

ie I divided both sides by $d\tau^2$ to get $(dt/d\tau)^2$

0celo7

No, to get $dt/d\tau$ you use the geodesic equation for $t$.

gonenc

@0celo7 it just gives me $t = c_1 \tau + c_2 $ for some constant

0celo7

5:10 PM

@gonenc Exactly.

Pick the constants accordingly.

gonenc

sh*t and since I'm free to choose whatever I want I can choose $c_1=1$

Genau

^this :D

Alles klar?

kristalklar

@0celo7 do you have the answer for c?

DanielSank

5:16 PM

Unusually high density of crappy answers today.

0celo7

@gonenc Yes.

5061s?

gonenc

@0celo7 sh*t I didn't compute the number

just a sec

or a minute depends on my ability to find my calculator

0celo7

If you don't get this, you're outta luck. I gotta play the Skyrims.

gonenc

@0celo7 perfectly fine by me thanks btw :)

0celo7

Did you get that?

gonenc

5:18 PM

@0celo7 I need to calculate the numerical value

0celo7

You have to use the frequency calculated in part (b).

That is the classical value, I believe.

gonenc

@0celo7 yup it is

0celo7

Don't forget the $c$s.

6.371x10^6m

gonenc

@0celo7 goddamn $c$ s!

0celo7

My solution is $\Delta\tau=2\pi\sqrt{\frac{R_1^3}{GM}-\frac{2R_1^2}{c^2}}$

I wonder if h Bar works in Steam overlay...

gonenc

5:21 PM

@0celo7 that is

0celo7

That is?

gonenc

steam overlay?

a question mark missing sorry :)

0celo7

You can supposedly use the Internet when fullscreening games using it

Games like Skyrim die when you alt tab out of them

Hello from Skyrim!

Sadly I can't post screenshots from within Skyrim Chrome because Steam can't access the fraps folder

gonenc

5:37 PM

@0celo7 I have $$d\tau= \sqrt{1-\frac{GM}{c^2r}} dt$$ integrating it gives me a solution as simple as $\tau = \sqrt{1-\frac{GM}{c^2r}} t $

no $2\pi$ s or whatnot. I am sensing I'm doing something terribly wrong

gonenc

5:50 PM

@0celo7 and 5061s seems a bit too much since the satellite doesn't move with that great of a velocity and the earth isn't such a big gravitational source

0celo7

7:19 PM

Ugh, Skyrim just suicided and brought the whole PC with it.

Kyle Kanos

-1

A: If conservation of energy was wrong, how would we know about it?

i blv that the law of conservation of energy z not true ...,,, i knw more than 1 example where we get 2 outputs ,, under same experimental conditions , and i knw a way of production of infinite electricity in space ,, i blvs we cn create aswel as destroy energy with our activities , it is possibl...

0celo7

@KyleKanos It is wrong...

Kyle Kanos

Only 'cuz hez txting n nto wrtg it out

0celo7

@gonenc What is your answer for a)?

I used it for c).

@KyleKanos I can't read that, and I'm 17.

Kyle Kanos

You've spend the last two weeks deciphering people's signatures and you're telling me you can't read texting?

0celo7

7:22 PM

It was like 3 days.

@KyleKanos Only because he's texting and ... it out?

not writing?

I don't text like that.

Kyle Kanos

Pretty much

Yes, but you're the odd one out of kids your age

0celo7

I don't know anyone who texts like that.

You watch too many TV shows about teenagers.

gonenc

@0celo7 $\tau= \sqrt{1+2\Phi} t$

0celo7

@gonenc There's more to it.

You need the angular part of the metric.

gonenc

I assumed the guy in the tower doesn't move with the earth so that part is zero

@0celo7 didn't you do so?

0celo7

7:25 PM

$\Delta\tau=\sqrt{1+2\Phi-R_1^2\sin^2\theta\omega^2}\Delta t$

I took the worldline to be $(t,R_1,\theta,\omega t)$.

$\omega =2\pi/\mathrm{day}$.

Kyle Kanos

@0celo7 I don't watch any TV really

gonenc

@0celo7 I see that's where the $2\pi$ comes

I've calculated as if the guy were hanging still.

0celo7

@gonenc $\phi$ is the longitudinal angle.

$\dot\phi=\omega$ is the angular velocity.

gonenc

@0celo7 I meant the $2\pi$ in your answer above.

0celo7

I forgot what something is called, one moment.

We need $2\pi$ because radians.

The angular velocity of the guy is just 1 rotation / 1 day, which in radians is $2\pi$.

gonenc

7:30 PM

@0celo7 What I meant was

0celo7

$2\pi$ rad / 1 day

gonenc

you said:

2 hours ago, by 0celo7

My solution is $\Delta\tau=2\pi\sqrt{\frac{R_1^3}{GM}-\frac{2R_1^2}{c^2}}$

I said

2 hours ago, by gonenc

@0celo7 I have $$d\tau= \sqrt{1-\frac{GM}{c^2r}} dt$$ integrating it gives me a solution as simple as $\tau = \sqrt{1-\frac{GM}{c^2r}} t $

0celo7

Oh, that's from the conversion from angular to other quantities.

You need the period there.

gonenc

and I was wondering where the $2\pi$ come from.

0celo7

The period is just $2\pi/\omega$.

What do you think $\Delta t$ is?

@gonenc Let's go slower. Do you see how I got my expression for $\Delta\tau$ above?

@KyleKanos Oh really...

gonenc

7:34 PM

$\Delta\tau=\sqrt{1+2\Phi-R_1^2\sin^2\theta\omega^2}\Delta t$ this one I do cause it is like obvious :)

you also considered thought that the guy was rotating with the earth imo?

0celo7

@gonenc Of course he's rotating with the Earth...

Kyle Kanos

@0celo7 Yep. Analog TV w/o cable = No TV watching

I occasionally watch movies

gonenc

@0celo7 well I assumed he wasn't :D

what did you calculate for $\tau$ for the guy in orbit

0celo7

@gonenc You thought he was walking perfectly against the Earth's rotation?

We should give this problem to @KyleKanos.

@KyleKanos You up for a basic GR problem?

Kyle Kanos

@0celo7 What's the query?

gonenc

7:37 PM

@0celo7 unrealistic but hey it made my life easier :P :D

0celo7

@gonenc Same, but with $R_1\rightarrow R_2$.

Wait...orbit?

Mine says "tall building".

@KyleKanos I'll get it to you.

gonenc

are we on the part c?

@0celo7 btw should we not clutter here and go to another room?

0celo7

It's fine.

No one else is here.

ACuriousMind

::whistles::

gonenc

@ACuriousMind what for was that :D

0celo7

7:40 PM

@ACuriousMind You're playing PoE.

ACuriousMind

@0celo7 Still able to make silly comments in here while doing that ;)

gonenc

2 hours ago, by gonenc

@0celo7 do you have the answer for c?

user54412

@ACuriousMind You too?

0celo7

I already put my answer for c.

Is PoE that good?

gonenc

@0celo7 well I was discussing c :D

0celo7

7:43 PM

For c, do the same as above, but this time $\omega$ is $\dot\phi$ calculated in b, which is just the classical value.

user54412

@0celo7 More grinding-oriented than my usual preferred games, but the leveling mechanism alone makes it pretty awesome.

ACuriousMind

@ChrisWhite Me too what? Able to make silly comments, or playing? :D

user54412

yes

gonenc

@0celo7 ok we are on the same page now :D

fine with everything :D

0celo7

@gonenc So you know where the $2\pi$ comes from?

ACuriousMind

7:44 PM

@ChrisWhite Alright, then, me too.

0celo7

@ACuriousMind But unable to take part in @gonenc's GR education I see.

user54412

@0celo7 He doesn't know GR, remember? :P

ACuriousMind

^that

0celo7

@ChrisWhite take part != teach

He could audit my class.

gonenc

@0celo7 lemme get a pencil and paper

0celo7

7:46 PM

@KyleKanos Do you know how to do that problem?

@gonenc It's from the conversion from angular frequency to period...

gonenc

@0celo7 No I meant for the whole calculation.

I know a bit of classical physics you know... :D

0celo7

You can't do that in your head? :/

ACuriousMind

@0celo7 I like it far better than I liked Diablo3

0celo7

It's a diagonal metric...

Kyle Kanos

@0celo7 Hmm. Maybe if I spent some time on it I could figure it out

ACuriousMind

7:47 PM

And from time to time grindy hack-and-slash is necessary

user54412

So I'm only half paying attention, but I'm confused why there's so much discussion of $2\pi$ -- all $2\pi$'s should just come out of the $\phi$ integrals naturally?

Kyle Kanos

Anyone notice the rather harsh user Recognize Evil as Waste?

0celo7

@ChrisWhite $2\pi$ is from $T=2\pi/\omega$.

ACuriousMind

@KyleKanos Yes, already voted to delete one of his...contributions

gonenc

@0celo7 I should've read the question first!!! God damn it

0celo7

7:52 PM

@gonenc ??

gonenc

yada yada yada completes one orbit

didn't read that part

user54412

"completes one orbit" means $\phi$ wraps around from 0 to $2\pi$ -- that $T = 2\pi/\omega$ is something that should just fall out of integrating $\mathrm{d\tau}$ along the geodesic, rather than something you should have to know already.

user54412

At least I think. Again, I'm not 100% paying attention.

0celo7

@ChrisWhite True.

gonenc

@ChrisWhite I know I just integrate from 0 to $t$ not to $2\pi/\omega$

0celo7

7:56 PM

Back to Skyrim ~~

gonenc

I'm fine just a missing a factor of 2, which I thought I've lost anyway :D

@0celo7 thanks have a nice game :D

when is the next chat event?

Kyle Kanos

It's every other Tuesday

HDE 226868

@gonenc Should be Tuesday.

Kyle Kanos

Not sure if that means this coming T or the past one though

gonenc

@HDE226868 ie 14th?

HDE 226868

8:01 PM

@KyleKanos It's the next one. I checked.

@gonenc Yes. . . . (?)

gonenc

@HDE226868 it could have been 21st

HDE 226868

I just know that I saw "in 2 days", and remembered that there are events on some Tuesdays, which I inevitably miss.

gonenc

@HDE226868 how do you check that

found it

chat.stackexchange.com/rooms/info/71/the-h-bar?tab=schedule

Kyle Kanos

chat.stackexchange.com/?tab=events&sort=event

HDE 226868

I actually just went to "all rooms" and checked the blinking notice about the next chat event, but actually checking the schedule works, too.

Hippalectryon

8:12 PM

Hello !

One of my friends has the following question at an oral exam : "A cylinder explodes when the pressure inside of it is P0, at what pressure would it explode if it was a sphere" ? I don't have a single clue how to solve that, does anyone have an idea ?

Kyle Kanos

Probably through relating the stresses caused by the pressures

Hippalectryon

I mean, this was the whole exercise, no additional info was given. I don't see how to solve it with just that.

Kyle Kanos

Well you might have to make assumptions, e.g. the radius of the cylinder is the same as the sphere

Hippalectryon

Even with that I don't really see how to do it

alarge

8:34 PM

@Hippalectryon Surface area to volume ratios?

Hippalectryon

@alarge But I doubt it's this simple. Shouldn't we analyse what happens in the cylinder's angles ? Isn't the pressure maximal there ?

alarge

@Hippalectryon Well if the course was on continuum mechanics I suppose that yes you might have to compute the stresses somehow. I've forgotten almost everything about CM so can't really help you there.

Hippalectryon

@alarge We didn't have any course on the mechanic of solids or on stresses specifically

alarge

@Hippalectryon So what course was this then, thermo?

Hippalectryon

It was competitive exams to get in some schools, it's over two years of studies

Not a specific course

alarge

8:41 PM

Ok. Well, I'd basically go the area/volume route, so I'd try to estimate the total stress by what is essentially surface tension. I think with the cylinder though you'd have two degrees of freedom unless you make more assumptions.

Hippalectryon

Mk

skill patrol

@Hippalectryon what does "P0" stand for?

Hippalectryon

@skillpatrol some pression

skill patrol

This may make a good question for the main site.

ACuriousMind

@skillpatrol It's homeworky, and probably not enough information for a unique answer

skill patrol

8:53 PM

True that.

user54412

I actually think there is a unique answer the original person asking the question had in mind.

user54412

Like, there should be a pure geometrical conversion from cylinder to sphere (of course ignoring the weak points at the end of the cylinder -- that's uninteresting)

user54412

Think of placing a bunch of particles on a curved surface, linking them together with springs

user54412

Consider a change $\delta A$ in the surface area. How does this require the springs lengths to change?

user54412

For a fixed material, the stress/strain properties of each spring are the same.

user54412

8:55 PM

In a sphere, all lengths grow together, since there is no preferred direction among those restricted to be tangent to the surface

alarge

@ChrisWhite The problem is that this is not uniquely determined: The cylinder can stretch upwards or sideways or a combination thereof. Otherwise writing down the Young-Laplace eqn should be easy.

skill patrol

It is an interesting question imo @ACuriousMind

Homeworky or not :-)

user54412

But in a cylinder there is an anisotropy -- bonds along a circumference would get the full burden of expanding, whereas those aligned along the cylinder's axis would not be stretched at all.

user54412

@alarge An infinite cylinder is how I understood the problem. All stretching is in radius, not longitudinal. Then the answer should be unique, right?

ACuriousMind

@skillpatrol I made no judgement about how interesting the question is - I said why it would not make a good question for the main site

alarge

8:57 PM

@ChrisWhite Well an infinite cylinder doesn't sound that interesting, but yes, it is trivial then.

I'll take that back: It's not trivial as the area would increase an infinite amount from a small dR.

skill patrol

@ACuriousMind is having a unique answer a requirement?

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