No account by Jasper anymore.

(deleted all again)

@JasperLoy I wanna listen to some song by you! Create, produce some new song and bring it here (in the room).

@DanielFischer have you ever considered to sing some opera and put yourself on youtube as Jasper?

Too much talk here and I need to finish some stuff.

OK. BBL

@Semiclassical Are you around?

yeah

Great. Something in my book is driving me nuts.

physics?

So, $Y = 3K(1 - 2\sigma)$. The constants involved are the Young modulus, Bulk modulus and the Poisson ratio respectively. I don't know why this is true; nobody could tell me why (and I for some reason didn't think of asking you). But let us assume for a moment this is true.

@Semiclassical Yep, what else? :P

lol.

I don't know off the top of my head either, I don't work with the various moduli much. But presumably it follows from definitions.

I tried to prove it but I couldn't. But that's the least of the woes and worries; shall I proceed with my question?

one moment

The Wiki page for the Poisson ratio includes an approximate result labelled as Lame's relation, which may be equivalent.

$dV/V = (1 - 2\sigma)d\ell/\ell$ : that's exactly what I am having trouble with.

I can believe the proof. But invert both sides and multiply by $F/A$.

You get $(1 - 2\sigma)K = Y$.

Where did $3$ go?

If I were to guess, the issue is that the last step isn't actually valid i.e. you're conflating some quantities implicitly. hmm

So I thought. But I don't see how.

Admittedly, I don't know these definitions that well. Elasticity of solids is something an engineer works with in much more detail than a physicist.

The $3$ should come out of derivation some $\ell^3$ term of so. But how, I don't see.

lame's relation is lame

The integral is too crude to detect finite-time blowup.

@Semiclassical $Y = \frac{F/A}{d\ell/\ell}$, $K = \frac{F/A}{dV/V}$ and $\sigma = -\frac{dr/r}{d\ell/\ell}$ if it helps.

Oh, we have an engineer. @0celo7.

I'm sick and stuck on a Riemannian geometry problem

what do you need

well, that's what you think the definitions should be :)

I haven't taken a materials course yet

@Semiclassical So does my textbook. Admittedly my textbook imagines a lot of false things.

hmm

@0celo7 Scroll up; I have a question.

There is no way this problem can be this hard

I must be missing something

@BalarkaSen I don't know anything about derivations in solid state physics

I tend to stay away from physics "proofs"

I am writing out $dV/V$, yes. I have a wire of length $\ell$ and radius $r$, so $V$ here is $\pi r^2 \ell$ (unlike wiki where they do it with a cube). But w/e

@0celo7 I would do too, if I could.

But I can't, so I don't.

why can't you

Wouldn't want to mess up my grades this time.

why do your grades depend on physics derivations

Because it's a course involving those computations, I presume?

the average PhD physicist probably doesn't even know what this is

Because I seem to be getting two different formulas from two different derivations.

long forgotten at least

And that would give me two different answers for a given question.

One the thrice of the other.

yes, there are some tricky threes in this stuff.

@0celo7 I wouldn't even call this physics. It's something far more likely to be seen by an engineer.

So I'd have to write "either this is the answer, or it is thrice this (or one-third of this, dunno)"

@Semiclassical but certainly never a proof or anything like that.

Which would give me a fine shiny 0.

Depends on the course, I suppose. One might be expected to be able to explain why the formulas are true.

But, again, never having taken a course in continuum mechanics...

balarka is like 15, why is he taking a course in continuum mechanics

I was actually supposed to take a course in that this semester

but I took quantum instead

Might have been a mistake

i'm a bit surprised to see this in a course, but i'm operating under certain assumptions

namely, my own experience in the american college/university system

we have a course on mathematical continuum mechanics, whatever that means

it probably covers this kind of thing.

That's an oxymoron.

here comes @BalarkaSen the math expert...

eh, not really. a mathematical formulation of this just comes down to expressing things properly in terms of tensors etc.

BTW, nice fact. Eagle flies in escape velocity.

Cool, eh?

how fast is that

11.2 km/s IIRC

eagles do not fly 11.2 km/s

take your bs outta here

c.f. the first section of en.wikipedia.org/wiki/Linear_elasticity

Tell that to the new guy who's teaching us Newtonian gravity

@BalarkaSen Does 11.2 km/s pass your common sense test

for an eagle

Yeah, that's...not terribly reasonable.

Ofc not. I made a joke.

...that's not a joke

It was if you were there when he said it.

I wonder if one can define Riemannian distance with smooth curves.

"You had to be there" isn't a very accessible basis for humour :)

Certainly one can homotope a piecewise smooth curve to a smooth curve.

Can one make it arbitrarily small?

@0celo7 Sure

@BalarkaSen You might look at that last wiki page i linked. it probably gives a sounder basis for understanding what's going on

Just approximate it arbitrarily close by smoothenings of the curve by bump functions near the non-smooth points.

of course, it also uses index notation and summation convention for most of what's there. so you might find that distateful :p

If you can make it fit inside a tubular nbhd, you can product a homotopy of arbitrarily small size.

Which you can always do by definition of "approximate arbitrarily close".

seems reasonable.

@Semiclassical yes, at a glance, that page is not quite accessible to me. I'll try to get anything out of it later.

sure. here's how I'd state the issue, btw.

@Semiclassical He also taught us electromagneticity in a Newtonian gravity class.

What you're interested in is the ratio of Young's modulus to the bulk modulus

I think I need to develop a way of comparing topologies induced by various Riem. metrics.

And he claims that $K.E. + P.E.$ is said to be "conserved" only when it's $0$ everywhere, but not when it's $1$ everywhere.

i.e. $$\frac{Y}{K}=\frac{\text{volumetric stress}/\text{volumetric strain}}{\text{tensile stress}/\text{tensile strain}}$$

I changed the way I draw integrals.

They look much better now.

now, if you can equate volumetric and tensile stress (as you did when you wrote your equations with that factor of $F/A$ in both) then that becomes $$\frac{Y}{K} = \frac{\text{tensile strain}}{\text{volumetric strain}}$$

Right.

but we concluded earlier that that should equal $1-2\sigma$

hence, a problem of 1/3.

Yes, so the issue should be in equating the volumetric stress w/ the tensile one.

But I don't see how so.

do you get a nobel for proving everyone wrong or do you have to be right about something

For that, I'd need to review the definitions of volumetric v. tensile stress

Admittedly we weren't told, even by the good teachers we have, of any particular difference between them that should explain why I have this $1/3$ issue.

the definitions are given here: eduresourcecollection.com/civil_sm_Stresses.php

tensile is at the top, volumetric at the bottom

it could be that the point is that, in the volumetric case, you're providing 3 tensile stresses at once

Right, but I can't come up with any rigorous formalization that'd tell me the quotient is $1/3$.

and therefore volumetric stress is three times the tensile stress in that case.

hmm

@Semiclassical I thought that's how it worked.

@BalarkaSen Physics is not rigorous.

That doesn't sound right. On a cross section, I apply tensile stress along only one axis.

@0celo7 ...okay, i'm really getting tired of that attitude. not being fully rigorous does not mean being inconsistent.

and while i'm not versed enough in linear elasticity to actually give the arguments, I presume that when one actually uses them (i.e. elasticity tensors and the like) none of this is particularly strange or inconsistent.

The other two tensile components of the volumetric stress are orthogonal to the cross section, so doesn't really do anything to it, does it?

@BalarkaSen Depends on the definition, perhaps. How does one define volumetric stress when the forces aren't distributed symmetrically?

@Semiclassical I didn't say it was inconsistent

I'm not really paying attention anyway.

@Semiclassical No idea actually, there isn't a fully rigorous definition in by book.

If it's the sum of the stresses along the three orthogonal axes, then if the stresses are equal then "volumetric stress = 3 tensile stress" is pretty sensible

Yeah.

I think that's really the issue: You're trying to get the book to make sense when it's not really trying :P

I always assumed it's the amount of perpendicular force which is applied on the given cross section.

That may depend on how you choose the cross-section, though.

I agree, admittedly I have always tried to hand-wave past that issue: most objects (wires, cylinders) have a canonical cross section associated to it :P

right.

Good point, though. That does clear up a lot of mess

Well, it does if I'm right :)

Let us be more of a physicist than a mathematician and assume you're right for now. When another issue comes, I'll ping you and ask you to come up with a better definition :D

lol

I think this part of the page on Poisson's ratio might contain the same reasoning though not transparently: en.wikipedia.org/wiki/Poisson%27s_ratio#Isotropic_materials

what i'm saying about volumetric stress amounts to equating it to $\sigma_{xx}+\sigma_{yy}+\sigma_{zz}$ i.e. the trace of the stress tensor

Mhm.

in fact, Lame's relation just corresponds to taking the trace of the following equation (stated there in index form): $\mathbf{\epsilon}=\frac{1}{E}\left[(1+\nu)\mathbf{\sigma}-(\text{tr }\mathbf{\sigma})1\right]$

...maybe. maybe i'm just being loopy :/

should've been a factor of $\nu$ in front of the trace in that last expression, woops

hmm, $K=\frac{\text{tr }\sigma}{\text{tr }\epsilon}$ is a nice statement @BalarkaSen

@SemiC No, wait a second. $dV/V = d\ell/\ell ( 1- 2\sigma)$. $F$ be volumetric stress. $\frac{F/A}{dV/V} = \frac{F/A}{d\ell/\ell} (1 - 2\sigma)^{-1}$. You're claiming $F$ = 3 x tensile stress (call it $F_\ell$)? That doesn't seem to give me the right thing. That says $1/3K =Y(1 - 2\sigma)^{-1}$.

Should have been $3K$ on the left.

hrm

uck

Well, shit.

lol

So $F_\ell = 3F$? That does make sense in a way: if you apply the same amount of force by the x-axis, say, and symmetrically through all the three axes, the internal reaction in the first would be thrice the internal reaction of the latter.

Stress, after all, is internal reaction, and not the amount of force you apply to it.

hmm

what's also annouying to me is it looks like i'd need to have $3K=\frac{\text{tr }\sigma}{\text{tr }\epsilon}$ which isn't nearly so cute :/

Oh well

The real answer being, "look at an engineering text"

Eh, I am more or less convinced with $F_\ell = 3F$. You're applying the same amount of force, but equally distributing it through all the three axes.

The internal reaction perpendicular to a given cross-section is then 1/3rd (aka $F/A = 1/3 F_\ell/A$).

This definition of volumetric stress seems to be a hotch-potch of your definition and my earlier definition. :P

Both the three tensile components and the cross-section survive.

K, I gotta go

@BalarkaSen seems plausible. Anyways talk to you later

@ForeverMozart got Spivak 2

looks like crap, but can't beat $20

@Semiclassical what are some physicists who are only famous for their books?

Jackson?

Unless you specialize in plasma you've only heard of him because of his book

but every physicist has heard of him

@Semiclassical would you be willing to listen to some riemannian geometry ramblings of mine?

@user1618033 i dont create math problems i just solve them

whilst, i create puzzles (logic is my focus)

meantime, i am getting accross an idle periode, my stoopitch mind getting rigid and i am more connected to a sensless uncreative software with that sour flavour of databases and computational-statistics

@balarka I just realized that when I said ratio of Young's modulus to bulk modulus earlier, I said it backwards i.e. I wrote Y/K on the LHS but K/Y on the RHS

hence why i had the factor of 3 backwards.

@Agawa001 I see.

Hey

Can anyone tell me what the partial derivative symbol $∂$ is called?

I've heard it being referred to as the del operator, @SirCumference.

What are $\Gamma_g$ and $\rho$, @ColdGolf?

@Kari One person told me it was "dee", another told me it was "doh"