@bolbteppa Yeah, but as $2e_{ij}dx^idx^j = ds'^2 - ds^2 = g_{ij}dx'^idx'^j-g_{ij}dx^idx^j = (g'_{ij}-g_{ij})dx^idx^j$, so we have $g'_{ij} = 2e_{ij} + g_{ij}$. This is to say that instead of looking at a change in coordinates, we look at a change in basis. If you want more details, you can look at the reference I cited (or I can give you more, as well; e.g. this as you were more interested in quantum stuff anyway)
@alarge Landau's derivation 3.1 assumes the stress tensor is a divergence, he derives it by taking an integral F = < - dE/dr > and applying Newton's third law. So 2. in your post is derived from the first sentence of 1. in your post. Non-uniqueness of the stress tensor is manifest in this way, but it may explain why elasticity is built on displacements du_i = x_i' - x_i and not coordinates!!!
Basically, I would say if you take definition 2. you are basically assuming a non-unique stress and are defining something that depends on changes of coordinates but focusing on g_mn which will give non-unique answers if you get me
I think that's a compelling reason to phrase strain in terms of displacements u_i = x_i' - x_i anyway, thus you'd naturally want to talk about changes in energy with respect to changes in the changes of displacement so that everything is gauge invariant, right?
Landau says the whole point is to express free energy in terms of the strain tensor at the start of section 4, so if F is a function of a gauge dependent term you almost definitely want to measure it in units that do not reflect this gauge dependency right? Thus you want dF/du not dF/dg right?
It looks to me like your equation 1 is related to Landau's equation 2.9, i.e. that your establishing of some form of uniqueness is just defining an average stress tensor as Landau does in 2.9
Landau assumes F = - <dE/dr> and integrates to derive a gauge-dependent stress tensor from equilibrium statistical mechanics (reducing them to Newton's laws), from this fundamental physics he shows it has to be non-unique. That's kind of hard to dispute conceptually, the gauge dependency seems rooted in fundamental physics and mathematics, though I'd like it to be unique (I thought it was!).
What Landau derives, 3.1, seems to be on the assumption of a small displacement du so it may only be unique in that case, the integrals not mattering at the endpoints apparently depends on this du assumption, just as the Einstein-Hilbert derivation holds assuming the endpoints do not matter, but the concept of stress is more general right? It's got to hold when the boundaries are fucked up, thus this gauge invariant definition seems more fundamental
Another reason I don't think it's a good idea to define free energy as a function of the geometry is that the geometry might stay the same but points will just move around, a hydrostatic compression, would you not have to start parametrizing points as functions of time to do that, but this is a theory of equilibrium so we want to ignore time right? Functions of displacement, du, ignore this and allow us to talk about distance between internal points changing relative to one another
So the definition of stress at the microscale is the thermodynamic force (change in average energy with respect to position) over a surface shifting between equilibrium states, thus gauge dependency may reflect the statistical mechanical nature of this concept?
Anything to do with free energy is quantum mechanical, the force in these discussions is the statistical mechanical force which is the change in average thermodynamic energy with respect to position, you simply cannot conflate thermodynamics and newtonian mechanics
It's entirely possible gauge invariance reflects the statistical nature of the force, that'd be a pretty cool explanation too
Also I think the boundary conditions settle the matter on strain being defined as a derivative with respect to a metric or a displacement, it's a small-displacement approximation just as the Einstein-Hilbert action is derived assuming boundary conditions don't matter :)
By posting an explanation that 2 is flawed can I get the bounty? :D You might want to take 1 as a definition if you want to pretend you are doing classical mechanics, but logically it doesn't make sense,
I told you exactly what it was, in Landau section 3 they derive your equation under the assumption of small displacements du not affecting the boundary conditions, check the section yourself, "By considering an infinite medium which is not deformed at infinity, ..." this is why they can define the stress tensor as a derivative with respect to du, This is exactly the same story as in the Einstein-Hilbert action, I've said it twice now :(
In the Einstein-Hilbert action they have to assume dg = 0 at the boundary right? Well the concept of stress can affect the boundary, of course it can deform boundaries, it's supposed to be flexible enough to handle those cases, why do you disagree with that?
I would love to give an Einstein-Hilbert definition of that haha but I don't see it as justifiable :( I've always ignored boundary conditions in those action integrals apart from in basic string theory, this is the first time I guess it's mattered to me haha but honestly I just think 2 is a special case
I swear up until tonight I had the complete wrong idea about elasticity
No that's not true, it's a combination of the boundary condition AND the du = 0 at the boundary that lets you derive this, look at the integration by parts step in the derivation of 3.1
You can't define the strain as a derivative of the free energy except in a special case, Landau has to invoke this du = 0 assumption at the boundary in order to define strain as a derivative of work/energy, right? It doesn't matter if you derive w.r.t. u or g. I actually don't know how you can write any of the equations in 2, but Landau shows why they are special case definitions anyway.
Yes, and of course the boundary integral is set equally zero in the def'n 2.
dF / de is Landau's result, anyway. And as e can be written as a function of g (or vica versa), surely we can just be looking at dF/dg
In any case, it's not like you get wrong results with dF/dg as far as I've tried using it; All the classic cases you get the correct equations. You also get, for microscopic stress, the only definition that is actually used from the family of eqs from defn 1 (you get that one of the integration contours is unique. This is the contour that hasn't been shown to give unphysical results)
Okay you're right about the boundary conditions, thanks for spotting that. Mistura's paper looks interesting, but I still don't agree with anything in 2. I don't know where you got them
I see Mistura says the classical definition of stress is as dF/du, refers to an old paper of his to justify re-writing it in terms of derivatives of the metric (this is a non-trivial result to me and apparently him anyway), and in this paper he derives this result on the assumption that dF is an integral of the strain tensor. Where did this assumption come from? Who said you could write the free energy as the integral of a strain tensor?
If we can clear this point up I think we will have a final answer :)
Landau seems to claim that to write this integral with a quantum statistical mechanical justification you must already assume the strain tensor exists as a gauge dependent quantity because of Stokes theorem. Mistura's earlier paper says he's modelling his idea on the EM tensor in classical field theory which is nice and what I thought you were supposed to do, but I don't see where this integral comes from.
Interestingly, Mistura refers to Landau's book to justify why "we start from the familiar definition of the stress tensor in the thermodynamic theory of elastic solids", but this definition is derived on the assumption of gauge dependency, I think it's circular
I am teaching myself tensor calculus.
In some of my calculations the term $\nabla_{[i}x^i \nabla_{j]}x^j$ keeps turning up. In 2 dimensions this is up to a constant the determinant of the $\nabla_{i}x^j$ matrix. When it is zero this means that the direction $y^i\nabla_{i}x^j$ is independent of ...
@bolbteppa Taken the definition f = div sigma, You end up with a gauge dependency on sigma. You can then derive the result that dF = sigma de. I think this latter result, i.e. that sigma = dF/de no longer has gauge dependency due to the way it's derived. In fact, dF/dg doesn't, but if you restrict the metric to be Cartesian and do the variation then, the gauge dependency reappears.
So I think it's a question of what to choose as your definition, and I think the one with no gauge dependency is cleaner.
@Paul I flagged the question about elastic collision suggesting to be closed, due to the too low level. I guess that other people did the same as I see that the question was flagged by other people too.
@alarge I think that writing free energy as the integral in your post, and the definition Mistura starts from, is actually a consequence of Landau's argument, thus Landau is more fundamental, there is no justification for starting from that integral. However, it might be justifiable to define it as the derivative of free energy w.r.t. g_mn as in your post, & in equation 18 Mistura 1987, that seems like the only hope at justifying this, but I think this also runs into trouble.
If we have to start from that derivative of free energy definition, it implies that writing your integral is a consequence of the derivative definition, just as it's a consequence from Landau's starting point (you derived it using math, Landau using physics). It's a nice idea because it syncs with the idea that thermodynamic quantities are just derivatives of free energy.
The problem is that it seems equivalent to the statistical mechanical definition of pressure as F = -<dE/dR> = - <dE/dV>dV/dr = P dS if we think of - <dE/dV> = - <dE/dg_mn> in some sense. If this is so then you are actually trying to use an analysis of F locally, at a single particle, to justify saying F is not a divergence, but Landau is integrating over the total surface to ensure we take all particles. That seems to be the source of the difference between Mistura and Landau.
@bolbteppa Your latex doesn't render with math.ucla.edu/~robjohn/math/mathjax.html, because you omitted the "$". I guess you just don't know about the MathJax bookmark everybody else here uses for latex rendering in chat.
Landau is also talking about local free energies, there's nothing problematic with Mistura's definition in this regard. What I did worry about a bit, as I mentioned in the question, is whether long range forces here would give you problems as these quantities would then not be additive. The thing is, though, that Mistura's formula seems to give all the right results and looks very similar to standard definitions in GR.
As for surface effects, you always neglect them in bulk thermodynamics, so that's not really an issue
I don't know what you mean, you told me that you know of a similar integral for pressure that mimics this integral for free energy $\delta \mathcal{F} = \int \sqrt{g}\sigma^{mn}\delta\varepsilon_{mn}\mathrm{d}^3x$, what do you mean?
Landau 3.1 derives the integral form by assuming the stress tensor comes from a divergence and is gauge dependent, can you show me how to get this equation without that assumption from quantum equilibrium statistical mechanics
@KyleKanos I've not read Griffiths, but everytime people come here and ask questions about it I get the impression it is not very good at explaining concepts
I have Gasiorowicz, and from what I remember it's quite short and to the point; Shut up and calculate. I've only glanced at Griffiths', but I think it tries more to build some intuition about the concepts.
@alarge you are waving your hands when you say we can just write some magical integral for the free energy because it mimics some random integral for pressure. What is your justification for this? The only thing you've offered so far is that Landau derived it in equation 3.1. But he derived it on the assumption that strain is gauge dependent. How do you justify writing this integral down without that assumption? Can you please lead me from thermodynamics to this integral?
I have read Landau, both those papers and the Schofield paper trying to help you and spent a day at this, you have never explained it to me once, I just told you the flaws in Mistura's approach which you haven't addressed yet, can you please just carefully, without any hand-waiving, explain it to me, I'm confident you will not be able to do it without some flawed assumption
I'm quite sure Landau did not derive the formula with the assumption that it was gauge dependent. It's just that he starts with f = div stress, then "derives" another formula for stress which happens to be gauge invariant, I think. You could just as well start from the other direction, I don't think there's anything wrong with it.
I can just say that the integral defines a thermodynamic quantity called the stress. And then taking the divergence, I note that it indeed satisfies the requirement that its divergence is force.
Yes Landau did, in section 2 he defined the stress tensor using force F as a gauge dependent quantity and spent a whole section using it deriving different things, only in section 3 did he derive your result using work F.dr
So you are just whipping out random thermodynamic quantities with absolutely no justification, are fine with labelling them as fundamental quantities, but don't like gauge invariance? That's infinitely worse than a gauge dependent function tbh
In section 2 Landau defined stress in terms of force F, you want to define it in terms of force F and work dr, in order to do one single step in the 3.1 computation you must assume the force is a divergence of some rank 2 tensor, you are immediately forced to admit that the force is gauge-dependent
div dF/de is force, yet there is no gauge dependence
I'm quite sure there is absolutely nothing wrong with the definition of stress I have. The issues, if any, probably arise due to long range forces, but it's not obvious how this happens (considering, especially, that the form does seem to give meaningful results regardless).
I know that there are nonabelian gauge theories and their supersymmetric extensions. Mathematically, gauge theories basing on the fact that one can introduce a fiber bundle with a Connection. From this the curvature (field strength) can be computed.
But what is when one has a non-smooth fiber bu...
@alarge you did not answer my point about the computation in 3.1, the second you write anything down in any step of that computation you are assuming gauge dependence, could you please address this
You want to skip all the logic and write down the end result out of the blue
I don't see where you assume that sigma should be gauge-dependent as you derive 3.1. Symmetric though, yes.
Suppose sigma were unique; Which step of that derivation would fail?
Taking sigma = dF/de, like Landau, does, I think, give you a unique stress tensor. I'm quite sure Landau in the later pages does indeed not care to add the extra terms for any to any of the stress tensors he derives
@alarge the second you write $F = \frac{\partial \sigma_{ik}}{\partial x_k}$ you are allowing $F$ to be gauge dependent because when we look at $F$ on it's own, as is done in section 2, applying Stokes theorem leads to gauge dependency. Just because you decided to use work $W = F.dr$ it does not somehow change this fact... There are fundamental statistical mechanical reasons for this also, I've already posted them but will again and illustrate how Mistura's derivation does not bypass them.
Landau did not "take" dF/de he derived it as a consequence of his more fundamental definition of stress, he was just expressing the differential work in the variables of area in 3.1. I'll ask again, look at the derivation of 3.1 and explain to me exactly why I can write $F = \frac{\partial \sigma_{ik}}{\partial x_k}$ without assuming gauge dependency...
@bolbteppa It's not required in the derivation of 3.1 that sigma be gauge dependent. The only thing you are assuming is that div sigma gives you the force (and if this were the only definition of sigma, it would indeed depend on the gauge, but you need not take this equation as the def'n of sigma for the derivation to work) and that it is symmetric. You could well derive 3.1 had you fixed the gauge, and in fact I believe that the form dF/de does just that.
I don't think this line of questioning is going forward, as I don't think there has to be anything wrong with the definition sigma through dF/de (and indeed, it'll give you all the right results, gauge invariant)
@alarge this is very basic vector analysis, if you write $F = \frac{\partial \sigma_{ik}}{\partial x_k}$ you are allowing $F$ to be gauge dependent because when we look at $F$ on it's own, as is done in section 2, applying Stokes theorem leads to gauge dependency. That is sheer basic mathematics, why are you denying it?
F can't depend on anything, you probably mean sigma? Yes, it would make it non-unique if you were to take the equation you postulate as the definition of sigma. But you don't use this as a definition as it's not a good one. Rather, you just take it as a consequence: If you have a sigma, then obviously this equation must hold true (because you get the forces through sigma).
I mean, it's actually just as simple as $\vec{F} = \Delta \sigma = \Delta (\sigma + w)$ such that = \Delta w= 0$, I don't know why I'm referring to Stokes theorem lol in electromagnetism we don't randomly throw away this fact, why do you want to do it here?
If sigma is nonunique, you can get completely nonphysical results. I'm not much of a continuum mechanics buff, but I suspect that you do posit extra constraints, such as torques or something.
Which make sigma unique
For example, as I mentioned: take the case of microscopic stress. If sigma is defined through momentum flux, like in Schofield/Henderson, you get a nonunique sigma: Integration contours connecting the particles are arbitrary.
I'd like to do what you want to do, but it doesn't make sense, I have spent the night trying to justify it but there's no way to do it, the justification Mistura tries to give literally does not justify it, but you don't appreciate any of these subtleties you just want to ignore the physical derivation of the strain tensor and pick something you like with no justification
I'm just asking you to justify that integral definition, it's very very simple, if you can do it then I'll be very happy, but I think it's impossible, I will not accept you waiving your hands :)
Now there are 2 common choise, the Irving-Kirkwood contour and the Harasima contour. IK goes straight from one aprticle to the other, Harasima goes through xy. Now if you were to compute sigma, then integrate it to get surface tension, you get that surface tension depends on the integration contour, a nonphysical result. It's been shown that Harasima is incorrect in some cases, but I've never seen a proof that IK would be incorrect anywhere.
I've already said that the integral defines stress. You have a stress-strain conjugate pair of thermodynamical variables, and when your system is inhomogeneous, you'll have to integrate over.
They took statistical mechanics as their basis for their derivation of that equation, you take absolutely nothing as your justification for it, but other times you circularly take L&L's derivation
I mean, it's actually just as simple as $\vec{F} = \Delta \sigma = \Delta (\sigma + w)$ such that = \Delta w= 0$, I don't know why I'm referring to Stokes theorem lol in electromagnetism we don't randomly throw away this fact, why do you want to do it here?
You keep saying that if you define sigma through div sigma = f that means that sigma is nonunique. Which is absolutely true. The thing is, though, we are not using this as the defintion.
Rather, you use sigma = dF/de as the definition and immediately see that this satisfies div sigma = f.
You keep trying to force div sigma = f as the definition of sigma. What I am saying is that any acceptable sigma has to satisfy this, of course, but that this is not the definition of sigma.
In fact, from what I remember from reading the book yesterday, L&L proceed to define the free energy for Hookean elasticity and then dF/de for sigma. Note how you can proceed from free energy to sigma, to showing that div sigma = f rather than the other way around.
Landau's eq. 3.1 is $\int dR dV = \int (F \cdot dr) dV$. We cannot to a thing to this equation. That is basic mathematics. However, in the special case that $F = \frac{\partial \sigma_{ik}}{\partial x_k}$ we can play with that first integral. However, by very elementary mathematics, we simply must allow the possibility that $F = \frac{\partial \sigma_{ik} + u}{\partial x_k}$ where $\frac{\partial u}{\partial x_k} = 0$.
A quantum field theorist would love to take the lazy route of throwing this fact away, but they can't, okay, it's a major major issue
I get sigma from dF/de, like L&L for the Hookean system. This sigma is fixed. But it is true that I could add a constant, say, to it and it would not change the force (but I suspect it would change the moments).
Show me mathematically how to get $\int dR dV = - \int \sigma_{ik} \delta u_{ik} dV$, without writing $\int dR dV = \int (F \cdot dr) dV$, L&L get $\int dR dV = \int (F \cdot dr) dV = ... = - \int \sigma_{ik} \delta u_{ik} dV$ which is gauge dependent
L&L do not assume sigma is fixed, they can't, neither does Schofield
Why wouldn't I write it? I'm telling you that any proper sigma will have the property that div sigma = f. The only thing I'm saying is that this is not the equation that defines sigma.
No there is not mathematics that disagrees with what I'm saying. The integral works out for ANY sigma such that div sigma = f. It just happens that my sigma = dF/de (a unique selection) also satisfies this requirement.
You are again using div sigma = f as the definition of sigma. I am well aware that if you were to do so, you would end up with a nonunique sigma. This is what your equations show.
In other words, you have defined a completely random integral $- \int \sigma_{ik} \delta u_{ik} dV$ out of the blue, it has absolutely no association to physics, nobody knows where it came from, but you want to say that working backwards in this calculation $\int dR dV = \int (F \cdot dr) dV = \int ( \frac{\partial \sigma_{ik}}{\partial x_k} \cdot dr) dV = ... = - \int \sigma_{ik} \delta u_{ik} dV$ we can define justifiably define $dR = - \sigma_{ik} \delta u_{ik}$ and pretend we're doing physics, because we are magically going to call the L.H.S. of this work for some unjustified reason
No. It's the same equation as in L&L except in curvilinear coordinates. Note that the stress tensor you get out of it will give you work and is thus physically grounded. But you just get a unique stress rather than going the f = div sigma route.
Okay, where did that come from? It came from this calculation: $\int dR dV = \int (F \cdot dr) dV = \int ( \frac{\partial \sigma_{ik}}{\partial x_k} \cdot dr) dV = ... = - \int \sigma_{ik} \delta u_{ik} dV$ Either we take $$\int dR dV $ as our conceptual starting point, or we take $- \int \sigma_{ik} \delta u_{ik} dV$ as our conceptual starting point. You seem to think, and actually said earlier, that starting from $- \int \sigma_{ik} \delta u_{ik} dV$ and reversing it will justify things, and apparently curvilinear coordinates are the reason why, correct?
Okay, well curvilinear coordinates do not eliminate this gauge problem :) Nothing in this computation $\int dR dV = \int (F \cdot dr) dV = \int ( \frac{\partial \sigma_{ik}}{\partial x_k} \cdot dr) dV = \int ( \frac{\partial ( \sigma_{ik} + u)}{\partial x_k} \cdot dr) dV = ... = - \int \sigma_{ik} \delta u_{ik} dV$ is affected by curvilinear coordinates, it doesn't matter whether you go forwards of backwards, \int ( \frac{\partial \sigma_{ik}}{\partial x_k} \cdot dr) dV = \int ( \frac{\partial ( \sigma_{ik} + u)}{\partial x_k} \cdot dr) dV
Yes I've looked at eveything, Mistura does not show this, he tries to but he is incorrect, he takes that integral as his definition just like you are and commits a logical fallacy, but his computations make sense because he is doing exactly what Landau does in his statistical mechanics book
He just reproduced a computation in Landau with different notation that still implies this gauge problem
Yes, the 1987 paper explicitly, earlier I gave you the derivation of pressure from equilibrium quantum statistical mechanics, $F = -<\tfrac{dE}{dr}> = - <\tfrac{dE}{dV}>\tfrac{dV}{dr} = P dS$, (Landau Stat Phys Vol. 5 eq. 12.1) Mistura is actually just computing $- <\tfrac{dE}{dV}>$ replacing volume with the intrinsic metric, $- <\tfrac{dE}{dV}> = - <\tfrac{dE}{dg_{mn}}>$ in fact Mistura's derivation just analyzes the P in $F = PdS$ using a partition function
His claim of uniqueness, the unique path, is actually just defining P uniquely, the force per unit area, the whole gauge problem comes AFTER all this, when you integrate this equation F = PdS over the volume and assume it's a divergence
Equilibrium statistical mechanics, yes, as in classical equilibrium classical mechanics. There's no need here to invoke anything from quantum mechanics.
It's not like statistical mechanics or thermodynamics didn't exist before quantum mechanics.
Well if you want to go on hearsay and someone elses opinions that's fine, please go back and answer Landau's point about the non-fundamental nature of entropy
Okay you're not making any sense, you want to use classical statistical mechanics, yet your latest paper is on "quantum stress fields", and again equation 5 in your paper is using the same unjustified definition out of complete thin air
Once you use that definition you assume what you want to prove
Again a quantum field theorist would love to take this lazy route, but they are not going to pretend it doesn't exist
The uniqueness discussion doesn't use quantum mechanical arguments, but does shed light as to why the dF/dg will pop out unique (I think this is also true in GR, for the Hilbert stress-energy tensor, so again, this shouldn't be an issue)
@bolbteppa I've given you several different ways to arrive at the equation. Take the integral from L&L, pop it into curvilinear coordinates to generalize.
Yes, so you want to define some fundamental concept that only exists classically when you take differences, but quantum mechanically exists irrespective of differences, then you want to tell me that in the quasi-classical limit of quantum mechanics we will reproduce this classical phenomenon, if what you were saying made any sense I should be able to reproduce fixed entropy values in the quasi-classical limit, but I can't, yet this all makes sense
@alarge Your only justification for reversing Landau's equation 3.1 has been that curvilinear coordinates "magically" make it work, but they don't affect anything, you've literally got no justification for your derivation anymore, none, other than the authority of random papers which logically don't make any sense. The paper you linked to uses a flawed definition, worse it actually refers to Mistura as it's motivation, but Mistura's paper is just reproducing Landau's calculations, no sense...
You are always presupposing that we know that the world is quantum. Classical statistical mechanics does not know that. It has no reason to believe observables do not commute.
If we get a quantum result that has a classical analogue then the quantum concept should literally reproduce the classical concept in the quasi-classical limit, if it doesn't then one of them is wrong, the very existence of entropy in quantum mechanics makes sense by dimensional analysis, in classical mechanics this dimensional analysis breaks down, nobody has mentioned commutativity...
This is why I recommend reading Landau, two major conceptual errors completely avoided :)
@bolbteppa That's not a very satisfying answer given that the approach seems to work pretty much everywhere. I don't see why you can't conceptually just generalize the equations and see where that takes you. Physics is a tool anyway, not the Truth (well, not before everything's nailed down, anyway, which I don't think will ever happen).
As for what the curvilinear coordinates do and why they do affect the end result, you can read the Phys Rev B paper. You asked why I trust a random paper; I suppose that over a random internet personality, I'd go with the paper anytime. Especially seeing the arguments for/against.
Basically, you want to treat elasticity as though it derives from classical particle mechanics or classical field theory, i.e. CM. I wanted to do this for the past year tbh. But I simply could not understand why they analyzed the whole thing with thermodynamics in Landau, made no sense. Only by thinking of elasticity as a macroscopic quantum equilibrium statistical mechanics of solids does any of this make sense, that's why you can employ thermodynamics, it doesn't exist in classical mechanics.
Okay well enjoy following these papers, I see I can't convince you to give up this ideological belief, but at the end of the day you are using classical mechanics and a flawed understanding of statistical mechanics to justify it, unswervingly want to ignore quantum mechanics for some strange reason,
are happy to take some random starting point for ideological reasons, anytime you're asked to justify your starting point you give a circular justification or an incorrect justification for throwing away gauge issues because of coordinates, also in this circular justification you actually still can't throw away the gauge invariance issue, let alone the fact that you fundamentally misinterpret the meaning of Mistura fixing his arc.
If this leads you to believe you've unified QFT & GR don't say I didn't warn you, I mean you're already trying to define quantum functions of geometries... It's a dangerous road dude, all because of some silly stress tensor :)
@ACuriousMind : I need your help. You are quick and you can check a trivial calculus. Please see the question . It's not a simple home exercise, I showed there different tricks how to solve such problems. But, you put the question on hold, and I was punished by God knows whom with a -1.
@Sofia I think your calculus there is correct. The downvote is probably from someone who found your solution to this homework-like question too explicit. (Answers that fully solve a homework question are generally discouraged)
@ACuriousMind No, and please help. There is a small mistake of lack of attention, in some place I wrote that $\int x^2 dx = x^2/2$. It doesn't change the general course of the proof, but I am afraid that he would take the result without checking it. I need to call his attention on that, and he will see the minus. And his will loose any trust in my calculus, he is a novice. THIS is the problem. Help me! Please!
@ACuriousMind of course the calculus is correct (I corrected that small item) and it is not too detailed. It teaches tricks, this is why I made it detailed. Please read it a bit. Help me!
@KyleKanos let's leave editing, you see my problem. Please help. I don't trust these young fellows that they follow the calculi equation after equation. Otherwise he would find and correct the mistake along - it's a small thing. But I don't trust that.
@ACuriousMind about reopening, this is another issue. The tricks I show there are useful, they are of general benefit.
@ACuriousMind : I don't need you to remove the minus, but to place a comment and explain that an answer should not be too explicit, can you do? So, the OP will see that not the calculus is the reason for the minus, but that it is considered too much explicit. This thing, can you do?
@Sofia Yes, I could leave that comment. It seems the minus is taken care of anyway, though - your answer is at a net score of 0 at the moment.
@Sofia Good or useful answers do not mean that the question is redeemed by them. I don't think we should reopen (or leave open) question just because the answers may be useful.
@ACuriousMind I cannot care less about the score. I have to leave him a note that there was a small mistake and I corrected it. If he would have seen the minus, he would have lost trust in whole the calculus. He doesn't have experience.
@ACuriousMind I will return to you, I want to leave my note, I repeat, I don't trust that these youths check calculi.
@Sofia If they do not, then it is their own fault. I don't really see what I can do about it (and they'll believe me as much or little as they would you if you said that the -1 was only because the answer is explicit).
@ACuriousMind the -1 was removed, and I sent my note. I assume that learning is easy for you, but not everybody has your qualities. I am afraid that under pressure of exams, they can take a calculus without checking. There is no fault here, you don't know details about the respective person. I heard many bad things about schools. I am in contact with different people, and what I hear from them about in which conditions they learn, is simply revolting. This is why I care so much.
@ACuriousMind if you knew how many questions, simply of revolting low level, I flag to the moderators. This one question was by far more normal.
What are good examples of Hamiltonian systems outside physics? I heard there are financial systems that can be described by a Lagrangian, and was interested to see some examples
@ChrisWhite Good question. It has only 3 questions tagged with it, too.
user54412
I don't really get this "if I can write a differential equation for it, it must be physics" mindset. I agree fully that most fields of study, from economics to biology, are woefully incompetent when it comes to math. But that's their loss. Physics doesn't have some monopoly rights to applying equations to nature.
@DavidZ : why am I punished with 23 points? Not only the question was put on hold, but also points were taken from me. Why? Or, let me ask otherwise, why only 23?
I am teaching myself tensor calculus.
In some of my calculations the term $\nabla_{[i}x^i \nabla_{j]}x^j$ keeps turning up. In 2 dimensions this is up to a constant the determinant of the $\nabla_{i}x^j$ matrix. When it is zero this means that the direction $y^i\nabla_{i}x^j$ is independent of ...
What are good examples of Hamiltonian systems outside physics? I heard there are financial systems that can be described by a Lagrangian, and was interested to see some examples
