2:28 AM
@ACuriousMind omg
the greatest coffee shop of all time
2
@ClaudioMenchinelli Let a,b refer to the particle number and i,j,k refer to Cartesian components. Then $[S_{1i},S_{2j}]=0$ and $[S_{aj},S_{ak}]=\mathrm i\hslash\varepsilon_{ijk}S_{ai}$ are your standard starting points. What you now want is \begin{align}\tag1[S_{1j}+S_{2j},S_{1k}S_{2k}]&=[S_{1j},S_{1k}]S_{2k}+S_{1k}[S_{2j},S_{2k}]\\\tag2&=\mathrm i\hslash\varepsilon_{ijk}\left(S_{1i}S_{2k}+S_{1k}S_{2i}\right)=0\end {align}
@ACuriousMind well maybe it is experimentalist terminology...
honk
where the zero at the end is because the bracketed term is obviously symmetric in i,k but is contracted with the levi-civita symbol
H O N K ~
@Sanjana I am not sure where you are even coming from, why you'd think that this even had the chance to work. When deriving Euler-Lagrange equations, the delicate relationship between positions and velocities had to be made clear. In Maxwell's, A is the position and F are velocities. Of course, there would be special cases that gives the same result, but those are the exception, not the rule.
@ClaudioMenchinelli Also, if you wanted to use the $\otimes$ notation, it should be $$[\vec S\otimes\mathbb I+\mathbb I\otimes\vec S,S_x\otimes S_x+S_y\otimes S_y+S_z\otimes S_z]$$ which looks horrible.
3:18 AM
@naturallyInconsistent No, here $A$ and $F$ are taken to be independent in the sense that when you vary $A$, you don't get to vary w.r.t. derivatives of $A$. They are completely decoupled "fields" in some sense. I mean to ask why does that special case work and when does those work?
This kind of thing also probably occurs in say Palatini formaluation of gravity, where the connection and metric are taken to be independent and the variation of these quantities separately give the EFE and the relation of Christoffel symbols and metric components.
This is also probably analogous to the worldine action of a spinless particle (with no restriction on mass), or Polyakov action in string theory.
3:50 AM
@Sanjana are you sure it is done that way, and not just the usual EL style partial derivatives (in the latter case you are meant to pretend that the varying of derivatives is independent)?
@naturallyInconsistent Yes. Atleast in the similar exercise you can check yourself also that if you take that to be the case, then you get Maxwell equation+ what we call the "definition" of field strength tensor in terms of the gauge fields
@SillyGoose Wow. When are you putting it up online :o ?
ok, I'll leave that to you, since I'm sure you know more than meow
4:14 AM
nah. I might be just fooling around anyway...

2 hours later…
6:04 AM
@Sanjana i wish i knew how to put pdfs online...but i would be happy to email them to you if you are interested :D
they're very elementary hehe

1 hour later…
7:08 AM
now, now, let's not star the honks
2
@SillyGoose Yeah. I am having some trouble in deriving even the basic classical EOM actually. Didn't think about it before.
Here's my email ID: [email protected]
@SillyGoose I also asked a question regarding CS theory action variation if you are interested.
0

Summary/TL;DR: I want a detailed calculation of the derivation of classical equations of motion from the Chern-Simons action using differential forms, using variational derivatives. I mentioned "using variational derivatives" because the Lagrangian density might be case as an exact form (which I ...

sent! then that would be perfect i think. i write down the variation of the action explicitly. however, my note is specific to non-abelian chern-simons on a trivial bundle (i think is the simplest case of chern-simons theory)
i use a little different notation than your post. the omission of brackets and etc. was quite annoying while i was writing up the note so i kept all the operations explicit in the variation of the action
let me know if it is helpful or not/if there are some unclear bits
8:21 AM
hi

2 hours later…
9:58 AM
@SillyGoose scribd
10:09 AM
physically, what does it mean here if $V_y, V_z = 0$? it would seem this means that the y and z spins cant couple to eachother, but how would it be possible that the x spins can couple but not the y and z?

2 hours later…
12:18 PM
@naturallyInconsistent this is exactly what I did, but I considered the generic $S_j \otimes S_j$

1 hour later…
1:26 PM
@Relativisticcucumber I think it's just a way to model a system in which the spins are naturally already aligned along one axis (parallel or antiparallel)
every article seems to start with your Hamiltonian, where $$\mathbf{B} = B\hat{z},\text{ }V_y = V_z = 0, \text{ }V_x = J = const$$
but I'm probably wrong, maybe it has a physical explanation
That Hamiltonian sure is scary :P

3 hours later…
4:06 PM
Hi, can you provide a quantitative answer as to why, if a rod of mass m is rotating around an axis with two masses attached to its sides and one of these masses falls off, the angular velocity of the system between before and after detachment remains the same?
It is not so intuitive to me...
@Bml Who claims the angular velocity "of the system" remains the same, and what even is "the system" after detachement - there are then two parts, presumably with different angular velocities?
@ACuriousMind The angular velocity of the system composed by the rod and the two masses is equal to the angular velocity of the system composed by the rod and one of the two masses, after the other mass detached. Right?
@ACuriousMind As far as I have heard, it should be analogous to the case where a man drops an object from the palm of his hand.
@Bml again - who claims this? I don't think this is true - what is true is that since no torque acts on this system, the total angular momentum should remain constant, but since both the mass and the rod+mass can move freely, their moments of inertia will change after the detachement
4:23 PM
@ACuriousMind My professor said: consider a rod (mass $m$) with two attached masses $m_1$ and $m_2$, rotating with angular velocity around an axis parallel to the axis of CM. If the mass $m_2$ falls off without any external impulse, then the angular velocity of the system composed by the rod and the two masses is equal to the angular velocity of the system composed by the rod and one of the two masses, after the other mass detached.
@ACuriousMind This occurs because the total angular momentum should remain constant: before detachement, the angular momentum is $(I_1 + I_2 + I_{rod}) \omega$, whereas after detachement the angular momentum is $(I_1+I_{rod}) \omega + m_2 v_2 r$, where r is the distance between one side of the rod and the axis of rotation.
@ACuriousMind He said that the moment of inertia after detachement will be less than before the detachement, but we have to add the spin angular momentum of the detached mass, who moves on a parabolic trajectory in a vertical plane. It is not so clear or intuitive to me.
of course, but I don't think this immediately implies the rotational velocity stays constant - consider conservation of linear momentum, which the particle now flying in a straight line away also has. Since the c.o.m. of the whole system (rod + both masses) has to stay still (no total linear momentum before detachment), the c.o.m. of rod + remaining mass has to have momentum in the opposite direction
that means the rod + remaining mass will move away from the former center of rotation/mass, and so you also have a linear component there - the rod doesn't just stay still and spin about the initial center of rotation forever
@Relativisticcucumber it doesn't mean anything physically, this model describes no real-world physical system - it is designed to be the exact analogue of the classical Ising model rather than to reflect an actual quantum system. The real-world model for magnetism is the Heisenberg model
@ACuriousMind Sorry, I do not understand this answer. What do you mean by 'a linear component'? I cannot understand why conservation of angular momentum actually results in angular velocity remaining constant before and after detachment, whereas your assertion seems to be contradictory. Which equation would you apply?
@Bml by "linear component" I mean a term analogous to the $m_2v_2r$ you wrote down there - there's also a $m_\text{rod} v_\text{rod} \times r_\text{rod}$ because the rod has to start moving in the opposite direction to conserve overall linear momentum
@ACuriousMind My professor said that it is analogous to the case of a man who drops an object from the palm of his hand (is it true?). I searched for it and I found this:
2

A man holds in his hands two equal masses with outstretched arms, standing on the center of a platform that rotates with a certain angular velocity. If you drop both of the masses without moving your arms, what happens to the angular velocity of the man and the angular velocity of the masses? I...

and so "the angular velocity" of the rod also becomes a bit of a strange term because a) it is also moving linearly and b) its new center of mass it not the total center of mass (the former center of rotation)
4:36 PM
@ACuriousMind OK, so how would you write the conservation equation for angular momentum before and after detachment?
@Bml Note that there the man drops both masses at the same time!
in that case, the center of mass of the rod/man after the detachment is still the same, and my argument doesn't apply - it is specifically the asymmetry of releasing only one of the masses that causes the problem
@ACuriousMind So, it is the case of a man who drops a single mass from his hand, right?
not exactly (since we there also have gravity) but yes
@ACuriousMind OK, I got it, but I have difficulties to write the equation of conservation of angular momentum in this case. Could you help me?
you just sum up all the angular momenta - where's the problem?
4:41 PM
@ACuriousMind $(I_1 + I_2 + I_{rod}) \omega = (I_1+I_{rod}) \omega' + m_2 v_2 r$ is not correct, right?
what is $\omega'$ and $I_\text{rod}$?
again, at least in my interpretation, the rod moves away from the total center of mass / initial center of rotation
so you must compute the new $I_\text{rod}$ taking that motion into account, and $\omega'$ won't be a constant, either
@ACuriousMind $\omega'$ is the angular velocity after the detachement, $I_1, I_2, I_{rod}$ are the moment of inertia of mass 1, mass 2 and rod about the axis of rotation, respectively.
also it's not really rotating around that point, it's rotating and translating - this is actually a pretty complex system if you want to write the momenta down explicitly
@Bml sure, but all of those things are now time-dependent since everything is moving away from the initial center of rotation!
@ACuriousMind In the problem presented by my professor, the system mass 1 + mass 2 + rod didn't rotate about the combined center of mass, but around an axis parallel to it. Is it relevant?
ohhhh
so the axis of rotation is parallel to the rod and the rod is kinda orbiting it, not spinning around itself?
4:46 PM
@ACuriousMind OK, so you are saying that immediately after detachment the angular velocity might instead be preserved?
actually I think that makes the situation even more complicated :P
or, rather, the question is then what makes the rod orbit around that other axis - you need a centripetal force there
@ACuriousMind The axis of rotation is parallel to the axis passing through the combined center of mass of the system and perpendicular to rod. In the original statement it was at a distance of $\ell/6$ from the beginning of the rod.
@ACuriousMind I'm copying the original statement, maybe I neglected some important details:
@ACuriousMind A homogeneous rod AB of length L and mass m is placed in equilibrium on a pin placed at the point O, located at a distance AO = L/6 from the end A. At the end A of the rod there is a body of negligible dimensions and mass m_A is welded, while at extreme B a body of negligible dimensions and mass m_B=m/6 is welded. The rod is horizontal.
a) Calculate the m_A value for which the rod remains in equilibrium under the action of the weight force;
The rod is then constrained to a vertical axis z passing through O, around which it can rotate with negligible friction. A constant moment
oh, yes, you neglected the extremely important detail that the rod is pinned to the point O :P
this changes the entire setup completely, since the rod isn't free to move (as I assumed)
@ACuriousMind Ah, I was sure I would have omitted something important. So, the linear component exerted on the rod + remaining mass system is no longer present, right?
yes, the pin prevents it from moving
in that case, the angular momentum conservation of the whole thing is just $I_\text{rod+1+2} \omega = I_\text{rod+1} \omega' + m_2v_2 \times r_2$ (note the cross product!)
4:59 PM
@ACuriousMind OK, but... Masses have been found for the system to be in equilibrium on the horizontal, the system is pivoted in O. Should the mass m_B fall for the other fixed values there could be no equilibrium in a stationary case, right? In the case where the system is in motion, why does detachment not affect if mass m_B is removed, since the system is rotating?
@ACuriousMind Why did you underline the "cross product" (sorry for not understanding)?
@Bml because you kept writing it without a cross product
the trick is that $v_2 \times r_2$ is actually constant for a particle moving in a straight line (a single free particle has conserved angular momentum), so this term has to be equal to the initial $I_2\omega$ when it detaches, and so $\omega = \omega'$
@ACuriousMind And this is what I do not understand. In my professor's solution, it must be $m_B \times{v_B} \times \vec{r_B} = m_b v_B r_B$. Why? We do not know which side the mass falls on...
I don't really know what you mean by either of the two sides of that equations
the mass isn't a vector, so $m\times v$ doesn't mean anything
perhaps I should clarify that I treat $v$ and $r$ as vectors, i.e. with the arrow notation I mean $m_2 \vec{v}_2\times \vec{r}_2$
just the ordinary angular momentum of the particle
I don't know what you mean by "which side the mass falls on" nor why that would be relevant
@ACuriousMind I said this because $m_B \times{v_B} \times \vec{r_B} = m_b v_B r_B$ implies that the angle between radius and velocity is 90°. Is this always true in this case? Why?
5:15 PM
no, it's absolutely not true
it's 90° at the moment of detachment
but not after that
but probably that's what your solution is trying to say - at the moment of detachment, the angle is 90° simply by how circular motion works
@ACuriousMind Moreover, in this case where the system is rotating, why does detachment not affect if mass m_B is removed, since the system is rotating? Isn't there a situation in which the angular velocity is affected by the fact that the mass m_B is detached?
17 mins ago, by ACuriousMind
the trick is that $v_2 \times r_2$ is actually constant for a particle moving in a straight line (a single free particle has conserved angular momentum), so this term has to be equal to the initial $I_2\omega$ when it detaches, and so $\omega = \omega'$
this argument does not depend on $m_B$ at all
@ACuriousMind I am not saying it is not right (because it is certainly correct), but I still think it is something counterintuitive, or at least not so intuitive.
intuition is overrated :P
@ACuriousMind Ahah, that's true.
@ACuriousMind More than anything else, I keep thinking of the analogy between spinning top in motion (not falling) and spinning top at rest (falling). This is why I imagined influences on angular velocity when a mass falls from a system....
Probably this analogy is wrong!
5:54 PM
@ACuriousMind Sorry to bother. The problem I have with this is that the angular momentum is a vector, so it has a direction, so we could have a negative contribute (-) if the mass falls on the side going in the direction in which the rigid body was rotating before the mass detached, a positive contribution (+) if the mass falls on the opposite side. Where am I going wrong?
I don't really understand what you mean by "the side in the direction in which the rigid body was rotating" but that doesn't matter: Any signs/orientations are part of the vectors $\vec v$ and $\vec r$ themselves, no?
angular momentum is always $\vec p \times \vec r$
and you always add all those
if stuff is going in opposite directions then one of the $\vec p$ will have an opposite sign to the other or whatever
@ACuriousMind But if they have opposite signs, why shouldn't the contribution of mass m_B be negative?
it's a vector!
there's no notion of it being "negative"
each thing here has an angular momentum vector $\vec{L}_i = m_i \vec{v}_i\times \vec{r}_i$
and total angular momentum is by definition the sum of all of these, $\vec{L} = \vec{L}_1 + \vec{L}_2 + \dots$
If $\vec{v}_1$ and $\vec{v}_2$ are opposite, you have $\vec{v}_1 = -\vec{v}_2$, but that doesn't change anything about these formulae as such
@ACuriousMind Wouldn't there be an extra minus sign? That is, we would have that $\omega = - \omega'$, right?
no?
how would that happen?
6:09 PM
@ACuriousMind Because $\vec{v}_1 = -\vec{v}_2$, or not?
@Bml can you point out where in the formula exactly you think that would matter? i.e. actually spell out the computation?
@ACuriousMind $(I_1 + I_2 + I_{rod}) \omega = (I_1+I_{rod}) \omega' + (- m_2 v_2 r)$
@Bml why did you put a minus there? $I_2 = m_2 v_2 r_2$. It would be $I_2 = - m_2 v_1 r_2$ (since $v_2 = -v_1$) but even then how does that lead to $\omega = -\omega'$?
also, you should see that this would be not only "unintuitive", but patently absurd: If $\omega = -\omega'$, that would mean that in the instant where the mass is released, the rod suddenly reverses its direction of rotation and rotates as fast as before - but in the opposite direction
@ACuriousMind To fall, does the mass m_B not need a small impulse, i.e. a force that is also transmitted to the rigid body by action and reaction? Just for curiosity.
@ACuriousMind Yes, you're right, it doesn't lead to $\omega = -\omega'$.
@Bml you're just disconnecting the mass from the rod, why would there need to be an impulse?
the linear velocity of the mass doesn't change
6:14 PM
@ACuriousMind Mass falls by itself, doesn't it?
I don't understand the question
@ACuriousMind We are not disconnecting the mass, the mass falls off and drops, or not?
sorry, I still don't know what you mean
@ACuriousMind "Disconnecting the mass" implies action by human beings, whereas "mass falls off" doesn't.
no, I didn't mean to imply that
for some reason - be it humans or magic or the inevitable decay of the universe that destroys all things - whatever fixes the mass to the rod stops working, and the mass flies off
the actual reason doesn't matter, and there's no forces here
the only forces here are the one between the remaining mass and the rod, and a brief wiggle when the pin stops the rod from flying off in the opposite direction
6:20 PM
@ACuriousMind So there is no reason for the mass to come off? Isn't it due to internal contact forces or something in particular?
I mean if anything it's due to contact forces stopping working
note that there's a force on the mass when it is on the rod - the centripetal force that keeps it rotating
after it's disconnected, it flies off in a straight line, no forces acting on it anymore
but really the reason doesn't matter (and if it mattered the exercise would need to specify how the mass is affixed to the rod, by a glue or a screw or whatever)
@ACuriousMind Isn't there also a moment of force exerted by the rod on the mass?
that is the centripetal force
@ACuriousMind So this is the one that makes the angular momentum contribution mvr?
I don't understand the question
angular momentum is $mv\times r$
you don't need a force to make it so
6:36 PM
@ACuriousMind So it is not due to the centripetal force, right?
how would momentum be "due" to any force?
momentum is due to velocity, not forces
@ACuriousMind Yes, I know, in fact I was talking about moment of force.
7:33 PM
@ACuriousMind One question: I should treat the rod + mass 1 + mass 2 system as an open system, right?
@ACuriousMind How well do you know ancient greek? I'm curious as to whether the definitions and postulates written in book 1 of Euclid's elements can be transcribed to some first order logic and set theory
but reading this(page 6-7) so far the definitions & postulates seem almost arbitrary in english
postulate 2 doesn't seem to make sense to me. 5 seems highly paraphrased from the parentheses.. darn, must I really learn ancient greek to appreciate whether there was any rigor in the defintions he gave
@Obliv There was none.
The modern conception of mathematics as a game of formal logic is really that - modern
@Bml I don't know what you mean by that in this case
@ACuriousMind AFAIK, an external torque is the only way that one can change the angular momentum of a closed system. But if one considers the system consisting of the rod and the two masses, one does not have a closed system. One has an open system -- a system in which mass is allowed to enter or leave. The mass m_B start on the rod, inside the system, but ends flying freely, outside the system. Yes?
@Bml if your system is the rod + the 2 masses, how is it open? One of the masses is moving, sure, but it's not leaving the system of "the rod and 2 masses"
But why do some people say that Euclid's 5th postulate is "wrong" or that leads to a paradox?
7:44 PM
and why are you talking about an external torque? The total angular momentum of this system is not changing, that is the entire point of the exercise
I feel like it's just an axiom for which one can axiomatize their geometry.. it's not a paradox afaik?
@ACuriousMind Why not? Aren't they two different systems?
@Bml why would they be?
The non-euclidean geometries exist when you relax the metric/discard the 5th postulate, you can't get them from the same set of axioms so how can that be a paradox?
@ACuriousMind You also said here -- there are two parts...
7:47 PM
@Bml My point there was that the system of "a rod and 2 masses" doesn't really have a single angular velocity after one mass flies away. I wasn't implying anything about it not being the same system or whatever
and I don't see how this is relevant to the problem at hand at all
@ACuriousMind I wanted to say that, if one is trying to account for angular momentum in an open system, one must consider the angular momentum carried in or out of the system when mass enters or leaves.
also are you trying to learn math from Wikipedia again instead of from an actual textbook :P
3rd paragraph
Well, I was just skimming that article (I have a textbook open in another tab don't worry lol)
@Bml that is correct, but entirely irrelevant to this situation since the total angular momentum is conserved - it's the same before and after the mass flies away
@ACuriousMind Because in the initial angular momentum we consider the total angular momentum of the system rod + 2 masses, whereas in the final angular momentum we have the angular momentum of the system "rod + remaining mass" + angular momentum of the system "mass flying away". Right?
7:51 PM
@Bml In both cases you're computing the total angular momentum of the system "rod + 2 masses". Just in one case both masses are attached to the rod, in the other they aren't. The system is still "a rod and 2 masses" in both cases. This doesn't have anything to do with open systems.
@Obliv the article directly clarifies that "Some of these paradoxes consist of results that seem to contradict the common intuition", then talks about non-Euclidean geometry as an example of that. What exactly is confusing about this?
do you disagree that non-Euclidean geometries are counterintuitive :P
@ACuriousMind OK, so the difference is that the formula for moment of inertia as $L = I \omega$ applies for rigid bodies rotating around a fixed point. A mass flying in straight line does not form a single rotating rigid body, so its total angular momentum will be computed not as $L = I \omega$ but as $L = \vec{r} \times \vec{p}$. Right?
those aren't really two different formulas
it's just that for a point mass, $I\vec \omega = \vec r\times \vec p$
@ACuriousMind No, but I disagree that Euclidean geometry is "wrong" or paradoxical :P I think, while informal, his axiomatization of geometry is perfectly valid depending on one's criteria. Since it's not written as a formal system, anyway, it's like comparing apples to oranges shrug
@Obliv that's not what that section is saying
they're just saying that non-Euclidean geometries are examples of geometries where the parallel postulate is false (because there there are more parallel lines than Euclid's 5th would allow)
@ACuriousMind OK, it's right. What I thought is that the angular momentum of the system does change, since the system separates (rod + mass remaining and mass falling off). The speed and angular momentum of the "rod + mass remaining" does not change - meaning that part of the momentum that is "with" the "rod+mass remaining". The momentum of the mass has gone away with the mass, and eventually been transferred to the Earth when they finally impacted. Isn't that an equally valid point of view?
8:03 PM
this is not a metaphysical claim about Euclidean geometry being "wrong" in some abstract sense, merely the basic diagnosis that the parallel postulate is false in non-Euclidean geometries in the logical sense - it is not a true statement there
@Bml no
there's no Earth here
and the argument for why the angular velocity of the rod + mass does not change is precisely because of conservation of the total angular momentum
@ACuriousMind So maybe it is a valid point of view if we talk about the man + object in his hand?
no
the earth is irrelevant even if it is there :P
@ACuriousMind Yes, the total angular momentum, but I did not understand why it is not right to consider two separate systems (in general, not only in this situation).
stop trying to overcomplicate the situation, it is very simple: The total angular momentum of the system "rod + 2 balls" is conserved for all time. When you write out what that means, i.e. once you realize that the angular momentum of the ball that flies off stays the same for all time, too, it follows directly that the angular momentum of "rod + ball" must stay constant, too
this is a consequence of figuring out that the angular momentum of a ball moving in a straight line doesn't change, not the result of some arbitrary partitioning of the system into new "systems" with their own conservation laws or whatever you're imagining here
@ACuriousMind OK, and it is conserved because of the impulsive forces created on the constraint, precisely considering with respect to the pole. Right?
8:08 PM
sorry, I don't know what an "impulsive force" is, nor what pole you're talking about
the entire argument rests solely on conservation of total angular momentum + you being supposed to have learned that something in uniform linear motion has constant angular momentum, too
you literally don't need to say the word "force" once in this entire chain of reasoning
@ACuriousMind The impulsive forces (constraint forces or other) acting on the constraint, the pole is the pin around which the system does rotate.
@ACuriousMind OK, so what is the real reason why angular momentum is conserved? Do no external torques act?
@Bml Where would there be any torque here?
@ACuriousMind The problem statement said that first an external torque was applied, then the engine was switched off and then the body was dropped.
@Bml As you said: the engine was switched off - so where's the torque supposed to come from now?
did you not realize that the meaning of this sequence of events was that the external torque was created by said engine?
@ACuriousMind There is no external torque, so that is why angular momentum is conserved, right? That was my question...
8:18 PM
@Bml Yes, there are no external torques (after the engine is switched off), so total angular momentum is conserved.
@ACuriousMind OK, thank you very much for your help!
8:37 PM
@ACuriousMind I see. Do you happen to know if there is a difference in Hilbert's formalist school and the logicist school of thought for foundations of mathematics?
The distinctions are all muddled imo, also intuitionist school just sounds like the school for people who don't seem to care :D
oh... there are many schools.. jeez