@GiorgiLagidze the point is again about infinitesimals! The $\alpha(t)$ is an infintesimal variation to $q(t)$, but by definition of what an extremal point of the action is, all infinitesimal variations of a solution $q(t)$ vanish!
> Again, if the motion of the heaven is the measure of all movements whatever in virtue of being alone continuous and regular and eternal, and if, in each kind, the measure is the minimum, and the minimum movement is the swiftest, then, clearly, the movement of the heaven must be the swiftest of all movements.
How was that guy the basis for all physics for a thousand years
the whole idea of this proof is first, author uses $a$ transformation(a is infinetisemal number) and then uses a = a(t) and clearly, a(t) is not 0 at $t_0$ and $t_1$. That's for sure :P jhwilson.com/blog/2022/Noether-classical
i will take a look, but i think a(t) doens't have to be 0 for proof to hold. if you got some path q(t) for which you know action is minimum, then action for q(t) + a(t) have to be bigger. no ?
Hi everyone. I have a problem with this. Consider a rigid, thermally insulated cylindrical container divided by a wall parallel to the cylinder bases into two parts A and B, each containing one mole of the same ideal gas. The wall prevents the exchange of particles and heat between the two parts.
Initially, the position of the wall is locked in the centre of the container and the system is at equilibrium with the two parts at the same volume V_A(0) = V_B(0), but different temperatures T_A(0) > T_B(0).
At a certain instant, the moving wall is left free to slide without friction, keeping parallel to the cylinder's base, and the system, after a rapid transient, reaches a new equilibrium condition. Determine the pressure values P_A(f) and P_B(f) in the two parts of the system once the new equilibrium condition has been reached.
At first glance, one would say that this situation obeys an adiabatic transformation, so one should use the relation $PV^\gamma = const$, but this is only valid for reversible processes. Does the physical situation described sound like a reversible process to you? Not really to me...
in variational calculus, we use variation such as $q(t) + \epsilon f(t)$ where $\epsilon$ is infinetisemal and it considers such paths where $f(t_0) = 0$ and $f(t_1) = 0$ and argues that every path's action other than true path has to be bigger. Ok. Now, on this article I shared, we use $a(t)$ variation(infinetisemal, sure), but since author doesn't discuss the initial/final points, that means his proof works for any case of $a(t)$ as long as it's infinetisemal
and I wanted to confirm if the theorem that action on non-true path will be bigger here even without considering the boundaries
the author simply didn't want to get into a discussion of what the symmetry might to do the boundary terms
the symmetry is in fact not required to respect the boundary conditions
i.e. there is no need to have $a(t_0) = 0$
the correct argument for that in the context of that blog post is just that $S[\sigma(q,\alpha)] = S[q(t)]$, so extremizing the former is the same as extremizing the latter for all $\alpha$
My point is that you can vary $q(t)$ (with that $q(t_0) = q(t_1) = 0$ condition!) in $S[\sigma(q,\alpha)]$ and the extremal points must be that same as those of $S[q]$
that's the whole point of the exercise here, shifting the Lagrangian by a total time derivative does not change the E-L equations
so $S[\sigma(q)]$ has the same E-L equations as $S[q]$
and so when we vary $S[\sigma(q)]$ (still with $q(t_0) = q(t_1) = 0$), the resulting E-L equations must be that of the original L, and in particular the term multiplying the $\dot{\alpha}(t)$ must vanish
you are in both cases just varying $q$ with the same boundary conditions, the value of $\alpha(t_0)$ doesn't enter into this at all
@GiorgiLagidze The rigorous proof is that the boundary terms only depend on the value of $q$ at $t_0$ and $t_1$, which are held fixed during a variation (a point which you have been quite insistent about!)
so such boundary terms cannot contribute anything to infinitesimal variations with fixed $q(t_0) = q(t_1) = q_0$ and therefore also not to E-L equations
this is a proof you should have done before you started trying to prove Noether's theorem, since this is the whole reason we allow the Lagrangian to vary by a total derivative - since it only contributes boundary terms to the action
I suspect it's mostly the effect that his physics are intertwined with his morals and ideas about what is "perfect" and so on, so if you want to believe in the latter you have to pretend to believe the former
IIRC Ptolemy partly based his model on Aristotle but in his book you don't really read such things
I think "working" astrologers probably didn't have time for that nonsense
Only the nonsense of astrology
Aristotle's proof that stars are spherical is that stars are fixed objects carried by the motion of the celestial sphere, and therefore do not need any method of locomotion, and as nature does not use any more than it needs, it would give them the simplest shape for this, the sphere
Otherwise they would have tiny legs
or presumably wings since they fly
I kinda wonder what it actually means when ancient astronomers mention the records of the egyptians and babylonians
I'm guessing they didn't pull them on Arxiv
Did they have to travel all the way there to look at them and construct whatever model they have in mind
Maybe Hipparcus would talk about that let's see
"Hipparchus obtained information from Alexandria as well as Babylon, but it is not known when or if he visited these places. "
@Slereah It was not uncommon for Roman elites, for instance, to have slaves that were scholars - many famous Romans mention in passing that their tutors were slaves who spoke multiple languages and knew a lot of texts
> Most people-all, in fact, who regard the whole heaven as finite-say it lies at the centre. But the Italian philosophers known as Pythagoreans take the contrary view. At the centre, they say, is fire, and the earth is one of the stars, creating night and day by its circular motion about the centre. They further construct another earth in opposition to ours to which they give the name counterearth. In all this they are not seeking for theories and causes to account for observed facts, but rather forcing their observations and trying to accommodate them to certain theories and opinions of th…
Is there anything like Lie-group/Lie-algebra correspondence but for superalgebra and supergroup...Does the notion of exponential map and all that naturally extend...what is the canonical reference on the representation theory of superalgebras/supergroups or other mathematical aspects of SUSY?
According to Aristotle, heliocentrism's position was mostly due to people thinking fire was more important than Earth and thus deserved a central place in the universe
> By these considerations some have been led to assert that the earth below us is infinite, saying, with Xenophanes of Colophon, that it has 'pushed its roots to infinity',-in order to save the trouble of seeking for the cause.
That's how flat Earth doesn't fall :p
> Anaximenes and Anaxagoras and Democritus give the flatness of the earth as the cause of its staying still. Thus, they say, it does not cut, but covers like a lid, the air beneath it. This seems to be the way of flat-shaped bodies: for even the wind can scarcely move them because of their power of resistance.
> It is clear, then, that the earth must be at the centre and immovable, not only for the reasons already given, but also because heavy bodies forcibly thrown quite straight upward return to the point from which they started, even if they are thrown to an infinite distance.
If Aristotle had stronger arms he could have discovered the Coriolis force
The maximum bosonic subalgebra of $\mathfrak{su}(2,2|4)$ is given to be the product of Lorentz group and dilatations $\mathfrak{so}(1,1)$ and the R-symmetry $\mathfrak{su}(4)$. Why not $\mathfrak{so}(4,2) \times \mathfrak{su}(4)$?
@WaveInPlace OK, but one must look for a way to find the volumes V_1 and V_2, and this can be done via the PV^\gamma relation, but, as I said, this only works for reversible processes. How can this process be reversible, since it is not controlled and no external force is exerted?
@naturallyInconsistent Which ones? I can impose that the difference in internal energy in the two compartments does not change, but then what? How do I find the final temperatures of the two subsystems?
@RyderRude, I feel like we've been here before. "I don't know of anyone between these two time periods" =/= "There was no one between these two time periods". Galieleo didn't work in a vacuum.
@Bml Sigh, you have $p_A V_0=Nk_BT_A$ and $p_BV_0=Nk_BT_B$ and you want to find $V_A$ and $V_B$ subjected to constraints $V_A+V_B=2V_0$ and ideal gas relations after equilibrium has been established. It must be doable.
@naturallyInconsistent Sorry, don't I have to find the final pressures (which are the same, at least I think so)? Do you think finding them is as straightforward as you say?
@Bml Well, the question is somewhat ambiguous. There is one single final equilibrium position, which is not ambiguous, but the question seems to be wanting to get a metastable intermediate position, which is where the ambiguity arises. I'm going to sleep. Like now.
@RyderRude, I don't necessarily disagree. In hindsight a lot of ideas have been pretty destructive, closing off avenues of discussion and preventing advancement.
I mean, unless you've done a bunch of other spring problems like this where you're supposed to consider gravity, I'd say it's a safe bet that you aren't supposed to do it here, either :P
is there a way to write "Every natural number is of the form $4a,4a+1,4a+2,4a+3$ for some $n \in \Bbb{N}$" where $\Bbb{N}$ includes zero, but rewritten without the zero?
@ACuriousMind Thanks. What I understand is: Conformal transformations on the sphere or Mobius transformations are locally isomorphic to Lorentz transformations, and the particular transformation corresponding to scaling is a boost in one particular direction, so the group is SO(1,1). Is that it?
@ACuriousMind But what does it mean at the algebra level? If I know $x^\mu \partial_\mu$ is the generator---how to make them look like that of $\mathfrak{so}(1,1)$?
yeah, I feel like if I got on a natural sleeping rhythm I wouldn't need it as much. I just am the antithesis of a morning person so i'm relying on it. How did you get over that feeling @ACuriousmind did it get better overtime
@ACuriousMind I said $\mathfrak{so}(4,2)$ because that's the conformal group. They took a direct product of Lorentz group and the dilation group; I am wondering what makes them leave out the SCT and why wont these stuff mix with each other?
yeah I'm pretty nocturnal as well. While in quarantine during the pandemic, I managed to completely roll over my sleep schedule to a normal time simply by sleeping later and later. Like going from sleeping at 12am to 2am, 4am, ..., 12pm, 2pm,...,8pm. I think I decided it wasn't worth trying to fix it by the 5am mark.
how does the argument for $|\Bbb{N}| = |\Bbb{Z}|$ bypass the diagonalization argument for uncountable sets
I understand this argument but I wonder why $\Bbb{Z}$ has no issue
@Sanjana I think they're just wrong; the bosonic part of $\mathfrak{su}(2,2\vert 4)$ should indeed just be $\mathfrak{su}(2,2)\cong \mathfrak{so}(1,3)$
nvm I think it's because u can define a bijection from the even natural numbers to the negative integers, then odd natural numbers to the positive integers +0
@ACuriousMind Just the Lorentz group? Why not the conformal group and the R symmetry group? They didn't "define" bosonic subalgebra clearly: is it clear what they are referring to? I thought that it is just bringing all the "bosonic" generators all together...
and re: the definition - you are reading the article, not me, how am I supposed to read their minds whether they meaning something different by "bosonic subalgebra" than anyone else :P
I think, inorganic chemistry could become real fun, if yet another electron would appear, with exactly the same (or very similar) properties, but being a different particle as the electron itself. So, this e' would distract e electromagnetically, but pauli exclusion would be non-existent among them.
So, hidrogen with e' would be very similar to ordinary hidrogen. But, in helium there would be He without e' (or with double e'), and also He with an e and e'. This latter would behave... I have no idea, how would it behave
i see. im just beginning interactions and tong is discussing that the formalism we are building now only holds for weakly coupled theories so i wanted to see more physically what the domain of this is. i think the examples he focuses on are yukawa and $\phi^4$
idk for some reason i really feel disoriented if i dont know the domain of the content in discussion. maybe this is smth i should learn to get over
just look at everything you're looking at as toy models - parts of this will recur in the formulation of the Standard Model, but almost everything in a slightly more complicated fashion
does anyone know of a good example to wrap my head around: 1) every hermitian matrix is diagonalizable. 2) decompose a hermitian matrix into the sum of two non-commuting hermitian matrices. 3) the two matrices you are summing cannot be simultaneously diagonalized. 4) the sum, however, is diagonalizable.
for some reason I have this faulty idea that if the sum of two matrices is diagonalizable then there should be a way to diagonalize sums of matrices that sum to the matrix in question
@SillyGoose What trouble do you have coming up with an example? The sum of any two self-adjoint operators is self-adjoint, so just sum your two favourite non-commuting operators
well i know that this is mathematically true but i guess i was hoping for a more illuminating example, but maybe what i am thinking of does not exist :P
or maybe a more fleshed out reason for why this should be the case
the first thing I look at when I look for examples is in 2 dimensions and in particular the Pauli matrices
@SillyGoose so what is mysterious about $\frac{1}{\sqrt{2}}(\sigma_x + \sigma_y)$ being diagonalizable, but not simultaneously so with $\sigma_x,\sigma_y$
that's just the statement that the spin operators along different axes don't commute (the sum is the spin operator along the 45° diagonal between x and y)
i think in my mind, the simplest mechanism for a sum of two matrices $A+B = C$ to itself be diagonalizable is if both matrices $A, B$ are simultaneously diagonalizable. but I am having trouble seeing the actual, general mechanism for the sum of two matrices to be diagonalizable
this does not, in general, impose any restrictions on the two matrices it is a sum of
in fact, it's the other way around: Every matrix is the sum of two diagonalizable matrices (just split it into its upper triangular and lower triangular parts), and it's a special occurrence when those two parts are simultaneously diagonalizable
