@Physicsismylife Please don't post your questions here directly after you asked them; interested people watch the main site anyway, and if everyone did it, the room would be flooded with new questions.
Tbh I doubt people who come to the chat to post their question read carefully through the rules beforehand, also it's not like there's an enormous flood of people posting their questions, it's a few a day max, and worse case the messages can be removed by a mod
Meta posts are relatively few and far between, I can't give you a satisfying answer as to why they're posted here, but either way neither of these things is really that big of a deal
Just on the solutions of the Dirac equation again, do the electron and positron solutions transform into each other under the time evolution of the Dirac spinors? This isn't a problem in $\Bbb C^2$ in qm since this is just a linear combination of two spin states for a single particle, but if we have a linear combination of electron/positron solutions, is this not an issue? My guess would be either LC's of $e^-$/$e^+$ solutions are non-physical, or the time evolution doesn't interconvert
them
which would be strange because I've seen that the mass term mixes all 4 components of the spinor
Guys , it is assumed in my book that fluids are incompressible i.e they’re density is constant even if we try to change it . My question is that if I wish to change the density of a fluid , how would I do it by applying pressure ? How can pressure change the density of a fluid ?
Well, you can't increase the mass of a liquid, you can reduce the volume of the container it is in (unless your fluid is incompressible, or the container isn't full)
If your fluid is incompressible and you have a container full of that fluid you won't be able to compress it
@Physicsismylife No, if both of those containers are the same volume and are full of that blue liquid and the liquid is incompressible then they will have exactly the same density.
Incompressibility is a (often good) assumption that simplifies the mathematics of fluid dynamics a lot. In real life all liquids are (at least within reason) compressible, yes.
@ACuriousMind Ah I see, thank you I will read through that
Also, and I'm 99% sure I know the answer to this but as someone fairly uninitiated in physics I feel I should just check, we use the phrase "positive/negative energy solution" and "positive/negative frequency solution" completely synonymously in physics, right? Because of the de Broglie relation between 4-momentum and 4-wavevector, $p^\mu=\hbar k^\mu$
Why is there H,N and CO attached to thesuffix of equilibrium constants?Particularly how to use these as suffixes?Nothing can be extracted from the reaction.There are many H,N and so on like C and O so how do I know which one to use as suffix here for getting the equilibrium constant.
The spin representations of $su(2)$ that we use in QM are identical to the angular momentum representations used in say the hydrogen atom right? And by extension the reason we can so easily talk about total angular momentum as being spin, orbital angular momentum or any combination of the two is because we're just taking tensor products of identical representations
"Identical" and "isomorphic" to me are two different things
sometimes we work in contexts where we consider things that are isomorphic identical, but if I have a spin-1 representation $V$ and some orbital-angular-momentum-1 representation $W$, I would say the total state space is $V\otimes W$ and it still matters which of the factors is spin and which is orbital for some questions, so I wouldn't call $V$ and $W$ identical, even if they are isomorphic as $\mathfrak{su}(2)$ representations
something that is slightly bothering me, is that there's the spin-1/2, 2-dimensional rep of $su(2)$, but with the invariant subspaces of the orbital angular momentum operators, none of them are two dimensional
maybe I'm just getting hung up on a label, but the spin reps are labelled by $s$ and the OAM reps are labelled by $j$, but $s$ takes half integer values and $j$ takes integer values
in the case of orbital momentum we don't start with some symmetry group and just allow random representations, we start with the representation of $[x,p] = \mathrm{i}\hbar$ and observe it decomposes into integer-valued angular momentum representations for the orbital angular momentum algebra $J_i = \epsilon_{ijk}x_jp_k$
there just are no half-integer values of orbital angular momentum in this decomposition of the standard CCR representation
So the irreps of $su(2)$ are labelled by $j=1/2$, but the ones that are the invariant subspaces of the OAM operators are only those for which $j\in\Bbb Z$?
in other words the spin-1 rep of $su(2)$ is the same rep as the $j=1$ rep of OAM
I'd be happy with that, if the QM Hilbert space only decomposes into invariant subspaces of the angular momentum operators which are only the integer labelled reps of $su(2)$
@Charlie yes, but I'm not sure why you call one of the algebras here "$\mathfrak{su}(2)$" and the other "OAM"
both algebras are $\mathfrak{su}(2)$, just one is the spin algebra "living alone" and the other the orbital angular momentum algebra sitting inside the larger algebra of observables generated by $x$ and $p$
The "larger algebra of observables generated by x and p", this is not just the Heisenberg algebra is it? This is the $C^*$-algebra thingy that is more complicated
@Charlie I essentially mean the algebra of polynomials in $x$ and $p$ there, whose representation is induced by the representation of the $x$ and $p$ of the Heisenberg algebra
I've never encountered algebras of polynomials like that
I still try to think way too much in terms of pictures :P it was a strategy that half-worked in chemistry but seems to backfire quite a lot in maths/physics
@Charlie don't think too abstract about it - "the representation of polynomials in $x$ and $p$ is induced by the representation of $x$ and $p$" just means that if you know how $x$ and acts and you know how $p$ acts, then you also know how $xp - px$ acts.
The harmonic oscillator Hamiltonian is a polynomial in the algebra of the form $\hat{H} = \hat{p}^2/2m + k^2 \hat{x}^2/2$ and the commutation relations between $\hat{x}$ and $\hat{p}$ matter, it can be written in different forms if you try to factor $\hat{H}$
In equilibrium, {\displaystyle \mathbf {v} _{1}}\mathbf{v}_1 and {\displaystyle \mathbf {v} _{2}}\mathbf{v}_2 are random and uncorrelated, therefore {\displaystyle {\overline {\mathbf {v} _{1}\cdot \mathbf {v} _{2}}}=0}{\displaystyle {\overline {\mathbf {v} _{1}\cdot \mathbf {v} _{2}}}=0}, and the relative speed is
In a Lie algebra you can't tak about $AB$ you can only talk about $[A,B] = AB - BA$ so the enveloping algebra is an extension where you can talk about things like $AB$ but e.g. $ABA$ is equivalent to $AAB - A[A,B]$ where $[A,B] = C$ is some other element in the Lie algebra
@Charlie what is true is that the standard CCR representation in 3d $L^2(\mathrm{R}^3)$ is the direct sum over all finite representations of integer angular momentum $\ell$ inside it, this is just the statement that the spherical harmonics form a basis
If $\mathbf{v}_1 \cdot \mathbf{v}_2$ wasn't zero on average, it would imply that the velocity of one half of the particles on average would depend on the velocity of the other half of the particles
@PrateekMourya You have to realize there are two independent random variables here - the velocity of particle one and the velocity of particle two, it is not the angle $\theta$ between that that is randomly distributed
if we are in 2d, then we can represent the directions of the particles by two angles $\theta_1$ and $\theta_2$ that are uniformly random since no direction is preferred. The dot product is then proportional to $\cos(\theta_1 - \theta_2)$. So you would have to evaluate $\int_0^{2\pi}\int_0^{2\pi} \cos(\theta_1 - \theta_2) \mathrm{d}\theta_1 \mathrm{d}\theta_2$.
this results in zero since you can switch to coordinates $\theta_1 - \theta_2, \theta_1 + \theta_2$ and then it's just the integral of the cosine over its entire period, which is zero
note that this does not mean the "average angle" between the particles is 90° - averages do not commute with other functions, the cosine of the expectation value is different from the expectation value of the cosine
E.g. if you compute the average angle as the average of $\theta_1 - \theta_2$, you just get 0.
(Fun fact, if you compute the average magnitude of the difference in angle as the average of $|\theta_1 - \theta_2|$, you get $\frac{8\pi^3}{3}$ according to WA, which seems a bit large)
@PrateekMourya what does that have to do with how the dot product is computed?
The random direction of one particle is $\theta_1$ (relative to some arbitrarily picked reference axis), the random direction of the other is $\theta_2$ (relative to the same axis). So the angle between them is $\theta_1 - \theta_2$ and that's what we have to use in the dot product
well, why are you doing statistical mechanics in the first place then if you're going to throw away the statistics and assume everything is average anyway? :P
@PrateekMourya No one here can know what you will or will not learn - we have no idea what you're studying or where (and even if we knew we might not be able to say)
My answer (posted below) was deleted from the following question: Calculating mass of unknown object using center of mass. The deleter's comment read, "I'm deleting this in accordance with our homework policy. Please do not give complete or near-complete answers to homework-like questions."
Answe...