Why don't the "Does this answer your question...?" comments get deleted when I press delete

they seem to return from the dead

with an edit symbol

do I smell mod intervention

@NiharKarve sounds more like a bug, does this happen reproducibly?

oh never mind about the edit symbol, that's there for all auto-generated dupe flags

@ACuriousMind I don't really have any concrete evidence, but I swear it's happened to me a few times

no biggie

@ACuriousMind may I suggest that you put this message in the room description

or some summary, there of

@user85795 it's already there - it's part of the guidelines

but that's not as visible as the room description

"Just ask" is misleading, in that sense

Perhaps, a link to the problem solving room for homework type and newly posted questions.

As an aside, having a meta feed that posts newly posted questions directly into the room seems like a double standard

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

if meta had a flood of questions, the room would be flooded with new questions.

sure

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

sure

Don't ask about asking, just ask sounds like an open invitation to ask whatever you want, whenever you want.

noobs don't care about the further guidelines

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 ?

Are you asking how you do it mathematically or how you do it in real life?

You can compress a fluid with, say, a piston for example

So , @Charlie which factor would change.volume or mass of liquid.

Volume shouldn’t change I think

Water can flow but how can you compress it

and make its volume less

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

@Charlie I feel in my book it is incompressible so that we can solve question easily

in real life , it is compressible right

Do you mean to say this @Charlie that in both the cases , density is different

@Charlie The Dirac time evolution mixes states of different chirality, not particles and anti-particles

@Charlie you can consider them in 3D

I talk a bit about chirality and antiparticles and their mixing in this answer of mine

@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.

Confusion in vertical equilibrium

It says online if flow of system is upwards , to balance it or to not let it go up.

It is vertical equilibrium

is it correct

Hi All.

@NiharKarve well, if they're deliberately anonymous, I think you shouldn't be trying to publicly de-anonymize them, right?

I understand the sleuthing impulse but the polite thing is to leave them be

@ACuriousMind you're right

I apologise

No worries :) But let me move this out of public sight nevertheless

Good idea

6 messages deleted

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

@Charlie depends on what you mean by "identical" :P

both spin angular momentum and orbital angular momentum are $\mathfrak{su}(2)$ representations, if that's what you mean

The only way of forming an equivalence I know is through invertible intertwiners

"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

yeah hmm

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

Spin 1/2 is 2-dimensional in the complex sense, though

as in, if I look at the hydrogen atom, the OAM with just an $s$-orbital and $j=0$ is 1 dimensional, but the next level with $j=1$ is three dimensional

oh

You can cast all other spins in complex form, if you want

In which case everything is $(2j + 1)$-dimensional

Or wait

Hm

Spin 1 particles are represented by two spin indexes

So 4-dimensional, I guess?

Although not quite since that's not an irrep

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

It's $2 \otimes 2 = 3 \oplus 1$

The complex version of spin-1 particles is symmetric 2x2 complex matrices

what does "complex version" mean here?

There's an isomorphism between the $3$ irrep of $SU(2)$ and the usual Minkowski space

oh that formula you wrote refers to the dimensions, I'd never really thought about it until now

Errr

@Charlie see physics.stackexchange.com/q/153369/50583

SU(2) so it's just Euclidian space

Or something

I guess it's hermitian and not symmetric

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

spin half and spin 1 are inequivalent irreducible representations of $\mathrm{su}(2)$

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 $su(2)$

I'd say you can say yes to that

@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

x and p is the Heisenberg algebra

$\{ \hat{x}, \hat{p}, \hat{I}\}$ to be exact

well, if the answer was no my follow up was is $su(2)$ a subalgebra of the Heisenberg algebra?

@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

there's a lot of dimensions

Kinetic energy is a polynomial in them right, $\hat{p}^2/2m$, but you never cared about the algebra of polynomials in $x$ and $p$ to work with them

@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.

hmm

and in principle you also know how $x^{10} p^4 + x^9 p x p x$ acts

The angular momentum operator obeys the rotation group algebra, so I guess su(2) is a subalgebra of the (2 x 3) + 1 dimensional Heisenberg algebra?

I've never given it that much thought, but I guess if asked I would have just said it's the composition of maps

@Charlie it is!

ye :p

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 $\hat{H}$

@Slereah no, the Heisenberg algebra is really just the $x_i$, the $p_i$ and the identity

there's no multiplication in it to form the polynomials

@ACuriousMind How do you get the commutator then

@Slereah it's a Lie algebra

But $\hat p^2$ doesn't make sense pre-representation right, assuming it's a lie algebra rep, there's no map composition there

Lie algebra have magic Lie brackets, not "commutators" :P

you need to form the universal enveloping algebra if you want to talk about the bracket as a commutator

let's not

it's been a long day

i'd like to second that

The Hamiltonian is a polynomial in the universal enveloping algebra right

I think I am one layer of abstraction away from some kind of major psychotic break

It just means the algebra generated by the Lie algebra basis vectors such that the commutator relations hold

Welcome to physics

@Charlie yes, on the pure Lie algebra level it does not exist

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

Am I right in saying that the direct sum of all of the integer $j$ reps we use in OAM isn't the entire Hilbert space

Please tell me why is this true

@Charlie depends on whether your particle has spin or not

You need to put your equation inside "$'s" @PrateekMourya

The Hilbert space can be many things

in general, for a single particle, the Hilbert space will be something like $\mathcal{H} \times C_{j}$

The "basic" Hilbert space of square integrable functions and the irrep of $j$

Oh I mean ignoring spin yeah

I forgot for a second there we have to tensor it on

You can take those individual Hilbert spaces and combine them in various ways

@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

Why v.v is zero

They are saying the average of the dot product of the two particles' velocities is zero

How?

if you put two particles in a box and let them bounce around, over a long period of time the average of the sum of their velocities will be zero

On average they move in directions completely orthogonal to one another, think of two pool balls colliding then scattering in orthogonal directions

How do get this that average angle is 90°

@ACuriousMind ah yeah of course

I tried int(vrev(dtheta))/int(d(theta))

that looks like python code no?

oh no nvm lol

Sorry i don't know mathjax

$$\frac{\int v_{rel}\text{ }d\theta}{\int d\theta}$$

Limts (π,0)

I got 2v

But only on 42° i got approximately √2

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

Is there any mathematical proof in stat mech?

I can then simply then move on and learn it in university in future

V.v is v^2 costheta

Inetgral of costheta dtheta over integral dtheta

On interval 0-π is zero

Is this proof ok?

@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

What if we assumed all to move at average speed?

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

Thanks acm

Thats most acceptable easy to understand maths

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.

@ACuriousMind from what theory did you said that dot product is proportional to theta1-theta2

Stat mech?

@PrateekMourya that the dot product is proportional to the cosine of the angle between the vectors is just basic geometry, is it not?

Ya but velocities are random

(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

In the integral then two velocity will be taken out

@PrateekMourya I don't understand what that's supposed to mean

the function whose average we have to proove zero is v1.v2.cos(theta1-theta2)

Rather than explicitly computing it, isn't the average dot product required to be 0 to not have a net movement or is that just bad intuition?

Ya but taking all particles to move at average speed still works and simplifies

@DanielUnderwood that's entirely the correct intuition but they asked for maths :P

Will that be a good assumption?

To suppose all to move at average speed ?

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

Sorry but i only know that all assumption of ideal gas are based on statistical mechanics

Not a bit of theory

Thats why i was asking if i will learn it in future

If they're statistically independent, can't we just use that

@PrateekMourya How are we supposed to know whether you will learn it in the future or not?

$$E[v_1 \cdot v_2] = \sum E[v_1^i] E[v_2^i]$$

Same distribution

I mean all the explanation to assumption i will learn in future when i will learn stat mech

and if it's angularly independant it should be zero, vector-wise

@Slereah I think in order to really prove that and not just argue with intuition you'd have to do essentially what I did

mb

@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)

I think i should leave it for now

Till i have enough knowledge

I am not asking if one knows what i will learn

I am asking that this particular derivation

Is it given in stat mechans with proper proof

Most texts will probably argue with intuition as the one you quoted and expect you to figure out the formal proof yourself if you need to

there is no one "statistical mechanics" reference we could talk about, there's just different texts

Where is the holy bible

Ok

Thanks

Hello guys.

Pls clear.. If mechanics book said moment about point does it mean point is fixed for rotation as in pin joint.

What if point is not fixed does object rotate about center of mass