« first day (3730 days earlier)      last day (41 days later) » 

9:35 AM
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
 
9:47 AM
@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
 
10:30 AM
@ACuriousMind may I suggest that you put this message in the room description
17 hours ago, by ACuriousMind
@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.
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
 
10:53 AM
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
 
11:27 AM
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
 
11:44 AM
@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, $p^\mu=\hbar k^\mu$
 
12:05 PM
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
 
123
12:26 PM
Hi All.
 
 
2 hours later…
1:56 PM
@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
 
 
1 hour later…
3:28 PM
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
 
3:39 PM
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
 
3:43 PM
oh that formula you wrote refers to the dimensions, I'd never really thought about it until now
 
Errr
 
SU(2) so it's just Euclidian space
The isomorphism is just $$\begin{pmatrix}
z & x + iy\\
x - iy & -z
\end{pmatrix} $$
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 $\mathrm{su}(2)$
 
3:51 PM
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?
 
4:00 PM
@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
 
4:04 PM
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 $\hat{H}$
 
4:06 PM
@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
 
4:07 PM
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
 
4:10 PM
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
 
4:11 PM
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
 
4:13 PM
They are saying the average of the dot product of the two particles' velocities is zero
 
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
 
4:16 PM
I mean some maths?
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
 
4:22 PM
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.
 
4:38 PM
@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
 
4:44 PM
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
 
4:48 PM
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
 
4:51 PM
@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
 
4:55 PM
Ok
Thanks
 
123
5:34 PM
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
 
 
1 hour later…
6:41 PM
0
Q: What constitutes a 'near complete' solution

user256872My 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...

 

« first day (3730 days earlier)      last day (41 days later) »