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12:28 AM
the thought occurs that the speed of sound is actually a random variable
due to the mean free path
2 hours later…
2:20 AM
@antimony I think it's not what you have in mind, but tandfonline.com/doi/abs/10.1088/0959-7174/4/4/001
2:41 AM
ooh interesting, thanks @fqq :)
2 hours later…
4:20 AM
is fundamental of physics by Halliday taught to physics freshman?
6 hours later…
10:15 AM
@NiharKarve I've tried Mathematica a lot of times for integrals and ODEs
It doesn't work that often tbh
I recommend this instead : sciencedirect.com/book/9780123849335/…
10:32 AM
Ah, I'd forgotten about Gradshteyn and Ryzhik
yeah it's really good
"The perturbation series in QFT is an “invention of the devil” and this is actually a good thing"
Good title
11:14 AM
In the free supersymmetric Wess-Zumino model, doesn't the fact that the supersymmetry algebra doesn't close correctly off shell mean that this isn't a supersymmetric field theory? I thought that a symmetry that only closes correctly on-shell was literally how we defined a trivial symmetry that isn't actually a symmetry since all infinitesimal variations on-shell vanish?
I see that you can add extra terms to the Lagrangian to correct this, maybe the thing I'm looking at is more of a toy model for introductory purposes
You can add terms to Wess-Zumino to get it to close off-shell, and the superfield perspective gives those terms automatically
oh I've not heard of a superfield
Wait is that $F$ and $F^*$ "auxiliary" fields that correct the non-closure?
ok that's interesting
When you compute the susy commutators on the fields you find you need to use the eom if you don't include them, but including them cancels those terms out, just by a basic degrees of freedom counting argument you can see one should be including them from the beginning
11:21 AM
does anyone know if it's worth attending a few presentations from the Strings 2021 Conference?
As far as I can tell, it's going to be all online this year
Is the wess-zumino model used to model an actual interacting that is being searched for in supersymmetry experiments at the lhc and stuff? or is it just a simple examples of a susy field theory
I'll watch anything that has ed witten talking about stuff in it, I think he's awesome
@Charlie no, the WZ model is simply one of the simplest possible theories with SUSY
it's like the free scalar field :P
@Charlie Yes, I was actually motivated after someone linked to Witten's presentation slides for Strings 20xx - it's not forbiddingly difficult, and I think they do aim to make the talks accessible to wider audience
Experimental stuff involves broken susy
Q: Why must SUSY be broken?

Federico CartaBackground One usually claims that supersymmetry must be spontaneously broken. The reasoning is roughly the following: Since $M^2=P^{\mu}P_{\mu}$ is a casimir operator of the supersymmetry algebra, all the particles in a supermultiplet will have the same mass. Therefore the electron and the sel...

11:24 AM
ed witten tops my list of favourite physicists, even above feynman don't @ me
I saw something or other about spontaneous SUSY breaking but seems complicated
luckily I think intro-SUSY is one of those fields that has an abundance of high-quality lecture notes
I like Bertolini (pdf warning)
Am I wrong for associating "non-linear sigma model" with the polyakov action approach in string theory?
is that just one context in which that's what it means or are those two things synonymous
"non-linear $\sigma$-model" just means that your fields are valued in a manifold, just like the worldsheet coordinates are valued in their 10 target space
it has nothing inheretly to do with string theory, the Polyakov/NG actions are simply examples of non-linear $\sigma$-models
oh ok that's interesting
so kind of like whenever you're considering a theory on a manifold that is being embedded in another (larger) manifold?
yes, it mostly comes up in contexts like "thing with shape X is constrained to move on Y", then your fields are the coordinates of the embedding $X\to Y$, most commonly it's just a particle moving on a manifold $Y$.
11:32 AM
pretty neat
A wave function is represented as $$\psi(x,0) = A(a^2-x^2)$$ for x in $[-a,a]$, and zero otherwise. Griffith then asks you to calculate $<p>$
and then states "You cant get it from ehrenfest's theorem"
why not?
( I am sorry if I interrerupted an ongoing discussion)
Is it maybe because you've defined a piecewise wavefunction? I.e. you've had to put in by hand that the wavefunction vanishes outside of a particular region
Your $\psi$ above satisfies the boundary conditions, when $x=a\rightarrow\psi=0$, but in general that function is non-zero outside of the closed region $[-a,a]$, so if you compute $\langle p\rangle$ in that state you're not accounting for the fact that you've manually set the WF equal to zero outside of the box
actually no sorry, even though that's true I don't think it's the reason ehrenfest's theorem don't apply idk
@satan29 how are you going to calculate the time derivative of $\langle x\rangle$ :p
@satan29 Unless you are also given the Hamiltonian with which to evolve this wavefunction, how would you apply Ehrenfest?
@NiharKarve oh wow
XD i thought theres some deep fundamental reason for the theorem to fail
the wayy griffiths put it in caught me off guard...
11:48 AM
Griffiths tells you to derive the value of the fine structure constant from first principles as an exercise in chapter 6
then reveals that its a joke on the next page
I think we've had "homework-like questions" for that exercise at least twice :P
I wonder if he can sleep in peace knowing how many undergraduate hours he's sucked away with that one
I think he's still alive :P
ah, sneaky edit
that was ... terrible wording
11:52 AM
no but wait
for my question,
do we neeed to calculate:
$$ \int_{-a}^{a} \psi^{*}\hat{p} \psi dx$$
what does $p$ mean if your function is in position space?
i meant p operator, sorry
then sure, that's the definition of the expectation value
also: note the bounds of the integral, maybe you won't have to fully evaluate the integral ;)
the wave function is even
is that what you are poining to?
the integral is coming out to be complex :(
no wait, zero
12:05 PM
a very similar issue is discussed here
12:18 PM
I see, thanks.
actually, in my case, i got an i multiplied by a real value, so I thought its complex... but it turns out that the real value is zero.
it may well be that the "real" value doesn't even exist, as discussed in that question you have to deal with the discontinuities properly
but in my case, the wave function is continuos right?
oh, you're right
sorry, in that case the linked question is pretty irrelevant
no, its still interesting
thanks, i gtg now. thanks @NiharKarve too!
2 hours later…
2:03 PM
hey chat.
I'm a math undergrad. Took advanced linear algebra and I know what an element of $V \otimes W$ is supposed to be but I still don't know what the hell physicists are talking about when they say that "a tensor is something that transforms like a tensor". Like... a tensor is an element of a vector space $V \cong \bigotimes\limits_{i=1}^n V_i$? What is that supposed to mean either way?
@LucasHenrique when physicists talk about "tensors", they usually do not mean the generic notion of mathematical tensor products, but they mean a tensor field on a manifold $M$, i.e. sections of $(TM)^{\otimes n}\otimes (T^\ast M)^{\otimes m}$
and just like you can define a vector (field) on a manifold as a tuple of numbers showing a particular transformation behaviour under coordinate changes, you can define a tensor field the same way
ok, so I need to finish my analysis on manifolds classes first. :p
at least to the point where you define a vector field, yes - there are like three or four different common definitions of vector fields, and the one via transformation behaviour is one of them
string theory definitely exists
so how could it not be real :P
2:20 PM
what the heck
wait, that's how you unify QFT and GR? What a stunning argument, you should get that published!
@LucasHenrique if you write something like $T_{\mu_1 \mu_2} = T(e_{\mu_1},e_{\mu_2})$ you can see saying a tensor transforms like... is basically just saying it changes under a change of basis in each argument as usual
3:07 PM
A: Why quantum mechanics?

Urs SchreiberI am late to this party here, but I can maybe advertize something pretty close to a derivation of quantum mechanics from pairing classical mechanics with its natural mathematical context, namely with Lie theory. I haven't had a chance yet to try the following on first-year students, but I am pret...

where has this answer been my whole life
@NiharKarve the longer you think about this answer, the more it's just a very convoluted way to say "the Poisson bracket looks like a commutator if you squint"
it's one of the better written motivations for the mathematical structure of geometric quantization, but in the end the main "wow" effect here is achieved via calling some arbitrary elements $\mathrm{i}$ and $\hbar$
The sad part is that all quantum theories aren't realised as "quantizations" of classical data (?)
@ACuriousMind I was just happy to find out about the extension of the Lie algebra of Hamiltonian vector fields, I never knew that
@NiharKarve depends on your field, really. In practice, all QFTs are quantizations (but not formally understood ones :P), but none of the finite-dimensional systems of quantum information are
3:22 PM
ah, interesting
but in practice no one does geometric quantization anyway unless the naive canonical approach produces garbage, which is rare
and where we would really like to have a proper quantization procedure (i.e. for field theories), we don't have one, either
I mean there is no known formal quantization procedure that produces well-defined quantum theories for arbitrary field theories
Could I just check real quick, the dimension of the gamma matrices being $2^{d/2}$ with the exponent rounded down, that $d$ is the dimension of the Clifford algebra, not the representation right?
Maybe that's somewhat obvious, but I wanted to check that we are actually dealing with different representations of different clifford algebras, not just different representations of the same one
@Charlie The $d$ is the number of spacetime dimensions and hence the number of gamma matrices
the dimension of the representation is $2^{\lfloor d/2\rfloor}$, that's what the size if the matrices tell you :P
And different dimensional spacetimes have associated to them a different clifford algebra
ok, just sanity checking
2 hours later…
5:22 PM
@ACuriousMind I have a question about this
I once read that geometric quantization does not work for the harmonic oscillator
am I remembering correctly?
@RyanUnger you're probably thinking of the fact that you need a metaplectic correction for some geometric quantizations to work correctly, I think the HO is one of these cases
6:18 PM
It's a sign that I have been doing physics for a while that I haven't thought that "Fock space" sounds dirty in forever
@Slereah XD
no matter how mature you are, you gotta atleast chuckle at that
there is no reason why quantum mechanics
quantum mechanics just be
if the universe was different, quantum mechanics would not be, but that math argument would be the same
3 hours later…
9:26 PM
intro physics lab sanity check. my students will be putting together the following setup next week:
to do that at home, they're using simple plastic pulleys with nails (taped to a flat table) as axles
for practical purposes, is there actually a difference between the pulleys versus just resting the string on the nails?
a few of my students seem to only have one of the pulleys, but if they have both nails then I think the setup still works
(it'd be even easier if I could just tell them to hammer in the nails somwhere but I probably can't)
you'll have more losses due to friction on the nails I'd say, but it's not a fundamental difference
@ACuriousMind this is actually just supposed to be an equilibrium scenario: how much does the middle mass make the string sag
so i think even that doesn't matter here
that said, friction does matter insofar as it makes it easier to reach equilibrium in the 'wrong' position. so not a loss so much as a source of uncertainty

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