what are some good books on thermodynamics my class will be using schroeder

any other recommendations?

I have commented multiple times that Schroeder is not good. There are some defenders too. You can search the multiple convo and multiple viewpoints up.

@nickbros123 Pippard is not covering remotely the similar topics. It is a good book to read, but it is very far away from a textbook.

@naturallyInconsistent yeah it's not a textbook; i think blundell and blundell may be worth a shot though, covers thermo, KTG, boltzmann and Maxwell stuff, near the end covers statistical stuff, what do u think

I didnt think Blundell & Blundell is great, but then again, I didn't pay too much attention there either. After all, I read the greats...

youre gonna say kittel....

why do i have to pay to view research articles pubs.aip.org/aapt/ajp/article-abstract/69/12/1280/529319/…

@nickbros123 Yes, I read the standard holy trinity, Kittel & Kroemer, Callen, Pippard, and more

Simultaneously, and whilst learning it for the first time.

Ill probably do a Mickey Mouse version of your treble with fermi, blundell and so on

hahha

I think Fermi's textbook is a bit small, isn't it?

@naturallyInconsistent yeah, it's just thermo

@nickbros123 When he wrote that text, there was just not that much in the syllabus. Nowadays, his book is covering way too little to be tolerable as a textbook.

His book is primarily useful to get a feel for the subject before a course. I will probably finish it on the long train journey from my home to my college

@naturallyInconsistent It's one fermi in width

lol

@naturallyInconsistent what makes it so unreadable/not good? just from reading the preface it seems nice in that it tries to provide a middle ground between thermo & stat. mech kind of easing you into both subjects?

guys, I have a question: let us suppose I initially have a free wave packet ($V = 0$ in the Hamiltonian) (Gaussian in my case, but this can be done in general, I did my computations under this assumption). Its representation in the momentum is space is given by the Fourier antitransform of $\psi(x,t = 0) = \psi(x,0)$.

the probability distribution associated with the momentum wavefunction will be constant in time since $[\hat{p},\hat{H}] = 0$

Now I want to determine the time evolution of the x-space wf $\psi(x,t)$. I know that $\phi(p,t=0) = \phi(p)$ is constant in time and I know, after performing some computations, that a plane wave, which is an eigenfunction of momentum and of the hamiltonian, evolves in time according to this law $$ \phi(p,t) = Ae^{\frac{i}{\hbar}(px-\frac{p^2t}{2m} )}$$

so the expression for $\psi(x,t)$ of my packet at some time $t$ can be obtained by making the time dependence of the single plane waves in the Fourier transform explicit, obtaining something like this: $$ \psi(x,t) = A \int_{\mathbb{R}}\psi(p)e^{i/\hbar(px-\frac{p^2 t}{2m})}dp$$

sorry not the wavefunction but probability density associated with it

My problem is that I can't interpret this result

I mean what's happening here, all these calculations are ok, but the meaning behind it is kind of fuzzy as of now

could someone explain to me why all of this is happening?

@ClaudioMenchinelli if by "dispersion" you mean that the position space wavefunction of a free particle "spreads out" if you leave it undisturbed, then that's correct

there isn't really some deep significance to this, it just shows that states that are well-localized at one point in time won't remain so forever

@ClaudioMenchinelli it's happening because of your calculation :) beyond that if you want a handwavy explanation for classical people I guess you could say that if you start with a well-localised distribution in space, the corresponding momentum distribution is spread out, so the real-space distribution spreads in all directions as time passes

@stringynonesense are you available?

@fqq But for a Gaussian wavepacket the uncertainty principle inequality becomes an equality so why does the dispersion in position become larger while the one in momentum is constant? Isn't this contradictory ?

Oh wait

I just realized that $\psi(x,t)$ has a gaussian shape only at $t = 0$ so for $t \ne 0$ the product will be greater than $\hbar/2$. It's the probability distribution that retains a gaussian shape at every instant

Lol, thanks to the Wikipedia page for letting me come to my senses

@JohnA. ya, I'm here

I would absolutely love it if you could post an answer to my post at the level of sakurai and napolitano

If you are not interested to do that, please tell me some improvements to my answer

But still in the scope that is intended (probably tangental mention of lie algebras, as its not usually discussed at this level)

Thanks so much for your time btw

ya, no problem

writing an answer now

Thanks!

@JohnA. added an answer. Hope it addresses your confusion. If not, let me know.

I saw and upvoted thanks!

Also happy to say you taught me something as well

Hi @PM 2Ring

Yes, so when we talk about "feeling something", in general we feel forces right? F = dp/dt

If someone pushes me with a constant force, and therefore I have a constant acceleration, i should be able to feel someone pushing me

Except with gravity, you don't feel a push or pull. Because every part of you is getting pulled identically, and no part of you is pushing on any other part.

I see, and what if gravity acts the same way on your entire body, but then the gravitational force is increasing more and more (for example due to the fact that the mass of the planet is increasing randomly), would you still feel this force?

i assume you would not feel this right?

Planet masses don't increase randomly. ;) Gravitation sources can only change smoothly. And changes to the gravitational field (that is, changes to spacetime curvature) propagate at lightspeed.

I know, but it's just an artificial example to illustrate a point

So you could feel the change in gravity, depending on how big & how fast it is.

according to your logic, if every point on your body accelerates the same way, you would not feel this, so that would imply that even if the acceleration increases of every point of your body in the same manner, you would not feel this. is that correct?

But in this scenario, we're assuming that the gravitational field is uniform. So there are no changes. And there are no tidal effects.

thanks i think i get it

@user46792 Yes, but the acceleration can't instantly change over your entire body

@user46792 Artificial examples can be useful, but they aren't allowed to break the rules. ;) In GR, mass-energy isn't allowed to just appear or disappear. See en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor

Is there any other "solid" reason for considering $B_{\mu \nu}$ and the dilaton $\phi$ in the DBI action other than including them just because they appear together in the closed string spectrum?

@Sanjana all the low-energy QFTs in string theory come from "amplitude matching" - both the 10d SUGRAs and the brane DBIs come from matching their amplitudes to stringy amplitudes; however there's an alternative path to the brane worldvolume actions via the Green-Schwarz mechanism

@ACuriousMind Setting the beta function calculated from the various superstring worldsheet theory gives the various SUGRAs. What should the DBI action give if such a thing is done?

@Sanjana I'm not sure I understand the question. This, too, is a "beta function computation", just for open strings ending on the D-brane. See any of the references on nLab for the actual arguments

@ACuriousMind Thanks for the references but why is the beta function computation using DBI action a string computation? It is an action for the brane and not the strings ending on them, right? The embedding coordinates are that of the brane and not the string, etc.

@Sanjana For a "space-filling brane" at p=9, it's just a QFT on spacetime like the usual SUGRAs; the other brane actions are dimensional reductions of this theory

@Sanjana if your problem is that the brane worldvolume theory is confined to the worldvolume: This is supposed to be an effective theory for the low-lying massless states of the open string, which is just the brane gauge field $A$ and a bunch of scalars (and their superpartners in superstring theory), which are both fields which are perfectly well "live" only on the brane, so you should expect the effective field theory to be given in terms of fields on the brane

finished reading this interesting paper about how complexity and novelty could appear and be detected. It's probably too simplistic, but nevertheless interesting. In case anyone wants to take a look here it is nature.com/articles/s41586-023-06600-9

i've made a summary but no one will donwload a 'xournalapp' file I think

i may do a gist-summary soon