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12:33 AM
@lucabtz haven't read it all, but it cites Zamolodchikov so it must be good :P
12:54 AM
@lucabtz aww, I dont think these are the Elliptic Hyperbolic group of functions that solve reciprocal square root integrals of quartics. Those form a nice little group that have been recently unified under a new banner.
1:43 AM
5 hours later…
6:13 AM
this article seems interesting: ams.org/journals/notices/202402/rnoti-p174.pdf
6:52 AM
@SillyGoose mews in opposition of AI
7:23 AM
I just found this place, the name's absolutely fantastic
@AlanFox86 hi hi, been seeing your questions quite a lot
7:52 AM
@naturallyInconsistent roars in opposition of AI
8:40 AM
@AlanFox86 they could equivalently call it 1 too
Dyson does not like fundamentalism and theories of everythings. He talks about the older generation of physicists who liked fundamentalism
and later he talks about the quantum-classical divide. he changed his mind about QM and completely agrees with Bohr's interpretation
in the earlier days, Dyson was on about consciousness and stuff. but he changed his mind to quantum classical divide
he says that the previous generation discovered qft and his generation cleaned up the math
9:16 AM
@RyderRude He is completely misinterpreting Bohr when he tries to pretend Bohr was somehow saying a classical theory can exist that does not need to be quantized as a way to try to say that gravity doesn't necessarily need to be quantized it can somehow exist only as a classical theory which doesn't need to be quantized
It's really amazing to see one of the pioneers of QED, quantizing the electromagnetic field, who talks about gravitational waves (i.e. a field-theoretic aspect of gravity that was basically predicted based on on the field theory analogy between free EM and free GR giving wave equations), to then say this gravitational field theory may not need to be quantized and misquoting Bohr to do this
Apart from that the rest of it is great
I have some confusion about a proof in Hall. So Hall presents the theorem that a Lie algebra homomorphism induces a Lie group homomorphism for simply connected Lie groups. to show this, he constructs a five step proof. the steps are 1) define $\Phi$ in a neighborhood of the identity 2) Define $\Phi$ along a path 3) Prove independence of the partition 4) Prove independence of the path 5) Prove that $\Phi$ is a homomorphism and is properly related to $\phi$. [...]
[...] I have a few issues with this proof. First, what is the logic of step 2? I know we want to show that there is a $\Phi$ and that it is related to the Lie algebra homomorphism. I assume that we should need to use the simply-connectedness somewhere, but I'm not seeing the idea behind this approach.
Here, $\Phi$ is the Lie group homomorphism, and $\phi$ is the Lie algebra homomorphism
@bolbteppa yeah i think Bohr wouldnt have agreed with that part. But Dyson also gives an experiment to show why gravitons can never be observed
@Relativisticcucumber The simply connectedness will probably be relevant in step 4, where you show that $\Phi$ only depends on the homotopy class of the path. Simply connectedness then says all paths have the same homotopy class, so $\Phi$ is independent of the path.
Yeah I'd like to see more about that graviton experiment
9:33 AM
@naturallyInconsistent oh i see
blu lu blu lu lu
@naturallyInconsistent H O N K
@naturallyInconsistent omg u reinforce schweg XD
9:48 AM
@ACuriousMind but why do we even want to define $\Phi$ along a path
@naturallyInconsistent MERP
im confused whats going on?
@Relativisticcucumber M E O W
@Relativisticcucumber Why not? We will use it to obtain that it is actually independent of path, which is a rather important result.
@Relativisticcucumber Meanwhile, the reasoning behind why we define $\Phi$ along paths is that relation between the Lie algebra and the Lie group again - remember the Lie algebra is all the left-invariant vector fields,
and you can take a neighbourhood of the identity and reach any neighbourhood in the group by moving it along the flow of those vector fields. So what you get is that for any point $g\in G$, you take a point $g_0$ in the neighbourhood of the identity, and connect it to $g$ by some path $\gamma$, then you "translate" $g$ along that path to get to $g_0$, apply your definition of $\Phi$ in the neighbourhood of the identity, then translate the result back along tha tpath
oh no the flows are back
9:58 AM
@Relativisticcucumber Actually, it is pretty nice. Lie groups are usually curved manifolds, but Lie algebras are the tangent space of the Lie groups at the identity element, i.e. flat. Working in flat spaces is a lot nicer.
I'm very much handwaving here, I don't know if Hall phrases the proof in those terms, but that's what's happening on a structural level
at 3:20 and then at 6:00, u can see the big questions
the question at 6:00 is big, but it's still imprecise
@ACuriousMind i feel i am having a hard time uniting these notions. so i see $\frac{d}{dt}e^{tX} \bigg|_{t = 0} = X$ and I know there are two defs of a Lie alg that Hall presents: 1) the collection matrices corresponding to one parameter subgroups for a group and 2) a vector space with a bracket where the bracket satisfies bilinearity, jacobi identity, and I think asymmetry? [...]
[...] but conceptually I am having a hard time uniting it with the concepts you mention here which remind me of our discussion we had related to topics in Diffe g when discussing some GR stuff
@naturallyInconsistent i do see that this comes from the derivative that i put in the last message but idk its not all coming together for me
the brain scientist answers the identity question in affirmative
@Relativisticcucumber What do you want to come together?
10:08 AM
@naturallyInconsistent I see what you mean but in mathematical terms i think the most natural connection on a Lie group is the Maurer-Cartan connection which is actually flat if im not mistaken (?) @ACuriousMind help
@lucabtz meow~
@naturallyInconsistent wof wof
@ACuriousMind wait what exactly do you mean by the left-invariant vector fields
@Relativisticcucumber oh i will reply to this
on a Lie group $G$ a vector field $X \in \mathfrak{X}(G)$ is left-invariant if $d \ell_g (X) = X$ for all $g \in G$
$\ell_g$ is the left translation i.e. $\ell_g(h) = g h$
@ACuriousMind Trying to figure out if you can do AQFT for classical spacetimes
From what I can see, if you try to take the direct limit of some sequence of open sets in a spacetime to a point, it will generally converge to a trivial algebra for the relativistic case
Do you think maybe this would not be the case in general for a Galilean spacetime, and maybe you could define the causality condition as some direct limit case
In which case you may not need to use open sets to talk about the spacelike separated regions
10:32 AM
@Relativisticcucumber all these formal theorems about compactness, connectedness, etc... yet none of it addresses the most basic stuff like where the basic axioms of a lie algebra come from, I recommend picking up a physics book on this stuff instead
Georgi basically derives the commutator from this matrix group element thing, Jacobi is very hard to find anywhere, from this perspective it comes from looking at associativity but it's also trivially satisfied because they are matrices
@bolbteppa i find most physics books do a big mess explaining Lie theory
@Relativisticcucumber A common definition of the Lie algebra is as the algebra of left invariant vector fields; it then turns out that these are already defined by their value at the identity (or anywhere else), which means the algebra is equivalently the tangent space at the identity
10:50 AM
Yeah no book on this is perfect but a near 500 page book on this that doesn't even prove the classification but wastes time on technicalities is just...
@Slereah How are you taking the direct limit? The algebras are a cosheaf, not a sheaf, i.e. they come only with inclusion maps $f : A(U)\to A(V)$ for $U\subset V$, not the restriction maps we use to define the stalk as the direct limit for a sheaf
@ACuriousMind I am seeing people talking about the algebras at a point as $$\bigcap_{\mathcal{O} \ni p} \mathfrak{A}(\mathcal{O})$$
Isn't that a direct limit
Apparently in the relativistic case it just restricts the algebra to $$\bigcap_{\mathcal{O} \ni p} \mathfrak{A}(\mathcal{O}) = c \mathbf{1}$$
hence not a fun algebra
at a point, there are only operator valued distributions and no operators
@Slereah I think that's an inverse limit and this is indeed the co-stalk
I don't think it is a particularly relativistic effect that this is empty/trivial
Isn't it
10:57 AM
it's more just a consequence of fields being operator-valued distributions, and distributions don't have values at points
if the intersection was non-trivial, you'd suddenly have operators localized to a point
I have read that in the classical case you do have operators at points in a non trivial way, at least for the point particle case
I mean some distributions do have values at points
and the reason that fields are distributions is, at least to me, not relativistic, because it's simply a consequence of the CCR containing a $\delta$-distribution
@Slereah yes, but the fields cannot be such distributions, because their product has to result in a $\delta$
you can't get the $\delta$ that lacks a value at a point from operations on distributions that have values everywhere
There is something going on in the classical case though
Like typically apparently the net of algebra is defined on thr spacelike hypersurface
do u mean the Poisson algebra?
I'd like to figure out of you can define this properly in the general context
11:03 AM
i would say operators do correspond to integration on R^3 rather than R^4, so it is okay that we no longer have distributions on subsets of 3-space
we only have distributions at points. subsets of three-space have operators
@Relativisticcucumber The $\mathrm{e}^{tX}$ is just the exponential map starting at the identity of the group.
The idea is now that you want cover the group with patches where you can present elements in this form (since then simply $\Phi(g) = \mathrm{e}^{t\phi(X)}$), and you do that by translating the neighbourhood of the identity that's an image of the exponential map around the group (along those paths!)
the challenging part is proving that this doesn't depends on "how" you translate it, and that's the step where you prove this is path-independent
note that only in the Heisenberg picture can we talk about the algebra on spacetime in the first place and the heisenberg picture is not mandatory
@ACuriousMind Why can we define the algebra at points if the spacetime is just R tho
in the usual picture, the algebra is on subsets of 3-space
In the case of a point particle
We generally don't define operators on some time open $(t_a, t_b)$
11:10 AM
this is because the CCR have $\delta ^{3}$, not $\delta ^{4}$. in case of a 0+1 field, the CCR does not have a delta @Slereah
I guess because there's no delta term in time in the Poisson algebra
the 0+1 field is just QM
@Slereah yes
Why does the other delta drop out, tho 🤔
and how would we define causality for this particle?
i dont think causality makes sense without space
Well just pick it as being in some Galilean manifold
It's some map from R to R4 but Galilean
Trying to figure out what's the nuance
I vaguely recall a paper saying that in the Galilean case there's a natural identification between the curve parametrization and the time function of the manifold
11:14 AM
we can re-formulate it on usual spacetime, but then position will no longer be the field amplitude
Maybe it relates
Yeah but you can get back position I think?
yes, as a parameter
it is a big re-formulation
But it is a possible one in the Galilean case
The relativistic case too I guess but it will not be as good
all kinematic groups have a notion of a plane of simultaneity, right?
from there, we can get a Hamiltonian formulation
and then the poisson bracket being zero follows for independent co ordinates
They have a notion of spacelikeness yeah
Well I guess not techniclly
The galilean case you only really have the null space
11:20 AM
all that matters is a Hamiltonian formulation. i think the Schrodinger field does have one. but it's constrained, I think
I'm guessing for the Galilean case you have a natural time direction from the structure which makes things go more smoothly
Although do you really
The natural structure involves a one-form, not a vector
You can pick a variety of timelike vectors presumably
And you can't ask for them to be orthogonal to the plane since the metric is degenerate
Actually isn't the lack of time deltas also true for the relativistic point particle
the field corresponding to a point particle in 3+1 should itself be a delta function, I think
I think it's just fundamentally due to the one dimensional aspect
yeah. if u formulate it in 0+1, then there's no deltas
it follows from.the phase space having only a discrete number of variables
but when we formulate it in 0+1, causality is hard to define
have u tried the Schrodinger field rather than the point particle
but the Schrodinger field has its own problems. the eqn is not Galilean invariant in the easy sense
@Slereah If you read e.g. Streater and Wightman, you'll notice that they have to introduce the notion that for a certain kind of theories, the Wightman functions do have sharp values in $t$, and only have to be smeared in space - this is not a generic feature of all Wightman theories, it's an additional property
I suspect this reflects in the algebraic approach as being able to talk about the operators associated to a hypersurface, but note that this is also not part of the minimal Haag-Kastler axioms
11:37 AM
I'll give S&W a look, thanks
note also that this "sharp in time" property is a requirement to be able to talk about the equal time CCR we often see
Is it impossible to do QCD using operator methods?
also, talking about time dependence of operators is only a Heisenberg picture thing. when u quantise, u only get operators on 3-space.
Yeah I'd like to untangle a bit how quantum theories work
I guess another question is
Is there something remarkable about the algebra of observables on the light cone, relativistically?
according to Rovelli, the Heisenberg picture is mandatory for some spacetimes like those with CCTs
11:41 AM
The equal time surfaces classically are essentially just the light cone
the immediate analogue of equal time CCR non relativistically are equal time CCR relativistically. this is just the Dirac quantisation postulate
but there may be something deep going with the light cone CCR
do u know a deep reason for general kinematic groups why space-like CCRs have to vanish?
i think this vanishing is strongly dependent on the dynamics too @Slereah
Hey, does anyone know a nice website to check out undergrad research / some research I can remotely understand as a 2nd year student?
I wanna get familiar with research (structure, subjects of interest, etc.)
@AlanFox86 I think that depends very much on what your interests so far are and what you've learned ("2nd year student" can mean very different things to different people); the best thing is probably to first look around at your university whether there are any undergrad research programs or positions
11:57 AM
i think the light cone CCR depends on the dynamics
it may not follow from.the kinematic group alone
@ACuriousMind I don't think I'm eligible for undergrad research until I get to 4th year at my uni. The only project to which I'm able to contribute rn is my Optics professor's Python module. It's quite great if you're into optics, but it's not really my thing, I'm more interested in mathematical Physics
do you really mean mathematical physics or do you just mean theory?
@AlanFox86 What area of theoretical physics is mostly done at your university? Do you not do a 3rd year research project?
@bolbteppa Have you looked much into quasistationary states (see L&L QM section 132) and their connection to decaying particles in QFT?
@AlanFox86 If you are interested in QFT, there are plenty of methods used in quantum optics that will be of interest to you. My 3rd year supervisor was in quantum optics on the theory-heavy side and it was probably one the most interesting projects I've ever done.
12:26 PM
@AlanFox86 In particular, are you meaning to say theoretical physics rather than mathematical physics?
Not enough to say much about it, I doubt it has any bearing on the stuff I'm guessing you're looking for, but there is something one could say about all this I'm sure
12:53 PM
also wait what am I even saying
The commutator doesn't even vanish at equal time in CM
Well, classical QM
Stupid idea
@Slereah what do u mean
Had some notion that the equal time would correspond to spacelike separation
But wrong on several accounts
Schrodinger field has $[\psi (x), i\psi ^{\dagger} (y)]=i\delta ^3(x-y)$ for equal time @Slereah
do u mean the point particle theoru
I mean the condition $[A(O_1), A(O_2)] = 0$
@bolbteppa There is some information here under equation 22 jetp.ras.ru/files/JETP_7_96.pdf
1:09 PM
But fundamentally the Galilean spacetime doesn't have a spacelike region
It's just a null region
> The cartoonist Luis Bagaría devoted many cartoons to Einstein in El Sol. In one of them, a boy asks his father:

> –Tell me, dad, is there anyone wiser than Einstein? –Yes, son. –Who? –The person who understands him.

> Another cartoon has the following caption: “Einstein says there are no straight lines, just curves.” In the drawing, a man yells at a passing woman: “Oh, what curves! Long live Einstein!”
@Slereah is the conclusion that the causality axiom holds vacuously
@Slereah you can't even phrase the axioms of AQFT in non-Lorentzian spacetimes, can you?
1:22 PM
yeah the equal time CCR are just from Dirac quantisation. this has little to do with causality maybe @Slereah
@ACuriousMind You might idk
I mean if the causality is true, the rest mostly doesn't depend that much on the properties of the Lorentz group
@ACuriousMind general kinematic groups have notions of spacelike regions
You can replace the unitary rep of Poincaré by Galilean
according to Slereah
and you still have two light cones
1:22 PM
@Slereah how do you define a causal structure on a non-Lorentzian spacetime?
@ACuriousMind By the time ordering of time slices pretty much?
I don't know if it works, but I can't see any big obstruction for it
what do you do e.g. with the spectral condition that the energy-momentum spectrum should be contained in the forward lightcone?
i think the speed limit only makes sense for minkowski spacetimes
there's no lightcones in Galilean space
also a lot of constructions explicitly use causal diamonds (intersections of a forwards and a backward lightcone)
@ACuriousMind There are
They are just very open
1:27 PM
I suspect you just have to drop all of that, meaning you're left with a local net of observables that obeys essentially no restrictions
The intersection of two causal diamonds would just be the whole region between t1 and t2
causality is an empty requirement, the spectral condition does nothing, you're just left with a net
I mean not that far
there's probably not a lot you can really say about that
There's no spacelike region but there's still a past and future light cone
and a null region
1:28 PM
what's the null region???
The region of equal time
oh, that's what you mean
I guess it's harder to define since they're not associated to eigenvalues of the metric
the forward cone is region just all of spacetime ahead
But they behave essentially the same way
1:29 PM
yeah, I really think this is too weak to have some kind of nice theory :P
I'm guessing you probably need some additional conditions yeah
yes, but there r some more interesting kinematic groups
or, really, any theory at all, there's no restrictions here
galilean is not interesting maybe
From what I've seen, when people do nets for non-relativistic QM, they do it only on space, which makes me think you probably need some condition relating to the time slicing for this to work
Something to indicate the sharpness there idk
but then how to motivate it
1:33 PM
i think the operators are always sharp in time, even relativistically
aqft axioms may be different, but in the Schrodinger picture, the operators of any quantum theory are sharp in time
Yeah I don't remember what's the relation between equal time commutators and the covariant ones
@ACuriousMind I really mean mathematical Physics as in the development of mathematical tools which can later be applied to Physical theories. Theoretical physics draws my attention too, though, and being in second year I still have the benefit of not having chosen a determined path yet
And no, we don't do a third year project in Spain, but we do have a TFG ("end-of-degree project", would be its literal translation). I have the opportunity to do some research in my 4th (last) year, but up until then I'm not supposed to do anything whatsoever other than studying for my exams, and I'd like to get involved some more
i think spacetimes with CCTs may not admit a Hamiltonian formulation. they may not satisfy equal time CCR
these spacetimes do not have initial value problems
@DIRAC1930 my 2nd year optica class covers exclusively electromagnetic optics, there's not much quantization involved. Typically, subjects get more interesting in 3rd year, but I'm nor wasting this summer doing nothing, that's why I'd like to spend ir reviewing resesrch I can begin to understand
@AlanFox86 In that case you probably will not be able to understand any contemporary research in mathematical physics unless you already know a lot of math (at least advanced knowledge of differential geomety or functional analysis (either operator theory or differential equations) or statistics/stochastics)
1:43 PM
I see, no, I don't know that much math haha
@Slereah this should depend on the Hamiltonian
I'm currently trying to learn as much as possible on my own I'm doing Linear Algebra now (I didn't like how it was taught last year so I'm doing a more personal review) and then I'll move onto real variable analysis (of course, I know analysis but again I'd like to know it more in depth)
2:16 PM
@AlanFox86 you took like one of the most involved fields of research
with linear algebra you may be able to understand quantum computing, which is a very active field of research
2:27 PM
Interesting! I'm also currently taking Quantum Physics so perhaps I could delve some more into that
2:39 PM
Please read this and explain all the subtle details to me, like why a $\Omega_3$ closed ensures locality, what the hell is going on with the indices, more generally what the ... is going on, thanks
@AlanFox86 ... you are so farrrrr removed from maths phys right now. You are much likelier to understand a bit of what theo phys might be playing with. You will have to learn a lot more maths to catch up for either case, but at least theo phys often goes to old maths.
> But symmetries also occur in nature; many flora and fauna show symmetry. (I shall leave it to the reader to imagine them.)
@ACuriousMind your heros names are in there ^
@bolbteppa no freely available copy :(
2:56 PM
It's of course based on this which has another 'derivation' which is simply insane
@bolbteppa idk unpaywall doesnt find free versions of this either
It's basically talking about this, how to derive that extra term $S_{\kappa}$:
Q: What is kappa symmetry?

DilatonOn page 180 David McMohan explains that to obtain a (spacetime) supersymmetric action for a GS superstring one has to add to the bosonic part $$ S_B = -\frac{1}{2\pi}\int d^2 \sigma \sqrt{h}h^{\alpha\beta}\partial_{\alpha}X^{\mu}\partial_{\beta}X_{\mu} $$ the fermionic part $$ S_1 = -\frac{1}...

oh this is about the Green Schwartz formalism
3:20 PM
> In this category-theoretic language, the quantum teleportation protocol is essentially just the zig-zag identity for dualizable objects
is this relation for trace of the inverse of a $2 \times 2$ matrix $$\operatorname{Tr}(A^{-1}) = \frac{\operatorname{Tr}(A)}{\operatorname{det}(A)}$$ always true?
my proof was based on bringing $A$ to uppertriangular form and then it seems to follow trivially
@lucabtz trace and determinant are both independent of basis, and so yes
@naturallyInconsistent yeah i know
but im surprised i never saw this formula before
i guess its because its only valid for $2 \times 2$ matrices
@lucabtz it aint that useful, is it useful?
@naturallyInconsistent i mean it can be useful
like i need it rn
also im in SL(2, C) so i even have tr(A^-1) = tr(A)
omg found out i can plot the space of monodromy data (or maybe just part of it idk) we were speaking about a while ago @ACuriousMind
3:58 PM
@lucabtz It's just based on the explicit formula for the inverse of a 2-by-2 matrix: you divide by the determinant, swap the entries on the main diagonal and negate the entries on the lesser diagonal
since swapping the entries on the main diagonal doesn't change the trace, your formula is true
@ACuriousMind oh yeah i mean i know how to get it, i just proved the formula too
actually i didnt use the explicit formula because i though could work in arbitrary dimensions, but of course it does not
yeah, it's a happy accident in 2d because there dividing the trace by the determinant yields the sum of the inverse eigenvalues since the determinant is just the product of two values
that is $a+b$ divided by $ab$ turns into $1/a + 1/b$, while dividing $a+b+c$ by $abc$ does not turn into $1/a + 1/b + 1/c$
i think u hav $tr(A)+tr(A^{-1})=log(det(e^A)) + log (det(e^{A^{-1}}))$
so this gives $log(det(e^A) det(e^{A^{-1}}))$
@ACuriousMind yeah i was using upper triangular matrices so i thought it would be general to any dimension, but then i realized that
did not realize that since its just 2d could as well use the formula for the inverse
this doesnt simplify
may simplify for unitary matrices
we hav $log(det(e^{A+A^{-1}}))$
4:12 PM
this is a plot of the monodromy data space I think
idk its a bit odd i feel like i made some mistakes
it doesn't look like much but tbh I have no idea how it should look :P
i have a typo in my code
it looked odd that nothing happened along the z axis in fact
this should be correct, i suppose at the center there is some kind of singularity, idk if its the coordinates that are singular or the space itself since its not a manifold
do u think qft people must know the standard model or are there other things to do
4:27 PM
@RyderRude i mean the standard model is a great example for a lot of qft concepts so id say its good to learn about it
it feels like a strange model to me especially with symmetry breaking
@lucabtz do all qft people learn qcd and weak force?
4:44 PM
You must learn the standard model or else
the standard cosmological model?
A basic course on it will be mostly yang-mills, ssb etc it's basically direct motivation for a lot of qft stuff
But it wont equip you for the madness of kappa :\
do they focus on non scattering things
My intro qft course starts with a cursory glance at the standard model
i think the idea is to get a concrete notion of what we mean by particle and field before diving into the maths
There was only one QFT module at all, and so it could not even cover QED. Instead, it focused upon teaching renormalisation.
4:52 PM
I don't think we ever did Yang-Mills here
i feel there is too much focus on scattering
We just did some QED scattering computations
If you want something non-scattery then don't read HEP style QFT books
There's no lack of other genres
i'd prefer to learn foundational things like positions of particles
but this is glossed over in favor of scattering
4:54 PM
I'm not sure theres's really any good book like that for QFT
It's more the realm of papers
It’s time for u to write your qft textbook
qft in curved spacetime doesnt focus on scattering, right?
It doesn't even focus that much on interacting systems, for the most part
At least for introductory books
@SillyGoose I'm really not one to write a QFT book :p
thats great. i love the bogoliubov stuff
there is the local gauge symmetry hack for the standard model lagrangian
this is in the intro books
but the actual lagrangian looks nothing like that
the GR lagrangian is pretty, for instance. but the SM lagrangian is kinda random
i see a bigger one too which is whole mess
5:10 PM
The other one is cheating a bit
It expands things and include ghost terms and renormalization terms
But if you included all this the GR lagrangian would be much longer too
thats what i hoped for
but the lagrangian in my link only has one $F_{\mu _\nu}$. does this cover all of the forces with indices suppressed?
It is shorthand
then it doesnt sound bad now if this is all of it
GR if you include the ghosts
And that's not including boundary terms and renormalizations
5:36 PM
can someone explain to me, if we choose dv=0, wouldnt then the other partial derivatives, that include V, also be Zero? i dont understand this step of calculation.
Total derivative versus partial derivative
does that read $$dV = \left(\frac{\partial V}{\partial T}\right)_P dT + \left(\frac{\partial V}{\partial P}\right)_T dP$$
it is a V
all the U's are V's in what I wrote?
there is no U.
5:41 PM
@Mad No, the relation $$\mathrm dV=\left(\frac{\partial V}{\partial T}\right)_p\mathrm dT+\left(\frac{\partial V}{\partial P}\right)_T\mathrm dP$$ is true for any choice of $\mathrm dT$ and $\mathrm dP$ giving arbitrary $\mathrm dV$. This means neither of the partial derivatives should be zero, especially in the general case.
In the steps below, when taking partial derivative , and setting something constant, the partial derivatives of those things set constant disappear.
by setting "something" constant what do u mean
it's true if $dV = 0$ then $$0 = \left(\frac{0}{\partial T}\right)_P dT + \left(\frac{0}{\partial P}\right)_T dP=0$$
Oh you're asking why $$dV=0, \left(\frac{\partial P}{\partial T}\right)_V = -\frac{\partial P}{\partial V}\frac{\partial V}{\partial T}$$ is nonzero
say, for example, we have $ dS=( \partial S/\partial T _V +\partial V/\partial T _p \partial S/\partial V _T ) dT +\partial V/\partial p _T \partial S/\partial V _T dp$ and now wanting to calculate $ \partial S/\partial T _V= \partial S/\partial T _V$
And those partial derivatives including V disappeared. So why does this not apply in this case?
or did they actually not disappear and they just put in $\partial S / \partial T_V $ as it is
my issue is, if the change in V is zero, how can we talk about for example the change of P if V changes?
because P doesn't just change w.r.t. to V but also w.r.t T implicitly
or I guess explicitly(?)
through the relation $-\frac{\partial V}{\partial T}\frac{\partial P}{\partial V}$
again, you speak of change in p, when v changes. and we said,v doesnt change.
thus nill.
5:51 PM
$-\frac{\partial P}{\partial T} \iff -\frac{\partial P}{\partial V}\frac{\partial V}{\partial T}$
nevermind that, can you look at the example i gave with dS?
but if dV=0 you are deviding by zero.
I'm gonna transcribe that because it's hard to read: $$dS = \left[\left(\frac{\partial S}{\partial T}\right)_V + \left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial S}{\partial V}\right)_T\right]dT+\left[\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial S}{\partial V}\right)_T\right]dP$$
is what you wrote
Does anyone know of a String Theory book written in the style of L&L that puts physicality at the forefront
@Mad no idea what calculating $\left(\frac{\partial S}{\partial T}\right)_V = \left(\frac{\partial S}{\partial T}\right)_V$ means
By that I mean, something that concentrates on the problems string theory solves
5:53 PM
Yes, it is then proceededed to calculate $ \frac{\partial S}{\partial T}_p - \frac{\partial S}{\partial T}_v = \frac{\partial V}{\partial T}_p \frac{\partial S}{\partial V}_T$
and i want to understand, from that equation, if you calculate $ \frac{\partial S}{\partial T}_V$ all the terms must disappear except the first. this means also the partial derivatives that include V
i think my question is perfectly clear and well formulated.
The result is solely that $\left(\frac{\partial S}{\partial T}\right)_p=\left(\frac{\partial S}{\partial T}\right)_V+\left(\frac{\partial S}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_p$
I agree!
But i am not talking about this one, i am talking about the other one while holding V constant.
It is not trying to calculate $\left(\frac{\partial S}{\partial T}\right)_V$, so you are simply misinterpreting what is being meant there.
But lets say we do want to do that!
what do u mean by calculating
5:57 PM
Given a function $f(x,y) = x^2 + y^2$ its differential is $df = 2x dx + 2y dy = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy $. Setting $df = 0$ for this function of two variables is an equation for a curve i.e. $2x dx + 2y dy = 0$ leads to $\frac{dy}{dx} = - \frac{x}{y}$ leads to the curve(s) $y = \pm \sqrt{C-x^2}$ or $x^2 + y^2 = C$ which is just an implicit equation for that curve.
Note $2x = \frac{\partial f}{\partial x} = - \frac{\partial f}{\partial y} \frac{dy}{dx} = - (2y) (-\frac{x}{y}) = 2x$.
Then you are going against the logical flow of the thing you are reading, and you cannot expect whatever excursions you are taking to be fruitful
So its not generally true.
It is a coincedence?
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