@DebanjanBiswas Gases are rather inefficient at radiating heat. If you have a container of gas with constant volume the gas will typically cool by conducting the heat into the walls of the container.
The EoM shouldn't be used in the Lagrangian (cf e.g. two body problem and effective potential). Why does the wiki article of GR Kepler problem use the EoM (conserved quantities) in the Schwartzschild metric to derive the radial EoM? Why does it work?
Can't remember off the top of my head, but I do remember getting into trouble before doing that in the non-relativistic Kepler problem
Is it not that you're not working with the Lagrangian, you're just working with $ds^2$ directly (whose root also happens to be part of the Lagrangian), which is simply a displacement (squared) in (curved) space-time, and treating some quantities as known and others as unknown, so you can use the equations of motion
We know $ds^2$ in one frame is $c^2 d \tau^2$, and in another frame it's Schwarzschild, so we have enough info to fix some variables in terms of others. If we did it with a non-relativistic Lagrangian, it would be like knowing $L$ in two frames $L' = L + \frac{df}{dt}$, where $L'$ (and maybe also $f$) was known explicitly, and we could use this equation to fix variables, and assume the eom
@bolbteppa This way it makes much more sense. In his book Rovelli imposes natural parametrization on the lagrangian $\mathcal{L}=-1$ (as $S=\int\mathcal{L}d\xi=-\int ds$) and then plugs the conserved quantities in there. Thinking about it he's just using the metric too
Is the case when we consider a potential barrier and a potential well the same, with the only difference that the potential, in the potential well is finite?
@RyderRude why do you think Descartes needs to construct an "objective reality"?
dualism just means that you believe in two distinct kinds of things, and Descartes believes in the mind, the soul - the first kind of substance - inhabiting matter, the body - the second kind of substance
there's "stuff that thinks", the mind, and "stuff that can be perceived", matter or "bodies"
Regarding the wavefunction in rel. QM, is it so much that we can't measure it due to $\Delta q \sim c \hbar/E$ where $E$ is the energy of the electron or that it doesn't exist?
@ACuriousMind so we are still saying that the "stuff that can be perceived" exists within our subjective experience? If yes, then idk y descartes is drawing a boundary between thoughts and other qualia. They're all qualia.
the modern jargon for Descartes is that he's a substance dualist: He really thinks there's "stuff that thinks but has no physical extent" and "stuff that has physical extent (and hence can be perceived) but doesn't think"
well, you have to read the whole passage about the demon and the existence of God to get to why we should trust an external world exists according to Descartes
@RyderRude what do you mean by "objective reality"
I don't really know of anyone that denies "objective reality" in the sense that there is something that serves as a substrate for our and other minds existence
the thing epistemology is concerned with is not "does objective reality exist?" it's "how and with what certainty can we know anything about that objective reality?"
dualism and monism aren't statements about epistemology or how we can know the external world, they're categories your philosophy gets sorted into depending on whether or not you think the mind is fundamentally different from "other stuff"
But that's wut i said. Descartes's dualism is a boundary between the outside stuff and the mental substance. I think Descartes inferred outside stuff based on behavior of his mental substance and some leaps of faith like : "other minds exist" @ACuriousMind
@RyderRude no, that's not what you said, and Descartes' dualism is not about a boundary between "outside" and "inside"
Descartes believes the minds of other people exist, too
like, he believes in souls
souls are different from matter because they think and cannot be perceived
this isn't about his subjective experience being different from objective reality, it is about stuff that thinks being different from stuff that doesn't think
@DIRAC1930 you're not measuring full "position", you're measuring whether or not a photon crossed a surface (a layer of cells in your eye), see physics.stackexchange.com/a/104565/50583
again, the dualist claim is just that two different kinds of things exists: Minds and Not-Minds. The inside/outside comes in when Descartes then also says that the mind can know itself in a much more certain way than it can know anything else.
but the aspect of inside/outside/certainty is not a core aspect of dualism; that Descartes argues the mind can know only itself with certainty is much more a rationalist and skeptic position; you can believe in that aspect of Descartes without believing in his dualist claim that the mind is fundamentally different from any other stuff that exists
the modern jargon for Descartes is that he's a substance dualist: He really thinks there's "stuff that thinks but has no physical extent" and "stuff that has physical extent (and hence can be perceived) but doesn't think"
@RyderRude again, no, that's you imposing what you want him to say because it makes more sense to you than what he's actually saying :P
just because you don't believe in disembodied souls as a fundamental ingredient of reality that doesn't mean Descartes isn't saying that
@DIRAC1930 note that uncertainty here refers to an actual uncertainty of a specific measurement process
this is not the same meaning as the "uncertainty" in the standard uncertainty principle, where the uncertainty is a theoretical quantity that's a property of a state (and not a measurement process!) and only depends on the expectation value of the operator in that state
it is very annoying that we use "uncertainty" both for experimental uncertainties and what is more properly called a standard deviation
So an example of the standard uncertainty principle (between momentum and position) is a localized particle (definite position) being constructed of a superposition of momentum eigenstates i.e. it is a property of the state. But I can in principle measure the momentum and the momentum space wavefunction will be an eigenvalue of the momentum operator with a definite eigenvalue $k$.
In the L&L 4 case, is he saying that if I do a measurement of the position of a state, the state will collapse to a state that does not correspond to a definite eigenstate with an eigenvalue that you have measured
The uncertainty in position measurement seems to imply that if there was a position space wavefunction, it wouldn't even give you the probability at a specific location because your measurement wouldn't even be able to tell you where it is
'For photons, the ultra-relativistic case always applies, and the expression (1.5) is therefore valid. This means that the coordinates of a photon are meaningful only in cases where the characteristic dimensions of the problem are large in comparison with the wavelength.'
Is this to do with your ruler argument
Is the last paragraph before section 5 on page 14 a better argument?
@DIRAC1930 Ultra-relativistic case for a massive particle is when $E$ is much larger than $mc^2$, so in $E^2=(cp)^2+m^2c^4$ you can ignore the last term and we get $E\approx pc$. For photons, this is exact i.e. it always holds, as they say.
It's not just photons, the arguments there show there is always some error in the relativistic particle position i.e. you can never measure it exactly, so the whole idea of a position-space wave function is just gone
You can still work in position space, but the interpretation in terms of a precise position is just gone, but in momentum space a technicality with free particles saves a physical interpretation
@DIRAC1930 oh, I just read back what you were saying and I'm sorry: You were correct, I thought you were talking about the existence of uncertainty in measurement in general, but it is correct that a lower bound on the uncertainty in measurement implies position wavefunctions are questionable
But isn't measurement the important thing because that's precisely what the wavefunction tells us? The X-P uncertainty is a statement about the states so that if we theoretically have a wavefunction and we measure the state, the wavefunction will give something meaningful. But with the measurement uncertainty in $X$ stated in L&L 4, even if we had a wavefunction, we wouldn't even be able to physically interpret it properly
There seems to be something very strange about the relationship between quantum field theory and quantum mechanics. It is bothering me; perhaps somebody can help.
I'll consider a free Klein-Gordon field. In standard treatments (e.g. Peskin & Schroeder and Schwartz) the one-particle momentum ei...
So for example, we can still use the quasi-classical appoximation $\psi \approx e^{iS/\hbar}$, and for a relativistic particle $S = p \dot{x} - H d \tau$ where $H = e(p^2+m^2)/2$ (assuming Dirac's constraint stuff, with $p^2+m^2=0$), so that the Schrodinger equation is $i \hbar \frac{\partial }{\partial \tau} \psi = \hat{H} \psi$, but imposing the constraint as an operator equation gives $\hat{H} \psi = 0$ which is just Klein-Gordon. So we're literally just doing normal QM in that sense
@ACuriousMind I can't find the video, but basically the person explains how in the 6-dim phase space of a neutron star, you have a position/momentum uncertainty relation from QM and as you add more matter to the star, your position space gets more defined and smaller, and momentum space gets larger.
So from this Schrodinger equation, the stationary state solutions are the Klein-Gordon $e^{ipx}$ solutions, and we can then package these stationary state solutions into a quantum field operator and discuss quantum field operators as usual, exact same thing as for the non-relativistic case
However the fact we get negative probabilities in the associated four-current density means the whole probability interpretation in position space is gone
This positive square root Hamiltonian thing is not serious, you can't work with square roots at best you get non-local nonsense, it's always squared, this square root amounts to working in the lightcone gauge, a special gauge choice, but you still end up squaring and getting the above stuff back (because the results should not fundamentally depend on a choice of gauge), so it's not an answer to any of this
@bolbteppa but wut i described is equivalent to usual QFT restricted to single particle case. Non-locality is only a problm if u want 2 use the strict position operator. It's not a problm as an approximate position operator in the regime where the McDonald's function is close enough to the delta function
@Obliv not sure what "theoretically measureable" means, I mean that there is no experiment in a non-gravitational theory for which it would matter what number you assign to "energy of the vacuum"
all that matters in practice are energy differences, this isn't a particularly quantum phenomenon
You can't restrict QFT to single particle problems without making approximations, it's a useful tool in some situations you are throwing away basic things if you do stuff like this, this is irrelevant to the question of what the basic principles say. Dirac's Hydrogen atom solution uses approximations like classical external $1/r$ fields
You'd need a theory of everything if you wanted to do it this way though, since you'd need to be able to represent the universe totally by some parameters
@DIRAC1930 is it not that you go back to $\Delta x \Delta p \approx \hbar$, and you realize if you send $\Delta x \to 0$ that there is no problem with sending $\Delta p \to \infty$, whereas (1.3) says there is a problem because of the least value there (ultimately due to the finite speed of light in the relativistic case)
@RyderRude I have trouble understanding this statement, but yes I'm a returning student. I am currently a math major because I have requirements to fulfill first.
@ACuriousMind i read in qft notes that the vacuum energy cud explain the cosmological constant becuz "gravity sees all energy". Is this not a mainstream view?
@bolbteppa But then under eqn 1.2, it says that you can measure $p$ as accurately as you want (in the rel regime) under a measurement time approaching infinity
What I said is done at one time (i.e. a "rapid" measurement as they discuss below (1.2), only possible non-relativistically), whereas this (1.2) thing is done over a time interval and relates to time-energy uncertainty, which is very different to the above uncertainty
Now, take all this complicated stuff, and remember books like Griffiths tell you that free particles with well-defined energy do not exist because the normalization integral diverges, and pretty much all of formal QM tries to do the same thing without a care in the world for any of these (physical) subtleties
So in the non-rel regime I can theoretically measure $p$ instantly (due to no limiting velocity) and infer the collapsed wave-function $\psi(x)$, but in the rel regime, I can no longer instantaneously measure the momentum with exact certainty and hence there will be an inherent uncertainty in the position?
Yes, unless the particle is free in which case the measurement occurs over a (theoretically) infinite time interval, which means every QFT interaction process can only talk about before and after an interaction (when things are free particles), but not what goes on during the interaction
But if I am measuring a single particle, won't the momentum be conserved over the entire course of the switching on and off of the interaction so I could determine the exact momentum over an infinite time?
When you actually perform a measurement, that's another interaction, which affects the state etc... all the above is trying to say is there is something to measure in the first place
This is a petition for the attention of moderator team, and the professors of Physics Stack Exchange, about the clousure of my post on Physics Stack Exchange about gravitational geons in our Solar Sytem, to be reopened. If I can I order better this list of reasonings and expand this with other to...
So am I assuming that I am measuring the asymptotic states over a very large time such that $\Delta p $ approaches $0$ and declaring these to be the asymptotic states?
@bolbteppa What did you mean by this 'Yes, unless the particle is free in which case the measurement occurs over a (theoretically) infinite time interval'?
that's because TeX thinks $\star$ is a binary operator - this is the same logic that produces the space around the minus sign in $a - b$ but not so much space in $\int_S -b$
depends very much on what you mean by this, on a very abstract level I'd say all physics is about information: Initial data/observational information in, predictions out
"But if quantum mechanics isn't physics in the usual sense -- if it's not about matter, or energy, or waves, or particles -- then what is it about? From my perspective, it's about information and probabilities and observables, and how they relate to each other."
"Many commenters seem to assume the pilfered quote is just my expression of the conventional wisdom. Well, it should be, and I wish it were! But as Avi Wigderson pointed out to me, the idea that quantum mechanics is about information rather than waves or particles is still extremely non-standard, and would have been considered insane fifteen years ago."
to me quantum mechanics seems like the same physics with a new logic, which leads to new physics
hm I guess I am wondering after all is said and done and we have 1) physical theories before quantum, and 2) the textbook theory of quantum mechanics, what makes 2) essentially different from 1)
it's not the "physical framework" that we attempt to model the world by system and states, at least to my understanding
the logic definitely seems different because we have new types of states
Aaronson is advocating for another approach, one I'm sympathetic to, where we focus on the idea that states assign probabilities to observables
in that formulation the difference between classical mechanics and quantum mechanics is the aspect of non-commutativity: That the probabilities of getting answers to questions like "Is the particle's position between $x$ and $y$?" and "Is the particle's position between $q$ and $p$?" change depending on the order in which you try to experimentally obtain answers to these questions
maybe I misunderstand this approach, but I am quite confused why we say things have a wave-particle duality. Just because you represent something by a wave function does not mean it is a (physical) wave, right?
personally I think the notion of "wave-particle duality" and the focus on wavefunctions has done a lot of damage to people's attempts to learn QM because they keep getting stuck in the places where there isn't really anything "wavy"
and to me the foundational experiment with which we teach QM should be Stern-Gerlach and not the double slit
although, my answer would not be much better :P though less inaccurate still: "a state is something a system occupies. a state is mathematically represented (as we always try to represent the world, through symbols; in particular mathematical symbols) as a vector in Hilbert space. it is not a single point object, it is not a wave. I don't know what it is physically, but this math gets pretty close :P."
but historically Stern-Gerlach just supported "old" Bohr-Sommerfeld theory and it was the double slit that vindicated Schrödinger's wave mechanics (and implicitly Heisenberg's matrix mechanics) as the new correct theory
so the double slit gets to be the thing we focus on
yeah idk it seems people really want to attribute something real to wave functions when i don't think it makes sense to do that but :P. or, at the very least the yearning to constantly interpret something meant to be abstract as real takes away from what that thing actually is
@SillyGoose in the more mathematical treatments of QM there is the $C^\ast$-algebraic approach (going back to von Neumann) where we just start with an abstract algebra of observables (not even operators on a Hilbert space!) and states are just linear functionals on the algebra, the idea being the value of the linear functional on the algebra is the expectation value of the algebra element for that state
this to me is the "informational" approach because we really just start with the base necessities; in the end it becomes equivalent to the standard treatment because you can show that every $C^\ast$-algebra is isomorphic to an algebra of operators on a Hilbert space and every state can be represented as a density matrix on some Hilbert space on which the algebra is represented
and together with Stone-von Neumann saying that there is really just the standard wavefunction representation of $[x,p] =\mathrm{i}$ on $L^2(\mathbb{R}$) we also recover why the wavefunction approach works equally well as a foundation
@SillyGoose It will not become the norm because it requires considerably more mathematical machinery
if you ignore all the subtleties you can just throw the hydrogen atom Schrödinger equation at people who barely know calculus in the wavefunction approach and they can go compute energy levels
meanwhile, in order to establish all the theorems about algebra representations you need to finally be able to do explicit computations like that in the algebraic approach you need a whole lot of functional analysis physics students typically don't learn
also are computers built to do linear algebraic things or something? I feel like everyone tries to reformulate their problems in linear algebraic terms, not in group theoretic or etc. terms.
@SillyGoose my biased view is yes, the algebraic approach is more elegant because it derives a lot of the usual Hilbert space setting without explicitly assuming it; however you can complain that the abstract notion of $C^\ast$-algebras is more or less designed to do that so that it isn't really a meaningful difference
well if I want to say construct an arbitrary element of SU(2) in my python code, I don't write a presentation of the group and construct it; I use the pauli spin matrices to construct said arbitrary element and exponentiate (I guess this is not quite solely linear algebra either)
okay i see. i was just wondering if I hadn't seen how people use computers compatibly with group theory(as an example) or if it is just easier to code in linear algebra
there are very general problems with presentations of groups, e.g. it's in general an undecidable problem to determine algorithmically whether two presentations are presentations of the same group
I mean for SU(2) specifically you could just write down some function that gives you an element of that group "directly", but the question is what parametrization would you choose?
it seems very natural to use as coordinates just the prefactors of the Pauli matrices and then exponentiate
because when you write $\mathrm{e}^{\mathrm{i}\vec n\cdot \vec \sigma}$, the length of $\vec n$ says how far we rotate and the direction of $\vec n$ is the axis of rotation
There seems to be something very strange about the relationship between quantum field theory and quantum mechanics. It is bothering me; perhaps somebody can help.
I'll consider a free Klein-Gordon field. In standard treatments (e.g. Peskin & Schroeder and Schwartz) the one-particle momentum ei...
