The global (non-gauge, physical, whatever) symmetries associated with non-Abelian gauge symmetries are only the discrete $\mathbb{Z}_n$ center symmetries so Noether's first theorem does not apply.
@Slereah oh yeah. I think, despite diffeomorphic invariance, the Poincaire grp still holds a special place becuz the metric still has the +---- signature
@ACuriousMind ooh
@Slereah For example, Diffomorphic invariance is there even in SR and Newton. But we cant really say that these theories do not have Galilean and Poincaire group built in
But yeah, symmetries do go away in curved spacetime
I guess we have to talk about local symmetries in case of curved spaceime. Now the Poincaire group shows up as an approximate symmetry that holds locally
Becuz the first derivative of metric can b made to vanish
So we get Poincaire group as sort-of a symmetry of GR as well
I'm just wondering what QM is trying to tell us about the fundamentals of the universe. If QM is fundamental, the observables it describes have to have something special about them
@MoreAnonymous why are u treating me like BJP agent it is all we grown up with even Congress government did support the cause because they believe in India as civilization state.
@RyderRude both classically and quantumly, every observable is an infinitesimal transformation and so generates a one-parameter group of transformations that are potential symmetries. They are symmetries exactly when the observable commutes with the Hamiltonian, i.e. when the observable is conserved.
@ACuriousMind oooh i got it. The Poincaire symmetry is the global symmetry that we "deform" to get to GR.... just like how we deform the global U(1) symmetry to get to electromagnetism
@ACuriousMind i agree that, on paper, we can hav non-symmetries as observables, both classically and quantum mechanically. But maybe the observables that r physically realised in nature have to do with symmetries
Or maybe the observables that r physically realised just have to do with the measurement devices we've been able to build
But i dont like QM's dependence on the measurement devices we've been able to build
If only there was a natural mechanism for energy/momentum/angular momentum, etc to get "measured" in nature
@ACuriousMind (ignore if your not interested but do you know the answer to this)?
Ugh ... One moment
What is the minimum potential energy required (to behave as a turning point) for an elastic collision between $2$ point particles $A$ and $B$ with velocity $v_A^{\mu}$ and $v_B^{\mu}$ in a relativistic classical mechanics? any thoughts would be welcome
I just dont know how to take QM seriously as a fundamental theory, if the Born rule-aspect of the dynamics is so reliant on the existence of man-made measurement devices
How can we get energy, momentum, angular momentum, etc measured using a more simple natural process. @ACuriousMind
tough luck; whenever whatever else we hope to be fundamental clashes with quantum theory, it is quantum theory that wins. So, I do not know how you want to have a fundamental understanding of anything if you do not accept it as being fundamental. And what flight of fantasy led you to think it is all about man-made measurements this time?
@RyderRude look at those conserved quantities. arent they pretty little things?
@RyderRude Who claims that natures "likes" to measure any observable?
Again, this is just the measurement problem
you seem to have decided that the measurement/collapse whatever must happen objectively in nature and you're demanding an explanation
but e.g. $\psi$-epistemic relative interpretations don't have this problem at all because they don't consider it an objective fact that any measurement happens at all
You are not alone in wanting a definite answer to the question of what the heck actually is going during a measurement, but again: This is the measurement problem, it has haunted us for a 100 years, and you will not find an answer by just trying to re-ask the same question in slightly different ways
I am of the firm belief that having a proper lecture set to cover just interpretations alone will be helpful. It will at least be helpful for weeding out really silly misinterpretations of interpretations.
@ACuriousMind sorry again. I said "measure" instead of "decoherence". I'm currently subscribing to the "objective collapse due to an unknown mechanism after decoherence" interpretation, so this is y I tend to use the terms interchangeably. But i really only mean to ask if nature prefers to decohere certain obserbables on its own. This question is answerable independent of interpretation about objective vs relative collapse
So I've managed to confuse myself.
We know if the energy equal to the potential energy then point at which the energy exceeds the potential energy it behaves like a turning point (slide 2). Usually in the collision of (say billiard) rigid balls we assume $V = \infty$.
I realized since it's behavi...
@naturallyInconsistent Yes, the (understandable) interpretational agnosticism ("Let's just say we do Copenhagen here and never return to the question of what that actually means") of most standard texts and courses produces a lot of damage in the long run
Hello people, just wanted to confirm that will the total wavefunction of a Helium atom be antisymmetric? I read this in a Atomic and Molecular physics book, and am confused that why is it so. Helium acts like a boson, and the wavefunction should be symmetric I feel.
@MoreAnonymous Part of the problem is that I have worked out a bit on why you are having this misinterpretation, yet I am also worried about reviving such an old dead horse to give a totally different answer.
That the atom as a whole "is a boson" means that wavefunctions over several such atoms should be symmetric under exchange of such atoms, but that doesn't tell you anything about what the wavefunction of a single atom is under exchange of its individual electrons
@MoreAnonymous You started with boost invariance when you wrote down that adding constant c to both the velocities of A and of B in the expression for KE of each, and that immediately leads to an absurdity. Nobody talks about turning points that way.
@MoreAnonymous And so what does "potential energy" mean in special relativity? Since energy is not invariant under Lorentz transformations, what does your question mean in that context?
@ACuriousMind Did I get this right--->> The wavefunction of individual helium is antisymmetric, but combination of this wavefunction when written for identical helium atoms will be symmetric?
Inspired by this question a normal extension would be to ask:
What is the minimum potential energy required (to behave as a turning point) for an elastic collision between $2$ point particles $A$ and $B$ with velocity $v_A^{\mu}$ and $v_B^{\mu}$ in a relativistic classical mechanics? (Assume numb...
@MoreAnonymous It is physically absurd. You did not even work out which scenario you are looking at a turning point for. There is no saving that entire line of argumentation. There is no sense in having turning points be invariant to transformations that way.
@MoreAnonymous Now, you've started this conversation by asking a one-liner in chat as if it was a simple question, and now I'm up to 2 full physics.SE posts I have to read to understand what you even meant. I feel tricked :P
(Also, I regularly look at bountied questions anyway)
@MoreAnonymous It should be quite easy to prove that after deriving the conditions you want the turning points to appear, there will be no more possible transformations of the form you are wanting to appear.
@MoreAnonymous As usual I don't really understand what you're talking about. For instance, "we" don't usually "set $V=\infty$ in elastic collisions", we usually don't talk about potentials at all when we talk about this kind of elastic collision.
@ACuriousMind we do. For example see hard sphere model
If the potential is 0 the particles go through each other right? So between an elastic collision and particles going through each other. There should be a minima I'm potential that's all I'm saying
@MoreAnonymous Look, are you talking about repulsive turning point or oscillations? Are you talking about 1D or higher? Are you talking about orbits? Elliptic or Hyperbolic? Are you talking about van der waals? Are you talking about ideal gas? Are you talking about rigid bodies?
@MoreAnonymous Then by statistical thermodynamics, because there will always be higher and higher energetic particles of ever lower probabilities, V has to be infinite. Is that problem solved?
@naturallyInconsistent repulsive turning points. All collisions are reduciable to a 1 d problem. These details of point or rigid object do not matter in this context.
@naturallyInconsistent interestingly this model is used for gases:
@MoreAnonymous They manifestly do. Even in linear collisions you have to take angular momentum into account, and the collisions are often only reducible to 2D, and only by throwing away some information can you do it in 1D
@MoreAnonymous So why did you not put this in your question statement?
@naturallyInconsistent ah i have not mentioned the spinning rigid bodies so u can ignore that
@naturallyInconsistent cause most of the physics i learned wasnt in University so i have difficulty gauging what to include or not. Secondly it only becomes relevant when people start claiming this is physically absurd
The hard spheres model arises from considering statistical thermodynamics. The moment you have T>0, the infinite KE limit is potentially possible. Unless you arbitrarily set that the spheres are not actually hard at high enough energies, the only natural choice is for V to be infinite too.
Relativistic thermodynamics is in a self-delusional state. It is very easy to just guess the Maxwell-Jüttner distribution, and it is also appearing in many textbooks, but anybody who has thought through about the subject understands that that particular distribution is physically nonsense.
Or rather, I should not say self-delusional, but only that people who seriously treat that subject is self-delusional. One must be honest and note that obvious glaring problems with the naïve treatment of that subject.
Physics, or reality, is just that much more horrible than we like to pretend it is.
@ACuriousMind Can we say that this GL(n) is an "active transformation" symmetry of GR, as opposed to diffeomorphosm invariance which is just a passive transformation symmetry of any co-ordinate invariant theory
I mean the minimum potential problem is a problem of mechanics imo. Yes, in reality the world isn't solely special relativistic. But generally there's are considered to be complete subjects
Here's something that i wanted to ask about transformations. In QM the observer is never in superposition. Does this mean that there is an asymmetriy between active and passive transformation in QM?
But diffeomorphism just give different labels to the same manifold. There is no notion of diffeomorphism transformations if we directly describe the theory using manifold objects without using any co-ordinates @Slereah
@Slereah oh. In that case, active diffeomorphisms seem equivalent to "local GL(n) transformations". I guess then the only special thing about GR is that it makes the Christoffel symbols a dynamical entity
@naturallyInconsistent because u apply one transformation and not a superposition or whatever of transformations. If the observer was in superposition and one superposition of him accelerated. How would u describe the system?
I'll take weak equivalence principle in the formulation as given on Wikipedia page:
The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception.
Consider a wave function $\Psi(r,t)$ and suppose that...
Let's take free Dirac field theory. This can b expressed in local Gauge invariant way if we wanted. This does not require us to make the $A^{\mu}$ dynamical
@RyderRude There is at least half a dozen alternatives to GR. It is just that it seems that there are no known deviation from GR that isn't quantum in nature. It is actually more a case that it is so difficult to do experiments involving GR than anything else.
@naturallyInconsistent also i will use language as i please especially when not crossing an social boundary. Secondly, this is only the second time i raised it.
@RyderRude With SR the problem is tremendously easier. It is more a case of understanding what the story is about and realising that the consequences follow from the basic postulates. It is also why we have relatively little difficulty with SR QFT, as opposed to GR.
The difference between GR and SR and the difference between GR and EM are kind of a different nature
SR and GR technically have the same structure as spaces, you could just define your spacetime in the same way
In that sense it is a "gauge theory", but it is not dynamical
GR and EM are both dynamical gauge theories but GR has the Special Thing
Diffeomorphism invariance, being geometric, being a Cartan connection, being natural, general covariance, the splitting of the Atiyah sequence, the soldering
I don't understand this part of the answer : "It is rather crucial to note that diffeomorphism invariance is not the same as gauged GL(n)-invariance - the former is a basic aspect of all "coordinate-invariant physics", while the latter essentially arises because the Ricci scalar in the Einstein-Hilbert action is analogous to the gauge-invariant Tr(F) terms in ordinary gauge theories."
Y is the Ricci Scalar term in the action necessary to have the GL(n) gauge invariance @Slereah @ACuriousMind
@Slereah this is what im saying. GR is special only in that it makes the connection dynamical using the Ricci scalar term. This is analogous with electromagnetism where they use the $F^{\mu \nu} F_{\mu \nu}$ term.
But in both cases, the free theory (SR and Dirac respectively) can be written in a way to have the same local gauge symmetry
Hello. Consider a device which shoots electrons through a slit. We have two situations 1) don’t measure the electron at any point 2) put a detector after the slit.
In this context, what does “the ensemble evolves unitarily, not the individual system” mean?
I had thought that one can assign the shot electron a pure state, and then manipulate it mathematically as per usual (apply the unitary time evolution operator on it and so on)
I'll take weak equivalence principle in the formulation as given on Wikipedia page:
The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception.
Consider a wave function $\Psi(r,t)$ and suppose that...
