The h Bar

DIRAC1930

00:06

So he is viewing this $\sin^2$ formula as being the probability as a function of time for the system to transition from a state with proper energy $H'_0$ to one with proper energy $H'$ and since this function tends to a delta function centered around $H'_0 -H'=0$, he makes the conclusion that as time increases, the system becomes more likely to transition to an energy that obeys $H'_0=H'$

So are we just measuring the kinetic energies of the particles?

bolbteppa

It's perturbation theory, you are assuming a basis of solutions of an exactly solvable problem, and viewing the action of the perturbation as bouncing the state of the system between the different states of the exactly solvable problem. If your exactly solvable problem is a system of free particles, you are measuring the energy of those free particles which is kinetic energy, however it doesn't have to be a system of free particles

DIRAC1930

Ah yes I forgot about that

bolbteppa

Making conclusions about large values of $t$ (or limits of $t$ rather) is also problematic because we are talking about the argument of a periodic sin function, only it's value which maximizes the value of the squared sin function is going to be relevant to the amplitude

DIRAC1930

At the time of measurement, the perturbation is still on. Why aren't we measuring total energy there?

bolbteppa

The form of the perturbation is basically irrelevant, if the perturbation is time-independent it's not going to do anything to any of this, if it's time-dependent it's going to add the extra $\omega$ thing to the sin (and an ugly integral over the $w$ to deal with, which is why they talk about periodic perturbations to ignore it), that's all. In terms of modelling a measurement as a classical system weakly interacting with some closed quantum system, the interaction is naturally time-independent

All you're talking about is squaring the coefficient in a mode expansion of a wave function at the end of the day, it's very primitive, in this case it's writing out the wave function as a en.wikipedia.org/wiki/… and squaring the coefficient to first order

I don't really see Mandelstam explaining why you can't take this $t \to \infty$ limit in his (6) and make some conclusion about it, instead what he says below is basically invoking what L&L do, but without going the step further of deriving (6) as they do which then lets you make the point he's trying to say, even though he's apparently got a problem with this derivation (note no references are given, yet comments about references are made, maybe it's a translation error? Doesn't look like it)

This is all so primitive I can't see why there's a problem with it, I don't like having to use perturbation theory to arrive at a conclusion like this, but I don't trust these other approaches and everything seems to make sense this way and it is model independent

This little thing about the paragraph below (6) reducing to L&L does not look like a good sign, but I'm not doing this properly so it may be off

DIRAC1930

00:42

Btw that periodic perturbation $F e^{\imath \omega t} +. F^* e^{-\imath \omega t}$ reduces to a normal time independent constant perturbation $\mathrm{Re} F$ in the limit $\omega\rightarrow 0$

DIRAC1930

00:53

I think he doesn't take the limit in (6) immediately because he is trying to interpret what the equation actually means first

ACuriousMind

01:12

@SillyGoose no, that doesn't look like the kind of thing I'm interested in

DIRAC1930

The thing that is confusing about L&L is that he seems to be dealing with two uncertainties simultaneously i.e. the one that is stated in the Mandelstam paper but also the one where the act of measurement is changing the momentum of the particle you are trying to measure

Also, I don't know why $v_x$ is an exact value when there is a $\Delta p_x$

bolbteppa

it's just $\Delta f = f(x + \Delta x) - f(x) = f'(x) \Delta x$ applied to $\Delta E = (\partial E/\partial P) \Delta P = v \Delta P$

DIRAC1930

But surely an uncertainty in P means that there will be an uncertainty in v since P=1/2 m v

bolbteppa

In a Taylor expansion, the derivatives are evaluated at some point, that's all it is, the $(x - x_0)^n = \Delta x^n$ terms encode the approximation

DIRAC1930

01:37

Ah thanks

Slereah

08:53

One thing I hate in GR is that weird thing where the less information you have in a bundle, the more dimensions it has :V

Silly Goose

why is there "no single expression for all physical systems" for the Lagrangian? (from wikipedia on Langrangians)

ACuriousMind

@Slereah what do you mean by "less information"? Do you think about e.g. smaller structure groups as having "more information" than larger ones?

Silly Goose

Is $\mathcal{L}$ just defined to work...as in produce the equations of motion as wikipedia proceeds to say?

Slereah

@ACuriousMind Technically a larger structure group I guess

ACuriousMind

@SillyGoose Well, for one, because there is no guarantee that every physical system even has a Lagrangian, arbitrary dissipative systems are notoriously hard to formulate via Lagrangian mechanics

Slereah

08:57

Like how conformal geometry uses those weird tractor bundles despite having less information

ACuriousMind

Not exactly sure but that sounds to me more like a case of "fewer constraints" rather than "less information"

Silly Goose

oh now what would we do when we cannot use Lagrangian mechanics :0

ACuriousMind

depends on why you cannot use it - I mean, to even state that there is no Lagrangian you already need to have some other description of the dynamics of the system

Slereah

@ACuriousMind Well you can't get an absolute length in the case of conformal geometry

although I'm not sure that pattern holds true for all structure groups

ACuriousMind

sure, so - completely handwavingly - you need to carry around a relative length everywhere

Slereah

09:00

pretty much

Silly Goose

does Hamiltonian mechanics face similar issues?

Slereah

the tractor bundle is basically like the tangent bundle, but instead of being over the manifold it's over the scale bundle

ACuriousMind

because you don't have a constraint that says everyone has to use the same length scale

Silly Goose

And so then does Quantum Mechanics also have trouble describing certain systems? If we cannot write down its Hamiltonian

Slereah

actual vectors are just tractors for which you have set a scale everywhere

You don't need to explain it though

I said that I hated it, not that I was confused :V

can't explain away the hate

ACuriousMind

09:02

@Slereah I understand, I'm just saying perhaps a change of view might alleviate the hate ;) When you have fewer constraints, you need more information to specify something uniquely

Slereah

It feels like it is somewhat artificial, though

Like this only happens because we have to define everything via vector spaces

Same way you can use RP^n or you can use R^n+1

ACuriousMind

@SillyGoose Not exactly. When I say "dissipative", you have to think of something like friction, or an open system from which stuff can "escape"

but of course such systems have, at least in principle, non-dissipative descriptions where you include the part to which stuff can escape in the system which then may be more amenable to Lagrangian or Hamiltonian descriptions

Silly Goose

hm interesting i shall look into this more next semester or over the summer >:D

now i must naively ask if there will ever be a ball theory

ACuriousMind

09:19

not sure what ball theory would be, but there is a hairy ball theorem

Silly Goose

xD

i mean to ask if one can make a theory like string theory except instead of strings use higher dimensional things instead (balls)

ACuriousMind

09:34

@SillyGoose that's called brane theory and there are some general results that 1d objects are a sweet spot

like, these branes appear as dynamical objects in string theory/M-theory already but they are not well-suited as fundamental objects in the way strings are

mostly I think it's just that the 2d worldsheets of strings are very special - 2d CFTs are very particular theories and you don't have analogues for higher-dimensional worldvolumes

Mr. Feynman

11:24

I caught a flu and since then I enjoy GR way more than QFT, is this a side effect of the flu?

Slereah

11:45

got the spacetime flu

bolbteppa

@Slereah What's a bundle

Mr. Feynman

are PS5 bundles fiber bundles?

Slereah

12:06

@bolbteppa The money I have to pay for my bills every month 😔

ACuriousMind

12:20

@Mr.Feynman clearly it's the other way around: You started enjoying GR more, and then you got sick as karmic punishment

Ryder Rude

Hi

@bolbteppa But a classial Spinor field wudnt make sense tho becuz spinors field isnt observable. This is y I'm dealing with the spinor field by transitioning from QFT to rel. QM or non-rel. QM

But my rel. QM roadmap is handvawy becuz idk if the steps work out mathematically. And the non. rel QM roadmap is fine but it is too indirect. We remove relativity and then add it back

bolbteppa

The only spinors are in the four-current, but the four-current obviously has to reduce to the classical four-current regardless of spinors implicit in it

This transition from QFT to rel QM doesn't exist, it's the same thing in a different language (unless you mean single particle rel QM which is just an incorrect theory)

Ryder Rude

12:37

@bolbteppa what is ur starting point in QFT? Is it not the operator field equations?

@bolbteppa what is ur starting point in QFT for this derivation? Is it not the coupled operator field equations?

bolbteppa

The non-relativistic limit in electromagnetism is crazy, you're taking the $c \to \infty$ limit, there's no magnetism in this limit

Ryder Rude

Yeah, I'm getting only a classical particle in Coulomb potential. We're adding relativity later to it

It is a very indirect route

bolbteppa

You can't add relativity by taking something with $c \to \infty$ and then ignoring that, if $c$ is not at $\infty$ then you've immediately already got all of EM

The standard approach is very simple, I would recommend learning the normal way first before thinking about all these dances like 'Coulomb plus blah equals Maxwell', these things usually have issues or technicalities or some issue

Ryder Rude

@bolbteppa I think the Ehrenfest theorme approach works out for free electromagnetism. But with the interacting theory, how does it make sense to take the expected value of the Dirac operator field?

Becuz we can observe electric fields, so it makes sense 2 take the expected value. But we cannot observe Spinor fields

ACuriousMind

you really need to stop calling the classical limit via coherent states/correspondence principle "Ehrenfest's theorem"

Ryder Rude

12:42

Yeah, I'm omitting the details :P

ACuriousMind

Ehrenfest's theorem does not produce a classical limit, this is a frequent misunderstanding. Again, the problem is $V(\langle x\rangle) \neq \langle V(x)\rangle$ and that's the hard part

Ryder Rude

But Ehrenfest theorem is involved. The coherent states only allow us to shift from $\langle V(x) \rangle$ to $V(\langle x \rangle) $

I'm just saying that Ehrenfest theorem is our starting point

Coherent states r the next step

ACuriousMind

@RyderRude yes, but it is the latter that is the true "quantum-to-classical" transition

bolbteppa

Ehrenfest is trivial, it's just a time derivative of an expected value, it basically doesn't matter

ACuriousMind

because that these two things are not equal is an essentially quantum phenomenon

@RyderRude and I'm saying that that's a terrible naming convention that fuels common misunderstandings of Ehrenfest's theorem

as bolbteppa says, it's barely even a theorem :P

Ryder Rude

12:45

Ehrenfest is only the starting point of the derivation. To get classical teory, we need specific states like coherent states along with ehrenfest theorem

@ACuriousMind ye, it's just Heisenberg equations basically

So we can say that Heisenberg's equations r our starting point

bolbteppa

The real magic is basically in understanding how to take a classical Lagrangian, write out it's Poisson brackets, promote them to commutators, and understanding the meaning of these operators, and what they act on, and why.

Or something equivalent if you don't make the Poisson brackets explicit (Which most sources don't)

ACuriousMind

that's quantization - the exact inverse of what Ryder is trying to do

Ryder Rude

@bolbteppa what is ur roadmap from the free Dirac QFT to the free charged particle SR.

The onlu roadmap i know is to switch to rel. QM first by dotting with $|x\rangle$

bolbteppa

Right, but you simply can't start from a quantum theory without the classical theory, it's just a mathematical game trying to do this, it's like missing the whole point of QM

Ryder Rude

If we started with Heisenberg's eqns, taking the expected value wudnt make sense for spinors

ACuriousMind

12:49

@bolbteppa I disagree with this claim

bolbteppa

Spinors don't exist classically, however we know spinors satisfy the Klein-Gordon equation, and we know the Klein-Gordon equation can be derived from quantizing the free particle SR action

ACuriousMind

if we really believe in quantum mechanics as a theory of the world, then we should be able to start with a quantum theory without already having to also specify some accompanying classical theory

the classical theory should just be some limit

Ryder Rude

@bolbteppa this is a mathematical relationship, but this derivation isnt justified physically. Physically to get from QM to CM, u gotta take expected values

Becuz QM is probabilistic

ACuriousMind

@RyderRude I also disagree with this claim; depending on your idea of how the classical world emerges from the quantum world, it's not obvious that the classical values are expected values - they could just be measurement results

bolbteppa

Absolutely not, if the classical theory does not exist, there is absolutely no way to measure the quantum theory in the first place, so the quantum theory does not exist. Without being able to measure the quantum theory via the classical theory, we just have nothing. Again, this is the unavoidable contradiction in QM that the main founders spent their lives trying to tell people to accept

ACuriousMind

12:52

this is a fundamental complication for idea of the classical limit - if you leave everything undisturbed you probably want to talk about expected values, but what if the correct "classical limit" of your situation is "everything becomes hopelessly entangled with the environment" and then do some sort of decoherence?

Ryder Rude

@bolbteppa u always talk about this orthodox interpretation of QM. U can just use many worlds or relational QM to solve measurement problem

This classical/quantum divide is not a mainstream idea anymore

@ACuriousMind yeah. Decoherence is also involved i guess

ACuriousMind

@bolbteppa I understand where you are coming from but I disagree: We can start to formulate QM purely as a theory about probability measures on observables with no claim about a classical/quantum distinction or some systems being "classical" or "quantum"

bolbteppa

@RyderRude These alternatives are not serious

> Before the book was released for sale, Heisenberg privately expressed regret for having used the term, due to its suggestion of the existence of other interpretations, that he considered to be "nonsense". en.wikipedia.org/wiki/…

ACuriousMind

(In fact, the book by Moretti I mentioned yesterday has a chapter showing that under reasonable assumptions such theories of probability are necessarily modeled by projection-valued measures on Hilbert spaces, bypassing entirely any arguments about quantization to arrive at the usual formulation ofQM)

I'm not talking about interpretational issues here: This does nothing to "solve" the measurement problem, the nature of the probabilities is still exactly as mysterious as usual. I'm just saying that the claim that we need a classical theory to establish a quantum theory is wrong on a technical level.

Ryder Rude

@bolbteppa i think the classical / quantum divide is not serious. How do you make this idea mathematical? How is microscopic physics supposed to fundamentally depend on macroscopic systems?

This is very convoluted math. Many worlds just uses the Schrodinger equation

But Born rule needs 2 b imposed by hand

bolbteppa

13:00

5

A: What does it mean that a free particle has no definite energy in quantum mechanics?

if you measure the energy of a free particle you will still get a value right? And won't this energy necessarily correspond to one of the infinite solutions for the free particle? No, when you measure the energy of a (not necessarily free) particle you find an interval of values due to the fini...

I sincerely doubt he's doing anything right if he's writing this waffle about a free particle

This stuff is so simple, it really is, and they have to state nonsense about energy not being definite etc, I just can't believe it

@RyderRude Where is the Schrodinger equation coming from in MWI? Pulling it out of thin air (unless you accept everything QM already says before misinterpreting it)

Ryder Rude

But physics is about modelling and not about where fundamental equations come from

bolbteppa

Where is the wave function coming from in WMI, these things are just random blind assumptions when you go to these ridiculous alternatives

Ryder Rude

The wavefunction is suppose to be the ontology of the universe. It takes the place of the manifold from classical mechanics @bolbteppa

All our observations are within that wavefunction

bolbteppa

If normal QM had an ounce of this kind of waffle, the entire world would have rejected it a century ago, there's no way people would give up classical mechanics for this kind of flimsy logic, it's because normal QM is solid that people had to accept it despite the implications, we'd all like to deny it and pretend the world is rational like these alternatives really want to say, but it's not convincing

@RyderRude By this logic, angels dancing on the head of a pin as just as cogent an explanation for the world as any other theory, and equally valid, you know in your bones this logic is wrong, why apply it to something more serious

Ryder Rude

@bolbteppa But the classical/quantum divide school is severely ill-defined. How r we supposed to judge it if it doesnt provide its math?

Besides, quantum entanglement has been confirmed on larger and larger systems

ACuriousMind

13:05

Okay, I see there is little use in trying to continue this discussion, we're just talking past each other.

bolbteppa

We're not just blindly modelling theories and playing games, there's serious reasons for doing what people do, all these alternatives are just picking and choosing things from normal QM and saying we can ignore the implications that normal QM implies

Ryder Rude

@bolbteppa Does the classical-quantum divide school say that Schrodinger equation wud stop applying for some large enough systems? And the dynamics wudnt be reversible in principle?

bolbteppa

@RyderRude The classical limit is $\psi \approx e^{iS/\hbar}$, where we then take $\hbar \to 0$ and the only way to get classical physics back is for $S = 0$ i.e. $S$ minimized i.e. classical mechanics exists completely again. It's not ill-defined, you literally derive the Schrodinger equation and every QM operator from this, etc...

Ryder Rude

@bolbteppa But then it's not a divide. It means that classical mechanics is obtained in an approximate limit

bolbteppa

When we take that limit, the wave function doesn't exist, we have $e^{0/0}$, there's your divide

Ryder Rude

13:11

How is this different from saying that everything isnt fundamentally relativistic becuz we can take the $c\rightarrow \infty $ limit @bolbteppa

bolbteppa

Everything is fundamentally relativistic (in a classical world), the non-relativistic limit is a very accurate approximation

Ryder Rude

Yeah. But same logic goes for QM and CM

bolbteppa

No it doesn't, this is why QM is so different and upsetting etc...

Ryder Rude

U can get stuff like 0/0 while doing $c\rightarrow \infty$ too

bolbteppa

@RyderRude Instead of assertive claims about very basic things in QM not being serious (and talking about alternatives...), you should at least try to learn what the canonical approach says before writing it off, at the very least realize this stuff was thought up by nobel prize winners nearly a century ago

Ryder Rude

13:14

It just means we're doing it the wrong way

@bolbteppa i want to know a prediction of this classical-quantum divide school. Does it say that, for large enough system, schrodinger eqn wud stop applying in principle and evolution wud b irreversible?

Mr. Feynman

Question (for students and former students): do you work out all the exercises in all books you read or select just a bunch of them?

bolbteppa

It's exactly what you'd expect, classical experiments on classical systems yield classical results to the classical accuracy of your experimental apparatus, there's no need for QM in such descriptions (i.e. $e^{0/0}$ stuff), if QM arises it's because of non-trivial quantum effects where the wave function starts to matter, it's basically tautological

@Mr.Feynman There's not enough time to try to do everything, and you may regret wasting time on unimportant stuff

Mr. Feynman

That's precisely the reason why I asked that. I realised that the number of sources is growing without bound and a lifetime wouldn't be enough to work out all of them...

Ryder Rude

@bolbteppa this classical/quantum divide school is not accepeted among Nobel winners. E.g. thooft is working on a cellular automata model to reduce everything to a classical theory.

If the solution was as simple as a classical/quantum divide, thooft wudnt b working on that theory

ACuriousMind

@Mr.Feynman I have never worked out an exercise unless it was assigned to me as coursework or I found it genuinely interesting :P

Ryder Rude

13:24

@ACuriousMind do u get confidence in ur understanding by reading alone?

Mr. Feynman

@ACuriousMind I tended to to avoid exercises and focus on the theory too but then I thought my problem solving skills could be affected from that

I mean, conceptual exercises do need theory, but it's not automatic that knowing theory one (I)'ll be able to solve them

On the other hand I can't really enjoy most of them unless the question is somewhat exciting

Oh, also @ACuriousMind there's no such thing as assignments in most italian universities :P

ACuriousMind

@RyderRude No, I get confidence by having to apply what I read, e.g. trying to explain it to someone else, or following an argument building on what I read and noticing whether it makes sense to me or not

Ryder Rude

@ACuriousMind that's very badass

Mr. Feynman

There was a German student following my QFT course last semester and he was surprised there were no assignments

ACuriousMind

@Mr.Feynman Hm. Well, to me the primary benefit of exercises was never the experience of me trying to solve them on my own, but that during the week they were assigned, I would discuss them with other students and then we'd discuss their solutions in class when they were returned after grading

I've learned much more from talking about the exercises than from just solving them

so I generally don't bother doing exercises just for myself

Mr. Feynman

13:33

Oh sure, that's why I don't really like doing them either

On the other hand some authors suggest that one shouldn't skip the exercises otherwise they will not really understand the material and I see why they're saying that

DIRAC1930

@bolbteppa Is he using $v_x$ to denote the mean/expectation value of $v_x$ (i.e. $\langle v_x \rangle$)which he is expanding around in the Taylor expansion?

Mr. Feynman

To be honest I enjoy doing Math exercises more as in some Physics problem even the questions are somehow mysterious :P

bolbteppa

It's all involving the eigenvalues

ACuriousMind

@Mr.Feynman well, it depends on what the nature of the exercise is and on how you read texts. if the text says "as shown in exercise 2.3" and you're the type of person who cannot continue reading until you've seen the most polished solution of that exercise, then of course you have to do the exercises :P

generally I'm fine with leaving gaps in my understanding and filling them later, I know others are not - no matter how often I recommend it to them ;)

Ryder Rude

What r ur favorite theories of physics and philosophy

DIRAC1930

13:37

But how can he even define an absolute value of $v_x$ when there is an error

Ryder Rude

Mine is quantum gravity for physics, and neutral monism for philosophy

bolbteppa

$E$ is an exactly measurable eigenvalue, $E = p^2/2m$, $\frac{\partial E}{\partial p} = p/m = v$

(Notice even here, if we start waffling about energy not existing, or energy living in an interval, we can't do a single thing...)

DIRAC1930

Thats true

ACuriousMind

@RyderRude Favourite physics is obvious: It's gauge theory and QFT, just look at my physics.SE tags.

DIRAC1930

I think it's best to think of the errors as being the difference in the proper energy and that of the total energy

ACuriousMind

13:41

I have no idea what a "favourite" philosophy is supposed to be

Ryder Rude

It means what philosophy do u find favorite for the ontology of the universe

ACuriousMind

Are you asking me what I actually believe in

Ryder Rude

@bolbteppa @Mr.Feynman @DIRAC1930 @Slereah What r ur favorite physics and philosophy theory

@ACuriousMind yeah. That wud b the favorite one.

ACuriousMind

@RyderRude I mean, you realize you've just used the word "favourite" in very different meanings for physics and philosophy, right?

for physics you just named a subfield, for philosophy a specific philosophical viewpoint

Ryder Rude

Yeah :P. Becuz most physics is tested. Philosophy is a belief

ACuriousMind

13:45

I don't know how you can say something like that after all the discussions about epistemology

bolbteppa

I would recommend reading Socrates and Descartes

ACuriousMind

nitpick: Can't read Socrates, only what Platon claimed Socrates said ;)

Ryder Rude

@ACuriousMind i mean that there is no way to settle the debate for philosophy of ontology. So it becomes a belief

Slereah

Vulcanology and confucianism, obviously

just love volcanoes and obeying the rituals of my ancestors

Ryder Rude

@ACuriousMind Y is gauge stuff ur favorite? Mathematical beauty or the real world application?

bolbteppa

13:48

The guy was so smart he even said this:

> “I may instance olive oil, which is mischievous to all plants, and generally most injurious to the hair of every animal with the exception of man, but beneficial to human hair and to the human body generally; and even in this application (so various and changeable is the nature of the benefit), that which is the greatest good to the outward parts of a man, is a very great evil to his inward parts: and for this reason physicians always forbid their patients the use of oil in their food, except in very small quantities, just enough to extinguish the disagreeable sensation of smell in meats …

ACuriousMind

@RyderRude I don't think we agree on what distinguishes "belief" and something that is "tested", but I'll just point out that that discussion itself is already philosophy, whatever. I guess my favourite philosophy is Camus' absurdism.

Slereah

Ryder Rude

Ooh this philosophy is like upgraded nihilism

I mean Camus absurdism

I dont think "meaning of universe" is a meaningful concept

So it's absurd to look 4 it

Like questions like "y is there something instead of nothing"

@Slereah omgg :P

ACuriousMind

@RyderRude It is an extremely layered subject - being discussed in intro to EM classes as well as being the foundation of some recent purely mathematical advances - with plenty of different ways to think about it, occurs in plenty of mathematical and physical contexts and it is an instructive example of the idea that different physical theories can describe the same system so it also can lead to discussion of the nature of models. It's just very rich.

Ryder Rude

Yeah, Gauge stuff is extremely successful for all forces for some reason

I had a very nice generalisation of gauge stuff

It is taking gauge stuff to the maximum

ACuriousMind

13:58

@RyderRude That's not really what I meant but that is easily explained by Weinberg's argument that massless vector bosons can only occur via gauge fields

Ryder Rude

Yeah. There's that too

But i was thinking about the extreme success of the other approach

@ACuriousMind the approach where we try to make global symmetries to local and discover the interaction term

It is not a motivated approach but it works very well

ACuriousMind

it's garbage and I hate it

Nihar Karve

physics.stackexchange.com/search?q=user%3A50583+fetish

ACuriousMind

@NiharKarve lol, that is exactly how I find that post when I need it

Nihar Karve

haha

Ryder Rude

14:01

@ACuriousMind yeah, there's no motivation 4 this approach other than "we have symmetry fetish"

Mr. Feynman

@RyderRude Philosophy-wise I'm so illitterate I'm not able to express any opinion. Physics-wise, it depends on the period. Naturally, if we ever get to a working theory of everything, that'll be my favourite :P

@ACuriousMind LOL

Ryder Rude

@ACuriousMind I had this very nice generalisation of the Gauge stuff. We try to make the dynamical equations invariant under all diffeomorphisms on the gigantic phase space of a classical field theory. So this way, we no longer restrict to canonical transformations. And then we also make the "residue" term dynamical.

ACuriousMind

oh no

you're not baiting me into another rant about diffeomorphisms

Ryder Rude

I mean just co-ordinate changes

Mr. Feynman

Oh, we have one of the three most popular topics in the chat going live

Ryder Rude

14:08

Imagine what force we wud discover if we made the residue term dynamical

Mr. Feynman

For the record, I'm talking about representation theory, spinors and diffeomorphism invariance

ACuriousMind

I've written up my current understanding of "diffeomorphisms as gauge transformations" here and I don't really have more to say

Ryder Rude

@ACuriousMind Ok i think i didnt phrase it well. The theory on phase space is obviously already diffeomomorphism invariant. We're NOT introducing anything new in that regard

We're just making the residue term dynamical

That's the new stuff we r introducing

Mr. Feynman

@ACuriousMind Weinberg QFT I?

Ryder Rude

Basically, my idea is "wut if u generalised the logic to come up with GR to the phase space manifold". To come up with GR, we turn the residue field, i.e. the connection, into dynamical

ACuriousMind

14:13

@Mr.Feynman yes, it should be in there when he develops the theory for spin-1 particles

DIRAC1930

I'm going to stop thinking about it now but the whole problem is essentially 'given that we can measure the exact eigenvalues $\epsilon,\epsilon'$, what are the fundamental bounds to which we can deduce the energies $E$ and $E'$? The answer is dependent on the time of interaction.

Ryder Rude

So wut if we made the residue term for the phase space diffeomorphism into a dynamic entity?

ACuriousMind

I don't really know if this argument is original to Weinberg but it's where I learned about it and what everyone always references

Ryder Rude

Ok here's wut im saying precisely : "wut if we defined a connection on phase space and made it dynamical"? @ACuriousMind

Mr. Feynman

@ACuriousMind I'll check it because I've been wondering some things about how the gauge field quantization is different from other fields where you just impose the commutation/anticommutation relations

That clearly has something to do with the fact that there is a local gauge symmetry in the classical theory already, that is a constraint but I have to expand on that

ACuriousMind

14:20

the high level argument is more or less this: A massless spin-1 particle has just 2 d.o.f. based on Wigner's classification of Poincaré representations. But a "spin-1 field" has to be a 4-vector, which has 4 d.o.f. It's simple to get rid of one of these d.o.f. for the massive particles, it's essentially just recognizing that there's a "spin-0 part" you can remove.

But for the massless case, you need to remove one additional d.o.f., and that's what a gauge theory is - a theory where one of your d.o.f. is "fake" and doesn't represent anything physical

Mr. Feynman

@ACuriousMind The last point I think is written in many books. What I think a lot of books fail to convey is the fact that the the electromagnetic field is different from e.g. the scalar or the Dirac field upon quantization because of this freedom

Ryder Rude

Ye, becuz u wud need to part ways with unitarity if u used the non-gauge invariant classical theory

I think this stuff also somewhat arises for spinors

Lemme think

Ofc its not gauge invariance anymore but its smthing similar for the dirac field

Mr. Feynman

@RyderRude as ACM asked some days ago, could you please avoid using so many abbreviations? It makes it difficult to read

Ryder Rude

Okay

Mr. Feynman

Thanks

Ryder Rude

14:27

Okay so the usual complex spinor field would have 8 degrees of freedom at each point of spacetime

Because it's four components, each being a complex number

But the Dirac equation restricts the degrees of freedom

Of this mathematical object

This stuff also aligns nicely with the degrees of freedom of the spin-1/2 field according to Wigner's classification

Remember that in the general solution of the Dirac equation, you only need to use four coefficients

So you get four creation-annihilation operators upon quantisation

ACuriousMind

no, this isn't the same at all, you're just confused about what a degree of freedom is in this context :P

Ryder Rude

@ACuriousMind But since Dirac yields you four creation-annihilation operators, this aligns nicely with the spin-1/2 representation which has two degrees of freedom

So u get two creation operators for particles, one for each spin

And two for anti particles

ACuriousMind

yes, but this removal of the d.o.f. is the same kind of removal as for the spin-0 part of the 4-vector

it's not the same as the removal due to a gauge freedom

Ryder Rude

Yeah, i did say that it's not the gauge stuff :P

But it's still degree of freedom stuff

ACuriousMind

in the Hamiltonian formalism, this is the difference between a first-class and a second-class constraint - only first-class constraints generate gauge transformations

but I'm sorry, yes, this is the same as for the massive 4-vector

Ryder Rude

14:46

@ACuriousMind the constraint classification is very interesting.

@ACuriousMind If the phase space has a dynamical metric defined on it, do the usual quantisation methods apply for that model?

ACuriousMind

why would the phase space have a metric?

Ryder Rude

I mean suppose theoretically

ACuriousMind

What I mean is that I don't know what that would imply for the physics

the usual phase space is just a symplectic manifold, a metric never enters

Ryder Rude

Oh. Since it's dynamical, it affects observables

ACuriousMind

@RyderRude how?

time evolution of observables, i.e. the dynamics, is $\partial_t f = \{ H, f\}$, nothing there depends on a metric on the phase space

Ryder Rude

14:48

Yeah, suppose we change that formula to accomodate for this metric

ACuriousMind

again, how?

Ryder Rude

Idk :P

I was just wondering if quantising this modified phase space theory could yield us a quantum gravity theory

ACuriousMind

so why do you ask me about quantization methods when you don't even know what classical theory you have here :P

Ryder Rude

@ACuriousMind i will try to find some "natural" coupling

Ooh i have an idea

It's very interesting

In GR, we couple the metric to the energy density. In my phase space theory, we couple this metric to the probability density on the phase space @ACuriousMind

So curvature wud b directly proportional to probability density

And then we quantise this stuff

ACuriousMind

I don't know what any of that means in technical terms

First and foremost, I think you misunderstand the nature of the phase space

remember that that the phase space consists of the cotangent bundle of the configuration space, it is not spacetime

so a metric on spacetime has nothing to do with a metric on phase space and vice versa

Ryder Rude

14:53

Yeah, im just taking inspiration from GR. My theory is purely theoretical

ACuriousMind

I haven't seen a theory yet

Slereah

the phase space isn't even the same category of manifold as a spacetime

ACuriousMind

what do you mean "couple the metric to the probability density"

Ryder Rude

I will try to come up with a precise coupling formula to the probability density

Slereah

you're gonna have to deal with symplectomorphisms instead of diffeomorphisms

ACuriousMind

14:54

the probability density also evolves via $\partial_t \rho = \{\rho, H\}$

Slereah

And spoiler symplectic structures are always flat

ACuriousMind

and in general a metric on phase space will not respect the symplectic structure

Ryder Rude

@ACuriousMind yeah, my idea is that this evolution is modified after the coupling

@ACuriousMind ok this answers one of my querries

It means my new theory is no longer symplectic

So usual quantisation wont make sense either

ACuriousMind

you can't just make up a new theory if you know nothing about the concepts you're trying to work with

Ryder Rude

I would need to come up with a quantisation recipe too

ACuriousMind

14:56

in fact, there are manifolds that are both Riemannian and symplectic (and complex), so-called Kähler manifolds and they're rather special

Ryder Rude

Oh

ACuriousMind

the Kähler potential on such manifolds plays a role in many supersymmetric theories

Ryder Rude

I have not studied Kahler manifolds

Slereah

but the symplectic structure is still boring because as I've said they are always flat

there's only one symplectic connection and you're already using it

ACuriousMind

in a sense it is the purpose of the symplectic structure to be boring ;)

Darboux' theorem says we're not losing much if we insist on always thinking in terms of coordinates with the canonical Poisson brackets

Ryder Rude

14:58

I'm just trying to couple a newly defined metric to the probability density on the phase space. It's okay if the structure is no longer symplectic

ACuriousMind

which is why the manifold/coordinate aspect of Hamiltonian mechanics is easier to sweep under the rug than in e.g. GR

Slereah

you can add torsion to it but idk if that does anything cool

ACuriousMind

@RyderRude what does "couple" mean

again, I think you are confused about what the phase space is

if your metric is a dynamical field of the theory, then the space of all metrics is a subspace of the configuration space, on whose cotangent bundle the Hamiltonian theory is built

a dynamical object on phase space itself is an oxymoron

Ryder Rude

But I can mathematically define a metric on the phase space

ACuriousMind

the points of phase space are the possible states of the system, if you have something dynamical "on it", that's not a phase space

Ryder Rude

15:00

It's just pure math

ACuriousMind

you can do whatever you want on a symplectic manifold, but I have no idea why you call this phase space or what this has to do with physics :P

Ryder Rude

@ACuriousMind yeah, this is just pure math. I'm extending the usual stuff

@ACuriousMind maybe it could give you something interesting in the end

Pure math explorationa end well sometimes

Imagine a dynamical metric on which the probbaility evolution depends

ACuriousMind

@RyderRude oh, you can think about whatever you want, I just don't like your implication that this is somehow a minor modification of what we already have

as in, you started this conversation by asking me how to quantize a "metric on phase space"

and it turns out not only don't you actually know what you meant clearly enough to think about quantizing it, you're essentially trying to invent new physics out of nowhere

bolbteppa

Maybe if you add bundles it will work :p

Ryder Rude

I know now that the usual quantisation wont work

ACuriousMind

15:04

you still don't even know what the classical theory here is!

how could you possibly talk about quantizing it

Ryder Rude

I have defined my observables tho

It's just a manifold on which there are functions

ACuriousMind

physical theories aren't defined by just saying a few vague words and hoping someone else will work out the math

Ryder Rude

I was jus enquiring if others have fit this stuff into usual quantisation

But it seems like this modification is major

But i want to say that my stuff isnt completely vauge. I have a manifold whose dimensions r the same as that of phase space. I have functions on it that are classical observables. I have a metric on it which is supposed to modify the usual hamiltonian evolution

The only vague stuff is the modification

of the time evolution

ACuriousMind

I don't think you appreciate how fundamental the time evolution being the Poisson bracket is to all aspects of Hamiltonian mechanics

Ryder Rude

Yeah, it conserves probability for one

ACuriousMind

15:10

Imagine I said "I'm just doing slightly different Newtonian mechanics. Everything is the same, I'm just modifying $F=ma$"

like, that's not a minor modification, that's just entirely new physics

Ryder Rude

In my new theory, I have a vague idea that probability will only need to be conserved locally

Just like energy is only conserved locally in GR

And the probbailistic interpretation would be local

So the entire stuff need not add to 1

I mean that if you do an experiment in a small area, u get back ur probabilistic predictions

ACuriousMind

talk is cheap, writing down a theory that's not immediately disproven by trivial observations is gold

bolbteppa

Again, I'd just recommend first getting through normal physics first :p

Ryder Rude

Lol

I'm just doing this for fun. I know it's wrong :P

There is hope to reproduce usual Hamiltonian mechanics with this as the coupling with the metric tends to zero

So this satisfies a basic tenet of physics

It can reproduce older theories

ACuriousMind

I don't find talking about limits of a theory you haven't even defined interesting

bolbteppa

15:13

What's missing are BRST ghosts and a path integral on this metric phase space

ACuriousMind

how could you possibly know anything about the limit of the "coupling with the metric" when you have steadfastly refused to actually specify how that coupling is supposed to work

Ryder Rude

It wil have a coupling constant that can tend to zero. But i shall try to come up with a formula

ACuriousMind

"I'm not telling you what I'm replacing $F=ma$ by but I promise you it will reduce to it in some limit"

Ryder Rude

Lol. I know this stuff is wrong. It's just for fun exploration :P

It's just for personal purposes

ACuriousMind

my problem isn't that you're probably wrong, my problem is that you're not even wrong

nothing here is specific enough to even be wrong

Mr. Feynman

15:16

@ACuriousMind Sounds like something someone standing in a dark hallway of a Physics department would say

Ryder Rude

I will try to write the math for this too. In the meantime, i shall also learn GR

ACuriousMind

@Mr.Feynman we didn't have any dark hallways

Ryder Rude

I will share it if i can get some good probability dynamics

ACuriousMind

well, unless you count the rather large tunnel system under the campus

Ryder Rude

U guys went to the same university?

ACuriousMind

15:18

no

Mr. Feynman

@ACuriousMind neither do I but it sounded just right to imagine :P

Ryder Rude

I think the metric i have is extremely crazy.. Because classical field theory has a continuum dimensional phase space. I will probably abandon this whole idea

It's just too stupid

I had already abandoned it. I just tried to discuss it here as its last funeral

to see if maybe there's anything interesting in this

Ryder Rude

15:41

I saw a cool idea of quantum gravity on phySE. It was about assigning a Hilbert space to each point of space and then defining a connection on it

But i think this idea is just, at best, an alternative formulation of QFT on curved spacetime

But there is already the usual formulation of QFT in curved spacetime

The usual formulation is nothing weird. You just have a pre-defined metric on spacetime, and then u translate it to Hamiltonian mechanics

The only difference from QG is that the metric affects the fields but the metric isnt affected in return

So it's like a one-sided interaction

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