The h Bar

bolbteppa

00:00

@danielunderwood what is $d \mu$?

danielunderwood

Well in that notation, I kind of wrote $\int F [q^\mu] d \mu$ due to taking the limit of $\sum F[q^i]$. It really would have been more sensible to write $\int F[q(x)] dx$, hence my question about index vs function

I don't think I slept last night because I was trying to figure out what $d \mu$ could possibly be

Though now I don't know why that limit should end up with the location being the index

bolbteppa

In non-relativistic mechanics, for say the Hydrogen atom problem, you use the fact that the angular part of the wave functions are eigenfunctions of the angular Laplacian, which is the casimir of the so(3) angular momentum algebra, and you use the fact that the eigenvalues of the Cartan generator are bounded above and below by the eigenvalues of the Casimir, i.e. $M$ takes $2L+1$ values. If $L = 0$ then $M$ has $1$ value, if $L = 1/2$ then $M$ takes two values, etc...

For an electron in the Hydrogen atom, $L = 1/2$, so you get spin wave functions. For the Klein-Gordon equation, $L = 0$ so you get one. This all arises because you are representing a connected but not simply connected group, however you need to specify which representation you're working with, the angular momentum algebra does that locally easily.

So that's how spin arises in the non-relativistic Schrodinger equation

@danielunderwood are you trying to transition from a classical point particle Lagrangian to a field Lagrangian?

danielunderwood

@bolbteppa Yeah. I can get to the field EL equations once I take $L = \int d^3x \mathcal{L}$ and $q_i \to \phi(x)$, but I have no idea why the label $i$ should go to position $x$

I'm also interested in the case that you can't write the Lagrangian as the integral of a density, but "integrable with compact support" goes a bit over my math knowledge

bolbteppa

Basically you started from $i$ independent degrees of freedom, and in 3 dimensional space you are now sending $i$ to $\infty$ to get an infinite number of independent degrees of freedom in 3-D space, in other words, you're filling 3-D space full of particles, it's going to end up with a particle at each point of space, i.e. an independent degree of freedom at each point of space, hence parametrized by $\mathbf{x}$

I think there's some lattice argument as well to keep things discrete but meh

Averaging over a countable number of lattices

danielunderwood

Ahh that makes sense. I wasn't really thinking of $i$ as indexing a position, but it does. Does this always result in position space or can you do it in configuration space like $\phi(q)$? I've always seen $\phi(x)$, but don't know why configuration space shouldn't be possible

Although I guess if you write your configuration space as some invertible $q^i(x)$, then it doesn't really matter

bolbteppa

00:12

Same thing

Basically $q_i \to q_{\mathbf{x}} = \phi(\mathbf{x})$

Your $q^i(\mathbf{x}) = \phi^i(\mathbf{x})$ looks like you have $i/n$ fields

Basically $q$ is just a way of saying $x$ where you assume arbitrary coordinates, even though $x$ can also mean arbitrary coordinates

@Semiclassical so if you followed that spin argument above, spin is baked into the cake of the Schrodinger equation

"Try to add the spin to a particle. If the logic of Bohmian mechanics – the wave function "is" a classical field and we should also add some classical values of a maximum set of commuting observables – were universally valid, it's clear that aside from the spinor-valued wave function (cup,cdown), we should also assume that Nature "objectively knows" about the classical bit of information that tells you whether the spin is "actually" up or down.

However, even the Bohmists realize that if every electron "objectively knew" whether its spin is up or down with respect to the z-axis, then the la…

Is more what I was wondering about, it seems like there's a conceptual issue, the math should spit it out fine, but whether that makes sense I'm not sure

Semiclassical

well, again, there's the whole thing about contextuality

quoting from that paper I linked earlier: " There is thus a sense in which, for the pilot-wave theory just as for ordinary QM, the measurement cannot be thought of as passively revealing some pre-existing quantity, but should instead be thought of as an active intervention which brings about the new, final state of the particle corresponding to the measurement outcome."

which Norsen contrasts with 'naive' theories for which "measurements simply passively reveal some preexisting value of the property being measured, without affecting the state of the particle.

bolbteppa

right let me check that, meant to sorry

Semiclassical

np

i'll note that, for better or worse, spin is different in pilot-wave theory than position in a key respect

for position measurements, there can be situations where it makes sense to say that a given measurement has revealed where a particle is (and that it makes sense to talk about where the particle was before then, even if you have no way of deducing it)

danielunderwood

Alright that makes more sense. Thinking about this also explains a bit more how you can have a massive field (I think). What you really have is a mass density inside the Lagrangian density, which gets integrated out to a mass correct?

Semiclassical

By contrast, in pilot-wave theory the spin is never given that status

bolbteppa

00:21

It sounds like contextuality means a measurement basically changes the state so you can never discover what it was like before measuring or something

Semiclassical

they've got a section on contextuality that i'm quoting from

so i'd look there

bolbteppa

Yeah, "but should instead be thought of as an active intervention which brings about the new, final state of the particle corresponding to the measurement outcome ... there is an even deeper sense in which, for the pilot-wave theory, the measurement cannot be thought of as passively revealing a pre-existing value"

Semiclassical

yeah

that continues for another page and a half

bolbteppa

@danielunderwood it depends, the Lagrangian becomes a Lagrangian density, the dimensional constants in there depend on the case at hand

danielunderwood

Ahh well I haven't really gotten to any real examples yet, so hopefully that will clear things up

And ugh quantummechanics.ucsd.edu/ph130a/130_notes/node452.html

> $F_{\mu \nu} F_{\mu \nu} = 2 (B^2 - E^2)$

bolbteppa

00:26

If I had read this I think I would have fallen for it

Before I understood spin and it's link to simple connectedness etc

I was thinking maybe it was the orientation of the coordinate system of the measurer for a while

"That is, the outcome of a measurement of a certain component of a particle’s spin depends, in the pilot-wave theory, not just on the initial state of the particle but also on the overall experimental context – i.e., on the particular “way” that the measurement is implemented."

Definitely would have fallen for that for a while anyway, yikes

Semiclassical

well, in the story Norsen is telling, it's a story about the orientation of the detector and where exactly the particle is passing through the detector

bolbteppa

I spent months trying to figure out if that made any sense

The whole Dirac Belt thing I thought was like a link between the two

Sheer nonsense of course

Semiclassical

but you're not in a position the measure said location, at least not if you want to then record where the particle will end up after exiting the detector

that said, I'm not entirely sure his story convinces me. what he's doing is basically a story about the Pauli equation, i.e. Schrodinger equation plus appropriate interaction

what I'd have liked to see is a story about the Levy-Leblond equation, i.e. the appropriate non-relativistic version of the Dirac equation

there I am not at all sure his story works the way I'd want it to

bolbteppa

I think if you replace Pauli equation with Hydrogen atom Schrodinger equation what he's saying would be fine

Think it was a bad idea to pick the Pauli equation but still, just because he's got Schrodinger in there, his logic should still hold

Semiclassical

well, if he's going to do Stern-Gerlach, then just doing $H=-\mu Bz\sigma_z$ is the most 'obvious' choice

but it feels like it undersells the problem a bit

bolbteppa

00:33

@danielunderwood those notes are not that great, e.g. the indices on $F_{\mu \nu} F_{\mu \nu}$ should be $F_{\mu \nu} F^{\mu \nu}$!

danielunderwood

Yeah that's what I said ugh for lol

Semiclassical

@bolbteppa ew, they actually had the former?

double plus ungood

danielunderwood

They seem to only use lower indices

Semiclassical

yikes

bolbteppa

Einstein summation convention!

danielunderwood

00:35

Except when they wrote $\frac{\partial}{\partial x_\nu}$ and similar. Though I'm sure that's not intended

I suppose it kind of works if you're just using Minkowski metric

bolbteppa

I don't like where this 'spin is different' thing is going in Bohm, really seems like there should be zero issues with copying spin exactly as in normal qm

Semiclassical

well, formally it's not that different

as I said, it's basically just a matter of writing down the probability current in that context

and that's a question for regular QM just as it is for Bohm

and, on the QM side, would the usual interpretation insist that all three spin components have meaningful values before you make a measurement?

bolbteppa

No, that's just a property of Lie algebra generators not being simultaneously diagonalizable

Semiclassical

That doesn't seem so different on the face of it from the Bohmian position that spin is contextual.

I'll note the quotation of Bohr which Bell employed (as noted in that paper): "“the impossibility of any sharp distinction between the behavior of atomic objects and the interaction with the measuring instruments which serve to define the conditions under which the phenomena appear.”"

bolbteppa

"Bob: No. The spin is rather a property of the wave function."

www-f1.ijs.si/~ramsak/km/dialog.pdf

Semiclassical

00:47

yeah, that seems like a very poor way of putting it

note the prior statement of Bob before that: "It is understood that all functions ($\Psi$ and its derivatives) are evaluated at the actual configuration."

which would seem to say that the wave function (and thus the spin itself) depends on the actual (hidden) configuration

bolbteppa

So both the casimir, total angular momentum squared, L^2 = L_x^2 + L_y^2 + L_z^2, and L_z, are simultaneously diagonalizable for say the Hydrogen atom, you get two quantum numbers L^2 \psi = L \psi, L_z \psi = M \psi, right. You can actually measure/observe these things, does that not hold in Bohmian mechanics?

Well I guess I should say they specify a wave function for which observables can be measured

Surely that can't change with Bohm

It's just math

Semiclassical

Sure. You can send such an electron through a Stern-Gerlach device with orientation +z and get a certain output. but of course that's a different scenario than if you used a device with orientation +x or +y, in both Bohm and in standard QM

bolbteppa

This seems to be a complicated issue in Bohmian mechanics anyway, seems tricky to figure out

Semiclassical

but this does raise a question for me, now that I think of it

suppose I run an electron through a stern-gerlach device with a particular orientation and find that it comes out as spin-up. how is it, in the bohm account, that I'm guaranteed to again get spin-up if I run said electron through another device with the same orientation?

I think I can see how it's resolved but I'm not sure

danielunderwood

Back to the Lagrangian stuff, is there a reason we consider $L(\varphi,\partial_\mu \varphi, t)$ instead of just $L(\varphi, \dot{\varphi}, t)$? I'd assume because not including the space derivatives would just result in a trivial homogeneous field, but didn't know if there's anything more interesting.

Semiclassical

00:57

(briefly: in figure 2, the possible trajectories are all oriented in the horizontal direction. however, if I take only the spin-up trajectories and run them through another S-N magnet pair in the same orientation, then necessarily they'll all enter the second detector at an angle. so I guess the story will be that, when a bohmian trajectory enters at an appropriate angle, it always leaves at that same angle)

bolbteppa

You consider $L(x_{\mu},\phi,\partial_{\mu}\phi)$

Semiclassical

(the appropriate angle being whatever angle the spin-up electrons exit with)

I'm sorta disappointed, in retrospect, that Norsen didn't address this point

he considers the effect of using a z-oriented SG device and then an x-oriented SG device, but not two z-oriented SG devices in sequence

danielunderwood

Ahh right I forgot the coordinates themselves. Though I suppose technically just $L$ wouldn't care about the space coordinates. But your EOMs come out to be in terms of the density, so it's kind of pointless to think of just $L$

Semiclassical

hence he doesn't really address what it means to be an eigenstate of Sz according to pilot-wave theory

(which seems like an awfully big gap...)

Cows

02:38

@bolbteppa so on the physics side I will be taking analog circuits, and thermal physics this fall, and thinking about taking nuclear and particle physics research methods course in the spring. I am really excited about returning to school lol

danielunderwood

03:19

Is there any use in taking the limit of the field Lagrangian for infinite fields so you end up with something like $L = \int d^6x \tilde{\mathcal{L}}$?

Basically the same procedure as particle -> field, but it seems to just result in a field with twice as many spatial dimensions?

Cows

05:48

coding .. .. . .. . . .. .

Secret

06:16

This is an interesting finding

What if decoherence is largely thermal in nature in the direction of increasing entropy, thus explaining why classically the memory overhead to infer a cause from an effect is harder than predicting an effect over a cause, and yet quantum computers have no such overhead?

> This holds even when the memory overhead is unbounded, resulting in quantum models with unbounded memory advantage.

Since in computer science, there is a tradeoff between CPU time and memory, this result may suggests we actually have unlimited memory in some form in retrodiction. But whether my impression is correct will require reading this paper in detail

dx.doi.org/10.1103/PhysRevLett.121.033603

probing vacuum friction, hmm...

Referring back to phys.org/news/2017-02-friction-vacuum.html, where the KISS principle in engineering is used to understand what happened

But we cannot always overlook these small things. They are small, but they are also interesting in its own rights

Correction: why classically the memory overhead to infer a cause from an effect is larger than predicting an effect over a cause

Fawad

07:28

There is pulley with masses m1 and m2 connected to ceiling. What does net force acting on pulley mean?

Secret

It means the tension acting on the pulley by the two masses

Fawad

@secret if there are two pulleys . 1st pulley is connected to masses m1 and m2 and second pulley is connected to m1 and pulled with F=m2g , both pulleys are connected to ceiling. Then (Q) net pulling forces in both the cases are same or different?

Secret

So you mean something like this?

Fawad

Secret

It should be identical, since the net force is given by $(m_2-m_1)g$ assuming the pulley is frictionless

Fawad

07:42

I didn’t get what you meant

Secret

In both systems, the forces that are pulling down are weights, correct?

Fawad

@Secret correct

Secret

yeah, thus the net force acting on the system is all the weights summed up with its directions taken account of

Fawad

So pulling forces doesn’t have anything to with tension in string right?

Secret

The tension only plays a role if the pulley has friction and you are drawing the free body diagram of individual masses

If you consider the whole system, the tension is internal and thus it cancels out by newton's 3rd law

(otherwise the string would have snapped already)

Anonymous

08:01

@Abcd The cringe is real....$g$ is not even a constant unlike $\pi^2$! Surely Ted was annoyed and understandably so. :P

Anonymous

The only people for whom those two would be equal, are engineers :D

Secret

> They seriously don't get that ignoring what they considered as garbage and not recycling it will only cause the garbage to pile up, until the garbage dump is full and all their services to suffer

> O well, then go ahead, cause this pile of garbage is only going to pile up. Ignoring the need to recycle is simply not an option, and nature will teach them this lesson because nobody can oppose nature

theconversation.com/…

for context

One of my favourite hobby of dealing with people that are inaction is to massively amplify their inaction so that in a very short time, the explosively growing inaction will explosively devour those people who are inactive, thus erasing them from the system

having said that, since I and many others do actively engage in recycling, I don't need to amplify our inaction because we have the opposite of that

The big challenge of course is how to write that into the algorithms that controls the market, so that the market knew how to punish inactive people

that might be just the way to solve all economic problems without governments

Fawad

08:23

Value of g will always have some error. But pi is a pure constant. You can’t compare someone so pure with impure

4

Secret

absolutely

user1732

actually, g has been defined to be exactly 9.80665 m/(s^2)

user1732

09:13

in Mathematics, Sep 26 '13 at 17:38, by robjohn

@JackM I am assuming $g=9.80665\,\text{m}/\text{s}^2$, that is the acceleration of gravity at the Earth's surface.

Yashas

10:33

-1

A: How do we know that Quasars actually produce collimated jets?

I just finished my 30 year book the first part of May of 2018. It tells about something that I have never heard mentioned by science. Anything that rotates, rotates in two opposite directions at the same time. Look at the front wheel on your bicycle. The right side rotates clockwise while the lef...

That answer contains a link to a document which contains non-mainstream physics concepts.

It also has several bits of religious stuff.

bolbteppa

12:34

@Cows Nice

Try get a summer internship for the following year early into it

@danielunderwood where does the $6$ comes from?

user2236

13:03

twice as many spatial dimensions as 3 is 6

Secret

Very strange philosophical thing from last night dream:

I am assuming everyone here knows what "doing nothing" means, right

user2236

OK?

bolbteppa

Yeah but why go from 3 to 6

Secret

Now, last night dream introduce something very strange. Think about not only you are asked to do nothing, but everyone is protesting, rioting, campaigning etc. in order to force others to do nothing

It's like maintaining the status quo on steroids

a "status quo activism" so to speak

user2236

like forcing people to "look the other way"?

Secret

13:12

No, its like, given a problem, the solution is to force people to do nothing about it

user2236

example please

Secret

If conservatism is advocate the preservation of tradition and changes are to be made slowly and carefully, and reactionary is to return the society to some state in the past, then whatever this dream is suggesting is some ideology between reactionary and conservatism, which basically the radicalisation of preserving whatever status quo that is in our society political structure and economics

user2236

sounds like a nightmare

danielunderwood

14:38

@bolbteppa Yeah I think it should only be 4 now. Or maybe better yet $d^3x dy$. I was thinking you take the field Lagrangian $\mathcal{L}[\phi_i(x), \partial_\mu \phi_i(x), x^\mu]$ and take the index $i \in \mathbb{N} \to y \in \mathbb{R}$ to get $\mathcal{L}[\phi(x, y), \partial_\mu \phi(x, y), \partial_j \phi(x, y), x^\mu, y^j]$ where $\mu$ represents $x$ dimensions and $j$ for $y$.

I was thinking to add 3 $y$ dimensions to be like particle to field, but skipped over the fact that there are 3 $q_i$ in the particle case if there aren't constraints.

I think what I've done is at least sensible, but don't know if it's useful in any way

Though I suppose what I've done is no different than having 4 degrees of freedom in the particle case and extending it to fields

John Rennie

The final data from the Planck mission has just been published

4

There's an article about it here

2

vzn

@Secret thx for sharing, vacuum friction sounds like a hot new/ trendy/ cutting edge research topic but also thermodynamics/ fluid related o_O

@user2236 sometimes the boundaries between dreams + nightmares seems quite "blurry"...

> For the time being, experimentally measuring the effect is not likely, since the energy involved is roughly three orders of magnitude smaller than what can be detected by today's most precise measurement techniques.

> "We will try to extend the successful model currently used to describe atom-light interactions to include the possibility of a changing mass," Sonnleitner said. "Of course this will only be a rather small correction, but it should help to complete the picture. It is never wrong to revisit, rethink and, if necessary, tweak an established theory."

bolbteppa

14:53

@danielunderwood doesn't look right

Section 1.2 gives an example nikhef.nl/~t45/ftip/Ch01-1.pdf

danielunderwood

15:13

Is there a certain part that doesn't look right? Note my $\phi_i$s are each separate fields rather than separations as in that document. Rather than trying to model a material, I've been taking the Lagrangian itself and expanding to different dependencies like $L(q, \dot{q}, t) \to L(q_i, \dot{q}_i, t) \to \mathcal{L}(\phi, \partial_\mu \phi, x^\mu) \to \mathcal{L}(\phi_i, \partial_\mu \phi_i, x^\mu)$ and I'm wondering what the next step to an infinite number of fields is.

I still get to the same result in the field case, but by making $L = T - V$ at the end instead of at the start

...I think

bolbteppa

What you just wrote is fine, the other post had $y^j$'s and $\partial_j \phi(x,y)$ for no reason, and not sure what the dimension thing you said meant

danielunderwood

Well say you have the last version with multiple fields $\mathcal{L}(\phi_i, \partial_\mu \phi_i, x^\mu)$ and you promote $i$ to a continuous index. If you call that continuous index $y$, it seems like you'd get the above. Or you could also stick it in with $x$, but it may be neater to keep the typical space dimensions separated.

Although I don't even know if it's sensible to talk about an infinite number of fields

bolbteppa

I have never seen that anyway

You're integrating over the arguments of the $\phi_i$, they are not the variables in the integral, so I don't see why you can't simply say $\mathcal{L}(\phi_{\lambda},\partial_{\mu} \phi_{\lambda},x^{\mu})$ for $\lambda \in \mathbb{R}$ etc

danielunderwood

15:53

Well you have the $d^3x$ that comes in from the fields and when you vary the action, you get $\delta S = \int d^4x \sum_i \left( \frac{\partial \mathcal{L}}{\partial \phi_i} \delta \phi_i + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_i)} \delta (\partial_\mu \phi_i) \right)$ then you take $\sum_i \to \int dy$, right?

You'd also have to go from the plain density $L = \int d^3x \mathcal{L}$ to some new density $L = \int d^3x dy \tilde{\mathcal{L}}$

Well not to do the part above, but to do the integration by parts to solve the variational problem

bolbteppa

16:24

Then if you varied $L$ in terms of $\tilde{\mathcal{L}}$ the same argument would lead to a new $\int d z$ etc, it you take the total derivative I guess you'd introduce the integral which is probably why you don't usually go near this stuff :p but I don't see why you should introduce $\int dy$ into $S$ itself

I'm not sure, maybe it's fine

Maybe this is a way to get to path integrals over function spaces

dmckee

@user1732 But if you use that value you are almost certainly wrong. Actual surface gravity varies in the third and fourth decimal place on distance scales of only kilometers in places.

Seriously, gravinometric surveys are a standard part of the prospecting toolkit.

Emilio Pisanty

18:52

Uhm ... yes. I guess I do. — dmckee ♦ 6 mins ago

@dmckee :-P

Emilio Pisanty

19:09

@dmckee also, I imagine you mean "gravimetric" :-P

a gravino sounds like the graviton's equivalent to a neutrino

ACuriousMind

@EmilioPisanty No, that's a gravitino

Emilio Pisanty

@ACuriousMind ah, figures

Rarita–Schwinger equation

In theoretical physics, the Rarita–Schwinger equation is the relativistic field equation of spin-3/2 fermions. It is similar to the Dirac equation for spin-1/2 fermions. This equation was first introduced by William Rarita and Julian Schwinger in 1941. In modern notation it can be written as: ( ϵ μ κ ρ ν γ 5 ...

hehe

hehehe

rarita

Adjective: rarita

feminine singular of rarito...

ACuriousMind

Heh

Semiclassical

19:26

hmm, more elementary propagator questions

for the two cases of an exact propagator in non-relativistic QM which I know (harmonic oscilator and free particle), the following is true: $|K(x,t;x',t')|^2$ is a function of $t,t'$ alone

Now, the fact that it's true for a harmonic oscillator means that it has to be true for the free particle (since for small times the former reduces to the latter)

so this isn't really that strong of an example

But, is there an obvious counterexample? i.e. a Hamiltonian for which $|K(x,t;x',t')|^2$ depends on $x,x'$

another example: the propagator for a linear potential is of the form $K(x,t;x',t')=A(t-t')e^{i S[x_{cl}(t)]}$ where $x_{cl}(t)$ is the classical trajectory from $x$ to $x'$. since $S[x_{cl}(t)]$ is real-valued, $|K|^2$ is again independent of $x,x'$

ah, Qmechanic had a nice answer which addresses this in passing: physics.stackexchange.com/a/81278/55641

dmckee

20:38

Several questions on the site today that (a) have been asked and well answered before, but (b) have titles very different from those of their predecessors.

Emilio Pisanty

21:42

The car hated driving by the tree in front of the Baker place. "It's a tire swing," they said, "it doesn't mean anything." Still, creepy.

Tweeted by ASmallFiction on July 13, 2018 at 4:39 AM

heh

Secret

23:58

To be investigated later: A molten causal network

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