The h Bar

2:30 AM

@Relativisticcucumber I'd make minor changes to straighten the logic flow. 1) After Einstein published 1905 SR pair of papers, Mink studied it and realised that Mink metric can summarise a lot of ingredients that would make SR tick. The symmetry group is Poincaré. 2) It is found out that if you simply start from a scalar action that is SR invariant, implying Poincaré invariant, then EL equations will spit out a theory for you that obeys SR. 3) These are subsets. SR actions are subset of Poincaré actions,

In particular, theories that arise from SR actions are subsets of all possible SR theories. But this guaranteed link means that if we are to search for SR theories to try, it is vastly easier to search from SR actions first. Other physicists will have a much less rejection rate because this path is so well-trotten.

@SillyGoose The actual answer to this is that, if you assert that you found an SR theory that you want other people to accept as being fundamental, and it is not found from an SR action, then you would have to have entire chapters worth of proving work to make others accept that your SR theory actually is SR.

@SillyGoose I think I just answered this. It is not some "finite list of assumptions". It is that everybody is going to be lazy. Students are lazy, teachers are lazy, the professors are the laziest of the entire lecture theatre... Like, even if you found an SR theory another way, e.g. Dirac guessing his equation first, he would just need to find an SR action that derives his equation, and viola, Dirac theory is necessarily SR. Note that it is an SR action, not the SR action.

Of course, Dirac also put some effort into proving directly that the equations and their solutions actually are SR, but that is beside the point.

nickbros123

4:50 AM

@naturallyInconsistent someone should make an Iceberg meme about this. Step 1: "everything is a point dipole", step 5: "wtf is even macroscopic energy" step 10:"reject macroscopic, return to microscopic" xD

naturallyInconsistent

5:40 AM

Correct

bolbteppa

6:48 AM

@Slereah Some great answers on both sides of the fence in that post now, also he put it on arxiv

Slereah

lots of answers for the Kerr part but what's going on with the Schwarzschild part

The Kerr part seems to be very extendible tho

Mr. Feynman

7:17 AM

What's with the recent Kerr talk?

bolbteppa

7:29 AM

17

Q: Why does Roy Kerr claim that the Kerr black hole does not contain a singularity?

In a preprint posted on the arXiv, Roy Kerr claims that there is a widespread misunderstanding related to the singularity inside the black hole that bears his name. Can anyone explain his argument in "simple terms"? Are there any pitfalls in his reasoning? Here's a link to the preprint: https...

Giorgi Lagidze

8:39 AM

Hi all. was just wondering why there's such a complex derivations of noether's theorem(such as this weizmann.ac.il/particle/perez/Courses/QMII19/…) . article assumes that there's a lagrangian symmetry. I mean, if there's a symmetry, euler lagrange equation's right side already is 0, which means LHS is conserved. What's the point of so complex derivations ?

bolbteppa

In the reddit, ChatGPT is contradicting Kerr...

Giorgi Lagidze

as another example, en.wikipedia.org/wiki/…, on the image, the author says that middle parts' actions are equal and only change contributing is buffering pieces, but on the buffering pieces, he assumes that only speed changes. why does he assume that ? i think he uses the fact that $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, if so, i don't see what's the point of this derivation as EL is simpler

bolbteppa

Derivation 1 says if you plug a symmetry transformation into the Lagrangian directly and it changes to first order as a total derivative (4 - 6), then this has to be equal to what you get when you abstractly vary the Lagrangian and invoke the equations of motion (7 - 8), where the equality is in (9-10) implying (11-12).

The second way is a poor explanation for the 'Noether' method, which is basically just equation (9) or (14) integrated, where in (9) or (14) the parameter is constant as it was all through the derivation, and you now notice that if it was local and did IBP you'd end up with the conserved current attached to the derivative of the local parameter.

Thus this is another way to find the conserved current, vary a Lagrangian, throw away every total derivative you can until you're left with stuff attached to the derivative of the local parameter, and you call that stuff the conserved current

Giorgi Lagidze

8:54 AM

@bolbteppa thank you <3. though, if lagrangian doesn't change when applying transformation, that this means: $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, which means LHS of lagrangian is conserved

this seems super simple. is this something wrong with this way ?

$\frac{d}{dt}(\frac{\partial L}{\partial \dot x_1}) = \frac{\partial L}{\partial x_1}$ <br>
$\frac{d}{dt}(\frac{\partial L}{\partial \dot x_2}) = \frac{\partial L}{\partial x_2}$ <br>

Let's multiply both sides by $\epsilon$(equation won't change for sure).

$\epsilon(\frac{d}{dt}(\frac{\partial L}{\partial \dot x_1})) = \epsilon(\frac{\partial L}{\partial x_1})$ <br>
$\epsilon(\frac{d}{dt}(\frac{\partial L}{\partial \dot x_2})) = \epsilon(\frac{\partial L}{\partial x_2})$ <br>

$\epsilon(\frac{d}{dt}(\frac{\partial L}{\partial \dot x_1} + \frac{\partial L}{\partial \dot x_2})) = \epsilon(\…

in the last equation, symmetry of lagrangian implies RHS is 0, which means left side is conserved

If you tell me that this way is not good because it doesn't help me find conserved quantity in all cases of problems and tell me that on the article I shared, that way helps to find conserved quantity in all cases, then sure, I would understand

bolbteppa

Not sure what you're doing, but $L$ depends on more variables than $x_1$ and $x_2$, when you added those two they don't sum to zero

Giorgi Lagidze

well, as many as it depends, my case can be used for as many as you want

bolbteppa

The Lagrangian can change, in (4) for example it changes, if it changes it just needs to change up to a total derivative

Giorgi Lagidze

if you understand how i got to: $\epsilon(\frac{d}{dt}(\frac{\partial L}{\partial \dot x_1} + \frac{\partial L}{\partial \dot x_2})) = \epsilon(\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2})$, then what I am saying from here is: RHS basically says how much L changes when you change both $x_1$ and $x_2$ by $\epsilon$ and if you got symmetry, RHS is 0, so LHS is conserved

I think, my proof is for the cases, when Lagrangian doesn't change at all.

@bolbteppa on the image(en.wikipedia.org/wiki/…), can you tell me what logic he uses such as on the buffering pieces, he assumes only speed changes ?

bolbteppa

Symmetry of the Lagrangian does not imply anything here necessarily has to vanish, if $L$ does not depend on $x_1$ and $x_2$ then the RHS will vanish

Giorgi Lagidze

9:02 AM

yes, i was referring to the cases when $L$ doesn't depend on position variables or if it does, it depends on them such as $U(x_1 - x_2)$ in which case, my proof is solid. but for the cases when $L$ changes by total time derivative of some function, my proof is not covering that case

bolbteppa

That picture is just trying to describe the case where $L = L(x,\dot{x}) = L(\dot{x})$ does not depend on $x$, so that $\delta S = \int L_{\dot{x}} \delta \dot{x} dt \approx L_{\dot{x}} \delta x |^2_1 = 0$ tells us that $p = L_{\dot{x}} = \partial L/\partial \dot{x}$ is conserved throughout the motion. In this case there is no 'total derivative' term in the variation of the action, i.e. the $J$ term is zero in (5) and (9),

we are intuitively showing the momentum at the end of (10) is conserved directly (i.e. I just integrated across assuming $L_{\dot{x}}$ is approximately constant), you need to argue like (10) to justify it properly over the whole region

You don't need to add your two equations in this case, they independently vanish, so the LHS involves two separately conserved quantities, whose sum is also conserved

Giorgi Lagidze

so if the picture is only describing the scenario when $L$ doesn't depend on $x$, that case is like super simple one

@bolbteppa how did you figure out that the picture uses the scenario of $L(\dot x)$ and not $L(\dot x, x)$ ?

maybe they use $L(\dot x, x)$, but assumes that since symmetry happens, $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2 = 0$

bolbteppa

The language is confusing but in the picture they say $\Delta S = \int \Delta L = \int \frac{\partial L}{\partial \dot{q}} \Delta q$ which is only possible if $L$ only depends on $\dot{x}$, otherwise they'd have to write $\Delta L = (L_x \delta x + L_{\dot{x}} \delta \dot{x} + L_t \delta t)$ in general for $L = L(t,x,\dot{x})$

Giorgi Lagidze

exactly. that's what was confusing me and I thought they skipped the part where they had to say: $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, but seems like that's not the case and it's just $L = L(\dot x)$

@bolbteppa so the proof on the image is pretty weak since it uses the super simple case :( would the same type of proof be possible(as on wikipedia picture) for $L(\dot x, x)$ ? because if I try that, I fail because to succeed and end up with conservation, I need to use $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, and if I use this, what's the point of such proof lol i could just directly use EL to show conservation.

bolbteppa

9:19 AM

No, if you try it with $L(x,\dot{x})$ it's going to completely fail, you can't fix it by adding those terms, also in the picture there is only an $x_1$ no $x_2$

Giorgi Lagidze

it's not going to fail only if $L$ is symmetric without even time derivative difference. if that's the case, the proof succeeds, but the proof becomes pointless as to use that proof, as i said, i need to use $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$ and if i use this, proof is pointless at all. makes sense ? @bolbteppa

I think I have to follow the proof on the first article I shared

bolbteppa

$\Delta S = \int \Delta L = \int (L_x \Delta x + L_{\dot{x}} \frac{\Delta x}{\Delta t} ) \Delta t = \int (L_x \Delta x \Delta t + L_{\dot{x}} \Delta x)$, what the hell is the first term supposed to mean,

Giorgi Lagidze

@bolbteppa $\delta S = \int_{t_0}^{t_1} \epsilon (\frac{\partial L}{\partial \dot x_1} \dot f(t) + \frac{\partial L}{\partial \dot x_2} \dot f(t) + \frac{\partial L}{\partial x_1} f(t) + \frac{\partial L}{\partial x_2}f(t))dt = 0$

bolbteppa

No it doesn't make sense, there is no $x_2$ anywhere, and even if we include an $x_2$ the Lagrangian now depends on $x_1$ and $x_2$ so they don't even add to give zero nor separately give zero

Giorgi Lagidze

$\delta S = \int_{t_0}^{t_1} \epsilon (\frac{\partial L}{\partial \dot x_1} \dot f(t) + \frac{\partial L}{\partial \dot x_2} \dot f(t) + \frac{\partial L}{\partial x_1} f(t) + \frac{\partial L}{\partial x_2}f(t))dt = 0$ if lagrangian is symmetric under $x_1(t) + \epsilon$ and $x_2(t) + \epsilon$ (symmetric such as it exactly stays the same, then 3rd and 4th member in the interal add up to 0, i.e only 1st and 2nd stay in the integral and you can get conserved quantity from that @bolbteppa

bolbteppa

9:25 AM

You can't use the same function $f$ for $x_1$ and $x_2$

I'm not sure what you're saying

Giorgi Lagidze

why not ? if you got 2 paths($x_1$ and $x_2$), we can choose f such as $f(t_1) = 0$ and $f(t_2) = 0$. this works for both paths.

bolbteppa

In $L = L(x_1,x_2;\dot{x}_1,\dot{x}_2)$, you vary $x_1$ and $x_2$ independently of one another in general, also, what have you got other than a big mess right now that doesn't mean anything

Giorgi Lagidze

I wouldn't agree. I can just vary them with the same function. @bolbteppa see here(physics.stackexchange.com/a/19865/366605)

bolbteppa

It's not this hard, these kinds of posts just confuse and mislead people

Giorgi Lagidze

my point was that there's no constraint why i can't vary $x_1$ and $x_2$ by the same $\epsilon f(t)$

bolbteppa

9:32 AM

Fine vary then by the same function, now what, you still have a big mess

Giorgi Lagidze

I don't in the case when Lagrangian is symmetric without total time derivative(i.e L = L'), because if so, $\delta S = \int_{t_0}^{t_1} \epsilon (\frac{\partial L}{\partial \dot x_1} \dot f(t) + \frac{\partial L}{\partial \dot x_2} \dot f(t) + \frac{\partial L}{\partial x_1} f(t) + \frac{\partial L}{\partial x_2}f(t))dt = 0$ (3rd and 4th members add up to 0) and I successfully get conserved quantity @bolbteppa

bolbteppa

What's conserved, all you have is some integral of complicated functions over the whole path ends up being zero

Giorgi Lagidze

no, we got $\int_{t_0}^{t_1} \epsilon (\frac{\partial L}{\partial \dot x_1} \dot f(t) + \frac{\partial L}{\partial \dot x_2} dt = 0$ .

you now split this in 3 integral parts for each interval(as on the wikipedia page) and you will notice that 2nd integral will be 0, i.e you're left with: $\delta S = \underbrace{\epsilon (\frac{\partial L}{\partial \dot x_1} + \frac{\partial L}{\partial \dot x_2}) f(t)\Bigr|_{t_0}^{t_0+\tau}}_{\text{At the beginning}} + \underbrace{\epsilon (\frac{\partial L}{\partial \dot x_1} + \frac{\partial L}{\partial \dot x_2}) f(t)\Bigr|_{t_1 - \tau}^{t_1}}_\text{At the end} = 0$

$\delta S = \epsilon (\frac{\partial L}{\partial \dot x_1} + \frac{\partial L}{\partial \dot x_2})(1-0) + \epsilon (\frac{\partial L}{\partial \dot x_1} + \frac{\partial L}{\partial \dot x_2})(0-1) = 0$

and you got conserved quantity :P

the proof works, but i need to use $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$

which i don't like because if i use that, it's the same as if i had directly used chat.stackexchange.com/transcript/message/64827821#64827821 which is simpler

@RyderRude hey mate. what's your thoughts ? basically, I wanted to use the same technique as in here(en.wikipedia.org/wiki/…), but for $L(x_1, x_2, \dot x_1, \dot x_2)$ and if I do, only way I succeed is by using $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$ and if i use that, the proof loses its point, because I could directly just use EL to show this without such complications.

bolbteppa

9:52 AM

I'm just not sure what you're trying to do, maybe it makes sense

Ryder Rude

@GiorgiLagidze i dont understand which proof on wiki u r following

Giorgi Lagidze

all good. @bolbteppa thanks for your time and listening and helping ^_^

@RyderRude en.wikipedia.org/wiki/… (this should directly get you to the image)

Ryder Rude

oh it's the one u linked

Giorgi Lagidze

it kind of uses feynman's technique

but the situation is for $L(\dot x)$ and not for $L(x, \dot x)$

Ryder Rude

this seems identical to Feynman's proof

Giorgi Lagidze

9:56 AM

yes, but they discuss the case when $L = L(\dot x)$.

if you now use the same technique for $L = L(x, \dot x)$, you will realize that you need to use the fact that $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, otherwise, you can't prove. and if you use this fact, then what was the point of using this feynman technique in the first place ? if $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, you could just directly use EL ( chat.stackexchange.com/transcript/message/64827821#64827821)

Ryder Rude

i dont think Feynman's technique is a general proof. it only works for the conservation of canonical momentum, which is the simplest case of Noether's theorem

Giorgi Lagidze

yes, i agree, but can you make it work for $L(x, \dot x)$ without using $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$ ?

bolbteppa

Maybe this will help, if $L = \frac{1}{2}(m_1 \dot{x}_1^2 + m_2 \dot{x}_2^2 - k_1 x_1^2 - k_2 x_2^2)$, from from $\frac{d}{dt} (m_1 \dot{x}_1) = k_1 x_1$ and $\frac{d}{dt}(m_2 \dot{x}_2) = k_2 x_2$, you're saying in the special case when $\frac{d}{dt}(m_1 \dot{x}_1 + m_2 \dot{x}_2) = k_1 x_1 + k_2 x_2 = 0$ i.e. when $x_2 = - (k_1/k_2) x_1$, then $m_1 \dot{x}_1 + m_2 \dot{x}_2$ is conserved. Okay, but what about when $x_2 \neq - (k_2/k_1) x_1$? There is no reason this has to be true in general

Ryder Rude

@GiorgiLagidze do u mean for L(x1, v1,x2,v2)?

Giorgi Lagidze

@RyderRude yes

@bolbteppa yes, true. it's just I wanted to use feynman's technique without using $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$

it seems impossible to me :P

Ryder Rude

10:05 AM

the symmetry transform here is L(x1,v1,x2,v2)---> L(x1+a, v1, x2+a, v2)

for the conservation of momentum, the case where Feynman's proof works

Giorgi Lagidze

@RyderRude and how do you prove with feynman's technique ?

bolbteppa

If $L$ depends on $x$, or $x_1$ and $x_2$, then at best you can only use it in some special directions, but you can't use it in general because $x_1$ and $x_2$ are in general more independent than to be constrained such that $L_{x_1} + L_{x_2} = 0$ has to hold. We're trying to find the completely general solution and general conservation laws not special cases

Ryder Rude

@bolbteppa he wants to assume invariance of the Lagrangian under the transformation i mentioned, and only then prove conservation

he is not trying to prove conservation for a generic Lagrangian

but he also wants to use Feynman's technique

Giorgi Lagidze

+1

bolbteppa

The problem is, you have two separate equations, $\frac{d}{dt} L_{\dot{x}_1} = L_{x_1}$ and $\frac{d}{dt} L_{\dot{x}_2} = L_{x_2}$, when you add them you are restricting the set of possible solutions and thus considering only a special case, not the most general solutions

Monty

10:09 AM

@ACuriousMind Well (1.12) defines the measure $\delta_{H(q,p)-E}$, as the text says it is defined through expectations of observables $\varphi$. I am just confused why you can straight away define it as the uniform measure over the level set $\{(q,p)~:~H(q,p)=E\}$ of the Hamiltonian. To me thats what it looks like they are trying to do.

Giorgi Lagidze

@RyderRude here(physics.stackexchange.com/a/19865/366605), why do you think that Ron only assumes that only speed changes on those paths ? :P Definitely, he uses the logic that $\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2} = 0$, but if you use that logic, there's no point to even start feynman's proof because EL simply is enough.

Ryder Rude

@GiorgiLagidze Feynman's proof is only for conservation of momentum. that's y the term $dL/dx$ isnt there, but this term is there in the general proof of Noether's theorem

Giorgi Lagidze

you can still get conservation of momentum even if dL/dx is there

Ryder Rude

u can see the diagram Feynman draws. it only involves an unchanging x and an infinite v

it is not the case of a general symmetry

Giorgi Lagidze

I think the wikipedia page doesn't use feynman's technique.

Ryder Rude

10:12 AM

as wiki notes, this proof is only for illustration. it's not a general proof

Giorgi Lagidze

forget feynman at all. my point is that, on wiki, can you use the same proof for $L(x, \dot x)$ ?

no right ?

Ryder Rude

wiki also doesnt consider the term dL/dx

it's the same proof

Giorgi Lagidze

exactly :P

if you don't have $x$ in the L, fine.

Ryder Rude

it's only an illustration of conservation of momentum

Giorgi Lagidze

and why can't i have illustration for $L(x, \dot x)$ case ?

Ryder Rude

10:15 AM

Noether's theorem considers all the terms in the variation

but it loses this illustration

y cant u just accept the mathematical proof :P

Giorgi Lagidze

yes. let's forget this conversation ever happened.

Ryder Rude

@GiorgiLagidze ok wiki says this : Roughly speaking, they contribute mostly through their "slanting"

so they r ignoring the dL/dx contribution

ACuriousMind

@Monty Ah, that's indeed because of the zero Lebesgue measure of $S(E)$ - you want to express the uniform measure on $S(E)$ as an integral w.r.t. the Lebesgue measure $\mathrm{d}q\mathrm{d}p$ in the ambient phase space

of course you could just abstractly say "let's take the uniform measure on $S(E)$", but as a physicist you want this integral representation so you can do (more or less justified :P) symbolic manipulations of it

bolbteppa

If you include $x$, then the wiki proof now goes as $\Delta S = \int \Delta L = \int (L_x \Delta x + L_{\dot{x}} \frac{\Delta x}{\Delta t} ) \Delta t = \int (L_x \Delta x \Delta t + L_{\dot{x}} \Delta x)$ where you now have this first term to deal with as well, if you just assume it's constant over some small interval and integrate then you get an extra $L_x \Delta x t$ which obviously quickly gives the wrong answer as it only depends on $t$ and linearly at that

and it's not conserved over the interval either it's some function of $t$ which is different at different times

Monty

@ACuriousMind gotcha

Ryder Rude

10:23 AM

in Neother's theorem proof, this dL/dx gets cancelled out due to the EL eqn

Giorgi Lagidze

@bolbteppa that's not the case. because you use $\epsilon f(t)$ . there's no way you end up with $t$ at all

the proof works as I said, but assumes that dL/dx1 + dL/dx2 = 0

anyways, gotta go. thanks

Ryder Rude

@GiorgiLagidze note that Ron is ignoring this term because the change is x is an infinitesimal while the change is v is infinite, so the former is negligible

Giorgi Lagidze

yes, Ron's proof is better than wikipedia

that was my point

bolbteppa

I explained why this only works for a special case where the variables are related in a special way, it does not hold in general so it's not a general conservation law

Giorgi Lagidze

I can't understand fully Ron's proof, so I gave up on it

Ryder Rude

10:31 AM

but this argument only works for translational symmetry. Noether's theorem is always more general

wiki is using the same proof but they also want to consider a general symmetry. so they just handwave and say "Roughly speaking, they contribute mostly through their "slanting"

this allows them to throw away the dL/dx term

but they onlh claim it to be an illustration, rather than a proof

Giorgi Lagidze

yeah, I don't get why they throw dL/dx out. is it because of translation invariance ?

bolbteppa

You might as well assume $x_1$ and $x_2$ are constants, then you get another useless special case conservation law

Ryder Rude

in case of translational symmetry, delta x is negligible because delta v is infinite in the diagram

Giorgi Lagidze

@RyderRude even on wikipedia ?

Ryder Rude

no, on wikipedia, they consider a general symmetry

so they dont have a rigorous reason to throw away that term

Giorgi Lagidze

10:34 AM

yes ! that's why they don't consider $L(x, \dot x)$ in the first place, but only $L(\dot x)$ :D

Ryder Rude

the diagram they've drawn is not a mere translation of the trajectory

it's not that they dont consider the dependence of L on x

Giorgi Lagidze

if they consider, why do they throw $dL/dx$ ?

Ryder Rude

it's that they assume delta x due to the infinitesimal symmetry transformation is negligible

the full change in the lagrangian wud hav been dL/dx times delta x

just like it's dL/dv times delta v

Giorgi Lagidze

okay, they assume delta x is negligible

Ryder Rude

yes.

Giorgi Lagidze

10:37 AM

why don't they assume that delta v is also negliglbe :D

Ryder Rude

they write : Roughly speaking, they contribute mostly through their "slanting"

@GiorgiLagidze idk :P

it's a bad proof

Giorgi Lagidze

well, it's horrible.

yeah, seems like this noether proof, I have to forget about illustrations because every single time, I saw anyone's illustration proof, it sucks unfortunatelly

the only way is pure mathematics and that's it i guess :P

Ryder Rude

yes, the original proof is a very short manipulation

stick to it until Hamiltonian mechanics proves it in one line

Giorgi Lagidze

yes, better to forget it now ! thanks so much. have to go for a couple of hours. have a good day

Ryder Rude

ok :)

Mr. Feynman

11:57 AM

Hello, does anyone know the name of this change of variables?

$b_i^\dagger\to e^{i\varphi_i}$

ACuriousMind

12:35 PM

@Mr.Feynman isn't this Jordan-Wigner

Mr. Feynman

1:20 PM

@ACuriousMind The context is that of Jordan Wigner, but the definition I have is a bit different. I'll check with some algebra after my siesta

By the way, the definition I have is that of the inverse transformation $$\sigma_i^x=(a_i^\dagger+a_i)\prod_{j=1}^{i-1}(a_j^\dagger+a_j)$$, something similar in for $y$ matrix and $z$ matrix is the same as wiki

ACuriousMind

that was a short siesta :P

Mr. Feynman

Lol, a couple of minutes and I'll go non-interacting

Oh ok, wiki writes it for the ladder operators, but still in the picture above there is something different: they are mapping $\b^\dagger$ to a "phase" operator

It would make sense if it were some kind of polar decomposition but it isn't

(And now I'll go non-interacting)

Ryder Rude

5 mins of seista can feel like hours of sleep

naturallyInconsistent

1:39 PM

I just learnt that siesta is the human norm throughout history. We are working way too much and getting paid way too little and paying too much and the almost doubling the amount of work done by basically everybody in the world only improved actual productivity by 30%, i.e. overworked for nothing much.

And there are just the usual devils to blame.

Absolutely psychotic

Ryder Rude

i wish superheros existed to sort this out

why on earth do we need so much productivity

the world shud work the same with same pay on half the productivity

one could have expected that sort of a world back when desk jobs started getting big

but government and rich people made it awful for everyone else anyway

Silly Goose

8:52 PM

The day of honking has arrived

Honk

So do you all think that physics will forever remain in pieces that each describe their respective regimes precisely

Relativisticcucumber

10:13 PM

@naturallyInconsistent i see -- thanks !

@SillyGoose going ryder rude style today

Mr. Feynman

10:44 PM

@Relativisticcucumber the impression is spot on, that question isn't receiving any feedback (sorry Ryder, I needed to make this joke :P)

John Zimmerman

11:16 PM

Consider a Cauchy foliation $\Omega_s(x)$ of a Lorentzian (1+1) manifold for space variable, $x$ and time variable $s$. Does anyone know how to interpret $\Omega_s(x)$ given that it satisfies a heat/diffusion equation, with Cauchy data being $\Omega_s(x)|_{s=0}$?

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