The h Bar - 2023-01-26

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3:50 AM

What are the solutions for no matter and just a cosmological constant called?

1 hour later…

5:00 AM

Lambdavacuum solution

In general relativity, a lambdavacuum solution is an exact solution to the Einstein field equation in which the only term in the stress–energy tensor is a cosmological constant term. This can be interpreted physically as a kind of classical approximation to a nonzero vacuum energy. These are discussed here as distinct from the vacuum solutions in which the cosmological constant is vanishing. Terminological note: this article concerns a standard concept, but there is apparently no standard term to denote this concept, so we have attempted to supply one for the benefit of Wikipedia. == De...

5:28 AM

@PM2Ring thanks

7 hours later…

12:53 PM

I want to ask a question regarding asynchronous variations (moving boundary) but I'm not sure whether Physics.SE or Math.SE is better suited for it

1:11 PM

The reason is that there are physical applications but I don't think the core of the question really has Physics content

2 hours later…

3:15 PM

I decided to post my question on Math.SE

Nonetheless I have a related question. I'm not sure I agree with Frankel here. You could make $\delta t$ variations also in the configuration space functional

user image

Is that right?

@Mr.Feynman I mean, I think all he's saying is you can vary a functional on $T^\ast M\times \mathbb{R}$ in the $\mathbb{R}$ parameter?

@ACuriousMind Yes, but you could also vary the $\mathbb{R}$ parameters in the Lagrangian case

The books seems to suggest this is a peculiarity of the Hamiltonian formalism, doesn't it?

3:30 PM

@Mr.Feynman no, that's the part about lifts of variations in $M$: In the Lagrangian case you're not varying a curve in $T^\ast M$ or $TM$ directly, you're varying a curve in $M$ and this induces a variation of the lift of this curve to $TM$. Since the lift depends on the choice of $t$ along the curve (via $\dot{q}$), you can't do an independent variation here

I've been lied to

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@ACuriousMind I was told that to make asynchronous variations instead of $\int_{t_0}^{t_1} L(q(t)+\lambda\eta(t),\dot{q}(t)+\lambda\dot{\eta}(t), t)dt$, the varied action should be $\int_{t_0+\lambda\delta t_0}^{t_1+\lambda\delta t_1} L(q(t)+\lambda\eta(t),\dot{q}(t)+\lambda\dot{\eta}(t), t)dt$

I mean, it might be you can do this in some approaches?

just not in the context your quote is about

The problem is that changing the extrema of the integrals is the only meaning I have for varying $t$

as it is an integrated variable and I don't know what else I could do

4:08 PM

I mean, althought $H$ is defined on $T^\ast M\times\mathbb{R}$, in Phase space $S[p,q]=\int(p\dot{q}-H)dt$ time is an integrated variable too, so I don't see what else a variation of the functional wrt to time could mean

@Mr.Feynman The action is a functional on a path $\gamma : I\to T^\ast M\times \mathbb{R}$, $S[\gamma] = \int_\gamma (p\mathrm{d}q - H\mathrm{d}t)$

when you write $\mathrm{d}q = \dot{q}\mathrm{d}t$ I think you've essentially already decided not to vary $t$

there's a reason that section talks about the Lie derivative of $p\mathrm{d}q-H\mathrm{d}t$, not about the version with $p\dot{q}$

@ACuriousMind In fact I was wondering why in phase space one would write $dq=\dot{q}dt$. I guess I was presented a weaker version of Hamilton's principle

Which somehow reconciles with the most general one if we vary the extrema in the way I wrote above

Or maybe even in that way it is not as general

now that I think of that, I should recover the asynchronous variations of "my version" as soon as I use $dq=\dot{q}dt$

Well, I'd like to spend more time on this issue and learn about the differences between this approach and mine but I guess I can't spend much hours rn

2 hours later…

6:50 PM

@DIRAC1930 have you tried to derive $\eta_{ijkl} = 0$? Going a bit haywire...

1 hour later…

8:14 PM

@bolbteppa I've stopped looking at where the equations come from and just accepted it for the time being

I think it may be helpful to try and derive the equations just starting out in $D=4$ without the $X^0$ term initially

Then you can compare it to similar constructions such as the one in Penrose etc.

The $D = 4$ case immediately follows from the $D = 5$ derivation on setting $x^0 = 0$ and just re-arranging the $X$ matrix rows/columns, the derivation is exactly the same as $D=5$, it goes exactly as I did it in one of the pictures I sent, the only loose end is really motivating the $\eta_{ij} = 0$ equation properly so that it's unavoidably obvious ( which I can do now), another loose end is really justifying the $\eta_0 = 0$ equation the way he justified it, but that's a smaller issue

(Another smaller loose end is motivating the re-arrangement of the matrix in even dimensions the way he motivates it, but you can see from the $D = 5$ matrix with $x^0 =0$ that the re-arrangement is an obvious thing to do, I posted it in one of the pics I sent)

8:38 PM

In the $D = 5$ case, we know $\eta_0 = \xi_0 x^0 + \xi_i x^i = 0$ involves none of the $x^{'i}$ and all the $x^i$, and each $\eta_i = \xi_0 x^{i'} - \xi_i x^0 + \xi_{ik} x^k = 0$ involves one $x^{i'}$ and is missing $x^i$, so an equation like $\eta_{ij}$ should be an equation for a plane involving $x^{i'}$ and $x^{j'}$ while missing $x^i$ and $x^{j}$. Thus $\eta_{ij}$ should be anti-symmetric, since $\eta_{ii}$ would involve removing only an $x^{i}$ and inserting an $x^{i'}$, so it should be $0$

Clearly $\eta_{ij}$, as a linear equation, should include the terms $2 \xi_{[i} x^{j']} = \xi_i x^{j'} - \xi_j x^{i'}$. These terms can be found in the relation $2 \xi_{[i} \eta_{j]} = \xi_i \eta_j - \xi_j \eta_i$. If you work this out you will get $2 \xi_{[i} \eta_{j]} = \xi_0 (\xi_{i} x^{j'} - \xi_j x^{i'}) + (\xi_i \xi_{jk} - \xi_j \xi_{ik}) x^k$. This does not involve $x^0$ so it's not the full equation yet, thus we add $\xi_{ij} \eta_0$ to both sides, which now gives $\eta_{ij} = 0$.

The exact same motivation leads to $0 = 3 \xi_{[ij} \eta_{k]} = \xi_{ij} \eta_k + \xi_{jk} \eta_i + \xi_{ki} \eta_j$ which is how he gets $\eta_{ijk} = 0$, where $x^0$ is in this result so we don't need to do anything more to it, if you try the next one this way it gets a bit weird, as do other things so far...

8:54 PM

Sorry I meant that Penrose starts off with the plane $T-Z=1$. It seems like this maybe equivalent to Cartan starting with $\eta_0 = \xi_1 X^1 + \xi_2 X^2=0$ is he was starting directly from the $D=4$ case.

Information about what exactly is motivating Cartan could come about from this

Why can I have two different conventions for complex conjugating Grassmann numbers?

Syncing the Penrose stuff with all this is another issue to deal with later, it will only confuse things right now, but yeah it needs to by lined up with all this eventually

The other thing is, if you start with the $D=4$ case and enforce real $X_0,\dots$ from the start, you can at least somewhat visualise null vectors on the light cone. With the simplest $D=3$ Euclidian case, it's impossible to visualise because all isotropic vectors are complex.

@bolbteppa In coordinates $$X^1=(T-Z),\,\,\,\,X^{1'}=(-T-Z),\,\,\,\,X^2=(X+\imath Y),\,\,\,\, X^{2'}=(X-\imath Y)$$

The hyperplane will be $$\eta_0 = \xi_1 (T-Z) + \xi_2 (X+\imath Y)=0$$

The other issue is that Cartan's construction is reducible for the Minkowski case

It's the $(1/2,0)\oplus(0,1/2)$ rep of the Lorentz group

9:14 PM

From this perspective, Penrose is starting from the $D = 5$ case with $x^0 = 0$ (i.e. the $D = 4$ case) and then setting $x^2 + y^2 + z^2 + w^2 = x^2 + y^2 + z^2 - t^2 = 0$ and then setting $t = 1$, and then using the fact that the $D = 4$ matrices split into two separate $2 \times 2$ matrices, and using all this to describe the $D = 3$ case before bringing it back to $D = 4$, also it seems to involve a pair of spinors not just one spinor, so there are a load of things to sort out

Yes you are right

I thought about ACM's answer and what I posted above and I finally understand that the two problems were completely different and unrelated, so thank you for helping, ACM. However, I have to say I have seen several Physics books perform a time variation on the action $S=\int Ldt$ to prove Noether theorem: I'm now wondering whether that is the phase space action in the lagrangian form, otherwise variating time would be problematic as in the above case

There's also the small matter of using real vectors to talk about things like $x^2 + y^2 + z^2 = 0$ which is only satisfied by $x = y = z = 0$, but let's not talk about this little issue...

That issue is not present in SR however

@Mr.Feynman this should be a sign not to read a book like that, over-complicating something as simple as the definition of the action/hamiltonian

9:20 PM

There are some interesting comments about the Electromagnetic fields in section 154

In 154 he's just talking about $F^{0i} = E^i$ being a vector and $F^{ij} = \epsilon^{ijk} B^k$ being a pseudo-vector (with the right signs)

How can we choose how we define complex conjugation for Grassmann numbers?

@bolbteppa I don't know, that's my hypothesis to make a rigorous sense of what I'm reading

Because thinking about the discussion I had with ACM and what I've read on Frankel, it makes sense one should avoid time variations in configuration space

$S = \int L dt = \int (p \dot{q} - H) dt$ is the action, that's all there is to it, you can treat it as the integral of a differential i.e. of a 1-form and write it as $S = \int p dq - H dt$, which makes it clear that $\frac{\partial S}{\partial t} = - H$, it's just a game to pretend you can make it any more rigorous than this, defining some space in which you end up doing the exact same thing I just said

@Mr.Feynman bolbteppa isn't a big fan of rigor for rigor's sake :P

9:27 PM

Or framing it in terms of a lie derivative when the usual approach is pretty much just deriving the lie derivative from first principles (without calling it that) instead of defining it in complicated notation

14

A: Grassmann Variables and Complex Conjugate

If $\eta$ is a complex Grassman variable then we require $\eta^*\eta=x$ to be a real (non-Grassmanian) variable. In particular it means that $$(\eta^*\eta)^*=x^*\overset{!}{=}x=\eta^*\eta$$ Write $\eta$ in terms of two real Grassman variables $\eta=a+ib$, then $$\eta^*\eta=(a-ib)(a+ib)=iab-iba$$ ...

@bolbteppa Your words are hurting me like a sword, my friend :P

It's a common mistake nearly everyone makes to think rigorous notation will save them, it's going from being a baby to a child when you outgrow this fad ;)

I think Classical Mechanics done rigorously is one of the most beautiful things out there :P

But why can we just change how we complex conjugate normally?

@ACuriousMind I'm sure you are though :P

What's your opinion about my above message regarding Noether theorem, ACM?

9:37 PM

@Mr.Feynman I am, but less because of some sort of aesthetic sense and more because the kind of abstraction and formalization rigorous approaches perform closely mirrors what I experience as "understanding" something

@ACuriousMind Well, that's my main reason too. I need a systematic approach to be sure to have a solid understanding (that's why without functional analysis in my background I feel like I don't really understand QM :P), then aesthetics are definitely an added value

Because the numbers you are applying complex conjugation to are not real/complex numbers, they are an entirely different type of number, so how we define complex conjugation is dependent on what we want it to satisfy (the thing that post begins with)

@bolbteppa Sometimes it's just about enjoying something :P

@Mr.Feynman you just need to read a good book which explains the physics from first principles, books that formalize all the math in terms of vector spaces and limits are focusing on the wrong thing, it's like trying to read e.g. the bible written in ye-olde Elizabethian grammar that satisfies the conventions of some grammar book instead of just reading a modern version and not getting confused by BOTH the concepts AND language

In the dirac eq. the $\Psi$ is called a spinor?

9:45 PM

just the concepts*

@bolbteppa I understand your point, but I'm already comfortable with that side of classical mechanics. So I'm reading the old versions of the Bible after reading the modern one :P

And I've already read Landau I, it was my first classical mechanics book

@imbAF A Dirac spinor or bispinor

Is it correct to say that the Dirac equation describes the time evolution of a spinor/bispinor?

The SE, describes the time evolution of the state of a quantum system

What does the Dirac eq describes, apart from parity

@Mr.Feynman I think the "correct" way to talk about that kind of time translation is to just look at that as $q(t)\mapsto q(t) + \epsilon \dot{q}(t)$. This corresponds to infinitesimal "time translation" because $q(t+\epsilon) = q(t) + \epsilon\dot{q}(t) + \mathcal{O}(\epsilon^2)$

Sure, that's the induced variation of coordinates and it's fine if we work on the lagrangian

Working on the action, though, you have some problems lifting the variations of the curve

what's the problem? $q(t)\mapsto q(t)+\epsilon \dot{q}(t)$ is a variation of a curve on $M$ that then induces a corresponding variation of a curve on $TM$

it's annoying to try to figure out how to phrase this globally, but locally this is well-defined

and since we only care about infinitesimal variations anyway, we don't need to figure out what this would mean globally :P

10:00 PM

@ACuriousMind That's also true. Then what kind of time variation can't I do in the configuration space action (which what we talked about this afternoon)?

7 hours ago, by ACuriousMind

@Mr.Feynman no, that's the part about lifts of variations in $M$: In the Lagrangian case you're not varying a curve in $T^\ast M$ or $TM$ directly, you're varying a curve in $M$ and this induces a variation of the lift of this curve to $TM$. Since the lift depends on the choice of $t$ along the curve (via $\dot{q}$), you can't do an independent variation here

also, note that your idea about varying the boundary terms is equivalent to just varing the time argument in the Lagrangian, since $\int_{t_i - \epsilon}^{t_f - \epsilon} f(t)\mathrm{d}t = \int_{t_i}^{t_f} f(t+\epsilon)\mathrm{d}t$

Kind of, because I had two different variations at the boundary

@Mr.Feynman the point is that you can conceivable vary $t$ for a curve $(q(t),p(t),t)$ and it stays an "allowed curve" since there is no a priori relation between $q(t)$ and $p(t)$

But that was a moving boundary problem in CV

Also with spinor indicies, is it just a definition that the spinor indicies take the form $(\sigma^m)_{\alpha \dot{\alpha}}$ and $(\bar{\sigma}^m)^{\dot{\beta} \beta }$?

10:03 PM

that is, in the Hamiltonian formalism the action is a functional on any curve in $T^\ast M$, but in the Lagrangian formalism, the action is not a functional on arbitrary curves in $TM$, it is only a functional on curves in $M$

it's essentially the first order formalism vs. the second order formalism

@ACuriousMind And if I do that in the case $(q(t),\dot{q}(t), t)$ I break the relation $\dot{q}=dq/dt$

yes

This is clear. What I'm saying now is: didn't we just vary time just above?

where :P

also, I think there's a very simple fix here to just say that the action becomes a functional on random curves in $TM$ for the purposes of this kind of variations

@ACuriousMind This ^ remains a time variation. Unless you're saying that we are just considering the induced variation on coordinates plus possibly the coordinates variations and lifting that

10:06 PM

that's not what it "is" in Lagrangian mechanics, but you can straightforwardly extend it to one

@ACuriousMind That was kind of what I'm asking

Or you just see it as the phase space action but written in terms of $L$

While I like rigor, I think you are overthinking a throwaway line in that book :P

Oh my, I'm definitely doing that

That's a bad habit of mine

E.g. I had three nights of bad sleep for a sentence in Zee

the way I read the quote, it is just making a point about the first-order vs second-order natures of the Hamiltonian and Lagrangian formalisms

the $q$ and $p$ of a path in Hamiltonian mechanics are truly independent in the sense that given $q(t)$, you have to solve the equations of motion to get $p(t)$

but in Lagrangian mechanics, you get $\dot{q}(t)$ simply by differentiating $q(t)$, i.e. you only need "half the path"

Sure, my 2021 version spent a lot of time on this concept, which I find very crucial to understand the difference between the two formalisms

10:11 PM

a lot of stuff people do with variations in Lagrangian mechanics can be justified by temporarily forgetting about this and pretending $\dot{q}(t)$ is truly independent

So you may wonder why I overthought so much on that line

as long as you restrict to the "true" paths again at the end, you're not doing anything wrong

@ACuriousMind Well, again, $S[q,\dot{q}]$ is not really different from $S[q,p]$, so that's what I mean and I agree

there is no such thing as $S[q,\dot{q}]$ :P

the action is a functional of the physical path

It's the name I gave to the concept you've proposed of momentarily abandoning the lifts

10:15 PM

i.e. it's a functional of $q(t)$ in the Lagrangian and of $(q(t),p(t))$ in the Hamiltonian case

@ACuriousMind Yes yes I know, I was referring to this ^

ah, okay

Now it's time to go back and read Chainsaw man. Thank you!

Someone recently said to me I could probably guess what that manga was about and I did not guess correctly which body part was the chainsaw :D

10:40 PM

@ACuriousMind I'm trying not to imagine what your guess was to get it wrong because the mc has a chainsaw in his head but also on his arms and potentially on the legs

oh it's also the arms? I have been lied to!

Yes, basically all the limbs... When they are attached to the body

I guessed the arms!

The characters are not always topologically connected

there will have to be Words about this

10:44 PM

I don't enjoy violent manga (but I somehow enjoyed violent videogames...?) And now I went from slice of life manga to a manga where people and demons are being cut in pieces

ペガサス Seiya

@Mr.Feynman dragon ball?

@ペガサスSeiya Chainsaw man. It's a bit more violent than DB :P

ペガサス Seiya

@Mr.Feynman how can you get more violent than a guy literally impaling your pectoral area?

I am studying the Dirac equation and how one can derive the Pauli Eq. and the Fine structure from it. The derivation starts, with the following assumption: $E \propto ma^2<<<m$ therefore, we only consider positive energies. Can anyone explain to me, how do we say that energy is positive, because of our initial assumption ?

I don't think "violent" is a particularly coherent category of media because violence is used in a lot of different ways and one can like some of these ways and disapprove of others

ペガサス Seiya

10:51 PM

@ACuriousMind he probably means physical violence in which case it's pretty clear cut IMO

@ペガサスSeiya Even restricted to physical violence what I said holds

ペガサス Seiya

I guess you're right. Physical violence in Dragon Ball is indeed different to physical violence in Chainsaw Man

a very basic distinction being whether the violence is being glorified or recognized for its horror

@ペガサスSeiya I'm not sure I want to describe some scenes here :P

Yes, I meant physical violence

ペガサス Seiya

@Mr.Feynman anime SE rooms exists, if you think offtopic may not be a good idea here

10:53 PM

one can dislike media that glorifies violence yet still enjoy very violent media that doesn't do that

I don't think there is anything really off topic here, but I'll join Anime.SE someday

ペガサス Seiya

@Mr.Feynman you can use the chat without joining but anyway, what scenes were you talking about? I was talking about Moro impaling Goku's chest completely

@ACuriousMind you know, that just sounds like a paradox

@ペガサスSeiya in what way?

I'm saying that violence can fulfill multiple different functions in media, and people can like some of these functions but not others

ペガサス Seiya

@ACuriousMind Maybe you can dislike physical violence in, say, a slice of life anime, but absolutely love it in Dragon Ball for example

I haven't seen a lot of Dragon Ball but my understanding is that its portrayal of violence is very cartoonish (not because it is literally a cartoon but because it does not portray it realistically, in particular with respect to permanent consequences)

11:00 PM

I mean, how you expect realistic portrayer of violence, when you have creatures that blow up galaxies xD

ペガサス Seiya

@ACuriousMind well, it's "realistic" in the sense that violence is well, very violent in that anime. Lots of blood, impaling people, breaking bones and what not. It is unrealistic in the sense that the characters, even after all of that, are still alive and well and better than ever

@imbAF energy is always positive, the assumption $E << m c^2$ affects the two component form of the Dirac equation in different ways

@ペガサスSeiya I won't talk about specific scenes online until I've reached the latest chapter :P

@bolbteppa it is $E \propto ma^2$ not c

ペガサス Seiya

@Mr.Feynman the latest for DBS manga?

11:02 PM

What's the $a$

No, CSM. I dislike DBS manga

ペガサス Seiya

@Mr.Feynman same. The manga is absolute dogwater right now

Anyone watches AOT here??

surely there are people who dislike any form of violence in principle, but I was more thinking of stuff like this:There's a lot of media that celebrates violence, without a lot of critical thought behind it. And there is media that shows similar kinds of violence, with similar levels of gore, but uses it for shock value (i.e. you're intended to be horrified) or reflects on it instead of celebrating it.

ペガサス Seiya

No

11:05 PM

@ACuriousMind I think the impact of violence also depends on the context. In a fighting anime where people die and get revived all the time, violent stuff is quite normal. If you had even just a punch in a everyday-life anime you'd be more upset

I would find it weird if someone dislikes "violence in slice of life" but likes it everywhere else; I would not find it weird if someone dislikes a lot of instances in slice of life because they don't think it adds anything worthwhile but likes it somewhere else where it is actually serving a function

ペガサス Seiya

Let's keep in mind though that violence, especially in anime, can go way beyond physical

I mean I don't think anyone experiencing itachi's Tsukuyomi would describe it as pleasant and it wasn't physical at all

You don't have to go to Tsukuyomi for that. Simple genjutsu by him was enough to fire up kakashi

@Mr.Feynman yes; when the violence is ultimately inconsequential, it doesn't carry a lot of weight either way

ペガサス Seiya

@imbAF Nah it was a Tsukuyomi. Kakashi's sharingan won't let him be trapped in a normal genjutsu

11:09 PM

first time Tsukuyomi was used, was vs sasuke. it's a mangekyou sharingan ability

He never activated it, against kakashi

ペガサス Seiya

Wasn't itachi being on a cross getting stabbed while time was slowed down, is basically what a Tsukuyomi is like?

I don't think a normal genjutsu does that

@ACuriousMind I do not like physical violence in fiction if there is explicit imagery and most of all if it is against "good" characters, it makes me feel bad. In the case of CSM I wouldn't say I enjoy violent scenes (except when I really want someone to die horribly), more like I bear them

A genjutsu is a genjutsu. The thing with tsukuyomi is that it deals a great deal of torture, for days in the genjutsu world and for sec in the real one

ペガサス Seiya

I prefer DBZ over DBS simply because it had more violence and blood

Other genjutsu, do not have the same affect or the same duration

ペガサス Seiya

11:12 PM

@imbAF pretty sure Shisui's does

Cuz he had mangekyou

it's a mangekyou dōjutsu

@Mr.Feynman I mean...that you feel bad for the suffering for the protagonists of a story can just mean that story has likeable protagonists :P

if stories consisted only of scenes we "liked" most of them would be very boring

ペガサス Seiya

@ACuriousMind its called fanservice

I feel you and I are operating on very different levels of discussing media :P

Some of the few animanga where violence is portraited properly and it is necessary and significant is Berserk and AOT..prolly Vinland Saga too

11:16 PM

@ACuriousMind Nah, some years ago I could get fazed even for random background characters dying horribly

ペガサス Seiya

@ACuriousMind obviously I'm superior :P

@imbAF Berserk is an anomaly. It is way too violent in the manga

It's medieval europe . aka land of Cavemans with intelligence to inflict pain

so pretty accurate imo

@Mr.Feynman I think what I'm trying to get at is that there is a big difference as to the intent behind showing that. Is the story doing that because it thinks gore is cool? Then you being horrified really is disliking it, you're not on board with what it's saying. Is it doing it to make a point about the world it's portraying or the characters perpetrating the violence? Then you being horrified may be the intended effect.

Being uncomfortable with any particular scene is not automatically the same as thinking the scene is bad or overall disliking the story.

@ACuriousMind Oh, that's a sure thing. It's the same thing that happens with character: you may hate a character, but that does not mean it is a bad character.

For this reason I always make a distinction between "hating the person" and "hating the character". In your case it would be between "hating the content scene" and "hating the scene"

11:33 PM

yes, exactly

ペガサス Seiya

@Mr.Feynman Some characters are written to be hated

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