The h Bar

Bml

00:06

@ACuriousMind OK, it was just a thought.

@ACuriousMind Do you mean here that @RyderRude's demonstration is no good because it includes non-affine transformation?

ACuriousMind

@Bml No, my point was that you need to add the condition about preserving constant relative velocities to rule those non-affine transformation out; it's correct that if you just talk about preserving the Galilean (pseudo-)metric, you get more than just the usual Galilean transformations

but in the end this is all the wrong way around anyway: We know the world is not Galilean, and the way the Galilean group arises from relativistic physics (the way the world actually is) is as a group contraction of the Lorentz group in the limit $c\to\infty$

it's an interesting academic exercise to try and figure out an axiomatization of physics that produces Galilean physics, but I'm not sure you realize how academic it is :P

Bml

@ACuriousMind So, if I have understood correctly, this means that: in a strictly Galilean group, one 'preserves' the metrics (time intervals and spatial distances) ALWAYS when switching from an inertial frame to a non-inertial frame? And, similarly, can it be shown that the metric is preserved in the transition from inertial to non-inertial system for any other Galilean transformation that does not belong to the strictly Galilean Group (10-dimension)? Correct me where I am wrong...

ACuriousMind

I don't know what "in a strictly Galilean group" means, we're not "in" any group in any sense

Bml

@ACuriousMind I mean: "considering the Galilean Group"

ACuriousMind

The Galilean group can be shown to be the full group of transformations that a) preserves the Galilean metric and b) preserves inertial observers (this is the condition on the relative velocity). If you omit b), you get a larger group that includes some (but not all) non-inertial transformations in addition to the Galilean group, e.g. the time-dependent translation RyderRude talked about

note that "non-inertial" here just means that this is a transformation that's not in the Galilean group

we haven't made any other definition of what (non-)inertial means

Bml

00:22

@ACuriousMind So, if we omit (b), that means we also consider non-affine transformations, right?

ACuriousMind

@Bml We should always "consider" all possible transformations, affine or not. But it turns out a) and b) together restrict us to the affine transformations, while indeed a) alone does not.

Silly Goose

01:54

@ACuriousMind what do you mean by the

This*

naturallyInconsistent

05:15

H O N K

user400188

Is there standard notation for replacing three vectors, $\mathbf{x},\mathbf{y},\mathbf{z}$ with just one vector $\mathbf{s}$, in physics? I need to integrate over all of space, but writing them all out seems cumbersome. The functions are not symetric so I don't want to use $\mathbf{r}$.

I.e. I want to write $\int f(\mathbf{x},\mathbf{y},\mathbf{z})d\mathbf{x}d\mathbf{y}d\mathbf{z}$ as something like $\int f(\mathbf{s})d\mathbf{s}$, without confusing any readers.

Wait, I'm being silly... I don't need to consider $x,y$ and $z$ as vectors anyway. Forget what I asked, sorry about that.

naturallyInconsistent

05:31

@Bml Unlike ACM, I'm seeing what it is you are trying to get at with $\phi$. However, note that ACM is correctly pointing out that in no part of your question here did you actual state any metric at all. Nor did you state what non-inertialness appears.

14 hours ago, by ACuriousMind

I feel you're getting stuck in a weirdly circular situation here where you simultaneously state you know that Galilean transformations are the unique transformations that preserve the metric and you simultaneously ask whether there are any others :P

I mean, ACM is kinda trying to tell you that, from the maths side, your question is not yet posed well enough to even be answerable. Whereas I am trying to tell you that, from the physics side, that you are not even considering enough physics to start answering your own question. At some point, you have to put in the effort to actually get to a point whereby your question can be answered.

One of the reasons why it is frustrating to argue with RyderRude, is that he would be bringing in situations that are extremely unhelpful to the discussion. For example, the x+o(t) style rigid time-dependent displacement is a non-inertial situation to consider, and it is simple, and so it is good to try something with it. However, it is totally obvious that such a case will not pose any challenge to your "preserving metrics" question at all.

Instead, the stereotypical university 1st year 1st module textbook example of a non-trivial non-inertial frame, is one that is rotating at a uniform angular velocity. It is not expressible in x+o(t). Then you will have to be confronted with how N2L starts acquiring new terms. Then you will think about what it means to measure things, because immediately it means that measuring forces is affected by this change.

Then you have to think about what it means when you say that something is at time t at position $\vec x$ in such a frame. What do you mean that you know where something is? If you are not willing nor able to start putting in the mental effort to consider such basic physical measurement issues, you are not able to start talking about non-inertial frames, Galilean group be damned! You have to either let go of all this, or give us our peace and quiet back.

That is, you do not have to refine your understanding enough to become sensible in both the maths side and the physics side. But if you even want an answer at all, you have to at least choose to meet the criteria on either, and then we can have a meaningful conversation.

@Semiclassical Have you considered that there is a niceness to having a plurality of tolerable interpretations? e.g. it is of course still great if we can find out, say, the particular mechanism by which carbon-based life on Earth evolved, but it is perfectly fine that we have a plethora of plausible alternatives. Similar to that, having a few "good enough" interpretations is much better than having no good interpretations at all.

Of course, we should discard bad interpretations like consciousness collapse or objective collapse interpretations (the latter of which is at variance with experimental results), because we are doing science, but it is not necessary that it is one-interpretation-rules-all.

Ryder Rude

06:14

@naturallyInconsistent rotating frames also preserve the space metric in an obvious way. the example i gave was a time dependent translation. a rotating frame just induces a time dependent rotation on the space slices

@naturallyInconsistent bml didnt ask about N2L. they asked about preservance of space and time metrics. we have already made the question well formulated if u hav read the chat

for preservance of N2L, u need Galilean transforms

Ryder Rude

06:31

in general, one can take a combination of time-dependent translations and time-dependent rotations to preserve the space metric. Galilean transforms r a special case of this

naturallyInconsistent

06:54

@RyderRude No, you have completely missed the point of the conversation. Everybody else in the conversation have had to continue the conversation despite you, not with you. It is actually rather interesting how you could fail to see the relevance of N2L when bml had repeatedly tried to steer the conversation towards the non-inertial. Please just leave us be.

Ryder Rude

07:24

i see the relevance of n2l. i mentioned n2l before u did. but bml 's main question is about preservance of metric, not of n2l

but i hav said many times that u need galilean transforms to preserve n2l, while more general transforms preserve the metric

as always, u just keep writing paragraphs that amount to nothing

Bml

@naturallyInconsistent The speech you made is certainly interesting, full of rigour and conceptually very stable. However, I never said I wanted a solution to my problem. My initial challenge was to try to understand if it was possible to preserve the characteristics of Galilean transformations by moving from an inertial frame to a non-inertial frame. We realised that this can be done, although doing it is difficult. @RyderRude proposed his version of a time-dependent transformation and was quite helpful to me, showing me how my problem could be approached. I make no claim to a total resolu…

naturallyInconsistent

@Bml When you say that

> My initial challenge was to try to understand if it was possible to preserve the characteristics of Galilean transformations by moving from an inertial frame to a non-inertial frame.

I had been trying to tell you that this is not the case. It is not that we are trying to preserve these characteristics from inertial frames to non-inertial frames. It is that we are defining what it means to measure stuff in an non-inertial frames, by bootstrapping them from knowledge we have accrued in inertial frames. That is what is physically meaningful to do to extend the region of validity of application.

> If you have been reading, we finally managed to pose the problem in a decent manner. The last two posts by @ACuriousMind answer my original question, they are the solution I was looking for.

I have been reading. I did not think that a mathematical resolution to your issue had been satisfactory to you, but it is good that you are somewhat satisfied, then.

Bml

07:47

@naturallyInconsistent Read the penultimate post by @ACuriousMind. It is not a mathematical resolution.

naturallyInconsistent

@Bml It is precisely that way of phrasing, and how it is ACM ending the convo rather than you, that made me not realise that you had been satisfied. It is, indeed, not a resolution at all, so how are you happy with that, I do not know. But as long as you are happy, that is fine.

barlop

10:50

Is it true that Physicist Richard Latter found that 4s is lower than 3d for neutral elements of the periodic table, until you reach scandium(Z=21), then 3d and 4s switch round, and 4s becomes at a higher level than 3d?

John Rennie

@barlop Other way round, surely? Atoms lighter than scandium had 4s > 3d and elements heavier than scandium have 4s < 3d.

barlop

atoms fill lower energy orbitals first then higher energy.. indeed google says "The Aufbau principle states that electrons fill lower-energy atomic orbitals before filling higher-energy ones"

Mr. Feynman

Do you have a clearer explanation of Carroll'a argument (concerning Birkhoff theorem and foliations) quoted in this question? Basically I'm asking the same thing as OP since I don't consider the accepted answer compelling enough

John Rennie

@barlop The levels are different in different atoms.

Mr. Feynman

(So as OP I'm asking only how he leaps to the conclusion)

John Rennie

11:00

The 3d and 4s orbitals in a hydrogen atom are not the same as the orbitals in a scandium atom.

barlop

and 3d and 4s in hydrogen aren't the same as in calcium(Z=20) I suppose

John Rennie

So there is no reason why the order cannot change as we move between different atoms

barlop

I think it may well change.. but the question is when..

Would you say the afbau rule is true of Hydrogen?

so 4s is lower than 3d

naturallyInconsistent

@barlop John just told you that it changes at Scandium.

barlop

but John said it's the other way around to what I have it

John Rennie

11:02

That's not a terribly exciting question. We can calculate the orbitals using a big computer and find out exactly where the 3d and 4s change order.

But I'm not sure what is interesting about that calculation.

barlop

what's the name of the calculation?

John Rennie

@barlop We know that 4s < 3d for the transition metals. Yes?

naturallyInconsistent

@barlop And John should be correct. It is trivial to note that in H atom, 4s is higher energy than 3d, so the statement that is correct is John's.

John Rennie

@barlop I guess it would be a Hartree-Fock self consistent field calculation.

But this is heavy maths!

barlop

hmm I have some HF code somewhere.. I might be able to adjust it

well a lot of code with heavy maths, has the maths done in libraries.

so you just call the function and it does the maths behind the scenes

John Rennie

11:04

True :-)

naturallyInconsistent

hmm, could you go beyond Hartree-Fock? Miahahaha

naturallyInconsistent

11:25

@Mr.Feynman Would it help you if I say that I think there is a typo in the quotation, namely that the statement should be "But $R_p$ also takes the set of vectors tangent to $S_r$ into itself, since these rotations leave the spheres invariant.", $S_r$ rather than $S_q$?

What that quotation is saying is that $V_r$ is a tangent space rotated into itself by $R_p$ and so is the tangent space at $r$ of $S_r$, so those two tangent spaces are the same.

John Rennie

@naturallyInconsistent In my day (40 years ago) we would start with the self consistent field calculation then so a configuration interaction calculation to include the effects of electron correlation. Assuming your basis was good enough the result would be within experimental error.

naturallyInconsistent

@JohnRennie Yes, we are still taught this. The current world record seems to still be about CI of 12 electrons?

John Rennie

Don't know. It was 40 years ago :-)

naturallyInconsistent

I think 40 years ago, probably the number of electron pairs thus taken is less than 4?

barlop

11:41

@JohnRennie I guess for neutral atoms before scandium, there is no 3d! 'cos no electrons are in 3d

so one wouldn't really compare energy levels of 4s compared to 3d before scandium?

John Rennie

The 3d and 4s states exist, they are just not populated.

barlop

ok thanks

John Rennie

It makes perfect sense to ask what the 3d and 4s energies are in hydrogen even though they are only populated in excited states.

naturallyInconsistent

@barlop The full spectrum of eigenfunctions necessary to fill up the Hilbert space always needs to exist.

Mr. Feynman

@naturallyInconsistent I suspected this might be the case too. The quotation is faithful to the book, which I own and it is not listed in the errata, so I got paranoid

naturallyInconsistent

13:18

@Mr.Feynman but the actual argument is somewhat trivial if you just reduce what it is saying to the familiar flat space 3D situation. It is just saying that for every point $r$ in the neighbourhood of any spherical surface $S$, there exists a $p$ such that $r$ and $p$ are only differing in radial length, and any rotation of $S=S_p$ about the axis passing through $p$ (and thus $r$), it would also rotate $S_r$ into itself. This axis is thus perpendicular to both $S_p$ and $S_r$

And because of this understanding, I deduce that that particular thing should be a typo. Sadly, it also trivially holds true for the original wording, so it isn't obvious if it is a typo or not.

Mr. Feynman

@naturallyInconsistent Yeah, in 3D it is so evident it's almost no use :P but for the time being I'll accept that's what Carroll meant. And even if the argument were flawed, the conclusion is correct regardless

naturallyInconsistent

@Mr.Feynman But the argument follows exactly the same in the GR case considered there. All that extra talk about orthogonal planes, that is just doing the exact same thing as what I was talking about. My argument is also the content of the answer accepted by OP.

I just think that this proof is needlessly complicated. I would just chalk up that such a representation must be possible whenever spherical symmetry exists.

Sanjana

16:15

Why don't $N$ D p-branes break $1/2^N$ or $N/2$ SUSYs? In some brane configurations like $D5$ branes compactified on a $T^5=T^4 \times S^1$ along the spatial worldvol. directions and a smeared $D1$ (along the $T^4$ directions) and compactified on the $S^1$, and $P$ units of $KK$ momenta along $S^1$, the amount of SUSY broken is not $1/2$ but $1/2^3=1/8$ (All of $D1, D5$ and $p$ are half BPS objects).

And there can be many of the $D5, D1$ and $P$s but still $1/8$th SUSY is broken. Is it the configuration or the different "type"s of 1/2 BPS objects, which is responsible for the enhanced SUSY breaking? Is it the configuration or the different "type"s of objects, which is responsible for the enhanced SUSY breaking?

Mr. Feynman

16:28

@naturallyInconsistent I know, right? Either you go full-rigorous or you just handwave your way using a simpler example

I don't like the idea of explaining it all by means of a wall of text :P

ACuriousMind

@SillyGoose I mean that, if we're being strict about it, there is no need to construct a fully consistent axiomatization of Galilean physics because we know it's merely a limit case of SR in reality. It may be interesting to do it, but it's not required for consistency of the eventual theory that describes reality.

I frequently make a similar point about quantization - it sure is interesting that there's a procedure that turns classical systems into quantum systems with some accuracy, but the only process that matters for our actual theory is the limit that turns quantum systems into classical systems.

@Sanjana you need to look at the boundary conditions these various objects impose, and which supersymmetries preserve them - don't forget that "a D-brane" is originally just a bunch of Dirichlet conditions and similarily KK-compactification imposes boundary conditions on spinors.

More or less these conditions are always of the type "periodic" or "anti-periodic" in certain directions, and you can choose a basis of supercharges so that half of them preserve the periodic conditions and half of them preserve the anti-periodic ones, which means imposing one choice of condition breaks half of the charges.

More generally preservation of some supersymmetry in compactifications is related to the existence of "parallel spinors", which is why CY or G2 manifolds are preferred as compact spaces, since those have such spinors

Sanjana

So I have to do the full thing. There's no shortcut? For the $D5-D1-P$ system I described above the book says $1-1/2^3=7/8$ SUSYs are preserved. Is there a general rule or something like that?

ACuriousMind

welcome to string theory :P

they always state random facts that took some original paper like 10 pages to derive :P

Sanjana

every time :)

Slereah

@ACuriousMind It is still a pretty active area of research

ACuriousMind

16:41

@Slereah I'm not saying it isn't!

Slereah

although tbh galilean mechanics is way worse than GR

Galilean is an ugly bastard group

Sanjana

@Slereah Nowadays people are also looking into the opposite limit $c \to \infty$. They have a fancy name for it: Carrollian limit!

Slereah

Yes I'm aware :p

Average classical mechanics course

Mr. Feynman

@ACuriousMind this means that by 2070 I'll finish reading Polchinski

Then in 2071 ST will be disproven

ACuriousMind

maybe by 2070 string theory is finally dead :P

I see we had the same thought

Mr. Feynman

16:51

I just realized that my message doesn't exclude the possibility of me being the one to end ST once and for all

At the time of writing it was like "I trashed my life doing this stuff"

Sanjana

It's fascinating to know that people had to use Computer algebra in the '70s to derive one SUGRA Lagrangian.

Obviously no book I saw has a derivation...

Manas Dogra

Hello everyone. Today I watched Oppenheimer and saw that making of the bomb that was dropped during WWII involved fairly advanced theoretical calculations. No book on nuclear physics includes those. Is there any book where I may find those calculations or any modern version of it for intellectual/academic purposes. I am sure the complete bomb making guide is not released publicly for good reason. I am interested in the theoretical side only.

Does anybody know any reference?

I just wanna know the basics.

Mr. Feynman

In what sense "derive" a lagrangian? :O

Sanjana

17:38

@Mr.Feynman Write some terms which are invariant under local SUSYs.

Found this

Slereah

17:59

Los Alamos Primer

The Los Alamos Primer is a printed version of the first five lectures on the principles of nuclear weapons given to new arrivals at the top-secret Los Alamos laboratory during the Manhattan Project. The five lectures were given by physicist Robert Serber in April 1943. The notes from the lectures which became the Primer were written by Edward Condon. == History == The Los Alamos Primer was composed from five lectures given by the physicist Robert Serber to the newcomers at the Los Alamos Laboratory in April 1943, at the start of the Manhattan Project. The aim of the project was to build the first...

Slereah

19:33

Reading Duhem's book on physics and it is a refreshing change

He's not the "Physics is only describing experiments" crowd

He KNOWS there's some underlying metaphysical truth

He will not shut up and calculate

ACuriousMind

booo

Slereah

It's pretty rare to see popular books like that in the 20th century

Logical positivism and its consequences have been a disaster for the human race

ACuriousMind

eh

I'm a bit of a fan

Slereah

It's nice to see some conviction once in a while :p

Plus I can only read the spiel about it so many times

ACuriousMind

oh, I agree deviant positions can be more entertaining to read about :P

Slereah

19:41

I wonder who's the most recent PHYSICS IS ACTUALLY ALL GOD guy

Tipler maybe

I know he has plenty of weird opinions

Slereah

20:23

Duhem explaining magnetism within the lens of various metaphysics

Aristotelian v. atomism v. mechanism etc

Slereah

20:35

People really thought vortexes would turn out to explain a lot of physics for like 3 centuries

idk if much ever came of that idea

Slereah

21:14

Duhem is attributing conservation of momentum to god

Claudio Menchinelli

guys I have some problems with the displacement operator: $$ D(u) = e^{-\frac{|u|^2}{2} e^{ua^{\dagger}}e^{ -u^{\ast}a }
}$$

I have to prove its unitarity but this is what I got

Claudio Menchinelli

21:52

Also, if $|\alpha \rangle $ is a coherent state (namely $D(\alpha) |0\rangle = |\alpha \rangle, \alpha \in \mathbb{C}$) how do I compute $\langle \alpha |\hat{p}|\alpha \rangle, \langle \alpha |\hat{x}|\alpha \rangle$ ?

if anyone's wondering I'm trying to see how my professor obtained the wavefunction for a coherent state

basically what he writes is the following equalities: $$\alpha \hat{a}^{\dagger}-\alpha^{\ast}\hat{a} = i(\sqrt{2}Im(\alpha)\frac{\hat{x}}{x_0}-\sqrt{2}Re(\alpha)\frac{\hat{p}}{p_0} = \frac{i}{\hbar}\left( \langle \alpha |\hat{p}|\alpha \rangle \hat{x} - \langle \alpha |\hat{x}|\alpha \rangle \hat{p} \right)$$

where $$x_0 = \sqrt{\frac{\hbar}{m\omega}}, p_0 = \frac{\hbar}{x_0}$$

well the first equality is ok, but the second one just doesn't make any sense, mainly because I don't know how to compute the mean values wrt the coherent state

ACuriousMind

@ClaudioMenchinelli This is literally the worst possible form to try to prove unitarity

just write it as $D(u) = \mathrm{exp}(ua^\dagger - u^\ast a)$ - the exponent is anti-Hermitian, so the exponential itself is unitary, no need to do any computation

(equivalence between the different forms should come via some version of BCH)

Claudio Menchinelli

22:07

wait is this a result I never heard of?

ACuriousMind

which one

Claudio Menchinelli

the operator $A = ua^{\dagger}....$ is antihermitian therefore $e^{A}$ is unitary

ACuriousMind

that's Stone's theorem

Claudio Menchinelli

which we never encountered lol

Sanjana

ACuriousMind

22:10

usually we write it as "for any self-adjoint operator $A$, $\mathrm{e}^{\mathrm{i}At}$ is unitary", but this includes as a special case that $\mathrm{e}^{A'}$ is unitary for $A'$ anti-self-adjoint (choose $A' = \mathrm{i}A$ and set $t=1$)

Sanjana

@ACuriousMind Got this from Polchinski. This apparently counts the number of SUSYs for a pair of branes. I can't understand how the $\beta^{\perp-1} \beta^{\perp'}$ could possibly a reflection in the DN and ND directions when $\beta^\perp$ is defined to be reflection in the D direction.

Claudio Menchinelli

@ACuriousMind Thanks, this makes things much easier, I might have to use this as a black box for now

ACuriousMind

@Sanjana sorry, I'm too tired to try to make sense of string theory :P

Sanjana

Hehe...okay :)

ACuriousMind

I'll have a look at it tomorrow

Sanjana

22:15

Thanks ^_^

ACuriousMind

@ClaudioMenchinelli you've probably seen a variant of this argument before - how did you prove that for self-adjoint Hamiltonian $H$ the time evolution operator $\mathrm{e}^{\mathrm{i}Ht}$ is unitary?

for some reason physicists often don't name this fact Stone's theorem

Claudio Menchinelli

@ACuriousMind we defined the time evolution operator as $1 -i \Omega dt, \Omega \text{ hermitian}$

ACuriousMind

::sigh::

Claudio Menchinelli

and if you stop at 1st order terms is $dt$ it's ok

hahaha

Claudio Menchinelli

22:34

@ACuriousMind I found it: my professor wrote at the start some 'useful' results, which were obviously left to the reader: $$ \langle \alpha |\hat{x}|\alpha \rangle_{t} = \sqrt{2}x_0 Re(\alpha e^{i \omega t}), \langle \alpha |\hat{p}|\alpha \rangle_t = \sqrt{2}m\omega x_0 Im(\alpha e^{-i\omega t})$$

I just need to derives those

ACuriousMind

I'm not sure what those have to do with unitarity of the displacement operator

Claudio Menchinelli

no no the equality

the final equality

ACuriousMind

that's just some specific equations for the time-dependent expectations values of the QHO

Claudio Menchinelli

for $t = 0$ you get the right result

ACuriousMind

nonono, that $t$ there is not my $t$

the $t$ in Stone's theorem is just a generic parameter, not time

Claudio Menchinelli

22:38

yeah yeah I'm talking about the above-mentioned formulas, stone's theorem is "okay", I read the wiki page. My bad

Oh I get it, I need to use the time evolution of coherent states

Mr. Feynman

@ACuriousMind I've never seen you get so heated up with a no :P

But apparently it was my fault

ACuriousMind

@Mr.Feynman and you call yourself an expert in ACM lore?

Mr. Feynman

I was about to announce my discovery!!!!

Claudio Menchinelli

@Mr.Feynman imagine hearing Stone's theorem for the first time in 3 years, we've been using these super powerful results as black boxes for so long I'm starting to feel ashamed of my life :P

Mr. Feynman

How should I read it though? A calm and polite sequence of "no" or a NO NO NO NO FUCK NOOOOO

ACuriousMind

22:44

also, that's not meant to sound "heated", it's supposed to be me frantically waving my hands as I try to prevent someone from going down the wrong path

Mr. Feynman

@ClaudioMenchinelli well, most QM courses never mention it is called Stone's theorem but it's always used nonetheless

@ACuriousMind ::takes note

Claudio Menchinelli

@Mr.Feynman lol

Mr. Feynman

You won't laugh when my plan of cloning ACM finally succeeds

Claudio Menchinelli

that seems preposterous

Mr. Feynman

I'm just fulfilling a wish

Sep 11 at 16:09, by ACuriousMind

I firmly believe the world needs more ACM

Claudio Menchinelli

22:50

that's one of the most ACM-like sentences I've ever read tbh, and I firmly agree with it

anyways I wanted to share this question with you guys math.stackexchange.com/q/4236005/1096913

do you think there's a way to complete the proof following the OP's steps

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