The h Bar

Relativisticcucumber

1:10 AM

@naturallyInconsistent hm ok

John Rennie

4:41 AM

@Relativisticcucumber I spent a happy three years using X-rays to measure reaction kinetics :-)

It is true that X-rays will damage your sample. For delicate samples like protein crystals this could be a problem.

You just have to find out how the X-ray dose is related to the damage and make sure you don't exceed the dose where damage starts to be significant.

In my case my materials were pretty resistant to X-ray damage so I was fine.

I did some work using Rutherford backscattering instead of X-rays and there the damage was really significant and we had to work with low beam currents and short measurement times.

Ryder Rude

7:15 AM

hi

Ryder Rude

7:34 AM

@ACuriousMind i didn't understand why it has to be a central extension, just that it happens to be a central extension for the Galilean group

e.g. if we represent the translation + rotation group on phase phase, it is not a central extension, right?

then why must it be central extension for the Poincaire group?

123

7:56 AM

Hello Everyone...

In SR Rindler Coordinates are constant proper acceleration. Does it mean he can travel faster than speed of light?

Slereah

No, constant acceleration in SR doesn't mean that the speed just adds linearly with time

ACuriousMind

@RyderRude Of course the answer there does not contain the full proof of all possible statements one can make here. The idea is this: We want nice position operators that transform "correctly" under our symmetry group, so we seek Hamiltonian actions of these groups on phase space.

In the case of the Galilei group, we get some amount of choice since it has non-trivial extensions, and this allows us, for instance, to choose the correct transformation behaviour so that the SE transforms as we want in the end.

In the case of the Poincaré group, there is no choice, the trivial 2nd cohomology essentially fixes the action on phase space, and the choiceless result we end up with are the Newton-Wigner operators, which are not quite the kind of operators we were looking for, but there's no other choice here, so there aren't any "true" position operators.

123

@Slereah Okay Is there any way by means of acceleration we can have speed more than speed of light?

I have study in lecture there are observer at which Rindler horizon speed of light looks stationary, and other observers which never catch/recieve light.

Slereah

8:16 AM

@123 No in SR any acceleration of a mass will always have results inferior to c

123

@Slereah I have study in Rindler coordinates some observer found speed of light to be stationary and some never receive light.

Ryder Rude

8:31 AM

@ACuriousMind oh. Valter Moretti once told me that the main problem with the quantisation of the relativistic particle theory with $H=\sqrt{p^2+m^2}$ was the violation of causality. He didn't say that there was anything wrong with the position operator there, or any mathematical consistency problems. Did he miss this fact? I am linking that post

ACuriousMind

@RyderRude Valter-Moretti is well-aware of the theory and problems with Newton-Wigner operators, he's written several answers on the topic.

Ryder Rude

yes

I meant his comments on this post physics.stackexchange.com/q/784700/156987

I am referring to this part :

> Thank you. What I understand is that nothing goes wrong in, e.g., the geometric quantisation of the phase space theory with H=p2+m2−−−−−−−√
. I mean we do get a consistent quantum theory with a position basis, which is fine mathematically but the theory isnt experimentally correct because of superluminal communication. Is this correct? –
Ryder Rude

Yes, in summary the problem is the one you wrote. - Valter Moretti

ACuriousMind

@RyderRude The problems with "causality" are the problems with position/localization - your "position basis" is implicitly based on Newton-Wigner if it is to be consistent.

Ryder Rude

oh

ACuriousMind

Valter says this explicitly e.g. in physics.stackexchange.com/a/770850/50583

Ryder Rude

8:39 AM

thanks

i was thinking that the causality problem was a separate thing. At least we get a well behaved operator here. It is the usual multiplication/derivative operator combo as we are representing $[x,p]=ih$ just like in non rel QM

@ACuriousMind why is the causality problem related to the locality problem

> A crucial point is that no notion of localization observable can be defined in terms of projection-valued measures (PVMs) i.e., standard selfadjoint observables, because they automatically would produce a superluminal probability detection spread. This is a non-trivial consequence of the fact that the states of single particle (but also all states of the field) have positive energy.

ACuriousMind

@RyderRude The only operators you could choose for position (and hence use to have a notion of "localization") are the Newton-Wigner operators. But (Hegerfeldt's theorem) it turns out if you do so, "local" initial conditions for your wavefunction can spread superluminally, so this "localization" cannot be actual position as that would then violate causality. Whether you call this problem a problem "with causality" or "with locality" is just words.

Ryder Rude

oh. Thanks

So it just means that the operator, while a good operator mathematically, is just unphysical for causality reasons

ACuriousMind

Well, at least the NW operators aren't position in the traditional sense, that doesn't necessarily mean they are "unphysical". There's a lot of discussion in the literature about what they represent instead.

Ryder Rude

But i just don't see the violation of causality to be a mathematical issue. It seems more like a physical issue because u can construct paradoxes using causality violation

Like, this operator predicts that the probabilities can go outside the light cone. It is just a prediction. It's not mathematically inconsistent as a quantum theory

Do u agree

@ACuriousMind oh

The paradoxes of causality violation arise when we consider multiple observers

But QM treats observations like a black box thing

Ryder Rude

9:12 AM

but the cohomology thing is mathematical in nature, while the superluminal spread of probability is more like a physical problem

the relation between these, i understand it, is that the trivial cohomology gives us only one choice for the representation, and it happens to be a causality violating one

so the cohomology issue is related to causality problem, but in an indirect way. Thanks

Ryder Rude

9:59 AM

Also, Hegerfeldt's theorem anticipates that this rep was going to violate causality

Mr. Feynman

10:18 AM

What makes Bloch's theorem more than a restriction to discrete values of the (continuous) translation operator?

naturallyInconsistent

@Mr.Feynman what are you even asking about?

ACuriousMind

yeah I don't really get the question either

Mr. Feynman

Mhhh, I keep finding full derivations of the Bloch theorem on solid state books, but it appears to be just the translation group/algebra whose knowledge is obvious(???) from elementary QM

Just with discrete translations

So I wondered if I was losing something that makes it any different

Let me know if I'm still too vague

ACuriousMind

Well, I would agree with you that "Bloch's theorem" is much simpler than its usual presentations suggest but you're saying it very confusingly :P

naturallyInconsistent

But the Bloch wavefunctions are continuous functions on the 1BZ; it aint trivial!

ACuriousMind

10:30 AM

Bloch's theorem is really just two statements: 1. The simultaneous eigenfunctions of the translation operators $T(n_i)$ that translate by $n_i$ by the i-th lattice vector are periodic functions on the BZ times $\mathrm{e}^{\mathrm{i}kr}$. 2. If the Hamiltonian commutes with all of the $T(n_i)$, then since all the $T(n_i)$ commute with each other, these periodic eigenfunctions form an energy eigenbasis of the space of states.

The second statement is obvious from elementary QM, the obviousness of the first statement depends on the person :P

Mr. Feynman

The way I interpret 1. is just as the eigenvalue problem of the translation operator

I'll consider a more specific related question (it was not a XY problem, ACM!!!)
$$\psi(x+a)=\mathrm{e}^{iqa}\psi(x)$$
From what I know that $q\in\mathbb{R}$ is a consequence of unitarity of translations, which is very well known in QM; some books suggest it's a consequence of periodic boundary conditions $\psi(x+Na)=\psi(x)$ without ever mentioning unitarity

ACuriousMind

I wouldn't call this "unitarity of translations", that's just the condition for $\psi$ to be an eigenstate of the translation by $a$

but you're right it's because the translations are unitary operators

naturallyInconsistent

@Mr.Feynman If $a$ is not in the set of lattice vectors, this does not need to hold. $q$ actually isn't any real valued vector; if you pick it some other irrational value than $k\pi/a$, especially for a finite crystal, this will fail. Also, your $N$ depends upon $q$, which, again, isn't trivial

Mr. Feynman

10:45 AM

Yes, for a lattice system it doesn't need to hold, as the translation invariance is only discrete. What I am saying is that the Bloch's theorem is "just" the eigenvalue problem of the group of translation operators, restricting to the subgroup of discrete translations, in the way I currently understand it.

I'm not claiming this in an arrogant way: if that is just what it is, I have the impression that solid state books present the derivation as a novelty and it confuses me a lot

ACuriousMind

Yes, Bloch's theorem is just solving for the simultaneous eigenstates of the lattice translation operators

naturallyInconsistent

Well, prior to Bloch's theorem, a generalisation of which is Floquet theorem to mathematicians, and Floquet theorem appeared earlier than Bloch's theorem, people didn't know that crystalline physics had a simple solution form. It was very surprising in that way.

ACuriousMind

it's historical that this is presented differently: Bloch found this in 1929, one year before Dirac's QM book (which was the first text that would probably have allowed you to make that statement in that way)

Mr. Feynman

Historically, of course it is

naturallyInconsistent

Other than that, and those caveats you have to be careful about, yes, this is not exactly complicated.

ACuriousMind

10:49 AM

(also I think the solid state books might not actually expect you to have all that much QM prerequisite knowledge...)

naturallyInconsistent

And I dont think it is presented as a novelty rather than as a "you had better be proficient in this; most of you will be working using this theorem"

@ACuriousMind this is definitely false. You only get to touch SSP after you have proven proficient, at least in exams, in QM and Stat Therm.

Mr. Feynman

That's probably very university dependent

naturallyInconsistent

I dont think so; at least, I have not heard of any university that does things any other way. It is a strict requirement, after all. All the textbooks also assume the same.

Mr. Feynman

In my uni SSP was the first course we took in the master's course and we only had QM in our bachelor's

Well, there was a stat mech course (that I took) but it was optional, while this one was mandatory

But if I had to explain why I felt like asking that is that the presentation gave me the same sensation as GR books explaining differential geometry as if it's something specific to physics :P

And at that point you start wondering "wait, am I losing something?"

naturallyInconsistent

11:07 AM

I'd have to point out that no textbook on SSP that I know of, would take the time to explain stat therm. Familiarity with Bose-Einstein and Fermi-Dirac distributions are just openly assumed. Same thing with stat therm texts assuming basic QM concepts, even though they would take care to discuss particle indistinguishability and not use much of QM maths.

But I understand you just want to point out that concepts in physics usually apply far wider than the specific places they get taught, and professors tend to present them as if they are only applicable in those specific places

Mr. Feynman

11:22 AM

@naturallyInconsistent Yes, that it the reason. It's not about pedantry, rather about the possibility of being misleading

Ryder Rude

11:50 AM

does quantisation (replacing functions with operators according to Dirac postulate) fix the inner product, or do we have to choose it

I think it at least "almost" fixes it, don't know about completely fixing it

e.g. when we do $[X,P]=ih$ we require the inner product such that these operators are Hermitian

So, if X is the multiplication operator, and P is the derivative, does the inner product get fixed to be $\int \psi ^* (x) \phi(x)dx$?

Slereah

I mean you're the one saying it's a multiplication operator and a derivative

You have some freedom in defining $x$ and $p$

Ryder Rude

yes. But it is always unitarity equivalent to multiplication and derivative

so assume we have exhausted that freedom

Slereah

Sure, but that's something you've gotta show

and it's only true for classical systems

Ryder Rude

oh and the notion of unitary equivalence depends on the inner product

@Slereah what do u mean

Slereah

Not true for QFT

Ryder Rude

11:58 AM

yes. U meant to say it is only true for finitely many dof

Silly Goose

@Slereah do you know a good reference that thoroughly goes through index notation (primarily for relativity applications)? I was looking at Wald which seemed good, but didn't go through things like what transposing looks like and etc.

Slereah

errrr

You can try this if you want to look in depth I guess?

link.springer.com/book/10.1007/978-3-662-12927-2

If you need to see fancy words like transvection

Ryder Rude

the reason im confused is the quantisation of relativistic fee particle theory. Suppose we quantise and the rep of momentum is the multiplication operator

and the inner product is fixed to be $\int \psi^* (p)\phi (p) dp$

then suddenly boosts are not unitary, as boosts take $\psi (p)$ to $\psi (\Lambda p)$

Does this mean there is freedom in choosing the inner product?

it's really weird. maybe boosts don't do what i wrote

in qft, there is freedom of choosing inner product when we are looking for unitary reps of the Poincaire group

but I'm not sure about Dirac quantisation postulate. The postulate doesn't say anything about the inner product

nvm it does say. It says something like "replace these with Hermitian operators satisfying the CCR"

So the inner product comes first, and the CCR comes later

Claudio

@SillyGoose I’m interested in this too, but I’m open to more introductory stuff, not necessarily something too heavy

123

12:23 PM

What is "$virtual$ $displacement$" ?

I still don't get the idea after watching a lot of videos on youtube, reading books, google threads etc...

Silly Goose

12:44 PM

what exactly are the conditions to obtain a unique $(\vec{E}, \vec{B})$ from the Maxwell equations?

Slereah

It's a differential equation, so just initial conditions of the EM field and its derivatives?

Wait it's first order in EB

So just initial conditions

Silly Goose

what about (spatial) boundary values?

(I am thinking of the differential formulation of the equations)

Slereah

If you're gonna ask about specific boundary value theorems for PDEs I'm not gonna know them on the top of my head I fear

Silly Goose

or i just am asking if temporal initial condition + spatial boundary values => unique E and B

Slereah

If you have the tempora initial conditions sure

Just have $E(x, t_0)$ and $B(x, t_0)$

123

1:27 PM

I found by definition virtual displacement is differential of transformation of generalized coordinate to cartesian exclude time term.

What exactly is virtual displacement?

Silly Goose

what is the proper way to write a yang-mills lagrangian without components?

is it $-\frac{1}{4} \text{tr}(F \stackrel{\wedge}{,} F)$?

123

$\Large{\delta{x_i} = \frac{\partial{x_i}}{\partial{q_{\alpha}}}dq_{\alpha}}$

Slereah

@SillyGoose $F \wedge \ast F$ in the trace

123

$\Large{dx_i = \sum_{\alpha = 1}^{n} \frac{\partial x_i}{\partial q_{\alpha}}dq_{\alpha} + \frac{\partial x_i}{\partial t} dt }$

Hello @RyderRude

Silly Goose

Hm so the field strength $F = dA$ is a $2$-form, so is the hodge dual $\star F$ also a $2$-form?

ACuriousMind

1:43 PM

@SillyGoose In 4 dimensions, yes

@SillyGoose There's like 5 different notations for this, I prefer the straightforward $\mathrm{tr}(F\wedge{\star}F)$

Silly Goose

is the trace taken only over the lie algebraic part of the resulting $4$-form?

Slereah

Yes

ACuriousMind

technically it's not even necessarily a trace :P

Slereah

That way the remaining thing is an $n$-form

which you can integrate

Silly Goose

Right okay so the Lagrangian actually should be a $4$-form over Minkowski space

Slereah

1:45 PM

that is how you do it yes

That's what $d^4x$ is

Silly Goose

@ACuriousMind what is it technically :P

ACuriousMind

@SillyGoose you can do a Yang-Mills-type Lagrangian with any ad-invariant bilinear form $K(F,F)$ on the space of curvature tensors, it's just that the most natural choice of ad-invariant bilinear form is the trace, i.e. $K(A,B) = \mathrm{tr}_{\mathrm{ad}}(A\wedge{\star}B)$ (the $\star$ is necessary for this to work in arbitrary dimensions)

naturallyInconsistent

@Mr.Feynman In GR that is extremely misleading. But I think Bloch's theorem is mostly only used in crystalline physics and so is ok to be slightly misleading

@123 It is one of those esoteric topics that always confuses students and doesnt really help understanding other stuff when you put in the effort to understand it. i.e. you can safely skip the topic.

Silly Goose

what concretely is the hodge dual of the electromagnetic field strength $F$ :P

123

@naturallyInconsistent Thanks. Is my definition correct for virtual displacement?

naturallyInconsistent

1:53 PM

@SillyGoose we usually prefer writing it as that the lagrangian density should be a 0-form. The difference, of course, is a Hodge dual, and you are free to pick either case.

@123 I don't care if it is correct or not. You had better ask someone else. And seeing how nobody wants to touch that topic, you had also better not tag people randomly

123

Okay.. :(

ACuriousMind

@SillyGoose It's the field strength tensor but with the electric and magnetic fields swapped

i.e. the temporal components are the B-field and the spatial components are the E-field in the dual

it does not have any particular physical interpretation different from the usual field strength

Silly Goose

i see

also so to write out the electromagnetic gauge potential over Minkowski space, would I write $A^\mu(x^\sigma) = (c^a T^\mu{}_a) dx^\sigma$?

this seems like quite bad notation if not wrong :P

ACuriousMind

yeah that's wrong :P

Silly Goose

or like heuristically, for each $\mu$, $A^\mu = g \otimes \omega$ where $g \in \mathbb{R}$ and $\omega \in \Omega^1$ for electromagnetism

ACuriousMind

2:05 PM

it's $A(x) = A_\mu(x)\mathrm{d}x^\mu = A_\mu^a(x) T^a \mathrm{d}x^\mu$

Ryder Rude

@123 hi

Silly Goose

@ACuriousMind and this is for one component of the electromagnetic 4-potential?

ACuriousMind

@SillyGoose no, this is the full potential - summation convention is implied

Mr. Feynman

@naturallyInconsistent To everyone their own, I guess. I'm more grounded in differential geometry than I am in solid state physics, so it's a matter of perspective :P

Ryder Rude

@123 i haven't studied this topic and i haven't found that i needed it for anything

ACuriousMind

2:07 PM

the $A(x)$ is the "abstract object", i.e. a Lie-algebra valued 1-form, the $A_\mu(x)$ are Lie algebra-valued functions in a coordinate patch, the $A_\mu^a(x)$ are real-valued functions in a coordinate patch, the $T^a$ are the algebra generators

Mr. Feynman

Who's @RyderisnotRude. ? Lmao

Silly Goose

@ACuriousMind hm so when someone writes $A^\mu(x)$, they are referring to the (raised) $A_\mu$ you have in your expression? This is like the (linear algebraic) coordinate of $A(x)$ in a basis $dx^\mu$?

ACuriousMind

Yes, $A^\mu$ is just the raised version of $A_\mu$

Ryder Rude

@Mr.Feynman it is user5019261819010018191

Mr. Feynman

Is that your antiuser

Ryder Rude

2:09 PM

no

ACuriousMind

@Mr.Feynman in that case the two should have annihilated when they met :P

naturallyInconsistent

@ACuriousMind just to be sure, there is no $T^a$ for EM, correct?

ACuriousMind

@naturallyInconsistent Well, there is a single one, which we conventionally call 1 (the Lie algebra $\mathbb{R}$ is generated by 1 :P) and then omit, if we want to put it into this framework

Mr. Feynman

@ACuriousMind What if they haven't? Moreover, are you implying that the scattering probability of a Ryder and an antiRyder is 1?

naturallyInconsistent

@ACuriousMind that was what miao miao wanted to type as miao miao sent that out

ACuriousMind

2:12 PM

@SillyGoose "component", not coordinate - a vector field $v$ has components $v^\mu$ relative to a coordinate basis by $v = v^\mu \partial_\mu$ and a 1-form $A$ has components $A = A_\mu \mathrm{d}x^\mu$

Mr. Feynman

What if I write $v_\mu\partial^\mu$? How does make you feel?
(Ignore this message, Silly Goose. Just trying to trigger ACM :P)

naturallyInconsistent

@Mr.Feynman that is clearly equal to $v^\mu\partial_\mu$ and so there is no difference

ACuriousMind

@Mr.Feynman uh...it doesn't make me feel anything in particular

Mr. Feynman

@naturallyInconsistent Of course, it's just "unnatural" to write it like that

naturallyInconsistent

technically, it isnt

Ryder Rude

2:16 PM

@Mr.Feynman this notation implies that metric tensor has been used

naturallyInconsistent

we have to write $\partial_\mu\partial^\mu$ quite often

Ryder Rude

It is different from 1 form

naturallyInconsistent

anyway, time to go have fun again

bye

Mr. Feynman

@naturallyInconsistent I don't mean that it's wrong or anything, I just mean that no one would in the context of expanding a vector :P

ACuriousMind

@Mr.Feynman much more unnatural would have been $A^\mu \mathrm{d}x_\mu$ :P

Mr. Feynman

2:17 PM

GOD NO AAAAAH

ACuriousMind

now that's just horrible

Slereah

@ACuriousMind It's not natural to put indices on $x$ anyway

Mr. Feynman

Do you realize you've done the same thing? :P

Slereah

That's a coordinate function!

ACuriousMind

@Slereah the index is on the atomic symbol $\mathrm{d}x$, it's not a combination of $\mathrm{d}$ and $x^\mu$!

Mr. Feynman

2:18 PM

@Slereah $(dx)^{\mu}$ what are you going to do now?

ACuriousMind

@Mr.Feynman yes, but my version is clearly worse on a spiritual level

Mr. Feynman

There are no ghosts anyway in QED

Ok, I've run out of cheap physics wit for today

Ryder Rude

@ACuriousMind this is $g_{\mu \nu} dx ^{\nu} A^{\mu}$

123

@RyderRude NP Thanks

Ryder Rude

nvm it doesn't make sense

Slereah

2:21 PM

@Mr.Feynman you can add ghosts to QED

But they don't couple to anything

Ryder Rude

$dx^{\mu}$ transforms like the components of a vector

Mr. Feynman

@Slereah Like the ones appearing in Gupta Bleuer iirc

I haven't done QFT in such a loooong time :P

I've actually ran out of interest for physics in the last few months, it's been horrible :'(

Ryder Rude

@Mr.Feynman what are u interested in rn

Mr. Feynman

Nothing

Ryder Rude

U r interested in nihilism

i haven't been studying physics too

ACuriousMind

2:32 PM

@Mr.Feynman are those months of forcing yourself to do it nevertheless or months of not doing physics? Because the first is a recipe for burnout, the second might just mean you need to start again to feel the interest again

Shing

3:14 PM

any intuitive way to see the Jordan's lemma?

Jordan's lemma

In complex analysis, Jordan's lemma is a result frequently used in conjunction with the residue theorem to evaluate contour integrals and improper integrals. The lemma is named after the French mathematician Camille Jordan. == Statement == Consider a complex-valued, continuous function f, defined on a semicircular contour C R = { R e i θ ∣ θ ∈ [ 0 , π ]...

Mr. Feynman

3:41 PM

@ACuriousMind first months of forcing myself that led to a burnout that led to months doing stuff with a very slow pace (in Italy you can take exams whenever you want). Before the burnout I would just do whatever I liked (be it a course I was taking or not) and focus on the exams at the right moment. In the current situation, I can't afford to "waste" time doing that and the only connection left to physics is taking the few remaining exams, but "constrained" study is repulsive to me right now

When I say "constrained" I mean from a course

Obliv

4:02 PM

What does $+Ze$ mean in the context of atoms?

@Relativisticcucumber my lab TA for this computer based experimentation class is also German, born, raised, and studied in Munich though.

was curious about University of Heidelberg he said it was a very good school for physics and philosophy too.

ACuriousMind

@Obliv $e$ is the elementary charge (the constant), $Z$ is the number of protons in the nucleus

Obliv

oh right! It looked so familiar it's been a while since I studied atoms

thank u

Debanjan Biswas

What will be linear and angular acceleration of an extended object if a single force acts upon it?

Obliv

@DebanjanBiswas depends on the geometry of the object and where the force acts

Debanjan Biswas

4:18 PM

Denote them by moment of inertia and angle created by the force with line passing through com

Obliv

nvm I figured it out. the photon has frequency $\nu$ but we're taking half of it to supply PE+KE for $|E_{total}|$

Debanjan Biswas

@Obliv?

Obliv

@DebanjanBiswas the component of the force orthogonal to the path from com to where the force is acting gives the object angular acceleration, the linear acceleration is supplied by the component of the force acting on the com

Debanjan Biswas

My actual question is how will the orthogonal force effect upon angular acceleration?

What will be it's magnitude?

Obliv

depends on the object. $\tau = I\alpha$ if you can determine the torque and the moment of inertia, just use 2nd law

then u can use normal $F=ma$ to determine the linear acceleration for the component of the force not orthogonal to some lever arm

Debanjan Biswas

4:25 PM

Will the object rotate around the axis passing through com?

@Obliv?

Obliv

@DebanjanBiswas the axis that is perpendicular/orthogonal to the line of force acting on the object to the com

Debanjan Biswas

I couldn't understand

Obliv

so for a disc, the force acts on a point on the disc, draw a line from that point to the center of mass/center of the disc that is perpendicular to a diameter through the com

Debanjan Biswas

So the axis passing through com?

Obliv

the axis passing com is too vague

the axis is aligned so that it's passing through com AND orthogonal to the lever arm

in that picture the force is acting on the disc inward/outward of the page so the axis of rotation is drawn to reflect that it's perpendicular to the lever arm

well also perpendicular to the force itself.

Debanjan Biswas

4:37 PM

Okay, now why the axis passes through com?

Obliv

basically $\tau = \vec{r}\times \vec{F}$ produces a vector that's orthogonal to both, just from the definition of vector product. (right hand rule)

@DebanjanBiswas idk, it's a mathematical derivation which I don't know off hand

but you can just think of intuitively. When you flip a coin, it flips about the middle of the coin, not about the edge

you'd need an attachment/hinge to make something rotate about a different axis besides com

Debanjan Biswas

Ugh, it's the thing that is bothering me for hours

I assumed that it will rotate about the axis passing through com

But didn't know the reason

Obliv

maybe this can help

Obliv

5:15 PM

what does it mean to "derive the speed of an electron in Bohr orbit n in terms of the speed of light"

I got $v = \frac{e^2}{2\epsilon_0 nh}, n = 1,2,\dots$

not sure how I'd write this in terms of $c$ though

do I literally just divide by $c$ to get fractional speed or something

the followup question is "is it justifiable to neglect relativistic effects in the development of the Bohr model" and I would say no but idk how I'd justify that

nvm I gotta use $\lambda = \frac{c}{\nu}$

and work that into the equation somewhere

naturallyInconsistent

@Obliv yes

Obliv

dang, it was so simple

I suck at interpreting things

naturallyInconsistent

you already got there

and then you backed off

Obliv

5:32 PM

So I said: We can divide $(2)$ by $c$ to get $\frac{v}{c} = \frac{e^2}{2c\epsilon_0nh} \approx \frac{1}{137n}$ which implies $v_n$ is on the order of magnitude of $c-1.37n\times10^2$ which is very close to $c$ for small $n$. Thus, we should not neglect relativistic effects.

naturallyInconsistent

@Obliv no. Go and compute gamma

Obliv

so $\gamma = \frac{1}{\sqrt{1-\frac{1}{137n}}}$

naturallyInconsistent

and...

Obliv

this is a non-negligible gamma since $n$ must go to $\infty$ for it to be $=1$

naturallyInconsistent

sigh

Obliv

5:37 PM

:(

naturallyInconsistent

$$\tag1\beta=\frac vc\approx\frac1{137n}$$ is clearly fastest when $n=1$. $$\tag2\gamma-1=2\sinh^2\frac{\arg\tanh\frac1{137}}2=2.664073635\times10^{-5}$$ means that $\gamma$ is really negligibly greater than 1, and thus we are justified to ignore relativistic effects in the Bohr model for Hydrogen atom.

Obliv

ooh..

That's surprising. I thought relativistic effects come into play a lot with atomic physics.

guess light is just that fast

naturallyInconsistent

If it comes into play a lot, then Dirac would not have conjectured that SR would have no rôle in quantum chemistry

There is an entire Wiki page for that

Relativistic quantum chemistry

Relativistic quantum chemistry combines relativistic mechanics with quantum chemistry to calculate elemental properties and structure, especially for the heavier elements of the periodic table. A prominent example is an explanation for the color of gold: due to relativistic effects, it is not silvery like most other metals. The term relativistic effects was developed in light of the history of quantum mechanics. Initially, quantum mechanics was developed without considering the theory of relativity. Relativistic effects are those discrepancies between values calculated by models that consid...

You can also compute via $$\tag3\gamma-1=\frac{\beta^2}{\sqrt{1-\beta^2}+1-\beta^2}$$ to get the exact same number, with all the digits kept accurately. Even on a simple handheld calculator

Bml

6:14 PM

Hello everyone. As is well known, the entropy change of a system for an irreversible transformation is not the same as that of the same system for a reversible transformation (just think of the case of adiabatics, where $\Delta S_{sys, rev} = 0$ and $\Delta S_{sys, irrev} > 0$).

Thus, in general, the same formulas cannot be applied. Instead, the entropy change of a system for an isochoric transformation and that for an isobaric transformation is the same in both the reversible and irreversible cases. Is there a precise explanation for this?

Obliv

7:41 PM

it's amazing that this simple formulation (basic newtonian orbital mechanics + discretization of energy via planck) agrees with experimental results for atomic radii and rydberg constant for balmer-rydberg series in spectroscopy

Obliv

8:06 PM

Why is planck's constant appearing everywhere though? Like, it was originally introduced to explain the data for thermal (blackbody) radiation and now it's a fundamental quantity in electron orbital angular momentum..

I guess if you quantize frequency, you end up quantizing energy and now orbital angular momentum.

sorry, $E=nh\nu$ was developed first, the bohr atom stuff came after.

I don't understand the continuum part of this diagram

imbAF

A lorentz invariant quantity is said to be lorentz covariant, because we know how it transforms under LT, which is that it doesn't transform ?

Obliv

oh I guess the E>0 electrons can have any random energy since they're not bound to the atom. nvm

Amit

@imbAF invariant is not covariant

imbAF

So isn't invariance included in covariance?

I mean, you know how it transforms under LT, so how is it not a trivial case?

Amit

covariant afaik refers to tensors, similar to contravariant, it means a certain transformation relation. invariant is used with regards to a scalar quantity. it just remains identical

imbAF

8:17 PM

I see

I got it

Amit there is one other thing I want to ask as well

Amit

ok

imbAF

I was reading some notes of mine about the Dirac equation, and, after deriving the equation, in a way that I wouldn't even call phenomenological, but w/e, somehow we were given the eq. At the end the following is said:
After realizing that the matrices involved are square ones, it is said:
nxn, where n=2 is not possible. Because there are only 3 matrices which are anti-commutative with each other, the pauli matrices.

I have no clue what is being said here

And how the fact that there are only 3 pauli matrices

is a consequence for n=2 not being accepted

afk 7 min

Amit

What? The Pauli matrices are 2x2 matrices...

Obliv

aint no way I've been saying "bro-glee" all this time when deBroglie is pronounced de-broy

I'm pretty sure every professor I've come to know has used bro-glee not broy

and how/why does glie translate to oy?

oh maybe it's that soft/guttural g sound so it's not actually broy but bro(glee) just extremely subtle

Silly Goose

de Broglie is french

it is probably more like de bro-hlee if I had to guess

Obliv

8:28 PM

yeah i'm trying to think of an example for that sound

Amit

@Obliv that's a very common thing with French surnames. Not only French... to refer to another famous Physicist and (past?) user on Phys.SE, few people understand how to pronounce 't Hooft's name correctly, and fewer still are capable of doing it even though they know... 🤣

Silly Goose

i think if someone is saying de-broy it is meant to sound like bro-hlee

the "gl" is like a huff sound

to my understanding

@imbAF do you have the original text of these statements

Obliv

oh I thought it'd be more like you start by making a g sound in your mouth then go to a y sound

Amit

lol

tongue gymnastics

imbAF

@Amit the n=2 represents the dimensions of $\alpha_i$ and the solution to the Dirac equation the vector wave function which would be of 2 dimensions in this case

But it is said that n=2 is not a possible solution

Hence we start with n=4, for which the solutions to the Dirac equation, are what we know as the spinors, if I am not mistaken

Silly Goose

8:36 PM

@imbAF it sounds like this note is making the (canonical, i think) argument that the dirac matrices are a representation of a clifford algebra. the "smallest" representation of said clifford algebra is in terms of four $4 \times 4$ matrices.

Amit

@imbAF Those are the Dirac matrices! Not Pauli

imbAF

I don't know I am translating exactly what it was written

Amit

@imbAF Well anyway, you're better off getting helped by someone else. My knowledge wrt the Dirac equation is superficial at best

imbAF

Ok

fqq

@Obliv de Broglie's name comes from Italian, in which 'gl' indicates this sound

Obliv

8:39 PM

Italian?

What if I had a car, weighing 800kg and moving at 20m/s, made entirely of electrons?

Does the DeBroglie wavelength change?

(obviously you can't make a car out of pure electrons, just assume it's an electron cloud of 800kg and still moving as a group at 20m/s)

Amit

@Obliv Should it? There's no dependence on charge I'm seeing in $E=hf$

Obliv

I ask because this last statement leads me to believe it's the type of particle that matters, not the mass and velocity alone?

@Amit I don't think it should, but the way they said "On the other hand, if an extremely tiny mass is involved, such as that of an electron, then it turns out the associated wave can be comparable in size to its parent particle."

Amit

yes, but it refers to mass only, not charge

Obliv

Okay I erroneously thought maybe this quality would be additive so if you have 800kg of electrons moving at 20 m/s you can still compare to the "parent particle"

but yes it's basically just saying if the object is small enough or fast enough, or both, you can have the wavelength comparable to the original object.

Bml

Would no one like to answer my question?

Silly Goose

8:54 PM

actually, why do we write the dirac equation in terms of dirac matrices? would it not be more sensible to write it down in terms of the image of a spin-1/2 irrep of $\mathfrak{so}(1,3)$?

Obliv

@Bml what is it ur asking exactly

thermodynamics is extremely context dependent

Silly Goose

I mean the only reason I can think of is that the Dirac matrices are far more compact. However, morally it feels better to write everything in terms of the appropriate representation

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