The h Bar

bolbteppa

02:54

@Charlie all of chapter 4 of Peskin and Schroeder, especially cross sections and actually getting an amplitude like the $\phi^4$ example they give

user434058

06:13

yesterday, by FakeMod

Is $(\mathbf v\cdot \nabla)\mathbf v$ same as $(\nabla \cdot \mathbf v)\mathbf v$? (I think not)

user434058

Also, is $$\mathbf v \times (\nabla\times \mathbf A)=\nabla (\mathbf A\cdot \mathbf v)- (\nabla \cdot \mathbf v)\mathbf A$$ or, is it $$\mathbf v \times (\nabla\times \mathbf A)=\nabla (\mathbf A\cdot \mathbf v)- (\mathbf v\cdot\nabla)\mathbf A$$ (I think the latter is correct)

John Rennie

06:28

Hyperphysics is now linking to us. Fame at last!

5

(sadly not to one of my answers)

John Rennie

06:46

@FakeMod isn't $(\nabla \cdot \mathbf v)\mathbf A = (\mathbf v\cdot\nabla)\mathbf A$ ?

JingleBells

07:04

BAM

John Rennie

Yes, but ACM's stars are for useful comments while mine are mostly for smutty jokes.

JingleBells

xDD

John Rennie

If the citation indices included puerile jokes I would be world famous.

JingleBells

07:28

ACuriousMind

@JohnRennie Are we even sure that they have more traffic than us and it's not hyperphysics that becomes more famous by being linked here? :P

John Rennie

@ACuriousMind :-)

Still, it's nice to see that the PSE is fulfilling the role we all hoped it would i.e. becoming a definitive repository of information about physics.

ACuriousMind

It's neat. We're also the top result in many google searches for physics topics, but I'm not sure how much of that is a bleeding effect from the rest of the stuff on the *stackexchange.com domains.

Hrishabh Nayal

08:10

@ACuriousMind So english isn't my first language and I tried to google what "bleeding effect" was this is what I found.

"The Bleeding Effect refers to a disorder wherein the genetic memories of one's ancestors begin to blend with the sufferer's own..."

what does it mean though?

ACuriousMind

@HrishabhNayal I meant 'bleed' in the sense of 3b here, i.e. that the popularity of other topics on the stackexchange domains may "spill over" into the popularity of its physics sites.

probably not the clearest way to put that, sorry

Slereah

08:29

Man, hyperphysics

that brings back memories

Hrishabh Nayal

@ACuriousMind oh no worries. It gave me a good laugh :-)

tbh I thought we were gonna get a hidden blade or something for being part of this community :-D

ACuriousMind

09:02

Hidden blades, you say? Now there's an innovative idea for solving meta disputes

user434058

09:22

@JohnRennie Yeah, but that's what I am not sure about. For example, this question talks about $(\mathbf v\cdot \nabla)\mathbf A$ as the directional gradient of $\mathbf A$ along $\mathbf v$. Something definitely different than $(\nabla\cdot \mathbf v)\mathbf A$.

JingleBells

'dmckee --- ex-moderator kitten' seems to have 1 message (since 2015) that has 8 stars :O I can't find it

ACuriousMind

@FakeMod How is it different? The gradient acts on $x$, (right?), so $v$ and $\nabla$ commute.

From the viewpoint of $\partial_x$, $v$ is a constant, so $\partial_x v f = v\partial_x f$.

user434058

09:43

@ACuriousMind In $(\mathbf v\cdot \nabla)\mathbf A$, this $(\mathbf v\cdot \nabla)$ acts as an operator on $\mathbf A$ (directional derivative), whereas in $(\nabla\cdot \mathbf v)\mathbf A$, it's just a constant $(\nabla\cdot \mathbf v)$ being multiplied with $\mathbf A$. That does make them sound different, though they can, after all, be the same, idk. That's why I am asking.

John Rennie

10:10

Ah, I see. I don't think $(\nabla\cdot \mathbf v)$ means $\mathrm{div}(\mathbf v)$ in this context.

user434058

@JohnRennie Then what else could it mean? 🤔

user434058

$$\Delta\nabla\Delta\nabla\Delta\nabla$$ Nice!

bolbteppa

10:30

\begin{align}
[\mathbf{v} \times (\nabla \times \mathbf{A})]_i &= \varepsilon_{ijk} v_j (\nabla \times \mathbf{A})_k \\
&= \varepsilon_{ijk} v_j \varepsilon_{k lm} \partial_l A_m = \varepsilon_{k ij} \varepsilon_{k lm} v_j \partial_l A_m = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl} ) v_j \partial_l A_m = v_j \partial_i A_j - v_j \partial_j A_i \\
&= \mathbf{v} \cdot (\partial_i \mathbf{A}) - (\mathbf{v} \cdot \nabla) A_i \\
\mathbf{v} \times (\nabla \times \mathbf{A}) &= \mathbf{v} \cdot (\nabla \mathbf{A}) - (\mathbf{v} \cdot \nabla) \mathbf{A}

user434058

11:11

@bolbteppa Thanks a lot! :-)

bolbteppa

The only issue is remembering $\varepsilon_{k ij} \varepsilon_{k lm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, just remember you need a formula anti-symmetric in $ij$ and $lm$ that reproduces $\varepsilon_{k ij} \varepsilon_{k ij} = 3! = 6$. An easier example is $\varepsilon_{k ij} \varepsilon_{k im}$, clearly the only condition is that setting $j=m$ and summing should give $3!$ so it has to be $\varepsilon_{k ij} \varepsilon_{k im} = 2 \delta_{jm} $.

Charlie

11:54

@bolbteppa ok thank you I'll definitely look into that

What actually motivates the form of the scalar field Lagrangian that produces the Klein Gordon equation?

In point particle mechanics we heuristically use $\mathcal L=T-V$, but how would we even begin to infer what form $\mathcal L$ would take in classical field theory?

ACuriousMind

@Charlie isn't the Klein-Gordon equation motivation enough for you - you write down the Lagrangian that produces the correct equation of motion

the Lagrangian is not an intrinsic feature of a system, it's just a tool to get the equations of motion. The one and only motivation for using a Lagrangian is that it delivers the correct e.o.m.

Charlie

Oh ok that's fair

ACuriousMind

$T-V$ is the heuristic because this produces - for velocity-independent potentials $V$ - the same e.o.m as $ma = \nabla V$

There's nothing magic about the Lagrangian that it "should" be $T-V$ except that this produces the same equation as Newton's laws

Slereah

Hell you can totally make systems that aren't even close to it

bolbteppa

If you're happy with the Klein-Gordon equation, e.g. as as being the mass-shell relation $p^2 = m^2$ in operator form applied to a wave function i.e. $(p^2 - m^2)\phi = 0$, you can then integrate $(\partial^{\mu} \partial_{\mu} \phi + m^2 \phi ) = 0$ against $\delta \phi$ so that $\int dV (\partial^{\mu} \partial_{\mu} \phi + m^2 \phi ) \delta \phi = 0$ is the minimum of some action. Bringing the $\delta$ out front gives the Lagrangian

Emilio Pisanty

12:09

28.7 kiloGhandis. — Daniel R. Collins May 1 '18 at 18:04

possibly the best comment in all of SE?

ACuriousMind

12:25

lol

Slereah

We all know about Gandhi

JingleBells

12:53

I'm scrapping all messages from 2018 to now that have over or equal to 3 stars and then I'mma feed that text data through an RNN to generate new similar messages. So soon I hope I'll be posting messages that will have many stars

It's so exciting to do random things that no one else has done with ml and ds

I might as well stumble upon a way to connect quantum mechanics with relativity

ACuriousMind

It's more likely you'll get gibberish because stars on a message are usually highly dependent on the context the message was posted in :P

Charlie

@JingleBells The infinite monkey theory suggests you might find it if you scrape the chatlogs

JingleBells

@ACuriousMind That's true, I'm looking at the text I'm extracting and it makes no sense alone, but still :P

@Charlie lol

I can't wait to see what gibberish comes out of the experiment

I hope it doesn't invent words

ACuriousMind

inventing words is a perfectly cromulent activity

JingleBells

@ACuriousMind Not when they look like this: bkksiU#(sjk:

ZeroTheHero

13:55

@ACuriousMind @DavidZ what to do with an edit by third party that should IMO be considered as colliding with the original intent?

physics.stackexchange.com/posts/575672/revisions

ACuriousMind

@ZeroTheHero where's the third party? Ritesh is the original author.

ZeroTheHero

v3 is an attempt to clarify the question but it actually introduced a number of elements that were originally and could have actually changed the contents

never mind...

I confused the author of the revisions.

how silly of me.

ACuriousMind

no worries

ZeroTheHero

sorry ‘bout that.

More Anonymous

@Slereah up for a GR question?

(hey)

Slereah

14:03

Sure

More Anonymous

0

Q: $2$'nd law of thermodynamics of an isolated system as it passes into a blackhole?

So I'm quite confused by this. Let's say I have $2$ friends $A$ and $B$ with metrics: $$ ds_A^2 = -dx_1^2 + dx_2^2$$ and $$ ds_B^2 = dx_2^2 - dx_1^2$$ Ideally I would imagine a symmetry in their dynamics say: $$ S_A(x_1,x_2) = S_B(x_2,x_1)$$ where $S$ is the action. Let's say I have an isolated s...

Slereah

If you're counting the evolution of entropy for an observer, you shouldn't use coordinate time

But proper time of the observer

Or at least a local time function, which can be $r$ if you're inside the horizon, yes

More Anonymous

I see ... thanks :)

Slereah

It is sometimes advised to define the time function via the entropy, even

The York time, as it is called

More Anonymous

Really? Where can read more about this York time?

I'll add this to my to read list

Slereah

14:15

It's in Rovelli I think

Also Carlip

BioPhysicist

14:35

2

A: How do I calculate the effort force exerted by my hands while holding a high plank?

Great response by John Alexiou above. At the end of the day, you can validate the equation with a bathroom scale. It’ll be a narrow stance. If both hands are on the scale, then it’s reading will equal to the force exerted on your hands.

How is this an answer to the question?

Superfast Jellyfish

Yeah I don’t think it answers OP’s question about wanting to calculate the force. It should be a comment on the answer it’s referring

BioPhysicist

That is what I thought as well

Superfast Jellyfish

15:25

When did the font for QnA get changed?

ACuriousMind

15:38

@SuperfastJellyfish Do you mean the change to serif font in '16?

Ah, no, I just looked at the main site :D

The font is still the same, but the line spacing is different. They're...doing things, see meta.stackexchange.com/q/353446/263383

Superfast Jellyfish

15:53

Ah okay. I noticed it when it was surprisingly easier for me to read on mobile.

user434058

How did they come up with the equation (11.4.14) [here](https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Book%3A_Variational_Principles_in_Classical_Mechanics_(Cline)/11%3A_Nonconservative_systems/11.04%3A_Reference_Frame_Undergoing_Rotation_and_Translation). The equation: $$\frac{\delta L}{\delta q}-\frac{\partial R}{\partial \dot{q}}=0
$$

user434058

Why is linking not working?

ACuriousMind

links don't work in multiline messages

user434058

@ACuriousMind Mine isn't multiline, I am on a phone.

ACuriousMind

@FakeMod Your equation is displaymode (because of double dollars), which begins a new line

user434058

15:56

@ACuriousMind Oh, never thought that 3rd party LaTeX bookmarklet could affect SE chat.

ACuriousMind

test

test $$ a = b $$

Hm, no, it's not the display mode

Robin Singh

16:16

I have heard that once we are able to make superconductors at room temp then we would be able to reduce laptops to palmtops. Can someone tell me how?

JingleBells

16:37

I seem to answer it interesting is pretty and was last college and barning about the place work on the queue. They are of the time works on the star bare's get an ever in a proportion and comminged something to do not even on a sige of the study of mathematical theorem :P

ACuriousMind

@JingleBells Please mark it as such when you post something from your neural nets

JingleBells

My bad

ACuriousMind

But I'm willing to bet that it got the ':P' from me :P

JingleBells

yes xD exactly what I was thinking

Here's what the RNN generated when I gave it a starting string "T":

The should be completely barned to the first pease of the community, have been placed of the problem, in the site of the sume of the site and the universe is the site of the site of the time to do the study of the day I was plose students do thing you like the study for models a real expranation of the comment to dees out of the sure of a sige of the sume of the site and the interest of the most today.

I have a posted that not offen is a termint to make a comple are sinitely generated lines and explain the formula with an…

It hurts my brain just to read it

delon

hey, this might be a really dumb question but is $$\frac{\partial}{\partial A^\mu}\left(\frac 1 2 m^2 A_\mu A^\mu\right)=\frac 1 2 m^2 A_\mu$$

ACuriousMind

16:43

it makes you appreciate how much better at this transformers like GPT are - even if they still often don't quite make sense, they at least manage coherence over one or two sentences

JingleBells

@ACuriousMind Sure, this is just a dumb RNN that I copy-pasted from the web. I could tweak the structure a bit and gather more training data, then it'll work better but I suppose it's no match for GPT

ACuriousMind

@delon you should use a different index to sum over in the brackets when $\mu$ is your free index on the derivative

and I don't think that's true, it should be $m^2A_\mu$ without the $1/2$

delon

@ACuriousMind ah gotcha, thank you!

user434058

BTW, (FWIW) I have satisfactorily convinced myself that the EM potential term in the Lagrangian isn't just out of thin air. Deriving it involves intelligent solution guessing, however, deriving it this way definitely makes it less obscure than just stating the potential and verifying it.

bolbteppa

In other words, $\partial_{A^{\mu}} (A_{\rho} A^{\rho}) = (\partial_{A^{\mu}} A_{\rho}) A^{\rho} + A_{\rho} (\partial_{A^{\mu}} A^{\rho}) = (\partial_{A^{\mu}} A^{\rho}) A_{\rho} + A_{\rho} (\partial_{A^{\mu}} A^{\rho}) = \delta^{\rho}_{\mu} A_{\rho} + A_{\rho} \delta^{\rho}_{\mu} = 2 A_{\mu}$

ACuriousMind

16:57

those aren't words ;P

delon

how did we get from $(\partial_{A^{\mu}} A_{\rho}) A^{\rho}$ to $(\partial_{A^{\mu}} A^{\rho}) A_{\rho}$?

bolbteppa

@FakeMod it's a very non-trivial assumption to say the interaction of an EM field with a point particle is given by $- \frac{e}{c} \int A_{\mu} dx^{\mu}$, this introduces gauge-invariance and introducing it in this form may be responsible for renormalization issues in classical EM.

$(\partial_{A^{\mu}} A_{\rho}) A^{\rho} = (\partial_{A^{\mu}} A^{\sigma}) \eta_{\rho \sigma} A^{\rho} = (\partial_{A^{\mu}} A^{\sigma}) A_{\sigma}$

delon

ah I see! thank you.

Superfast Jellyfish

17:26

@JingleBells that was painful to read. Quite funny though. “models a real expranation of the comment” gave me a chuckle.

Abhay Hegde

18:03

Is it just me or the line height of the posts is increased?

ACuriousMind

2 hours ago, by ACuriousMind

The font is still the same, but the line spacing is different. They're...doing things, see https://meta.stackexchange.com/q/353446/263383

Abhay Hegde

Ah okay, forgot it would be rolled out today

ACuriousMind

I think I'm largely okay with it, but I find it weird that the preview of the question text in the list of questions still has the old line spacing

Abhay Hegde

Do you all like it? I feel it would have been nicer to include more gap between paragraphs.

@ACuriousMind True, that remains unchanged. Also, I was hoping they would use a font like Helvetica/Roboto for easier reading.

ACuriousMind

I'm not at the point where I can like it because it just looks different than before and when I look at my old posts I just can't stop thinking "Those don't look like I wrote them!"

@AbhayHegde We explicitly don't use a sans-serif typeface on physics.SE because it blends in better with rendered MathJax, see physics.meta.stackexchange.com/q/7419/50583

Charlie

18:09

I feel like the line/paragraph gap difference isn't big enough now

Abhay Hegde

@ACuriousMind I see. Anyway, if anyone is so bothered, their preferred font (in fact, the whole CSS) is just one plugin away in Chrome or Firefox.

user434058

19:02

Is the following constraint holonomic (I think yes, but Goldstein says otherwise): $$y\mathrm dx+(f(t)-x)\mathrm dy=0$$ where $f(t)$ is a differentiable function dependent on time (only).

user434058

Integrated version of that constraint: $$x=f(t)+cy$$

ACuriousMind

@FakeMod What's your definition of "holonomic"? The usual definition of a holonomic constraint is that it can be written as $f(q,t) = 0$.

user434058

Which conforms with the definition of a holonomic constraint.

user434058

@ACuriousMind Mine is, isn't it?

ACuriousMind

you wrote down an equality of 1-forms and I don't know how you got your "integrated version" from that

user434058

19:05

(FWIW, it's Derivation 6 of Goldstein's chapter 1)

user434058

@ACuriousMind I integrated that ODE.

ACuriousMind

where's the ODE?

user434058

Below I present the integration process:

user434058

@ACuriousMind Oh! Now I understand my blunder. I assumed $x$ and $y$ to be independent of $t$. Silly me ;-)

ACuriousMind

I don't know what you're doing but Qmechanic gives the proper condition for such an equality of 1-forms to be equivalent to a holonomic constraint here

user434058

19:09

@ACuriousMind Was I wrong in assuming that x and y are independent with respect to t?

user434058

(I think yes)

ACuriousMind

I will repeat that I don't know what you're doing and hence cannot comment on what you're doing wrong :P

What you wrote down is an equality of 1-forms, a priori $x$ and $y$ are just coordinates on configuration space and not dependent on $t$

user434058

@ACuriousMind Oh, I was hoping to integrate the constraint equatio to obtain an equation of the form $g(x,y,t)=0$. So I went on integrating without acknowledging the dependence of $x$ and $y$ on $t$. Does this make sense?

ACuriousMind

well, it makes sense to me why you'd think that way but you cannot just "integrate" the equality of 1-forms

What are you integrating them along? They're 1-forms, you cannot generically integrate them without choosing a path

user434058

@ACuriousMind Alright, I need to know what 1-forms are. Hitting the wiki... But, as a general advice, what should I read to clear up these mathematical fundamentals? I mean, I thought Goldstein might be self sufficient, but now I see these terms being thrown at me by you and Qmechanic (in his answer), how am I supposed to know them, and from where?

ACuriousMind

19:14

Qmechanic's post gives the proper condition for "integrating" such an equation

user434058

Btw, thanks :-)

user434058

Dammit, I give up. JEE was so much kinder :P

ACuriousMind

@FakeMod I don't know if you're supposed to know them, but learning about them is part of differential geometry

user434058

@ACuriousMind So what are the prerequisites for classical mechanics?

ACuriousMind

And knowing about $p$-forms is crucial to make rigorous arguments about such things like the equation involving "infinitesimals" like $\mathrm{d}x$ up there. If you don't care about rigor you don't need to learn about them

user434058

19:19

I suppose multivariable calculus, vector calculus, linear algebra, differential geometry, topology(?). What else?

ACuriousMind

@FakeMod impossible to answer since you need very different things for hand-wavy Newtonian mechanics than you do for advanced Hamiltonian mechanics!

julien surel

@ACuriousMind, I have a couple questions regarding Witten's paper "quantum gauge theories in two dimensions" which we discussed a few days ago. Is there anything known about the dimension of moduli space of flat connections on a surface (the one obtained mod action of gauge group)?

user434058

@ACuriousMind So what should be the best way? First get all the prerequisites solid, or just keep on learning them as you keep on encountering them (as in this case)?

ACuriousMind

physics doesn't really work like giving you a list of math you need to know to understand a specific subject - there's often a way to muddle through with knowing much less that "usual", and there's also a way of throwing much more math at it than the average

I'm often partial to the latter approach but that doesn't mean it's the only one :P

user434058

@ACuriousMind No, I care about rigour :-) And I like the latter way. Anyways, enough physics for today, see you later. Thanks.

ACuriousMind

19:27

@juliensurel as you can see at the beginning of section 3 of that paper, it's $3g-3$ for $g$ the genus of the underlying surface

julien surel

@thanks, I had just looked at section 2.

ACuriousMind

@juliensurel you can see this by thinking about a flat connection being essentially determined by its holonomies about homotopically non-equivalent loops. In general, the moduli space of flat connections is the space of maps $\pi_1(\Sigma) \to G$ modulo configuration by elements of $G$

it's not that straightforward to see though, here an exposition on it

julien surel

21:07

@ACuriousMind, I actually realized I knew the answer. I was aware of the computation before.

Section 2 of that paper was just like magic, I understood the overall structure, but physical arguments were like magic I didn't understand why they actually worked. They gave a TQFT argument which made more sense. Apparently they promise a different mathematical proof in section 4 which I'm going to look at.

Charlie

21:26

During second quantisation, when we "promote" $\phi$ and $\pi$, the field and the momentum density to operators, are talking about effectively an entire field of operators?

As in, an operator defined at every point on spacetime

ACuriousMind

yes

Charlie

ok nice ty

ACuriousMind

(although mathematically that doesn't quite work and we should really talk about operator-valued distributions but we usually ignore that :P)

Charlie

Anything even slightly resembling rigour in my journey into qft will be ignored for now so it's ok :P

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