The h Bar

Prateek Mourya

3:23 AM

A sphere rolls without slipping is rolled on a hill rises up to a height h

How can we reason here that it stops rolling and moving simultaneously

RewCie

4:57 AM

Nihar Karve

5:16 AM

The only search results Google shows for 'comtupar engynar' verbatim are the transcripts of this room

RewCie

6:42 AM

I'm a comtupar systems engynar, I come here.

That's it.

Soon you'll see my picture between them... I'll soon becom a Comtupar Systems Engynar

🤣😂🤣😂🤣😂🤣😂🤣😂🤣😂🤣😂🤣😂🤣😂

Prateek Mourya

7:11 AM

Is there any stack exchange for tourism stack exchange

John Rennie

@PrateekMourya yes, give me a moment ...

Prateek Mourya

7:45 AM

Thanks

satan 29

7:56 AM

Hello people, I have a confusion from lagrangian mechanics. Consider the lagrangian:

$L = (0.5)m\dot{x}^2-U(x)$. The lagrange equation :

$\dfrac{\partial{L}}{\partial{x}}=\dfrac{d}{dt}\dfrac{\partial{L}}{\partial{\dot{x}}}$.

Evaluating this, we get:

$m \dot x \frac{d\dot x}{dx} - \frac{dU(x)}{dx} = F_{x} - \frac{d}{dt}\left(\frac{dU(x)}{d\dot x}\right)$.

To get the familiar result $F_{x}=- \frac{dU(x)}{dx}$, the equation seems to suggest $\frac{d\dot x}{dx}=0$ and $\frac{d}{dt}\left(\frac{dU(x)}{d\dot x}\right)=0$.

123

Yo...

satan 29

hi

ZeroTheHero

8:19 AM

@satan29 : $\partial L/\partial \dot{x}}= m\dot{x}$. $d/dt (\partial L/\partial \dot{x})=m\ddot{x}$

satan 29

how?

ZeroTheHero

what do you mean how? It is clear that $\partial L/\partial \dot{x}$ is $m\dot{x}$ since this is a partial derivative and $U(x)$ does not depend explicitly on $\dot{x}$...

I mean... if you have $F= my^2/2 + U(x)$ and you take $\partial F/\partial y$ the answer is clearly $m y$.

satan 29

@ZeroTheHero this is obvious ofcourse

@ZeroTheHero but this is the crux of my doubt

just because U is not explicitly a function of $\dot{x}$ , is it valid to conclude hat the partial deriv is zero?

ZeroTheHero

its partial w/r to $\dot{x}$ is 0 yes.

satan 29

for instance consider a sitution where $\dot{x}$ = ax+b

ZeroTheHero

8:31 AM

ah! hold on.

satan 29

then we can use the chain rule, and we can see that the deriv wont be zero

so how is that we say it is zero in general?

ZeroTheHero

Here you have a function $L(y,x)$ with $y=\dot{x}$. The partial derivative gives you the change in $y$ only, even if y depends implicitly on $x$.

this is the difference between a partial and a total derivative.

i.e. $dL=\partial L/\partial y dy +\partial L/\partial x dx$...

so the $\partial L/\partial y$ is the coefficient of $dy$ with $x$ remaining constant.

in other words, in a partial derivative, you treat every variable as constant except the one you’re taking the partial w/r to.

satan 29

@ZeroTheHero hmmmmm this seems to be it

ZeroTheHero

I recommend you have a look at the wonderful book of Mary Boas... Boas ML. Mathematical Methods in the Physical Sciences.

this is really well done in there... the bit on partial derivatives.

and the distinctions between these implicit and explicit functions of chained variables.

satan 29

I see. thanks a lot for your time!

ZeroTheHero

8:36 AM

it’s quite a subtle point...

moreover what is doubly tricky is that in Lagrangian mechanics $\dot{x}$ and $x$ are considered independent variables..

good luck but don’t get discouraged... you can check the MathSE and there are plenty of question of this type on partial derivatives.

Ratman

9:54 AM

does anyone know a good resource for t' hooft double line formalism? It is the first time I see it. It would be interesting if it is also explained with a group theory approach

Ratman

11:20 AM

never mind, I found the nlab site. And I also found that maybe I should study a bit more of group theory for SU(3)

John Rennie

12:13 PM

So that's the end for Arecibo. It seems a shame.

2

Nihar Karve

12:30 PM

"57 years of astronomical discoveries" - has its lifespan been longer than other telescopes of comparable scale?

Charlie

12:56 PM

I'm failing to see how, if we define a creation operator in QFT as: $$a_1^\dagger=\int d^3k\text{ }\exp{\frac{-(\mathbf k-\mathbf k_1)^2}{4\sigma^2}}a^\dagger(\mathbf k),$$ which is done in a few derivations of the LSZ-reduction formula that I've seen so far, why the multiplication of $a^\dagger(\mathbf k)$ by a Gaussian necessarily creates particles that are of Gaussian form

ACuriousMind

@Charlie What do you mean by "particles that are of Gaußian form"?

Charlie

ah I mean relatively localised in position/momentum space

at least in the limit of some $t$

ACuriousMind

$\int f(k)a^\dagger(k)\mathrm{d}^n k$ creates a state whose momentum wavefunction is (proportional to) $f(k)$

Charlie

If we act on the vacuum we'll get something like $$\int d^3k \text{ }\exp{\frac{-(k-k_1)^2}{4\sigma^2}}|k\rangle$$

oh

ACuriousMind

precisely because you get the state you wrote down there - the momentum wavefunction is $\psi(k') = \langle k'\vert \psi\rangle$ and if you plug in that state there for $\lvert \psi\rangle$ you getr $\psi(k') = f(k')$.

Charlie

1:01 PM

ahhh I see

ty!

RewCie

3:11 PM

@satan29 No, lagrangian mech, assumes $x, \dot x$ to be independent! It's an assumption!

The Pride Flag is homophobic because the lines on it are straight

Legit...

ACuriousMind

@RewCie @satan29 Don't listen to this, it's not an "assumption", it's a consequence of what "the Lagrangian" actually is a function of. See physics.stackexchange.com/q/885/50583, physics.stackexchange.com/q/307794/50583 and their many linked questions.

Nihar Karve

3:41 PM

er, why does a source say that QCD is completely perturbative?

I would have thought the complete opposite

ACuriousMind

the source is wrong, QCD lattice computations are certainly not perturbative :P

Slereah

Do they converge, though

ACuriousMind

most lattice computations "converge", the question is usually whether you can show they actually converge to the right thing

@NiharKarve is the source talking about high-energy QCD, perhaps?

the QCD coupling runs, and at typical hep-th collider settings it's perfectly treatable by perturbative methods

it's in the low-energy regime where it becomes strongly coupled and hence you need non-perturbative methods

Sir Cumference

@Charlie you can tell a lot about a person by whether they put integrands before the $dx$ :P

Nihar Karve

@ACuriousMind Yeah, that makes sense

Actually, when would a theory be "completely perturbative"?

satan 29

3:59 PM

@ACuriousMind thanks a lot.

ACuriousMind

@NiharKarve Well, "completely" is a weird qualifier there

All QFT perturbation series are inherently asymptotic, meaning there'll always be a maximum order in perturbation theory beyond which it's not useful to continue

Charlie

4:15 PM

Mathematically, "functional integration in field theories" is just applying the tools of calculus of variations to multivariable functions, right?

ACuriousMind

@Charlie No, "functional integration" is math-speak for "path integrals"

Charlie

oh

ohh the integrands found in functional integration is the CoV for multivariable functions

i guess that's obvious

Nihar Karve

@ACuriousMind I wonder why they used "completely". Thanks anyway, ACM you're a claß act

skullpatrol

5:45 PM

@RewCie but, the rainbow it is a part of is not straight

Illegit...

RewCie

6:22 PM

@ACuriousMind I guess you both need to check this Why is velocity considered to be independent from position in the Lagrangian formulation of Classical Mechanics?

If you don't assume it, the chain rule will never let cancel $\dot q$ cancel $q$

ACuriousMind

@RewCie The answers there seem to make the same points as the answers in the physics.SE questions I already linked to, what in particular do you think I should pay attention to there? No one there just claims it's an "assumption", they all explain why the Lagrangian as a function on initial data to the e.o.m. should have $q$ and $\dot{q}$ as independent input variables.

Prateek Mourya

What is point of application of resultant of forces

RewCie

@ACuriousMind The question claims it's an "assumption":

Answer seems following it too

JMac

7:19 PM

@RewCie Neither that quora question or any of the answers use the word assumption...

Charlie

This is similar to the fact that $x$ and $y$ coordinates on $\Bbb R^2$ are independent but if we constrain the coordinates to a circle of radius $r$ (by analogy, a particular solution to the EL equations is a constrained curve through $TM$) then the coordinates aren't independent, $x=\sqrt({r^2-y^2})$?

skullpatrol

7:55 PM

ymmv

ZeroTheHero

8:11 PM

@ACuriousMind Can I chime in in this? It is clear from a variational perspective that the velocity $\dot{q}$ is related to $\Delta q/\Delta t$ only “on the solution trajectory”. But it is possible to obtain the Lagrange EOM using - say - the principle of virtual work. Is it so obvious in the context of virtual work to argue that $\dot{q}$ should be independent of $q$?

or rather *that $\dot{q}$ corresponds to the actual velocity only on the solution trajectory

ACuriousMind

@ZeroTheHero The argument that because the e.o.m. are second order the quantities $q$ and $\dot{q}$ must in general be independent in the "space of initial conditions" doesn't seem to me to depend on whether you're looking at virtual work or not

ZeroTheHero

8:33 PM

Let me rephrase this. In many derivation that use virtual work (Goldstein does this also) one reaches an equation of the type $(d/dt)(\partial T}{\partial \dot{q})-\partial T/\partial q - Q=0$ (I’m omitting indices on $q$ and $Q$.

$T$ is the kinetic energy $m\dot{q}^2/2$.

so in such a case with $T=m\dot{q}^2/2$, is $\dot{q}$ not “the velocity on the solution trajectory”?

I mean: the variational formulation is much clearer in this sense...

because yes in this case the variational procedure explores all possible $q$ and $\dot{q}$ to find the best trajectory.

(the trajectory that makes the action extremal)

hmmm... now that I think about it the argument about eom being 2nd order still applies.

ACuriousMind

@ZeroTheHero I really think this isn't tied to any specific equations - whenever we have partial derivatives w.r.t. $q$ or $\dot{q}$, we're not talking about a path $q(t)$, we're talking about some sort of coordinate for a space. This space is the space of possible states/initial conditions of the system, and there $q$ and $\dot{q}$ are independent coordinates because the e.o.m. are second-order

ZeroTheHero

yes this is really the best way to think about this.

ACuriousMind

I think a lot of confusion about this point - path vs. coordinates on the space of initial conditions - would be resolved if we all committed to using e.g. $(q,v)$ as the coordinates and $\dot{q}$ only in the context of path, but the convention to use $\dot{q}$ in both cases seems to be pretty much set at this point :P (alas, my proposed convention would probably also would destroy some of the "act as if they're fractions" applications of the chain rule...)

ZeroTheHero

8:59 PM

agreed this is not terribly clear in the literature. Actually it was @satan29 ‘s query that send me back to textbooks.

Charlie

If the equations of motion weren't second order why would that affect whether $q$ and $\dot q$ are independent on $TM$?

ACuriousMind

@Charlie It would not but we would probably not formulate Lagrangian mechanics on $TM$ if they had a different order :P

Charlie

oh ok

ACuriousMind

the way to argue that Lagrangian mechanics lives on the tangent bundle if you don't want to just postulate it is to argue that the infinitesimal virtual displacements in the principle of virtual work are essentially tangent vectors, see physics.stackexchange.com/a/129817/50583

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