The h Bar

user 85795

00:14

Not seriously at all.

SnoopyKid

00:36

@naturallyInconsistent it's ok I got the answer for my previous question. Is it appropriate if we use probability formula in order to understand whether a dog breed is likely to attack us? Can we use probability formula instead of crime per capita in order to understand if some place is more dangerous than others?

user 85795

00:54

Sure, why not?

123

01:18

Hello @naturallyInconsistent , In perfectly elastic collision we do two experiments where masses $m_1$ and $m_2$ and initial velocities $U_1$ and $U_2$ before collision are same. But in experiment-1 we use different material objects than experiment-2.

We know change in momentum in perfectly elastic collision depends on masses and initial velocities $\Delta{P} = 2\Big(\frac{m_1m_2}{m_1+m_2}\Big)(U_1 - U_2)$

My question is that in experiment-1 and experiment-2 (both uses different materials objects) during collision the magnitude of force always same, or it can possibly be different. because $\Delta{P_{before}} = - \Delta{P_{after}}$. It is predictable. Same response of change in momentum in both experiments can result of different forces.

Obliv

01:51

Why is the otto cycle depicted as a 4 step cycle in PV diagrams

isn't it technically like 3 cuz after the mechanical work is done, the mixture gets replaced by the valves

before you complain it's Schroeder, I googled otto cycle and all the diagrams look like this

the exhaust step seems fictitious

I guess we're separating the steps so that we can evaluate the mechanical work done as its own quantity

yes otherwise we'd be overestimating how much actual mechanical work is output since we'd be considering the work done in replacing the mixture to be part of it(?)

SnoopyKid

02:12

@user85795 how?

123

03:53

If i am in closed elevator under free fall. And i throw a ball upward with velocity 2m/s. Does velocity of the ball remain constant in my frame of reference?

Obliv

@123 i dont have a good reasoning but i'd say yes because you're both in free fall so you're effectively weightless

until it hits the ceiling of the elevator, it'll just be going with the imparted velocity is my final guess lol

0

Q: Derivation of Lighthill Equation

I'm trying to understand the meaning of Lighthill equation. Based on my text book (sound and source of sound, Dowling and Williams 1983), it is derived from combining mass conservation and momentum conservation equations. $$\frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x_i}\left(\rho ...

aeroacoustics is a cool subject I hope I can take a class that covers this stuff some day

@naturallyInconsistent I wanna learn more about navier-stokes equations & fluid dynamics :D

Aeroacoustics

Aeroacoustics is a branch of acoustics that studies noise generation via either turbulent fluid motion or aerodynamic forces interacting with surfaces. Noise generation can also be associated with periodically varying flows. A notable example of this phenomenon is the Aeolian tones produced by wind blowing over fixed objects. Although no complete scientific theory of the generation of noise by aerodynamic flows has been established, most practical aeroacoustic analysis relies upon the so-called aeroacoustic analogy, proposed by Sir James Lighthill in the 1950s while at the University of Manchester...

I wonder if I can consider an air & vaporized gasoline mixture to be an ideal gas.. Shrug

naturallyInconsistent

04:42

@Obliv This is correctly treated by Schroeder. Everybody understands that you really are pulling in new air and expelling exhaust. However, we can only analyse, with equilibrium thermodynamics, closed loop cycles. So some way to theoretically close the loop is always being done. It is also sensible to be considering stuff like constant pressure intake and expulsion, and seeing how the energy and entropy changes that way, etc

Obliv

so you can't close the loop with a different adiabat after ignition step?

guess not, that'd make no sense in how an adiabat is defined

naturallyInconsistent

@Obliv if you have actually seen a real experimental plot of these things, you will realise that it makes no difference. The experiments are nowhere near as precisely defined as we show it on the plots

Obliv

with how $\Delta U = Q + W$ and $Q = 0$ there's only 1 way to get $dU = dW \implies \frac{f}{2}NkdT = -PdV$ anyway so yeah you'd need an extra step

can't have same starting point and 2 different adiabats going to different pressures

@naturallyInconsistent I think because I'm learning 1800s thermo in the absolute most barebones and basic sense, I find it incredibly impressive how far we've come with engineering like I was watching a mechanic take apart a toyota engine to replace a short block and it was mesmerizing

an absolute marvel of engineering like 300k miles in and the engine was still capable of more. If not for a small defect in piston rings it could have gone who knows how much longer

naturallyInconsistent

@Obliv no, what you are getting is all there is. (Except, of course, from where you are getting it, lol.) The issue is that you are not seeing how the theory doesnt actually map to experiments.

again, a good book will emphasise it extremely strongly what is the link between theory and experiment in thermodynamics.

Obliv

well I mean I can't design such an engine yet but I might have the underlying principles(?) lol

experiment & practice is another beast. Eh maybe it's not so bad. Probably problem specific stuff

it's all about the scope of the problem I guess

naturallyInconsistent

04:55

No

There is a gigantic assumption in thermodynamics theory. Every good book emphasises it.

And it is also a persistent source of beginner confusion on adiabatic processes.

Obliv

you mean the assumptions of quasistatic and adiabatic processes?

naturallyInconsistent

yes

just the former

not the latter

No, it is extremely in need of emphasis. Your question right then falls under it.

Obliv

actually I would have enjoyed if he went more in detail about quasistatic processes and what it means for it to not be

I remember this section

naturallyInconsistent

@Obliv that is done in incredible detail in Callen.

Obliv

cuts off but it says "You've created "extra" entropy, because you added extra energy to the gas--more than was needed to accomplish the change in volume."

Ok ill bookmark that

this measly paragraph didn't fully illuminate what is meant tbh

had to spend a lot of time visualizing and thinking about it

esp cuz in my mind I think the box would have increased T from the excessive compression by the piston so the $dV$ and $dP$ would be accordingly

like impulse > work done and entropy increases due to the impulse but the work done doesn't reflect it? idk im sleepy

naturallyInconsistent

05:11

come back when wakey

this is very difficult physics, dont do it sneeppyyy

123

05:38

$\Delta{KE} = \frac{1}{2}mV_f^2 - \frac{1}{2}mV_i^2$

$\Delta{KE} = \frac{1}{2}m(V_f^2 - V_i^2)$

$\Delta{KE} = m(V_f - V_i)\Big(\frac{V_i + V_f}{2}\Big)$

$\frac{\Delta{KE}}{t} = m \Big(\frac{V_f - V_i}{t}\Big)\Big(\frac{V_i + V_f}{2}\Big)$ dividing "t" obs

$\frac{\Delta{KE}}{t} = ma\overline{V}$

$\frac{\Delta{KE}}{t} = F\overline{V}$

$F = \frac{\Delta{KE}}{\overline{V}\times t}$

Why we never use this relation of force and kinetic energy

Hello @JohnRennie Sir

$\Delta{KE} = Ft\overline{V}$

$\Delta{KE} = J \overline{V}$

naturallyInconsistent

05:59

@123 because the step that replaced $\frac{v_f-v_i}t$ with $a$ assumed constant acceleration, in which case the result is obvious and trivial and the standard formulæ already have it, so that there is no point in stating this any more.

123

Hello... :65278135 . Oooh i see. It means in change in KE, we can have variable acceleration

123

06:18

We have thermal energy, but we don't have thermal force. Why?

naturallyInconsistent

06:52

Because that is pressure. And in non-equilibrium thermodynamics we have entropic "force". we are trying to shield beginners from extremely complicated and difficult topics when we teach them

123

Ookay.. Thanks

@naturallyInconsistent I am struggling in finding the equation of motion using law of momentum. I want to make animation of this. But without knowing equations for each object i can't animate. Pls give me hint how to find equations for both objects in elastic and inelastic collision.

naturallyInconsistent

That is most commonly in textbooks. What are you missing?

and dont tag me on that. You should be asking in general and let someone else answer

123

Yes i know three equations of motion. but in LoCM we don't know forces or acceleration or time of collision. without knowing force/acceleration/time i can't reverse the objects after collision.

All three equation of motions have uniform acceleration.

naturallyInconsistent

What even is the scenario you are trying to work on? If you dont tell people, they cannot figure out what your problem is

123

07:09

Yes. I am trying. i found $s = \Big(\frac{V_i + V_f}{2}\Big)t$.

naturallyInconsistent

I am asking for a description of what it is you want to have solved. I am not asking for what you have arrived at.

Slereah

08:07

This paper claims that quantum computing does not actually obey either quantum logic or linear logic but basic logic

Ugh

I don't know enough about logic to judge

Ryder Rude

09:53

what is right context, left context and structure?

Slereah

Structure is about structural rules, which are rules not involving any logical operator

The right/left context is about distributive rules regarding context, I'm not sure what it is

Ryder Rude

oh

Slereah

Ah, apparently S is specifically weakening and contraction

Ryder Rude

i feel we will never be able to classify all of logic

it is essentially derived from human language, but human language is just too much mystery

@Slereah oh

do u think that first order logic can, in principle, do everything that any language can?

Slereah

idk

Ryder Rude

10:06

it would be nice if that held

but liar sentences are not allowed. it says they are allowed in "fixed point logic"

Slereah

L and R are apparently whether the structural rules allow for contextual information on the left and right of the sequent

The parts of the sequent that are not transformed during the computation

Ryder Rude

oh

what would be an example english sentence that this logic can represent (let's say B+L logic)

Slereah

10:23

they are not very generous in explicit examples

Ryder Rude

oh

lucabtz

11:28

can someone help me with something

I have some doubts about 1103.3919

in particular (35) has clearly a wrong sign using their definition of W and the expressions above for the poisson bracket

Slereah

Maybe related to these rules idk

lucabtz

however using the parametrization (37) in terms of the Darboux coordinates $\alpha$ and $\beta$ and computing the poisson bracket as $$\{F, G\} = \partial_\alpha F \partial_\beta G - \partial_\alpha G \partial_\beta F$$ i get exactly (35) even though the sign seems to be wrong

my doubt is that i dont understand the skein relations and the knot theory stuff so well

maybe they just got the sign in the expression above (35) wrong and it should be $\{A, B\} = C^- - C^+$?

however in 1404.5188 they use again the same sign for that and correct (35) with a minus sign

the correction is eq (4.29)

these are different people though

it is also possible that the mistake in Nekrasov Rosly Shatashvili is just that $\{\alpha, \beta\}$ is actually -1 rather than 1

123

13:56

In perfectly elastic collision between Earth and ball. The final velocity of the ball $V_b = 2\Big(\frac{M_E}{M_E + m_b}\Big)U_E + \Big(\frac{m_b - M_E}{M_E + m_b}\Big)U_b$

$M_E >>>> m_b$

$V_b = 2U_E + U_b$ .If frame of reference outside the Earth. It is head-on elastic collision. Direction of Earth taken to be positive.

lucabtz

also i dont understand the red marked sentence

the symplectic form alone does not define a poisson structure, I would also need to specify an hamiltonian to define a Poisson structure no?

123

$V_b = U_b$ . If frame of reference at the Earth surface then $U_E = 0$

How to justify $2U_E$ in the final velocity of the ball when frame of reference taken from outside the Earth.

Also when frame of reference taken at the Earth surface final velocity of the ball and initial velocity of the remains same. The results shows ball rebound at same height. But this results doesn't make sense when frame of reference outside the Earth. $2U_E$ means ball gain twice the speed of Earth.

lucabtz

oh maybe the hamiltonian is the one generating gauge transformations?

lucabtz

14:27

no i think it is not

lucabtz

14:40

oh wait i was just confused i dont need an hamiltonian

Obliv

18:05

@123 should be $-U_b$ I think

based on what you wrote $\left(\frac{m_b-M_E}{M_E+m_b}\right) \approx -1$

@123 You gotta take the velocity of both objects, convert to center of mass frame and rewrite the velocities there

get the velocities of the objects in the COM frame after the collision, and then convert back to their respective frames

i.e. $m_1v_1 + m_2v_2 = m_1v_1'+m_2v_2'$ and in COM frame $v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_{1_{cm}}=v_1-v_{cm}$ etc

using $M_E\gg m_b$ you can do the necessary approximations to get your final ball velocity

Guys is there any way I can get rid of this constant in the work done through this adiabat

For the work done through the adiabat (assuming the process is quasistatic): we integrate the pressure or temperature over the volume change given that $VT^{f/2}=V^{\gamma}P=K\text{(a constant)}$, $$W=-\int_{V_2}^{V_1}PdV=-\int_{V_2}^{V_1}\frac{\text{K}}{V^{\gamma}}dV=-K\int_{V_2}^{V_1}V^{-\gamma}dV = -K\left(\frac{V^{-\gamma+1}}{-\gamma+1}\Big|_{V_2}^{V_1}\right)$$ $$=-K\left(\frac{V_1^{-\gamma+1}}{-\gamma+1}-\frac{V_2^{-\gamma+1}}{-\gamma+1}\right)$$

nvm I think i found it

bolbteppa

18:32

@lucabtz Very hard paper, are you following the rest of it

Obliv

i had a dream about the kronecker delta function I can't remember why

OH I was trying to figure out why it was useful because in my sleep deprived state last night I wanted to know what it was since I always see it being used. Based off the wiki it's just like the dirac delta function except it's 1 when the indices are equal

but in my dream I used it wrong, combing matrices with billions of entries for some reason

lucabtz

19:36

@bolbteppa oh in the end I think I found the mistake

There is of course a sign mistake in that paper but I'm good now

Anyhow the preceding part to what I'm speaking about seems all pretty standard. The parts following isn't what I'm interested in right now so I haven't read it

bolbteppa

20:07

If you want to give a dumbed down explanation of the previous sections, I am all ears, I tried to read the backstory to (1) and gave up trying to find it in ref (2) which I have given up on before, this whole subject is like this, pulling matrices and ideas out of thin air all the time

I've never even seen a simple example of how to find lax matrices, I don't know how people can study this stuff

lucabtz

20:21

@bolbteppa you mean quantum lax operators?

I think the Fadeev notes are quite good

The reference 1 in the paper

bolbteppa

Lax pair

In mathematics, in the theory of integrable systems, a Lax pair is a pair of time-dependent matrices or operators that satisfy a corresponding differential equation, called the Lax equation. Lax pairs were introduced by Peter Lax to discuss solitons in continuous media. The inverse scattering transform makes use of the Lax equations to solve such systems. == Definition == A Lax pair is a pair of matrices or operators L ( t ) , P ( t ) {\displaystyle L(t),P(t)} dependent on time and acting...

Even just that example is crazy

lucabtz

Yeah I know what's a lax pair

But there is a difference between classical lax pairs and the lax operators which appear in quantum systems such as spin chains

For classical field theories such as KdV anyhow I don't think there is a general way to find the lax pair

You just guess it and see it works

bolbteppa

@Slereah en.wikipedia.org/wiki/Belinski%E2%80%93Zakharov_transform

lucabtz

Also a generic equation won't have a lax representation in general

0xVikas

21:06

When we talk about an infinite conducting plane in electrostatics, should we assume it has a small but finite thickness or no thickness at all? How would the charges be distributed in both the cases when there is a point charge near the plane? I think In case of a small thickness, the negative charges will appear on the side of the charge and positive charges on the opposite side. Is this correct? What if the thickness is 0?

0xVikas

21:18

I'm trying to find the solution to the classic method of images problem with charge above a conducting plane, but without the plane being grounded, any ideas on this?

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