This question comes to my mind once in a while what worthy in solving questions with ideal conditions i mean massless frictionless pulley and string...

Please answer this

Anyone here?

@sheltonBenjamin hi :-)

Sir please answer the above question as it is bothering me

sciencealert.com/…

@sheltonBenjamin what question?

This question comes to my mind once in a while what worthy in solving questions with ideal conditions i mean massless frictionless pulley and string...

In real life, if you're working as an engineer or physicist, the problems are never as simple and well defined as the ones you solve for the JEE.

But you use the same skills to solve them that you developed by solving the simpler problems.

Ya that's what bothers me is how can we face difficult problem only when we are given such ideal condition problem

The reason we give students the simplified problems is to help them develop the skills they will need when they're solving problms for real.

How we deal with Friction pulley?

With complicated problems there are often so many things to consider that it easy to get lost in all the detail and be unable to do the calculation. But if you're already experienced with problems where there is no friction then you know roughly what the solution is going to look like.

i.e. it's going to be a bit like the frictionless problem but with some differences caused by the introduction of friction.

So it's much easier to solve a hard problem if you already know how to solve similar simpler problems.

Ok sir it will take a bit of time for my mind to accept it but i am happy that one of my fears will end now

Everything new you learn is based on stuff you already know and skills you have already acquired.

Hippocampus

try to think of it as learning to walk before attempting to run @sheltonBenjamin

Room temperature (15°C) superconductivity achieved under lab conditions quantamagazine.org/…

Why is there no bound state of two quarks?

@ManasDogra There are bound states of quarks and antiquarks of course i.e. mesons.

A bound state of two quarks would have a net colour charge and therefore an infinite energy.

Why does a net colour charge give infinite energy?

Because the colour force does not decrease with distance. Suppose you start with a proton and pull out one quark. You have to exert a force F on the quark to pull it out, but this force does not decrease with distance so the work done i.e. W = ∫F dx to remove one quark to infinity is infinite.

What actually happens is the energy stored in the interaction between the quarks becomes so high that a quark-antiquark pair is created from the vacuum. Instead of a separate diquark and quark you end up with a hadron and a meson.

Thank you @JohnRennie

hello fellow fizyssitsts

What is wrong with the following reasoning?: $$|\phi_3\rangle\cdot\left(\langle\phi_1|\phi_2\rangle\right)\underset{\text{commutativity}}{=}\left(\langle\phi_1|\phi_2\rangle\right)\cdot|\phi_3\rangle\underset{\text{associativity}}{=}\langle\phi_1|\left(|\phi_2\rangle|\phi_3\rangle\right)?$$

or maybe this is allowed, and the last equality has a tensor product in it, $\langle\phi_1|\left(|\phi_2\rangle\otimes|\phi_3\rangle\right)$?

@Charlie The equality is true if the last expression is meant as a tensor product, but what "associativity" do you think you're using there?

just matrix associativity

maybe I'm abusing Dirac notation

In $(\langle 1\vert 2\rangle) \cdot \lvert 3\rangle$, the operation between 1 and 2 is an inner product, and the operation between the result and 3 is a scalar multiplication. Where's the matrix multiplication?

hmm

good point

by which rule is the last equality correct in that case? if associativity is off the table

Well, you first need to really define what you mean by the r.h.s. and then it's true by definition :P

if the thing in brackets is a tensor product, then you need to say what it means to write $\langle 1\vert$ next to such a product. It's of course natural to say this means taking the inner product with the first factor, but if you write out what that means it's just the middle expression

When you say "true by definition", what definition are we using?

is that just a standard definition in linear algebra?

I should really just take the time to read a proper linear algebra textbook at this point

would probably save a lot of grief

@Charlie It's just (partial) contraction of a tensor - think about how you'd write this operation in index notation: $1_\mu 2^\mu 3^\nu$.

and now in this notation you could talk about associativity

ok I see what you mean ty :)

why, may i ask is there no direct link between the stackexchange chatrooms and the stackoverflow chatrooms?

ok, thanks

are flags seen by moderators from all the servers?

I can only see the flags on chat.stackexchange.com. I don't even have an account on the other servers.

Wow!what a research

google.com/amp/s/www.thehindu.com/sci-tech/science/…

@FadedGiant thanks

Only organization in the world which has rich experience in research in sun

I am happy if you guys share your valuable views on it

@skullpatrol Heyo, I've enrolled in CS50 x2020. It' FrEaKiNg CoOl!

@ACuriousMind — Help me make sense of this comment: “I suspect that when dimensionally reducing from (4+1)𝑑 to (3+1)𝑑 space, there's a magical conspiracy such that after the initial positive wavefront passes, its infinite-duration negative wake is continuously exactly canceled by successive positive wavefronts arriving from sources that are ever further-away in the fourth spatial dimension”.

You told me the other day that my interpretation of an “upside down” wavefront on the light cone (in respect to the strictly positive “wake” inside the cone) wasn’t the right way to think about it. But isn’t it exactly what tparker argues in this comment? An initial positive wavefront followed by a negative wake?

Congrats! @Azmuth Have fun :-)

@skullpatrol I've downloaded all the reference books too! :-)

Along with CLRS and Donald Knuth, Art of Programming!

will B phun

Coolio.

@skullpatrol You've enrolled too?

@RobertoValente I said that you interpreting the Green's function as the shape of the wave was wrong. I don't really know what the actual shape of the wave looks like.

Nope.

@skullpatrol I'm doing week 1 assignments....

I can complete whole course in 15 days..

Easy...

Nice.

@skullpatrol You've completed Donald Knuth?

not all of it

@skullpatrol Chapter 1?

lol, are you testing me, pal?

:-)

@skullpatrol Yes. I wanted to know if you are an professional expert or not

I'm nobody.

@skullpatrol Me too, same here. No difference.

@skullpatrol Have you completed CLRS?

Don't forget to use Concrete Mathematics as your backup.

@skullpatrol Yep. I've that too... I read that everyday for a week and then I left that, now, I'll pick that up back again until the course ends.

You should go through that first,

@skullpatrol Have you read The Alchemist by Paulo?

@skullpatrol Okay...

@skullpatrol Completing that?

Yes, complete it. Throughly.

@skullpatrol Okay... I'll

It's 670 Pages long :\

@skullpatrol But concrete Mathematics has all mathematics!

you'll need it

I like their notation of definite integrals... $$ \sum_{a}^b f(x) \, \delta x \equiv \int \limits_a^b f(x) \, dx$$

For The Art of Computer Programming you'll need it.

@skullpatrol O.o Okay...

@skullpatrol Are you a professional coder?

@skullpatrol I thought he was a coder...

Computer Scientist

@skullpatrol Wow. So, you've ever tried writing books on Computer Sc?

I gotta go pal

Oh okay!

Cya

Have a great day ahead :)

you to pal

Thanks :D

:-)

@ACuriousMind Heyo ACM, you there?

@Azmuth Yes, why? (Please just tell me whatever you want to tell or ask me directly instead of just checking if I'm here)

@ACuriousMind You've a Bachleors in Computer Science?

No.

Oh okay...

Any other Professional Coders here?

We live in a 4D space where we travel through the 4th dimension (time) at 1 second per second, right? So what would the 5th dimension be? Will a 5D world means we are at all places at all times?

No. :P

@JingleBells Lol

1 second per second

Depends on your model of string theory...

@ACuriousMind No to which question :D?

@Azmuth I am a professional programmer, I just don't have a degree in CS.

5th dimension is a tiny one..

@ACuriousMind Have you completed Donald Knuth?

@JingleBells To yours :P A fifth dimension is usually just another spatial dimension, since more than one time dimension messes everything up

and we don't really have a notion of dimensions that aren't time or space

Can I possibly imagine the 5th dimension plz :P

Yes, ezy

@Azmuth I don't know what that means. I haven't read Art of Computer Programming if that's what you mean

@ACuriousMind Yes, that's what I meant... Then how you learnt programming?

@JingleBells there are neat animations of a 3d section of a rotating tesseract (=4d cube), that's technically 4+1 = 5 dimensions

@JingleBells $ds^2 = \sum_{1=1}^5 \sigma_i \,dx_i^2$

$|\sigma_i| = 1$

Tessaract

probably

Hmm, why isn't 3D 2D + time?

@JingleBells it can be

@JingleBells You can make your own model, but it has to work

@Azmuth I have no idea what you're talking about but I appreciate the random equations that I barely understand.

@JingleBells You can design your own model of universe 3D = 2D + time... But, if it doesn't works, it's useless... simple!

@Azmuth We had classes in school and once you know one language it's not so hard to pick up other ones. I had to do some data analysis in Python and some simulations in C when I studied physics and I also did some programming for fun (that never got anywhere useful but taught me plenty while I was figuring out how to do stuff like display a triangle on a screen :P)

@ACuriousMind So Homer Simpson is living in a 3D world? I mean, I'm confused by what people refer to when saying 3D, 2D, 4D...

If 3D = 2D + Time, then why do people say we live in a 3D world, and not 4D?

@JingleBells sometimes they mean the spatial dimensions, sometimes they mean the spacetime dimensions, it depends on the context

Got it

I learned to play boogie woogie on the piano so well, it's really cooool

@JingleBells the equations don't make any more sense to me either :P

@ACuriousMind Oh okay, We had programming classes, Python, C++, AI, ML, Web Dev, CGI.... But you know what happened? I passed out from school and those subjects came as different... In high school we had C++ only... I was pretty able to do it but Python sounds a better language and I'm unlearning C++ and relearning Python

@ACuriousMind In 4D, $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$ Is that difficult to understand the length of spacetime?

ofc $c=1 m/s$

@Azmuth Once you've tried enough languages you'll come to realize that most languages aren't "better" or "worse", they are just more or less suited for particular use cases and styles.

@Azmuth and what does that have to do with JingleBells asking about imagining the 5th dimension?

@ACuriousMind Honestly, Python is more useful for AI/ML thingy stuff than C++/ C bec they have got some awesome libraries and Tensorflow stuff...

@ACuriousMind geometry! That's how dimensions are constructed in Mathematics.

I'm well aware of that but how was your cryptic metric formula intended to help JingleBells' imagination?

@ACuriousMind You can construct every part of spacetime using metric formula...

That's not an answer to my question but if you're just going to say random facts instead of having a conservation I'll drop it :P

lolololol!

Riemann Geometry! stonks

In linear algebra transformations, it's required that the origin point stays at (0, 0). Is there a branch of math that "moves" the origin point, something like this? :D

@JingleBells that's called an affine transformation

oh wow, this exists :D

Does anyone know How to automatically put ; after end of line in vim automatically?

Use Visual Studio Code instead of vim. Who uses vim? Are you insane? :-)

sell these to make Easy Money during the pandemic :-)

@JohnRennie whoa, will we have a three-way holy war between vim, emacs and VS code now? :P

I like VS code (and its extensions) a lot, but I feel it's not quite what the core vim demographic is looking for...

Why can we say $\partial_iP^\mu = 0 $ due to homogeneity?

I used command line editors in the days before GUIs existed (on an IBM OS/370). The experience left a permanent scar on my soul! :-)

$P$ is the four-momentum?

YEs

Sorry, should have mentioned

It's kind of just a glancing comment in the notes

of what is that the 4-momentum?

Yes it is the four momentum

I got that, but the 4-momentum of what?

A photon

then I don't get what that has to do with "homogeneity" at all

Or a massive particle

The lecture notes just say it quite generally

@ACuriousMind lol!

that wasn't a joke

what's funny about it?

@skullpatrol I've been laughing for straight 15 min...

@ACuriousMind No not that, that one^

skull's!

sorry, wrong tag

then why did you ping me?

@ACuriousMind Mistakenly mouse got overscrolled

@SirCumference Heyo

Any1 on9?

@JohnRennie Vim is classic, I just wanted to give it a console kinda hacker style... I'm doing cool thing, very less people get it.

During the motion of a vertical pendulum

In equilibrium position

If the suddenly the velocity of the bob is increased

Then can we say because mechanical energy of system is increased

Therefore system has tendency to go up the level

By vertical pendulum i meant this

What does "go up the level" mean @sheltonBenjamin?

Increase the potential energy

The way you've drawn your diagram seems to show the pendulum going in a circle, so that it and the string trace out a cone shape, is that intentional? Or are you talking about a pendulum swinging in the plane?

Yes it is conical locus could this be explanation for this phenomenon of it moving up

Bob goes in circles

it's not about energy, it's about the centripetal force balancing with the gravitational force

Uh, it seems related to the fact that centripetal force is a function of velocity, and the change in the centripetal force changes the position of equilibrium for the bob

Anybody knows what the heck is the "variation of a scalar field"? I think it might be the gradient, but maybe it has some different meaning in QFT.

Any chance it's talking about the variation of the action of the scalar field?

@Charlie Wow, thanks! It is a trace. Next question: what is the variation of the action. Maybe the change of S for infinitezimal changes of the path?

@Charlie Btw, the field is constant in time. That makes the action probably significantly simpler.

@ACuriousMind Please wait with the closure, I am thinking with full capacity

"path" doesn't make so much sense for a scalar field

If the field is constant in time surely that is trivial no?

Ok. I know basic vector calculus but my skills are quite enthusiast-level in QFT

do you mean the lagrangian doesn't have explicit time-dependence?

@Charlie No, I mean that $\phi(\underline{x},t)=\phi(\underline{x})$

What likely includes that also the lagrangian has no time-dependence

that seems odd, what book are you reading from?

I would probably not shoot for qft if you only just understand basic vector calculus

do you know regular qm?

Yes

Idk, in my limited experience of qft I've never seen a time independent scalar field used

@Charlie Here is an example: physics.stackexchange.com/q/506020/32426

@Charlie I am trying to calculate energy densities in SI units of laboratory-sized Higgs-field in various configurations :-)

How to prove that tension is same even though string isn't massless

@peterh-ReinstateMonica Why are you trying to do that? Note that in QFT you can't just get the value of a function by plugging in some expectation value - the Higgs field does not have a definite value, and while we know it's VEV $\langle \phi\rangle$, this doesn't mean you can compute an energy density as $E(\langle \phi\rangle)$ - you rather need to compute the expectation value of $E(\phi)$ as an operator.

additionally, since the zero point of energy is arbitrary in non-gravitational energies, you'd need to compute some sort of difference for the numerical value to have any sort of meaning

@Azmuth :D

@ACuriousMind pardon the mix up, pal.

@skullpatrol now that's a smart investment

Yup.

Does anyone know a book for practice of coriolis force and other forces in rotating frame

Like questions

@ACuriousMind Uhm, I see Peskin & Schroeder is too hard for me yet. What if $\langle \phi(\underline{x},t) \rangle$ is known and we calculate with basic quantum uncertainity principle? <- sorry if it is complete nonsense, I wrote this... intuitively ;-)

Please tell

What are you using now?

@peterh-ReinstateMonica I'm not sure what you mean by that. The problem is that the expectation value of a function is not the same as the function of the expectation value - already in basic QM you have that $\langle x\rangle^2\neq \langle x^2\rangle$ in general.

@skullpatrol I'm not sure why you're apologizing for something you didn't have anything to do with?

@ACuriousMind If $\Delta r$ is macro-sized, does it help?

@ACuriousMind In the Feymann formalism, the value of a (scalar) quantum field is a real number, as I read here: physics.stackexchange.com/a/48033/32426

@peterh-ReinstateMonica sure, but you compute expectation values there by the path integral over all possible field histories, not by plugging in some specific value of the field.

@ACuriousMind Ok. Imagine I have the $\Delta r$ sided cube. It is static in time, so $\langle \phi \rangle$ depends only on $\underline{x}$. Does it not make the count of the all possible field histories to 1?

no, that's not at all how the path integral formalism works

You can't just declare that the field is static. Even if the expectation value or the solution to the e.o.m. is static, the path integral integrates over all histories, not just solutions to the e.o.m.

This is that thing that I need to integrate to all possible pathes, weighted by the action?

@sheltonBenjamin Taylor Classical Mechanics?

the intuition behind the coriolis force is it's the "force" that makes everything spin around you, from your perspective

what does it mean by bounded distribution of mass

does it mean mass distribution is finite?

what is continuous distribution of mass?

does it mean for eg earth is approximately round and suppose a single particle isolated outside earth then every particle exert force on that particle and vice versa ?

I don't see the phrase "bounded distribution of mass"

Or "mass distribution is finite"

A continuous distribution of mass would be the limit of a discrete distribution of mass as the size of each of the discrete piece approaches zero. Similar to the difference between the integers and the real numbers.