The h Bar - 2017-07-20 (page 2 of 4)

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3:00 PM

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Research effort: Watched a video on Normal forces in elevator (khanAcademy), read pseudo force stuff from 2 books.

Attempt: N- mg = ma where a = g/10

Anonymous

What is N?

Where am I going wrong and can someone please explain why this pseudo force technique works?

@Blue Normal force = Tension in string of Spring Balance = N

@AccidentalFourierTransform But that only works for $L^1$...so what I said about Sobolev spaces is the best you're gonna get in the $L^2$ case. The $L^2$ Fourier transform is terrible...

It's not even defined by an integral.

AccidentalFourierTransform

what!? the $L^2$ FT is the best thing ever

Anonymous

Spring balance reading is twice of string tension....

3:03 PM

@AccidentalFourierTransform You must be a physicist...

@Blue ?

@Blue I found doing the open courses (lectures + P-sets + reading ) from MIT are more helping than walking through QM textbooks alone.

AccidentalFourierTransform

$f_n\to f$, like a true physicist >_>

@AccidentalFourierTransform what?

@JohnRennie Can you help me with the question and explain why the steps work?

AccidentalFourierTransform

3:04 PM

though we usually say "we introduce a small photon mass to make the integral converge"

oh no

Anonymous

@Shing I'm doing the same thing ! However the MIT OCW for QM suggests reading from liboff...so :P

@Blue They have two QM I . each stands on its own, I think.

@AccidentalFourierTransform I have a post on PSE from back in the old days where I computed $1+2+3+4+\cdots$ by introducing a frequency parameter on some metal plates and taking it to zero in the end.

Pretty awful stuff

AccidentalFourierTransform

gool ol Casimir

3:05 PM

@Blue speaking of textbook, I found Shanker is quite confusing from time to time (and a very little bit dated). but the P-sets are quite helpful.

Anyone?

AccidentalFourierTransform

pretty awful indeed

@Abcd suppose the elevator was stationary, then you have a simple Atwood machine. You have a gravitational acceleration $g$ and it's all straightforward to calculate. Yes?

@AccidentalFourierTransform In any case, the $L^2$ FT is defined by density on $L^1\cap L^2$, via mollifications, or density on $\mathscr S$. Something like that.

@JohnRennie Yes.

3:06 PM

@Blue Personally, I found reading Townsend's Quantum Physics quite helpful.

It's not a straightforward integral, although there is the formula $$\hat f(\xi)=\mathrm{l.i.m.}_{R\to\infty}\int_{|x|\le R}f(x)e^{i(x,\xi)}\,dx.$$

AccidentalFourierTransform

@0celoñe7 yeah, that is what I meant by $f_n\to f$ :-P

also, MIT's introductory QM by French are great too, but a bit dated. (I have read some parts of it though)

AccidentalFourierTransform

you take a family of $L^1$ functions $f$ such that $f_n\to f$ in $L^2$

and you define $\hat f\equiv \lim_{n\to\infty}\hat f_n$

its great, dude

@Abcd OK. In the rest frame of the elevator the apparent gravitational force is $g$ + the acceleration of the elevator. So if you're standing in the elevator the apparent gravitational force is $1.1g$. So just solve the Atwood machine with the gravitational acceleration set to $1.1g$ instead of $g$.

Anonymous

3:08 PM

@Shing Yeah. I'm using 4 books (which are suggested here) including Shankar and also doing the exercises in the books along with the problem sets. As I am starting early I want to do all the stuff thoroughly :)

Anonymous

@Shing Oh, I'll check out that book then. Thanks for the suggestion

@Blue No problem.

AccidentalFourierTransform

@0celoñe7 physics.stackexchange.com/questions/105635/… have fun

@JohnRennie Why 1.1g?

Anonymous

@Shing This one: amazon.com/Introduction-Quantum-Physics-M-I-T-Introductory/dp/… ?

Anonymous

3:10 PM

Lol, at first I thought that book would be written in French :P

@Blue yeah. It introduces some very nice experiments.

but the math methods are a bit elementary.

@Abcd Consider what we mean when we say the gravitational acceleration is $g$. It means that if I am stationary and drop an object then that object will accelerate downwards away from me at $g$. So far so good?

@JohnRennie yes

Anonymous

@Shing I see. By the way, do you know any MOOC for undergraduate level electromagnetic theory?

Anonymous

The MIT one is graduate level

3:12 PM

@Abcd OK suppose I drop the object, then I start accelerating upwards at $0.1g$. What is the total acceleration of the object away from me? Hint, acceleration is a vector and adds like vectors do.

@Blue I know one introduction to EM (i.o.w. no field theory)

@Blue here: youtube.com/playlist?list=PLyQSN7X0ro2314mKyUiOILaOC2hk6Pc3j

former MIT open course

@JohnRennie 9.9 m/s^2

Anonymous

@Shing Oh, but that one is like too basic. We already did that much in high school. I'm looking for an MOOC equivalent of Griffiths "Introduction to Electrodynamics"

@Blue mine starts on Saturday :P

@Abcd are you taking $g = 10$?

3:15 PM

@Blue no idea then, but the problem-set is another thing... they are much harder than the lectures

ik it is an unusual day of the week to open college :P

@JohnRennie Yes

@Blue if you are looking for field theory, then you may want to check out Susskind.

@Yashas why don't they simply start on Monday?

@Sid mby they want students to spend one day at the hostel?

Anonymous

3:16 PM

@Shing I know I know. I solved many of Walter Lewin's problem sets during my high school. They are nice.

@Abcd OK, $10 - 0.1 \times 10$ isn't equal to $9.9$, but let's gloss over that for the moment.

Anonymous

@Shing Oh. Checking.

@JohnRennie 9.1 m/s^2

@Abcd Suppose I drop the object, so it starts accelerating downwards at $g$, then I jump off a cliff so I also start accelerating downwards at $g$. I and the object would fall at the same rate. Yes?

Anonymous

@Yashas Wow. Good luck! Are you already in Hyderabad?

3:17 PM

@Blue not this Saturday :P

@JohnRennie Yes

Anonymous

@Yashas heh :P

Anonymous

Next?

@Blue Personally, I prefer Purcell much more than Griffiths. They are roughly at the same level. but Purcell doesn't cover field theory at all. :(

@Blue yea

it's my last Friday at home :'( for few months

3:19 PM

@Abcd OK, so if I accelerate downwards it reduces the acceleration of the object relative to me. That's because I'm in effect chasing the object downwards so it doesn't go away from me as fast. Does that make sense?

@JohnRennie Yes

Anonymous

@Yashas We used to live in Guwahati. Now I've shifted to my Kolkata house with my mom. Father is staying back in Guwahati. Next year he is planning to take up a new job in Kolkata only. Don't worry. You'll enjoy in a new place. Personally I wanted to stay away from home for a while. :D

I miss my pets more than anything else :'(

I was already JEEPrepSick :(

@Abcd But the elevator isn't accelerating downwards. It's accelerating upwards. And if I accelerate upwards I'm going in the opposite direction to the object so it accelerates away from me even faster. Still OK?

I still feel like solving JEE questions -,-

Anonymous

3:21 PM

@Yashas Call them via Skype everyday :D!

Anonymous

lol

Anonymous

@Yashas Oh, not agaaaaaaaaaaaaiiiiiiiiiiiiiiiiiiiiinnnnnnnnnnnnnnnnnnnnnnnn!!!!

@Yashas Rank?

@Abcd classified

@Yashas what does that imply?

3:22 PM

@Blue I actually studied biomolecules after JEE :D

@Abcd secret

studied some basic bio too

@JohnRennie No. Not clear.

Anonymous

I sold off all my JEE books today morning. :P Got only Rs. 2500 for books worth over Rs. 50,000

WTTTH

you had books worth 50k?!

@Blue I sold off some of my books. Got 2900. :P

?!!?!?!?!!

3:23 PM

@Blue Are you serious - over Rs. 50000

I gave away some others to a relative

I did not sell anything

Anonymous

@Yashas Yeah, including fiitjee materials and all

Anonymous

@Abcd Yep

becaz for some mysterious reason I have some personal attachment to books

I still have my 3rd grade library book

and many others

3:24 PM

@JohnRennie Talking in terms of lift will help me understand better probably

Anonymous

@Yashas I didn't keep them. Otherwise I would suffer from nostalgia XD

:38899818 omg I wasn't talking about such books

(removed)

my library book was some rough book like thing

I used to take notes when I used to visit the library

user image

Anonymous

3:26 PM

However, the good thing about university books is that I can borrow all of them from the library and don't have to buy too many new books

@Blue But you can't highlight/ mark in them. That's a con.

@Abcd: suppose I'm stationary and I drop the stone, so it accelerates downwards at $g$. At the same time you start accelerating upwards away from me at $0.1g$. The question is, how fast is the stone accelerating away from you?

Anonymous

@Abcd Well, yeah. But I usually make my own short notes from the books. So I don't need to highlight.

@Blue Alright, I am obsessed with highlighting :p

@Blue Here we have a much cheaper price than the US, for the exact same book. (some are international edition though... those sucks)

I bought A.Zee's Einstein Gravity for like $28 USD

Anonymous

3:28 PM

@Yashas I spoke with Ravindra Kumar from TIFR today. Nice guy. He told me how to go about learning the extra topics in Physics which are not included in the ECE syllabus. Also that you can apply for the VSRP program from your third year onwards.

@JohnRennie so much patience? you drew a detailed human whose parts are in proportion and has shoulders :P

@Blue Where did u meet him?

TIFR reminds me of olympiads

Anonymous

@Yashas I just called him by phone

TIFR conducts the 2nd stage of Physics, Chemistry, Astronomy and Biology olympiad

@JohnRennie Very simple. g+0.1g , intuitively

@Yashas I'm not sure I got the proportion right. The body looks too long and the head looks too small. Oh well :-)

Anonymous

3:29 PM

TIFR is awesome *_*

@Abcd Yes, exactly.

Anonymous

@Yashas yeah

@JohnRennie Then?

Anonymous

@JohnRennie You drew that on MS Paint? :O

@Abcd So if you're standing in an elevator accelerating upwards at $0.1g$ and you drop a stone it will accelerate downwards away from you at $1.1g$. Agreed?

@Blue Google Draw

2

Anonymous

3:31 PM

@JohnRennie aw :) You are an artist :D

2

@Blue Google Draw is free and really good for simple diagrams. Most of the diagrams in my answers on this site were done in Google Draw.

2

@JohnRennie yes

@Abcd Good. We're nearly there :-) Now suppose you're standing on a planet slightly heavier than Earth so the surface gravity is $1.1g$. When you drop an object how fast will it accelerate away from you?

Anonymous

@Shing Could you just check this MOOC once? Does this along with the em field lectures look sufficient for UG level?

We probably ought to remove those last three stars. They kind of clutter up the star board and I'm not sure Google Draw is so great it needs to be on the star board.

3:34 PM

@JohnRennie 1.1g?

Anonymous

@Shing Where do you live?

@Abcd Yes. The point is that being in an elevator on Earth accelerating upwards at $0.1g$ is just like being on a planet where the gravity is $1.1g$. In both cases when you drop something it accelerates downwards away from you at $1.1g$.

hat trick stars

So to solve your problem just treat it as if it was stationary in a gravitational field of 1.1g.

@JohnRennie So N = m*1.1g?

3:38 PM

@Abcd well with an Atwood machine the system isn't in equilibrium. The lighter mass is being pulled up by the string and the heavier mass is pulling other end of the string down. So calculating the string tension isn't straightforward.

:38900223 No.

@JohnRennie Let's consider the blocks as a system

@Abcd: have you learned how to solve Atwood machines?

@JohnRennie Nope.

@Abcd You need to study Atwood machines first. Learn how to do a simple Atwood machine without accelerating elevators being involved.

See the Wikipedia article I linked or Googling will find you lots of articles and videos on Atwood machines.

@JohnRennie But I have done such questions with pulleys and there I used to consider two blocks as a system because thread doesn't break and acceleration is same

3:43 PM

Ah, OK, so you're familiar with the basic idea but you probably didn't realise it was the type of system that engineers call an Atwood machine.

@JohnRennie yes

@Abcd So you know the tension in the string is given by: $$T = \frac{2 g m_1 m_2}{m_1 + m_2} $$

Anonymous

@Yashas So, what are you doing nowadays ? Completed Shankar's lectures?

California institute of technology, physics videos: the mechanical universe

@Blue^

@JohnRennie no

3:46 PM

@Blue I have been doing some CS work recently

@Abcd Ah, I thought you said you'd done this type of problem ...

Anonymous

@Yashas Like?

Anonymous

I did the MATLAB mooc from coursera

@Blue misc stuff

@JohnRennie yes but didn't learn any formulas

Anonymous

3:47 PM

Coursera seems awesome. I plan to cover all the CS courses there

@blue and I deactivated fb :D

Anonymous

I recently discovered that I can audit all the courses for free :P

@Blue have you watched those videos?

Anonymous

@Yashas Oh. Why?

@Blue I keep fighting with creationists, flat earthers, anti-science, anti-vaxxers, etc, etc...

Anonymous

3:48 PM

@skullpatrol No. I am looking at them. Are they rigorous?

waste of time

@Abcd OK. The point is that if you work through a stationary (not in an accelerating elevator) Atwood machine the tension in the string is given by the equation I quoted.

Anonymous

@Yashas lol. XD Use FB to connect to interesting people instead

@JohnRennie mmkay

And we've already agreed that the accelerating elevator just increases the effective value of $g$ from 10 m/s² to 11 m/s² i.e. from $g$ to $1.1g$.

So just use the same equation that you get for a stationary Atwood machine but replace $g$ by 11 m/s².

AccidentalFourierTransform

3:50 PM

@Yashas have you ever changed your mind because of something you read on FB? because if you didn't, you should really consider the possibility that they won't change their minds either, i.e., arguing with them is wasting your and their time.

4

There is a rigours textbook for engineers that goes with them @Blue

Then the force on the pulley is just twice the tension in the string.

Anonymous

@Yashas I got to know Somak Raychaudhury (IUCAA) and many more such people on FB. FB is very helpful for building contacts. Also, you can follow the Physics/CS pages you like. You'll get to know lots of great people.

@AccidentalFourierTransform I realized that

AccidentalFourierTransform

well there you go :-P

3:52 PM

Mar 29 at 4:59, by Yashas Samaga

ignorants have to be ignored

AccidentalFourierTransform

OTOH, I have changed my mind several times when discussing random topics here, so there's that

Anonymous

Basically in FB you attract the crowd that you indulge in. If you quarrel with trolls you'll invite more trolls. XD

nah

most of the pages I have liked are science, math, animals, etc. stuff

@Abcd: I get $T=22$ N as well.

every science and math page has some ignorants who talk nonsense

3:54 PM

So the force on the pulley is $F = 2T = 44$ N

idk why they even like science pages if they are anti-science

Anonymous

@Yashas It's your mistake if you go ahead to argue with them. :P

ironically, they use a computer which is a product of science

@Yashas that's getting a bit crude for the chat room. We (intermittently) have ladies present.

fixed :-)

3:55 PM

:-)

Anonymous

ladies?

Yes

We need more!

RIP Maryam

:'(

she died too early :(

I think she was 40

ik ik

3:56 PM

...who?

AccidentalFourierTransform

so, $\hat a$

is it a hat or a caret?

@Sid Maryam Mirzakhani

Lemme just google that

@Sid She was the first woman to win the fields medal.

Oh, that lady.

I had forgotten the name. yeah, I have heard about her genius

3:57 PM

Everyone dies to early. Did you know anyone who felt they had lived too long? But it's certainly true that some people die much too early.

It's a "pretentious little hat" according to Weigand ;P

AccidentalFourierTransform

lol

I can use that

Thanks @JohnRennie

I think there's a footnote in his QFT script where he says "We'll omit the pretentious little hats from now on"

@Abcd is 44N right?

3:59 PM

@AccidentalFourierTransform Go after Forrest Fenn's treasure

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