The h Bar - 2020-06-12

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2:08 AM

@SujalMotagi I go for C

If you have a car which is uniformly accelerated so initial it's velocity is zero and then reaches to certain value X. Here speed changes continously but the direction of car is same.

4 hours later…

user434058

5:50 AM

@SujalMotagi The answer should be (a). The most common example of such a scenario is uniform circular motion.

user434058

@Yuvraj But how does that justify the velocity being the same. Velocity depends on both, the direction and the magnitude which is equal to the speed. I believe that (c) is not the correct option.

6:10 AM

@FakeMod then how would you define the speed would be constant?

user434058

@Yuvraj the derivative of velocity is the acceleration, not the derivative of speed. So the speed can stay constant even if there's a non zero acceleration. And as I said, uniform circular motion is a great example for this scenario.

user434058

Yo @rob what's up! :-)

@FakeMod Uniform acceleration" that doesn't change direction means linear motion, not circular. In circular motion, only the magnitude of the acceleration does not change. The direction of acceleration always points towards the circle center, which is a different direction at each point on the circle

user434058

@Yuvraj It's written in the question "uniform magnitude acceleration". They're only talking about the magnitude.

Ah! My mistake I haven't read the question correctly.

Then answer is A.

6:18 AM

@Yuvraj it's one of those deliberately tricky questions :-)

@JohnRennie I love this

hello humans

I'm not sure I approve of questions tryin to catch students out by being deliberately trick. I guess it's fir since it checks if the student read the question carefully, but it still seems unnecessary to me.

user434058

@JohnRennie It's just trying to mess with intricacies, which is good, but there should be some "explicit-ness" about the conditions while writing the question.

@JohnRennie I remember one question, of two disk rotating by satan

6:25 AM

@Yuvraj yes?

user434058

@Yuvraj that question is an abomination! :P

He has asked the wether angular velocity about the axis is same as angular velocity about the point on axis.

@JohnRennie

I feel it is same

What you feel?

7:13 AM

@FakeMod Searching for Academia memes shows anime memes because there's an anime called My hero academia.

2 hours later…

9:26 AM

Is $sin^2a = sina * sina$?

yes

coolio

why not $(sina)^2$ instead of $sin^2a$?

note that $\sin^{-1}(x)\ne\frac{1}{\sin(x)}$ though

it is an example of slightly poorly chosen notation

@Charlie hmm weird

$\sin^{-1}(x)$ is the inverse $\sin$ of $x$, $\arcsin(x)$

9:31 AM

$sin3a = 3sina - 4sin^3a$. Can it be $sin3a = sina(3 - 4sin^2a)$?

looks fine to me

the power on the sine behaves in the usual way, you just have to be careful that the -1 means something different, other than that all power orders of sine behave like you would expect

=> $sin3a = sina(3 - 4sina(sina))$ lol I don't know where i'm going with this

other than the missing bracket that also looks valid

2 hours later…

11:11 AM

hi

hello

I'm about to install Linux on my main work laptop. Wish me luck! :-)

godspeed

The installer claims it will leave existing partitions untouched. We'll see :-)

11:39 AM

Is there a reasonable free alternative to mathematica? I'd like to learn to use a program like that but I can't get it through my uni

ah I just found a nice list online nvm

@Charlie depends what you want to do

Matlab is usually the alternative

I'm not really sure what I want to do, just hear it talked about a lot and figured I might as well at least try it

I'll check out matlab ty

oh sweet I actually can get matlab free with my uni

12:08 PM

user image

3

12:22 PM

category theory? lol I feel like there's a theory for everything

what is there to theorize about categories. Just go to amazon, click categories and there stuff like electronics, ebooks etc... ....

Question no.1 I am self taught

@Slereah Best memes post ever.

@JingleBells How are you Santa. Any present for me or any good quotes?

@Slereah How to stop it?

@Questionno.1Iamselftaught Sorry, you've been a naughty boy my elf tells me

@Questionno.1Iamselftaught don't go to university

Question no.1 I am self taught

@JingleBells best suggestions ever! I doubt that you are not real Santa=You must be real santa!

Hohoho

Question no.1 I am self taught

@Slereah Matlab sucks it needs money.

12:28 PM

I need money too

Category Theory seems to me to be another successful attempt to unnecessarily overcomplicate simple stuff

Question no.1 I am self taught

@Slereah Go program the whole calculator by yourself.

@JingleBells Easy to say when you never have to deal with the topics involved

Question no.1 I am self taught

My book teaches my Numerical analysis and whole book is devoted to program things from scratch lol.

slereah what area of physics do you work in that requires a lot of category theory?

Question no.1 I am self taught

12:30 PM

Seleteah I though you profile picture is a cat lying on bed but when I zoomed it up it is completely different

or is it something that's used all over the place

@Questionno.1Iamselftaught I thought it was a white crab

Question no.1 I am self taught

@JingleBells you as a atheist Santa believe physics so you should be thinking it is schrodinger cat

I think hbar is fun place to chat. A place you can see a santa who knows electrical engineering,biology, chemistry...etc but math is lame

But why does hbar has all users with profile picture of cartoon lol

I am happy to answer questions here rather than Mathematics chat.

@JingleBells Don't give up on a dream.

Keep sleeping.

12:50 PM

^ finally someone who speaks my language here

I'm fluent in nonsensian

@Charlie Category theory is used in the axiomatization of QFT

also string theory and such

oh right

Category theory is useful if you have to deal a lot with equivalence of different mathematical objects

1:10 PM

I'll get there one day

Question no.1 I am self taught

Category theory is theory that tells category and string theory is a theory with lots of strings super string theory is super string theory of sub string theory.

good night guys. I see stars when I rub my eyes.

night :)

2 hours later…

Captain Bohemian

3:23 PM

is the dimension of any Riemann group $O(p,q)$ of fixed $p+q$ the same?

I think the dimension of $O(0,4)$ is 6, and the dimension of $O(1,3)$ is 6, so the dimension of both $O(0,4)$ and $O(1,3)$ is 6.

but I am not sure about other examples.

@CaptainBohemian Yes, the dimension of the orthogonal group is independent of the signature

user434058

3:38 PM

Will $$\frac{\partial L}{\partial \dot q_i}=p_i$$ hold even if the potential $V$ depends on $\dot q_i$?

That's essentially the definition of $p_i$

But the $p_i$ there is the canonical momentum, which is not necessarily the kinematic momentum $m\dot{x}_i$.

user434058

@ACuriousMind ah, yes.

user434058

@ACuriousMind So changing the dependency of $V$ on $\dot q_i$ will also change $p_i$, right?

sure

user434058

@ACuriousMind so, why is $\{p_i,q_i\}=0$ true? To be precise, what are $x$ and $y$ in the following equality $$\frac{\partial p_i}{\partial x}\frac{\partial q_i}{\partial y}-\frac{\partial p_i}{\partial y}\frac{\partial q_i}{\partial x}=0$$

3:50 PM

no idea, there are no "x" and "y" in Hamiltonian mechanics

user434058

Canonical coordinates

In mathematics and classical mechanics, canonical coordinates are sets of coordinates on phase space which can be used to describe a physical system at any given point in time. Canonical coordinates are used in the Hamiltonian formulation of classical mechanics. A closely related concept also appears in quantum mechanics; see the Stone–von Neumann theorem and canonical commutation relations for details. As Hamiltonian mechanics is generalized by symplectic geometry and canonical transformations are generalized by contact transformations, so the 19th century definition of canonical coordinates in...

the Poisson bracket is $\{f,g\} = \sum_i \frac{\partial f}{\partial q^i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q^i}$

user434058

Alright, so how is $\{p_i,q_i\}=0$ true?

@FakeMod it's not

It's 1, just like the page you just linked says

Is space vomit, poop and pee thrown into space towards Earth so it can burn and no problem?

user434058

3:52 PM

@ACuriousMind Ah, I messed up Kronecker delta. Inverted its definition :-)

user434058

@ACuriousMind What if I wanted to prove that $\{q_i,p_i,\}=1$? How should I go about doing that?

just plug it in?

user434058

@ACuriousMind yeah but what should be there in the denominator?

user434058

What should I take as the dependent variables?

I don't understand the question - I just wrote down the definition of the bracket, just set $f = q^i, g = p_i$

user434058

3:57 PM

@ACuriousMind So whenever I say something of the sort $\{f,g\}$ in classical mechanics, does it mean that I have always have to partially differetiate stuff with $p_i$ and $q^i$ (momentum and coordinate, respectively)? Why are these both special, or is it just the definition of Poisson brackets?

Hamiltonian mechanics is physics on the phase space and the phase space is spanned by the $q^i$ and $p_i$ by definition. They're "special" because a tuple $(q^i, p_i)$ fully specifies the state of the system, just like a tuple $(q^i, \dot{q}^i)$ would

user434058

@ACuriousMind So $q_i$ and $\dot q_i$ are independent, right?

user434058

@ACuriousMind thanks, makes sense.

@FakeMod Yes. I talk about the dependences between $q^i, \dot{q}^i, p_i$ at some length here

user434058

@ACuriousMind Thanks, but your answer is out of my "mental" reach as of now. Will surely take a look at it once I am acquainted with the pre-requisites.

5:30 PM

user image

There is a meme war between commutative algebraists and category theorists apparently

@FakeMod The $x$ and $y$ variables you're used to would be denoted $q_1$ and $q_2$ in generalized coordinate notation, examples of the functions $f = f(q,p)$ and $g = g(q,p)$ could be say the Hamiltonian $f(q,p) = H(q,p) = p_1^2/2m_1 + V(q_1,q_2)$ and the '$y$' coordinate $g(q,p) = q_2$, or the i'th component of angular momentum $g(q,p) = (\mathbf{q} \times \mathbf{p})_i$ or something.

You should derive the Poisson bracket by simply time differentiating a random function $f = f(q,p)$ and then using Hamilton's equations, and it's obvious what the P.B. is saying

2 hours later…

Captain Bohemian

7:07 PM

@ACuriousMind So the n-1-dimensional deDitter space really has dimension n-1 by the construction O(1,n)/O(1, n-1) for n>1!

because $\frac{n^2-n}{2}-\frac{(n-1)^2-(n-1)}{2}=n-1$.

user434058

8:10 PM

@bolbteppa Thanks for the clarification, but the catch is that I encountered all this stuff when trying to build the prerequisites for Hamiltonian mechanics (I haven't learnt it yet), so any explanation which needs a prior knowledge of Hamiltonian mechanics is hard for me to understand :-)

8:42 PM

Okay, I'm out of practice. I want some Poisson-distributed random numbers in my quick-n-dirty Numpy/Python script. I don't remember how I did this before. There's cookie-cutter for uniform, gaussian, exponential, and some others, but I don't think Poisson. Hints?

Does the atlas of a manifold contain the transition functions? If not, is the information about the transition maps stored anywhere or do we just construct them from the chart maps as needed?

It's not really "stored"

It depends how you define the manifold

so we don't define a structure that contains all possible transition functions in the same way we define an atlas the contains all the chart maps?

If you define the manifold via some external mean, then you can reconstruct the transition function from the atlas

But you can also define the manifold intrinsically, from the transition functions and the atlas

in both cases the atlas contains all the information necessary to define transition maps right?

8:54 PM

Well, in the second case, you have the transition maps

hmm ok

If you have the original space, like say, as a set

Then it's pretty obvious what the transition functions are

It's just the composition of maps

ah yeah that's kind of what I was thinking

tyvm

Yeah, a chart is just a homeomorphism from an open subset of a topological space to Euclidean space, an atlas is just a collection of charts which cover that topological space, in the case each chart maps into the same Euclidean space it's a manifold, and this immediately implies the existence of transition functions which pass from Euclidean space through the intersections of the charts then back to Euclidean space due to the fact we're using homeomorphisms (into the same Euclidean space)

9:16 PM

Ok I'm starting to get the hang of it thanks guys

Then if your transition function is differentiable, noting it's just a map from Eucliden space into Euclidean space, we can say the manifold is differentiable because the homeomorphism carries the Euclidean space structure onto the manifold continuously/bijectively, so you're now talking about differentiation on weird curved spaces using only Euclidean space

ah defining differentiability on the manifold is just where i'm at in the video lectures

Frederic Schuller has a really nice balance between maths and physics

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