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02:00 - 23:0023:00 - 00:00

2:50 AM
@ACuriousMind I want to reason spacetime and black holes with you. Can you spare a few minutes tonight?
@JohnRennie the movie is online. . . you can watch it at 3 bucks a pop. :)
@JohnRennie kudos to Ed for the courage, and dedication, and for learning a new skill.
 
3:45 AM
I want to discuss GR, black holes and some associated math and solutions to things
If anyone is free, willing and capable of sustaining such a conversation with at least minimal competence please ping me
 
4:18 AM
@JakeRose Yes, something applying a frictional force takes momentum away from whatever it acts on; and gains that momentum itself. That's an oversimplification, of course, because momentum is a vector, not a scalar. But essentially, that's Newton's third law. If I'm driving my car, and I apply the brakes, then the friction between the road and the wheels both pulls the wheels back and pushes the road forwards. But because the earth is much heavier than my car, you don't really notice the
effect on the earth when I brake.
@JohnRennie Where do you think little mathematicians come from?
 
> He also has said that he wanted to dispel the common conception of mathematicians as depicted in films like “A Beautiful Mind.”
A lot of mathematians say that...
2
Q: The Fifth Element or What Elements Are Correlated With CS Success and Does It Matter?

user4333After a long search a colleague says that the only 'predictor' of computing ability seems to be Autism, and that only with a low correlation (60%, whatever that means). But in our experience of teaching students with various disabilities, including mild forms of Asperger's, we find that even this...

...so, controversial.
 
Anonymous
4:46 AM
@DawoodibnKareem Mitosis :)
 
vzn
> Frenkel told Italian website Oggi Scienza that the film’s challenge was finding a way to “unite the beautiful body and mathematics,” and the “beauty of a woman’s body” served as a metaphor for the task.
ok time to come clean, whos seen it... sounds even more "artistic" than tarkovsky eh? o_O :P
 
@vzn I have the link if you are interested
 
vzn
hmmm reminds me of Ikkyu my favorite japanese zen master =D
 
I was going to check it out but I can't front out 3 bucks yet. I will wait for the reviews first
 
vzn
Ikkyū (一休宗純, Ikkyū Sōjun, 1394–1481) was an eccentric, iconoclastic Japanese Zen Buddhist monk and poet. He had a great impact on the infusion of Japanese art and literature with Zen attitudes and ideals. == Biography == === Childhood === Ikkyū was born in 1394 in a small suburb of Kyoto. It is generally held that he was the son of Emperor Go-Komatsu and a low-ranking court noblewoman. His mother was forced to flee to Saga (嵯峨), where Ikkyū was raised by servants. At the age of five, Ikkyū was separated from his mother and placed in a Rinzai Zen temple in Kyoto called Ankoku-ji, as an a...
 
5:49 AM
I should really buy that camera so I can just make videos of math
I really should . . . . . .
 
6:04 AM
Guys I'm not joking! I want to have a serious GR conversation. I have the skills . . .
I will hang around for another hour if anyone is down to talk
 
6:30 AM
@JohnRennie Good Morning :)
 
6:52 AM
@Cows what do you want to ask about GR?
 
@JohnRennie Are you around ?
 
@Tanuj I generally don't like diverting off into separate chat rooms. It's just too many rooms to keep an eye on. I prefer to stick to this or the problem solving room.
 
okay
noted
@JohnRennie Is there a standard method to find the partial fraction of the following type of fractions ? (Without actually solving it completely )
 
Anonymous
Integrate both sides...
 
@Blue I'm not sure how that would help ? I'll still have to solve for the partial fractions right ?
 
7:05 AM
@Tanuj doing integrals is mostly a matter of practice and I'm badly out of practice since I haven't hand integrated anything for decades. I'm not really the person to ask.
Having said that, just integrate the right hand side and see what you get.
On the left side the denominator obviously factors to (x-1)(x+2). It would be worth seeing if the numerator also factors.
 
@JohnRennie $$Ax-\frac{B}{x-1}-\frac{C}{x+2}$$
 
I would guess the factoring is related to the way you're expected to solve the problem
So the right hand side is $$ \frac{Ax(x-1)(x-2) - B(x+2) - C(x-1)}{(x-1)(x+2)} $$
 
you mean $$ \frac{Ax(x-1)(x-2) - B(x+2) - C(x-1)}{(x-1)(x+2)} $$
 
$$ \frac{Ax(x-1)(x+2) - B(x+2) - C(x-1)}{(x-1)(x+2)} $$
 
yea
 
7:13 AM
so... what are we having here...
 
So multiply out the top of the fraction ...
 
@JohnRennie and then compare with the left side ? Well i know this , but isn't there another way ? Without actually multiplying and comparing so much stuff ?
 
I can't think of a quicker way to do it
 
hmm..neither can I . If i go by that method , its surely gonna consume some time
 
are A,B,C functions of x or just constants?
 
7:16 AM
Constants
 
Actually, I think integrate is the way to go (quotient rule of differentiation is just too messy.) Note that since A,B,C are constants, we can integrate everything on the RHS term by term. Meanwhile, the denominator of the LHS obviously look like it can be broken up into (x-1)(x-2). Thus after integrating you should get:
 
@Secret I get this method , thanks !
 
$$\frac{2x^3+3x^2+x-3}{x^2+x-2} + K = Ax - \frac{B}{(x-1)} - \frac{C}{(x+2)}$$
now simplify the LHS to get:
 
Anonymous
@Tanuj Plug in $x=1,-2,0$ on both sides, to get the values of $A,B,C$
 
@Blue I know this ! lol . I was wondering if there was a better method and d/dx had do to something with the question ! But okay.
Thanks so much guys :) @JohnRennie @Secret @Blue
 
7:24 AM
$\frac{2x^3+3x^2+x-3}{(x-1)(x+2)} + K = Ax - \frac{B}{(x-1)} - \frac{C}{(x+2)}$
multiply both sides by (x-1)(x+2) to get:
$(2x^3+3x^2+x-3) + K(x-1)(x+2) = Ax(x-1)(x+2) - B(x+2) - C(x-1)$
Now plug in x=1,-2 for each case and B,C should pop out, after that you should be able to get A
 
Yea , @Secret thanks
 
(Meanwhile, I need to think about whether I should include that arbitrary integration constant K in here..)
 
Anonymous
It's okay to include $K$ here. There are $4$ different powers of $x$, whose coefficients you can compare anyway, to get values of the $4$ constants.
 
right, and we happen to have a nice result where A is the only constant associate with $x^3$ thus A has to be 2
sometimes when I do these questions, I sometimes wonder what is the geometric interpretation when we plug roots into the expression and then seeing a lot of terms just vanishes...
It will be cool if the above problem can be viewed geometrically, but I guess I need better background on polynomials in general to understand how
Let me just make a note of this in my notebook...
 
Anonymous
That would be an interesting question, yes
 
Anonymous
7:30 AM
Normally teachers justify this step by multiplying by $(x-1)$ or $(x+2)$ on both sides and considering the limit
 
Anonymous
The point is that polynomials are continuous but whenever you have something like $(x-a)$ in the denominator, it blows up at that point (similar to poles in complex analysis)
 
7:43 AM
indeed, and I am suspecting multiplying the (x-a) somehow removes the pole?
 
Anonymous
Not really. We have to sort of "redefine" the function
 
Anonymous
@Secret This is relevant
 
right, so you can redefine a function which matches the original one, but also continuous at where those singularities of the original function is
 
Anonymous
Yup
 
Anonymous
So all these steps are justifed. After all we just need to find the constants such that RHS=LHS holds for all $x$. We don't really bother about the singularities as long as they are removable
 
Anonymous
7:54 AM
But yeah, there are some interesting things going on in the background which most high school teachers wouldn't explain
 
So I am doing some quick analysis on this, and I noticed that the roots themselves will not move when the A,B,C,K changes. So the roots have the special property that only some components of the superposition contribute to the overall outcome
e.g. At x=1, the behaviour of the sum depends only on the fixed polynomial 2x^3+3x^2+x-3 and the B variable, thus explaining why we can extract the B since B is the only contribution to the sum at that point
In particular, let P(x) = Ax(x-1)(x+2) - B(x+2) - C(x-1). Then we knew that P(x) = (2x^3+3x^2+x-3) + K(x-1)(x+2). Thus P(1) = 2^1^3+3^1^2+1-3 = B(x+2). This means, at x=1, only the linear components of B contribute to 2x^3+3x^2+x-3, thus allowing it to be quickly isolated
ah sorry, I forgot the K, I mean 2x^3+3x^2+x-3 - 2K+Kx = B(x+2)
 
8:20 AM
@JohnRennie sorry for the late response I was gone for a bit. I had a whole cascade of questions. I am caught up in a phone conversation but I will be asking the questions here quite soon (perhaps in the morning since it is about 12:19 here ). I will ping you once I compose the questions.
 
@Cows I'm only around for another half hour, then I'll be out till this afternoon (UTC)
 
So, relate this back to the removable singularity stuff, at least for rational functions, it seems that given a superposition of N rational functions, the sum will have N degrees of
freedom. However, as we approach the removable singularities, because the sum blew up (which is contributed by some of the rational functions actually blowing up there), some degrees of freedom were being removed, e.g. for the above question on partial fractions, at x=1, there are only two degrees of freedom left since the rest blew up there
So in other words, what we are exploiting in solving partial fractions is that the degrees of freedom can be less than N at the removable singularities, which allow us to get the coefficients easily
I wonder if a more generalised result holds for all superposition of complex functions with poles...
The finite dimension analogue will be when solving matrix equations, some of the rows have zeros in them, thus result in that row to have less degrees of freedom in giving the result in the RHS.
 
 
2 hours later…
10:53 AM
Oh wait
I'm dumb
3
Obviously the adjunction of two manifolds is gonna be smooth
Because the weird junction only happens on a region for which there's no overlap with any other chart
 
 
2 hours later…
1:20 PM
@Slereah rookie mistake
Don't say nothing on hbar that you're not ok getting on the star board
 
I'm not ashamed of my dumbness
 
1:31 PM
Is it possible to define the tangent bundle of a $C^0$ manifold by using weak derivatives
Piecewise smooth manifold, even
 
@Blue do you have a minute?
 
Anonymous
Just ask...
 
@Blue okay so I did this integration question and I'm not sure why but it's evaluating to something undefined
 
Why would you integrate it
You introduce an extra constant of integration doing that
Factor the denominator then differentiate then just equate both sides...
 
1:46 PM
Hello
 
Or factor the denominator, partial fractions, then differentiate
 
Anonymous
The quotient quotient rule isn't really great to use
 
Anonymous
Plus integration is faster
 
Kya koi hai?
 
You introduce an extra constant doing that
 
Anonymous
1:47 PM
You can find the value of the extra constant
 
Factor the denominator into $(x - 1)(x + 2) = x^2 - x + 2x - 2 = x^2 + x - 2$ so the quotient rule is easier
No you can't find the exact value of the extra constant
If you do partial fractions first then differentiate easier it's fastest probably
 
Anonymous
@bolbteppa There are 4 powers of $x$, which you can compare. Four equation and four constants, so it can be solved
 
Long division then partial fractions
 
That's the long division done
 
Anonymous
1:53 PM
@bolbteppa Gosh, really. You want to do that long division? After integration you can easily plug in and $x=1,x=-2$ to get $B,C$. I can already see that $A=2$.
 
Anonymous
And integrating can be done mentally for that as it is of the form $C/x^n$
 
@Blue kindly have a look ibb.co/kbK2vx
 
Anonymous
@Tanuj Please send me a rotated image. It's difficult to keep my head rotated for long
 
$\dfrac{2x^3+3x^2+x-3}{x^2+x-2} = 2x+1+\frac{4x+1}{(x-1)(x+2)} = 2x+1 + \frac{3}{x+2}+\frac{1}{x-1}$
 
Lol okay I'll try
 
1:57 PM
Partial fractions on the last bit done here
It's madness to integrate it and then play with an extra constant of integration :p
 
Anonymous
It takes no more than 2 minutes.
 
$\frac{d}{dx}[\dfrac{2x^3+3x^2+x-3}{x^2+x-2}] = \frac{d}{dx}[2x+1+\frac{4x+1}{(x-1)(x+2)}] = \frac{d}{dx}[2x+1 + \frac{3}{x+2}+\frac{1}{x-1}] = 2 - \frac{3}{(x+2)^2}-\frac{1}{(x-1)^2}$
 
@Blue rotated straight now ibb.co/kbK2vx
 
Anonymous
Still tilted
 
@Blue what's wrong with me? It should work now ibb.co/b422TH
 
Anonymous
2:08 PM
$1+e^{-t}=m\implies -e^{-t}dt=dm$
 
Anonymous
Minus sign missing
 
Oh yea wait
 
@blue it's still coming $$\frac{(ln(e^{-x^2}))^{\frac{3}{2}}}{1+e^{-x^2}}$$ from x=0 to x= infinity
 
'Am I taking care of you? I have a thesis to write!' 'My parents are at their house; you visited last--' 'No, no, explain like you're five.'
2
:D
 
Anonymous
2:25 PM
In your solution I can't spot any step where you are doing any integration @Tanuj
 
Anonymous
I just see some substitutions going on
 
Anonymous
@Slereah lol
 
@Blue Wow I just realized that, what should I do? There is no standard integral I can spot or anything simple I can integrate
 
Anonymous
Use standard properties of definite integrals
 
I dont think any property can be used here
It's 0 to infinity
 
 
1 hour later…
3:43 PM
Hi all could somebody elucidate something in the wikipedia article for the RG group?
In theoretical physics, the renormalization group (RG) refers to a mathematical apparatus that allows systematic investigation of the changes of a physical system as viewed at different distance scales. In particle physics, it reflects the changes in the underlying force laws (codified in a quantum field theory) as the energy scale at which physical processes occur varies, energy/momentum and resolution distance scales being effectively conjugate under the uncertainty principle (cf. Compton wavelength). A change in scale is called a scale transformation. The renormalization group is intimately...
So the problem I have is basically the first two sentences. I'm aware of the particle physics/QFT explanation which feels very natural to me, but the first sentence confuses me:
"The RG refers to a mathematical apparatus that allows systematic investigation of the changes of a physical system as viewed at different distance scales."
I really cannot for the life of me see how one and the other are interrelated, it just seems something completely different to me.
 
@user55789 What are "one" and "the other" in your sentence?
 
The QFT variant of the RG group and the generic theoretical physics variant, i.e. changes in the underlying force laws as the energy scale varies (QFT) and changes in a system as the distance varies.
 
...doesn't the second sentence explain that? "[...] energy/momentum and resolution distance scales being effectively conjugate"
The more energetic the things you probe the system with, the smaller the distances you can resolve.
 
4:06 PM
I think I found another difficulty with polyhedral wormholes
I suspect that the tubular neighbourhoods are not trivial to do for polygons
 
Tubular neighbourhoods?
 
In mathematics, a tubular neighborhood of a submanifold of a smooth manifold is an open set around it resembling the normal bundle. The idea behind a tubular neighborhood can be explained in a simple example. Consider a smooth curve in the plane without self-intersections. On each point on the curve draw a line perpendicular to the curve. Unless the curve is straight, these lines will intersect among themselves in a rather complicated fashion. However, if one looks only in a narrow band around the curve, the portions of the lines in that band will not intersect, and will cover the entire band without...
If it's smooth you can just pick normal vectors to the surface
But for a square
who knows
 
I don't understand the idea of a tubular neighborhood
Seems very trivial
 
If it's trivial what do you not understand
 
Why even define it if it's just a bunch of normals to a manifold
 
4:15 PM
well for a start it's not
It doesn't have to be normal
It's just that if you have a metric, you can define it uniquely that way
 
Hmm
So it's defining the things that, when you have a metric allowing for the concept of normalicy, becomes the set of normals to e.g. a curve
 
a tubular neighbourhood is basically "thickening" a surface
so that it looks like $S \times (-1,1)$
for instance the circle and the annulus
 
In classical diff geom there are concepts of envelopes, evolutes and involutes, seems like it's generalizing these old ideas but I'm not sure
 
(plus you gotta show that it is an injection locally, since globally the normal curves can intersect)
 
4:22 PM
I think basically you can show that it's all good by using the convex normal neighbourhoods along the surface
It's pretty hard to find much info on non-smooth things in general
Math people love smoothness
Especially if it's not even $C^1$
 
5:06 PM
@Blue are you around ?
 
Anonymous
@Tanuj Not really, I'm a bit busy. I have a test on Monday. I didn't get time to see your integration problem. Maybe try asking on Math SE. (The general rule for those problems is to use the properties of definite integration. One hint is that $1/(e^t+1)+1/(e^{-1}+1)=1$)
 
@Blue I have one more question to ask , not really a question , just a concept , I'm sure it wouldn't take longer than a minute
 
Anonymous
Ok
 
And the answer provided is
I don't get this $|A-\labdaI| = 0$ , how ?
 
Anonymous
Ugh, so they are using the Cayley Hamilton theorem without even teaching it to you?
 
5:13 PM
Actually I haven't been taught anything of use by my coaching , but I'm trying to papers of other institutes now
 
Anonymous
Just google around Cayley Hamilton's theorem. There are some youtube videos on this too. This isn't really in syllabus afaik, but makes some problems easier to solve.
 
@Blue I know you don't have time to teach me this right now , could you suggest me some good source which should be easy for me to understand it ?
@Blue Okay , any recommended video though ?
 
Anonymous
@Tanuj I didn't watch any of those. But this seems to be okay, looking at the title. It doesn't prove the theorem though.
 
Where did you learn it from ?
 
Anonymous
Or this
 
Anonymous
5:17 PM
@Tanuj I think some textbook like Arihant had it. I don't remember now. You'll learn it properly when you study linear algebra in uni.
 
ah okay , see you around man :) good luck for the exam !
 
Anonymous
Thanks
 
Anonymous
See you
 
6:02 PM
Greetings, nitsua
what brings you to the h-bar
 
6:18 PM
Anyone help please ... If I fill up top half of a capacitor with one dielectric and bottom half with another , will the capacitance of capacitor be affected by the change in their height , not width ...?
 
Anonymous
7:01 PM
@NehalSamee That doesn't make much sense. Which part are you calling the "top half"?
 
Anonymous
And how are you defining your height, width, etc (of capacitor)
 
Anonymous
Better draw a picture and post it
 
Isn't waiting for someone to arrive the longest wait ever
 
@CooperCape Only if his name is Godot
 
Anonymous
@CooperCape That's why I'm always the last one to reach any event or meeting...
 
Anonymous
7:13 PM
Waiting for Godot ( GOD-oh) is a play by Samuel Beckett, in which two characters, Vladimir and Estragon, wait for the arrival of someone named Godot who never arrives, and while waiting they engage in a variety of discussions and encounter three other characters. Waiting for Godot is Beckett's translation of his own original French play, En attendant Godot, and is subtitled (in English only) "a tragicomedy in two acts". The original French text was composed between 9 October 1948 and 29 January 1949. The premiere was on 5 January 1953 in the Théâtre de Babylone, Paris. The English language version...
 
Anonymous
Hehe
 
Anonymous
This is new
 
7:23 PM
@Blue No, it is in fact a rather old play :P
 
and philosophically very deep
 
Anonymous
@ACuriousMind Yeah, but new for me :P
 
Anonymous
It's on youtube and I'm watching it
 
Anonymous
lol
 
Anonymous
7:33 PM
Okay...I can't tolerate this for 2hrs XD
 
27 mins ago, by CooperCape
Isn't waiting for someone to arrive the longest wait ever
:P
 
@Blue why are you such a legend ;)
Also that Godot reference is probably way before what I would count as my generation
 
yoho, how's it been going
 
@CooperCape Yes, I get it, I'm old :P
 
8:21 PM
can some1 help me?
 
8:44 PM
Suppose a particle moving with ω velocity and $\dot ω $ accelaration
how to derive
$x(t)=Rcos(ωt +1/2α^2)$
i can only derive Rcosθ
 
9:06 PM
@JohnRennie I just got a camera. I want to do a few videos (somewhat related to the question) then make a video of my question. It would be much easier. I also want to compute a few highly trivial things on camera before asking the questions so that it makes more sense (also so I am taken a bit more seriously)
 
9:49 PM
can some1 explain me a pdf?
its simple for you
on circular motion how did he derive the components
of velocity and accelaration
on polar coordinates
i can understand the cartesian on
 
@ManolisLyviakis take a derivative of $\mathbf{r}(t) = r(t) \hat{e}_r(t)$
 
Heading out to record first video
 
yes i did that
 
What equation is giving you trouble in the pic
 
isnt $r(t)=||r(t)||e_r$
 
9:59 PM
Yes
 
since the trajectory vector is the radius in the direction of the radius
the derivative of that is $Re_θ$
 
Using vector notation, $\overline{r}(t) = r(t) \hat{e}_r(t)$
 
ye
i mean its zero*
but then on the component $a_r$
 
No, the derivative is equation 3 of the pdf
 
isnt it the second derivative of r(t)
i dont get the accelaration components
acceleration*
 
10:02 PM
Do you understand how to get equation (3)
 
yes its the derivative $\frac{dRe_r}{dt}$
hmm.. but isnt $e_r$ a function of t
 
Yeah it is
 
so its $0e_r +Re_θ$
ohhh
i get it now
 
$\frac{d}{dt} \mathbf{r}(t) = \frac{d}{dt}[r(t) \hat{e}_r(t)] = \frac{dr}{dt} \hat{e}_r(t) + r(t) \frac{d \hat{e}_r}{dt}$, why do you think $\frac{dr}{dt} = 0$?
Acceleration is equation (4), you just take the derivative of equation (3), they give an equation in between (3) and (4) with no label which is what you get when you take a derivative of (3)
 
i thought since the component of velocity
is zero
 
10:07 PM
No
 
then the derivative of the same component is the component in acceleration
if $u=(0,Re_u)$
 
$\mathbf{r}(t) = x(t) \hat{i} + y(t) \hat{j} = \sqrt{x(t)^2 + y(t)^2} \hat{e}_r(t) = r(t) \hat{e}_r(t)$
 
i would diff that and get velocity where is my fault
ohhh yes
 
Kleppner's Mechanics chapter 1 explains this better than the pdf
 
but stil am i wrong that $u=(0,Re_θ)$
 
10:09 PM
What do you mean
 
what is the velocity vector
since the one component is zero
$u_r=0$ and $u_θ=R \dotθ $
so $u=(0,R\dotθ)$
and i thought to diff that to get velocity
acceleration*
which book you said says it better?
 
If $\overline{r}(t) = r(t) \hat{e}_r(t) = (r,0)$ then $\overline{v}(t) = \frac{d}{dt} \overline{r}(t) = \frac{d}{dt} [r(t) \hat{e}_r(t)] = \frac{dr}{dt} \hat{e}_r(t) + r(t) \frac{d \hat{e}_r(t)}{dt} = \dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_{\theta} = (\dot{r},r\dot{\theta})$ not $(0,r\dot{\theta})$
 
ohh i thought dr/dt is zero
 
Compare this to equation (3) in your pic, it's the same, here is the pdf of the pic:
 
i got how he derived the equations i think
he used the general formulas and applied the physical interpatation
im just asking if dr/dt is zero
 
10:17 PM
No $dr/dt$ is not zero in general
Maybe in some specific example
 
hm.. yes
thats maybe why i couldnt get it
somehow i thought it was zero
i thought r=R the radius
 
Like if a particle is moving around on a circle, $dr/dt = 0$, but in general no
 
ye if it moves on a circle
 
You can use polar coordinates and a polar basis to describe any particle in the plane more or less
 
ye nice
now on the circle
since $u=(0,R\dot θ)$ why cant diff that to get the acceleration?
 
10:20 PM
That is the velocity on the circle right
 
sorry**
meant acceleration
and if i diff each component i get
 
If that is the velocity for a particle on a circle, you can differentiate it to find the acceleration of the particle on that circle, however you need to differentiate both the components and the basis elements
 
but the first component is zero which means it stays zero
 
Differentiate $u = (0,R,\dot{\theta}) = 0 \hat{e}_r + r \dot{\theta} \hat{e}_{\theta}$
 
ye
ohh the second part will become a component of $e_r$
hahaha that is so strange
 
10:24 PM
$\overline{a} = \frac{d}{dt} \overline{u} = \frac{d}{dt} [0 \hat{e}_r + r \dot{\theta} \hat{e}_{\theta}] = \frac{d}{dt} [r \dot{\theta} \hat{e}_{\theta}] = \dot{r} \dot{\theta} \hat{e}_{\theta} + r \ddot{\theta} \hat{e}_{\theta} + r \dot{\theta} \frac{d}{dt} \hat{e}_{\theta} = ...$
 
yes
ok i think i got it!!
thank youuu
 
Cool
 
but its stange
having a vector (0,x)
and the derivative will be (y,k) :p
 
Basically, the basis is moving as well, so you get crazy behavior
 
10:26 PM
So you only use it in nice cases
The Kleppner book derives these results in like 3 ways because it's so strange :p
 
what is that book
can u show it
 
also isnt the angle between acceleretion with the placement vector 0
 
The position vector is $\overline{r} = r \hat{e}_r$, the acceleration vector is $\overline{a} = ??? \hat{e}_r + ?S?? \hat{e}_{\theta}$, so no
 
ohh i though it was point always to the centre
 
10:31 PM
If $r$ is constant, then it will point to the center, but it will still move around as well
It has to use both bases or else it will go in one direction right
 
if u u mean is constant will point to the centre
 
0
Q: Do polyhedral wormholes make sense?

SlereahVisser defines some class of wormholes with polyhedral mouthes, as a limit of smoothed polyhedrons as the radius of the edges go to zero. Does this limit actually make sense, as an actual spacetime? That is, can I remove two polyhedrons from Minkowski space and identify them in a way that makes s...

plz halp
 
@bolbteppa in you equation there is no square but in the pdf there is$(\dotθ)^2$ on component of the acceleration
 
what equation
In the acceleration, I didn't write out the derivatives, I wrote = .... and you can see the pdf for the final answer :p
 
10:46 PM
when you took the derivative of $0e_r+r\dot θ e_θ$
it doesnt match
i should get a component of $e_r$ with a square
 
Look at what I wrote, compare it to (4), and it should make sense
$\overline{a} = \frac{d}{dt} \overline{u} = \frac{d}{dt} [0 \hat{e}_r + r \dot{\theta} \hat{e}_{\theta}] = \frac{d}{dt} [r \dot{\theta} \hat{e}_{\theta}] = \dot{r} \dot{\theta} \hat{e}_{\theta} + r \ddot{\theta} \hat{e}_{\theta} + r \dot{\theta} \frac{d}{dt} \hat{e}_{\theta} = ...$
Look at $\dot{r} \dot{\theta} \hat{e}_{\theta} + r \ddot{\theta} \hat{e}_{\theta} + r \dot{\theta} \frac{d}{dt} \hat{e}_{\theta}$, the first term is $0$, the middle term is fine, what is the last term?
 
$-r\dotθ e_r$
 
No
 
10:55 PM
oooooooooooooooooooooooooooooooooooooooooh
dayumn in respect with the angle the derivative is just a sign change
in respect with t there is another θ
 
Yeah
 
hahaha im soo bad.
 
It's counter-intuitive
 
im a math undergrad and i havent got used to them
we do heavy analysis and stuff
but not all these coordinate calculus
ill right everything in respect to t so i dont get confused
θ(t)
 
Yeah include $(t)$ as much as possible
 
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