3:25 AM
@Danu NB: Numerical Recipes that ChrisWhite promotes there is actually copyrighted information. That is, you cannot legally use the subroutines in any publicly or privately available software without the recipient having a legal license to use NR.

user54412
@KyleKanos fortunately for us all, those authors have such awful, opaque coding style no one would ever want to directly use their code

@ChrisWhite Debatable. I've actually seen their routines in public code (which goes to show no one gives two sh*ts about software licenses)

3 hours later…
6:14 AM
@KyleKanos this is true. My advisor uses their code
Come to think of it, some of it may have snuck into our released software (!!) so I should check that... sigh more to do

1 hour later…
7:23 AM
@KyleKanos Luckily, I'm so bad nobody else will ever want to use my code

7:46 AM
@KyleKanos: Oh, I didn't know that... So, if I've included it in some software I've written for my computer, that's technically a breach?

8:00 AM
Hah
The Taylor seires thing I did does give the constant, I was just too stupid to realize @Qmechanic @JamalS

8:57 AM
@Danu: All I did in Mathematica was find the solutions, and then take a Taylor series around infinity, so I was surprised it didn't work for you

2 hours later…
10:28 AM
0

I contacted stack exchange several times because I had to make a duplicate account and I want to merge them. I have not still received any response after abunch of emails acros about a week. Mt problem is that when I created my google account I misspelled it. I do remember the password, but not...

10:55 AM
@JamalS If you bought a code license, or the package that contains the book and a code license (and the code on a CD, IIRC), then you can use it for whatever you want on your own computers. If you haven't bought a code license, then you can't use it.
see this
16

I took a look at the clip and my take on it that the object of the exercise was not to slow down the tank, but to move it sideways and land in the lake a half a mile away. Hannibal says, "rotate the main gun to 82º" which I take to be sticking out sideways. Background Information The facts I co...

Maybe someone feels like turning that into MathJax... I don't, not right now

One should think that one of the 16 people voting that up would have bothered to do so...

Yeah, you'd think. But apparently not :-(

2 hours later…
12:47 PM
Footnote from my QFT lecture notes: "Whenever we don't know what to do, and Fourier transformation is not the answer, we insert a $1\!\!1$."

1:37 PM
@ACuriousMind: I'm surprised you haven't stabbed your lecturer yet!

1:51 PM
I...actually found that note funny :D
And this is not the same one I ranted about
@JamalS Also, stabbing lecturers would look bad on my resume

@ACuriousMind: Yes, doesn't show sufficient creativity or ingenuity

2:35 PM
@JamalS As far as I know, purchasing the book does not actually grant you the license. So technically, if you did not purchase the license and are using it, that is a breach. If you have the license, then you can use it in whatever codes you want so long as it is only you using it!

2:58 PM
0

I'm taking a philosophy of time travel class. In one of the lectures, the teacher was discussing problems with the Many Worlds interpretation. He talked about how since anything that can possibly happen happens in some timeline, in one timeline there is a person that comes out of what appears to ...

Is this on- or off-topic?
How "non-mainstream" is it to ask about quantum interpretations and time travel?
(Pretty non-mainstream, if you ask me...)

Well the non-mainstream closure reason does say, ...specific questions evaluating new theories in the context of established science are usually allowed
This could be construed to fall under that
But I'm with you that it's not really mainstream (mostly due to the time-travel aspect)

3:24 PM
0

Wave Equation Variables and Interpretation was put on hold as too broad, but edited the same day. Now it's been on hold for a few days. Is it still too broad?

1 hour later…
4:34 PM
0

I think everyone should have seen something like a bowl of hot water moving by itself on a flat surface(seems like it is moving by itself but maybe there are some external force applied to it when it moves). What is the explanation of this phenomenon? Hypotheses : Since the bowl is heavy, for ...

Do soup bowls move for anyone else?!
I'm quite confused by this post

Answer: Magnets

Yeah, as hot water is known to replace supercooled electromagnets in our next big collider, I should've thought of that.

Yeah, I've got no idea
maybe the table is at an angle

I should think OP tested if a bowl of cold water moves, too, before so boldly stating it about the hot one

what type of bowl I wonder?

4:42 PM
I've seen the effect but only when there is water on the surface and the bowl is hot enough to make steam.

and how much does it move?

Also, the question is formulated as if hot bowls of soups moving around is one of the more usual occurences in their day

Then it lifts up on a cushion of steam and slides a bit.

If it's only a jump, it could be that the expanding glass of the table drags whatever part of the bowl is most stuck to it along for a very short ride

What kind of paper-thin bowls do you people eat soup out of? All my bowls are far too heavy for me to imagine them just sliding on the table, water or not
I would be very bewildered by my household items moving around that way

4:45 PM
Well when I've seen it happen with a steam cushion the weight of the bowl doesn't mater as much as the seal the bowl makes against the table surface. It needs a pretty smooth lip all the way around to trap the steam under it.

he posted a picture of the bowl

@Jim I see. Looks like mine. Now I need a glass table :D
Also, is your answer there really an answer?

Partially

@ACuriousMind Probably not. There are probably a few days it can happen.

it does say that what he describes doesn't happen
which is as much as we could say for a bowl sliding across a table of its own will
the rest is of the not-an-answer variety

4:52 PM
it's easiest to reproduce by putting a hot bowl or coffee cup or inverted tumbler (pint glass) on a smooth table with isopropyl alcohol on the surface of the table. The alcohol already has a higher vapor pressure and it evaporates at a lower temperature.

@BrandonEnright I really don't think that is what is happening in this case

ah I see the asker included this "I have heard that water may form a thin layer between the bowl and the surface, but l'm not sure whether it is the case or not."
I've only seen sliding on a cushion of vapor.

5:29 PM
@ChrisWhite Uh-oh. I may have spoke too soon:
0

I ofter build with lego and use this mechanism: It simply converts the rotational movement to making the stick between the two blocks go back and forward. What is this mechanism called?

@KyleKanos: Is that even on-topic? It is even less about physics concepts than the dreaded homework questions.

Chris's comment that I linked to asks if questions about would be on topic here

1 hour later…
6:44 PM
Hey guys, I'm wondering if this question would be a good one to ask on the main site: "Why is the Lane-Emden equation written as $$\frac{1}{\xi^2} \frac{d}{d\xi} \left( \xi^2 \frac{d\theta}{d\xi} \right) = -\theta^n$$ instead of something like $$\frac{d\theta}{d^2\xi} = -\frac{\theta^n \xi}{2}$$?"

@El'endiaStarman It could become one, if you told us why you think it should be the latter. It seems to me these two are not equivalent - did you neglect the product rule in resolving the differential on the LHS?

IMO, it's not really a good question. The Lane-Emden equation is the result of taking the polytropic model and inserting it into the hydrostatic equilibrium equation.

@ACuriousMind I initially was quite confused as to why you wouldn't just multiply out the $1/\xi^2$ and $\xi^2$. From what I've understood, $d/d\xi$ is acting as a function here (I suppose that all derivatives are functions, but they're usually not written that way). So, why not just take the derivative with respect to $\xi$ and simplify it?

Then some change of variables, obviously.
Note also that you are assuming that $\theta\neq\theta(\xi)$
And how do you get $d\theta/d^2\xi$? Should that be $d^2\theta/d\xi^2$?

@El'endiaStarman Derivatives aren't functions, they're derivatives!

6:53 PM
@ACuriousMind Then why was it ($d/d\xi$) written as if it was?

$d/d\xi$ is operating on the term in the parenthesis

Okay, so then, $\frac{d}{d\xi}(\xi^2)$, for example, should be $2\xi$, right?

Yes

Hmm. Maybe I forgot to take into account the product rule.
$$\frac{d}{d\xi} \left( \xi^2 \frac{d\theta}{d\xi} \right) = 2\xi \frac{d\theta}{d\xi} + \xi^2 \frac{d\theta}{d^2\xi}$$
Then multiply by $1/\xi^2$ and you get...
$$\frac{2}{\xi} \frac{d\theta}{d\xi} + \frac{d\theta}{d^2\xi}$$

You're doing your differentials wrong:
$$\frac{d}{d\xi}\left(\xi^2\frac{d\theta}{d\xi}\right)=2\xi\frac{d\theta}{d\xi} + \xi^2 \frac{d^2\theta}{d\xi^2}$$

7:02 PM
@El'endiaStarman I'd echo what ACuriousMind said, that it could be a pretty good question if you explain why you think it should be the "other" way in the question
But all this chat discussion may be leading you to an even better question

@KyleKanos Ahh, an error in notation. $\frac{d^2\theta}{d\xi^2}$. Excellent. Thanks for the correction.
@DavidZ I did my math wrong. Which kinda sucks considering I'm a math major. :P

heh, well to be fair mathematicians actually don't do much math :-P
(by which I mean they don't do as much raw manipulation of symbols as one might think)

Okay, then that means that one could write the Lane-Emden equation as: $$\frac{d^2\theta}{d\xi^2} = -\frac{2}{\xi} \frac{d\theta}{d\xi} - \theta^n,$$ correct?
Which, by the way, I would say is much more clearly a second order ODE, and poses less risk of confusion.

That is a correct formulation

I mean, really, $\theta'' = -\frac{2}{\xi}\theta' - \theta^n$.

7:09 PM
For $n=1$, this formulation is the way to go because it can be solved with power series

@KyleKanos Hmm. Could you explain, please?

Let $\theta(\xi)=\sum_m a_m\xi^m$ where $a_m$ are the coefficients. Sticking this into the Lane-Emden eq'n leads to a recursion relation for the $a_m$'s.
With enough work, you'll find $\theta(\xi)={\rm sinc}(\xi)$

@KyleKanos Coefficients of what?

Of the power series
$\theta(\xi)=\sum_ma_m\xi^m=a_0\xi^0+a_1\xi^1+a_2\xi^2+\cdots=a_0+a_1\xi+a_2\xi^‌​2+\cdots$

"method of Frobenius" I think that was?

7:19 PM
Google-fu says you are correct:
In mathematics, the Method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a second-order ordinary differential equation of the form with and in the vicinity of the regular singular point . We can divide by to obtain a differential equation of the form which will not be solvable with regular power series methods if either p(z)/z or q(z)/z2 are not analytic at z = 0. The Frobenius method enables us to create a power series solution to such a differential equation, provided that p(z) and q(z) are themselves analytic at 0 or, being analytic...

@KyleKanos What are the $a_m$?
Or how do we find them?

By applying boundary conditions for some
The remaining you actually just take derivatives and set $\xi=0$ to find the remaining values
Eventually you'll find a recursion relation between the $a_{m+2}$ and $a_m$ coefficients

@El'endiaStarman arbitrary coefficients - you just plug into the equation and differentiate. So on the left side you get $2 a_2 + 6 a_3 \xi + 12 a_4 \xi^2 + \ldots$

7:48 PM
@KyleKanos Okay. Boundary conditions were mentioned. For the problem I have to solve with numerical methods, one boundary condition is $\theta(0) = 1$ and the other is $\left( \frac{d\theta}{d\xi} \right)_{\xi=0} = 0$.
I don't quite understand the last one though. Is it saying that the derivative of $\theta$ at $\xi=0$ is $0$?

Yes, that's what it's saying

Okay. I think some of my difficulty in really getting this, understanding it, is that $\xi$ is just something weird.

Don't think of it that way
It's a coordinate

So I don't have a good mental picture.

Well, a scaled coordinate: $r=\alpha\xi$
where $\alpha$ depends on some constants (e.g., central density, Newton's universal G)

7:52 PM
And $$\alpha = \left( \frac{n+1}{4\pi G} \kappa \rho_c^{(1-n)/n} \right)^{1/2}.$$

Those look like the constants I was thinking of

What's the purpose of $n$?
@KyleKanos I'm currently in a Numerical Analysis class taught by a professor who does lots of physics work. Hence, his problems tend to be physics based, and in particular, this one is in the context of stellar structure. How the pressure/density varies inside a star, I believe.
I just copied that straight from the problem statement.

$n$ is just the power of the polytrope:
A polytropic process is a thermodynamic process that obeys the relation: where p is the pressure, v is specific volume, n, the polytropic index, is any real number, and C is a constant. The polytropic process equation is particularly useful for characterizing expansion and compression processes which include heat transfer. This equation can accurately characterize a very wide range of thermodynamic processes, that range from n=0 to n= which covers, n=0 (isobaric), n=1 (isothermal), n=γ (isentropic), n= (isochoric) processes and all values of n in between. Hence the equation is polytropic in the...
It relates the pressure and volume to some constant
(in this case, $v=\rho/\langle m\rangle$)

> The following derivation is taken from Christians.[1]
Lol.

Well, to be fair, the Christians did a lot for humanity: hospitals, education, peer review, Pasteurization, printing press, and so on. In this case, however, it's a persons surname :)

8:04 PM
@ACuriousMind heyy :D I just realised you've started a blog already, great! you finally got around to it ha?

@Phonon Hah, I meant to tell you when you joined chat today :D

I have a blog
I just don't use it

Seems you stalk my profile regularly enough that there's no need ;)

^ OoOoOoOoOoOoOoOo

@ACuriousMind lol

8:05 PM
Actually, I have two blogs

@ACuriousMind really happy you finally did it ;D

My academic one and the Fortran in a C World one that I link in my profile. Neither is used

@ACuriousMind I ll have a read at the two posts already later tonight, after work
@KyleKanos had no idea, you've stopped blogging then?
@ACuriousMind Gauging thoughts :pppp

@Phonon A Curious Mind, was, unfortunately, already taken^^

haha :D

8:10 PM
@Phonon Yeah, I just ran out of ideas to write about so I stopped

@KyleKanos :(

I'm over it. It's not a big deal, somethings I'm just not good at

Blaspheme! You're the best at everything!

Like the intuition for the Rubik's cube. I solve it rather rigidly (piece A goes there, piece B goes there) rather than trying to solve it smoothly (e.g., if I move this piece here, then rotate, I've got these 3 adjacent pieces together)

8:26 PM
@ACuriousMind one can use latex in comments? (your blog)

@Phonon Uhh...I guess so? You have to write latex directly after the initial dollar, though.

@ACuriousMind kk, testing it with a first comment

This is one of my favorite problems in physics:
-1

There is a common fact about projectile motion. Let me remind you with an example: Assume that you are aiming at a hanging banana on a tree. If there were no gravitational force, the bullet would follow a linear path at a time "t". But since it does exist such kind of force, a parabolic path will...

HEllo

8:41 PM
Hello

I want to see as far as I can around my hand with a mirror for armwrestling. I want to be be able to see both arund the side and somewhat over the top of my hand . would a homemade parabolic mirror lke this be able wrap around very far on a hand? youtube.com/watch?v=_8sd9UgjXLE

That sounds like a lot of effort just for armwrestling

@ACuriousMind done, lol it says "waiting moderation" :D

1

An applied physics question: How can I setup a mirror all around an arm wrestling table so I can see all angles of what I'm doing? The arm wresting table looks like this With the wrestlers on table looks like

@KyleKanos Their is no longer answers
There was at one point
Don

8:46 PM
It was deleted because it wasn't really an answer

Oh
@Jim Yep but is worth it if it is works. Can't think a simpler way with little money spent and being as simple.

I just don't get the "why" aspect

@KyleKanos Bravo! you referred him to his own question. That is a special kind of kindness right there

It wasn't meant for him
It was meant as reference
(e.g., for you)

How can I approximate if given parabola mirror would reflect ( wrap around ) around my hand enough

8:49 PM
@ChrisOkyen a simpler way for what?
Or rather, why?

The why - Arm wrestling is a lot of form and you can't always see what your fingers are doing without moving your head which is annoying and can put you in bad positon

Mount a camera and watch the video like a football team does when they practice?
otherwise, wouldn't this be not allowed by pro arm wrestling regs?

I am doing it during practice
If I took a bendable surface and rigged a reflecive material or painted a reflective material on it , would that work?

It could work, but you'd be the one who could tell, not us :/

Camera is the way to go then. That way you see your form while you're concentrating properly. You aren't looking at a mirror and distracting yourself (which might make your form worsen compare to when you aren't watching the mirror)

8:54 PM
We basically are stopping when the person I am pulling wants to show me my form or his form when we are practicing

Right, but if you capture it on camera, you can do the wrestle and go back after to watch your form. Just like the football teams do it

@Jim Your are right I just thought it would spend less time being done with it during instead of going back. But with a camera it is recorded so you can review :)

@ACuriousMind btw send me an email to that address, so I can tell you bout some typos here and there (don't wanna spam the chat here).

2 hours later…
10:30 PM
Rereading (the debates on) the questions of 't Hooft is so fascinating. physics.stackexchange.com/users/11205/g-t-hooft
Those were the days...
@DavidZ Lol.

@Danu idd, same for Peter Shor's

1 hour later…
11:41 PM
@Phonon Sry for delayed reply, I will - tomorrow. For now, I'm going to bed.