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00:11
@obe type it up texpaste.com
obe
obe
I'd prefer to write it first.
I'll type it tomorrow.
NSFW btw, Mr. Ban Bot
just a great show
00:31
@0celo7 haha, the question mark at the end is sooooo bad.
People seem to now routinely message me bad question titles.
I wonder whether this is a position of honor or pain for me...
00:42
@DanielSank honor
01:04
@FenderLesPaul I'm down for tonight
 
5 hours later…
05:58
@JohnRennie Yo.
Seems like the tearing issue is complex, as you say.
The usual example of cold welding is metals
The velcro example is particularly interesting.
The idea is that if you separate a metal along an atomically smooth surfec you can rejoin it
05:59
@JohnRennie I thought the presence/absence of oxide formation after the break was the key.
and the interatomic forces will bond the two surfaces back together
The surfaces have to be smooth on the atomic scale
I see. I suppose both clean and smooth surface are needed for cold welding.
Yes
Either roughness or contaminents will stop the rejoining
In practice, if you break metal then the metal surfaces deform plastically while breaking
06:00
Makes sense.
Sure
and even in a vacuum they won't rejoin
because they are no longer smooth
I get it.
Paper is a composite
The fibres are held together by hydrogen bonds between them
but mostly by a binder that's added to the mix
...and I imagine some amount of "things being wound around other things" like with velcro?
The cellulose fibres certainly tangle
but they aren't hooked
06:02
Allright.
Sure sure, no hooks like velcro.
and they aren't particularly rigid
0
Q: Similar/Same question on two SE sites

user35952If I have question in Computational Physics, I sometimes feel more comfortable putting up this question in compsci.SE, but nevertheless the response at that site is extremely slow and poor (since its a beta version and there are few users). Therefore, sometimes I am compelled to post the same que...

Without the binder paper would be flexible like cloth
Ah.
Ok.
In the old days the binder was starchy gunk from boiling the cellulose fibres
This dries out to a hard resin like consistency
06:04
Ok.
These days I'd guess some synthetic binder is used
So the trouble is that there's no mechanism for the torn paper to rejoin
It isn't smooth, and doesn't have hooks
But what happens when it tears?
The constituent parts sit all mixed together in binder when the paper is whole, yes?
Yes
When it tears you (a) fracture the binder away from the fibres and clay particles
(b) pull the (flexible) fibres apart
(c) bits of clay break off and fall away as dust
and the whole lot deforms during tearing
06:07
I think perhaps I'm puzzled by something which I can explain in a simpler way.
Suppose I have a many-stranded rope.
If I pull on it hard enough, I may get the strands to "disengage" and thus break the rope.
However, if I push the broken ends together, it doesn't spontaneously rejoin.
There's no binder here.
A rope is held together by friction between the fibres
Right.
The fibres are mostly aligned so there is a high surface area of contact between each fibre
The design of the rope also means the fibres are pressed together even when there is no load on the rope
06:09
Yes indeed.
When the rope is in tension the fibres are pressed together even more tightly
To rejoin two broken ends you'd have to weave all the dangling fibres back together
so they all lined up again
Right.
Possible in principle but hard in practice
06:10
Somehow, the pulling goes from a rather low entropy situation to a higher entropy one.
There are more ways for a rope to be broken than joined.
Though it can be done
Rope splicing in ropework is the forming of a semi-permanent joint between two ropes or two parts of the same rope by partly untwisting and then interweaving their strands. Splices can be used to form a stopper at the end of a line, to form a loop or an eye in a rope, or for joining two ropes together. Splices are preferred to knotted rope, since while a knot typically reduces the strength by 20–40%, a splice is capable of attaining a rope's full strength. However, splicing usually results in a thickening of the line and, if subsequently removed, leaves a distortion of the rope. Most types of splices...
^ Certainly.
The question in my mind is why simply pulling on the rope can go from the nicely lined up situation to the haphazard broken one.
I realize this question sounds somewhat naive. Perhaps I am over-thinking the issue.
I don't think fundamental arguments like entropy will help much here
Perhaps not.
If you have a carefully ordered system like rope then tearing it is only going to disorder it
but with such a complex process I'm not sure thermodynamic arguments help
06:13
I didn't mean to invoke thermodynamics. By entropy I just mean "situation with may possible realizations" versus "situation with less possible realizations".
In that case I agree
Anyway, it's still interesting: suppose my rope is actually a set of little magnetic pills.
If I pull on it, I might get a perfectly single-file string of magnets.
Still joined, but different configuration.
With a stranded rope this doesn't happen.
Magnets have a long range force (long compared to molecular scales)
so they can self organise on macro scales
Yeah, good point.
When the rope breaks, it doesn't seem to deform much at the break point.
It seems that at a particular spot the strands move relative to one another and disengage.
The individual fibres are heavily deformed
but of course they are very small
so it isn't obvious
06:16
::Looks for rope to break::
Interesting, I did knot know that.
Interesting, it seems that when a metal cable breaks it is due to strands actually breaking, not strands moving relative to one another.
I wonder if that's a generic property or due to the stands being very long.
I would guess strands also fracture to some extent in rope breaking
I have to get back to work now. Hope this helped ...
It did. Thanks
 
2 hours later…
08:31
Hello
@0celo7 What do I know
 
2 hours later…
10:38
Hello
11:27
Hello
@Slereah what that metric is called
It looks like a FRW metric?
But like
different factors depending on the coordinate
It has a name
11:44
It's a homogeneous spacetime, certainly
isn't FRW valid for all homogeneous spacetimes?
it's named after a dude
did you check the wiki article on all exact solutions
Kasner
The Kasner metric, developed by and named for the American mathematician Edward Kasner, is an exact solution to Einstein's theory of general relativity. It describes an anisotropic universe without matter (i.e., it is a vacuum solution). It can be written in any spacetime dimension and has strong connections with the study of gravitational chaos. == Metric and conditions == The metric in spacetime dimensions is , and contains constants , called the Kasner exponents. The metric describes a spacetime whose equal-time slices are spatially flat, however space is expanding or contracting at different...
how come I found that instantly
your GR knowledge is lacking
;w;
Never read up on that metric, I'm afraid
Never did that much cosmology
it's in MTW
::shakes head::
everything is in MTW
12:03
So you should know everything
"In addition to introducing the term "googol", he is known also for the Kasner metric and the Kasner polygon."
Kasner is full of trivia
In geometry, the midpoint polygon of a polygon P is the polygon whose vertices are the midpoints of the edges of P. It is sometimes called the Kasner polygon after Edward Kasner, who termed it the inscribed polygon "for brevity". == Examples == === Triangle === The midpoint polygon of a triangle is called the medial triangle. It shares the same centroid and medians with the original triangle. The perimeter of the medial triangle equals the semiperimeter of the original triangle, and the area is one quarter of the area of the original triangle. This can be proven by the midpoint theorem ...
he invented the triforce
Lololo
It's sad that ACM wouldn't get the reference
12:26
@Slereah : the FLRW metric starts with the assumption of homogeneity and isotropy of space. The Kasner metric is talking about inhomogeneous vacuum.
So I saw, yes
And inhomogenous vacuum is what a gravitational field is.
As you will know from your raisin-cake analogy, space expands between the galaxies but not within. So every galaxy is surrounded by a "halo" of inhomogeneous vacuum. And inhomegeneous vacuum is what a gravitational field is. Now my oh my, why can't we find them pesky WIMPs?
12:59
clicks "hide posts" from JD
Aaah
you should try it
It is refreshing
His icon is all tiny on the user list now
kinda puts things into perspective
We can totally talk about GR now
or anything else without mention of Einstein and The Evidence :-)
have you read his book meant for high schoolers?
where he combines SR & GR
according to him they should be presented that way
I have glanced at it
Here is a GR thing
Did you know
The Kerr metric isn't the generic metric for vacuum axisymmetric static spacetimes!
Unlike the Schwarzschild spacetime, it is not unique
It will depend on the quadrupolar moment of the source, too
Which is 0 for Kerr
Nobody mentions it because it is an awful awful metric
13:23
that diff eq test was easy
@Slereah I think Wald mentions that
Straumann certainly does
@Slereah indeed
@skillpatrol huh?
@Slereah link?
I wonder if JD can see posts from people who ignore him
Relativity: The Special and the General Theory began as a short paper and was eventually published as a book written by Albert Einstein with the aim of giving: . . . an exact insight into the theory of relativity to those readers who, from a general scientific and philosophical point of view, are interested in the theory, but who are not conversant with the mathematical apparatus of theoretical physics. It was first published in German in 1916 and later translated into English in 1920. It is divided into 3 parts, the first dealing with special relativity, the second dealing with general relativity...
thought you meant JD
@Huy hello
Huy
Huy
hi @0celo7
hi pal @Huy long time no see :-)
Huy
Huy
hi @skillpatrol
13:35
tsk tsk
what
Huy
Huy
@0celo7: u good at some basic non-euclidean geometry?
probably not
what is the question
Huy
Huy
Let's look at the upper half plane model of hyperbolic space only knowing that points are $(x,y) \in \mathbb{R}^2$ with $y > 0$ and lines are either half-circles with midpoint on the $x$-axis or rays parallel to the $y$-axis. Define the distance of two points $P, Q$ as follows:
If they have the same $x$-coordinate, define $d(P,Q) = |\log \frac{Q_y}{P_y}|$. Otherwise, look at the line (half-circle) through $P$ and $Q$ and its endpoints $A, B$ on the $x$-axis ($A$ on the left of $B$). Then, define $d(P,Q) = |\log \frac{k(A,Q) k(B,P)}{k(A,P) k(B,Q)}|$ where $k(A,Q)$ is the Euclidean length of the arc $AQ$.
Now take a Euclidean line $g: y = mx + b$ and a point $P$ that does not lie on $g$. I want to compute the distance from $P$ to $g$. Apparently, if I take the intersection of $g$ with the $x$-axis as the midpoint of a circle running through $P$, the arc on that circle between $P$ and $g$ realizes the minimal distance and therefore is the distance between $P$ and $g$. How do I see that it is this arc and if I don't know it beforehand, how can I find out?
^^
13:39
idk what the metric is for the hyperbolic space
or what it even is
never needed it
Huy
Huy
no knowledge of Riemannian geometry
required
what is the hyperbolic space
Huy
Huy
it says it in the question
:D
hmm
is this the poincare half plane
Huy
Huy
yes
13:41
the geodesics on it are circles
Huy
Huy
but we don't know anything about geodesics etc
just what's given in the question
ok, I guess I know what the hyperbolic space is
dude
if you know the geodesics are circles
then show the poincare half plane is the metric
and solve using proper math
Huy
Huy
is the metric
?
what's that supposed to mean
that $\frac{\mathrm{d}x^2+\mathrm{d}y^2}{y^2}$ is the metric
physics books name spaces by their metrics
Huy
Huy
@0celo7: yeah I know about Riemannian geometry
but this is an exercise for math/physics freshmen
and they don't
and some funny guy thought it would be a useful exercise whereas I think it's not very insightful to just introduce these notions without justification
so I need to figure out how to explain what I asked using just what's given and no other knowledge of Riemannian geometry
wtf is that
Geodesics of hyperbolic space are all weird shit that diverge
Parallel lines diverge at infinity
@Huy doing last minute homework, will look at later
maybe
can't promise, have lots of stuff to do
It's a bit annoying because there is very few good sources on TS metrics
I don't think I ever saw it in a book outside of Stephani
@0celo7 Regarding oil on mars: .... dieselsweeties.com
13:54
There's plenty of valuable shit in space
Yet we don't give a shit
Space industry is currently not v. profitable
Check.
Maybe we finally get fusion and need HE-3
Though I guess practical fusion is 30 years away from today
Practical fusion is always 30 years away
It has been 30 years away since the 50's
I know. :)
So anyway
I have this Notion
Of doing a QFT thing
The idea is to solve a basic QFT thing
Free scalar field
with every QFT method
Well every formalism
Shouldn't be too difficult except for AQFT mb
I tried looking at KG in AQFT once
I am still lost
14:12
back
hey hey
@Huy once again, I'd be at the bottom of the pack at a European university
Heisenberg and Schrodinger QFT aren't too difficult
Path integral's okay, although I should probably try to do it RIGOROUSLY
With Wick rotations
Bohm I know exists
Not sure what other formalisms exists for QFT
There's CQFT
dunno much about it
also a bunch of QM formalisms which I do not know if they apply for QFT
@Huy working on it now
@Huy so the minimal distance is the arc from $P$ to the intersection of $g$ and $y=0$?
that does not even make sense :O
that point is not in the hyperbolic space
::scratches head::
how are $A$ and $B$ in it either
I thought the Poincare half-plane does not include the x axis
@Huy halp
I got a $\log 0$
the worst log
14:25
oh the intersection is the midpoint
I thought it was the endpoint
log 0 avoided
but now my picture is ruined :(
what
now the picture does not make sense
ok, picture obtained
@Huy call the intersection of the circle and $g$ the point $X$
is the Euclidean distance from $P$ to the midpoint the same as from $X$ to the midpoint
::resists urge to parameterize a geodesic and do a minimization problem::
yes the distance has to be the same
ok, new terminology
$A,B$ are the intersections of the circle with the x-axis
$X$ is the midpoint, $Y$ is the intersection of the circle with $g$
so we need to show that $d(P,Y)$ is a minimum
Huy
Huy
yes
sorry was playing some fifa
let $e(.,.)$ be the Euclidean distance (along a line)
Huy
Huy
it should also be the unique minimum actually
I'm betting $e(P,X)=e(X,Y)$ has something to do with this
we can calculate the arc lengths using $e(P,X)$ as the radius of the circle
maybe?
Huy
Huy
what do you mean? is calculating the arc lengths a problem?
14:38
no, just thinking
Huy
Huy
@0celo7: https://www.dropbox.com/s/qhspyvfn2bpki54/l01.pdf?dl=0
let's look at the picture in 3b (bottom diagram on the second page)
is that the solution
Huy
Huy
the Euclidean line going to the top right is the curve $C$ given, and $P_2$ is the point given. the exercise is to find all points which have the same distance from $C$ as $P_2$
the bold line is the solution
if I'm given the solution it's very obvious that this solution is indeed such a set
a set?
Huy
Huy
but only if you know that the shortest distance is the arc of the circle with midpoint on x-axis
14:41
I thought we were looking for a point
Huy
Huy
you have $P_2$ given and the curve $C$ on the RHS (not labelled in the sketch). we want the shortest distance from $P_2$ to $C$
the exercise itself is something else, but that's a part of it required to solve it
and I want to see why it has to be the arc of the circle with midpoint in $E$
maybe write down the $k$s or something
something glorious might cancel
Huy
Huy
@0celo7: is it obvious from the definitions I gave before that hyperbolic lines (half circles / rays) minimize distance ?
yes
Huy
Huy
why?
14:45
but why that particular circle is the question, no?
Huy
Huy
yes
but the other thing would be nice to see too
@Huy well I knew that from solving the geodesics equation for this space long ago
Huy
Huy
yes, that's how I did it too
I thought that was a given in this problem
Huy
Huy
no
I only have the "metric" $d(P,Q)$ as defined earlier
14:46
yeah, that's a given here
Huy
Huy
and I know that hyperbolic lines are rays parallel to $y$ axis and half circles with midpoint on $x$ axis
no clue how to derive that off the top of my head
Huy
Huy
how do you see it from that?
ok
ok then let's see if we can figure out an answer to either of those two things
:D
no geodesics?
Huy
Huy
what do you want to do with geodesics?
14:49
ok, are we given that the geodesics are circles
Huy
Huy
I mean I know the geodesics of hyperbolic plane are these lines
and geodesics are locally minimizing
but the freshmen don't know that
0
Q: Net reputation from removed posts

AniketI have a thing that is bugging me for the time being. I just discovered that there is an option called "show removed posts". And when I enabled it, I saw (-2) rep corresponding to several of my posts.... check out the Link here. Yes I know that when a post edited by me gets deleted or removed, I ...

Huy
Huy
all they know is "these objects are lines and we define a notion of distance as follows"
ok so we need to derive $d(P,Q)$
Huy
Huy
no, we are given $d(P,Q)$ as earlier
unless you would like to use a different formula
but how would you derive $d(P,Q)$ without the Riemannian metric?
14:50
do we get it for the y direction at least?
Huy
Huy
If they have the same $x$-coordinate, define $d(P,Q) = |\log \frac{Q_y}{P_y}|$. Otherwise, look at the line (half-circle) through $P$ and $Q$ and its endpoints $A, B$ on the $x$-axis ($A$ on the left of $B$). Then, define $d(P,Q) = |\log \frac{k(A,Q) k(B,P)}{k(A,P) k(B,Q)}|$ where $k(A,Q)$ is the Euclidean length of the arc $AQ$.
this is what we get
dude you're confusing me :O
Huy
Huy
why?
I thought we wanted to derive d(P,Q) now
Huy
Huy
you're confusing me
no I want to understand why from the definition of $d(P,Q)$ as above it follows that
a) half-circles and rays perpendicular to $x$-axis minimize distance
b) why the shortest distance from a point to a Euclidean line is given by the arc as discussed before
14:53
ok I calculated $d(P,Q)$ explicitly
and...the radius fell out everywhere
it only depends on the various angles
Huy
Huy
ah
that's helpful
but that means the circle is irrelvant :O
or...not?
Huy
Huy
:O
lemme upload a pic
Huy
Huy
no it just means that distance is being preserved as we stretch in "circular direction" if you know what i mean ?
14:55
I have class in a few minutes
Anyone have any ideas about the question following this message?
Huy
Huy
like if you have 2 points on a circle and then draw a larger circle with the same midpoint, and draw two euclidean lines through the midpoint and the 2 points, the intersection of those lines with the bigger circle will give 2 further points and they will be as far apart as the initial 2 points
wait
Perhaps @ACuriousMind, @bolbteppa or anyone else?
class is about to start, but I'm so close I think
Huy
Huy
14:56
@Danu: you should ping Ted and Mike, they surely can help you
@Huy is this the only configuration for which all the rs are equal
Huy
Huy
@Danu: even DanielFischer might be able to
@0celo7: what do you mean by that?
@Huy I felt that it might be seen as rude by them
@Huy the radius falls out if each $r$ is equal
since I don't really talk to them often
14:57
see the 4 ks on the left
Huy
Huy
@Danu: then you should try to be online at some time when Ted is, and then ask after you say hi :D
but if the midpoint of the circle is not on $g$
then the $r$s are not equal
Huy
Huy
ah
Maybe you can solve it @Huy? :P
because right now the line $XY$ is a radius
and so is $PX$
Huy
Huy
14:58
@Danu: are the limits different?
but if $X$ is not on $g$, $XY$ will no longer be the same as $PX$?
@Huy I was hoping that they're not, somehow
but I think they are
gtg
think about $f:\mathbb R\to \mathbb R; f(y)=x$

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