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Secret
Secret
4:32 AM
I wonder if bubble implosion is related to sonoluminescence
 
rob
rob
5:18 AM
@Secret I thought that sonoluminescence was all about imploding bubbles.
 
 
2 hours later…
Nehal Samee
Nehal Samee
7:05 AM
@JohnRennie ... How are the facilities of graduation for overseas students in UK ?
 
John Rennie
John Rennie
@NehalSamee I don't know
@NehalSamee What exactly do you mean by facilities of graduation?
What aspect of the UK universities are you asking about?
 
Nehal Samee
Nehal Samee
@JohnRennie I mean ... scholarships
 
John Rennie
John Rennie
@NehalSamee I suspect an overseas student would be very unlikely to get any form of scholarship
I think UK universities tend to charge around £20K a year for overseas students
 
Nehal Samee
Nehal Samee
@JohnRennie like ... Cambridge ?
 
Secret
Secret
@rob it is (for bubbles in solution under exposure to acoustic waves of high intensity), and it's exact mechanism is not determined yet. The bubble implosion they proposed here however is something more general and produce several MeV photon outputs
 
John Rennie
John Rennie
7:10 AM
 
Novalium Company
Novalium Company
7:25 AM
Morning.
 
Dawood ibn Kareem
Dawood ibn Kareem
That's an insane amount of money.
Dawood hastily cancels his plans to study medicine at Cambridge over the next six years
 
Captain Bohemian
Captain Bohemian
in my country, going to public university is a very cheap expense compared to most countries. So almost everyone in current generation here goes to university.
medicine college is excluded, because I have never been interested in going to medicine college.
 
Abcd
Abcd
7:46 AM
@Sid From JR's link did you calculate the total fees for Indian engineering student
 
Sid
Sid
@Abcd , 25 lakhs per year.
 
Abcd
Abcd
@Sid including personal expenses?
 
Sid
Sid
Yeah, if you can live on minimum requirements.
 
Abcd
Abcd
@Sid and whats the fees of the top college of India for 1 year?
 
Sid
Sid
@Abcd 2 lakhs per year(without personal expenses)
If you add personal expenses, then probably 2.5 lakhs.
 
Abcd
Abcd
7:51 AM
@Sid Such a big difference :/
 
Sid
Sid
Well, if you live in a developed nation, then obviously the cost of living is high
 
Slereah
Slereah
8:21 AM
I think we can estimate when the Stack Exchange chats were implemented
Because one of the error page has this :
 
Mithrandir24601
Mithrandir24601
8:40 AM
@Slereah Or you could just find the first chat room, then find the date of the first message (which was the 20th October 2010), which is left as an exercise for the reader :P
 
Slereah
Slereah
2010 is a bit late to still be using lolcats
for shame
 
Mithrandir24601
Mithrandir24601
Hmm... Turns out that the first chatroom didn't have the first message...
 
Akash. B
Akash. B
guys I have aquestion
how fireflies produce photons?
through black body radiation?
 
Mithrandir24601
Mithrandir24601
@Akash.B According to Wiki, some sort of chemical reaction
 
Slereah
Slereah
9:18 AM
 
 
1 hour later…
bolbteppa
bolbteppa
10:33 AM
Who needs gravitinos when you can have violins
 
John Rennie
John Rennie
10:47 AM
The opening introduction is good ...
 
bolbteppa
bolbteppa
Yeah it's cool
 
Mithrandir24601
Mithrandir24601
11:00 AM
In a similar theme note, I found a brilliant quote that I put on the description of the Classical Channel (QC chat) - “A classical computation is like a solo voice—one line of pure tones succeeding each other. A quantum computation is like a symphony—many lines of tones interfering with one another.” ― Seth Lloyd, Programming the Universe
 
Slereah
Slereah
11:18 AM
What happens when the wavefunction collapse
 
John Rennie
John Rennie
You wave smelling salts under its nose?
 
Slereah
Slereah
12:12 PM
Looking for metaphores in physics for too long is a bad idea
Modern physics is mostly unlike anything we encounter
It is metaphores-proof
 
lobotomized_sheep_99
lobotomized_sheep_99
12:23 PM
i have some basic qm question
does [p,x] = ihbar always hold?
or does it depend on the wavefunction, i.e. on the system
 
gateprep
gateprep
what are interstitial solids?
pls help
lookws up on wikipedia though
want the answer in simple language
@Lozansky
@lobotomized_sheep_99
pls answer
 
Secret
Secret
1:09 PM
There is no such thing as topological disconstitutionalism
there also does not exists anything call Megan rescronomism
Most importantly:
$$\int^{\int^{\int}}IIT(x)dc =0$$
 
Slereah
Slereah
"Famous like Sean Carroll and Lubos Motl" is tantamount to "famous like Led Zeppelin and the rock band I was in when I was fourteen". — AccidentalFourierTransform 19 hours ago
My sides
 
Secret
Secret
and therefore. If the establisharanism want to put a break on the sconstitutionalisalioation, clearly Led Zeppelin's Zeppelin does not work
This is perfect opportunity because it does not exists
no longer exists
and will not exists
trouble is the stargate will not listen to our commands, thus calling that unsoliciated is extremely approporiatr
but clearly, they do not understand, for they only have the evolutionary stage of bacteria. no.. even calling that bacteria is a massive insult to bacteria as a highly noble species
and therefore, whatever nisha90's advice clearly does nothing to that stubborn stargate in order to force the destination coordinates to stablise
which is an absolute disgrace to that famous franchise
That wormhole is mot terribly stable either
Motl will be very proud when his LQG rivals all drop into the black hole and become one with the singularitu
Anyway...
What is the current status of the penetration...?"
::Still 8 hours to go::
Great, keep that going, and we will take over from that point
It's time to teach that unfairness a lesson...
What's the current status of the retraction
::At 30% progress. Madam::
Good, good. Continue the monitoting
We have been preparing this for almost a decade
Nobody will stop us from changing the rules forever
 
Semiclassical
Semiclassical
1:34 PM
@Slereah comparing Sean Carroll to Led Zeppelin is weird to me
 
Slereah
Slereah
yeah he's not quite that famous, either
 
Semiclassical
Semiclassical
right
Not sure what band would have the right level of fame for a good comparison tho
there's also the complication of 'internet famous'
Both Carroll and Motl are prominent in the online domain, but offline they're not anywhere near as influential.
 
Slereah
Slereah
He is the Jimmy Buffet of physics
 
Semiclassical
Semiclassical
lol, that'll do
 
Slereah
Slereah
Led Zeppelin would be more like
Kaku I guess
 
Semiclassical
Semiclassical
1:40 PM
eh, I'd put Led Zeppelin at the level of Hawking
 
Mithrandir24601
Mithrandir24601
@Semiclassical Umm, Carroll's published a lot, he's a professor at Caltech, he's written a rather well known textbook on GR. Motl hasn't published anything in over a decade and is known for essentially being forced to leave Harvard. I wouldn't say that Motl is anywhere near as famous as Carroll
 
Semiclassical
Semiclassical
Hmm, fair enough.
I'd still say that, in both cases, their prominence on the internet isn't proportionate to their influence offline. But the contrast is far sharper for Motl than it is for Carroll.
 
Slereah
Slereah
Hawking is the Beatles
Obviously
 
Semiclassical
Semiclassical
hah
What does that make Penrose?
The Moody Blues?
 
Slereah
Slereah
Einstein is Elvis
 
Semiclassical
Semiclassical
1:48 PM
Makes sense.
 
Mithrandir24601
Mithrandir24601
@Semiclassical Seems very possible - I do neither GR or ST, so the above is the extent of my knowledge :P
 
Semiclassical
Semiclassical
lol
 
Slereah
Slereah
Me I am the guy putting GUITAR LESSONS fliers at your local bakery
 
bolbteppa
bolbteppa
I'd say Einstein is the Beatles, and Hawking is Michael Jackson, Zeppelin maybe aren't that well known to non-music fans!
 
Semiclassical
Semiclassical
uh
 
bolbteppa
bolbteppa
1:56 PM
Or Elvis yeah
 
Semiclassical
Semiclassical
p sure if you know classic rock then you know Led Zeppelin
and while not everyone is a classic rock person, that's a pretty solid segment
 
bolbteppa
bolbteppa
Maybe Penrose is Zeppelin :p
 
Lozansky
Lozansky
Euler is Mozart
 
bolbteppa
bolbteppa
If you know enough about music/physics to know the categories and some examples from each, he'd be a name that would come up quick enough
Gauss is Beethoven
 
Lozansky
Lozansky
Oh I guess Euler is Beethoven
 
bolbteppa
bolbteppa
2:00 PM
Who nearly went blind
 
Lozansky
Lozansky
Euler
If that was a question lol
 
bolbteppa
bolbteppa
Mozart was born before though :p
 
Lozansky
Lozansky
Does ordering matter? :P
 
Semiclassical
Semiclassical
@bolbteppa I feel like that's a good comparison
 
bolbteppa
bolbteppa
There's a lot of (intellectual) inbreeding amongst these guys
Academic genealogy of theoretical physicists
The following is an academic genealogy of theoretical physicists and is constructed by following the pedigree of thesis advisors. If an advisor did not exist, or if the field of physics is unrelated, an academic genealogical link can be constructed by using the university from which the theoretical physicist graduated. An academic genealogy tree lists the physicists' PhD (or in some cases BA/MA) date and school, if known. Nobel Prize winners are indicated by †. If physicists are advised by mathematicians, their genealogy can be readily traced using the Mathematics Genealogy Project. For the meaning...
You can build direct chains of advisors between them, I had no idea e.g. Gauss had a phd etc, thought they all were just magician geniuses
 
Lozansky
Lozansky
2:17 PM
What's the physical reasoning behind having a wave function $\neq 0$ in the region of a potential barrier $U>E$ whilst at the same time total reflection occurs?
It's easy to show mathematically
But it still feels a bit unintuitive to me
 
Semiclassical
Semiclassical
have you seen frustrated total internal reflection?
 
Lozansky
Lozansky
I think so
Some sort of tunneling phenomenon
But I don't remember the details
 
Secret
Secret
details in the page
 
Semiclassical
Semiclassical
suppose you've got a wave incident on an interface, e.g. light passing through glass and reaching an edge with air outside
 
Lozansky
Lozansky
With another medium inside the glass?
 
Semiclassical
Semiclassical
2:22 PM
nah
 
Lozansky
Lozansky
OK
 
Semiclassical
Semiclassical
just glass-air
 
Lozansky
Lozansky
Sure
 
Semiclassical
Semiclassical
if you have a steep enough angle of incidence, you'll get total internal reflection and no transmission
 
Lozansky
Lozansky
Yep
 
Semiclassical
Semiclassical
2:23 PM
however: if you put another piece of glass close enough nearby, then you can recover that transmitted ray
like that
the significance is that, when you write down the appropriate waveforms
you'll find that the waveform is nonzero outside of the glass. it's of the form $e^{-\lambda x}$, i.e. exponentially decaying. (maybe it's really $e^{-\lambda x}\sin kx$, I forget)
 
Lozansky
Lozansky
It's the first
 
Semiclassical
Semiclassical
ok
main point is that it's not strictly zero outside of the glass. but it's so small once you go any appreciable distance away from the glass that it can usually be ignored.
 
Lozansky
Lozansky
I guess I find it a bit troubling that $\lambda$ is real
 
Secret
Secret
 
Lozansky
Lozansky
So it's not really a wave, is it?
 
Semiclassical
Semiclassical
2:30 PM
@Secret No.
 
Secret
Secret
isn't that getting dimmer as it go further?
as expected for an evanescent wave?
 
Semiclassical
Semiclassical
the decay I'm talking about is in the air gap between the two pieces of glass.
 
Secret
Secret
O, oops
 
Semiclassical
Semiclassical
what you're seeing there is presumably just the absorption of light by the prism
it's not as visible in the first piece of glass, because the intensity is so large initially
@Lozansky does it satisfy the wave equation?
 
Lozansky
Lozansky
Yes of course
 
Semiclassical
Semiclassical
2:32 PM
Then it's a wave. It might usually not be the relevant waveform for usual boundary conditions, but it is a wave
 
Lozansky
Lozansky
Yes but it gives an imaginary wave number (and therefore momentum)?
Inside the barrier
Which is expected I guess since the kinetic energy is negative?
 
Semiclassical
Semiclassical
Right.
So in principle there's a finite probability of finding a quantum particle at a classically forbidden location.
Of course, finite can still be very small
 
Lozansky
Lozansky
Yes but the reflection is identically one?
 
Semiclassical
Semiclassical
Yes, but: Suppose you had the barrier only extending a finite distance
Then the transmission coefficient would be nonzero.
 
Lozansky
Lozansky
That is true
 
Semiclassical
Semiclassical
2:36 PM
The point is that, in order for there to be transmission, the particle needs to get all the way through the barrier i.e. to +infty
 
Lozansky
Lozansky
Okay yeah that makes a lot of sense
I feel dumb now lol
 
Semiclassical
Semiclassical
eh, QM takes getting used to
plus I'm not sure how practical the following thought experiment is:
Suppose you've got a particle in a double-well potential, with total energy below the barrier
in that case, the particle would be classically trapped in one of the two wells. but in QM the particle is able to tunnel between the two wells
you could, in principle, ask for the probability of finding the particle within the barrier. this would be small but finite.
I'm not sure how practical a thought experiment this is, though, since I'm not sure how you'd measure the position of a particle at a given time in this setting
 
Lozansky
Lozansky
Maybe you could measure the flux of escaping particles if the barriers are bombarded at extremely high rates
 
Semiclassical
Semiclassical
maybe? I'm not going to pretend to have a definite answer
typically one confines the question to the simpler case of: does an incident particle transmit to +infty, or does it get reflected back to -infty?
 
Lozansky
Lozansky
I'm gonna preface every answer on my exam like that:P
 
Semiclassical
Semiclassical
2:44 PM
lol
 
Lozansky
Lozansky
Especially if there is a question on bonding/antibonding states
 
Semiclassical
Semiclassical
hah
 
Lozansky
Lozansky
That is just.. confusing
 
Semiclassical
Semiclassical
one thing to also note is that the usual potentials in QM are real
which has the implication that they're not acting as absorbers or emitters
i.e. if a particle comes in then it must either reflect or transmit
non-real potentials are used if you want to include the possibility that the particle can be absorbed by the barrier, or if the barrier can spontaneously emit a particle
but that's even more confusing and I don't deal with it
 
Lozansky
Lozansky
Sounds messed up
 
Semiclassical
Semiclassical
2:47 PM
I think you may need to do it if you want to model inelastic scattering
 
Lozansky
Lozansky
What is the mathematical condition for absorption?
 
Semiclassical
Semiclassical
reflection + transmission = 1, I'd say
if a particle is incident on a barrier, then eventually the particle will either bounce back out or transmit through
 
Lozansky
Lozansky
Then it isn't absorbed?
 
ACuriousMind
ACuriousMind
@EmilioPisanty I'm afraid I can't add the recommendation banner to the post you bountied because the system considers the bounty notice a banner and won't let me add another one. I'm searching for a way to circumvent this, but it doesn't look good.
 
Semiclassical
Semiclassical
right
 
Lozansky
Lozansky
2:52 PM
You mean $R+T <1$?
 
Semiclassical
Semiclassical
@Lozansky for absorption, yes
 
Lozansky
Lozansky
Mhm
I'm thinking if this could be modelled in Python
Not that I have the time for it now but
 
Semiclassical
Semiclassical
though the tricky thing is if you had both emission and absorption
 
ACuriousMind
ACuriousMind
Lesson: Next time only bounty after the banner has been added, I guess
 
Lozansky
Lozansky
Could maybe split the cases and then superposition them :P
 
Semiclassical
Semiclassical
2:53 PM
since if you had emission at the same rate as absorption then you might assume that neither are present
no idea
 
Lozansky
Lozansky
Fair enough
 
Semiclassical
Semiclassical
@Lozansky I think you could definitely do some calculations in this vein
 
Mithrandir24601
Mithrandir24601
@Semiclassical emission + absorption makes more sense than just absorption - use field theory ideas as per quantum optics where it's represented by creation and annihilation operators might be an idea. An alternative (in context) would be to model the material as a lattice of 2-level systems interacting with the magnetic vector potential, A. It depends on exactly what you want to do... Normally, on a more abstract level, absorption is an exponential thing
 
Semiclassical
Semiclassical
@Mithrandir24601 nice.
i had in mind something like an alpha particle scattering off of a nucleus
with the possibility of the nucleus absorbing the alpha particle to become a different nucleus
but that seems like it could get confusing fast
 
Mithrandir24601
Mithrandir24601
@Semiclassical Sounds like scattering theory? I think I heard something similar in the context of PT-symmetry, but this would have been a few months ago, so I could very easily be misremembering stuff
 
Semiclassical
Semiclassical
2:59 PM
sure.
 
Mithrandir24601
Mithrandir24601
(as in, S-matrix stuff)
 
Semiclassical
Semiclassical
i also have pt-symmetry in the back of my mind
typically that shows up in systems where there's a notion of balanced gain-loss
But open quantum systems are confusing
 
Mithrandir24601
Mithrandir24601
@Semiclassical You can say that again...
 
 
1 hour later…
Slereah
Slereah
4:04 PM
Causal boundary theory is so awful
The hell is a future hull lattice
They talk about maximally naked points
Can't get nakeder
 
Secret
Secret
It will be more interesting if there is future null lattice
There world is terribly lack of null
 
Emilio Pisanty
Emilio Pisanty
4:20 PM
@ACuriousMind what was the solution?
 
ACuriousMind
ACuriousMind
@EmilioPisanty Another mod not being as afraid as I was of accidentally removing the bounty and clicking on "remove post notice" :P
That just didn't do anything except apparently allow to add an actual notice
 
Emilio Pisanty
Emilio Pisanty
@ACuriousMind well, it removed the bounty =P
 
ACuriousMind
ACuriousMind
Wait, I just looked at that question a few minutes ago and saw the bounty and the notice there!
@EmilioPisanty Also, it didn't remove the bounty, just the notice, right? You awarded the bounty.
 
Emilio Pisanty
Emilio Pisanty
@ACuriousMind =P
Just kidding
 
ACuriousMind
ACuriousMind
Oh, you... :D
 
Emilio Pisanty
Emilio Pisanty
4:33 PM
It worked fine
I awarded the bounty
If I'd gotten your message in time I would've told you to hold on a bit and awarded it directly
 
rob
rob
You guys and your hijinks amuse me.
 
Secret
Secret
4:51 PM
I hope someone can install something so that next time gateprep ask a question, it get automatically moved into the problem solving room
otherwise, having his questions here is a massive category error
 
ACuriousMind
ACuriousMind
@Secret Discussing questions of any kind in here is not forbidden. The problem solving room is the preferred location for lengthy discussions about solutions of exercises, but that doesn't mean we have some sort of blanket ban on such discussion here. If you have a problem with what someone is posting here, please try to take that up with them directly and respectfully.
2
 
Secret
Secret
@gateprep I will say this one last time: Be patient and if someone want to answer, they will, stop multi pinging people
 
Slereah
Slereah
5:08 PM
"Quantum Gauge Theories: A True Ghost Story"
Good title
 
Semiclassical
Semiclassical
@Slereah "Quantum Hauntology"
 
ACuriousMind
ACuriousMind
Apr 30 at 17:40, by ACuriousMind
hauntology is where you chase the ghosts, cohauntology is where the ghosts chase you
 
Slereah
Slereah
So for the first homology group, I can see how the group is done
The group product is the composition of curves and such
But how does it work for higher homology groups
How do you do a composition of spheres like that
I assume it involves the boundaries of the simplicial decomposition of it or something
 
bolbteppa
bolbteppa
5:28 PM
Homotopy groups are composition of curves no?
 
Semiclassical
Semiclassical
First homotopy group
 
ACuriousMind
ACuriousMind
@Slereah Since you're talking about spheres, do you really mean homology or rather homotopy?
 
Slereah
Slereah
@ACuriousMind Aren't they related?
 
ACuriousMind
ACuriousMind
@Slereah Yes, but they are far from the same.
The precise relation is given by Hurewicz
 
Slereah
Slereah
I assume homotopy, then
How do you take the product of two spheres in $\pi_2$?
it's easy enough to see for $\pi_1$ since curves have a beginning and an end
but for spheres it's trickier
Do you just identify some edges of the simplicial decomposition?
 
ACuriousMind
ACuriousMind
5:36 PM
@Slereah Well, intuitively, you do the same thing as for the loops in $\pi_1$: Formally, you also do the same thing as for loops, but just in the first component.
 
Semiclassical
Semiclassical
In the case of first homotopy/homology, it's just abelianization
but all higher homotopy groups are abelian, so that's not enough in the higher cases
 
Slereah
Slereah
@ACuriousMind That's not very intuitive!
 
ACuriousMind
ACuriousMind
You can define the composition of two loops $f ,g: [0,1] \to X$ by $(f\cdot g)(t) = f(2t)$ if $t < 1/2$ and $g(2t - 1)$ otherwise.
And you define the composition of two maps $[0,1]^n \to X$ by simply doing that in the first component
 
Balarka Sen
Balarka Sen
If you have two (pointed) maps $f : S^n \to X$ and $g : S^n \to X$ consider the map $S^n \vee S^b \to X$ given by $f$ on the first component and $g$ on the second component of the wedge. Then precompose with $S^n \to S^n \vee S^n$ given by squashing the equator $S^{n-1} \subset S^n$
That's the product.
 
Slereah
Slereah
@ACuriousMind So two spheres composed together are just linked at a single point?
 
ACuriousMind
ACuriousMind
5:40 PM
@Slereah Well, the image of the product map is just the union of the images of the individual maps. Whether the spheres that are the images only touch at the base point or intersect depends on the specific spheres
 
Slereah
Slereah
@ACuriousMind Sure, but in every case, they touch in at least one point
and in some cases only one
is that correct
 
Balarka Sen
Balarka Sen
Correct, that's the basepoint of the homotopy group.
 
Slereah
Slereah
alright, thanks
 
ACuriousMind
ACuriousMind
Yes - you require them all to have a base point in the definition of the homotopy group, after all
 
Slereah
Slereah
And... what's the inverse of a sphere
Just the sphere with the coordinates backward?
 
Balarka Sen
Balarka Sen
5:41 PM
Antipodal map composed with it, yes.
Higher homology groups have a similar interpretation, but it's more complicated. If $M$ is a manifold, $H_n(M)$ is the set of all maps $X^n \to M$ from $n$-pseudomanifolds $X$ modded out by $(n+1)$-pseudomanifold cobordism. The group structure comes from just taking two maps $X^n \to M$ and $Y^n \to M$ from two pseudomanifolds and considering the disjoint union $X^n \sqcup Y^n \to M$.
 
ACuriousMind
ACuriousMind
@Slereah The elements of the homotopy group are not spheres, they are maps from the n-sphere to the space. You should not think of them merely as their images - the image of any element and its inverse are exactly the same.
 
Slereah
Slereah
I am aware, yes
 
Balarka Sen
Balarka Sen
So in a sense higher homology groups are like higher homotopy groups, but you probe your space with much more than just spheres.
You probe it with pseudomanifolds.
 
Slereah
Slereah
Errr, is pseudomanifolds the ones with degenerate points
 
Balarka Sen
Balarka Sen
Manifold away from a codimension > 1 singular locus
 
CooperCape
CooperCape
5:49 PM
ohai
 
Slereah
Slereah
6:27 PM
also btw
when people say homology and cohomology wrt manifolds
Do they mean simplicial homology and de rham cohomology
or are there several that apply there as well
 
ACuriousMind
ACuriousMind
@Slereah They...mean whatever they mean. All of singular, simplicial, cellular and deRham (co)homology work on smooth manifolds
 
Slereah
Slereah
Are they equivalent in some way
 
ACuriousMind
ACuriousMind
Yes - all these (co)homologies are isomorphic when they exist.
 
Slereah
Slereah
Alright
 
Slereah
Slereah
6:49 PM
Hm, i think the common past $\downarrow U$ is just $\cap_{p \in U} I^-(p)$
 
Abcd
Abcd
7:02 PM
Just a conceptual question, not a hw problem:
 
Slereah
Slereah
Looks like that rectangle moved
and changed color
 
Abcd
Abcd
If I pull the greenish blue block towards right, pause and then release, will the light blue block move towards right too?
Its obvious that the greensih blue block will move towards left
PS: Arrow denotes spring.
 
Slereah
Slereah
That greenish blue is turquoise, you philistine
Or is it cyan
i'm not sure
 
ACuriousMind
ACuriousMind
Depends on whether the force the spring exerts on it is enough to overcome static friction or not.
 
Abcd
Abcd
Assume that the surface is smooth (= frictionless) @ACM
 
ACuriousMind
ACuriousMind
7:05 PM
If by smooth you mean "assume no friction", then yes, the block will move
 
Abcd
Abcd
Why?
 
ACuriousMind
ACuriousMind
Because the spring, being extended by you pulling the other block away, will exert a force on it.
 
Abcd
Abcd
But that force will be on greenish blue block
and not the light blue one
 
ACuriousMind
ACuriousMind
Why would it? The spring will try to shorten, exerting a force on both blocks
 
Abcd
Abcd
How would force on light blue block even develop?
 
ACuriousMind
ACuriousMind
7:09 PM
As I said, since you're extending the spring, it will exert its spring force on both blocks. Why would the spring only exert a force at one end but not the other?
 
Abcd
Abcd
in Problem Solving Strategies, 7 mins ago, by Blue
The left portion is still intact and hasn't elongated yet
in Problem Solving Strategies, 6 mins ago, by Blue
In that situation if you release the right block then the left block should not move
 
ACuriousMind
ACuriousMind
What, we're assuming no friction but we are considering the time the extension takes to propagate through the string?
That strikes me as a rather inconsistent way of abstracting the situation
 
Abcd
Abcd
But spring force is always opposite to the direction of extension @ACuriousMind
SO spring force is towards left
 
ACuriousMind
ACuriousMind
No, that's not how springs work
 
Abcd
Abcd
:( :"( :/
 
ACuriousMind
ACuriousMind
7:14 PM
When the spring is extended beyond its equilibrium length, it tries to contract, exerting a force on both its ends towards its center
 
Abcd
Abcd
I see.
Thanks.
 
Lozansky
Lozansky
Newton III?
 
Abcd
Abcd
@ACuriousMind What if I had shifted the cyan block towards left?
Pause. Release.
Will the blue block again move (towards right?)?
 
ACuriousMind
ACuriousMind
I don't see why it would matter which block you move, unless there is a difference between them you have not yet mentioned
 
Abcd
Abcd
@ACuriousMind In case 1 also we had shifted the cyan block, again we are shifting the cyan block but in the opposite direction i.e. left
 
ACuriousMind
ACuriousMind
7:20 PM
Oh, sure, you're compressing the spring, so the spring will again exert a force on both blocks
(I didn't really pay any attention to the colors because the blocks seem to be interchangable :P)
 
Abcd
Abcd
So will the light blue block move towards right or not?
 
ACuriousMind
ACuriousMind
Why would it move towards the right?
I suggest you get an actual spring and play around with it a bit
 
bolbteppa
bolbteppa
8:15 PM
Superstring theory is weird and cool
Of course you get two versions, as to whether you want superpartners for the $\sigma^a$ coordinates or for the fields $X^{\mu}$
Even just a few pages of GSW has required a lot of susy and of course not susy in most susy books which is 4-D susy but 2-D susy
 
bolbteppa
bolbteppa
8:38 PM
The real madness, however, is pdf page 53-54 of this theory.tifr.res.in/~mukhi/Physics/mukhi-stringtheory.pdf which seems to say the bosonic string theory action with general relativity $R$ incorporated is such that because the lowest massless excitations contain a graviton one can expand $g_{\mu} = \eta_{\mu \nu} + h_{\mu \nu}$ and throw away all above-quadratic terms in $h$ and still describe gravity...?
 
Slereah
Slereah
down voter please undo!!!! — user192234 1 min ago
UNDOOOO
 
bolbteppa
bolbteppa
Great question
Who wouldn't want a Zee SUSTRI Nut(?)
sustri is no good as an abbreviation
 
Semiclassical
Semiclassical
8:55 PM
SUS$\Delta$
 
bolbteppa
bolbteppa
SuΔ
This is why it's so group-y
 
skull
skull
you may find this interesting @Semiclassical though incredibly long
 
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