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5:00 PM
I had it ;)
@PhyMan Could you help?
 
yeah no probs
 
Could you tell how angle affects tension?
 
realise that the horizontal component of tension is same throughout the rope
 
@AccidentalFourierTransform I'm horribly afraid that if you did me you'd have to write something like "[something abuot experiments with htree to siz typos per paragraph]".
 
@PhyMan Yes.
 
5:06 PM
@SwapnilDas Since the horizontale components are the same, you can use that to find the tension at the end points using the angles
 
yes..
Do we compare then?
 
@HritikNarayan His class was fun—and he brought cookies and milk to the final—but I wasn't mathematically sophisticated enough get the full effect.
 
@SwapnilDas And also, you can divide the rope into 2 parts,AC and BC and draw the FBD for each of them
 
Hmm yes.
C is COM?
 
5:10 PM
the bottommost point
@SwapnilDas No, the one in figure
They willhave their own weights, horizontal tension at bottom and the tension at top
@SwapnilDas No, the one in figure
 
ohk.
Is there any qualitative explanation to this problem?
@AccidentalFourierTransform Using this property?
^^ That proportionality thing, It gave me a feel.
 
@SwapnilDas It is quite qualitative :/
 
Is it correct?
 
Intuitively, since A is higher, it would feel a higher force since it is pulling a larger amount of rope upwards
 
@JohnRennie Ahem.
 
5:16 PM
@PhyMan Hmm yes.
 
What?:-)
 
Last time I checked, "scumbag" was not a nice thing to say about another user.
 
@JohnRennie I agree. What I don't understand is why this wasn't so obvious from the start, i.e. at the time when people could have said "you know, train wrecks are kinda bad, let's not put the train on the wreck track"
so, like I said, I'm completely mystified about what happened to Britain after I left
 
There is a hard core who are quite unconcerned about the bad effects (economic and social) that will result from Brexit. They're mostly people who are financially secure and not fussed about anyone else.
 
@JohnRennie yeah, but surely that doesn't add to 17 million people, right?
 
5:22 PM
If you add together the group I've just described with the group who simply don't understand the consequences of their actions I reckon that comes to 17 million.
 
that I can believe
I would star that but let's keep politics out of starboard
btw @JohnRennie you got anything on this one?
3
Q: Can gravitational wave induce electromagnetic waves?

voidMoving charged particle produces a magnetic field. If a gravitational wave passes by stationary charged particle and makes it oscillate together with the space, would that movement produce an EM wave?

 
@EmilioPisanty that's a tough one. It's like the question whether a charge falling freely under a gravitational force emits radiation. In both cases the proper acceleration of the charge is zero.
 
'cause I reckon that if you put a quadrupolar charge distribution on the dots of the standard image
they'd produce quadrupole EM radiation
(right?)
@JohnRennie but then, yeah, precisely that
 
Yes, if you arranged the charges like that and made them accelerate, e.g. with an external field, then they would emit EM radiation. But in that case the proper acceleration of the charges wouldn't be zero.
I saw the question, thought about it, and decided I simply didn't know the answer.
 
@JohnRennie yeah, but if I'm sat over to the right, at a location that's close enough to the ring that the roundtrip between the ring and me is shorter than the oscillation, the distance between me and the bits of the ring would change, right?
which means that the electric field I experience would also change, no?
 
5:30 PM
Well no, because you'd move as well in response to the change in the spacetime geometry
 
@JohnRennie yeah, but there's no way that the distances between me and every particle in the ring will stay constant, right?
that sounds like a vastly overconstrained system
I mean, the distance between the north and south poles of the ring oscillates, right?
 
Agreed. But I would have to sit down and think about exactly what is going on, and it's too late in the day for that. I started work 13 hours ago and now just want to crash.
 
@JohnRennie yeah, agreed
 
A bit like the British economy will in two years :-)
 
@JohnRennie d'you reckon it'll take that long for the uncertainty to set in?
I mean, tariffs are pretty bad for an economy
but uncertainty is worse
 
5:34 PM
It's already started. Growth is down and inflation is up.
 
@JohnRennie that's pretty bad
 
The UK had the highest growth in the EU and now it's slipped down to, erm, not sure but well behind the other big EU economies.
 
is that real-economy growth down, or mostly just the City taking the hit?
 
Don't know offhand - I'm trying not to think about it :-)
In fact I'd be pretty well insulated as I'm mostly retired and if the company I work part time for folded it would sting a bit but not in a life changing way.
But that doesn't mean I'm happy about the UK becoming a shittier place.
 
@JohnRennie yeah, it's that
though the City getting taken down a notch in preeminence within the UK economy, in favour of people who produce stuff instead of move money around, wouldn't be all that bad
 
5:42 PM
@EmilioPisanty It's certainly true that the size of the City has unbalanced the economy to a considerable extent. The trouble is that the UK manufacturing industry is in a poor state. We have the lowest productivity of all the big EU economies.
 
@JohnRennie I agree that that's a problem
 
If the City said thanks, but no thanks and moved to Hamburg there's no way manufacturing could pick up the slack.
It would be pretty disastrous.
 
agreed
 
And lots of the big manufacturing companies are multinationals and would probably move too. There's already talk of Airbus and Rolls Royce moving manufacturing to the continent.
No, it's an all round rubbish state of affairs.
 
@dmckee Hahahaha
 
5:46 PM
@JohnRennie yeah, please stop
 
@EmilioPisanty time to go back to my book I think. And maybe a glass of something cool and refreshing to go with it :-)
 
@JohnRennie what, is it blisteringly hot over there?
it was 28°C at 2am here a couple of nights ago
there's no escape
 
@EmilioPisanty That's sad
@EmilioPisanty Would you rather have that or rain all day, though?
 
Zee
is three hours of physics a day too little for a grad student?
 
@EmilioPisanty 28? What a nice cold Winter night :)
 
5:55 PM
@Zee I'm not a physics student and I end up spending that much time, so I don't know :P
 
@0celo7 The 1080.
 
Zee
@HritikNarayan do you feel exhausted after that time? Because I do, and am wondering if am working hard enough
 
@Zee I usually enjoy it too much to notice but it is mentally draining at times. It's different for different people, I guess! There's no metric as such. What are you working on, though? :)
 
Anonymous
@Zee I guess 3 hours of physics per day is too less even for a high school student. My opinion though. :) I guess if you utilize the remaining time productively then it doesn't matter.
 
Anonymous
Anyhow, I don't think you should be counting "hours" if you enjoy what you are studying.
 
6:10 PM
Obviously the other hours will be going on math for physics ;)
 
Anonymous
@bolbteppa lol, yeah I agree. Maths is the DEVIL. :P
 
Anonymous
I started learning Classical Mechanics and realized that I have to study Multivariable Calculus thoroughly or else I won't be able to proceed!
 
[Portal within a portal]
The way to think about it (ignoring GR effects, I don't know how to put a metric in here...) is to consider the two mouths are identified by some map
Suppose in mouth 1 I have points labelled by postions (x1,y1,z1) and in mouth 2 I have points labelled (x2,y2,z2)
We can identify them by doing x1=x2, y1=y2, z1=z2
That means, anything that touches a point x1,y1,z2 on the spherical surface of one of the mouth, must be mapped to its corresponding position on another mouth
Now:
As we move the mouths closer to brought them to intersect
one portion of mouth 2 will touch and then penetrate into part of mouth 1
But we have said earlier that as soon anything touches one of the mouths, it must map into the corresponding position of the other
So when the left side of the blue mouth enters the right side of the red mouth, that portion has to be mapped to the right side of the blue mouth
and and the same time, the right portion of the red mouth will get mapped to the left portion of the red motuh since it enters the right potion of the blue mouth
So as the portals continue to move closer, the mapped hemispheres continues to grow.
Now at this point, one interesting question to ask is, what happens when you enter one of these hemipsheres. well we can always follow the mapping. for example
1. Suppose I enter the blue hemisphere that is produced by the two portals intersecting on the right.
2. If the portals are not intersecting, you will expect to emerge in the right end of the red mouth.
3. Howeever, because the portals intersect, the right end of the red motuh is instantly mapped to the left end of the blue mouth, which because of the intersection, instantly mapped to the left end of the red mouth.
4. Therefore the overall result is that if you enter that blue hemisphere, you must emerge at the red and vise versa
Now, at the limiting case where the two portals coincide, the two hemispheres grew completely and enveloped the region where the mouths shoudl be, so you end up with something that look like a dipole,
where if you enter the red, you will emerge at the antipodal location of the blue
Now at this point, the portals can be pushed no further, because when you follow each point with the mappings, you found they are all fixed points
in particular, the centre of the portals must map to the centre of the other, so when they collide, they can go no further as thy are both mapped to the same spot
The fixed point of this mapping thus (expected to) create a physical effect as if two rigid bodies bump into each other, like a ball hitting a wall and can go no further
 
6:30 PM
@JohnRennie Not sure that's very nice.
 
Well, that's how classical portals should work ( I think....), I am not sure if we put actual wormholes in there, there will be proeblems
@JohnRennie That's what I tried to say many hours ago regarding the wormhole discussion. If GR does not allow such argument, I am not sure how exactly it will fail other than perhaps the spacetime in the vicinity of the two mouths will become so warped that it lead to divergences...
If we instead have the two mouths sliding past each other like so, then we should expect regions I and IV to grow with their intersections and the shrunk again after they passes by each other
 
@DanielSank I'd agree that I don't quite see how that's Nice. @JohnRennie, I can edit that for you, if you can give me an alternative wording. (A wording I chose probably wouldn't look right beside someone else's name.)
 
@SevenSidedDie "This ganger-schlamper!"
 
6:46 PM
If I have a semiconductor on top of an oxide on top of a metallic gate, and I apply a voltage V1 to the gate, what can I then say about the voltage at the semiconductor-oxide interface, V2? Given the thickness of the layers, and the dielectric constants
I thought that the dielectric would simply translate this voltage with some linear proportionality, but perhaps the derivatives also come in
(oxide being the dielectric)
 
7:09 PM
Nice
Experimental results of the Pound-Rebka experiment
 
@SevenSidedDie maybe scumbag is worse in American slang. On the UK side of the pond it's a cartoon insult i.e. not one that would be taken seriously. If it is more offensive than I thought please just delete it.
 
@JohnRennie It's somewhat a cartoon insult in NA too. It's just that being just joking doesn't really help. (And the user is being a bit of a pain there, so it's easy enough to read as serious.)
 
@SevenSidedDie Hmm, well, I'm not sure I see it as offensive but in the interests of world peace I'm happy for it to be deleted.
 
7:36 PM
@BernardoMeurer if you're giving me a windows key I need it today so I can get the .iso
 
8:28 PM
how do you calculate the information density of a system?
 
@heather What do you mean by "information density"?
 
well, i'm reading papers on different systems of encoding information in dna, and they all refer to their encoding system's information density.
so...how much information it gets across, in how much space, i suppose, as defined by shannon, i think.
 
@heather I suspect that they just mean the Shannon entropy of their system.
 
okay...let me look at that wikipedia article.
oh, dear, this seems to be well over my head.
 
@heather Well, what are you reading these papers for? Do you really need to understand the technical meaning of information density for that? Also note that that this sort of information/entropy is often not uniquely determined and depends e.g. for languages/string on what you consider the "letters"
 
8:43 PM
i am interested in being able to understand how to calculate the information density of an encoding system, yes.
from what i can tell, the equation seems to be dependent on the "probability" of a message, though I don't know what that means, really - what's the probability of someone saying "Hello world!"? how do you even calculate that?
 
Well, I'm not an expert, but for a start, I think there is no such thing as "information density of an encoding system"
What you can compute is the entropy of a given message in some encoding, but not that of the "encoding system"
 
9:07 PM
@heather That's kind of the point. You have to have some model that gives a probability distribution over the possible messages.
Given the probability distribution, you can calculate the average information content of a message chosen according to that distribution. The definition of information content is such that a uniform distribution over $2^n$ possible messages corresponds to $n$ bits of information (which should be intuitive).
Then the information density is simply the ratio of information content to length of the message. You can calculate this in various ways, like the average information density for all possible messages of a given length, or the asymptotic average information density for long messages, or so on.
As an example, it's been estimated (by Shannon I think?) that English text has an information density of something like 1.2 bits per character.
 
9:25 PM
o/
 
@DavidZ there's also meant to be a tight anticorrelation between the information density of a language and how fast its speakers speak
I.e. so in most languages the information density of the spoken language is about constant
But then that does sound like the kind of slightly-too-convenient fact that one should look up before parroting further
 
@Danu \o
 
@BalarkaSen you wouldn't happen to know how to work with multiplicative sequences, would you?
 
Multiplicative spectral sequences?
 
No, polynomial sequences
a la Hirzebruch
Signature theorem and such
 
9:39 PM
I don't think so.
 
:(
 
@ACuriousMind bits per unit volume
 
Any time you see a newspaper headline in the form of a question, you know the answer is "no". Any time you see a Stack Exchange question beginning "Is it ethical to", you know the question has nothing to do with ethics. — David Richerby yesterday
that's some immortal words there
 

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