Hungary

I've not thought about that.

@Mew I will have to think about this.

forbes.com/sites/victorhwang/2014/10/21/uber-will-lower-gdp/…

Economics is applied maths

oh that's a shgame

@MAFIA36790 He claims to want to be a nuclear engineer, that's pretty applied :P

<link to Purity xkcd>

:P

@BernardMeurer I'm open about it. Any particular reason you don't want to discuss it in public?

Caption: Turning the isomorphism class members into something that resemble Pigpen cipher

Jewish math?

I'm pretty sure that's a blueprint from Contact

I've seen that code before

@Secret what are you up to these days, anyway

MAFIA: lol, I just simplified all a s and b s into stick diagrams that only highlight their connectivity in the tables. 0celo7: Still group theory, halfway through covering groups

No, I mean are you doing grad school, or what?

I have my PhD proposals submitted, but results will not be out until December

Good luck.

My honours graduation will take place next week. Other than that, I sometimes blame myself for procrastinating from getting a driver license too much

What are you doing phd in?

Time?

Huh. I guess Australia is backwards.

no i meant are you doing your phd in time

time research

like joan

damn people at work are always taking my Swingline stapler and never returning it. That's my stapler

it's milton. From Office Space. I thought for sure more people would know that movie

Joan's research interested me because like most, I wonder about the nature of time and try to better understand it

@Secret this isn't the time to talk about time, we just don't have the time

@Jim Well its obvious, unless we cause it to go loop de loop

Can we please stop starring PO sales pitches. The entire star list is taken up by these. We get it. Somebody strongly agrees with the pro-PO sentiments and wants everyone to read them all. But some of these messages are very mundane/not star-worthy. I have no strong opinions for or against PO, but this starring of literally everything said in favour of them is starting to get very annoying

Is it really necessary to star every single message by ArnoldNeumaier and Dimensio1n0? Please stop it, I'll clear out most of those stars in a bit.

...it also wasn't necessary to star both Jim's and my post :P

^ that

Damn, I hate physics papers :P It appears I have to locate a math monograph in order to actually see how this computation is done :/

Why would you bother introducing all the machinery if you're not going to do any of the computations explicitly, thus never actually using the machinery?!

@ACuriousMind we all know who it is

@ACuriousMind Which one?

@0celo7 Compact manifolds with special holonomy by Joyce.

It appears our library has it, but not for lending, I have to physically sit in the library and read it :/

@ACuriousMind Lol, what the heck?

Pirate it

@0celo7 Yeah, that's an odd thing they do here: Many of the more specialized books are only for use in the library itself

@ACuriousMind Maybe it's to prevent the situation the like geometry people have here

I think that's because they don't want to buy enough copies so that more than one person could borrow it

My advisor has 30 library books

all the good GR books >:(

@ACuriousMind Does your library normally have multiple copies o.o

@0celo7 Of the books you can take with you, yes, often

wow

I think

We usually have one copy, and then they have scans available

I don't know really, I've rarely used it to get physical books

Hlw guys

What's the reference frame for an object having a velocity of 10m/s ?

@ffahim What do you mean?

Great, of course google books has exactly the wrong part of the book :P

@ACuriousMind Are you seriously anti-libgen

or are you just not wanting to admit using it

Suppose I am saying ""hey look the car's velocity is 10m/s .."". Ami I the reference frame now?

@ACuriousMind

@ACuriousMind If it makes you feel better, I can have the book delivered to my office from the library :)

@ffahim I'm sorry, I don't understand what you are trying to ask. A reference frame is a coordinate system, not a person or object.

Or it's initial position is its reference frame?@ACuriousMind

@0celo7 Yeah, that's a great consolation

@ACuriousMind How far of a walk is it?

Coordinate ? .. I say my school is 2miles away from my house eastwards.. isn't my house reference frame?@ACuriousMind

@0celo7 The library? I need about 20-25 mins to get to it

Is it cold in there? Ours is always freezing

Not that I'd recall

@ffahim No, that's not what "reference frame" means. You might say that your house is your point of reference (or the origin of your coordinate system) if you're giving the direction of things in terms of how far they are from your house, but it's not a "reference frame".

@ACuriousMind oh I see.. sorry.. :-) now what would the answer of my first question pls?

Which one?

I say the car is moving at 10m/s

Am I the reference point here?

What is the meaning of life?

@ffahim Uh...that's not something one can tell from that statement.

@ACuriousMind ..

@ACuriousMind As a rule, we always clear out stars any time it looks like somebody is indiscriminately starring one person's messages or starring one side of a debate (or both sides). Stars are for highlights.

Then what can we say about its reference point.. as its traveling relative to a reference point ? @ACuriousMind

@DavidZ Yep, that's what I thought. I left two messages that had more stars than the rest, but it appears someone else got rid of them.

@ACuriousMind .. sorry I am bothering u.. :-).

If you could help me then pls tell me the main problem

@ffahim you're being awfully impatient here

@ffahim I still don't understand the question. The statement "The car is travelling at 10 m/s" is incomplete, it is missing precisely the information with respect to what it is moving with that speed. You seem to think that you can somehow deduce that, but you can't.

Well, I don't know what instrument measures velocity.. however suppose the reading of apparatus gives us 10m/s for that car.. but how it can measure?

Definitely it requires reference point right?

@ACuriousMind

@ffahim Sure. Which point it is is a property of the measurement device.

Can't I say the car's reference point is the initial point from where it started ?@ACuriousMind

Sure you can

Ohh thanks.. tell me another thing pls@ACuriousMind

It's about Newton 's first law

The first law says an object at rest will be at rest and an moving object will keep moving in the constant speed unless any force F applied

But what about an object which was accelerating constantly.. and let's suppose no additional force was applied or even if the force was applied the net force will be 0 , in this case won't it always be accelerating?

@ACuriousMind

@ffahim An object can't accelerate and have zero net force acting on it.

Ohh wait I repeat again

@ACuriousMind the 5 kg block of stone was at rest. I applied 10 N force.. now it will start moving at acceleration of 2m/s^2 .. in its way of moving there will not be any friction,gravity and any other force

Now?

Doesn't Newton 's first law say so?

I'm afraid I don't follow you.

Sorry.. what wrong did I say? @ACuriousMind

As long as you exert the force on the stone, it will accelerate. As soon as you stop exerting the force, it will stop accelerating and continue with whatever velocity it had at that point

@ffahim You're not saying wrong things, you're just not making much sense to me - I can't tell what is confusing you and what you're trying to ask

There might be a language barrier here, I'm not sure

@ACuriousMind hang on.. let's go term by term

moving = moving with velocity v will keep moving with that velocity unless a force is applied

an accelerating object doesn't have constant velocity, which is what ACM was talking about

As it has the acceleration of 2m/s^2. It will keep moving at the acceleration of 2m/s^2 . By not to apply force I meant that the 10N will remain but no further force will be applied @ACuriousMind @AndrasDeak

Hope it's clear now

if 10N remains, then a force is applied:)

that's what is confusing to us

@ffahim Yes, okay. What is your question about that?

Okay as no other force will act then it will keep accelerating at 2m/s^2 , right?

@ACuriousMind

yes

Thanks ^1/0

Great. The result that's supposedly in this citation is nowhere in sight -.-

@ACuriousMind now you know how GR people feel when told to check HE

@Kaumudi He's just bought a new iPhone. Existential crises are a frequent result of contemplating how much money you just spent on jewellery :-)

@Jim : High noon stapler showdown in bureaucratic public office :)

@MAFIA36790 : Thx. Same 2 u.

@JohnRennie Good point

A new iPhone is a pretty good deal when it comes to jewelry

Nine problems on the topology problem set

good lord

@0celo7 As an Android user it's obligatory to have a go at iPhone users at the least excuse :-)

@obe Why ? Android is way better that anything else.

@Ocelo7 What Kind of Problems ?

Chill guys, there's no such this as better in this context. It just depends on which you like more.

I like Android more because Apple is crap ;-)

Yes?

Boo :-)

Yes. That's an app I wrote myself!

My memory is terrible. That's not old age - my memory has always been terrible. So yes, I schedule everything. Otherwise I would forget it.

@JohnRennie you're still on your OnePlus?

@EmilioPisanty Yes, and it's still frikking awesome (dude) !!

@JohnRennie good to know it takes well to aging

it's a model 3, I recall?

also, how long've you had it?

I'm growing increasingly annoyed with my Honor 7

I'm going to be distracted for a few minutes while I bid for a laptop on ebay ...

ebay.co.uk/itm/322309423718

If anyone's interested.

Being a dude I'll fix the screen myself.

I'm hoping to get this for £100 or so ...

BOOM!! Won it for £103. Whoo hoo. That's a 3GHz Core i7 in that baby :-)

Anyway, where were we? Phones!

@EmilioPisanty: I think the OnePlus 3 is by some way the best mid range phone available at the moment. However they are allegedly about to release a minor upgrade with a slightly faster CPU called the OnePlus 3 Pro.

@JohnRennie that's good to know

You might want to wait for the Pro model, or you could wait and buy the 3 at a reduced price.

I'll probably start phone shopping in earnest in a couple of months

Ideal timing as the Pro version should be out by then.

Actually I don't really need another laptop, but it was too good a bargain to pass up. Does anyone want a laptop? :-)

@JohnRennie proof?

@Kaumudi: did you want to ask anything further about that scheduler?

@0celo7 At the risk of taking this seriously. My view is that I want complete control over all my computers, including phones and tablets. And Android allows me that control. Apple's view is that they know what's best and you can't do whatever you want with their phones.

There isn't a right and wrong side to this, it's just a matter of what users prefer. Lots of people like Apple's approach which is why the company is awash with (your) money.

@JohnRennie oooof, yeah, but then you're just hostage to your OEM's security updating schedule

which often just reads "never"

but then once you start digging, you just end up at trusting trust.

@EmilioPisanty: I generally buy Nexus devices because they get the updates first. This is the first non-Nexus phone I've bought for ages. OnePlus don't have the greatest reputation for shipping updates fast, though they are good at bug/security fixes.

But if the phone doesn't get updated to Nougat I'm not fussed because the current OS is very good.

yeah, but then that's an argument for Nexus, not for android

it's not about updates to the newest .0 version

it's about security fixes

and you have huge chunks of the android ecosystem sitting there with months- or years-old bugs unpatched

because the OEMs just can't be bothered to push the update

The argument for android is that, for example, if I want to open a terminal and start grubbing around in the OS then Android will let me. Apple will never, NEVER, allow that sort of access.

@JohnRennie is this a rooted android or box-standard?

At the moment the phone is unrooted. But even unrooted Android allows far more control than Apple does.

@JohnRennie oh, for sure. It's just that the added control comes at the unfortunate price of depending on an intermediary for basic security fixes.

Oh, and now that @S007 is here: please, please stop misusing the term "mathematical physics"

↑ that is not an apology

that is digging down on your use of the term

@EmilioPisanty the vast majority of manufacturers issue fixes promptly if you buy an unlocked phone. If you go to a phone company and buy a locked phone, well, that's your choice.

in a public forum that does not contain said apology

@S007 That is irrelevant

@JohnRennie I don't think anyone needs such accesss

@S007 what this one?

@JohnRennie They do know better than me

@S007 do you really not get the distinction? You just posted a comment digging in on the position that you think it's right to use the term as you previously used it on that thread.

@0celo7 It's a matter of personal preference. I want such access and I don't want a bloated, arrogant, rich multinational telling me otherwise.

@S007 Yes, it's your comments going forward that are the issue.

Your latest comment just de-rails the conversation yet again.

@0celo7 but don't worry, I'm sure Apple will put your money to good use i.e. stash it offshore and reap the interest.

@JohnRennie tbh it's not like the alternative is actually different

Not that I'm being tendentious you understand :-)

@EmilioPisanty Quiet you! I'm on a roll here :-)

@JohnRennie ok, can do. I was going to link to a search for "google taxes", but I'll drop it.

:-P

:-)

@JohnRennie You missed the part where I think taxes are theft anyway :)

Good for apple I say

@0celo7 Um, er, that's your money Apple are taking and stashing offshore.

@0celo7 how would this work: $\int \int_{D} 1 dA = A(D)$ the constant function $f(x,y) = 1$ integrated over a region $D$ gives the area of $D$. It'd be integrated like $\int_a^b \int_c^d f(x,y) dy dx$ like so. Then, you just subtract the upper bound minus the lower bound and multiply that by $b-a$?

@S007 I am not saying you should delete any comments.

oh okay I thought this wasn't correct because it seemed too simple (I thought of a region whose boundaries were defined by a function). The bounds, for a complicated region, could be something like $\int_{g_1(x)}^{g_2(x)} \int_{h_1(y)}^{h_2(y)} dy~dx$ where $g_1(x) = x^2 + x - \sin(x), x \in [0,3]$ or something

@0celo7 nvm

Insisting that "mathematical physics" has anything to do with numerical problem-solving of set-piece exercises just makes you sound childish. Sure, the term is occasionally used by textbooks where the audience is not yet ready to understand what mathematical physics actually is, but that doesn't mean you should use the term in its for-children version. Instead of digging down on for-children language, understand and use the for-grownups versions.

@S007 In terms of immediates, you are still in time to replace this one with a comment that's actually constructive.

@S007 well, and here I was trying to help you

@JohnRennie I'm referring to the "stashing offshore" part.

Who keeps starring everything?

You?

@ACuriousMind ?

I'm on mobile.

@S007 He was trying to help you by helping you sound professional. Insisting that "mathematical physics" has anything to do with solving high school or undergraduate exercises makes you sound unprofessional and ignorant of the field your proposal is supposed to cater to.

@0celo7 What?

@ACuriousMind Undergraduate exercises? I will point out that Hilbert's classical text on mathematical physics contains quite a lot of "undergraduate" material in the form special functions.

@0celo7 Add a "physics" in front of "exercises" if you're in the mood to nitpick :P

Add a further "typical" if you're still not satisfied :P

Typical undergrad QM exercises, sure.

If they're particularly nasty PDE, they're math phys according to Hilbert.

@ACuriousMind @S007 pretty much what ACuriousMind said.

In any case, trying to tear into PSE in that thread is not going to accomplish much.

Robert Cartaino explained that Stack Exchange is not interested in investing in type-based sites

and that is precisely the kind of site you're trying to argue for

Can't we just let this drop?

@ACuriousMind Also, HE is undergrad level at some universities.

I'm sad the proposal got the boot because I think it would have been a useful site. But that's life. Let's all just move on.

It's somewhat mathematical

@0celo7 Your point being?

@JohnRennie I still think there's some useful info to be had from Robert Cartaino. He mentions that there have been failed experiments on sites built around types of questions, but didn't explain what they are and how they failed. I think those are useful references to have around.

@S007 Oh, OK, so your last comment there was one last angry swipe? Good to know, then.

@0celo7 did u do ur lab report

oh it's not wednesday nvm

@ACuriousMind Well, are the paths in the path integral continuous?

Huh. I was chasing up my mysteriously obtained Area 51 reputation, and it turns out to come from this old question, about the previous go-round

discuss.area51.stackexchange.com/questions/20534/…

that proposal is here: web.archive.org/web/20150402085037/http://…

@ACuriousMind If so, how is that enforced in the usual treatment?

originally here area51.stackexchange.com/proposals/77677/…

@0celo7 I have no idea what that has to do with anything, but yes

@ACuriousMind wait for it

So how does one prove this?

@0celo7 If you look at the typical limiting procedure physicists describe, you'll see that the path produced is continuous (you'll also see that you should worry about what its derivative is supposed to be). I'm not sure what you mean by "enforced".

@0celo7 Well, in the rigorous approach, you define the integral to be over the Wiener space of continuous paths. I'm not sure what you want to "prove" this from.

@ACuriousMind is the path just a uniform limit of polygonal paths?

@0celo7 I believe so, yes

@JohnRennie would you believe me if told you I wear gold chains

@0celo7 do you also wear a v-neck and aviators indoors?

@mafia what does it mean to find the moment about an axis?

@Obliv I like v-necks, yes.

But not aviators

@MAFIA36790 rekt lol

In my book it defines a moment of a particle about an axis as its mass * the directed distance from that axis

Oh I didn't really understand why they did this except maybe for torque problems or something

oh also to determine center of mass apparently. You get the moment w.r.t. y-axis and divide by total mass to get $\bar{x}$

I never understood center of mass.

@Jim It's complicated :)

@0celo7 $(\bar{x},\bar{y})M = \sum (x_i,y_i)m_i$ my non-standard definition of center of mass.

left side is center of mass * total mass = sum of each mass moment

ugh I ruined my large coffee by putting 4 pink sweeteners in it. I never put sugar in coffee before and thought that since they were small packets they wouldn't be very sweet..

yeah it was fine with just milk but idk I just felt like changing it up :( @mafia

usually i get unsweetened iced tea because it's not so strong but today is going to be long (~8 hrs of studying)

Man, gotta love Science Alert:

> Now, the team led by French physicist Guillermo Ballesteros from the University of Paris-Saclay says...

because googling is sometimes just too hard

@BernardMeurer I missed the removed part. So I guess that will have to do.

@Jim The deleted part was a failed attempt to write "Jim" :P

(Don't ask me how one screws that up :D)

uh would it be correct to say that the center of mass of an object is the location at which all moments at the boundaries w.r.t. this location are equal?

@ACuriousMind what's Rigor in German

@0celo7 In the context of "mathematical rigor" it would be something like (mathematische) Präzision or Exaktheit.

@acuriousmind ist mein erklarung nicht falsche?

@Obliv What?

my center of mass statement

I don't know what a "moment at a boundary" is

uh like imagine a circle, the moment at a boundary w.r.t. the center would (in my definition) mean the mass at the edge of the circle multiplied by the radius

no that's not right

There is no "mass at the edge": The mass at the boundary is zero (for almost all shapes, the boundary will have zero measure, so integrating the density over the boundary just gives zero)

it's the sum of the moments in every direction w.r.t. that center

should be the same

@Obliv I don't know what a moment of inertia w.r.t. to a point is, either.

Moments of inertia are a with respect to an axis, and I don't think that they in general have something to do with the center of mass, although one typically chooses axes passing through the center of mass.

@acuriousmind Not sure if this will be coherent or be logical but I will say it to determine so: I meant that the center of mass defines an infinite amount of axes (in 3D) that if you wanted to get the moment of a point some distance from the com, you'd multiply by distance from the point and the mass. The sum of these moments in this one direction should be equal to the sum of the moments in some other direction.

I'm just thinking of the case in 2D where you find the com of, say, a stick. The sum of the moments in one direction is equal to the moments in the other direction (otherwise it would not balance?). I though to extend this idea to 3D+ with that definition above

in this way, it's almost like the irregular shape's com can theoretically be modeled as a sphere with uniform density

@ACuriousMind average over all axes

where the sum of the moments in any direction of the sphere is equal to that of the irregular shape

@Jim Thanks a lot, I appreciate it

@Obliv The definition of the center of mass $r_0$ is $\int \rho(r) (r - r_0)\mathrm{d}V = 0$. Moments can not "sum to zero" because moments are not signed - they are always non-negative.

You are perhaps trying to say that when you balance an object on its center of mass, then the torque about the center of mass will be zero. That would be correct, but is a completely different statement.

@acuriousmind I don't get what that integral is. $\int r\rho(r)dV$ is this equivalent to $\int \int_D r\rho(r) dA$? So you're saying the entire moment about the origin of the coordinate system (assuming we're still using the sphere idea) is equal to the entire moment about the center of mass of the sphere $r_0$?

Wait

You're not talking about momenta of inertia!

I'm sorry

isn't that the same thing?

Should've read your definition

@Obliv no, the moment of inertia is $\int r^2\rho(r)\mathrm{d}V$.

oh okay yeah it's defined in the next section.

umm

I just joined in: are those cylindrical coordinates?

@AndrasDeak Yes

ah, good, thanks:)

the many occurences of "sphere" made me wary

@Obliv For your notion of moment, you are correct: The center of mass is where, if you put the origin of your coordinate system there, all moments of mass vanish.

@acuriousmind well, aren't they still there just equal in all directions?