The h Bar

Good question

12:57 AM

Why does it emit so much radiowaves?

First, remember that all stars turn. They all have angular momentum, right?

Yes.

In a supernova, the core collapses into a neutron star and the outer gas is ejected away.

The neutron star is much, much smaller than the original core. It's gone from a decently big object to something the size of Manhattan.

But remember: angular momentum is always conserved.

A bit of the angular momentum is lost when the star's outer gases were ejected, but most of it remains in the neutron star. The neutron star has only a tiny fraction of its progenitor's radius, and thus its moment of inertia is reduced.

That means it's going to rotate hella fast.

Okay.

But why are pulsars special, then?

What ya mean, "special"?

1:01 AM

They emit much radio waves. And they have a different name.

Well, they rotate up to 716 times a second, if I recall. They also emit radiation with an extremely predictable period — so much so, that they can be more accurate for keeping track of time than atomic clocks

@SirCumference Neat conclusion.

@BalarkaSen Thanks :D

Okay, but what is the difference between a pulsar and a neutron star, then?

@PhysicsGuy A pulsar is a neutron star rotating extremely fast

1:03 AM

Don't all neutron stars rotate fast due to your explanation?

Their short rotation periods cause them to have regular pulsation periods

That's not an answer....

?

Sorry, didn't see that

Well I'd also say that pulsars specifically emit high amounts of radiation

Yes, but why?

Well that's a more intricate question, and I'm afraid I don't know the answer to that

1:07 AM

Ok, but thanks, you helped me.

np :)

Any other questions?

You mentioned dark matter

No, thanks, that's a field where I worked on, too.

Oh really?

You an undergrad?

Yes, but very shortly.

What year?

1:08 AM

Yes.

@PhysicsGuy ?

@SirCumference Oh really? --->Yes Are you an undergrad?---->Yes What year?-----> I don't know anymore.

What do you mean, you don't know anymore?

I don't exactly remember when I worked with it, it wasn't officially anyhow.

Oh, ok

1:10 AM

But I understand it.

I think.

@SirCumference I'm no physicist, but rotating test charges emit electromagnetic radiation in Maxwell's theory of electromagnetism - maybe something analogous happens here?

@BalarkaSen I don't think there's any connection.

I think it has something to do with he surface of the star.

Shrug

In EM, it is about field dynamics and stuff.

Goddammit, I'm missing the conversation

SE isn't notifying me when people ping me

So I don't hear anything or see a desktop notification.

1:14 AM

strange

@PhysicsGuy Keep in mind that not all neutron stars are made in supernovae

Really?

Yup. Some are made if a magnesium-oxygen-neon white dwarf goes beyond the Chandrasekhar limit.

You mean, like, they collapse into themselves due to having a big mass?

A carbon-oxygen white dwarf would ignite fusion at that point, since carbon isn't very heavy. But a magnesium-oxygen-neon white dwarf has much heavier elements. It would be more energetically favorable for electron capture to happen at that point.

1:16 AM

Ah, at least I know what the Chandrasekhar limit is, because it has something to do with black holes, which have something to do with GR, which has something to do with mathematical physics.

@PhysicsGuy Actually, the Chandrasekhar limit is maximum stable mass of a white dwarf. At that point, it either explodes as a supernova or becomes a neutron star, depending on the composition

The Tolmon-Oppenheimer-Volkoff limit is the maximum stable mass of a neutron star. At that point, GR comes into play and it becomes a black hole.

I think you're thinking of the latter.

@SirCumference Anyway, when electron capture happens, you're left with a bunch of degenerate neutrons.

Thus, a neutron star is born.

@Kenshin Holy crap, I just realized

It's "Q and A Exchange", as in Q&A

Not qanda exchange

How can strings in string theory be simultaneously one dimensional and have a property of being open or closed

wouldn't you have to have two dimensions to have an open or closed string

user228700

1:34 AM

Hi, everyone :-) I have a "quick" question about the domain and range of a composite function.

user228700

This is how my textbook has defined composition of functions:

user228700

> "Let $f: X \rightarrow Y_1$ and $g: Y_2 \rightarrow Z$ be two functions and $D$ is the set of values of $x$ such that if $x \in X$, then $f(x) \in Y_2$. If $D \ne \phi$, then the function $h$ defined on $D$ by $h(x) = g[f(x)]$ is called composite function of $g$ and $f$ and is denoted by $g \circ f$"

user228700

My question is this: it is given that $D$ is the set of values of $x$ such that $x \in X$ and $f(x) \in Y_2$. Doesn't this mean that the domain of the function $h(x)$ is only a subset of $X$?

Yes

user228700

Why is this allowed? Shouldn't we include all of them to call it a proper function? I guess what I'm asking is whether doing this is still okay.

1:39 AM

It's when the composition makes sense. It's not mentioned anywhere but also, $Y_2 \subset Y_1$.

@BalarkaSen BTW, you an undergrad?

$g \circ f$ is not a function on all of $X$, just on a subset of $X$. That's what they call $D$.

@SirCumference something like that

@BalarkaSen Huh?

user228700

@BalarkaSen And this is how it's defined? It won't work for all of them? But I thought that the whole range of the inner function had to be equal to the domain of the outer one.

It's a yes-or-no question

1:41 AM

Strictly speaking, that'd be a no.

So what are ya then?

Think about $f(x)=x^2-1$ and $g(x) = \sqrt{x}$. $f$ has a domain of $\unicode{x211D}$, but $g(f(x))$ has a domain of $(-\infty, -1)(1, \infty)$

@Kaumudi.H Yeah, you're right. Strictly speaking it's not $g \circ f$, but $g \circ f'$ where $f'$ is the restriction of $f$ to $D$.

But you don't want to be so rigorous in general.

@Natecat Bad example, $\sqrt{-}$ is not a function.

It does if we take $\sqrt{x}$ to mean the positive square root of x, which is implied by the fact that I treated it as a function

user228700

@BalarkaSen Riight...

user228700

1:46 AM

@Nate: Thanks for the example :-)

user228700

@BalarkaSen But why should the domain of $g$ be a subset of the range of $f$?

Sorry, my internet died.

What did you mean by "Shouldn't we include all of them to call it a proper function?"

@Natecat Ah, got it.

Sorry if that was a bit rude

1:51 AM

@Kaumudi.H If you read the paragraph you quoted, you'll see that $D$ is the set of $x's$ in $X$ such that $f(x) \in Y_2$. But $f(x)$ lives in $Y_1$, so $Y_2$ better be a subset of $Y_1$, no?

Eh, maybe actually they just intersect. I don't like the way your book writes stuff.

user228700

@BalarkaSen But doesn't $Y_2$ contain other stuff as well? I too feel that they only intersect...

Yeah, I am inclined to agree.

But in any case, everything still makes sense.

user228700

Yeah, I guess. Last question...I think. My textbook has written this:

@SirCumference It's a bit complicated :P I do math, that's all.

Are you an undergrad?

@BalarkaSen Yep

user228700

1:56 AM

Question. How to write subset/equal to in LaTeX?

So wait, you're in high school?

\subseteq

user228700

Thanks!

user228700

> "The domain of $g \circ f$ is $D$, which is a subset of $X$ (domain of $f$). Range of $g \circ f$ is a subset of the range of $g$. If $D=X$, then $f(x) \subseteq Y_2$."

user228700

@Nate: Yep, thanks :-)

1:58 AM

Did you really mean $f(x)$?

user228700

Last statement? Yep.

The notation your book is following is all crap then.

user228700

:-/ Uh huh..?

In any case, they mean the range of $f$ is contained in $Y_2$.

user228700

But what's with $D$ being equal to $X$ and all? In that case shouldn't the range of $f$ just be equal to $Y_2$?

2:01 AM

The codomain of $f$ is $Y_1$, remember. Unless you restrict it to $D$, it doesn't get inside $Y_2$.

But if $X$ is already equal to $D$, no need to restrict. $f$ already fits inside $Y_2$.

Also, no, it need not be equal to $Y_2$. What makes you think this?

user228700

What? Your first statement seems contradictory to your last or maybe I don't understand.

user228700

What did u mean by "Unless you restrict it to $D$ it doesn't get inside $Y_2$"?

If you restrict $f$ to $D$, the range goes inside $Y_2$ aka the range becomes a subset of $Y_2$.

Yeah, based on your definiton paragraph it does seem that if $X = D$ then $Y_1=Y_2$

user228700

^^ That's what I'm asking about. But my textbook has written that if $X=D$, then $y_1$ in only a subset of $y_2$, which it already is, from the very beginning!

2:05 AM

Well, that need not be true.

Oh I see, if the range of $f$ doesn't encompass the entire domain of $g$

Right, @Natecat

@Kaumudi.H Where did it say $Y_1$ is a subset of $Y_2$??

user228700

@Natecat Ohhh, yeah, that makes sense. I was using twisted logic.

@Kaumudi.H Yeah, that definiton is phrased strangely imo

Let's all agree that your book likes to write crap

crap as in crap notation/phrasing

user228700

2:08 AM

@BalarkaSen If $D=X$, then $f(X) \subseteq Y_2$

Yep

user228700

What?

But $Y_1$ is not $f(X)$

$Y_1$ just contains $f(X)$ :)

user228700

U mean that $Y_1$ is the codomain, not the range?

Exactly right.

user228700

2:10 AM

But why does $D$ have to be equal to $X$ for that?

Otherwise you just have $f(D) \subset Y_2$.

user228700

Ah, okay...

$f(X)$ is a huge thing, which hits a bit of $Y_2$ but also a lot of $Y_1 - Y_2$

It's helpful to draw a Venn diagram to understand these

user228700

Gosh, okay, I think I've understood.

user228700

Thanks very much @Balarka @Nate :-)

2:13 AM

Sure. You should get a better book to read set theory from, IMO.

user228700

I'm not reading set theory per se. This is the beginning of Calculus.

strange

user228700

Why?

nobody really cares about composing awkward functions in calculus.

user228700

I see :-/

2:36 AM

Question: A wine bottle has high pressure fluid pushing on a cork. The cork should fly off with an acceleration and momentum that depends on that pressure value.

Then in another case you have a wine bottle with 2 corks. They should also feel that same accelerate and momentum each. Where does the extra energy / momentum come from?

They will feel the same force but the pressure is dissipating twice as fast, so they will experience the force for half the time

Thank you. But how do i work out the time part of it?

Does the duration change because there is twice the surface area of cork or something?

Are you just trying to find the impulse?

i think so yes, i want to know the velocity and model it in software

in a very simplified pipe system

mcnallyinstitute.com/13-html/13-12.htm

Oops that wasn't super relevant

2:52 AM

it was interesting though thanks

i'm starting to feel maybe my problem is not modelling time properly in my simulation

Actually it could be used.

in my virtual pipe system, the openings are all 1 square meter and 1 kg of fluid is pushed around etc..

so I don't need to calculate different sized openings etc. yet

You could use the head substitution to get the flowrate at any instant of time

i couldn't find the meaning of head substitution

Ignore the last one, it doesn't use SI units

2:56 AM

so if you had a Y shaped pipe laying flat, and you pushed fluid up the middle with pressure = 10, how do you calculate the resulting momentum on the two ends?

with that

what are the units of pressure

anything is fine, like 1 pascal or 1 N/m^2

Wait the momentum of the fluid or of objects being pushed in the pipe

say the Y shape is full, you push on the 'stork' section of the Y and try to calculate the resulting velocity of the fluid coming out the two prongs

It would be the velocity of the fluid in the stork times the ratio between the cross sectional areas of the stork and the two outputs, $V*(A_s/(A_{o1}+A_{o2}))$

3:03 AM

ahh so it would yes

but there are situations where the stork has a lot of static pressure (say a water column feeding in) and you need to work out the prongs given only this pressure value

Because unless you are compressing the fluid in the pipe, the flowrate has to be the same on both sides

yes ok that is simple

that's really helped me go in the right direction after days of wondering

thanks

i think my main problem was trying to solve each prong velocity independently, just from the pressure value at that prong. Instead I need to treat the whole pipe network as a single thing and work it out simultaneously.

What are you trying to work out on the prongs specifically

flow rate or pressure

Also is that static pressure dependent on the flow at all or can we treat it as constant

i really just want flow rate so I can calculate momentum and handle collisions of water parcels

I can already calculate static pressure

not sure about dynamic pressure

the formula looks simple but seems to complicate it for me

if a water parcel is traveling at 5 m/s then each frame I will try to move that parcel 5 m along the pipe, obeying T intesections etc.

Velocity won't nessecarily remain constant in the piping network

For example, the Venturi effect

3:15 AM

i want to avoid any suction and venturi for now, even though unrealistic

Hagen–Poiseuille equation

In nonideal fluid dynamics, the Hagen–Poiseuille equation, also known as the Hagen–Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section. It can be successfully applied to air flow in lung alveoli, for the flow through a drinking straw or through a hypodermic needle. It was experimentally derived independently by Jean Léonard Marie Poiseuille in 1838 and Gotthilf Heinrich Ludwig Hagen, and published by Poiseuille in 1840 and 1846...

I did come across that one but was unable to find a way to apply it to the fork problem

It's basically this: Take an upside down T shape, fill it with fluid so it is 10 m deep and stationary (somehow). What is the momentum of the fluid as it pushes left and right at the bottom.

It seems to depend on whether both sides are open at the bottom or only one side.

If the right hand part of the upside down T is closed, then the left side does get that full force of 10 cubic meters of water. But if both sides are open then they both should only get half that force. But how to calculate that?

3:31 AM

They both get the same amount of force

10 cubic meters of water

ok but impulse and momentum are different right?

head isn't dependent on shape

Impulse is the derivative of momentum with relation to time

is it possible to calculate the final momentum without stepping through time?

without a time value?

I don't really understand what you mean by final momentum

It would depend on the amount of water you had

Just assume a column 10 cubic meters high

3:34 AM

ok

the horizontal pipe on the ground is still empty

That tells me nothing about the momentum

everything stationary

but in real life you could try to predict that right?

and say the water will come out at X velocity each side

roughly at least, so surely that can be formulated

There isn't any pressure in there though

yes there is static pressure from the water column, so the bottom of that column is feeling a pressure value from all that water above

3:35 AM

And obviously it would depend on the amount of water, volumetrically, and the sizes of the pipes

just say 1 m cubed for ease

Once the valve is release the water is in freefall and there is no longer any pressure

say there are valves at the base of the upside down T that are holding the fluid back

Yes

you suddenly open them, then need to calculate momentum

3:37 AM

That's what I infered

There would be no pressure in the fluid

just afterwards?

If the system was closed once the water was at rest it would have head obviouslky

yeh

are you saying you suddenly have to pressure once the valve opens, because the fluid is able to enter the open spaces?

*valves i mean

suddenly don't have yeah

*suddenly have no pressure

ohh

3:39 AM

So my intuition says it would just act based on gravity

Would you say there is no way to calculate momentum for each prong given just the pressure values before the valves get released?

I am still confused by what you mean by momentum, the water isn't a solid object, it won't all have the same velocity

oh just velocity is fine, i am kind of breaking it into parcels in this simulation

it's in a grid of 1 m ^ 2

each square can hold between 0 a 1 cubic meter of water

static pressure is first calculated for each cell, which is fairly easy

then I try to calculate velocity for each cell, but that's where i'm struggling

Even just for an L shaped water column, I'm not sure what the acceleration should be at the bottom due to static pressure

I thought I could just make acceleration = whatever the static pressure is for that cell, but that fails when you have forks

I don't see how you can't apply the Haden-Poiseuille equation

Because I am using simple units for length, density = 1 etc., that equation just says delta velocity = delta pressure

but that doesn't seem to work for a fork

3:48 AM

Why not

at the base of the fork, each prong is getting a force of 10 units

but in real life the resulting velocity of fluid should be half each, not full each

Incorrect

otherwise energy is not conserved

Well no

because if you keep the pressure constant and add a fork, you have more energy avaliable

oh really

3:54 AM

Think of pressure as potential energy

but what about a champagne bottle with 10 corks, would they all fly off with the same momentum?

No because the pressure wouldn't stay constant

you mean it would drop very quickly at the start of the firing?

Well yes

It would drop 10x quicker with 10 corks

but to solve that you would have to step through time somehow right? I mean you'd have to calculate the time taken for the impulse etc. Or is there an easier way?

can you just divide by surface area maybe, 10 corks so 10 times the area

3:57 AM

Well you have to tell me the problem you are actually trying to model

because you have given me like 10 different problems so far and you keep changing it

in this pipe system i may want that cork-like behavior

Can we just talk about a wine bottle with a cork then?

sure

i think the problem is I am wanting the fluid to act a bit like a solid in this system, so it's a bit vague sorry

OH NO!!!

FDUCK

Thinking back, I realize on my test I said "core collapse supernova" instead of "pair instability supernova"

damn it all

I would say why is the cork example different to the fluid example though. Why does fluid come out at high velocity yet the corks seem to 'share' that energy?

4:04 AM

In one there is an infinite reservoir of pressure in the other there isn't

So if you had a very large container and a couple of small corks, they'd both come out at 'full speed' because the container remains pretty much full.

assuming both corks are at the bottom suffering from static pressure

They don't need to be at the bottom but yeah

ok thanks again for trying to understand where i'm coming from

vzn

Scientists detect a quantum crystal of electrons and “watch” it melt / MIT news

↑ (yet another) sub QM effect? o_O

user228700

4:38 AM

Hi, again! I'm having some trouble with piece-wise defined functions.

user228700

I've been asked to define the "pieces", so to speak, of the function $f(x) = | |x-3| - 2|$

user228700

I removed the inner mod first to get:

user228700

$f(x) = * |x-1| ; 0\le x < 3$

user228700

OR $|x-5| ; 3 \le x \le 4$

user228700

Splitting the mod again, in the first case (arfgh, that is a - sign, not = and now I can't even edit it), we have $(1-x)$ if $0 \le x \le 1$ and $(x-1)$ if $ 1 < x <3$

user228700

4:46 AM

Is this correct? 'Cause the textbook has a different answer. Well, slightly different. Their answer is that we have $(1-x)$ if $0 \le x < 1$ and $(x-1)$ if $1 \le x <3$

user228700

Arhjgui. Sorry. The textbook's answer is that we have $(1-x)$ for $0 \le x < 1$ and $(x-1)$ for $1 \le x <3$.

user228700

Notice that the $\le$ and $<$ signs are different. That's the only difference between my answer and my textbook's answer but that difference is still there. Can anybody be so kind as to tell my why..?

5:10 AM

@Kaumudi.H Hmmm. When in doubt, simplify.

Let g(x) = abs(x-3). That has a cut at x==3.

You have f(x) = abs(g(x) - 2), which will have cuts where g(x)==2

... which is at x==1 and x==5. Plus something funky will still happen at x==3.

@Kaumudi.H I fixed some of your typos

Anyway for nested absolute values I prefer to work outside in, though it probably doesn't matter much in this case

user228700

@DavidZ Oh, thanks very much :-)

Remember that Markdown (** and the like) doesn't work within MathJax

user228700

@rob Umm, yes. So..?

user228700

@DavidZ Oh, right :-/

5:16 AM

@Kaumudi.H Your answer and the book's only differ with respect to $x=1$. (makes sense?) Plug in $x=1$ and see what you get in each case.

user228700

@DavidZ Yes, and I realise that the difference doesn't matter anyway because in either case, it gives me the same answer.

@Kaumudi.H Just thinking through it out loud, plus getting interrupted.

user228700

@rob Sorry :-P

@rob oh, sorry

user228700

Anyhoo, okay, that's that. Thanks, guys :-)

5:21 AM

@DavidZ Not interrupted by you, sorry if I gave you that impression --- our new cat and I have a disagreement about the purpose of the kitchen table.

5:32 AM

@rob ah, gotcha.

P.K.Harshith

Just this one question and I'll be off ...

How did planets start rotating

Q: How do the Planets and Sun get their initial spin?

@P.K.Harshith I've always heard that the most plausible explanation is the ice-skater-arms-in effect. A fairly tiny rotational asymmetry in the proto-solar nebula, once the matter condenses into the compact sun and planets, requires a lot of spinning to conserve angular momentum. But I haven't worked through the arithmetic to see if that's reasonable.

8

How do the Planets and Sun get their initial rotation? Why do Venus and Mercury rotate so slowly compared to other planets? Why does Venus rotate in a different direction to Mercury, Earth and Mars?

@P.K.Harshith But note David Hammen's comment on the accepted answer to that question: since a forming solar system ejects some nebular matter, including the angular momentum associated with that matter, applying conservation of angular momentum to the entire system probably isn't appropriate.

user228700

5:52 AM

@rob Aww, cats!

@Kaumudi.H After several years of attrition I had gotten us down to one cat ... this year we took in two more that were abandoned at my wife's vet clinic.

user228700

"Attrition" is an...interesting word to use in this context, don't u think? :-P

@Kaumudi.H I think our maximum was four in 2001 or 2004. Mean life has been about ten years.

Actually our only cat at the beginning of last year was the first one that we adopted together in 1999 or 2000.

user228700

Boy, that's a lot of cats!