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12:58 AM
Can degenerate matter be in the form of a liquid?
 
1:29 AM
@SirCumference Degenerate matter is a phase in its own right. Some of it's properties are fluid like---it fills the potential to a isoenergetic level give or take the sharpness of the distribution---but you shouldn't be trying to bin it as on of the usual states of matter: it's its own thing.
Pointedly the properties of this stuff arise explicitly from the quantum nature of the governing physics.
 
@dmckee As I thought. The weird thing though is that metallic hydrogen is said to be a form of degenerate hydrogen, but the article then goes on to say
> At high pressure and temperatures, metallic hydrogen might exist as a liquid rather than a solid
What on Earth?
 
1:47 AM
I can't make those description agree with the way I usually think of degenerate matter, but I suspect this may be because I don't understand what they are going on about rather than because they've made some kind of horrible mistake.
 
@dmckee Well, ACM and I just saw Wikipedia make no sense in its Big Rip article...
It was flat out wrong
And I've agreed with your description of degenerate gases up until this point, so now I'm just unsure whether to trust Wikipedia
 
@dmckee @anonymous @ACuriousMind @DHMO Thank you for your continued help over the past few days in preparing me for my exam today. It is much appreciated, hope to pay all of you a beer some day.
 
2:44 AM
Man, I love getting downvoted with no explanation...
 
 
2 hours later…
4:18 AM
@0celo7 How's your eye?
Oh wait, you got 4 days left...
 
4:42 AM
@0celo7 Has a surgical eye patch
But he's feeling better
 
5:14 AM
Wait really?
 
14
Q: Isn't D'Alembert's wave equation enough to see that Galilean transformations are wrong?

Francesco BoccardoThe D'Alembert equation for mechanical waves was written in 1750: $$\frac{\partial^2u}{\partial x^2}=\dfrac{1}{v^2}\dfrac{\partial^2u}{\partial t^2}$$ (in 1D, $v$ being the propagation speed of the wave) It is not invariant under a Galilean transformation. Why was nobody shocked about this a...

Let's just take a moment to appreciate that an actual good question made it to the HNQ list
 
1
Q: Estimating value of Gaussian integral

Bernardo MeurerHow to show that the area $A$ of the region $$\{(x,y)\in\mathbb R^2 : |x|\le y\le e^{-x^2}\}$$ is such that $$0 < A = 2\int_{0}^{\beta}{(e^{-t^2}-t)dt} < 1$$ in which $\beta>0$ satisfies $e^{-\beta^2}=\beta$

Anyone?
 
5:31 AM
@BernardoMeurer I would appreciate butter chicken or pizza more :P How was your exam ? :D
@BernardoMeurer Have a look at this en.wikipedia.org/wiki/Gaussian_integral
The Gaussian integral, also known as the Euler–Poisson integral is the integral of the Gaussian function e−x2 over the entire real line. It is named after the German mathematician and physicist Carl Friedrich Gauss. The integral is: ∫ − ∞ + ∞ e − x 2 d x = ...
 
@anonymous I think it went well, we'll see in a few days
This was the exam
 
obe
@BernardoMeurer so he did surgery? or what?
 
@BernardoMeurer For estimating the integral first think of two functions between whom the value of $e^{-x^2}$ lies.
 
I think expanding the Taylor polinomial to n=4 works
 
@BernardoMeurer But then you need to know the solution of $e^{-x^2}=x$. Isn't it ?
I don't think you can solve that equation manually..
Though you can use something like Wolfram Alpha :P
 
5:43 AM
@anonymous I have no clue :P
 
obe
nice ignore @BernardoMeurer
 
@BernardoMeurer The value of $e^{-x}$ is less than $e^{-x^2}$ in 0 to 1. You can use that fact to estimate by the sandwich method !
 
@obe It's 6 in the morning, give me a break
Idk what he did, he was talking to me through Skype before
 
obe
why are you awake lol
 
Can't sleep
 
6:35 AM
@SirCumference In a de Sitter universe the distance of the cosmological horizon is related to the (exponential) expansion rate. For our universe this will end up at about 18 Glyrs I think. In a Big Rip the exponent of the expansion rate increases with time so the event horizon moves inwards with time.
That's why the size of the observable universe decreases with time.
 
@JohnRennie Do you know the event horizon's distance as a function of the scale factor, or as a function of time?
 
@BernardoMeurer bom dia @anonymous @JohnRennie hi
 
@SirCumference Remember this is for a de Sitter universe. The scale factor is something like $$a(t) = e^{kt}$$ for some constant $k$.
 
And also: in a Big Rip universe, the event horizon actually moves inward? Doesn't it remain stationary in a universe with $w = -1$?
 
The distance is related to the value of the constant $k$, but I've forgotten exactly how - I would need to go away and look it up.
 
6:40 AM
Jog my memory, it's early in the morning
 
In a Big Rip the constant $k$ isn't constant. It increases with time, and that's why the event horizon radius decreases with time.
 
Is a de Sitter universe necessarily dominated by phantom energy?
Or does it merely ignore ordinary matter?
 
A de Sitter universe has a cosmological constant.
But no matter.
 
right
 
Assuming dark energy behaves lie a cosmological constant our universe will tend towards de Sitter as the matter gets diluted away by the expansion.
 
6:42 AM
Yeah, that I've heard
 
That's why we will end up with a de Sitter horizon in the infinite time limit.
 
But a de Sitter universe doesn't necessarily have phantom energy, right?
 
Technically a de Sitter universe has only a cosmological constant. A universe with phantom energy wouldn't be de Sitter.
 
OK, so its equation of state is -1, right?
 
Yes
 
6:44 AM
So why bring up a de Sitter universe when talking about the Big Rip?
 
In practice a universe with phantom energy looks very like a de Sitter universe except that the constant $k$ increases with time.
As far as I know there isn't a name for a universe that is like de Sitter but with a time dependent $k$.
Anyway, Wikipedia is right and you and ACM were wrong - it doesn't happen often :-)
 
@JohnRennie Well, I didn't really know :P
I guess that makes me a loser here in any regard
 
A "loser", that's a bit harsh ...
 
@JohnRennie So question, is the scale factor as a function of time equal to $e^{kt}$ in a Big Rip universe, just that $\dot{k} > 0$?
 
Yes.
 
6:48 AM
Exactly that or similar?
 
I suppose strictly you'd write $a(t) = e^{k(t)t}$
 
Ok cool. Last-ish question, can we know $\dot{k}$ as a function of the equation of state?
 
But I can't remember the form of $k(t)$ offhand.
When you solve the Friedmann equations they give you $k(t)$
 
I'm an idiot
Especially considering we've done stuff like this multiple times
 
Somebody did the hard work for you :-)
 
6:52 AM
Oh right, last question. Do you know the event horizon's distance as a function of the scale factor?
Wait, I already asked that. Nvm
 
In a de Sitter universe the horizon distance is constant and independent of the scale factor.
 
Or better, do you know a proof for it? I asked a question about this on Astro and I might as well post an answer to it
@JohnRennie I meant in a Rip universe
 
@SirCumference I can't remember what the solution lookls like for a Rip universe. Sorry.
 
All right, thanks though
 
It shouldn't be hard to Google though
 
6:56 AM
On a side note, I've posted this question
1
Q: Can degenerate matter be in the form of a liquid?

Sir CumferenceThis title probably makes no sense, and that's the problem I'm having. As far as I know, degenerate matter is an entirely different phase of matter altogether — it may have fluidic properties, but it's very different from the traditional states of matter. It only occurs under extremely dense or l...

Just to solve the confusion Wikipedia's giving me
 
Well the degenerate bit is the electron gas not the matter as a whole. It isn't obvious that the electrons being degenerate forces the matter as a whole to be either liquid or solid.
 
@JohnRennie Oh right, I had a few extra questions about a shrinking observable universe.
Does the event horizon's distance reach zero in a finite time?
 
In a Big Rip? If so then yes, that's what causes the Rip.
 
Not the scale factor reaching infinity?
 
That happens at the same time. The expansion rate becomes infinite.
 
7:06 AM
Yeah, I haven't found the relationship between $a$ and the horizon's distance yet but I trust that.
 
Really you ought to go read up on the Friedmann solution for phantom energy. I don't think it's particularly hard. I'll have a dig around and see if I can find a good article on it.
 
@JohnRennie Thanks. As I mentioned before I missed the class when we went over the Big Rip in depth.
 
7:41 AM
@JohnRennie Are you free for a bit ? I need to ask something about young's double slit experiment...
 
@anonymous I'm working at the moment I'm afraid. I'll be done in around 15 minutes.
 
@JohnRennie Sure. Could you ping me when you are done ? :)
I'll be here for an hour or so :)
 
Well, imma try to sleep
'Night
 
@anonymous Finished more quickly than I thought. What did you want to ask?
 
I can't understand why
The maxima points are oscillating up and down
Should'nt the amplitude of each point be fixed?
According to the path difference...
@JohnRennie
 
7:57 AM
The incident wave is $\psi = A_0e^{-i(kx-\omega t)}$ so it is a function of both time and position.
If we conside the wave at a fixed position we get $\psi = Ae^{i\omega t}$
Where the constant $A = A_0e^{-ikx}$.
 
Oh I see, but then why don't the amplitude of the nodes on the screen change? There are some points as you can see on the screen where the amplitude is always 0!
 
And when we talk about the amplitude of the wave we normally mean this constant $A$. So we say the amplitude is constant even thought the amplitude of the wave oscillates with time.
At the bright spots $A$ is large and at the dark spots $A$ is small. What we see is actually the average intensity, which is proportional to $A^2$.
 
I think I got it now. Suppose the waves are $A_1\sin(wt)$ and $A_2\sin(wt-\phi)$ the resultant is $(A_1+A_2\cos(\phi))(\sin(wt))-(A_2\sin(\phi))\cos(wt)=R\sin(wt-\phi)$. So $R$ is the outline of the maximum possible amplitude of each point, right? However with time the amplitude oscillates between the maximum possible amplitude at that point and 0?
 
Yes. Well, the amplitude oscillates between $A$ and $-A$. The intensity oscillates between $A^2$ and zero.
 
@JohnRennie But what is the physical interpretation of $-A$ amplitude on a screen ? Does it imply a darker region than $0$ amplitude positions ?
 
8:07 AM
hi John & Anonymouse ... does $ 1 radian = 1 * r $ ?
 
@Hey-men-whatsup What is r?
 
I'm watching this youtube.com/watch?v=wcfkDuFpbiM&t=195s, but couldn't understand
radius
 
@Hey-men-whatsup Basically Angle In Radians= (Arc Length in any unit )/(Radius in the same unit as Arc Length)
 
I know what 'arc length' is.... , but the arc length itself is $r$ , so $1 rad = 1 r$?
 
1 radian = (r/r) radian
 
8:10 AM
assuming the term $pi$ never exist
hmmm that's one ..??
 
@Hey-men-whatsup Read this fully en.wikipedia.org/wiki/Radian
 
l mean $1 rad$ is plain 1
 
@Hey-men-whatsup Ya
 
@anonymous $A$ is the field strength i.e. the amplitude of the electric field associated with the light wave. What our eyes detect is the intensity i.e. joules per second and that is proportional to amplitude squared.
 
so $1 rad$ means $area$ between the $3 radiuses $ right?
 
8:13 AM
@JohnRennie Yay! I got it now :) That was the part I was confused about. Our eyes detect $A^2$ and not $A$. Thank you !
@Hey-men-whatsup I can't understand what you are asking....
How does area come into play ?
 
@Hey-men-whatsup Remember that a radian is dimensionless. $1 \times r$ will have dimensions of length so it can't be a radian.
 
so If a kid comes to me and said what is 1 rad, I will say the violet part right?
 
@Hey-men-whatsup Say that it is the angle between the red line and the x axis :)
 
Ah ok thanks
thank you so much.... all ya .. $angle$ , now I get it!
 
9:22 AM
@Oswald Hi! I like that profile picture of yours. It reminds me of one of my favorite cartoons in my childhood :)
 
@anonymous hehe.. My favorite too.. I'm too nostalgic about it :)
 
10:21 AM
Hi guys. A very stupid question... we know that the 4-momentum operator $P_\mu = i \partial_\mu $ is hermitian when acting on operators. Is that true even when I take the hermitian conjugate of the Lagrangian term $ i \partial_\mu\psi \sigma^\mu \bar\psi$ ? ( I wrote it in 2-component notation and $\psi$ is a well spinor)
I think that are two different things...
 
user228700
11:08 AM
Hi, everyone :-)
 
Hi :-)
 
user228700
How's it going?
 
Quiet this morning.
I've been playing with calculating the trajectories of objects falling into a black hole. I was thinking of writing it up as an answer but I'm having second thoughts.
 
user228700
@JohnRennie I know that you don't mean "quiet" literally but of course it's quiet, you don't have little children at home :-)
 
user228700
Ooh, that sounds...interesting and difficult.
 
11:12 AM
"little children" = younger sisters? :-)
 
user228700
Yes :-)
 
user228700
They're a handful.
 
I'm sure they'd be delighted to know you described them as "little children" :-)
 
user228700
Lol, definitely. They turned 10 last year, they would certainly be very pleased.
 
In my (admittedly limited) experience girls tend to regard themselves as grown up from somewhere around age 5 :-)
At least, they start telling you what to do from about that age!
 
user228700
11:15 AM
Hmph.
 
:-)
Admit it - it's true!
 
user228700
Certainly not! Well, little girls do try to act grown up from a relatively young age, but the best of us don't boss around other people!
 
:: John looks massively sceptical ::
 
user228700
...and I am not experienced in this, I'm afraid. Growing up, I was such a tomboy. Nobody even expected me to grow out my hair or anything!
 
user228700
Evidence:
 
11:19 AM
@DavidZ I think all of the "your question is opinion based" stuff was cleaned up by someone before you got to see this question. I noticed there was one last trace and I've edited and removed it. Shall we clean up our two comments and wait to see if I ever get a round tuit and do my high-rep/low-rep anaysis?
 
Evidence?
 
user228700
.__. Gimme one moment.
 
user228700
 
Indeed, though the pink trousers are a bit of a giveaway :-)
 
user228700
They're red!
 
11:24 AM
Whatever :-)
 
user228700
:-) That is me when I was about 3 and I am not ashamed to admit that I am still a little tomboy inside. If u want, I will show you one more picture from a few years ago, when I climbed a very large tree.
 
It's OK I believe you :-)
 
user228700
Thanks, it may have taken me ages to find it :-P
 
Does that have anything with you wanting to do physics?
 
user228700
Hmm?
 
11:26 AM
I'm being stereotypist I guess. Assuming that physics is a subject preferred by boys.
Or in this case tomboys.
 
how could she write from right to left, even the projection doesn't make sense to me youtube.com/watch?v=qTbDQ9gkKJg
 
user228700
@JohnRennie Oh, right. Well, no. I am sorry to disappoint but I'm afraid that my liking for physics doesn't have anything to do with that.
 
@Hey-men-whatsup Wow, yes, she's having to do mirror writing so it looks OK to us.
 
user228700
@JohnRennie I have noticed that out of my all my friends among girls, not one likes physics but I have absolutely no idea what that's about.
 
@Hey-men-whatsup Unless of course they have reversed the image in the video. I note she is writing with her left hand, and if the video had been reversed a right handed person would appear to be left handed.
 
user228700
11:29 AM
@Hey-men-whatsup There is at least one comment about this in every single one of her videos. She flips the video.
 
Then I should go to a doctor... there's something wrong with my visualization
or it's either distracted.. well
 
@Kaumudi.H that's a clever idea. The idea of writing ona transparent screen in front of her is nice, and flipping the video makes it easy for everyone involved.
 
user228700
@JohnRennie Indeed, it is! Unfortunately, if you scroll through the comments section, most of them will be along the lines of "You're so beautiful, you distracted me from the content of the video".
 
user228700
(I have watched several of her videos.)
 
Male nerds find any woman who will talk to them beautiful. Or am I being excessively cynical here? :-)
 
user228700
11:34 AM
:-) @JohnR: It is safe to say that gender might not have very much to do with passions. I am not a neurologist, it may well turn out that indeed, there is some dependence there but so far as I can tell, being a girl meself, there is no connection.
 
I bet most of the commentators there didn't even passed calculus...lol
 
Have you ever heard of Baroness Mary Archer?
 
user228700
Uh, no, I had not...
 
user228700
Why did you ask?
 
When I was a student she was a fellow at Cambridge studying various topics related to electrochemistry. She used to tutor physical chemistry undergrads - including me.
 
user228700
11:36 AM
Ohh!
 
At the time I was an impressionable 18 year old, and we were all utterly spellbound.
 
user228700
:-) I see...
 
user228700
Well, in great colleges here, the boys:girls ratio is ridiculously high--7:1, I think.
 
To be fair, the ratio was very high at Cambridge in the 70s, though I believe that girls now outnumber boys in at least some parts of higher education in the UK.
 
user228700
...I am talking of the present, Sir, here in India.
 
user228700
11:41 AM
One of my friends is doing an integrated course in Mathematics from one of the best colleges here, CMI.
 
Indeed. I guess my point (if I have a point) is that your daughters are likely to find things very different.
 
user228700
In her class of 60, she is the only girl.
 
user228700
@JohnRennie Hmm, yes. My point is that if I get into the college that I hope to, it will be boys, boys everywhere, especially in the course I hope to do.
 
What course are you applying for? I assume it's physics though I'm not sure you've ever said so.
 
user228700
Again, it depends on what rank I get in my exams but I hope to do an integrated course in physics and electronic engineering.
 
11:48 AM
Electronic engineering? Is that something you'd like to do after university?
 
user228700
If I'm unable to get enough marks to do that one, I won't be able to do physics at first--just engineering and then I'll have to see where it goes from there.
 
user228700
@JohnRennie I...erm, I'm not sure, but I don't know what type of engineering I want to do--there is no point in deciding this now anyway, because it all depends on my rank, but I do know that this integrated course exists so...
 
user228700
I know that it seems weird that the course will depend on my rank but that is how it works over here. One of my friends is now doing biology in IIT, just because it's the only course he got offered over there. He didn't even take biology in high school!
 
I suspect anyone motivated enough to study for a physics course would find any science interesting to do.
 
user228700
Well, yes, but he wasn't even interested in biology--he wanted to do computer engineering!
 
11:53 AM
From a young age I wanted to study physics and discover the unified field theory. But then I ended up doing physical chemistry instead because I found it more interesting. Then went to work as a colloid scientist.
There's not much call for general relativity or quantum field theory in colloid science :-)
 
user228700
Your story makes sense because you were led by passion. In his case, he was forced to take Bio simply because he couldn't do anything else. He didn't want to do any of the sciences at first--just engineering.
 
I guess my point is that if you're forced into studying a subject you didn't intend to you can probably find something interesting in it if you look hard enough.
 
user228700
And now he's doing a 5 year course in Bio. I am using his case to illustrate the point that the system is weird but in this particular case, he does seem quite happy, yes, so no problem :-)
 
You might end up specialising in electrodynamics :-)
 
user228700
For all I know, yes. Geez, I sure hope not :-P
 
11:56 AM
Whoever starred my male nerds comment, I'm not sure it's good to have that on the star board.
 
user228700
U know it wasn't me!
 
While it may be true I still don't think we should be highlighting it.
Oswald?
 
IMO, you suit science
here we are in engineering stand all day along in the middle of the day
get burnt by sun
 
Hii @JohnRennie
May I ask a question
 
@user123733 Yes, of course.
 
12:09 PM
Let there be two blocks m$_1$ and m$_2$ both are attched with a spring and have a velocity v$_1$ and v$_2$ respectively . They are attached to same spring.
 
Aha, I've seen this. You asked a question about it yesterday?
 
With c.o.m frame how the both blocks are doing shm
@JohnRennie yes
But I could not understand it
0
Q: Basic Concepts on Blocks and spring

user123733 Let there be two blocks $m_1$ and $m_2$ both are attched with a spring and have a velocity $v_1$ and $v_2$ respectively . They are attached to same spring. Now I have some doubts related to it. With respect to centre of mass, how these blocks are doing SHM. And in ground frame can we use ...

@JohnRennie can you help me in this
 
In the the centre of mass frame the total momentum is zero. So $mv_1 = -mv_2$. Yes?
 
Yes
 
Now suppose we differentiate both sides wrt $t$. $dv/dt = a$ so we get:
 
12:15 PM
Ok
 
$m_1a_1 = -m_2a_2$.
 
Yes
 
And $ma$ is just the force because Newton's second law tells us that $F = ma$. So the forces on the two masses have to be equal and opposite.
 
hEY @JohnRennie. I have a question and hope you can help me in it. Don't ignore or refuse please.
 
Which is kind of obvious because if the forces weren't equal and opposite there would be a net force and the whole system would move, so we wouldn't be in the COM frame.
 
12:18 PM
@JohnRennie yes
Yes it is obvious
 
And the force is given by $kx$, where $k$ is the spring constant and $x$ the extension of the spring.
 
@JohnRennie ????
 
@supersaiyan give me a moment, I'm just doing a problem with @user123733
 
@JohnRennie okay ,which is directly proportional to x
So shm
 
okay. Hope you will notify me when its done. @JohnRennie
 
12:21 PM
Yes. It's a little more complicated than that because if mass 1 moves a distance $x_1$ and mass 2 moves $x_2$ then these are realted to the total spring extension by $x_1 + x_2 = x$.
And for the COM to be constant you can easily show $x_1 = -\frac{m_2}{m_1}x_2$
 
Ohh i see
@JohnRennie you mean c.o.m is at rest
 
@user123733 yes.
 
But it is not necessary
 
It depends on how you want to approach the problem.
 
If we take the situation when it is moving
 
12:24 PM
I'd probably just consider one of the masses and write an equation for its distance from the centre of mass.
@user123733 Making the system move just means adding a constant velocity to both objects so it doesn't affect their acceleration.
 
Ok @JohnRennie
 
Is it a bit clearer now?
 
Yes
 
I forget exactly how the reduced mass creeps in ...
But I think it's obvious once you start doing the calculation.
@supersaiyan: what did you want to ask?
 
I think we can use it ,but if we have to find KE using reduced mass ,then what is velocity @JohnRennie
 
12:28 PM
@user123733 I can't remember. Try working through the calculation and it will probably be obvious.
 
Okay ,Thanks
 
i want to ask something related to spinning of a coin @JohnRennie
 
ah.. the question was "Why does a coin fall when it stop rotating" ?? I mean it can remain still on its base (Does it have to do anything with C.O.M./C.O.G.)
@JohnRennie
 
I honestly can't remember how the mechanics of a spinning coin work out, but as I recall it behaves like a gyroscope.
So it falls over when it stops rotating for the same reason a spinning top falls over when it stops rotating.
 
12:35 PM
And why does a spinning top falls when it stop rotating???? :P :)
 
You need to read up on gyroscopes.
There isn't a simple answer because gyroscopes are quite complicated things and their mechanics is somewhat non-intuitive.
 
Ah okay... So I think I should say the same to my friends who have asked me that question... :P
I also have another question btw..
 
Yes?
 
Everybody know that internal forces can't move a system right...
Then how do we move when we walk/run etc..
??
 
Internal forces can't change the position of the centre of mass of a system, but they can make it rotate.
 
12:40 PM
No... but what happens when I walk for example...(I m not rotating there ) and AFAIk there is no external force
 
In this case the system is you and the Earth. When you start walking you exert a force on the ground (the surface of the Earth) and the ground exerts an equal and opposite force on you.
So you move in one direction and the Earth moves in the opposite direction.
But because the Earth weighs much, much more than you the Earth moves by an undetectable small amount and you move by a large amount.
 
Okay... (So, I move forward because of friction) ??
Which earth applies on me??
 
Yes, the friction between the soles of your shoes and the surface of the Earth.
If you were standing on ice and tried to run you wouldn't get anywhere because your feet would just skid on the ice.
 
Ah... Now I m understanding things properly...So, I won't get more acceleration than what friction could provide me
??
 
Yes.
 
12:43 PM
Oh... that one was baffling..
Wait a minute let me remember my next question..
 
I need to head off soon, so just one more question!
 
Jim
@JohnRennie Relevant addendum: Living in a frequently icy region has taught me that you won't so much skid if you tried to run as you would slip and fall down hilariously. Also true with just standing still (stupid static friction being mostly necessary for stationary balance)
 
Yes that's fine.. I have only one @JohnRennie
That was kind of funny mention @Jim.. :P :P
Okay @JohnRennie.. So this time the question is based on neutral and stable equilibrium..
 
Jim
@supersaiyan This is the goal I try to achieve each day
 
@supersaiyan OK
 
12:48 PM
I will provide you the whole para from my book and what they say baffles me... please wait a minute
@Jim.. :) :)
See that underlined thing @JohnRennie
 
OK ...
What's the question?
 
The question is what is the diffrence b/w stable and neutral equilibrium??
I will prefer exammples of these two situations as
in book... they means same to me
 
Suppose you put a ball on your desk (I'm assuming your desk is flat). If you move the ball sideways it just moves sideways and stay at its new position. Yes?
 
yes it does...
 
That's neutral equilibrium, because if you move the ball it doesn't try to move back towards its starting point or move away from its starting point.
Now suppose you put the ball inside a bowl, so it settles to the bottom of the bowl.
 
12:57 PM
hmm....
 
If you move the ball away from the bottom of the bowl and let go the ball will start to move back towards the bottom of the bowl i.e. back towards its starting point. That's stable equilibrium.
 
It looks fine but I think the first one is unstable equilibrium as the ball have no tendency to gain its equilibrium position
??
 
No. Unstable equilibrium means if you miove the ball it starts to accelerate away from its starting point.
For example suppose you turn your bowl upside down and balance the ball on top.
 
oH.. LIKE planets do when they are in mutual gravitational attraction??
 
If you get the ball exactly at the top of the ball it will stay there, but if you move it even slightly it will start rolling down the side of the bowl away from its starting point.
That's unstable equilibrium.
 
1:01 PM
yes... I understand it now... One more thing, I have got something
Wait
2
Q: Why don't things balance after they stop spinning?

TdonutWhen I spin a penny, I spin it on its "edge" (circumference). So why is it that when the penny inevitably stops spinning, it never comes to rest on its edge, instead falling onto one of its faces? It is not a problem with the "thinness" of the penny; its possible to balance a penny on its edge in...

 
Floris' answer seems good to me. What is it you are unsure about?
 
nope.. I didn't knew that this question has been asked on PSE.. I just recently found it
 
So, I thought I should mention it
:P
 
I have to go now. I'm usually around in the mornings (British time) if you want to ask anything.
 
1:04 PM
Okay... Thanks a lot.. maybe we meet again with I have some intresting questions
 
1:25 PM
2
Q: Is spacetime supersymmetry realized at TeV scale? and more questions on supersymmetry

mounitha Is spacetime supersymmetry realized at TeV scale? If so, what is the mechanism of supersymmetry breaking? Does supersymmetry stabilize the electroweak scale, preventing high quantum corrections? Does the lightest supersymmetric particle (LSP or Lightest Supersymmetric Particle) comprise dark ...

Too broad?
 
@JohnRennie It's HERE
I have an amplifier now!
 
I see 51 things in the close queue. All >3kers, please review!
7
 
@ACuriousMind give me 3k rep :P
EVERYONE LIKE MY ONE NOT CRAPPY QUESTION
 
1:42 PM
@Qmechanic Yes; I closed it and left a comment
 
1:57 PM
Question, is $\dot{H} > 0$ if the universe is dominated by phantom energy?
 
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