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4:24 AM
I am so hungry.
 
5:02 AM
A cup of warm water is suspended in a large pot of water held at a steady boil.
Will the water in the cup ever boil?
Answer is no
When the cup of water is placed in the boiling water, the cup is cooler than the surrounding water and heat will flow into it. Eventually, the water in the cup will increase to 100 degree C if any more energy goes into the cup, then the cup will begin boiling.

But at this point, the boiling water in the pot and the 100 degree C water in the cup are at the same temperature. Since there is no temperature difference, there will be no more heat flow into the cup, so it will never boil.
I got this question from fb from a friend
I think answer is yes
is that explanation saying it takes more than 100 degree c to boil?
assuming atmospheric pressure is same as on earth
sorry i understand it now lol
sorry for asking dumb question
 
5:25 AM
@SpecterProphet it depends what you mean by "boil". The water in the cup won't form bubbles of steam because forming a bubble requires a temperature greater than 100°C.
 
in other words after boiling point it takes more energy for it to boil since water in pot has heat source it is boiling but water in glass doesn't bave one and it reaches only 100 degree since we know that the heat flow between two bodies will stop once they are at the same temperature
vaugely saying
I have some mistake in statement -_-
 
When you put a pan on the stove the burner heats the metal to more than 100°C, and this creates the bubbles that you see when water boils in a pan.
But the water itself is at 100°C so it can only heat the cup to 100°C and no higher.
That's why we don't get bubbling inside the cup.
 
looks like there are two types of boiling
and u mean nucleate boiling
 
@SpecterProphet yes, I mean bubble nucleation.
 
5:40 AM
I think both definition applies
The water outside the cup receives more heat from the heater and starts boiling but as the cup and its surroundings are in equilibrium there is no heat being transferred into the cup this means that the water in the cup is at 100 degrees but it doesn't get more energy to turn into steam and hence it does not boil
I was thinking about phase change than bubble
 
This is why I said it depends on what you mean by "boil". The water in the cup reaches 100°C and turns into steam. So in that sense it does boil.
But for many people, especially non-scientists, "boiling" means "bubbling".
 
so a scientist will say yes and non scientist will say no?
 
Probably ...
 
The surrounding water is boiling doesn't that mean steam bubbles present can't they be at a temp greater than 100°
 
and it just takes 100 degree c for it to boil not additional energy required to change it's phase if boiling means chf?
 
5:55 AM
@Protein there is a lot of water and very few bubbles, and the water keeps the temperature at 100°C. The steam inside a bubble will be greater than 100°C, but his won't make any significant contribution to heating the cup.
@SpecterProphet The water to steam change at the surface of the water occurs at 100°C and doesn't need any excess temperature. So the water in the cup will all turn to steam despite its temperature never exceeding 100°C.
 
It would be boiling without bubbling?
 
Yes
 
Heat is needed to increase temperature and change the phase of water in the cup.
Heat transfer can only happen if there is temperature difference.
The reason the water in the pot can boil is because there is temperature difference between the water and the heat source. Since there is also temperature difference between the water in pot and cup, heat transfer into cup happens to increase the temperature of the water in cup to its saturation temperature. However since the water in the pot is always boiling, its temperature is constant so when the water in the cup reached the saturation temper
But I am wrong since 100 degree Celsius is when water change to vapor so answer it yes it is boiling.
 
Vapour pressure of water equals atmospheric pressure . So water is boiling at surface but what about bulk of water is boiling!
 
lol didn't thought it will become this confusing
Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the temperature at which the vapour pressure of the liquid is equal to the pressure exerted on the liquid by the surrounding atmosphere
OK i take Wikipedia definition and accept it boils
 
6:21 AM
Hi everyone !
Is it possible to have notification like we get from any other apps like whatsapp on stack chat whenever someone ping me ?
 
 
2 hours later…
7:52 AM
@JohnRennie well it seems I am right latent heat makes it have phase change
getting 100 is okay but u need to try to exceed in order to have latent heat and phase change occurs
100 degree c is limit i think
I think i am sloppy but u might get it
 
8:12 AM
ok I was wrong
really depends on boiling def
 
 
2 hours later…
10:13 AM
It's like in Planck scale, space-time has radically different structure from what it is at macroscopic scale.
 
suppose I submerge a cube in water
will the pressure in bottom side of cube be greater than top?
I think yes
yes it is lol
today I am having dumb question day sorry for interrupting
 
10:48 AM
What if I have some data, let's say I'm trying to answer if education and IQ affect earnings, and I want to analyze this data and put in a linear regression model to predict earnings based on IQ and education. My confusion is, what if the data is not linear or polynomial? What if it's a mess but there are still patterns that the linear plane algorithm can't capture?
How do I even figure out if plotting all of the independent variables will form a line or a polynomial curve? I mean, with one dependent and one independent variable it's easy because you can plot it and see, but in a situation with multiple independent variables... how do I figure out if the relationship is linear or something like this? How do I figure out if I should use regression models?
 
How nice if the shop put computers for everyone to use freely.
 
0
Q: What could be introduced to provide a source of propulsion through time?

user271517This Ape has been suspended in a 4 dimensional anti-higgs field time sphere, where as the only source of gravity interfering with this chimp is the interior materials and the chimp itself. What could be introduced to provide a source of propulsion?

 
11:16 AM
What is the positive frequency condition in complex fields?
 
can i ask a doubt?
for further details see the room
 
yesterday, by ACuriousMind
@ronakjain Please don't ping random people to answer your questions; if someone here can answer and wants to, they will
 
@Slereah understood
so are you answering?
 
'fraid not
 
yuvrai can you?
 
11:26 AM
cf. above
 
"if someone here can answer and wants to, they will"
 
Stop asking random people
4
 
well i was asking so that i can go if no one's answering
 
if someone answers they can @ you and you will get a notification
aside from that there is a specific chat room for asking homework type questions
 
also try to do your damn homework
if you didn't try anything it's not particularly enticing to help you
 
11:30 AM
@Slereah i have done all 48 questions
just have doubt in this one
 
people often answer questions here very kindly but not everyone wants to drop what they're doing and grind through classical mechanics calculations
 
i underrstand
 
I vaguely and irrationally resent how useful WebPlotDigitizer is.
2
 
Jim
@Slereah Love it! The IT department is always coming down to my labs to try to help and always surprised to see that I've already built all the tech stuff I need. Normally, I only call them when it's something bureaucratic, like all the networked computers need to be re-imaged or something. It's like they think scientists can't figure out computers
.... let me backtrack a bit. Some scientists can't figure out computers
 
hmmm, kindly.
 
11:44 AM
@Slereah "Excel"
why does that program exist
 
Jim
@Charlie Preach! I'm a "teacher" and I still cringe every time I'm asked to drop what I'm doing and jump into a random physics problem
@SirCumference standby for xkcd reference
that's why. So church groups can break the laws of complexity
 
@Jim well...that's a little more specific than i expected to relate to
 
Jim
he's got one for everything
 
@Jim I've had to deal with the right side
Did some work for like
Companies that didn't have any programmer
And had a huge horrible excel script for their project management
 
Jim
lol, yeah. And you know they ain't commenting it cause it's excel
 
11:49 AM
for me it was just using excel in my classical mech lab
i guess i didn't have it that bad all things considered
still a bad program to be used so much
 
Jim
Excel is great. People just need to know when not to use it too
 
imo it's just very limited and e.g. getting graphs to look the way you want is overly time consuming
basic features like plotting functions with a continuous domain aren't an option
 
@Jim they weren't even using variable names
Just column numbers
 
Jim
most people don't use it for plotting. They use it for the ability to make records and have changes to those records update immediately without thinking
 
welp that is the correct way of using it
 
Jim
11:54 AM
@Slereah Jim shudders
 
unfortunately it's been like 20% of the use i've been assigned in my life
 
Jim
you have my sympathy
@Slereah that's kind of a "you're trapped" situation. I get students coming by all the time with excel sheets saying "what do these numbers mean?" And I get to say "I don't know because you didn't label anything. Go away until you can give me anything to work with". But if you're IT, people don't accept that answer. That sucks
 
Well you can reverse engineer it
But it is annoying
 
Jim
I'll bet
 
12:12 PM
0
Q: Comments to point out errors in OP's understanding

BioPhysicistDue to recent questions about the comment policy, and because I have seen this being done often, I am here with another question about comments. In the question The Law of Density at Extremes (that has been deleted, but access to the question isn't important here) a user commented A primary purp...

 
 
2 hours later…
2:26 PM
i haf a new pic yay
 
Hi physics!
Is anyone here well-versed in big O notation
I have a question in the context of the Lieb-Liniger model
But the context may not matter much since it's just an equation
 
2:42 PM
@1010011010 hi account
 
Did you reply because I said hi physics or should I ask the question :p?
 
@1010011010 I replied cuz hi physics triggered me, i'm a human not a subject xD sorry I'm not qualified to help you with this
I can help you set up a Minecraft server tho
 
Interestingly I already have one running on an old linux laptop
 
is ur name binary?
btw you should directly ask the question and not ask to ask
 
It's binary for 666
 
3:17 PM
AH great the thing uploaded
Fair, so I'm looking at some of these equations and I think that my big O is not correct
So some explanation about the equations
L is system size, epsilon is a small parameter of order 1/L
theta_jl is a phase of the difference of two parameters lambda_j and lambda_l
The set {l^0} is of order N
I'm assuming N = L in this calculation for simplicity
the limit "limtd" means N and L both go to infinity but N/L remains constant (thermodynamic limit)
The set {l^+} as well as {l^-} has cardinality "n" which is assumed to be a finite number in this limit
rho is defined as the difference of two subsequent lambda's (or mu's in case of underline rho)
 
If no-one in here knows you could also ask in the mathematics SE chat here
 
Thanks! I might just go ahead and ask there as well
Anyway back to my question,
The problem is that F_j = epsilon_j / L rho_j
epsilon is of order 1/L, 1/L is also of order 1/L (duh) and rho_j is of order 1 since rho_j^{-1} = L(lambda_{j+1} - lambda_j)
So it would seem that displacement F is of order 1/L^2, while its definition would seem to suggest otherwise
On the other hand, in the thermodynamic limit, 1/L sum_{{l^0}} (.) tends to int_{-q,q} dlambda rho(lambda) (.), which becomes a single term
Consistency thus imposes that each summation must be considered as one 'term' (not really, but it must really be considered as one element)
Which is turn implies that 1/L sum_{l^0} (.) is one term of weight 1, since the Cardinality of set {l^0} is indeed N giving the whole sum excluding the 1/L prefactor a weight N, and thus the whole sum including the prefactor weight 1
NB: "q" is the value of the largest instance of lambda in the continuous limit
Now F_j would appear to be of order 1/L^2 but this would not be consistent with the traditional derivations
oh I also forgot to mention that j is assumed to be in the set {j^0}
and that mu_j and lambda_j have a 1/L spacing in the limit
Regardless, looking at the last line (43e), theta_jl - theta_{underlinej underlinel} requires either of these phases to be expanded in the variable of the other set {lambda} or {mu} (in this case mu since F is defined in terms of the variable in thsi set)
 
3:36 PM
Just fyi it's a bit hard to read what you've written because it's not in mathjax, you can use mathjax in the chat by clicking here.
 
In this case, the derivation appears to hit a snag and most authors are just blatantly inconsistent with how each of the terms is expanded
Thanks Charlie, I think it's too late to edit my older messages
 
all good
 
3:59 PM
is a spin-vector just a spinor?
 
4:13 PM
by spin vector you mean a spin state $|\psi\rangle=\alpha|\uparrow\rangle+\beta|\downarrow\rangle$? this might answer your question @CaptainBohemian
 
@Charlie "I think that I had very much come around to the view that massless particles and fields were to be regarded as more fundamental than massive ones. My own reasons for believing this were probably very much bound up with the idea that 2-spinors should be regarded as more fundamental than Minkowski world-vectors, since it is a null vector (the 4-momentum of a massless particle), rather than a timelike one, which arises naturally from a single spin-vector of the 2-spinor formalism."
 
ah if that's qft I will tap out
 
When you read the wikipedia article on spinors, where do you get confused?
 
I forget if I have ever read the wikipedia article on spinors.
I learnt spinors from textbooks and my MSc advisor, not Wikipedia.
 
Yet you are still asking about that difference
 
4:22 PM
and some papers subsequently.
I am just asking what the spin-vector in the above text means. I am sometimes not very sure about some terminologies.
 
The spin vector is a vector operator built out of the Pauli matrices
 
@CaptainBohemian It seems very clear to me from context that "spin-vector" in the passage you quoted refers to the things the 2-spinor formalism uses.
This is different from the ordinary usage of the word, which is what bolbteppa just said
 
I think what that passage is saying is that $X = \begin{bmatrix} t + z & x - i y \\ x + i y & t - z \end{bmatrix} = t I + x^i \sigma_i$ is the spin vector, and $X^T X = t^2 - x^2 - y^2 - z^2 = 0$ arises naturally
I don't think it's $X^T$ it's something, the adjugate I think, the non-relativistic case is just $XX = 0$
 
4:39 PM
In MTW, it's just been said that the electrstatic equation of maxwell is: $$\vec \nabla \cdot \vec E =4\pi \rho,$$ where does this $4\pi$ come from? Is this a different unit system or something? I couldn't see any reference to it in my search
wait
Gaussian units, got it
 
ah that's also helpful thank you
 
@Charlie hi, can I ask a question? It can be regarded as both a physics/ physical chemistry question.
 
I can try
 
4:42 PM
Its based on Le-chateliers Principle
 
ok I faintly remember that
 
Why does increasing pressure cause more ice to melt?
 
my immediate response would be that if this isn't the expected result it has something to do with water/ice having flipped densities from what you'd expect
 
By LCP, if you increase the pressure , the system tries to decrease the pressure. What i dont understand is then why a reduction in volume (shifting to water) is how the system achieves it?
 
My reasoning would be that water is less dense, so if you increase the pressure the equilibrium is going to shift towards the product will higher density, which would be water
If you increase the pressure, the system tries to produce something with a smaller volume, this leaves more room for air in your container which would decrease the pressure
 
4:46 PM
oh
 
wait, but is it really appropriate to say that the air in the container will decrease pressure?
 
not sure what you mean
 
@Charlie yes, i just read that. "It can do that by making itself fit into a smaller volume". The reasoning behind this statement is all I am after.
 
If the ice has a lower density, the same number of water molecules in ice form occupy a greater volume than in liquid water form. So if you melt ice the volume taken up by those water molecules will decrease
If the water in the container then takes up less space by melting, this is equivalent to increasing the size of the container that the gas can exist in, lowering the pressure
 
4:54 PM
ah, i see it now
Thanks for your help!
 
no problem :)
 
 
2 hours later…
7:06 PM
ahh i love my new prof pic
 
7:28 PM
I can't find reference to it online, but I've just been introduced to the "wedge product", and it seems like given the hilbert space of two-particle fermion states, $\mathcal H_{\text{fermion}}$, we could write this as: $$\mathcal H_{\text{fermion}}=\{|\psi_1\rangle\wedge|\psi_2\rangle\},$$ does this work?
My reason for thinking this just comes from the following in MTW: "Wedge product is merely an antisymmetrized tensor product: given two vectors $\vec u$ and $\vec v$, their wedge product, the 'bivector' $\vec u\wedge \vec v$, is defined by $$\vec u\wedge \vec v\equiv \vec u\otimes v-\vec v\otimes \vec u"$$
 
@Charlie The "official" name is exterior product, and it is just the antisymmetrized tensor product, yes.
 
pretty neat
ty
I can't see any reference to a "symmetrized tensor product", is there a name for this? $u\vee v=u\otimes v + v\otimes u$?
 
7:44 PM
@Charlie It's the symmetric product
 
ah so we would have $u\odot v=u\otimes v+v\otimes u$
 
but the symbol the Wiki article uses is less common than the $\wedge$ for the antisymmetric variant, so you shouldn't bank on people recognizing it
Physicists usually don't write these (anti-)symmetrizations with operators, but with brackets on the indices of tensors
I.e. $(u\wedge v)_{ij} = u_{[i}v_{j]}$ and with round braces for symmetrization
 
oh interesting ok
 
8:11 PM
Suppose I want to find the COM of a large object. Suppose also I can break that big object down into smaller shapes for which I can compute the COMs. If I have the COMs for the individual sub parts, can I just take the average of all those subparts' COMs to get the overall COM?
 
@StanShunpike You could try deriving whether that's true or not yourself! It's just a bit of algebra
I.e. take your expression for the average of the sub-COMs and see whether it's equal to the expression for the overall COM or not
 
@ACuriousMind ok super i'll do that :)
@ACuriousMind I tried for some rectangular boxes and it worked
 
Is the notation $V^{\wedge N}$ in common use? Analogous to $V^{\otimes N}$
 
but i wasn't sure if it was generalizable to other shapes
 
 
2 hours later…
10:35 PM
@Charlie the wedge product is visualized in those pdf's as well
 
sorry which ones?
 
The ones I sent yesterday on visualizing forms
 
ahh got it thanks
 
1
Q: (Almost) ten years of Physics Stack Exchange

Emilio PisantyThis is something we should flag early if we want to do something in time for when it happens: the tenth anniversary of Physics Stack Exchange will happen in three months. Specifically, the private beta started on 2 November 2010 and the public beta started on 9 November 2010. So: do we want to d...

 
 
1 hour later…
11:57 PM
This exterior calculus stuff seems to just be tensor calculus on strictly anti-symmetric tensors
 

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