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2:23 AM
::grump:: I mostly loath the Hot Network Questions, but today it got me a badge I've been gunning for. Humbug!
 
2:34 AM
@dmckee Dmckee, can I ask ya a question?
 
Just did.
 
K thx bai
No seriously though.
I was thinking, main-sequence stars prevent collapse because they undergo nucleosynthesis, and their radiation pressure counters gravity.
Gas planets don't undergo nuclear fusion. Yet they are very massive, with little density.
Why don't their gravities compress them more?
I know their cores are supported by electron degeneracy pressure, but what about the rest?
 
Virial theorem.
 
Er...what?
Explain
 
The Virial theorem puts a limit on their compression. Potential energy increase must go into heating the gas in the short term, that holds it up.
Then they radiate and are able to compress a little more, but because they are pretty cool and small compared to starts that is a slow process.
It's actually the same thing that sets the size of stars.
 
2:38 AM
Wouldn't an increase in heat lead to less density?
 
But stars get into an equilibrium with the fusion process and hold the same size for longer.
@SirCumference Tries to. It's an equilibrium.
 
So...what prevents the further decrease in density?
Just gravity?
 
@SirCumference Yeah. What else?
 
Maybe they release the heat as radiation?
 
To lift some of the mass is to extract kinetic energy from it leading to a cooler average temperature which means it stops trying to expand. There is an equilibrium.
 
2:40 AM
I'm just guessing here.
 
@SirCumference They do, but that is a low process compared to finding the Virial equilibrium.
 
So where would they get the heat to expand?
 
Radiation causes the Virial equilibrium to change slowly over time. But of course, the cooler and denser they get the more slowly that process proceeds.
@SirCumference They don't without some kind of input. But they can't contract too much because that would warm them. There is an equilibrium.
I know I sound like a broken record with the equilibrium thing, but that's because that is all of the physics.
 
All right. So compressing too much would cause them to heat up, allowing them to expand before cooling again?
Basically "contract > heat up > expand > cool down"?
And repeat
 
@SirCumference Your trying to have there be steps that are followed in sequence, but an equilibrium is a state of constant balance with two processes evenly opposed.
 
2:43 AM
Ah.
 
So if you perturb the system you get some of that cycle, but it damps down right away.
 
Why wouldn't temperature alone be enough to prevent a star's collapse?
Because it also needs radiation pressure
 
Like the equilibrium between ice and iced tea. There is a slow overall melting, but the system is buffered against small perturbation in temperature.
@SirCumference Stars start shining a considerable time before they start fusing. They start out of equilibrium, warm as they contract (shine when they get hot enough) and eventually halt the collapse with energy from fusion.
Near stars, find their equilibrium before they start fusing and never get hot enough.
I do a BOtE calculation in mechanics about the sun.
If you magically turned off fusion, no one but neutrino physicist would notice for thousands of years.
 
All right, so why wouldn't a star just remain in equilibrium?
 
Because the Virial heating would compensate for a long time.
@SirCumference Well, eventually the core gets hydrogen depleted. Then it has to contract to warm up some gas further out enough to continue.
 
2:48 AM
Yes, but that's late in a star's life cycle
 
Eventually the center gets hot enough to burn hydrogen rapidly, and the dynamics change: red giant time.
 
I'm talking about early protostars.
 
@SirCumference They warm up as they compress, and shrink as the radiate. Until the fusion input is enough to compensate for the radiative losses.
 
Shouldn't they just remain in equilibrium, like gas giants?
 
@SirCumference They do. Meaning the collapse slows and is regulated by radiation.
But they're bigger than gas giants and get hotter, so they shrink faster.
 
2:51 AM
Er...sorry, so protostars emit radiation, contract, and eventually begin nucleosynthesis?
Sorry if I'm not understanding it right away
 
@SirCumference Everything hot radiates.
 
Blackbody, right?
 
Yeah.
 
As the system loses energy, it tries to cool, but that lets it contract which keeps the temperature up.
 
2:52 AM
So it will lose energy through radiation?
 
That's what gas giants are doing, too. But the planets do it slowly, and proto-stars do it relatively fast. Very fast for very massive proto-stars.
When the core gets hot enough, fusion begins. The core of planets never get that hot, which is how we know they are not stars.
 
So protostars begin in equilibrium, emit radiation, lose energy, contract, and eventually become hot enough to begin nuclear fusion?
 
why don't you read a book on it.
 
They begin out of equilibrium, reach an initial equilibrium that is pretty hot, emit radiation (which is energy loss, they're not separate) ...
 
Yes. Yes.
All right, now it makes sense.
Thanks!
 
vzn
3:13 AM
@ArnoldNeumaier hi saw you pop in. am a big fan after being tipped off to your work by TK & cited you heavily in this blog awhile back. wondering, have you seen the Couder et al oil drop experiments? any thoughts on those esp wrt your own research on semiclassical/ alternative theories?
 
@dmckee All right, one last question. Sorry...
If a gas giant could keep gaining more and more mass, would it begin nucleosynthesis, or would electron capture happen in the core, making the core neutron-degenerate?
 
Unless you're postulating some unusual composition the only difference between a big gas giant and a small star is mass. Add more mass and you make it a red dwarf.
 
3:40 AM
Hmm...couldn't an increase in mass cause stronger gravity, which would make electron capture favorable?
I mean, the core of gas giants is supported by electron degeneracy pressure
 
 
2 hours later…
5:18 AM
@AlfredCentauri Thank you dear Alfred because of your attention. I have simply done that by my questions, answers and comments and I never write that phrase in them. You had to read all of conversation and then accuse me. Do you know what is in my mind?
 
5:45 AM
-1
Q: Why should there be a -ve vote for questions?

Avinash MGEveryone have questions and most of them may be stupid but we must encourage them right? Even some of Newton's questions were pointless.So the negative votes may cause the curious minds to be unhappy and feel that they are dumb. So why should a question have negative vote in StackExchange?

 
6:04 AM
@PhysicsMeta Well, I put on my grumpy curmudgeon personality for that one.
 
6:42 AM
@Slereah Thanks for answering!
 
Answering what
 
what meant by angular distributions in measuring angular momentum
 
7:23 AM
@DeNiSkA See:
0
A: Transverse simple harmonic wave travelling in a string

John RennieWhen deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment sm...

@DeNiSkA I find it easier to use Google Draw. The vast majority of the diagrams in my answers were created using Google Draw. And it's free :-)
 
7:43 AM
@JohnRennie TikZ or bust ;)
 
@JohnRennie :3
Just trying to spread my knowledge and passion (somewhat like the plague)
 
8:26 AM
Bullshit (also bullcrap in the US) is a common English expletive which may be shortened to the euphemism bull or the initialism BS. In British English, "bollocks" is a comparable expletive. It is mostly a slang profanity term meaning "nonsense", especially in a rebuking response to communication or actions viewed as deceiving, misleading, disingenuous, unfair or false. As with many expletives, the term can be used as an interjection or as many other parts of speech, and can carry a wide variety of meanings. It can be used either as a noun or as a verb as in the question "are you bullshitting me...
 
@skillpatrol and your point is?
 
9:19 AM
no point @JohnRennie just an interesting history article :-)
especially the part: Brandolini's Law: "The amount of energy needed to refute bullshit is an order of magnitude bigger than to produce it."
so not only does one waste time, but a lot more energy
Hi @JohnDuffield
 
9:35 AM
@Danu I'm with you
 
9:59 AM
@vzn what do you mean by TK?
 
10:10 AM
@ArnoldNeumaier I would guess TK refers to Thomas Klimpel
 
10:20 AM
Hi @ArnoldNeumaier!
 
10:58 AM
Hi! Thomas Klimpel doesn't seem to have a blog to which vzn referred tonight.
 
11:17 AM
Hi, does anyone here know a little about orbital mechanics and/or tidal forces?
 
@NoahP that depends. Orbital mechanics can get very complicated very quickly. What do you want to know?
 
Hopefully this won't be the case! @JohnRennie If you had an object of a mass on the order of $10^10$ kg in orbit, how would you calculate the resulting tidal forces from this in comparison to those of the moon?
 
Suppose you are at a distance $r$ from a spherically symmetric mass $M$. The gravitational acceleration, i.e. the force per unit mass, is the usual Newtonian expression: $a = GM/r^2$.
 
Okay
 
Suppose you now move a small distance $dr$ away from the mass, then the acceleration becomes: $a = GM/(r + dr)^2$, so the acceleration is reduced.
 
11:27 AM
Yep
 
4 hours ago, by John Rennie
@DeNiSkA I find it easier to use Google Draw. The vast majority of the diagrams in my answers were created using Google Draw. And it's free :-)
 
This is the cause of the tidal force. Over the distance $dr$ the acceleration changes. If you crunch through the algebra and make the approximation that $dr$ is much less than $r$ then you get the equation for the change in the acceleration:
 
@JohnRennie but how do you post it after drawing, do we to download the image/file?
 
$da = -2GMdr/r^3$
Note that the tidal force falls away as $r^3$ rather than $r^2$.
 
okay
How can this be implemented to calculate the tidal effect of a mass X on the earth?
 
11:30 AM
@DeNiSkA once I've drawn the diagram I sceen grab it, paste it into Paintshop Pro and crop it to make the bitmap I upload here.
@NoahP Put $r$ in as the distance from the mass to the centre of the Earth, and set $dr$ to the diameter of the Earth. That gives you the difference in the force between the side of the Earth nearest the object and the side farthest away.
 
::*point to be noted*::
 
@JohnRennie with M being the mass of the object?
 
@NoahP Yes.
Though the equation I gave is only valid if $dr << r$ and that probably won't be true unless the orbital distance is very great.
 
How does 'very great' translate into numbers?
@JohnRennie I'm presuming all distances to be in km?
 
@NoahP Life will be easiest if you stick to SI units, so use metres for distance.
You want $r >= 10dr$ for the equation to be valid. You can still calculate the tidal forces if this isn't the case, but the equation will be more complicated.
 
11:37 AM
Okay, so mass would still be kg? Thanks!
 
11:47 AM
@JohnRennie I calculated it in both metres and kilometres
The kilometres answer is coming out correct
Having to multiply metres answer by 1000000
 
12:03 PM
0
Q: How do I make this comment?

Anubhav GoelHi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. While reviewing I have to make this comment certain times. How do I make it? Do I have to keep its copy so...

 
12:32 PM
any idea what will be the condition so rim will not "jump"
 
12:43 PM
@NoahP somewhere in your equation you've got a quantity expressed in km. If you use metres for everything the equation will give the correct answer.
 
1:05 PM
@skillpatrol : hi. I didn't mean to join the room this morning, my google chrome just displayed the last page I was on.
 
And I have to go now I'm afraid.
Bye.
 
cya later
:-)
 
1:54 PM
Is this a proof of Poincare duality or just a physicist derivation somehow having flaws, it's soooo nice img.ctrlv.in/img/16/04/27/5720b56ec8b65.png ?
 
2:36 PM
1
Q: Limit of sequence of continued fractions

RFZUsing fixed point theorem evaluate limit of sequence of continued fractions: $$2,\quad 2+\dfrac{1}{2},\quad 2+\dfrac{1}{2 + \dfrac{1}{2}}, \cdots$$ How to prove it?

That's the kind of shit I want
 
3:08 PM
@dmckee I rather like your grumpy curmudgeon personality :)
 
I don't care about the history
I want the convergence!
And sum
 
that converges to $1+\sqrt{2}$
 
Well yes but I don't care for that one
 
the proof is trivial using the fact that $\sqrt{2}=1+\frac{1}{\sqrt{2}+1}$
 
3:20 PM
Mine is $a_{i+1} = -1 + \frac{1}{4a_i}$
With $a_0 = -1$
 
it is not exactly a continued fraction
 
vzn
@ArnoldNeumaier (welcome back to this chat room/ site after your long absence. hope you find it to have a different spirit...) TK has a blog but hasnt written on physics yet. was referring to my own blog eg superclassical/ emergent QM developments which cites you & Couder/ Bush/ Wolchover et al.
 
Well, generalized continued fraction
I think
Whatever that is I could use the convergence and sum
 
convergence of the sequence or of the sum?
 
Both, hopefully!
But also the sum
$\sum a_i$
I think it may be Important
 
3:32 PM
I doubt that the sequence converges...
 
Hm
Unfortunate
First terms are like
$-1, -\frac{3}{4}, -\frac{2}{3}, -\frac{5}{8}, -\frac{2}{5}, -\frac{5}{12} ...$
 
no
 
Isn't it
 
$-1,-5/4,...$
 
I might have written the sequence expression incorrectly
 
3:34 PM
as you have written it, it is monotonically decreasing
 
I guess it's more $a_{i+1} = -1 - \frac{1}{4a_i}$
 
and decreases of more than 1 each time
 
I suspect it converges to 0
Which would certainly be nice
Not quite sure what the sum would be though
 
the convergence should be provable easily by showing it is monotonically increasing and bounded
 
3:49 PM
It is not monotonic, though
 
no?
 
One term is -2/5, the following is -5/12
-0.4 and -0.41
 
then it will probably neither converge
 
You can determine the possible limits by plugging in the limit $a$ for $a_{i+1}$ and $a_i$, giving $a^2 = - a -0.25 \implies a^2 + a + 0.25 = 0$, which is just $(a+0.5)^2 = 0$. So either your series converges to $a=-0.5$, or it doesn't converge at all.
 
3:53 PM
I guess that's not a good sign for having the sum converge
 
Either way, it's not a sequence going to zero, so the series doesn't converge.
 
Although
I suspect that the real series I'll get in the end will be like $\sum \frac{a_i}{2^i}$
I should do the full calculations once I got some free time
 
W.alpha suggests it converges to $-1/2$
the sequence
but maybe it oscillates forever
who knows
 
It is related to that path integral thing I've been investigating
btw does a term like $$\approx e^{i (\int dx \phi_a(\vec x) + \int dx \phi_b(\vec x))^2 }$$ sound like a reasonable form for the transition amplitude between two field configurations
I think there will be a $\Delta t$ somewhere too
That is just a vague hunch, tho
 
don't know...
 
4:02 PM
Also
 
why the form should be so explicit?
 
When I did path integrals for non-relativistic QM, I had the odd relation of the form $dx \approx dt^2$
Or somesuch
 
?
 
A heuristic due to the non-differentiability of curves
but this shouldn't happen here, I think, because the Heisenberg uncertainty isn't on $x$ and $p$, right?
 
that's too much heuristics for me, sorry
 
4:05 PM
Man what kind of physicist are you
 
I am none
 
Do you wait for mathematicians to prove a thing before using it? :p
No, just charge!
 
@Slereah Is that a (very crude) way to express that "a typical path in the path integral has Hausdorff dimension 2"?
 
yes.
it works pretty well for lattice stuff
And it's a great way to find the Ito integral
Hm
I wonder where I got that heuristic
I kinda forgot it
Probably from Demichev
There's a demonstration you can do with the RMS of Wiener processes, but an even cruder one is $\Delta x \Delta p \approx dx \frac{dx}{dt} \rightarrow dx^2 \approx \hbar dt$
 
::twitches::
 
4:14 PM
Oh wait I guess that's not quite how you'd write $\Delta p$
Something along those lines, anyway
Yeah that's how you do PHYSICS
None of that rigor
 
::weeps::
 
::pukes::
 
IIRC the serious demonstration uses the property that $\forall c \geq 0, X(t) = \frac{1}{\sqrt{c}} X(ct)$
It has been a while though
$\text{(Left as an exercize for the reader)}$
 
4:52 PM
Hi @JohnRennie ?
 
@NoahP Hi Noah
 
I'm a little confused over the tidal forces we discussed
 
Go on ...
 
What does the formula $-2GMdR/r^3$ actually quantify?
 
A tidal force isn't something that acts at a point, like the gravitational force.
If I have some object of some length d then if there is a tidal force there will be a difference between the gravitational forces acting at the two ends of the object.
The result is that the object is either stretched or compressed depending on how it is orientated.
The stretching/compressing force depends on the length of the object. If you double the length of the object you double the magnitude of the stretching/compressing force. OK so far?
 
4:57 PM
Yes, all makes sense
 
The equation $F_t = -2GMd/r^3$ is the net stretching force on an object of length $d$ at a distance $r$ from whatever mass is generating the gravitational field.
 
If I were to calculate the force between the earth and moon using G((m1*m2)/r^2) at the closest point to the moon and the furthest point, would the difference between these be equal to the tidal force?
 
Yes
 
despite the fact that the centres of mass would not be located at those points?
 
We generally assume that the size of the objects is small compared to their separation. This is certainly true for the Earth and the Moon. In that case all of the Earth is approximately at the same distance from the Moon.
 
5:02 PM
Ok, so the effect is negligible?
 
What effect?
 
Of the centre of mass not being located at the distance from the moon we are using
 
5:43 PM
@NoahP Not negligible or we wouldn't both with the tidal force, but well approximated by the first correction (which is the tidal force). That is, the second order effect is assumed negligible.
 
user116211
6:00 PM
> @JohnDuffield : I can give you both a correct answer in simple terms and the fairy tale, together with references to an explanation how the fairy tale is related to the real thing!
 
6:19 PM
@vzn thanks! I read you blog post and your answer at physicsoverflow.org/24612 . I know of Khrennikov's work and some other stuff but don't like it since it is too far removed from the standard quantum mechanical shut-up-and-calculate procedures. I don't think anything has a future that introduces stuff beyond what is needed in this most common approach.
I also stopped formally publishing on quantum foundations since it attracts the wrong kind of people. So I just put my insights primarily into articles visible from my home page and into discussions at PhysicsForums such as physicsforums.com/posts/5456567
 
@ArnoldNeumaier Hear, hear! Quantum foundations is a tricky field to be active in, I imagine ;)
 
I was also very disappointed of Bohmian mechanics. It explains nothing without taking the whole universe into account (hence is useless for small, open systems)., and spoils the kinematic symmetry of quantum mechanics under linear canonical transformations by giving position a special role.
 
vzn
@ArnoldNeumaier hi, ah... so you gave up on writing on interpretations, or finding/ researching them also? :|
 
@Danu Not tricky. But the standards are too low; so publishing makes you look in the eyes of others being in the low quality pool.
 
vzn
6:35 PM
@ArnoldNeumaier many authors/ scientists have talked about how difficult it is to publish in the area, because of career implications/ limitations; bell wrote of his own struggles, and a new book by gisin mentions it in the intro etc...
am not a huge/ strict adherent/ proponent of bohmian theory myself but think it was "a/ some step in the right direction".
 
@vzn I have since 2007 my own thermal interpretation mat.univie.ac.at/~neum/physfaq/cei and polished it a bit with time. It will be a tiny part of a book on quantum mechanics that I am supposed to finish by next year.
 
@ArnoldNeumaier Well, that sounds pretty tricky to me ;)
@ArnoldNeumaier Well wasn't the whole point of it to give up locality?
 
@vzn yes, for the reasons I just mentioned.
 
vzn
@ArnoldNeumaier yes, think thermal interpretation has some validity, have delved into it some... am glad youre not giving up on that at least...
 
I got "the joy of cats" in the mail
No cats at all
 
vzn
6:40 PM
@ArnoldNeumaier but (alas) it sounds almost as if youre giving up on "the truth"™ and sticking to "what is politically/ academically achievable...
 
It's all category theory
 
@Danu No. I restore locality by taking a quantum field theory point of view and giving expectations and correlation functions the status of beables. This matches with its use in thermodynamics and statistical mechanics, hence the name of my interpretation.
 
It will get a harsh review!
 
@ArnoldNeumaier I'm talking about the Bohmian perspective.
 
@vzn The truth doesn't depend on whether it appears in a journal or online.
 
vzn
6:41 PM
@ArnoldNeumaier exactly my point...
 
(I don't know anything about your own interpretation, so I cannot comment)
 
vzn
do you think there is some way to experimentally distinguish thermal QM from std QM, any thoughts on that? that would certainly draw some attn...
 
@vzn The whole point of the interpretation is that it reproduces all the well-tested predictions without introducing the philosophical problems of taking the Born rule as an axiom. Stop trying to read your "mainstream physics has it all wrong" belief into what he's saying.
It's really getting tiring.
 
@Danu ah; they add to quantum physics a useless true position that cannot be measured and depends nonlocally on the wave function. I don't want to introduce unmeasurable stuff with a bizarre intuition just to superficially justify the traditional postulates
 
vzn
@ACuriousMind ?!? are you talking about thermal QM? what do you know about it? plz dont put extreme words in my mouth, setup strawmen etc :(
 
6:44 PM
@ACuriousMind Don't feed...
@ArnoldNeumaier "bizarre intuition"---I guess the whole area is just about disagreements as to what these words mean
 
@Slereah It will be only a tiny part of the book, which is mainly about quantum mechanics in terms of symmetries.
 
vzn
ACM maybe you are not aware, thermal QM is not strictly "mainstream" physics (quite intentionally so/ by design)...
 
@vzn of course not. My main principle is to map only what physicists actually do in the applications of quantum mechanics.
 
@ACuriousMind is the snake lemma more general than the "derivation" of the long exact (homology) sequence?
 
I was talking about "The joy of cats"
 
6:46 PM
Or is that just it?
 
I was disappointed to find no cat photos at all
But instead nothing but functors!
 
@Slereah Slightly unfunny :P
Functors are more reliable than cats, anyways :)
 
Well a mildly amusing title deserves a mildly amusing joke
 
vzn
@ArnoldNeumaier but if its different than "std" QM, then it must have some limit cases that can be "pushed"... this was the rationale/ raison d'etre of the early applied Bell tests by Aspect et al etc
 
He could have called it "Go funct yourself"
 
6:48 PM
haha
 
So from general vibe I am guessing there is a Crank Alarm
 
Better
 
Who is the culprit
 
@Danu I can't recall any application of it besides constructing the long exact sequence, but the lemma itself has nothing to do with (co)homology in particular.
 
@ACuriousMind OK. We constructed the long exact sequence today in top2---it was tiresome :P
...but no snake lemma in a more abstract sense.
 
6:51 PM
@vzn My thermal interpretation is an interpretation of standard quantum mechanics - nothing is different except for an improved way of talking about it that removes the standard problems in the interpretation. The standard postulates are far too idealized to describe the actual practice of QM; so it is no surprise that taking them absolute produces conflict.
 
vzn
@ArnoldNeumaier have read some of your papers & plan to go back & review now, and basically agree with/ support the underlying rationale/ idea, but if the (std QM) "postulates are too idealized" then one would expect some discrepancy with experiment at some pt, or some way to contrive/ construct/ push such a "shearing"...
 
@Danu For me it is enough that I myself found Bohmian mechanics bizarre, and completely useless. I don't need pilot waves to se that an instrument sitting at x cannot be influenced by the part of the wave dunction far away from x, so that that part can be discarded without problems.
 
vzn
(anyway this is very interesting & hope to revisit at length later, have to go)
 
@ArnoldNeumaier =) Fair enough.
What other parts of physics interest you?
 
@vzn I don't expect discrepancy with experiment since i see lots of discrepancy between the standard postulates and how QM is applied in the shut-up-and-calculate mode. The latter is the stuff that has been found in agreement with experment, not the postulates. Thus the postulates can be altered, whereas the shut-up-and-calculate part must be preserved - exactly.
@vzn I am not often here; so don't expect to find me again....
 
6:58 PM
@ArnoldNeumaier :-(
The h bar can be really nice!
 
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