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12:01 AM
I should have known better, of course, I'd met about twenty physicists-by-training with advanced degrees at various adult functions my parents dragged me to. Exactly two were still doing physics, and most were not even in obviously technical fields.
@dmckee Thats a nice prospect for the physics students in here
But I was such a snot nosed little twat that I figured that was a sign that they were defficient somehow.
@0celo7 The market for BS's is rather more traditional. You can more fairly smoothly into engineering or into tech.
Taking the grad school plunge is entering your future into a lottery for the pleasure of pure learning, and a big hit in accumulated compensation.
I'm not a physics student
I wonder if the misery in this chat had anything to do with that
@0celo7 You keep saying that.
@dmckee I do, because it's true
I've done a good job of convincing myself of it
I really don't like physics
It's math I now have to rid myself of
12:10 AM
@dmckee Heh.
@dmckee You have to know people.
@dmckee Not your fault. Academia spreads that lie around and seems to be proud of it.
@DanielSank And the people I knew had retired, so they had little swing.
@dmckee ah
@0celo7 @0celo7: I posted an answer to your question at MSE.
12:26 AM
You missed the part in the OP where $p$ and $q$ are nonzero
@0celo7 You are absolutely right; I missed that.
You meant to say non-negative, not non-zero.
Anyway, yes, with that additional condition I think what you're claiming is likely to be true.
You're right
Strange, I distinctly remember thinking "nonnegative" when I typed that
(I get those two mixed up a lot)
@WillO I hope it is!
Well, it's much less of a blunder than my missing the fact that there was any condition in the first place!
I'd like an answer that doesn't use big machinery of homological algebra though
I'm afraid if I ask on MO they'll tell me to use a triple filtered Ext functor filter
or something stupid
12:30 AM
And my (geometric analysis) advisor will tell me to ask an algebraist
More likely they'll tell you it's off topic because the answer is obvious --- but it's not 100% obvious to me.
This Ricci flow thing WILL kill me
Are there any other conditions on the $K^{\bullet,\bullet}$? E.g. are they projective?
I don't know what that means
Or better yet, are they vector spaces?
12:34 AM
Indeed they are
Then it's just about got to be true.
I eagerly await your proof :D
I have no proof, but here is how a proof will go:
Break up your double complex into a direct sum of much simpler double complexes, all concentrated in just two total degrees. If all the $K^{\bullet,bullet}$'s are vector spaces, you can do this.
Concentrated in just two total degrees?
Do the same for $L$. Then you can break up your map $f$ into pieces that take one piece of $K$ to one piece of $L$ and analyze these....
Yes.... Total degree means p+q
12:38 AM
I still don't know what you mean by that
So---- pick your favorite $p$.
Your life would be much easier if all the non-zero terms of K were one-dimensional vector spaces that live in K^{a,b}'s where a+b is either p or p+1
I...don't know
Call those complexes "simple". (That is not standard terminology.)
You can write your complex as a direct sum of simple complexes.
I could be wrong, but I believe this will lead to a proof
12:41 AM
I'm sorry, I don't see what that has to do with vector spaces
I am invoking a theorem that says that if the K^{pq} are all vector spaces, then your double complex is a direct sum of simple complexes
Can you describe this in detail please?
I don't know much homological algebra at all
I wish I had a way to draw a picture here. Let me go draw one elsewhere and I'll be right back with it.
okay---is there a canonical place to post a picture, or is it possible to post one into this chat window? or should i put it elsewhere for you?
12:52 AM
see the upload button at the bottom
imgur wants me to create an account. try panix.com/~wotw/ocelo.gif
There you see four different double complexes (all non-pictured spaces are zero).
what kind of imgur are you using
These are even simpler than they look, because all of these vector spaces are 1-dimensional.
Theorem: Any bounded double complex is a direct sum of double complexes of this sort (and their obvious longer generalizations).
I'm not sure what "uhhh" means :)
12:56 AM
I don't know what this picture is telling me
You mean you don't understand what I'm saying, or you don't understand why it might be useful?
Okay....the first is probably a prerequisite for the second....
What is this picture supposed to be
@WillO Usually, yeah
It is not one picture. It is four completely separate pictures.
12:58 AM
oh, that's a helpful hint
The first picture is a double complex in which all but four entries are zero, and the other four entries are one-dimensional.
The second (to the right) is a double complex in which all but three entries are zero and the other four are one-dimensional.
The third is another double complex with just four non-zero entries. The fourth has just five, and other "staircase-like" double complexes can have six, seven, eight, etc.
If you knew that your double complex was one of these, it would be very easy to calculate with. It's easy to compute the horizontal, vertical, and total homologies.
But in the third picture
don't you get things from three total degrees
You don't know that, but you do know something along those lines: Your double complex is a direct sum of complexes that are this simple.
nevermind that
My total degrees are fixed along 45 degree lines that run NW to SE
1:01 AM
@WillO now that I do not see
you might be in the habit of indexing differently
you do not see.....what?
not all of the thingies in the complex are the same vector space
I do not see that $K^{\bullet,\bullet}$ is the sum of such complexes.
But they are all direct sums of one-dimensional vector spaces.
I am not claiming that it's obvious that your complex is a direct sum of such things. I am claiming only that it's true.
Ok, let us suppose that this is true
I am also claiming that I've often found it a useful thing to know, and I suspect it will be useful here, though I make no absolute guarantees.
1:03 AM
why is it useful?
Because it breaks big complicated problems up into much smaller uncomplicated problems (though at the cost of moving the complication over to how the solutions to the smaller problems fit together)
In the other direction, if you want to look for a counterexample....you can compute, from the structure of the double complex, maps $H^q(K^p,{\bullet} \rightarrow H^{q-1}(K^{p+2},\bullet}$$. You've got (by assumption) that the homologies are the same when you replace $K$ by $L$, but not necessarily that these maps between homologies are the same when you replace $K$ by $L$.
These are important invariants, and if they differ in some example, then that's a good candidate for a counterexample.
Okay, what you want is true. I have a proof, but it will take a little time to write it up.
But in brief (and recognizing that you might or might not know all these words):
Let C be the mapping cone of f. Then the first spectral sequence for the homology of C collapses at E^2. The boundedness condition says that the spectral sequence converges, and voila.
1:50 AM
6 hours ago, by dmckee
A lot of what goes on in these discussions is epistemology. Peolpe what to know what the quantum world "really is". Well, so do I, but I've had to come to terms with the lack of cog-wheels in there.
5 hours ago, by ACuriousMind
I get that it is somewhat dissatisfying that physics doesn't answer every "why" question you can think of, but that's, after all, not what it's supposed to do. It's supossed to predict nature, to give us a model we can work with.
One way to work out the why is allowing two "whys" of the same nature to interact with each other

As an example, before spectroscopy involving polarised light, we only knew something is an enantiomer because it interacts with another enantiomer
In the end at such level, we don't know what the "whys" are and why they act this way, but from these indirect experiments, we can identify what "why" we are dealing with

Another example is the stern gerlac experiment which told us an electron has a angular momentum like quantity that spans a 2 dimensional hilbert space. The maths then organise these findings into a systematic fashion so that prediction can be made from it. But what it "actually" is we don't know, other than it is some label of the representation of an more generalised notion of changing the orientation when looking at th
So I argue that most of the time, because most things cannot be directly probed, most of the whys have to be inferred from the hows, just like how daniel mentioned about the electromagnetic vector potential is a nice bookeeping device to predict motions of charged objects under electromagnetism
@Bass You nailed what I am trying to say back there
> Those "harmfully wrong" answers can be utterly confusing to students if one does not recognize them as wrong / non-mainstream. You read the answer and you try to understand and learn the (wrong) stuff in it. Then some time later, somewhere else, you read contradicting (but correct) content.
2 hours later…
3:41 AM
I'm annoyed that there's a featured question asking what a photon is.
It's good that the question "what is a photon" is getting attention, but it's less good that it's a duplicate which got away with being a duplicate because it 1) has a bounty, and 2) is (unnecessarily) long.
Also, @JohnDuffield, a question not having an answer is not an excuse to post a duplicate question. Instead, edit the original, put a bounty on the original stating that you want a better answer, campaign in chat for the original to get attention, or do nothing.
Don't post duplicates because you feel like it.
1 hour later…
5:02 AM
@PhysicsGuy You have to use lattice QCD to study glueballs because the perturbative approach doesn't converge. That's true for all QCD bound states and doesn't signify anything special about glueballs.
@jumbo Hey
@JohnRennie Isn't it late there?
@BernardMeurer Early not late :-) It's six a.m.
@JohnRennie Any servers turning evil?
I start work around 05:30 checking which of the 600 or so servers I look after have died overnight.
So far no problems.
It's the Windows update this week - it starts the second Tuesday in the month - and when 600 servers update and restart something usually goes wrong. But so far no problems.
Keeping my fingers crossed :)
5:07 AM
Most of the servers are configured to update on Saturday night.
We do that so I have all of Sunday to fix any problems.
Sunday morning (at 05:30!!!) is usually a bit stressful.
The Velvet Underground - that's practically musical archeology
I really like Velvet Underground
@0celo7 That's a GPIB / IEEE 488 connector - see radio-electronics.com/info/t_and_m/gpib/ieee488-connectors.php
@JohnRennie Holy shit how did you know that?
He asked me and I guessed it must've been some weird serial connector from the early 2000's
5:16 AM
Back in the stone age IEEE busses were quite common for hooking equipment together. We used to use them for running measuring kit from our BBC micros.
There you go, there are advantages to being old :-)
Could you daisy chain?
USB not being able to daisy chain sucks balls
RIP FireWire
They were actually a very good way to integrate stuff. Do a Google on it some time and you'll see how well designed the protocol was.
But it never caught on outside specialist applications and it was always expensive
@PhysicsGuy a virtual particle is a mathematical fiction. It's used as a description of a term in a perturbative calculation. the interaction described by the virtual particles is real, but the particles themselves are not.
@PhysicsGuy but in some circumstances the interaction is dominated by a single VP term to such an extent that the VP is nearly real. Push things a bit further and you get a real particle not a virtual one. This is what dmckee was referring to when he talked about boosting VPs to be on shell.
So the distinction between a real and virtual particle can sometimes be so fine that it's almost just a matter of semantics to say whether the VP is real or not.
5:44 AM
Is it correct to say that a real particle in the context of perturbation theory is an interaction term that dominates in the system?
@Secret No. Well, I suppose a virtual particle becomes real in the limit of infinite lifetime.
Which was what dmckee meant.
@Secret have a look at this Wikipedia article. This is quite a nice description of the difference.
In physics, particularly in quantum field theory, configurations of a physical system that satisfy classical equations of motion are called on shell, and those that do not are called off shell. In quantum field theory, virtual particles are termed off shell (mass-shell in this case) because they don't satisfy the Einstein energy-momentum relationship; real exchange particles do satisfy this relation and are termed on shell (mass-shell). In classical mechanics for instance, in the action formulation, extremal solutions to the variational principle are on shell and the Euler–Lagrange equations give...
It's when you get particles that are right on the borderline between on shell and off shell that the distinction becomes fuzzy.
6:03 AM
Hmm, so if I had a field whose interaction is "90% on shell (not sure if there's a way to formalise this, but what I have in mind is it is located very close to one of the mass hyperbola)", then I will find its interaction will be 90% like that of "classical" relativistic field while the rest of the 10% will be more quantum like such that conservation laws don't hold?
Suppose we consider some scattering process where some particles come in from infinity and other different particles head out to infinity. The way QFT describes this is by using creation operators $a^\dagger$ and annihilation operators $a$.
So the incoming particles get annihilated by $a$ an the outgoing particles get created by $a^\dagger$. The task is to calculate the probabilities for these operators.
@EmilioPisanty because it's too old. You can only cast VLQ flags on posts posted in the past week, I believe.
For an interacting field we have no way to calculate these operators precisely, but we can express them as a sum of various combinations of the free field operators.
So we end up with a sum of lots of different combinations of the free field $a^\dagger$ and $a$.
Each combination describes the creation and annihilation of the corresponding particles, so each term is like various different combinations of particles being created and annihilated. And these are the virtual particles.
NB I'd be cautious about starring these statements until ACM has had a chance to check them. This is my understanding but I would not swear under oath that this is a correct interpretation.
I see, Btw, it wasn't me who star them
I was it, because I found it wonderful, but now I am staring only the first sentence to find it laterf.
6:12 AM
So the point is that the real interaction is expressed as a combination of operators that look like they are creating particles, but they aren't really. That's why we say virtual particles don't really exist.
As we approach the mass hyperboloid we find the real operators are increasing dominated by a single term, and that's what we mean by the virtual particles becoming more and more real. But it isn't meaningful to say a particle is 90% real.
@JohnRennie These ops are playing not on a normal wave function, but on a Fock space. It is essentially an union of the 0 to infinite dimensional wave functions. Until this could I understand the things in the QFT books. What is not clear: do these operators have eigenfunctions? Hamiltonians, whose eigenvalues would be the observable energies?
Note that the Fock space only exists for free fields so it is an idealised structure. We don't know the equivalent of a Fock space for an interacting quantum field theory.
A creation operator creates a plane wave, but exactly what that plane wave is I have to confess I don't know. From the little I understood of ACM's attempts to explain this to me it is not simply the free particle solution to the Schrodinger equation.
@ACuriousMind: is this anything near the truth, or am I as usual wildly misunderstanding QFT?
12 mins ago, by John Rennie
Suppose we consider some scattering process where some particles come in from infinity and other different particles head out to infinity. The way QFT describes this is by using creation operators $a^\dagger$ and annihilation operators $a$.
@JohnRennie As I know, the creation operator creates a free particle, which is a plane wave with a beautiful $e^{j\omega t}$-like wavefunction.
@peterh See:
Jul 28 at 17:48, by ACuriousMind
The thing the creation operator creates is in no precise sense a plane wave because it is not a function of spacetime
and the associated discussion.
Uhh, it is harder as I thought...
6:22 AM
@peterh that's what I think too :-)
Basically there's no shortcut. To understand this stuff you have to sit down and learn QFT properly.
@JohnRennie I am whining the "simplicity" of the old eigenvalue-based QM
But there's such a high barrier to getting started. Every time I sit down with my QFT for Beginners books I stall in the first chapter due to my poor grasp of the maths required.
@peterh I suspect the simplicity of the Schrodinger approach to QM is actually not helpful if you want to go on and study it in depth. It encourages you to oversimplify then think that's all there is to QM.
@JohnRennie It uses a lot of abbrevations with a relative simple meaning, but it is not clear to me, what they really mean. It would be clear probably after I learned a lot.
6:47 AM
PSA: we should fix the homework policy
::looking for DavidZ's post::
7:12 AM
Before that discussion about creaton operator creating a plane wave, I was almost tempted to make a PSE question asking why all physical models can always approximate something either by integrals or series of some idealised things (It seemed to be unique to physics, as chemistry and biology and maths don't have such widespread use of such things) an why experimetnal result always seemed to agree with these approximations

The only case where I ever seen that some formalism becomes "discontinous" is when it requires a leap of the conceptual framework, such as you cannot derive any relativis
To elaborate what I am trying to say, recall how back in undergraduate physics when solving electric fields of a configuration of charges, all you need to do is integrate all the charge densities.

Another example is entropy, which is basically a counting problem in some sense

We also have quantum states which are superposition of some base state, vector spaces made of basis vectors

Even on the abstract level of groups, one can define some continous map from one transformation to another

And in differential geometry, you have continous deformation of something
Putting the above paragraph in "psedomathematics" form. why is the problem solving approach

$\text{Actual thing}=\lim_{n\rightarrow \infty} f(\text{Idealised things}_n)$ works so well...?
NB: "Pseudomathematics" sentences is a (desperate?) attempt used by me in order to communicate a thought that I cannot communicate coherently to most people due to the lack of vocabuary in my disposal. Therefore I sometimes "hijack" the properties of some mathematical object and use analogy to reduce their rigor into something such that it helps me to describe what I am trying to say
These sentences are known to cause mathematician to cringe due to the vagueness involve in them
The tendency for me to hijack the properties of technical terms and objects to communicate my thoughts is one reason I can easily make physics jokes
On the other hand, it is one reason why I easily fall into misconceptions in some cases because two different concepts became too similar under this framework and thus they start to mix together
8:00 AM
in theory salon, 1 min ago, by Thomas Klimpel
Surprises of the Faraday Cage, by Lloyd N. Trefethen. Surprising indeed!
And now for my third (pint¹) mug of coffee. Some mornings ...
¹ 0.568 litres
8:23 AM
> sinusoidal oscillation in one direction corresponds to exponential decay in the direction at right angles in the complex plane
i think we have seen something similar in the WKB approximation
probably because momentum and position operators are related by a fourier transform
Therefore, compelx mommentum becomes an exponentially decaying term
>L1. There are gaps out there. If you find something fundamental that nobody seems to have figured out, there’s a chance that, in fact, nobody has.

L2. Analogies are powerful. I would never have pursued this problem had I not been determined to understand the mathematical relationship between the Faraday cage and the trapezoidal rule.

L3. Referees can be useful. Thank you, anonymous man or woman who told us the Faraday cage section in our trapezoidal rule manuscript wasn’t convincing! We removed those embarrassing pages, and proper understanding came months later.
Anyone who think they have got something alogn the lines of L1, should check carefully with their collaborators on whether every steps that lead to that conclusion is not laden with mistakes.

Just as all the good ol' LIGO and LHC guys check their results very carefully in order to not have things like the "OPERA superluminal neutrino faulty cable incident"
It seems chapman and co. are doign a good job on that
As for L2, let us go to the next paragraph...
Analogies are one of the ways to related two seemly unrelated things together. Once again, if we "hijack" mathematics to describe this property you will get a picture that look like this:

Imagine there's a space (set+topology) consists of concepts as its elements. There's some concept A and some concept B somewhere in this space (we don't know where, because we only have a topology not a coordinate chart, we only know it is there).

Now imagine an analogy is like some kind of map f that maps A, B into some C such that they are 'somewhere in between in properties to' A and B. This map has a
8:42 AM
However, analogies can be dangerous, because not everything is always being truthful to you. for example, while coloumbs law and newtonian gravitation might look mathematically similar, it does not mean they are related. As our experience in physical intuition grows as we solve more problems, then we became better at seeing through these illlusions
Back in my chemistry demonstrator training session, we are required to watch this old documentary, which talks about the issues of pet theories
Since then, it caused me to became more aware of my gazillions of pet theories living in my mind. Thus my questions become sharper (although may not look so because my communication skill is incoherent, as most h barer noticed)
A long time back in my high school period, it is actually a biology teacher, not a physics teacher that taught me and my classmates the principle of solving problems by elimination
this method set up a chain of logical reasoning in biology, best reputed for its memorisation, so that it decomposes into something as systematic as physics
over time, the idea grows as the concepts in mathematics added to the mix as I study
Eventually, after reading susskind, the concept cemented itself into a perculier way of asking questions
and studying became complicated with the constant need to fight my gazillion of pet theories
One recent example had occured yesterday, involving the discussion about photons. Some long time ago in the history of h bar, I heard from some users on a pet theory that talks about photons being a bound state of electrons and positrons

Fast forward to the present day, for some unknown reason, the curiosity on that grew. The alamr system then said there is a high chance if left unchecked, it will become a pet theory and corrupt my thoguth process
Because of the concept space analogy mentioned above, combined with the increase in awareness of the effects of pet theories after watching A Private Universe, and then things become a bit "chunky" after finished reading susskind, the following action is carried out
> For a person who is not fully sucumbed to his/her pet theory, the best way to combat it is to ask that pet theory on what are the predictions, and then check if these predictions matches what we know in real life experiments and theories. If the pet theory fail the test, then it is incorrect
So in order to combat that "photon as a kind of positronium" pet theory that is suddenyl recalled for reasons unknown, the following question was directed to acuriousmind
But first, the pet model need to be generalise as much at it can to reach the limit of the scope of "predictions" it can offer, so that a question that can bring it down become most effective by blocking all possible loopholes that can make the theory survive
Therefore, "photon as a kind of positronium" generalised into something more plausible as "photon as a composite of massless electric charges so that it is overall neutral"
The next step is to ask this pet model what are its predictions: We then get "nth order multipole moment may be detected"
However this is not enough to kill the pet theory thus
18 hours ago, by Secret
What would instead would have been observed if photons somehow has a composite structure?
18 hours ago, by ACuriousMind
@Secret You're asking the question the wrong way around: Our current theories have the photon as non-composite, and QED matches experiment perfectly. What exactly you expect for a composite photon would heavily depend on what it is composed of
After some discussions in making that question use to bring down the pet theory clearer, we reached the answer
18 hours ago, by ACuriousMind
Considering that glueballs are massive, though, I think you're right that massless composite particles can't occur in ordinary QFT
17 hours ago, by ACuriousMind
@Secret What "vacuum solution"? A massive photon would just wreck the inverse square law, and make the force decay exponentially
17 hours ago, by ACuriousMind
Yeah, I'm not convinced that there are consistent QFTs with composite massless particles
17 hours ago, by Herr_Mitesch
well, of course a photon could have intrinsic higher multipole moments, which would couple it to itself in more complex ways, but those are non-renormalizable anyway, I guess.
The answer is more than enough to break the foundation of that pet theory thus it is no more
9:10 AM
One more thing, a possible explanation on why I seemed to always ask questions the wrong way around (in addition to that incoherency), might be because I use the answer of the question to falsify something, therefore, when I ask a question, it often has a purpose of using the answer to eliminate one of the many possible pathways in the thought process, so that the thoght process will stay in the correct path and lead to the correct conclusion.
This unusual thought process of mine (and others) might have something to do with that logic chain approach by our biology high school teacher
but so far it serve me well
Another interesting observation (probably might be just me, but maybe some others as well). When people ask questions where the answers help knock down his/her pet theories, unusual topics in the mainstream will be discussed. This often have the effect of making the overall discussion meaningful

Put this in the concept space framework, knocking down pet theory appraoches often result in us exploring rarely explored regions of concept space, and hence fresh insights or just a highly meaningful discussion
An interesting consequence of my thought process flow in a way similar to that concept space framework is the tendency to think about problems of this class in context where such questions are not comonly conceived about
To hear the shape of a drum is to infer information about the shape of the drumhead from the sound it makes, i.e., from the list of overtones, via the use of mathematical theory. "Can One Hear the Shape of a Drum?" was the title of an article by Mark Kac in the American Mathematical Monthly in 1966, but the phrasing of the title is due to Lipman Bers. These questions can be traced back all the way to Hermann Weyl. For the 1966 paper that made the question famous, Kac was given the Lester R. Ford Award in 1967 and the Chauvenet Prize in 1968. The frequencies at which a drumhead can vibrate depends...
This has been said to be one reason why when I do mathematical proofs, I often end up proving the converse without realising
hence the reason why people often said I think "the wrong away around"
9:28 AM
@JohnRennie I don't believe in virtual particles. Let's say in scattering theory, by definition virtual particles are inner vertices of a diagram and real ones the asymptotic states. So, the definition of a real particle is that it propagates in free space without interactions. Why does a free space propagation suddenly make a particle real? I don't believe in real particles either, since pretty much everything is always a quasiparticle. :)
Particle for me is a choice of a convenient 'basis'. It is usually
wanted that it resembles classical analogies though, but that does not make it any more real.
But then again, I don't know QFT that well...
@MikaelKuisma Hi Mikael. Did you look at my subsequent posts where I try to explain what virtual particles are in terms of a sum of creation and annihilation operators?
@JohnRennie I did now. I agree on that, because that was math. But somehow it seems that I always have quite strong opinions on interpretations.
I'm sort of interested into this QFT thing now. I think I will on some weekend sit down, and derive some lattice QED variant and code that with Matlab etc.
But QFT, it is such math. Everything is infinite etc.
10:21 AM
@JohnRennie It is near the truth
@ACuriousMind Phew :-)
Although it doesn't contain the hard part of how to actually get the amplitude from that ;)
I wonder if it worth writing up in a more coherent form as an answer to one of the many "what are virtual particles?" questions on the site.
@DanielSank the part here that worries me is that I was the only one of four reviewers who voted to close this obvious duplicate. I'm less baffled that this question was posted than I am how it was not immediately closed. Come to think of it, not tagging it quantum-mechanics or special-relativity probably was a conscious decision to not give the gold badge holders the opportunity to close it single-handedly :P
hey guys
10:33 AM
hey guy
(or gal)
wow.....44.2k reputation :O
you're a physics stackexchange legend
Just been here for a while and got too much free time on my hands :P
@ACuriousMind that question is a good example of the havoc wreaked by the hot network questions list. Bosoneando's answer is the only one that makes a serious attempt to answer the question and it's lagging at 11 upvotes. The top voted answer isn't an answer at all.
If Bosoneando extended the answer to show how we get experimental observations like $E=h\nu$ I think it would be verging on the awesome.
Hi K. T. Did you want to ask something or is this a social call? :-)
10:37 AM
Hey John...I didn't want to ask anything in particular just checking the chat out :D
@JohnRennie Well, JanLalinsky's answer is also a better attempt to answer the question, I think, but yeah, the good answers are at the bottom of the list because the average HNQ visitor can't understand them :P
@peterh it now has the ACM seal of not total disapproval :-)
17 mins ago, by ACuriousMind
@JohnRennie It is near the truth
@ACuriousMind I don't know, I wouldn't have applied those tags either.
Actually, I do have a question. But it's a bit broad and not really worthy of a SE post
Out of curiosity, you can't re-tag it and then VTC as duplicate, can you?
10:41 AM
@DavidZ nope, the dupe hammer acts on the initial tags
could I ask it anyway? it's regarding qft I guess
@DavidZ no, I've tried that before :-)
You get a message telling you to bog off.
@K.T. anything goes here, pretty much
Which has puzzled me before at other questions because Qmechanic added the QM tag but I just cast an ordinary dupe vote
@K.T. ask away. The worst that will happen is that we don't answer.
10:42 AM
Sweet @david @john
@ACuriousMind Ah, yeah, I thought so. I vaguely remembered hearing about that in the mod chat room or on meta some time ago.
@DavidZ Well, I would have chosen either the QM or QFT tag at least, seeing how the photon is a quantum mechanical object.
Okay so....If you had to explain the Higgs mechanism in quite simple terms (I know, sort sort of contradictory) then how would you do so?
Q: Entropy - State Function but not conserved?

B. CroydonBy the second law of Thermodynamics we can define the state function Entropy up to an additive constant via $ \Delta S := \int{\frac{\delta Q_{rev}}{T}} $ At the same time the second law of TD gives us the inequality: $dS := \frac{\delta Q_{rev}}{T} > \frac{\delta Q_{irrev}}{T}$ Hence when t...

Entropy is a nonconserved state function. How to explain that in nitty gritty details I am not sure
Q: Bug in closing as duplicate?

John RennieIf you have a gold badge in a tag then you can immediately close as a duplicate any question that has that tag. So for example I have a gold badge in [general-relativity] and [newtonian-mechanics] and should be able to close as duplicates any questions tagged with these. However I just VTC'd as ...

10:43 AM
May 24 at 22:15, by knzhou
Hey everybody, I'm trying to explain the Higgs mechanism to high schoolers, without using any 'fancy physics'.
@ACuriousMind I'd definitely go for QFT. I think on this site is for nonrelativistic QM.
@K.T. probably like this
@ACuriousMind how do I can only see the first line of that post
@DavidZ will check it out thanks!
@K.T. click on the blue text "May 24 etc" - it's a link to the original message
@K.T. Ah, I linked an old chat message to show you we've talked about that before here and didn't really come up with a good thing. To see the rest of the chat click on the date
However, now different people are here, maybe someone has a good idea
oh cool got it thanks curious and john :D
10:46 AM
Ah, well if we're targeting high schoolers, I would not do it the way I posted.
@K.T. the best explanation I've seen is by Matt Strassler but this is probably above high school level.
Oh, well that's what I am
Matt strassler!! I love his blog! Thanks John
Then trouble is that it's impossible to give a simpler explanation without it being wildly inaccurate.
Have a look at that article and see how you get on.
yeah exactly...will do, thanks!
@K.T. He's written some absolutely awesome articles. I wish I could write anything half as good.
10:49 AM
yeah, he has! And he's extremely nice too! He went to the trouble of personally answering some questions I had for him (through facebook), and answering them very thoroughly
@DavidZ Amusingly, even if the initial tags have been deleted the dupehammer is still possible.
hey guys sorry while reading one of the answers I don't quite understand something....When describing the higgs field potential, the writer writes "that Phi is a complex valued variable"
oh sorry nevermind I get it ignore my previous message
It still amaze me how the observable spin has a dependence on orientation in spacetime. I wonder if there are other observables that has no classical analog (e.g. can only be described by livinig in a hilbert space) that has such orientation dependent property?
11:07 AM
@DavidZ @dmckee @Qmechanic What do I do if I notice a consistent pattern that certain users vote to leave open obvious homework-questions and very clear duplicates? Especially voting to leave open the obvious duplicates is very disruptive to the site in my opinion, since people tend to stumble upon the newer questions, scattering good answers to the same question all over the place
@DanielSank : no.
@ACuriousMind : it isn't an obvious duplicate. It's just a question that you can't answer.
@JohnRennie : that's sometimes how it is even when HNQ isn't involved.
11:24 AM
@ACuriousMind Flag one of the user's posts with a custom mod flag explaining the situation.
@DavidZ : people who've taken a lot of time and trouble to answer a question aren't going to be too happy that it's been closed and the bounty removed. Especially when the top answer on the alleged duplicate is yours. And there was I saying this is physics stack exchange at it best.
Yes, I'm aware of that.
@DavidZ : but are you aware that your answer is wrong?
11:45 AM
@ACuriousMind Yeah, I've seen some of that.
@DavodZ : see Arnold's answer and his reference to The Nature of Light: What Is a Photon? edited by Chandrasekhar Roychoudhuri and Rajarshi Roy. I know the former. This is worth a read : arxiv.org/abs/0803.2596 . There is no n in E=hf. There is no evidence that a photon is some “infinite set of component sinusoidal waves”.
12:07 PM
@BernardMeurer Our IT person knew instantly
@JohnRennie physics.stackexchange.com/questions/273578/… check comments - the OP wants the Q's duplicate status to be removed - he has edited the Q
Hey @da
DavidZ* I read your answer it was super helpful thanks!
but I still have gaps in my understanding about how it works, in the physical sense
For example where do condensates come into play in the higgs mechanism? And also I've heard that a very similar thing happens in superconductors, but I don't understand how
12:52 PM
@K.T. You'd have to ask someone with more of a condensed matter background to understand how it relates to superconductors and condensates
oh ok

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