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John Duffield
John Duffield
4:00 PM
@WilliamBulmer : some people think a photon is thing that has energy. I prefer to think of them as a form of energy, taking my cue from Einstein: "If a body gives off the energy L in the form of radiation..."
 
William Bulmer
William Bulmer
@JohnDuffield hmmm...so is it appropriate to think of a photon more as a quantum of energy, rather than a distinct oscillation of a field?
 
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Testing theories to the breaking point is definitely cool @JohnRennie :-)
 
John Rennie
John Rennie
My old university too :-) One of my friends was doing a PhD in astronomy, though back in the 80s it was amazingly primitive compared to today.
 
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Do you go back to visit regularly?
 
John Rennie
John Rennie
No. I went back to Cambridge a couple of times in the years soon after I left, but that was to visit friends rather than the university. Most of my friends were students too, so within a few years of leaving almost everyone I knew in Cambridge had also left.
 
John Duffield
John Duffield
4:09 PM
@WilliamBulmer : neither sound quite right to me. I would say a photon is a singleton electromagnetic wave. If has an E=hf or E=hc\λ wave nature. A neutrino conveys energy, but it isn't a photon. People also say a photon is an excitation of the photon field, but I don't see any difference between that and a singleton electromagnetic wave myself. Which can also be described as a "pulse" of four-potential.
 
John Rennie
John Rennie
Though some years later I went back to interview students for jobs at my company (Unilever) which was a bit surreal :-)
 
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Nice.
 
John Rennie
John Rennie
It's weird. When you're a student in an interview you think half the questions I'm being asked are bollocks. When you're an interviewer you think half the questions I'm asking are bollocks :-)
 
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
:-)
 
Secret
Secret
@JohnRennie (Joke) Either it means there are exactly a certain set that has 1/2 of the full set of questions are bollocks, or that there is perfect correlation when it comes to what interviewer and students think about the proportion of questions being bollocks
 
Secret
Secret
4:22 PM
And no, since interviewers and students are too macroscopic that decoherence time is negligible, I cannot extend this joke using the concept of entanglement that the interviewers and students are maximally entangled such that the probability that they both think questions are bollocks are exactly half
 
John Rennie
John Rennie
When a big multinational comes to Cambridge to interview they aren't really probing to see how clever you are. It's a reasonable bet that if you're doing a PhD at Cambridge you're pretty clever.

What they are really looking for are good managers i.e. people who can assess things quickly and clearly and are good at organising their thoughts. These are the people who are going to be running large teams ina few years.

So the questions aren't specifically about your work - in any case the interviewers won't know enough about it to judge properly. The questions are intended to see how you thi…
 
Secret
Secret
true
 
John Rennie
John Rennie
They obviously thought I fitted the bill since they offered me a job, but I quickly discovered I preferred doing science to managing it. So as soon as I could afford to leave I did :-)
 
Secret
Secret
Yeah, managing people is hard work, I would rather do more research in the field or the lab or in the society than to deal with boring admin stuff
 
John Rennie
John Rennie
Hell is other people
(c) John Rennie 2016
Present company excluded of course - well, some of you :-)
 
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ
4:31 PM
Gee, thanks. (I think.)
 
John Rennie
John Rennie
@WilliamBulmer: defining what particles are in QM is surprisingly hard. QFT offers a different perspective, but even in QFT it's hard to point to anything and say here, this is a particle.
 
Mats Granvik
Mats Granvik
How much non zero must a commutator (matrix associated with a zeta function) be in order for there to be a uncertainty principle associated with it?
Would this example commutator matrix be non zero enough?
25
Q: Primes approximated by eigenvalues?

Mats GranvikConsider the matrix starting: $$\displaystyle T = -\begin{bmatrix} +1&+1&+1&+1&+1&+1&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \end{bmatrix}$$ defined by either the recurrence: $$\displaysty...

prime-numbers eigenvalues-eigenvectors arithmetic-functions
Anyone?
 
John Rennie
John Rennie
Hi Mats. Friday afternoon (European time) is a bad time to ask as most of our hard core theorists have finished for the week. The chat room is likely to be sparsely populated from now until Monday afternoon.
 
Mats Granvik
Mats Granvik
ok
 
Secret
Secret
@MatsGranvik have you also tried the maths chat room?
https://chat.stackexchange.com/rooms/36/mathematics
 
Mats Granvik
Mats Granvik
4:40 PM
@Secret yes that is where I usually hang out.
 
Secret
Secret
ok
 
Physics Meta
Physics Meta
5:09 PM
0
Q: The fine line between astronomy and physics

heatherSo, a few questions recently came up on the feed: What is Venus's core made of How can core of Mercury be a molten liquid They made me wonder: when does something belong on the astronomy.SE and when does something belong on the physics.SE (this could also apply for chemistry, electrical engin...

discussion
 
0celo7
0celo7
5:28 PM
Hello
 
Mats Granvik
Mats Granvik
@0celo7 hi
 
0celo7
0celo7
What is this
 
dmckee
dmckee
6:04 PM
If we're in the business of making canonical questions, we might want one about the classical versus quantum prediction for an electron's radiation when bound in an atom.
As in
and various duplicates from earlier.
0
Q: Why don't electrons moving around in a orbit produce electromagnetic waves in their natural state?

coderIf moving electrons produce changing electric field, and if changing electric field produces magnetic field, every electron must produce an electromagnetic wave. This means an atom in its natural state must emit light or other waves in electromagnetic spectrum. But why doesn't this happen?

quantum-mechanics electromagnetism atomic-physics
 
0celo7
0celo7
@ACuriousMind I take it you never checked that the energy functional is continuous in the path space topology?
 
William Bulmer
William Bulmer
6:21 PM
@JohnRennie (@JohnDuffield also), I think part of the problem is that the idea of a particle--per-se is somewhat outdated. It's more like it's been superceded by the notion of a particle field, which is different, though, right? I might be wrong, and I have not taken QFT. So, you can't really speak of an electron being exactly somewhere without running into all kind of theoretical issues. It's more like, in the statistical sense, there are n measurements of particles in a particular state
right?
But in QM (which, yes, has been supplanted by QFT), you CAN model an electron/electrons, by positing a Hamiltonian in O(n) dimensions. But, I think my issue is that I don't know how you would do that with a photon. I think that might be an ongoing line of research
 
0celo7
0celo7
@JohnRennie 1100C: glowing ceramic
 
William Bulmer
William Bulmer
You know what? While I am here, I wonder if somebody could help me with a concept. I have read that the Lagrangian of L - T, isn't strictly universal, and certainly is not accurate for the relativistic case.
That implies that there is some deeper definition of the Lagrangian, which implies L - T in on-relativistic case. What is it? I have read that it can be derived from the Hamiltonian, but I also get the sense that that isn't exactly the total energy of the system in the general case, either (for example, when friction is involved)
 
0celo7
0celo7
to you mean T-V
 
William Bulmer
William Bulmer
Yeah, sorry
I meant that
What is the physical reasoning that would enable one to arrive at T - V, and its other forms?
What I really hate, is when I get the cop-out answer of "because T - V works". My response is, well, yeah, but it didn't just come from nowhere, expecially if it can be derived from a deeper principle
 
0celo7
0celo7
I don't think it can be derived
the action has no particular physical interpretation
 
William Bulmer
William Bulmer
6:35 PM
@0celo7 Well, to be fair, neither can the Schrodinger equation, really
But is still comes from somewhere
Or is it more like, arriving at its form is a matter of trial and error? I guess, in the case of friction, you could always model it as a closed system where all the energy (and momentum) from the frictional force enters a kind of bath, and stays there
I should say reservoir
So, what if you had system A, B, and C. System A is the one experiencing friction. System B is the reservoir, and system C absorbs all of the equal and opposite momentum and energy from C as a result of A's interaction with B
Maybe that doesn't work...
No, this should be really simple. A interacts with B and C. Then, we model T - V for the entire system of A, B, and C. Then, using that, we calculate a new Lagrangian only for the system of A and B.
Is that the idea?
So, T-V would be extremized for A, B, and C, where there is a frictional force that "leaks" momentum and energy out to C. Then, we simply say that the Lagrangian for A and B is just the Lagrangian for A, B, and C.
So, you calculate a new set of Lagrange's equations based off minimizing the Lagrangian for A, B, and C, then you "forget" about C.
Err, I meant minimizing the integral of the Lagrangian. Sorry for spamming.
 
Balarka Sen
Balarka Sen
7:32 PM
@WilliamBulmer With the limited physics I know, T - V intuitively means to me the amount of energy of the system required for doing all the motion in the system (that's why you're subtracting the potential energy). In particular, I think of it as something which captures the dynamics of the system.
But then I don't know anything about physics, so you should take that as a grain of salt.
 
DanielSank
DanielSank
7:47 PM
@Danu yes and this is the problem I have with CuriousOne. Lots of comments which are not subject to votes and/or edits.
 
vzn
vzn
8:08 PM
@DanielSank did you get my msg? DZ says aug23 ok can you plz confirm
 
0celo7
0celo7
8:57 PM
@Danu Are there Riemann surfaces with strictly negative curvature?
 
0celo7
0celo7
9:36 PM
@ACuriousMind Good evening
 
ACuriousMind
ACuriousMind
You're essentially asking http://physics.stackexchange.com/q/15899/50583. Both the Lagrangian and the Hamiltonian formalism sacrifice direct physical intuition for mathematical strength. I don't think anyone doubts that Newton's $F=ma$ is more directly physical meaningful than the E-L equations or Hamilton's e.o.m.
But the Lagrangian and Hamiltonian formalism enable much more powerful general reasoning, like Noether's and Liouville's theorems, than the direct formulation through Newton's laws.
@0celo7 Evening
 
0celo7
0celo7
@ACuriousMind Does a space with compact universal cover have a finite fundamental group?
@ACuriousMind Ah, it must be true.
 
ACuriousMind
ACuriousMind
@0celo7 If the space itself is compact, yes. I'm not sure if it can happen that a non-compact space has a compact cover.
 
0celo7
0celo7
@ACuriousMind Yeah, the space is compact.
@ACuriousMind Probably not, the covering map has to be continuous and surjective.
 
ACuriousMind
ACuriousMind
Ah, yes
 
0celo7
0celo7
9:54 PM
@ACuriousMind So the way you prove that is that the universal cover is the ordered pairs $(X,[\gamma])$
You can't have a finite subcover unless there are finitely many $[\gamma]$s, right?
 
ACuriousMind
ACuriousMind
@0celo7 Can't you just say that if the fundamental group is infinite, the fibers are infinite, so taking the preimage of an open cover consisting of sets such that their preimages don't intersect in the universal cover gives you a cover of the universal cover that can't possibly have a finite subcover
 
0celo7
0celo7
That works too.
 
ACuriousMind
ACuriousMind
the double meaning of "cover" is unfortunate in this context :P
 
0celo7
0celo7
@ACuriousMind Is the sphere bundle of a compact manifold compact?
 
ACuriousMind
ACuriousMind
I think so
 
0celo7
0celo7
9:58 PM
How do you even check that
It would be compact in the topology of $TM$, right?
Milnor says "the space of unit vectors is compact"
Well, that's just the set of sections of the sphere bundle
 
ACuriousMind
ACuriousMind
For each open cover of the bundle, you choose a finite trivializing cover $U_i$ of the base, project the intersections of the open cover with $\pi^{-1}(U_i)$ onto the sphere (still open because the projections of a product are open), choose a finite subcover, which gives you a finite cover of the bundle when taking the preimages
Hm
That may not be it :P
 
0celo7
0celo7
@ACuriousMind I had that in mind, too.
But Milnor does not talk about bundles, so you know I have to reject this proof :P
 
ACuriousMind
ACuriousMind
@0celo7 Yeah, but I don't see right nowhow that fails for e.g. $\mathbb{R}\times S^1$.
Ah
It doesn't work
Taking the preimages does not give you back sets that were in the open cover
Since the preimage of $U\subset S^1$ is $\mathbb{R}\times U$, but the original cover might have contained not that but a small set that projects down to $U$.
So there must be another way
That the trivial bundle over a compact manifold is compact is clear
 
0celo7
0celo7
@ACuriousMind The trivial bundle?
 
ACuriousMind
ACuriousMind
@0celo7 Just $M\times S^n$.
 
0celo7
0celo7
10:08 PM
oh
@ACuriousMind Yes, by Tynchonoff
 
ACuriousMind
ACuriousMind
the trivial bundle is the one that is globally isomorphic to the product instead of only locally
 
0celo7
0celo7
Or however you spell that shit
@ACuriousMind I know what a trivial bundle is!
I didn't know which one you meant
i.e. which fibers
 
ACuriousMind
ACuriousMind
Googling tells me that indeed any fiber bundle with compact fibers and compact base is compact
 
0celo7
0celo7
Don't make me get Steenrod please
I think this should be elementary
well, maybe we can think of this another way @ACuriousMind
 
ACuriousMind
ACuriousMind
10
Q: Fiber bundle is compact if base and fiber are

Stefan HamckeI want to show that the total space $E$ is compact if the fiber $F$ and the base space $B$ are compact. Let $\pi$ denote the fiber projection. Since every point in $B$ has an open neighborhood $U$ whose preimage $\pi^{-1}(U)$ is homeomorphic to $U\times F$ via $\phi_U=(\pi,\beta_U)$, there are f...

general-topology compactness fiber-bundles
 
0celo7
0celo7
10:11 PM
What we need is that $\mathrm{Ric}(U,U)$ has a minimum, where $U$ is the set of unit vectors
Clearly the unit vectors at each $p\in M$ is compact
 
ACuriousMind
ACuriousMind
I'm not going to think about a topological property with damn Riemannian tools! :P
 
0celo7
0celo7
So $\mathrm{Ric}(U_p,U_p)$ takes a minimum on each tangent space
Then consider the function $p\mapsto \min \mathrm{Ric}(U_p,U_p)$
this should also take a minimum
plausible?
@ACuriousMind this is Riemannian geometry though :/
@ACuriousMind How difficult is Part 4 of Milnor
 
0celo7
0celo7
10:36 PM
@ACuriousMind How does the proof of Corollary 20.3 work on page 109?
It's too vague
 
0celo7
0celo7
10:48 PM
Never mind
 
ACuriousMind
ACuriousMind
11:17 PM
@0celo7 k
 
0celo7
0celo7
@ACuriousMind I am highly upset by what's on the next page
 
ACuriousMind
ACuriousMind
lol
 
0celo7
0celo7
Milnor uses the isometry invariance of the Riemann tensor
It is not proved anywhere
 
ACuriousMind
ACuriousMind
The point where math books highly upset you may be the point where you reevaluate your life choices ;)
 
0celo7
0celo7
No one ever proves it!
I would not know about it had I not read Lee
Is is so obvious?
No one even states it
They just use it
@ACuriousMind Physics books highly upset me and I now I hate physics
 
ACuriousMind
ACuriousMind
11:20 PM
I found this blog post that seems to prove it :)
 
0celo7
0celo7
I fear the same might happen to mathematics
@ACuriousMind that blog was not around in 1963, I assume
 
ACuriousMind
ACuriousMind
@0celo7 Did you even click the link? :P
 
0celo7
0celo7
btw, a blog takes a lot of damn time
@ACuriousMind I hovered
Seriously though, Milnor never mentions it
Then just uses it on the sly
same thing in Jost, Petersen
do Carmo
 
ACuriousMind
ACuriousMind
@0celo7 I guess I kinda expected more of a reaction of me linking your own blog to you...
 
0celo7
0celo7
@ACuriousMind I know the proof, but that does not mean I'm any less highly upset by this situation
don't you hate it when you know you found a cool result in Riemannian geometry while looking for something else, but the result could be in any one of eight books
crapola
Oh, it was in Petersen.
@ACuriousMind Do you not find it strange/worrying that they use isometry invariance with wild abandon?
Is it obviously true?
@ACuriousMind Can a self-adjoint linear map $V\to V$ have a kernel?
Probably, it would seem $0$ is self-adjoint.
 
ACuriousMind
ACuriousMind
11:32 PM
@0celo7 Exactly
 
0celo7
0celo7
@ACuriousMind I need a response to my isometry question or I will melt.
Literally.
@ACuriousMind Ah, but self-adjointness means the eigenvectors are orthogonal.
 
ACuriousMind
ACuriousMind
@0celo7 No, because it is very natural that this statement should be true, since the Riemann tensor is determined by the metric and the metric is by definition isometry invariant. It's a statement like "composition of homomorphisms is a homomorphism" - another thing which is almost always left to the reader.
@0celo7 No, it means that there exists an orthogonal eigenbasis
 
0celo7
0celo7
@ACuriousMind Is that not what I just said
 
ACuriousMind
ACuriousMind
Every vector is an eigenvector of 0, but certainly not all vectors are orthogonal
 
0celo7
0celo7
Oh god what if the eigenvalues are not real
CRISIS
 
ACuriousMind
ACuriousMind
11:36 PM
@0celo7 The eigenvalues of a self-adjoint operator are always real.
 
0celo7
0celo7
@ACuriousMind I've never heard of that before
 
ACuriousMind
ACuriousMind
@0celo7 It is part of the spectral theorem, and trivial to prove if you know that there is an orthogonal eigenbasis.
 
0celo7
0celo7
@ACuriousMind It takes way more work to prove than your homomorphism thing
And there is a subtlety with pushforwards
 
ACuriousMind
ACuriousMind
@0celo7 Oh, depending on the category it can get pretty ugly to show that the composition of morphisms is still a morphism :P
 
0celo7
0celo7
@ACuriousMind I thought the definition of category was that morphisms exist, can be composed, and this is associative
@ACuriousMind What
How am I supposed to know that there is an orthogonal eigenbasis
 
ACuriousMind
ACuriousMind
11:39 PM
@0celo7 ...yes. I mean that when you are given a category as "X as objects and maps between Xs with property Y as morphisms" you will very rarely see someone explicitly check that indeed the composition of two maps that fulfill Y still fulfills Y
@0celo7 ...because that's the spectral theorem?
 
0celo7
0celo7
@ACuriousMind There has to be an elementary way of proving this
 
ACuriousMind
ACuriousMind
@0celo7 ??? The spectral theorem is rather elementary, at least in the finite-dimensional setting
 
0celo7
0celo7
It's easy to show that any Hermitian matrix has real eigenvalues
@ACuriousMind What? It's seriously advanced
 
ACuriousMind
ACuriousMind
@0celo7 ??? What is the difference between Hermitian and self-adjoint
I have no idea what you are talking about
 
0celo7
0celo7
@ACuriousMind Beats me, is there one?
 
ACuriousMind
ACuriousMind
11:41 PM
@0celo7 Not in the finite-dimensional setting. In the infinite-dimensional setting it depends on your definition of "Hermitian"
 
0celo7
0celo7
Ok, so I believe that the eigenvalues are real
@ACuriousMind I'm working on a Riemannian manifold
So...
what do I do next?
Does the existence of real eigenvalues imply the existence of eigenvectors?
Probably
 
ACuriousMind
ACuriousMind
@0celo7 That makes no sense at all to me. "Self-adjoint" is a statement about operators on a complex vector space with inner product
Where does the manifold enter?
 
0celo7
0celo7
Self-adjoint works on any vector space with inner product, bub
$(v,Tw)=(T^*v,w)$
self adjoint means $T^*=T$.
 
ACuriousMind
ACuriousMind
...okay, strike the complex if you insist. What's your point?
 
0celo7
0celo7
...what are we talking about?
oh
 
ACuriousMind
ACuriousMind
11:45 PM
@0celo7 Why you replied to my statement about the setting with "I'm working on a manifold"?!
 
0celo7
0celo7
I have this thingie on $T_pM$, and I need an orthonormal eigenbasis
on $T_pM$
 
ACuriousMind
ACuriousMind
Yeah, so $T_p M$ is a finite-dimensional inner product space, why does it matter that it comes from a manifold?
 
0celo7
0celo7
Who said it does?
 
ACuriousMind
ACuriousMind
You. Unless you decided to randomly tell me that you're working on a manifold when this had nothing to do with what I said
 
0celo7
0celo7
So in the REAL setting, how do I prove $T=T^*\implies$ real spectrum?
Do I have to complexify?
 
ACuriousMind
ACuriousMind
11:46 PM
@0celo7 Uh. How could an operator on a real space possibly have non-real eigenvalues?
 
0celo7
0celo7
@ACuriousMind Oh for the love of god man
Then prove it has $\mathrm{dim}\,V$ eigenvalues
counting multiples
 
ACuriousMind
ACuriousMind
...I said "spectral theorem" like three times now. What is your issue with it?
 
0celo7
0celo7
never heard of it
I want to do it without
 
ACuriousMind
ACuriousMind
It's a fundamental result in linear algebra! I guarantee you that every book you have read in the recent past definitely assumes you know it
 
0celo7
0celo7
if I first complexify, it should work
@ACuriousMind Doesn't change the fact that I've never heard of it.
 
ACuriousMind
ACuriousMind
11:49 PM
Yes, but I don't know why you would want to avoid learning about such an important result
 
0celo7
0celo7
Too many other things to read!
Alright, if I complexify I can show the eigenvalues are real.
This shows that the eigenvectors are real
So I can decomplexify
now, how to show that these are orthogonal?
Screw it
I can only understand this in bra-ket notation
 
ACuriousMind
ACuriousMind
@0celo7 That is the spectral theorem. You are literally asking to show the spectral theorem without such "advanced" techniques as are needed to prove the spectral theorem.
 
0celo7
0celo7
$\langle b|H|a\rangle=a\langle b|a\rangle=b\langle b|a\rangle$
Hmm, so if $a\ne b$ we get it.
Ok, what about the case of degenerate eigenvalues...
or 0
No, that's the same issue
@ACuriousMind What do I do?
Wait a moment
Did I not explain to Secret how to do this like three days ago?
 
ACuriousMind
ACuriousMind
@0celo7 I have no idea what you are doing.
 
0celo7
0celo7
@ACuriousMind Calm down
 
ACuriousMind
ACuriousMind
11:55 PM
You said to me you wanted to do something without the spectral theorem because it is too advanced and now you are trying to prove the spectral theorem
That is highly upsetting to me :P
 
0celo7
0celo7
@ACuriousMind I will prove it without your damn Spectral Theorem, ok?
watch me
 
ACuriousMind
ACuriousMind
btw, how do you know the eigenvalues/vectors exist in the first place?
 
0celo7
0celo7
Eigenvalues exist because of fundamental theorem of algebra
Which I proved in GP
Vectors, just solve the equation $Hv=\lambda v$
 
ACuriousMind
ACuriousMind
How do you know that equation has a solution?
 
0celo7
0celo7
Are you trolling
 
ACuriousMind
ACuriousMind
11:57 PM
No
There are real matrices that do not have eigenvectors on their real vector space
 
0celo7
0celo7
We've complexified, @ACuriousMind
wait, are you saying the eigenvalue can be real but the eigenvector complex
 
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