« first day (2877 days earlier)      last day (2043 days later) » 
00:00 - 17:0017:00 - 23:00

12:10 AM
I've never even heard of The Mouse that Roared
Sounds interesting though
 
12:31 AM
Fast is done, just had three bagels. It tastes so good ;-;
 
 
1 hour later…
2:01 AM
Good night/morning/evening for everyone!

Is the Wronskian a example of second order tensor?
 
 
3 hours later…
4:40 AM
@JackClerk The Wronskian is a determinant isn't it? So it's a scalar.
 
4:52 AM
@JohnRennie how are you? Yeah, but it's a multilinear map
 
5:06 AM
@JackClerk You mean the matrix that the Wronskian is the determinant of?
 
@SirCumference I mean the Wronskian Determinant:

W(f,g)
My point is:

Consider the whole world of multilinear functions,right?
So, by the construction of Tensor Product vector space we earn something really special:

These multilinear functions are isomorfic to tensors (elements of tensor product).
Then multilinear functions are Tensors
Now I'm asking if my statements are correct (at least in conceptual/intuitive level)
 
Sorry, I don't know jack about tensors. But you could try asking in Math's chat too
 
Actually it seems that the authors in physics books (like General Relativity, Robert Wald),define tensors just to call everything multilinear in the text a Tensor.
It seems also that, the components of a multilinear function and a tensor are the same and then by the tensor product you can write the multilinear function in a very compact form....
 
5:24 AM
Last night dream, went to a drink shop and ordered a mixed drink called United Nations:
The lady (black curly haired, dark skinned, 160 cm something in height) first poured some blue lemonade into a small shot cup. Meanwhile, two other staff were placing whole papaya and a half cut giant mango into a blender and blend it so it becomes puree
After that, a quarter of the shop is capped under a plastic transparent box, as the lady then place some kind of mask like object on a cup, turned it on and the whole place become very foggy with liquid nitrogen vapor
After the fog cleared, the drink is completed, with 7 colored layers of liquid. 3 potatoe fries were then placed next to the straw followed by 5 slices of steak
As I drink it, the color easily mixed and then the whole drink became translucent yellowish, and then some areas becomes chocolate brown, as well discs of instant noodles floating up of it followed by some chicken soup and pieces of chicken breast
It has a weird sweet sour taste
 
 
1 hour later…
6:53 AM
Ok I have also finished reading this. So basically the way it works is that given a very large entangled state, choose POVMs such that local measurements were performed on each subsystem (thus will not break the entanglement) and then the final result can be made to one if it is the target state with the required quantum correlations. Thus in a way, even a single copy behaves like many copies in parallel as the correlations were exploited to ensure only for a target state will give the required
success
So in a way, quantum correlations represents an extra computational resources besides the traditional size of the system and number of replicates (hence time of computation)
I think it is safe to say that large entangled state essentially behaves like a generalised form of parellel computing with each subsystem holding infomation of the whole system
 
 
1 hour later…
8:24 AM
Hi, a little question too small for a full "thread" on the physics.SE front page
I stumbled upon this notation
...where the effective velocity $v^\text{eff}$ itself depends on the local distribution of quasi-particles $p(. ; x; t)$ through the "dressed" functions..."
where $p(\theta,x,t)$ normally had a $\theta$ instead of the "dot"
Is this normal notation?
I can't help but to say, this dot means nothing to me
 
8:37 AM
Huh. It's not very typical but I believe I've seen that used a few times to indicate a parameter which doesn't matter in that particular context.
But it might also be a typo or something.
Maybe if you link to where you found it, we could be more specific.
 
It's ArXiv 1711 00873 bottom of page 2, left column between (3) and (4)
Hold on while I get a link
 
(to the abstract page!)
Thanks
 
arxiv.org/abs/1711.00873 would be better
but yeah
it works
 
Saves 1 click off your mouse's lifetime ;)
 
8:41 AM
I am much more interested in not unexpectedly downloading (potentially large) PDF files than about saving mouse clicks
as are many other people
Anyway, I don't see any reason for the dot. Maybe it means something to someone who is familiar with that particular topic, but my guess would be just a typo, and it should be $\theta$.
 
Anonymous
10
Q: What is the meaning of expressions of the type $f(\cdot)$ (function (dot))?

MathAsFunSimple question, fully expressed in the Title line. Is the dot within the parenthesis intended to mean, "any possible function"?

 
8:57 AM
Hello my droogs
 
Anonymous
Morning :)
 
@Blue Cheers, yes I guess that makes sense in this context
 
Anonymous
I am...playing truant today
 
Anonymous
(more like got bored and came back home :P)
 
9:17 AM
@Slereah mornin'
 
How's life
 
> Once I created the interfinite, I will be able to break down the very foundation of all creation and noncreation: Hoerarchy and their stupid rules
nonsense!
Hierarchy are there for a reason
 
@Slereah sitting waiting to get a flu jab... Great times...
 
@Mithrandir24601 Beats having the flu
Two years ago I was the only one who had the flu shot at a family Christmas
Everyone got the flu but me
Dodged a bullet there
 
@Slereah most definitely going to agree there - I got it really bad last year
 
9:36 AM
a
The above diagram is of Coulomb's torsion balance from yesterday's question
0
Q: Coulomb's law: Torsion balance experiment

faheemahmed400I have read about Coulomb's torsion balance experiment for the first time from an online pdf article. I interpret it as follows: Black ball $-$ neutral Red ball $-$ positive charge Blue ball $-$ negative charge At point $P$, there is no spring force and the positive charge is...

User Peter says the following:
The problem is that the equilibrium will not be stable. It is like balancing a pencil on its tip. If it begins to move away from the upright postion gravity will pull it further away. In this case if the charge on the torsion balance moves away from the fixed charge, even as a result of microscopic vibrations, the electrostatic attraction will reduce, and the torque from the torsion balance will pull it further away, reducing the electrostatic attraction even further, etc, etc. There is also the problem that air is not a perfect insulator, so charge will leak from one to the other. — Peter 19 hours ago
It says the equilibrium won't be stable if we use opposite charges.
 
@faheemahmed400 I think Peter has misunderstood the experiment. In Coulomb's experiment the charges are the same and that is stable.
 
What if we use opposite charges? Why it will not work?
In my question, I use opp charges
 
Because the torsion force varies linearly with distance while the electrostatic attraction varies as $1/r^2$.
 
Then how does same charges (repulsion) work?
 
Suppose we find the distance where the two forces are exactly equal. Now move the ball on the torsion balance closer to the static ball by a very small amount. The attractive electrostatic force will increase more then the repulsive torsion force increases. So if you move the ball closer than the equilibrium distance it is attracted towards the static ball and moves until it hits it.
 
9:51 AM
I see.
 
10:03 AM
Is it like this for same charges? Suppose we find the distance where the two forces are exactly equal. Now move the ball on the torsion balance away from the static ball by a very small amount. The repulsive electrostatic force will be less than the attractive torsion force.
Do I understand correctly sir?
 
@faheemahmed400 What you need to do is graph the force for the two cases. You'll find that for same charges there is a minimum in the force-distance curve, and this is a stable point. For the opposite charges there is no minimum so there is no stable point.
 
Ok, tks... Let me see
 
10:45 AM
While vzn covered the groundbreaking ones, I felt this one need to deserve more attention:nature.com/articles/s41586-018-0503-6
So with the jump able to be timed, what else previous unmeasurable is now possible to be reassessed
 
@John Rennie: I am totally confused in the graph. Can you show me a rough sketch graph.
 
The most upvoted answer and the answer that was selected by the OP states (paraphrasing here) that "water is blue because it absorbs red light." If like me you disagree with this notion please offer feedback to the poster of the question directly :)
 
@pentane whoever asked that doesn't seem to have seen the sea on a grey and rainy day :P
 
 
1 hour later…
12:13 PM
2
Q: Are snide remarks allowed here?

just curiousI asked a question about dots on jerseys and got the following comment: Funny thing is, I saw the photo on a news website and a headline about how the Orioles were wearing dots (admittedly, I didn't read that article, which would have brought me to my answer, but not everyone knows the headqua...

 
12:57 PM
That's what the 'unconstructive' flag is for
 
 
1 hour later…
2:01 PM
Hmm...
everything so far is quite straightforward, except the surreal trajectory case. How on earth pointer 1 even click if it is mode 2 that bounce to detector 1 thus it should be pointer 2 that should have received a kick?
is it some entanglement occurred between mode 2 and pointer 2, thus after being detected by detector 1, the entanglement result in 1 to be clicked?
how does measurement at dectector 1 "back perturbs" the outcome of pointer 2 to pointer 1?
 
They talk in some more specificity about that setup in the appendix at the end
 
>
and that, by the time modes 1 and 2 cross, the pointer
barely moved. Then Bohmian mechanics predicts that
if detector 1 clicks, the particle went through mode 2
(as if there were no pointer); however, if one waits long
enough for the pointer to eventually move by more than
it’s spread, then one finds that it is pointer 1 that moves.
Accordingly, in the case of “slow pointers” the pointer
indicates where the Bohmian particle was not. This is
surprising, at least to physicists used to quantum theor
 
(I haven't actually done any of the thinking about that setup, mind.)
 
but it is mode 2 that bump into detector 1 due to the pointer being perturbed after the trajectories bounce off each other, so I don't get how pointer 1 can even be affected at all
unless detector 1 entangles itself with mode 2 after the particle hits it, thus modifying the trajectories, causing pointer 1 to be moved instead, in a countefactual fashion
 
I think the main thing is that, while the particle only moves along one of the modes, the wavefunction itself goes through both
 
 
1 hour later…
3:12 PM
Back in Victorian times it used to be said that self abuse was a terrible thing because it sapped the vital forces of young men. I can't help feeling that these days that's true of Bohmian mechanics.
5
 
but what about old men?
 
@danielunderwood I wouldn't know :-)
 
But even when the particle is in the blue trajectory while the kick at the red charge happened, there is no particle to kick the red charge. Even after it switched to the red trajectory before hitting the screen, there is still no particle that is travelling along the start of the red trajectory. Even if the pilot wave goes through both, the particle was never in the top plane, and I don't suppose the pilot wave can cause the kick itself?
 
Sure it can.
There's really nothing else that could provide the kick, given how Bohmian mechanics works
Stuff like potentials etc. act on the wavefunction, and the wavefunction acts on the particle through the guidance equation
 
3:22 PM
So the pilot wave kicked the red charge because the particle hopped into the red trajectory just before hitting the screen?
 
vzn
Jul 12 '17 at 14:52, by John Rennie
@ACuriousMind it's the only way I've stayed alive this long despite a lifetime of dissipation and self abuse.
 
@Semiclassical re Bohm contradicting the most basic claim of QM that there are 'no paths', even the big names like jstor.org/stable/20117507 (front page quotes it) are ok with QM being based on that and Bohm completely refuting it
 
@vzn that was posted by my PR man not me
 
Mostly, though i'd qualify the latter by saying that the particle doesn't 'choose' at the last moment; whether it'll end up being guided along in the support of the red or blue packet is determined by its initial location.
 
vzn
lol JR with a PR man. sounds irredeemably superficial :P
 
3:27 PM
@bolbteppa Haven't seen that review. One interesting sociological thing I'll note is that the reviewer and the reviewees all represent different Bohmian 'traditions'
 
So a surrealistic trajectory is one where there is an initial Bohmian location, which determines the particle to follow a trajectory made of red and blue trajectories pasted together. As the particle follow this trajectory and just before it hop from blue to red, the pilot wave then kicks the red charge, thus producing the surreal outcome?

That is, the initial location determines the above events happening in that order?
 
Bohm/Hiley are one tradition, with an eye towards the quantum potential as a key part of the story
 
@JohnRennie To prove 4 vectors A,B,C,D are not coplanar is it OK to prove AB, BC, BD are not colinear?
 
I think he is like one of the major names in the modern papers on the subject and is just reviewing older books, first two pages have his motivational comments on why he got interested in it
 
Bell is another, though I'm doing a disservice to him in 'boxing' him into the Bohmian story. (He was sympathetic to it, but by no means committed to it.)
And finally Goldstein is a representative of the philosophers who've taken up the Bohmian position
@bolbteppa here's the arxiv version: arxiv.org/abs/quant-ph/9512027
(so as to not worry about paywalls)
 
3:30 PM
@Abcd I'm not sure I understand this. When we talk about colinearity doesn't that normally refer of a set of points. It means the points all lie on the same line.
 
@JohnRennie no there are colinear vectors too
 
@Semiclassical thanks
 
@JohnRennie google.co.in/…
 
colinear vectors = parallel vectors, I imagine?
 
@Semiclassical ya
 
3:33 PM
oh, terminology, how you torment us:
3
Q: Difference between collinear vectors and parallel vectors?

devang singhiI can't understand the difference between the two. The definitions are as written in textbook: Parallel vectors are vectors which have same or parallel support. They can have equal or unequal magnitudes and their directions may be same or opposite. Two vectors are collinear if they have the s...

 
@Abcd ah OK. But I still don't understand. Does AB mean A + B?
 
@JohnRennie $\vec{AB} = \vec {B}- \vec A$
@JohnRennie Have you studied vector algebra in uni? Just asking...
 
Anonymous
@Abcd No, it's far from sufficient. Just non-parallelism doesn't imply non-coplanarity.
 
Well consider the four vectors from the centre of a square to the four corners. These are coplanar, but the differences BA, BC etc are not colinear.
 
Anonymous
I'd show something like $\mathbf{DA} \times \mathbf{DB} \cdot \mathbf{DC} = 0$
 
3:36 PM
@Blue Sorry
 
@Blue Equivalently, one can express that as a determinant
 
Anonymous
@JohnRennie I suppose by $A$ abcd is referring to the position vector $\vec{OA}$
 
vzn
@Secret thx hadnt seen that, nice find, Gisin is one of the worlds foremost applied Bell experimenters along with a deep theoretician, neat to see him ref Bohm
 
Anonymous
And by $\vec{AB}$ he means $\vec{B}-\vec{A}$
 
@Blue Please tell me/ link me to some nice article on Cayley Hamiltonian.
 
3:38 PM
also, I hate interuptions, whether it is shitty wifi or a whole barrage of linear mpa questions
my experience told me almost every conversation between me and some third party that get interrupted never continue
 
@Secret what are you talking about?
 
@vzn one of the points Gisin makes in his discussion at the end is one I endorse: Regardless of whatever conceptual satisfaction/clarity one obtains from the Bohmian POV, it's been embarrassingly unproductive as a source of new experiments
 
@Abcd I very much doubt Secret meant us talking about vectors was interrupting him.
 
Anonymous
lol
 
@Abcd well, just somehow recalling about some professional conversations in conferences where I and some guy had a good conversation, the conversation then get interrupted due to one of his old friend visiting. After that the broken conversation just stays like that and never continued
 
Anonymous
3:41 PM
@Abcd Just read Wikipedia and solve the past problems
 
Amazon is trying to sell me pouches of coffee grounds to be used like chewing tobacco...what an interesting world we live in
 
@JohnRennie oh I thought I disturbed him/her!
 
Anonymous
Do all the past 45 or whatever papers you have
 
It is very irritating when we could have get to the nitty gritty of a concept
 
@Abcd or it possibly :-)
 
3:42 PM
lol
 
> "From all we have seen so far, one should, first of all, recognize that Bohmian mechanics is deeply consistent and provides a nice and explicit existence proof of a deterministic nonlocal hidden variables model. Moreover, the ontology of Bohmian mechanics is pretty straightforward: the set of Bohmian positions is the real stuff.
This is especially attractive to philosopher. Understandably so. But what about physicists mostly interested in research? What new physics did Bohmian mechanics teach us in the last 60 years? Here, I believe fair to answer: not enough! Understandably disappointing."
 
Because of that observation, I often have the bad habit of not letting go of speakers easily when asking them questions, going as far to remove all possible distraction from the venue
 
That to me is the real embarrassment of the Bohmian PoV: not that it's inconsistent (I don't think it is) or incomplete (which is seemingly the case in the relativistic setting) but that it's seemingly never managed to actually be productive
 
@Blue I got scared by seeing it at first. Does it need stuff like eigenvectors??
 
as even one possible distraction, and the fragile quantum state that is our conversations, goes down into the drain into the eigenstate of an unfinsihed and unfinsihable conversation
 
Anonymous
3:45 PM
@Abcd It makes a lot of problems much simpler to approach. But I don't know if you will get time to study eigen-stuff.
 
Anonymous
For the time being just remember that a matrix satisfies its characteristic polynomial
 
(My calling it 'incomplete' in the relativistic setting is a gloss which shouldn't be taken too literally. i don't totally understand that aspect of what people have done)
 
@Blue I am asking does Caley Hamilton need eigen stuff in detail or just the meaning of it?
 
My sense is the most fatal flaw of BM is that it uses the notion of a wave function without every justifying why it's legitimate to even use such complicated objects, all the rest is just icing on the cake
 
Anonymous
@Abcd No, it's not necessary
 
3:46 PM
@Blue wow okay!
 
I am sometimes wondering, just why as a Bohmian particle I keep bumping into these surreal trajectory initial conditions where 30 mins ago the conversation go coast, and then 30 mins later it just breaks when one of their friends suddenly pops out out of nowhere and irreversibly derail the conversation
 
You need eigenstuff to understand the proof of Cayley-Hamilton, I think
But to use it, you really don't
 
Anonymous
I doubt he needs the proof to solve high school problems :P I didn't know it till I took a linear algebra course
 
Anonymous
But yes
 
Group theory is very confusing
 
3:50 PM
ya
 
I think one can say that because the Lorentz group is not connected, and because it breaks up into connected components all bijective to the component connected to the identity, with P/T/PT transformations getting us between them, one can reduce the analysis of Lorentz invariant theories to considering tensors which transform only under representations of the component connected to the identity,
 
Anonymous
(Honestly speaking tho, I think they should introduce linear algebra much earlier in school - like grade 8 or 9....much more intuitive than $3\times 3$ M A T R I X M U L T I P L I C A T I O N!)
 
@Blue does it take that much time to go over eigenvalues/vectors? Isn't it just pretty much that an eigenvector is the vector that maintains direction but scales by the eigenvector under a linear transformation? Then you have $A \mathbf{v} = \lambda \mathbf{v} \implies |A - \lambda I| = 0$ and $\mathbf{v}$ is found by $(A - \lambda I) \mathbf{v} = 0$?
 
I've been dealing with Wigner d-matrices lately, and it's been reminding me of everything I've forgotten about group theory (and everything I never actually knew about rep theory)
 
but because it's not simply connected the representations are multi-valued which is non-physical we must actually consider representations of the simply connected cover of this connected component. $SL(2,C)$ is the connected cover, a double cover, and so by constructing irreducible tensors from the vectors that $SL(2,C)$ acts on we have constructed irreducible representations of the Lorentz group connected to the identity, and using P/T transformations can get to the whole group.
 
3:50 PM
Of course actually knowing why those work would take some time
 
However complex conjugates provide an inequivalent rep of the component connected to the identity also, and we refer to vectors transforming under $A \in SL(2,C)$ as undotted spinors and under $A^* \in SL(2,C)$ as dotted spinors to distinguish.
 
> Ray's Dad5 months ago
Why doesn't Workgroup Bohmian Mechanics answer any of the posted questions?
lol, didn't you knew that:
$$\hat{H} \lvert\text{not answering questions}\rangle = \lvert\text{not answering questions}\rangle$$
 
Although I guess everything is easier once you know it too
 
We can show that inversion of a four-vector acted on by $SO(3,1)$ is represented in $SL(2,C)$ theory by sending a vector transforming under $A \in SL(2,C)$ to a vector transforming under $A^* \in SL(2,C)$, but we know that Lorentz transformations of the real world are those for which time increases (hmm) but still allow for parity transformations, i.e. involve 2 of the 4 components (potential nonsense on time increasing there),
so that mixed tensors involving at least one dotted and undotted spinor should correspond to real world objects
 
It's the most stubborn eigenstate in existence with an energy of null
 
3:52 PM
hmmm
 
Irreducible reps of the Lorentz group are found by taking the Levi-civita tensor, which is the invariant tensor of $SL(2,C)$ analogous to $\eta$ being the invariant tensor for $SO(3,1)$, and showing contractions on symmetric mixed tensors of $SL(2,C)$ are zero when you contract two indices of the same dot type, so that symmetric mixed tensors of $SL(2,C)$ with $k$ undotted and $l$ dotted indices are representations of the Lorentz group of what we call spin $(k+l)/2$
There's another confusing point about taking 'bispinors', apparently because of the sign indeterminacy of spinors (because one can use both $A$ and $-A$ to transform a given mixed tensor of $SL(2,C)$, this apparently basically gives you a proof of spin-statistics) one is forced to consider pairs of spinors of half-integral spin to remove this sign issue so you don't end up with double measurements or something
 
vzn
@Semiclassical feel theres something of a catch22 playing out. a theory needs adherents to be pushed fwd and it wont gain adherents unless there are signs of advances. but it also seems, the genuine advances are not being recognized. a classical explanation of the 2slit experiment is revolutionary. think the Bush-Couder theory is being ignored right now. but believe this "state of affairs" cant continue for "long"...
 
::Is wondering how on earth we hop from Bohmian stuff to rep stuff. Surreeeeaaalll metatrajectory anyone::
 
@vzn I'll agree on the first two sentences.
 
::No to find the detector that corresponds to the rep theory hop::
 
3:59 PM
But there's really not a logical contradiction. A theory that doesn't attract adherents doesn't have people working on it.
 
10 mins ago, by bolbteppa
Group theory is very confusing
wut, you call that a detector?
hmm...
 
And whatever conceptual problems there are in QM---well, just because it doesn't make sense doesn't mean it's not useful. And QM, above all else, has been an exceptionally useful and productive theory
 
vzn
@Semiclassical ultimately physics theories have an element of fashion (which has a fundamental element of irrationality). anyway there is an old zen-like yogi berra quote this reminds me of. nobody goes to that restaurant anymore because its too popular ...
 
Initial position, pilotwave set up trajectory X = blue -> red
Sequence of event deterministically encoded for that initial condition:
 
I think Bohmian mechanics would at least trick (if not convince :p) more people into studying it if it paid attention to basic group theory
 
4:01 PM
@bolbteppa It's rather straightforward if one phrases it in the proper mathematical terms: 1. Wigner's theorem requires that symmetry groups are quantumly represented as groups of certain ray transformations. This is what one calls a projective representation. 2. It is a mathematical fact that all projective representations of a group are obtained as linear representations of central extensions of the group.
3. For semi-simple Lie groups, this reduces to all linear representations of the universal cover. 4. Complex linear representations of the universal cover, in turn, are the same as all complex linear representations of the Lie algebra. Nowhere do confusing notions of "multivaluedness" enter. Dotted and undotted spinors are physically simply left- and right-handed Weyl spinors, which are distinct representations of the algebra (indeed conjugates of each other).
 
So, is the situation like as shown:
Initial location set up the following series of events in the surreal trajectory
blue -> trigger red charge nonlocally via the pilot wave -> red -> hit screen
 
Furthermore, you can purely mathematically show that the finite-dimensional complex linear representations of $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}$ are equivalent to the representations of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, which are easy to figure out.
 
Taking the non-relativistic Schrodinger equation is basically working with the Galilean group and they could at least try replace the group with the Lorentz group to fix the theory
 
@bolbteppa For better or worse, the people who advertise Bohmian stuff tend to be more motivated by the philosophical aspects
So the kind of people who find Bohmian mechanics interesting to work on, tend not to be the same people who find group theory very motivating
 
Sure, one requires a thorough grasp of group and representation theory, but there's no mystery here. It's math that has been well-understood for over half a century. I'll never understand why physicists are so enamored with confusing approaches through invariant tensors and indices.
 
4:03 PM
@bolbteppa Eh. That's not really the problem. I mean, if you want to tell the story of one Dirac particle in a Bohmian way, that's actually doable in a Lorentz-invariant way
The problem is that the Bohmian approach is so tied up with the notion of an instantaneous configuration
 
@ACuriousMind yeah I can see how what I said corresponds to 1, 2, 3, would like to avoid Lie algebras completely in 4 which I think the 'invariant tensor' argument does and can be generalized
 
Why would you avoid algebras? They're very physical from the Hamiltonian viewpoint: They are the generators ( = charges!) of the symmetry transformations.
 
Though calling it spin $(k+l)/2$ needs to be justified from this pov and I think you need algebras to do that
 
Which isn't a problem in the non-relativistic setting---Galilean relativity has absolute simultaneity---but is a big issue once you go to the relativistic multi-particle case
 
@vzn They will not see it because they are currently too caught up in rep theory stuff
 
4:06 PM
The true rotation algebra $\mathfrak{so}(3)\cong \mathfrak{su}(2)$ simply embeds diagonally in the $\mathfrak{su}(2)^2$ we are getting the $k$ and $l$ from.
 
You can talk about an 'instantaneous configuration' from the point of view of one of the particles, but how do you do that when you've got more than one particle?
 
@enumaris your best bet is to wait until the day is over and the chat gets it daily reset back to the initial condition
 
(There are technical ways around it, but there's seemingly no way to do so that doesn't do violence to the relativity of simultaneity. So I think one gloss is that the Bohmian setup is one which needs absolute simultaneity in a conceptual sense. Which...ugh.)
 
One question is - okay we are happy considering representations of $SL(2,C)$, why do we care that the inequivalent complex conjugate representations also exist, let alone considering mixed tensors transforming under them? It seems the parity argument I mentioned above is the only way to actually motivate doing it, but it's still a bit out of thin air
 
@user2646 and no that is not allowed. One more thing: The eigenstate NULL will not go away until the next system reset
 
4:10 PM
@bolbteppa Why wouldn't we care? Every single one of these representations yields a projective representation of $\mathrm{SO}(3,1)$, which is what we're physically interested in.
 
In other news...
 
quoting from an email exchange I had with a Bohmian person:
>" Re: Minkowski spacetime, in general it is the case that relativistic formulations of pilot-wave (type) theories -- such as the hypersurface bohm-dirac model you mention -- require a dynamically privileged foliation of spacetime (into, e.g., spacelike hypersurfaces). Depending on one's taste, one could either think of this as a perverse reversion to pre-relativistic Galilean spacetime, or as some additional structure (beyond the Minkowski structure) that turns out to be needed in light of Bell's theorem. But however you want to
 
Hamilton Howard "Albert" Fish (May 19, 1870 – January 16, 1936) was an American serial killer. He was also known as the Gray Man, the Werewolf of Wysteria, the Brooklyn Vampire, the Moon Maniac, and The Boogey Man. A child rapist and cannibal, he boasted that he "had children in every state", and at one time stated the number was about 100. However, it is not known whether he was referring to rapes or cannibalization, nor is it known if the statement was truthful. He was a suspect in at least five murders during his lifetime. Fish confessed to three murders that police were able to trace to a known...
 
@Secret I read the name, start to chuckle, and then I read the rest of it
 
@Secret sorry, I'm not sure what you're referring to?
 
4:15 PM
@enumaris well er.. you said "hmm" in the middle of the rep theory mire, so that makes me thinking, are you confused by the flow of the conversation like I do
 
Anonymous
 
hmmm
 
Anonymous
Right, it's 'hmmm', not 'hmm'. There's a difference, mind :P
 
@ACuriousMind Well, one might think we shouldn't care that complex conjugates provide an inequivalent representation because we can transform vectors under a given $A \in SL(2,C)$, and since the complex conjugate $A^*$ also lives in $SL(2,C)$ it's just some distinct element of $SL(2,C)$ like all the others, so why can't we just ignore complex conjugates and treat them as a different element like we do for every other element of $SL(2,C)$?
 
@Secret ah...I just say hmmm when I join :D
and also when I'm here
 
4:18 PM
My group theory frustration: I'm still bothered by the fact that I don't know why $\langle j\,\,{m},j\,{-m}|00\rangle =(-1)^{j-m}/\sqrt{2j+1}$
 
and also at other random times
Most of the time I gloss over the math/physics in here :P
 
I can calculate to to my satisfaction
But I keep feeling like there ought to be a better answer
 
So maybe there is an insanely physical reason for the fact that complex conjugates provide an inequivalent representation - e.g. this argument about parity transforming undotted to dotted spinors, or perhaps the whole $(-1)$ sign issue forcing us to work with bispinors in order to remove the $-1$'s that may arise, hmm
 
(I mean, the fact that it's + when m=j is conventional. But the rest isn't.)
 
@bolbteppa I don't understand what your argument is supposed to be. The resulting representations are not isomorphic, so they induce two different projective representations of the Lorentz group, i.e. two different kinds of spinors - namely, the left- and the right-handed Weyl spinors.
 
4:22 PM
About the smartest line of thinking I had was that $e^{-i \pi J_y}|j\,m \rangle = (-1)^{j-m}|j\,{-m}\rangle$
 
@Semiclassical there's a really interesting discussion in Landau volume 1 about how the Galilean invariance inherently forces one to consider instantaneous potentials to model interactions, indeed this is the only motivation one needs for replacing the Galilean group with the Lorentz group since instantaneous interactions are not observed in nature, this is one reason I find it so shocking people take BM seriously since it promotes a non-relativistic equation to an axiom
 
Eh. I don't think that's the right issue: if you're doing a relativistic theory, you'd presumably use a relativistic wave equation. And Bohmian mechanics, in the sense of the dirac-bohm stuff, does do that
 
@ACuriousMind I guess my point is - so what if we have two inequivalent representations, i.e. is there some theoretical reason forcing us to use them (e.g. the parity argument, or the 'pair of spinors is needed to remove the $-1$'s), of course they end up being useful but besides that... and are there other inequivalent reps?
 
vzn
@Secret who is caught up?
 
Not sure about whether the guidance equation is inherently non-relativistic---I haven't thought about it
 
vzn
4:29 PM
@Semiclassical sounds like Tenev-Horstemeyer spacetime fabric o_O
 
@vzn everyone else that does not have a red or pink shaded avatar or you
That whole conversation obliviated questions out of existence, meh, perhaps it means I should get osme sleep
I am starting to act like a madperson
 
vzn
@Secret lol youre talking about this room? bohmian mech has come a long way it isnt dismissed out of hand in here any more & there is 1 "other" credible adherent now + newly minted phd :P o_O
 
@enumaris I've always just assumed those times were you getting stuck at work
 
the crux of the problem is that, for better or worse, the Bohmian approach (in the sense of having particle velocities determined via the guidance equation) rests upon whether it makes sense to talk about instantaneous configurations
 
My version is "that's odd..."
 
4:32 PM
@danielunderwood not necessarily
sometimes it is
 
@bolbteppa Nobody "forces" you to use them. There's tons of representations that probably never occur, e.g. $(10,37/2)$. But for left-handed and right-handed Weyl spinors it turns out you need both to describe a massive spinor because the evolution equation couples the two parts.
 
in the non-relativistic setting, there's nothing too controversial about that.
 
done
 
vzn
@Semiclassical think people need to stretch their brains over relativistic vs nonrelativistic in a way its all a (anthropological) compartmentalization bias
 
but in the relativistic setting, the dirac-bohm story only works if you add extra structure beyond just 'minkowski spacetime'. which is not very appealing to most people
 
vzn
4:34 PM
@Secret ?!?
 
1 hour ago, by Secret
So a surrealistic trajectory is one where there is an initial Bohmian location, which determines the particle to follow a trajectory made of red and blue trajectories pasted together. As the particle follow this trajectory and just before it hop from blue to red, the pilot wave then kicks the red charge, thus producing the surreal outcome?

That is, the initial location determines the above events happening in that order?
 
vzn
@Semiclassical need to learn more but think minkowski spacetime isnt even full GR is it?
 
@vzn no, but it is the setting for the most familiar kinds of QFT
and in particular it's what you use for QED
 
@ACuriousMind one apparently can work some of that logic backwards and derive the Dirac equation based on the fact one is forced to couple dotted and undotted spinors, and apparently including parity is the (or at least one) reason why
 
vzn
@Semiclassical technically speaking a black hole cant even be modelled in minkowski spacetime right? o_O :P
 
4:36 PM
and crazily this thinking apparently lets you derive equations for arbitrary spin getting Maxwell, KG, Rarita, all of them
 
Yes, well, one would like to be able to talk about an electron in a relativistically sensible way without invoking black holes
 
Minkowski spacetime is just flat spacetime
i.e. no gravitational fields
anywhere
 
vzn
yeah thought so thx
 
It's a good approximation for weak gravitational fields
 
@vzn but anyway, both my brain and the chat currently is too tied up in turtles all the way down to see that so whatever, I am going to sleep
 
4:37 PM
like very weak gravitational fields :P
 
@enumaris which is thankfully true for scattering experiments
 
vzn
@Semiclassical lol reminds me of this ran across it not too long ago (think Its Not Totally Crazy™) en.wikipedia.org/wiki/Black_hole_electron
 
@vzn If Minkowski spacetime is just flat spacetime, and every spacetime is locally approximated by Minkowski spacetime, and black holes are impossible in Minkowski spacetime - how are black holes possible?
 
If my eyes aren't real how are things real?
 
@bolbteppa I imagine there's a simple out, but I don't actually know enough about black holes to see it
 
vzn
4:39 PM
@bolbteppa what is the sound of one hand clapping?™ :P (also reminds me of zenos paradox)
 
even black holes are locally minkowski
 
"I'm ignorant of the answer to the question" != "The question is a paradox"
 
mind=blown
 
(I am most definitely ignorant of the answer.)
 
I mean the simple answer is "Locally Minkowski" != "Minkowski"
or maybe "everywhere locally minkowski" != " Globally Minkowski"
 
4:41 PM
Right.
A sphere is locally Euclidean, but it's nevertheless not flat.
 
yes indeedy
Every manifold is locally $\mathbb{R}^n$
 
vzn
but feel you guys are illustrating my point that there may be cognitive biases about spacetime stretching that are artificial human constructs also interfering with freethinking on it...
 
But not every manifold is $\mathbb{R}^n$
 
@bolbteppa Ah, that's also not actually mysterious: Parity is in the pin group, but not the spin group. If you want a representation of the pin group, you find that the Dirac spinors are already irreducible, i.e. there is no representation of parity on a solitary Weyl spinor.
 
And I feel like you (to steal an old philosophical line) are like a bird in flight who assumes that you could fly all the easier if only there wasn't all that air in the way.
 
4:43 PM
well if the bird goes into space, it would be a bird astronaught
how awesome would that be
 
vzn
anyway "agreed" fundamental spacetime understanding is crucial to advancing new theory... think there are surprises in store/ on intermediate horizon...
 
It would be, right until its wings fail to find any air to push against
 
I indeed left out a step in my explanation of representations earlier: Since the Lorentz group is not simply-connected, you always need to represent the $\mathbb{Z}_2^2$ part that is parity and time reversal in addition to seeking representations of the cover
 
At that point it'd be a bird asteroid
 
The reason is that the "semi-simple" in the theorem about the universal cover sneakily includes the stipulation that the group be connected.
 
4:44 PM
How hard would it be for a bird to attain orbit?
 
(Asteroid? Meteoroid? Meteorite? I dunno)
 
vzn
thinking of every almost-now-cliche scene in a movie where astronaut gets sucked out into space like gerald butler movie last summer, almost comical imdb.com/title/tt1981128
 
I will say that, if you take the Bohmian stuff seriously when it comes to relativity, it does seem to make 'spacetime structure' a more complex question than in the standard story of QFT
which, if that ended up being useful, would be a Good Thing
But I'm not in a position to either advance or assess that, so w/e
 
vzn
21st century Daedalus :P
 
@Semiclassical be the change you want to see in the world
 
4:49 PM
lol
 
@ACuriousMind So the component connected to the identity has $\det(\Lambda) = +1$ and $\text{sgn } \Lambda^0_{~0} = + 1$, and reps of the double cover of this get you dotted or undotted spinors, but then including parity so that you can transform into the $\det(\Lambda) = - 1, \text{sgn } \Lambda^0_{~0} = + 1$ component I guess requires considering another spinor to represent this second component, and the Dirac equation is about relating the spinors in both part
 
vzn
@Semiclassical you have a point but so does zen :) :P
@Semiclassical try the Tenev-Horstemeyer paper, (nearly) everyone else in here hates/ trashes it, join the club™... what kind of phd physicist are you anyway if you dont have something knowledgable to say about GR/ spacetime structure etc, or blithely casually dismiss a new theory? :P
 
one who hasn't actually taken a GR class and therefore prefers to acknowledge his own ignorance
I can certainly have instincts and intuitions. But I have no way to ground them in what's actually knownn.
 
vzn
lol hasnt taken a GR class either, not completely ignorant either :P
 
@Semiclassical If only more people were so humble...
2
 
4:57 PM
tbf, it's hard to consistently acknowledge one's own ignorance
 
<--- taken 2 courses in grad level GR as well as diff geo, and done a lot of independent studying in GR...but still GR is a mystery
 
vzn
see, all physicists agree QM + GR have to be unified, but apparently without giving up anything we think right now... yeah right hows that working out for you all :P
 
it's not hard for me to say it in this context. But are there contexts where I'd blithely act as though I actually understand it? Probably.
 
Another point, the position-space wave function of a photon does not even exist, how could BM explain things like this when they call it a real thing
 
I'd gladly get rid of the standard model
just drop kick it out of existance
unfortunately, it agrees with experiment...=/
But don't take GR away, GR is so pretty
2
 
4:59 PM
Normal QM leaves you freedom to play with things, no problem when fundamental concepts make no sense :p
 
@bolbteppa the problem is that the photon is as essentially relativistic as you get
 
photons in black holes
 
which means that it's almost inevitably going to create headaches for Bohmian mechanics
 
photons in black holes in an expanding universe
 
00:00 - 17:0017:00 - 23:00

« first day (2877 days earlier)      last day (2043 days later) »