The h Bar

shashank

16:11

@JohnRennie Sure sir, just drop in a message whenever you're online .

John Rennie

Don't you just hate it when someone posts a better answer than yours? :-)

@shashank Hi, are you still around?

Sanya

@JohnRennie you did that to me so often :'<

shashank

Yes sir.

John Rennie

@Sanya Oops :-)

@shashank gyroscopes?

shashank

I would like to know the forces responsible for the precession of a gyroscope.

John Rennie

16:21

Ah I see you've just posted a question:

0

Q: Rotation of a disk

If I keep a rotating disc along the x-y plane and apply a force along the same plane in line with the disc's centre of mass, assuming no friction and no gravity, will the disc move along the direction of the force or in some other direction?

newtonian-mechanics

shashank

Yes sir, it's related to gyroscopes.

John Rennie

The answer to your question is that the disk will move in a straight line in the direction of the force.

shashank

In that case how is it any different from the rotating disc of a gyroscope?

John Rennie

Generally speaking gyroscopes only behave weirdly in response to a torque i.e. a force that isn't in line with their pivot point.

In your question the torque is zero. Torque is force times perpendicular distance, and since you have the force in line with the centre of mass the perpendicular distance is zero.

shashank

Yes sir I get that but then how does the disc move in a direction perpendicular to the force ?

For the disc to precess shouldn't the disc get some velocity in that direction ?

John Rennie

16:29

I think it's hard to explain this in a chat room. I suspect that to understand it would require us to sit down together with a large piece of paper and draw diagrams.

Did you look at the book Daniel Sank mentioned?

shashank

Yes sir I saw that image but I understood why the gyroscope doesn't fall but I failed to understand the precession part.

I understood that the disk twists about the z axis but this twisting doesn't explain precession.

John Rennie

I don't think I can help - sorry. I think it's just too hard to explain without diagrams.

shashank

No problem sir thanks a lot for your time.

Just one last thing .

John Rennie

Yes?

shashank

Does it have something to do with the speed of the atoms or is there some force that guides them?

John Rennie

16:38

You make it sound mysterious and it isn't. The direction of the torque isn't the same as the direction of the forces on the wheel, and that's the root cause.

shashank

Thanks a lot sir .

John Rennie

Suppose you start with the gyroscope horizontal. It's easy to think the force must be downwards, because gravity pulls downwards, but actually the force on the top of the wheel is outwards and the force on the bottom of the wheel is inwards.

Draw a diagram showing the gyroscope side on and you'll see what I mean.

shashank

Yes sir I got it .

John Rennie

And now I'm going to slump into my armchair and read about dogs

shashank

Thanks a lot sir and happy reading .

ACuriousMind

16:50

@JohnRennie Have you got a dog or are you just reading about them?

John Rennie

No, I've never owned a dog nor wanted to. I'm a typical nerd really and I find dogs intolerably touchy-feely. I quite like cats, though not enough to want to pay to feed one.

The book is about the evolutionary history of dogs rather than about how great they are as pets.

ACuriousMind

Heh, that matches my feelings about dogs quite well ;) I grew up with cats, though, and definitely want to keep one again when I live in a place that's suitable

John Rennie

The Book was recommended by New Scientist, and I find their book reviews are generally pretty good.

I've just read and very much enjoyed Coyote America by Dan Flores:

amazon.com/Coyote-America-Natural-Supernatural-History/dp/…

and that's sort of related since all the canines are genetically very similar - dogs, wolves and coyotes can all interbreed - in principle, it would be a brave coyote who'd sleep with a wolf :-)

Emilio Pisanty

17:15

@JohnRennie That's apparently been a thing for some time

Coywolf

Coywolf (sometimes called woyote) is an informal term for a canid hybrid descended from coyotes and gray wolves. Hybridization between the two species is facilitated by the fact that they diverged relatively recently around 6,000–117,000 years ago. Genomic studies indicate that nearly all North American gray wolf populations possess some degree of admixture with coyotes following a geographic cline, with the lowest levels occurring in Alaska, and the highest in Quebec. == Description == Hybrids of any combination tend to be larger than coyotes, and show behaviors intermediate between coyotes and...

Really putting biologists on their toes there regarding that whole "species" concept they keep using

Works really well except when it doesn't

@JohnRennie On that one I'm curious to know if this place comes up

Coyoacán

Coyoacán ([koʝoaˈkan]) refers to one of the 16 boroughs (delegaciones) of the Federal District of Mexico City as well as the former village which is now the borough’s “historic center.” The name comes from Nahuatl and most likely means “place of coyotes,” when the Aztecs named a pre-Hispanic village on the southern shore of Lake Texcoco which was dominated by the Tepanec people. Against Aztec domination, these people welcomed Hernán Cortés and the Spanish, who used the area as a headquarters during the Spanish conquest of the Aztec Empire and made it the first capital of New Spain between 1521...

0celo7

18:02

@JohnRennie wtf

@ACuriousMind Proof I'm not a nerd.

I love dogs

@ACuriousMind So are position eigenstates not physical

ACuriousMind

18:15

@0celo7 Please ask a more precise question.

We've talked enough about this that I know you can.

vzn

Proton radius mystery deepens as deuterium measurement comes up short / physicsworld

peterh

@ACuriousMind There are foundations and other volunteer things working on the re-socialisation of stray cats. In Germany it is a much smaller problem, but if you can live in a home where you can get a cat temporarily, you could have your cat - for a time - while you can also help one to find her final human.

Emilio Pisanty

@vzn Yeah, that whole thing feels like premature scandal to me

it may be worth asking on the site though

vzn

@EmilioPisanty why so? saw some interesting discussion in transcript on it... not really my area but do like any/ all "BSM" hints... aka "outside the box" etc

Emilio Pisanty

@vzn Lots of talk of "proton radius" and not so much talk of "actual physical observables", at least in public-facing popsci texts

vzn

18:19

@EmilioPisanty (sigh) am a bit tired of all the criticism of "popsci" articles that are well grounded in recent scientific findings. :(

Emilio Pisanty

@vzn (btw would appreciate not editing messages with pings in them, it sort of buts into stravinsky rather loudly at the moment)

ACuriousMind

2 days ago, by John Rennie

@ACuriousMind http://arxiv.org/abs/1502.05314

The "radius" is essentially a Lamb shift measurement

At least in that publication, I'm not sure if all of these experiments do it that way

Emilio Pisanty

@vzn actually, you know, I'm sorry I said anything. You can keep your tiredness to yourself and we'll all be happier.

vzn

@EmilioPisanty oh, so my complaints are less relevant here than yours. ok!

Emilio Pisanty

@vzn No. This is not a conversation I have time for at the moment.

vzn

18:22

@EmilioPisanty lol yeah whatever my time is valuable too have a nice day

@ACuriousMind the consternation seems to be over new muon measurements. tricky experiments to set up etc (reading from transcript etc). but agree with others its quite notable that low-energy/ low-cost experiments are challenging SM to some )( degree

2 days ago, by ACuriousMind

But as I said at the beginning, I'm a bit puzzled that we should expect the proton radius to be constant, after all the protonic states in a muonic atom should also differ from the electronic atom, but everyone seems to think it's obvious that the proton "radius" should be constant across all systems.

not an expert but wondering if a constant radius a prediction of SM? etc

Bernard Meurer

18:37

@dmckee Halp! x86_64 Assembly is making me cry

0celo7

@ACuriousMind enough about what

My prof said that physical states are normalizable

position eigenstates are not, so are they not physical?

Secret

@0celo7 Is there a name for a topology formed by stacking along the y direction uncountably infinite copies of $\mathbb{R}$ (which we knew is an ordered set) placed in the x direction to form something that looks like $\mathbb{R}^2$ except it is not dense in the y direction?

0celo7

That's not a topology you're creating, it's a set.

And I'm not sure what you're describing

Is it $\Bbb Q\times\Bbb R$?

ACuriousMind

@0celo7 yes

Secret

It is similar, except I want the y direction to be some unordered set of some infinite cardinality, and I am trying to figure whether it is dense in general in the y direction?

ACuriousMind

18:53

@Secret What does "dense in the y-direction" mean?

Subsets are dense, or they are not.

They can't be "dense in a direction"

Secret

The rough analogy is an uncountably infinite generalisation of arranging a horizontal pile of toothpicks to tile a square, and I am wondeirng whether in general there will be gaps if I move infinitesimally from one y position to the next. I am not sure how to make my statement more precise

ACuriousMind

...just take an uncountable disjoint union of the reals with themselves if you want something with "gaps"

Secret

ok that makes sense

dmckee

@vzn We're not really able to do ab initio predictions about baryons from the standard model, though lattice calculation are getting better that way.

Bernard Meurer

@dmckee imgur.com/a/uvj30

What am I doing wrong?

dmckee

18:58

But quasi-elastic scattering results suggest that the state of a proton in a light nucleus is similar to that of a free proton. Heavy nuclei are another matter entirely.

Bernard Meurer

I'm trying to print "Thanks Dan!"

Secret

@ACuriousMind a follow up question is then, now that we have an uncountable disjoint union of reals to form the conceptual 'square'. What we need to do next in order to make this set $\mathbb{R}^2$ which is known to be a dense set. Are these extra sets to be added into the union in general measure zero, or are actually empty sets because any intersection of two components in the disjoint union is empty?

dmckee

I never learned x86_64 and I don't know Gtk, so I"m not in a good position to help.

Bernard Meurer

@dmckee Oh god even you don't know it

I am lost

dmckee

I learned enough x86 to satisfy my numerical methods in physics prof that I understood what goes on at the on-the-metal level and then thankfully forgot it.

I prefer the calm, orthogonal, purity of m68k and similar architectures.

ACuriousMind

19:01

@Secret "Dense" is a property of subsets of a topological space. Saying "$\mathbb{R}^2$ is known to be a dense set" doesn't make any sense. On the level of sets, the disjoint union $\bigsqcup_{i\in\mathbb{R}}\mathbb{R}$ and $\mathbb{R}\times\mathbb{R}$ have a bijection. You don't need to "add" any "extra sets".

The disjoint union and the product differ in their natural topology, but that is all.

Secret

ok noted

peterh

Funny comment from the internet: "Early this year, LIGO peole reported first direct evidence of the gravitational waves. It was based on the change in path length amounting to very tiny fraction of the size of a proton. As the size of proton itself is in trouble, the discovery of GWs is in deep trouble."

0celo7

19:18

@Secret you should take a topology course

@ACuriousMind You were right about my topology prof, Morwen Thistlethwaite

He's an elf

Emilio Pisanty

@ACuriousMind Yeah. But it's still surprisingly awkard to biject $(0,1)\cup(1,2)$ with $(0,2)$.

ACuriousMind

@EmilioPisanty Hm? In this case, the bijection would be that we send the copy of $\mathbb{R}$ indexed by $x$ to $\{x\}\times\mathbb{R}$

Emilio Pisanty

@ACuriousMind Yeah, that's easier. Mine was only a vaguely related comment.

0celo7

@EmilioPisanty Biject each one to $\Bbb R$, so you get $\Bbb R^2$, then biject that to $\Bbb R$.

Emilio Pisanty

@0celo7 that's... not quite how bijection works. If you biject each to $\mathbb R$, what you get is the disjoint union $\mathbb R\dot{\cup}\mathbb R$.

0celo7

19:27

I know

Emilio Pisanty

and, in any case, you would still have to biject $\mathbb R^2$ back to $\mathbb R$, which is also an awkward construction (i.e. way overpowered for the task at hand).

Maybe this will show what I mean more clearly: show an explicit bijection $(0,1)\leftrightarrow [0,1)$.

@DanielSank Lucky you. I need to convince a referee that $\sum_{jk}\hat{x}_j \frac{\partial A_k}{\partial x_j} \hat{p}_k$, with $\vec A$ in the radiation gauge, is hermitian.

0celo7

@EmilioPisanty Hmm, not very fun.

I clearly made a dumb mistake earlier

Emilio Pisanty

@0celo7 Point is, it looks like it should be easy, right? They're both essentially the same, and they obviously have the same cardinality (since e.g. $(0,1)\subset[0,1)\cong [1/2,1)\subset(0,1)$), so there should be some bijection.

It's just surprisingly hard to come up with a natural one.

0celo7

Yes, also unsurprisingly uninteresting.

Emilio Pisanty

@0celo7 To each their own. But unless you can actually do it explicitly, don't diss it.

0celo7

19:36

I can't.

But I also can't weave a basket underwater.

I don't lose any sleep over that.

@ACuriousMind I should not have revealed I am taking topology and QM, now you won't want to help with them

Emilio Pisanty

@DanielSank Also, this is why you don't say Letter in the arXiv preprint ;-). That's how you get into e.g. arxiv.org/pdf/1403.0414v1.pdf ("In this Letter...") → nature.com/articles/srep11473 . Not terrible, but not amazing either.

ACuriousMind

@0celo7 what?

0celo7

@ACuriousMind if I ask you something about QM or topology, you'll write it off as homework and not respond

ACuriousMind

19:51

I think I'm going to write anything off you ask me if you keep making unfounded assumptions about me :P

0celo7

Seems fair

@ACuriousMind I have a legitimate question

Is he saying that the coordinates of $|\alpha\rangle$ in the new basis are the same as the coordinates of $|\alpha\rangle'$ in the old one?

Emilio Pisanty

20:24

@DanielSank I had a brief look at the preprint, and it looks interesting enough for a PRL to me. If you have a cool beyond-the-RWA effect, with experiment to match it, then you've definitely got my ear. On the other hand, it does feel like some restructuring would do it good.

Happy to discuss it later, probably when we're both online and I've had a closer look at the text.

DanielSank

@EmilioPisanty oh jeez

@EmilioPisanty Wait, what?

Emilio Pisanty

@DanielSank I know. Goes on to say "and therefore the rest of the calculations are probably wrong".

DanielSank

@EmilioPisanty That would be most welcome.

Emilio Pisanty

@DanielSank arxiv posting says "In this Letter...", and the paper ends up in Sci. Rep.

DanielSank

@EmilioPisanty Is that good or bad?

Emilio Pisanty

20:31

@DanielSank I mean, whatever works for getting your paper out, right?

DanielSank

@EmilioPisanty That would be most welcome.

Emilio Pisanty

But I do feel PRL to Sci Rep is a step down in terms of the audience that will see your paper and the impression that hiring/grant review/etc committees will get from the bibliometrics

DanielSank

I don't even know what Sci Rep is.

I'm not good about this sort of thing.

Emilio Pisanty

@DanielSank It's a Nature group journal

they review for technical correctness

but not on any interest criteria

DanielSank

I generally think official publication channels offer almost nothing to the scientific community anymore. The referee process is a very important thing, but I dislike that it's closed to outsiders.

Emilio Pisanty

20:34

yeah, your field is probably happier that way than mine

here people do browse PRL and PRA for new finds much more often than they do arXiv

and few people would casually browse Sci Rep for interesting finds

DanielSank

@EmilioPisanty Not my field. Mostly I think it's not being under the thumb of academic funding channels.

Emilio Pisanty

@DanielSank yeah, that sounds like a good place to be

Anyways, it's just something I do personally. If my initial submission is to PRL, I'd rather keep that fact to myself until the chips are down. All it takes is a bit of care when polishing the preprint. CVs of failures are probably a good thing but they have their place.

DanielSank

@EmilioPisanty That's cute.

I like that.

I'll add a section to my CV called "failures".

Emilio Pisanty

@DanielSank Yeah, it went viral some months ago. Hence the Meta-Failures at the end

vzn

21:03

@DanielSank suggest working with "physical id's" (labels/ numbering scheme/ serial codes) of the chips/ wafers... those probably already exist in some form right? one could create a numbering scheme also

ACuriousMind

@0celo7 I'm...not sure what that excerpt is trying to say

0celo7

@ACuriousMind I realized you're missing context.

$U$ is the unitary transformation operator from the basis $|a_i\rangle$ to $|b_i\rangle$.

i.e. $|b_i\rangle=U|a_i\rangle$.

peterh

0

Q: Rotation of a disk

If I keep a rotating disc along the x-y plane and apply a force along the same plane in line with the disc's centre of mass, assuming no friction and no gravity, will the disc move along the direction of the force or in some other direction?

newtonian-mechanics

I think the question is exact... in the VtC as "unclear".

0celo7

@ACuriousMind But I was able to show that the components of $|\alpha\rangle$ wrt. $b$ are those of $U^\dagger|\alpha\rangle$ wrt. $a$.

So I think that's what he meant.

DanielSank

21:20

@vzn I was asking @JohnRennie a more particular question.

vzn

@DanielSank yeah, read his ideas, sound good (assuming your db has constraint logic); you said you wanted good db insert efficiency, that can sometimes be improved by working with keys/ ids that are not (auto)generated by the database but are app-oriented instead.

Emilio Pisanty

4

vzn

22:39

Synopsis: Pentaquark Discovery Confirmed / APS

DanielSank

23:56

@EmilioPisanty Is that the paper you're trying to publish?

I'm not seeing the hermitian operator you described earlier?

Transcript for