The h Bar

user400188

12:00 AM

I guess I have to remove caps on distributive law as well for consistencies sake.

DanielSank

@user400188 Yeah. It's really interesting how often titles and random words in posts here are capitalized.

There must be some reason that our brains want us to do it.

user400188

When writing a book or something; its a standard to capitalize the title.

we leave words like "the" and "and" lowercase; but the main words have capitals at the front.

Looking directly ahead in my room right now I see "A Game of Thrones" in front of me.

DanielSank

For titles, yes, but there are many cases of words within the question text being capitalized as well.

(and for what it's worth, we decided not to use capitalization for titles on this site. No real reason, I don't think, but we had to pick something and stay consistent)

(Actually, there one real reason is that many titles are actually full sentences, e.g. "How do I compute the...?" which seems like it should use normal English capitalization.)

user400188

Good point

Anyway: I just went through the question and removed all the unnecessary capitals. The only one that remains is in the Wikipedia link.

DanielSank

^ hahaha, good point.

Anyway, I don't think I can help in that particular question.

user400188

12:13 AM

Yeah. The reason I asked it here actually was because I despaired when I realized most of the question on that topic in SE go unanswered.

DanielSank

@user400188 A bounty might help.

user400188

I will have to wait for at least a day to assign one.

Also, I do not wish to use a bounty to excuse a badly worded question. So I must ask; it this question well worded enough for physics SE?

DanielSank

@user400188 I was having trouble following it, but I'm not familiar with the formalism.

I will read it again now.

I don't remember what all those triangle symbols mean.

I know they're AND and OR but I forget which is which.

user400188

$\land$ is AND, $\lor$ is OR

$\bot$ is False and T is True

DanielSank

...but I think the resolution of the problem is that you're making some kind of assumption that a wavefunction can be thought of as saying a particle is "here or here or here or..." which is not true.

user400188

12:19 AM

we are not assigning locations to the particle; we are assigning regions the particle may be in.

for instance; it would be fine to say a particle is within the location (the entire universe). Similarly we can say it is somewhere within a location of [here, here] so long as that location is larger than the uncertainty in position of the particle.

DanielSank

Perhaps I misread the question.

You're invoking the uncertainty principle.

user400188

If its worded well enough you shouldn't be able to.

DanielSank

So you are thinking about quantum mechanics.

user400188

Indeed.

DanielSank

Right, so then you must remember that the information available to you about a particle does not come in the form of arbitrarily constructed probability distributions!

Quantum states are, by their very nature, a way of talking about the information available in a quantum system.

In fact, I heard Zoller say just that quantum mechanics is a theory of information.

user400188

12:29 AM

For the record: if I were to ask a bounty, I would add a request at the end asking that to answer this question, a person does not simply disprove or reject the proof I reference. Instead they should give an explanation as to why the distribution law fails.

DanielSank

Make sure that's clearly asked in the question, not the bounty message!

user400188

Well the question is titled "Why does the distributive law fail in quantum logic?" As opposed to "Why does this Wikipedia proof work?"

I was hesitant to add more info the the question because it was already so long too..

DanielSank

Well if you want to make it more concise, which you should, remove the irrelevant statements like this one:

> I spent many days considering this result...

That is completely irrelevant to the question you're asking.

user400188

Good point.

DanielSank

@user400188 It's a common mistake. Many posts here tell a narrative story of the author's mental life instead of getting to the point :)

user400188

12:38 AM

On math SE its can be a bit different. When you include just a bit of narrative, it can give some clue to the level of knowledge of the asker.

For instance a coder may talk about the situation they are in which alerts you they are not a mathematician.

DanielSank

What does $T$ mean in $p\land(q\lor r)=T$

user400188

T just stands for true

note that the = here should be read as $A\leftrightarrow B$ is $(A\land B)\lor ([\lnot A]\land[\lnot B])$

DanielSank

@user400188 That was not clear to me at all. Definitely explicitly say that $T \equiv \text{TRUE}$.

I once read very good advice about asking questions:

user400188

Ok. Ill put it near the start of the OP

DanielSank

Regarding your audience: Assume zero knowledge but infinite intelligence

6

In other words, start from the beginning and define everything carefully, but also know that your audience can understand whatever you tell them if it's clear (i.e. don't repeat yourself).

user400188

12:43 AM

Hmm, thats not a bad quote there.

DanielSank

@user400188 It's so very true and useful.

user400188

Anyway. I need to head off now. Thank you for the help rewording the question @DanielSank

0celo7

1:42 AM

@heather I recommend lots of thing s

Not all are serious

user400188

@vzn You are the second person to have misread the OP as assigning exact positions to the particle. Because of this, I fear I have badly worded my question. Could you point out the line which conveyed this to you? I would like to reword that part.

vzn

2:11 AM

@user400188 think its not a misreading. you talk about "particle in [x,y]" vs "particle in [y, z]". you have to try to convert your question to "observables" in QM. QM formalism tells you what is "observable".

Mithrandir24601

2:25 AM

@DanielSank I did (not that long ago) go through a large chunk of algebra of deriving the optical Bloch equations without using the RWA. It's definitely not an answer, but it might make a decent start to one...

vzn

3:04 AM

@Mithrandir24601 uh, not following. typo? "without" using RWA? what does that say about RWA? nothing, right?

0celo7

3:33 AM

@heather More hints: $|\int f|\le \int |f|$ and $\int f\le \int g$ if $f\le g$.

You now have all the tools, just figure out how to use them.

user123733

4:36 AM

Good morning

Can anybody help me in this

Yashas Samaga

What are your thoughts about the problem?

Stable equilibrium = the loop tries to return to the original position when displaced

Unstable equilibrium = the loop does not try to return to the original position when displaced

You need to do trial and error

check how the torque acts

The loop has a magnetic moment

You can find the direction of the magnetic moment using the right hand rule

The torque is given by M x B

user123733

And force is IL x B

In a option there is a torque

And F=0

So it is in equilibrium

But of what type

Yashas Samaga

if there is a net torque then it is not in equilibrium

user123733

Ohh

Yashas Samaga

if the loop tries to rotate, then that is not the answer becaz it is not in equilibrium

select the ones which are in rotational equilibrium first

then give a small push to the loop

if the loop tries to come back to its original position, then it is in stable equilibrium

if it tries to go away in the displaced direction, it is in unstable equilibrium

the loop has a magnetic moment, you can use U = - M.B too

negative max = stable

positive max = unstable

user123733

4:43 AM

In option (b) net force is not zero

Yashas Samaga

then it is not in equlibrium

what?

loops always have zero net force

in a uniform field

user123733

Why

Yashas Samaga

assume that is a wire

user123733

Ok

Yashas Samaga

Think of biot savart's law

you have dl x r

user123733

4:45 AM

Yes

Yashas Samaga

if you think deeper, you'll realize that that wire feels a force as if it was a straight wire from its ends

user123733

Yeah

Yashas Samaga

user123733

We can replace it with a straight wire

Yashas Samaga

the randomly shaped wire feels the same force as the straight wire

user123733

4:46 AM

Yes

Yashas Samaga

they are equivalent

so if you have a loop, you are going to end up with a wire of zero length

for the random wire in the picture I uploaded, you had two end points wher eyou could draw another wire

how are you going to do it in a loop?

user123733

But if we consider force at each side then

Yashas Samaga

if this is hard to understand, consider force at each side, opposite sides will cancel out

the effective length is zero

user123733

Oh yes

Got it

Yashas Samaga

the torque needn't cancel out though

to solve that problem, you first have to eliminate the options where there is a net torque

user123733

4:48 AM

In option (b) both force and torque are zero

Joshua Lin

but they are homotopic!

user123733

What is homotopic

Yashas Samaga

what I said is an example of homotopic

Homotopy

In topology, two continuous functions from one topological space to another are called homotopic (from Greek ὁμός homós "same, similar" and τόπος tópos "place") if one can be "continuously deformed" into the other, such a deformation being called a homotopy between the two functions. A notable use of homotopy is the definition of homotopy groups and cohomotopy groups, important invariants in algebraic topology. In practice, there are technical difficulties in using homotopies with certain spaces. Algebraic topologists work with compactly generated spaces, CW complexes, or spectra. == Formal... ==

user123733

@YashasSamaga i see

Got the concept

Thanks @YashasSamaga

Yashas Samaga

for the options where the torque is zero, imagine displacing the loop

user123733

4:50 AM

Yes

Yashas Samaga

*rotating or causing an angular displacement

the one which tries to return to it's original position is the one in stable equilibrium