The h Bar

user228700

5:02 AM

Sort of. There are days when it doesn't happen, but most days, I experience excessive dreaming during the night and wake up exhausted. Then, without warning, I'll be reading my book or writing something or anything and bam, I'll literally start dreaming and fall asleep.

John Rennie

That sounds like you're working too hard and getting tired, rather than narcolepsy!

user228700

:/ Either way, it's not helping. I was actually gonna ask u about it; do u get good sleep? (Since u wake up like me-I've been waking up close to 5 for awhile now. In the middle I made some changes, but yeah)

John Rennie

I wake up around 5, but I go to bed about 9 so I still get my 8 hours sleep.

user228700

Some symptoms do match tho...I start dreaming immediately after I fall asleep, even if it's only for a few seconds (when it happens during the day ie.)

user228700

@JohnRennie OK...

John Rennie

5:07 AM

@Kaumudi that's entirely normal for dozing, which is what you're describing. When sleep is light or disturbed you always get very vivid dreams.

user228700

Hm...

user228700

Sigh. It's been this way for the past 2 months now.

user228700

I dunno what to do about it. I'm so frustrated I could cry aaah!

user228700

Anyhoo.

John Rennie

Sounds like stress to me ...

user228700

5:08 AM

Perhaps...yeah, anyway, do read what I wrote about PE when u have the time, pls...

user228700

@S007: U don't attend school?

John Rennie

@Kaumudi In the equation $KE=h\nu -W- \Delta E$ the value of $\Delta E$ isn't fixed. It can be zero.

There will be some photoelectrons that hit a nucleus and bounce straight back with little or no energy loss.

user228700

OK, so what does $W$ give, then?

John Rennie

So it doesn't make sense to include the $Delta E$ in the work function. The work function is a constant defined by the energy difference between the Fermi energy and the vacuum.

Secret

This week, newscientist give us the SMASH model. To be tested in 10 years time

user228700

5:17 AM

The ionization energy of the electron in the conduction band that is bound tightest to the nucleus? That's what my textbook says...

user228700

@JohnRennie OK, got too technical for me right there. OK, thank you :-) I will accept this.

John Rennie

The conduction electrons in metals are delocalised and not bound to any particular nucleus.

They occupy energy bands. Have you come across the idea of the band structure of solids?

user228700

Yes...so no ionization energy?

user228700

@JohnRennie Yeah, I have, but it's not given in too much detail...

John Rennie

Well it's kind of like an ionisation energy.

In an atom you have oribital with two electrons per orbital, and the energy of the electrons is the energy needed to pull the electron out of the orbital and remove it to infinity.

In a solid the orbitals broaden out into bands with many electrons in each band.

The topmost (highest energy) band in metals is the conduction band.

user228700

5:21 AM

Hm, OK...

John Rennie

And the work funcion is the energy needed to pull an electron out of this band and remove it to infinity.

So it's actually a very similar idea to an atom.

However the band covers a range of energies, so the electrons in the band have a range of energies.

user228700

Ah, OK...

John Rennie

Within the band there will be an electron with the highest energy of all the electrons in the band, and this energy is called the Fermi energy.

The work function is the energy needed to pull this topmost electron out of the band and remove it to infinity. That's why the work function is related to the Fermi energy.

user228700

Oh, OK...

John Rennie

So I suppose it is kind of an ionisation energy ...

user228700

5:25 AM

Hm, yes...

user228700

OK, thank you :-)

John Rennie

That graph you showed of KE against number of electrons is inyeresting.

user228700

OK..?

John Rennie

It wil be telling us something about the scattering processes going on when the initial photoelectron is created.

user228700

Brb.

John Rennie

5:27 AM

But I'm not an expert on photoelectron emission, and to be honest it isn't that interesting.

Secret

6:08 AM

To teach a robot, you must show it and understand how it learns

Teach cars how to drive with GTA V

user116211

6:47 AM

26

Q: My advisor refuses to write me a recommendation for my PhD application unless I apply to his lab

I got my master’s degree in China and had worked for two years as a visiting student in the US. Although my master thesis’s supervisor promised to write me a recommendation letter, my visiting student advisor refused. He is the corresponding author of all my publications. What's more, research ex...

user116211

Yeh, some profs are jerks.

Secret

7:06 AM

If they said they saw everything seemed to be lined up with the axis, and that the axis itself lined up with the motion of our solar system, perhaps that might be reflecting we are measuring the CMB from a frame that is moving wrt it (i.e. the solar system)?

John Rennie

The axis of evil is one of those interesting areas that might just be leading up to a paradigm shift. If it's real then the cosmological standard model has a problem, and if it's not real then there's something we don't understand about the way we're doing measurements of the CMB.

Secret

Actually do we currently have enough data to show whether our solar system is moving relativistically wrt the CMB?

John Rennie

@Secret Yes, the observation of the CMB dipole proves that.

Did you really mean relativistically or did you mean relative to the CMB?

ffahim

@JohnRennie is it ever possible for an object in this earth to move spontaneously?

Secret

I need to double check the speed of the local group again, cause I am not sure if that is approching light speed

ok nvm 600 km/s, it is not very relativistic

John Rennie

7:14 AM

@Secret we certainly aren't moving relative to the CMB at anything like the speed of light.

ffahim

@JohnRennie @Secret pls guys let me know.. pls answer my question..

John Rennie

I forget the velocity but it's a tiny fraction of the speed of light. There may still be significant relativistic corrections of course.

@ffahim wouldn't that contradict conservation of momentum?

Not to mention Newton's first law.

ffahim

Cool then how can we imagine an object of constant velocity?? Since it gets that velocity from the rest state.. and it refers to acceleration..

@JohnRennie

user116211

@ffahim Why are you thinking about the initial condition?

user116211

We need force to accelerate. But that is irrelevant here as you are talking about uniform motion now.

user116211

7:17 AM

And not how the motion emanated.

Secret

Suppose the axis of evil is real and thus suggesting the universe is somehow anisotropic, how will these revised models of cosmology affect other related branches of physics such as astrophysics and particle physics. In particular what extra things will be included in GR models?

John Rennie

@ffahim You talk about a rest state as if it was something special, but all motion is relative. Every object is at rest in its own rest frame.

@Secret don't know.

user116211

@Secret, some halp in group theory?

Secret

@MAFIA36790 I am not very deep into it, but I'll try just ask

user116211

Well, I was reading about isomorphism....

ffahim

7:19 AM

Yeah I know all motions are relative .. Since earth moves but. Think about an object moving 5m/s .. in the uniform velocity ... how it's velocity won't increase ... ? @JohnRennie @MAFIA36790

user116211

user116211

@secret, read the example.

John Rennie

@ffahim what is the object moving at 5 m/s relative to?

i.e. what reference point are you choosing when measuring the velocity of the object?

ffahim

Think about any reference frame

Well I the reference point was in rest

user116211

The author adds the line:

user116211

7:21 AM

> Thus if the first row and column of the tables for $\Delta$ and $\nabla$ are headed by $a$ and the second row and column by $b$, then each entry in the table for $\nabla$ is not the same as the diagonally opposite entry of the table for $\Delta.$

user116211

the example is pretty clear.

user116211

We have ten isomorphic classes.

ffahim

I said u not to consider the earth's speed ..

John Rennie

@ffahim that doesn't mean anything. There is no absolute definition of at rest.

Do you mean at rest relative to the Earth?

user116211

But I didn't get that statement above; what did the author want to mean by that statement and how is it helpful in the discussion @secret?

John Rennie

7:22 AM

If so that's a perfectly good reference frame to use.

ffahim

I just said to forget about earth 's. Speed .. think about the earth is in rest @JohnRennie :-)

Secret

MAFIA: I hope you are not very urgent on it as I am still reading your pics

ffahim

Hmm I meant so @JohnRennie

user116211

@Secret no; I have understood the example; it was easy. But I couldn't comprehend the statement. I'm reading the other pages. But that statement still stings me. Yeh, take time.

ffahim

@JohnRennie .. r u there???

user116211

7:25 AM

And sorry for the low resolution @Secret.

John Rennie

@ffahim I don't understand what you are asking. If we have some object moving at 5 m/s relative to the Earth then that's fine, but what is special about that?

Swapnil Das

Hello people. I've a question. To what extent is rigor important in physics?

John Rennie

Lots of things move relative to the Earth ...

@SwapnilDas you'll need to explain what you mean by rigor. Do you mean should our theories be mathematically rigorous?

Swapnil Das

@JohnRennie Yes!

user116211

74

Q: The Role of Rigor

The purpose of this question is to ask about the role of mathematical rigor in physics. In order to formulate a question that can be answered, and not just discussed, I divided this large issue into five specific questions. 1) What are the most important and the oldest insights (notions, results...

mathematical-physics mathematics research-level

ffahim

7:31 AM

The reference frame A is at rest .. (neglecting earth's motion) now the objective was at 5m/s velocity reference to A .. as u said nothing can gain velocity from rest position unless force F was applied.. then simply that moving object must accelerate.. isn't that?@JohnRennie

user116211

and links therein.

Swapnil Das

@MAFIA36790 Thanks.

John Rennie

@ffahim Well, where did that moving object come from? For example Mars is moving relative to the Earth. If the object came from Mars then it be moving relative to the Earth.

Do you mean the object started at rest relative to the earth, and later we observe that it is moving relative to the Earth?

ffahim

No , I repeat again.. just forget about earth 's motion.. the reference frame was at rest .. and the object to be referred was also at rest.. now the object started moving at constant velocity relative to the reference frame. But to make the object getting that 5 m/s I definitely required F force.. that means there is an acceleration.. but it keeps moving at 5m/s .. how can it be possible? @JohnRennie

John Rennie

@ffahim Well I am currently at rest (sitting at my desk). In a few minutes I will walk into the kitchen to make a coffee and I will be in motion. Then when I walk back to my desk and sit down I will be at rest again. I changed my velocity by apply a force to the ground.

ffahim

7:41 AM

Now assume at my way.. u have a box and was at rest.. now u pushed it and after that it keeps moving at a Constant velocity of Xm/s .. (forget gravity,friction,earth 's rotation). Does that make sense? @JohnRennie

John Rennie

@ffahim Yes. While I was applying the force F to the box it accelerated at a = F/m. As soon as I stopped applying a force the box stopped accelerating, so it moved at a constant velocity in a straight line.

ffahim

You were not supposed to move along the box..

John Rennie

@ffahim but you said u have a box and was at rest.. now u pushed it

If I pushed the box then it moved.

ffahim

Yes I said that. U pushed it and not moving along the box.. just a push nothing else.. (U were told to neglect friction gravity and earth 's rotation)

Now it's all about Newton 's first law

John Rennie

Well if I applied a force F to the box for a time t the momentum of the box changed by Ft, so the velocity of the box is v = FT/m. Once I stop applying the force the box moves in accordance with Netwon's first law.

ffahim

7:50 AM

As soon as it will not have any touch of ur hands that doesn't mean that it would stop accelerating. Since no other additional force dismissing ur force F

user116211

@ffahim Who will provide the force then to make it accelerate now that the hands are not touching?

ffahim

Hmm no force so what?? The applied force won't disappear unless any opposite force acts.. I mentioned in my question to neglect any additional oppressing force like friction gravity

John Rennie

@ffahim yes, once I let go of the box no force is being applied to it, so it moves in accordance with Newton's first law.

user116211

@ffahim I'm not getting that.

John Rennie

You say The applied force won't disappear unless any opposite force acts but I don't understand why you say that.

user116211

7:53 AM

There won't be any force.

ffahim

Yes @MAFIA36790 @JohnRennie

I meant net force will neither be 0 nor decreasing

Since John's force will still work .. however @JohnRennie

Now answer my very first question

John Rennie

@ffahim I don't understand what you are asking.

user116211

The hands are not applying force any more. So, there will be zero force now. No change in momentum after the agent, in this context hands, are withdrawn.

ffahim

No @MAFIA36790. It's Newton 's first law since no other additional force will work on it.. there will be just John's force

user116211

Well, if there is some John's force (I'm not following the discussion), there will be change in momentum; but the force from the hands is zero and that force has nothing to do with the dynamics now.

user116211

8:00 AM

@Secret, any progress?

Secret

@MAFIA36790 Short answer, that statment basically means J is not an endomorphism, otherwise you should get identical tables as when you start

long answer: Still compling

user116211

okay!

Secret

because endomorphism is a isomorphism within the same structure

user116211

@Secret you mean automorphism?

user116211

yes, you meant that indeed.

Secret

8:03 AM

oops, yeah

maps are relatively hard to "visualise" thus I tend to muddle them up

user116211

$J\ne I_E$ so, why would we get the same table?

ffahim

Guys .. one last question I was having doubt ... is earth 's velocity 0m/s ?

user116211

with respect to whom? What reference frame?

ffahim

I meant average

Ummm just earth moves round right? That led me to this question@MAFIA36790

As earth moves round then relative to sun the Earth 's velocity should be 0m/s .. is it? @MAFIA36790

John Rennie

8:16 AM

-2

A: Time dilation in the dark

It follows from Einstein's 1905 (false) constant-speed-of-light postulate that a clock runs slower than a clock in a different reference frame only insofar as the measurements are done in the second clock's system. The presence or absence of light is irrelevant but still the effect is illusory in...

Sigh

user116211

Check who the poster is.

user116211

Just ignore his posts.

John Rennie

Yes, it's Valev again. Maybe he feels the urge to jump in now that Voldefield is less active.

user116211

Hmm. I respect Voldefield; he is different from all those other craps.

user116211

He is a victim of abusive Sith Lords here.

user116211

8:21 AM

Anyways, so what are your plans, @johnR for the Halloween?

John Rennie

Spending it at home. I've bought a big tub of sweets for the trick or treaters. Do children do trick or treat in India?

user116211

@JohnRennie Hmm, no it's not prevalent in India. Did they perform such custom in British Raj days? No idea.

John Rennie

I think trick or treat originated in the US, so I doubt children did it in the days of the Raj. Not in India and not in the UK either!

user116211

ahh!

John Rennie

Trick-or-treating

Trick-or-treating is a Halloween custom for children in many countries. Children in costumes travel from house to house asking for treats such as candy (or, in some cultures, money) with the phrase "Trick or treat". The "trick" is a (usually idle) threat to perform mischief on the homeowners or their property if no treat is given to them. It typically happens during the evening of October 31. Some homeowners signal that they are willing to hand out treats, for example by putting up Halloween decorations outside their door. Others might employ an honor system and simply leave treats on their porch...

user116211

8:25 AM

I remember Charlie and the Chocolate Factory.

John Rennie

> Almost all pre-1940 uses of the term "trick-or-treat" are from the western United States and Canada.[29] Trick-or-treating spread from the western United States eastward, stalled by sugar rationing that began in April 1942 during World War II and did not end until June 1947.

user116211

Willy Wonka's father used to burn all the chocolates he got in Halloween.

John Rennie

I don't remember going trick or treating as a child in the UK in the 1970s.

user116211

When did it originate in UK exactly?

user116211

I see it started in US and Canada.

user116211

8:28 AM

Do the Germans celebrate? I will ask ACM.

Secret

Actually I am kinda confused by the samples, how do two elements with a composition become just one element. If we say the triangles are either I or J, then after the evaluation there will be still two letters left, how are the rules on how the value of e.g. ab is assigned in the 4 tables?

user116211

@JohnRennie There is a nice Google Doodle in the UK Google (should be in US too); it's not in India :(

user116211

@Secret what thing is bothering you? I'm seeing them fine.

Secret

$a\nabla b$=ab (if $\nabla=I_E$) = aa (if $\nabla=J$)

so I still get two letters thus something else need to be done to map them into E

user116211

@Secret No one said $\nabla = I_E\,.$

Secret

8:34 AM

But near the beginning he said "Since the only two permutations are ... there are at most two such compositions"

user116211

Yes.

Secret

and the triangles are compositions

user116211

There is $\Delta, \nabla$ and there is $I_E, J\,.$

Mew

Sup dawgs

user116211

They are not same things.

user116211

8:36 AM

Since it is a mapping onto itself, so for the permutation there are only two mappings viz., $I_E$ and $J\,.$

user116211

When you are talking about $I_E,$ then $\Delta = \nabla$ comes in consideration.

user116211

When you are talking about $J$, then $\Delta \ne \nabla$ comes in consideration.

user116211

For the first one, there are four isomorphic classes.

user116211

for the second one, there are six isomorphic classes.

user116211

So, till now, it looks, to me at least, fine except that statement I pointed out earlier.

Mew

8:39 AM

Boring snoring

a triangle is a triangle

Secret

Ah in that case it makes sense now. Pointing out that statement is to illustrate how when J act on the $\nabla$ table, it swaps around the a and b in a diagonal fashion. Therefore, before $\nabla$ and $\triangle$ are being evaluated (e.g. listing out explicitly the members of the isomorphism class) one has to keep in mind the relationships of the two table as there is a relation being imposed by the map J

So for example:

For the bottom table, the entries marked $a^2$, you obtain an equation $J(a\nabla a)=a \triangle a$Therefore, you need to find an element that the triangles evaluated to in order to make it hold for all entries (and similarly for the inverse map, which is also J itself)

this will give you for choices, thus 4 members in the $\nabla = \triangle$ isomorphism class

user116211

reading...

Secret

Therefore that statement about diagonals is trying to highlight that given completely unspecified tables of the triangles, the two maps J and I does very different things to them, thus hinting the existence of two different classes, and the diagonal remark illustrates the specific way J did to the elements

user228700

@JohnRennie OK :-P

John Rennie

You're back! BRB was a slightly misleading claim :-)

Anonymous

8:50 AM

@Kaumudi I do

user228700

@JohnRennie I'm sorry :-P I needed to get some things done.

user228700

@S007 Then how come u're on here during the morning/early afternoon time? ._.

Anonymous

mobile

Anonymous

:-D

user228700

:-P I see.

Anonymous

8:52 AM

@Kaumudi you go to any coaching institute ?

Secret

9:08 AM

@MAFIA36790 I just cleaned up my muffled understanding: The reason that yellow statement is important is because it informs you how J correlate some entries in the cayley table

That is, the map J impose a constraint (especially for the automorphic case) to the entries of the table thus you only have 2 free parameters for the $\nabla = \triangle$ class, for example

Note how for the diagonals, the elements are always distinct

If the map is $I_{E}$ you will see the entries correlate in a different way

Sorry for taking so long to get it right, I am just so bad at tracking morphisms

user116211

9:56 AM

Hey Master @Loong!

Loong

@MAFIA36790 hello

user228700

10:07 AM

@S007 Nope.

user228700

U attend FIITJEE, yes?

Anonymous

@Kaumudi No, I only give tests at fiitjee

Anonymous

Actually I joined resonance test series...that is better than fiitjee...will give from january

user228700

Ah, so u don't attend any coaching institution?

Anonymous

I use Etoos Video Lectures...I used to take classroom coaching from Fiitjee but the teachers were not good

Anonymous

10:10 AM

Etoos is much better

Anonymous

and resonance test series too

user228700

Oh, OK :-) Yes, Etoos is good.

Anonymous

you give any test series ?

user228700

Yes, resonance.

Anonymous

oh....you will give the full tests from january toh ?

Anonymous

10:11 AM

I will

user228700

Yes, most probably.

Anonymous

good :)....btw how was your jee this year ? rank ?

user228700

Are u from CBSE?

Anonymous

icse

user228700

Ah, OK...

Anonymous

10:12 AM

jee rank ?

Anonymous

you are a dropper, right ?

user228700

I'd rather not divulge into my marks this year, but I must say that it is entirely different from writing tests at home/coaching center.

Anonymous

I saw the paper

Anonymous

It was too lengthy

user228700

Hm.

Anonymous

10:14 AM

especially physics

Anonymous

:-P

Anonymous

those paragraphs

Anonymous

@Kaumudi anyway all the best for your prep :)

user228700

Hm, yes. ICSE-now I understand why u're allowed ur phone :-P

Anonymous

@Kaumudi hehe, yeah :-D

user228700

10:15 AM

ATB :-)

heather

11:19 AM

@0celo7, sorry! I wasn't on by the time you commented for more analysis. ='(

Mew

do you want to be a physicist heather?

heather

@Mew, I don't know. Probably something over in that area of careers, yeah. Physicist, engineer, computer scientist, all three at once =D

@Mew, how about you/what do you do?

Mew

na

i work in finance

heather

I see

cool

Mew

i used to want to be a physicist

but then changed

still interested in physics though

obviously

heather

11:24 AM

right

Mew

mastering physics allows you to master nearly anything

user116211

Well, @Secret, let me take the first two tables for the two compositions respectively:

user116211

$$\begin{array}{|c|c|} \hline \Delta &a& b\\ a& \hline a&a\\ \hline b&b&b\\ \hline \end{array}$$

user116211

$$\begin{array}{|c|c|} \hline \nabla &a& b\\ a& \hline b&b\\ \hline b&a&a\\ \hline \end{array}$$

user116211

For $J$ :

user116211

11:31 AM

$$\begin{array}{|c|c|} \hline \nabla &J(a)&J( b)\\ J(a)& \hline a&a\\ \hline J(b)&b&b\\ \hline \end{array}$$

user116211

Did the author mean for the second table above, the element is not equal to the diagonally opposite element of the first table? I'm not seeing that; the first element of the second table is equal to the diagonally opposite element in the first ie. $b\,.$

user116211

However, if I observe the third table, the first element of the third table i.e., $a$ is not equal to the diagonally opposite element in the first table i.e., $b\,.$

user116211

So, I think the author is talking about the first and third table.

user116211

But I'm not noticing that in the statement; he explicitly wrote "...if the first row and column of the tables for $\nabla$ are headed by $a$ and the second row and column by $b$."

Swapnil Das

@S007 What's the cost of ETOOS grade 12?

heather

11:45 AM

@SwapnilDas, hello

Andras Deak

@0celo7 so how'd it go?:D

and does your phone require all this information, or are you using a buttload of apps?

also, hi

heather

@AndrasDeak, still have to buckle down and do the rewrite of the code =P

Andras Deak

@heather that's how these things work;)

have you pinned down the logistics yet?

ACuriousMind

@MAFIA36790 Costume parties are common, trick-or-treating isn't.

user116211

Ah! So you wore costumes? Captain Spock?

ACuriousMind

11:53 AM

It's noon of the 31th here right now, the parties are this evening

user116211

okay!

heather

@AndrasDeak, not quite yet, though I've got some general ideas.

0celo7

@AndrasDeak how'd what go?

heather

@0celo7, hello, and sorry about last night!

Secret

Suppose you apply $J$ on this table

\begin{array}{|c|c|} \hline \Delta &a& b\\ a& \hline a&a\\ \hline b&b&b\\ \hline \end{array}

you will get

\begin{array}{|c|c|} \hline \nabla &J(a)=b& J(b)=a\\ J(a)=b& \hline b&b\\ \hline J(b)=a&a&a\\ \hline \end{array}

However (unless I am missing something), it seems this is the only way to produce the distinct diagonal element relationship the author claimed

user116211

12:04 PM

@Secret Which table did you use for $\nabla\,?$

Secret

but if that's the case, then what the author claimed in terms of the two tables (with a,b as the row and column indicators) will not have that diagonal relation he claimed

\begin{array}{|c|c|} \hline \nabla &a& b\\ a& \hline a&a\\ \hline b&b&b\\ \hline \end{array}

This is the $\nabla$ table according to the isomorphism J

rob

@ACuriousMind I was surprised that you unilaterally closed this question:

0

Q: Is it possible to use natural units (particle physics units) with boost?

I would like to use boost to handle my units in my c++ code, and I was wondering if it was possible to use natural units? That is, can I add a mass and a momentum and set it equal to an energy without having to multiply by the necessary c's everywhere?

particle-physics units unit-conversion

user116211

@Secret But then $\nabla = \Delta\,.$

heather

have a good day everyone =)

the bus stop is calling my name =P

ACuriousMind

@rob Do you think it should have stayed open, or just that I shouldn't have closed it single-handedly?

user116211

12:08 PM

I'm talking about $\nabla\ne \Delta$ as I'm considering $J$ not $I_E\,.$

ACuriousMind

I thought that a) the policy that we don't advise how to use specific software is uncontroversial and b) the question very clearly asks how to use a specific C library, so I went ahead and closed it.

Mew

Gday Arnold

Secret

The $\Delta$ table we start with is part of the $\nabla=\Delta$ class. Let's try something different... (and yes I am using J there)

rob

@ACuriousMind I thought it was borderline, but could have been brought around to a physics context by a good software-agnostic answer

like the one I was writing :-)

user116211

Welcome in The h Bar @ArnoldNeumaier.

user116211

12:10 PM

@Secret You shouldn't be as $J$ is used when $\Delta \ne \nabla\,.$

ACuriousMind

@rob I see - if you then also edit the question to maybe make it less focused on boost, I guess that will be fine

rob

@ACuriousMind A reasonable response. Cheers.

Secret

The particular choice of elements for the $\Delta$ table above evalautes into the $\nabla\neq \Delta$ class. The author said that the difference between the two class is either J is an automorphism or J is not

Arnold Neumaier

@DavidZ @DavidZ Nobody at PhysicsOverflow has the bad intentions that you mentioned in your comment. We want to be a graduate+ level physics site continuing the no longer existent Theroretical Physics SE site, and this should be possible in a peaceful coexistence with Physics SE.

Secret

so both classes have to do with J, and not involving I

ACuriousMind

12:13 PM

@rob I don't really care, but would you prefer that I reopen it myself?

Mew

@ArnoldNeumaier, how do you feel about a site for physics problems that allows juniors to ask questions too?

Secret

o wait a sec... I made a carelss mistakes in the previous table

ffahim

@Mew it would be better for me...

Secret

nope, not helping, ...

Arnold Neumaier

@DavidZ We also don't impose any rules on spin-off sites using the PO software.

Secret

12:18 PM

Suppose you apply $J$ on this table

\begin{array}{|c|c|} \hline \Delta &a& b\\ a& \hline a&a\\ \hline b&a&b\\ \hline \end{array}

you will get

\begin{array}{|c|c|} \hline \nabla &J(a)=b& J(b)=a\\ J(a)=b& \hline a&a\\ \hline J(b)=a&a&b\\ \hline \end{array}

Rearranging:

\begin{array}{|c|c|} \hline \nabla &a& b\\ a& \hline b&a\\ \hline b&a&a\\ \hline \end{array}

Ok this one works and is in the $\Delta\neq \nabla$ class

this one does have the diagonal relationship the author claimed

user116211

@Secret yes, that was my example above.

user116211

@Secret yes.

John Rennie

@ArnoldNeumaier your contributions to the Physics Stack Exchange speak volumes for your good intentions, but I don't have the same confidence in Dilaton.

Secret

@MAFIA36790 no, look closely, there are 3 a s in here

user116211

I'm unsure of what the author really meant.

Arnold Neumaier

12:20 PM

If it fills a perceived gap, it would be fine. It would be wholly complementary to PhysicsOverflow at all, since there we are not catering for elementary physics, and we could transfer the occasional elementary questions to there.

user116211

@Secret ohh, yes.

user116211

@Secret I'm not seeing such composition in any of the four tables listed in the page.

Secret

because that one is in the $\Delta \neq \nabla$ class, which he had not show (at least for that page

user116211

Hmm; okay; these are the four compositions that are possible.

Secret

That one is an isomorphism, but not an automorphism (else diagonal entries have to oppose each other)

user116211

12:22 PM

For $I_E,$ they have four isomorphic classes each having a single element from the four tables.

user116211

And for $J,$ they have six isomorphic classes taking two different tables from the four as the two elements in each of the classes.

Arnold Neumaier

@JohnRennie @JohnRennie: Well, Dilaton was ostracized here and banned for some time, simply for positing links to PO. This is the only ''crime'' against Physics SE you can accuse him. He would surely be happy if that silent animosity between PO and Physics SE would come to an end.

John Rennie

@ArnoldNeumaier Dilaton made personal and derogatory comments about me on an unrelated public web site. That sort of thing doesn't inspire confidence in people.

Mew

Wow things are getting personal

Secret

How do he get 2 elements each for the 6 isomorphic class. Are

\begin{array}{|c|c|} \hline \Delta &a& b\\ a& \hline a&a\\ \hline b&a&b\\ \hline \end{array}

and

\begin{array}{|c|c|} \hline \Delta &a& b\\ a& \hline b&a\\ \hline b&a&a\\ \hline \end{array}
the same thing?

is bad at class counting

user116211

12:29 PM

@Secret Why should they be same? They are evidently different $\Delta$s.

Secret

But then, I end up with 8 tables for the supposingly remaining 6 classes (By permutating the position of the b, and 4x multiplicate caused by interchange the role of a and b in the table), not to mention a table with all a s and a table with all b s?

user116211

Hmm.

user116211

that's tiring.

user116211

See:

user116211

When you are using $J,$ you then have $\Delta \ne\nabla\,.$

user116211

12:31 PM

That means each class must have $\{\Delta, \nabla\}\,.$

Arnold Neumaier

@DavidZ PO isn't going to host any new site - it is lending software support to others who want to cater for subjects that generate less traffic than deemed necessary for SE. But they are otherwise completely independent sites.

user116211

Now, there are four tables: $1,2,3,4\,.$

user116211

Take $\Delta\equiv 1;$ then $\{1,2\}, \{1,3\},\{1,4\}\,.$

Andras Deak

@0celo7 check my reply:P The reddit password recovery...

user116211

Similarly, for $2,$ you have $\{2,3\},\{2,4\}\,.$

user116211

12:34 PM

Lastly for $3,$ you have $\{3,4\}\,.$

ACuriousMind

@AndrasDeak He has this habit where he accesses chat on the phone, can't click on arrows to see what they refer to and then demands everyone explains what messages they replied to :P

user116211

So, in total six isomorphic classes when $\Delta\ne \nabla\,.$

Arnold Neumaier

@JohnRennie Hmm; I didn't know. But you and David Z is making here derogatory comments about Dilaton that on the same token would prove that one shouldn't have confidence in you, too!??

Jim

@BernardMeurer Yeah, I don't really use my email to contact people from PSE. I'm willing to make an exception, but I'd have to know the basic subject of the conversation to decide that

user116211

Did you get it @Secret?

John Rennie

12:35 PM

@ArnoldNeumaier I don't recall making any comments about Dilaton, derogatory or otherwise. My reply to you is the first time I have mentioned him.

Andras Deak

@ACuriousMind oh thanks for the heads-up, let me ignore him in the future:P

(SO chat app sucks, I only use it with a desktop user agent)

Secret

I can see how you use permutation arguments and box counting and it makes sense. However it is still a bit abstract thus sometime later I might sit down and try to write them all out myself to have a better look at them

But otherwise, I think I have solved your query?

That is, I can see how there are 6 now, but I 'd like to have a closer look at them later

user116211

Hmm, I fear I still didn't get the statement.

user116211

1 hour ago, by MAFIA36790

So, I think the author is talking about the first and third table.

Secret

Well for the table with 3 a s, there's clearly that diagonal claim by the author in it, that is, getting table 3 from table 1 and then rearrange the column so they have the same ordering as table 1

In that table, a in $\Delta$ is opposed by b in $\nabla$

user116211

12:39 PM

I'll now not think about the statement and proceed forward.

Secret

ok then

0celo7

@AndrasDeak on mobile

Andras Deak

OK:D

Arnold Neumaier

@heather Hosting has to be done independently by any site using PO software. Thus PO is not involved at all, except giving initially advice on how to set up the software.

0celo7

@ACuriousMind Parties on a week day? Don't people have better things to do?

ACuriousMind

12:43 PM

@0celo7 The first of November is a holiday (Allerheiligen /All Saints' Day) here.

user116211

@Secret I mean I will read it later on and think more about it.

user116211

Thanks @secret.

Secret

ok

Jim

@ACuriousMind so people treat that as a legit holiday? Wow, it's pretty much forgotten around here

Arnold Neumaier

@JohnRennie Sorry then for mixing up you and David Z. Maybe it was partly because I remember you having called PO a failure here in the h-bar. it fails only if you think success is counted solely by the number of new questions and answers per day. That was why TP SE ''failed'' although those contributing had cthought of it as a success.

user116211

12:46 PM

@Jim They are Germans, after all. They must enjoy holidays, I suppose.

ACuriousMind

@Jim Yes, it's an official holiday. I'm not sure how many people actually celebrate it except for the older religious people, though.

Andras Deak

Isn't it the day of the dead there?

0celo7

@ACuriousMind I forgot about that. Ok, acceptable.

ACuriousMind

@AndrasDeak Yes, many people visit graves and it's a "silent holiday", meaning most forms of public music and dancing are forbidden (yes, that's a thing here)

0celo7

One forgets about such things in atheist America...wait

Andras Deak

12:50 PM

I imagine rebel ballet groups holding guerilla practices in their bunkers....

user116211

Do you visit cemeteries @0celo7?

user116211

Hop you are not atheist.

Andras Deak

atheists visit cemeteries too, you know

ACuriousMind

@AndrasDeak Protesters more-or-less regularly protest this arrangement by, well, dancing. I recall a clever group who all put on headphones and danced in front of a church - since there wasn't any music to be heard, they didn't technically violate the "silent" part

Andras Deak

@ACuriousMind wow. :D

user116211

12:52 PM

@AndrasDeak ohh.

0celo7

@MAFIA36790 the family one in Indiana, once

And Arlington, ofc

ACuriousMind

But I think most people don't particularly care much about this law one way or the other

Andras Deak

I see:)