The h Bar

12:08 AM

I need someone to check something for me

If you take the Sun, which has a polytropic index of 3, and you know its mass, radius and density, what's the pressure as given by $P = K \rho^{(n+1)/n}$

user218912

you need to know $K$

Sir Cumference

Yes

I'm asking someone here to calculate it for me, so I know whether I got it right or not

user218912

what's the formula for $K$ again?

Sir Cumference

www2.astro.psu.edu/users/rbc/a534/lec15.pdf

astro.princeton.edu/~gk/A403/polytrop.pdf

DanielSank

@ACuriousMind ...and yet Fourier transforming with regulators is done all the time, notably in the real analysis proofs that the Fourier integral converges.

heather

12:21 AM

@DanielSank, looking at this question, what does it mean to "dress" a qubit? I looked around on the google but couldn't find anything.

user218912

@SirCumference I think you have to find $K$ through experiment.

DanielSank

@heather That's a stupid bit of jargon. When people talk about "dressed" things (usually in quantum mechanics), they're talking about coupling that thing to something else.

So, if I connect a qubit to a resonator, I can say I "dressed" the qubit.

Now, usually you'll see the phrase "dressed states". That has a very specific meaning.

You see, a quantum object has certain states which have the nice property that they sort of don't change as time goes on. Those states also have definite values of energy.

0celo7

@ACuriousMind Norm topology?

DanielSank

They are called "energy eigenstates" (so much jargon, I know).

When you couple the qubit to something else, the energy eigenstates change. The new ones are called "dressed states".

I'm so sorry that physicists make up so much dumb jargon. I really am.

user218912

jargon is important.

heather

12:24 AM

@DanielSank, let me guess: eigenstates because an eigenvector doesn't move off of it's span, and they don't move off their energy "span"?

DanielSank

@heather o_O

You know linear algebra?

heather

And I mean, the jargon is pretty cool: it means I can sound like a complete nerd in front of my friends. =)

And about linear algebra:

0celo7

@DanielSank she knows more than me

heather

Sort of, I taught myself some of it recently so I knew what in the universe all these papers and videos were talking about. Don't ask me to explain vector spaces, but I can multiply matrices (barely).

@0celo7, I sincerely doubt that's true.

user218912

hmm 0celo7 can explain vector spaces but doesn't know how to multiply matrices apparently.

user218912

12:26 AM

so you're equal.

Sir Cumference

@bl00 No, it's calculated

heather

Sorry, but another question: what is the advantage/point to connecting a qubit to something else? (Coupling it)

ACuriousMind

@DanielSank Uh, but there we lift the "regulator" in the end. We don't make the Fourier transform converge by regulating it, we prove it converges by that method.

DanielSank

@heather You're a precocious little bugger, aren't you?

heather

@DanielSank, uh...googling precocious

oh. um.

"little bugger"? =)

first time I've heard that one

DanielSank

12:27 AM

@heather Remember our previous conversation about how a computer works?

0celo7

mfw heather is 7 feet tall

heather

@DanielSank, yes

DanielSank

The memory elements have to interact somehow in order for information processing to happen.

heather

@0celo7, and no, more like barely 5 ft

@DanielSank, right

0celo7

aww

you're probably taller than my sister

DanielSank

12:28 AM

Well ok, do the bits in a computer's memory interact directly?

heather

Hmm, no? I don't think

(not completely sure)

DanielSank

They do not. Remember, they both go into the input of the logic gate.

heather

Oh, yeah, right

DanielSank

Those gates are in the CPU.

heather

central processing unit

DanielSank

12:31 AM

@heather Yes.

Anyway, the point is, some times you want a third element to do the processing of information contained in two bits.

0celo7

Ok

user218912

@SirCumference where is the formula for it?

user218912

it's not in the pdf you linked

DanielSank

(In a normal computer, this is true all the time. Memory bits never directly interact).

0celo7

She definitely knows more about computers than I do

user218912

12:32 AM

I remember seeing a formula for it.

heather

@DanielSank, but sometimes in a quantum computer qubits interact...like when transmitting information through each other...? (that's the reason why the 1D laser system isn't efficient, right? Or am I mixing things up now?) And then entanglement, right?

0celo7

@ACuriousMind Did you even look at my final solution? It's ok if you didn't, just wondering.

I think my $\epsilon$-nbhd proof is really nice.

ACuriousMind

@0celo7 Didn't look at it, busy with other stuff

0celo7

@ACuriousMind If you ever want to see it, let me know. I'm tweaking it.

DanielSank

@heather Well, many quantum computer designs do use direct interaction between qubits.

You asked why anyone would want to connect a qubit to something else so I was just giving reasons you might want to do that.

Also, @heather, one of the hardest things in quantum computing is measuring the qubits' states. You definitely need to connect the qubits to something else in order to measure them!

(That's my expertise, by the way, state measurement)

heather

12:40 AM

I guess I never thought of a logic gate as connecting a qubit to something else. So then would putting a qubit through a logic gate be considered "dressing" a qubit? Or measuring a qubit would be dressing a qubit?

DanielSank

Well, remember, that logic gate has to be built of some physical thing!

So of course the bits are interacting with something else.

heather

Yeah =)

I've read about measuring qubits being the hardest part. It makes sense - how do you get the hidden information out, while only being able to read zero or one.

DanielSank

Now, in quantum computing, it's often the case that we don't need an actual physical element to do logic. We use the physical interaction between qubits to get logical operations.

This is not always the case, but it often is.

heather

Huh. I guess this goes back to using quantum mechanical effects to do computing, right?

Quantum computers are able to use those effects, so they don't always have to use gates that are separate from the qubits.

@DanielSank, and I guess the other thing I'm wondering is, why would an electromagnetic field make a difference in "the longevity of a usable signal-to-noise ratio of the qubit"? Would it have to do with sort of keeping the qubits in place, like with the laser system?

DanielSank

12:59 AM

@heather It's not a quantum thing. You can make classical computers where the bits interact directly too. We just don't usually do it that way for engineering reasons.

heather

Oh, I see.

DanielSank

@heather I lost you there. Where's that quote from? The question you linked?

heather

@DanielSank, yeah. Sorry, should've mentioned that.

The quote's from the question I linked earlier, here

DanielSank

@heather Ah.

Just ignore it.

heather

Oh, okay.

DanielSank

1:02 AM

The OP there is saying stuff that's not right.

They're confused.

...and a quick glance at the paper they're referring to suggests to me that the result in the paper is rather technical.

heather

Oh, okay. Thanks so much! Every time I talk to you you clear up so many things I misunderstood. =)

DanielSank

@heather :D

heather

=D

DanielSank

@heather When people say things like that, it's one of the happiest feelings.

When I was in college I gave a two-part lesson in our undergrad journal club on the Fourier transform. In the journal club, we had feedback forms to help each other get better at presenting. The feedback I got from the Fourier transform thing was like "Made everything seem so much simpler". I was so happy that I still have those little slips of paper from more than ten years ago.

heather

Wow!

You must have explained it really well!

0celo7

1:22 AM

@DanielSank I want an undergrad journal club :(

I want to explain balls to some one

DanielSank

@heather I wrote a 40 page document >.<

I was obsessed.

Keshav Srinivasan

2:12 AM

Can someone look at my question here? I've received six answers but none of them satisfactory:

3

Q: How do we prove that the angular velocity vector is equal to a limit involving the rotation vector?

The angular velocity vector of a rigid body is defined as $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$. But I'd like to show that that's equivalent to how most people intuitively think of angular velocity. Euler's theorem of rotation states that any rigid body motion with one point f...

user228700

Hi everyone :-)

user228700

Anybody knows geometry? I'm figuring out circlres and could use some help getting my head around "imaginary circles"...

0celo7

@Kaumudi I know some geometry sometimes

Keshav Srinivasan

2:40 AM

@Kaumudi Yeah, I know geometry. What's your question?

user116211

Welcome in the h bar @KeshavSrinivasan.

user116211

@KeshavSrinivasan The bounty is still active. So, you have to wait.

Keshav Srinivasan

2:59 AM

@MAFIA36790 Yeah, I'm just posting it in chat so more people see it.

user116211

sure.

Keshav Srinivasan

@MAFIA36790 So do you have any thoughts on it?

user116211

No.

Keshav Srinivasan

@MAFIA36790 Do you at least understand the question? I'm getting the impression that a lot of people are misunderstanding it.

user116211

My main suggestion would be to explicitly mention why the answers have failed to answer your question; what they lack in your question body.

Keshav Srinivasan

3:04 AM

@MAFIA36790 OK

rob

3:43 AM

@Kaumudi Do you mean the unit circle in the complex plane?

0celo7

@rob no, she means a circle in $\Bbb C^2$ or something

it's weird

user116211

Got the first typo in Hoffman.

rob

@0celo7 I don't remember that notation ... is that the set of functions with a well-behaved second derivative?

@0celo7 Or ordered pairs of complex numbers?

0celo7

@rob $\Bbb C^2$ is to $\Bbb C$ as $\Bbb R^2$ is to $\Bbb R$

@rob yes

user116211

-2

A: Confusion about P=VI and V=IR

thanks your answer are not nice https://socratic.org/questions/if-we-double-both-the-current-and-voltage-in-a-circuit-while-keeping-its-resista visit this website

user116211

3:50 AM

I want to flag it as spam.

user116211

nonsense.

user116211

@rob, What flag should I use? not an answer or spam?

user116211

Maybe shouldn't flag, just downvote.

user228700

@rob No, I mean a circle with an imaginary radius o.O

user218912

@MAFIA36790 lol that's clearly spam, should flag.

user116211

4:01 AM

I have removed the part your answer not nice

user228700

5:02 AM

@JohnRennie: Morning! :-)

John Rennie

Morning :-)

0celo7

should I go to sleep?

ok fine I will

bye

user116211

@0celo7 o/

user228700

Sir (u know who u are :-P), are u familiar w/ the term pseudo rotation?

user116211

Hmm, the Uncertainty Principle is way more general than the Fourier conjugacy.

user116211

5:09 AM

I think I will jot down something on this.

user116211

@yuggib o/

yuggib

\o

user228700

Anybody familiar with the term pseudorotation?

user228700

(With regard to Chemistry)

John Rennie

@Kaumudi Yes, sort of ...

user228700

5:19 AM

My textbook has written

user218912

HI

user228700

"The equatorial bond lengths and axial bond kengths are different. But NMR studies suggest that all five atoms in $PF_5$ are equivalent because of pseudo rotations"

user228700

@bl00: Hello :-)

user228700

What do they mean by this..?

user228700

For reference, $PF_5$:

user228700

5:23 AM

John Rennie

Pseudorotation is when atoms in the molecule swap places. In this case it means the axial fluorines can exchange with the equatorial ones.

If this happens fast enough all the fluorines experience an average environment that is a mixture of axial and equatorial.

user228700

When u say "environment", do u mean repulsions from bond pairs/lone pairs and all?

John Rennie

For NMR to see all the fluorine atoms as equal means the swapping must take place at a rate faster than the frequencies used in NMR.

@Kaumudi environment means everything. If two atoms are swapping places quickly enough you can't tell the difference between them.

user228700

OK...I think I understand.

user228700

May I ask u a question about bond angles and bond lengths and things like that..? (ie. Are u familiar with these topics..?)

John Rennie

5:30 AM

Yes of course.

user228700

Okay. Again, my textbook says:

user228700

John Rennie

OK

user228700

What do they mean by "The bridging I-Cl bonds are appreciably longer suggesting delocalized bonding rather than simple hydrogen bridges formed by coordinate bonds"

John Rennie

By coordinate bonds I'd guess they means the sort of bonds formed in complexes like transition metal complexes. Are you familiar with these?

user228700

5:36 AM

Yep! Those types of bonds in which a single atom contributes both electrons to a bond b/w the two atoms.

John Rennie

OK so I think a coordinate bond would mean you has a $Cl_2$ molecule in the middle, and it was forming a complex with two $ICl_2$ molecules.

That sort of bond would be rather weak and rather long.

user228700

@JohnRennie $Cl_2$ molecule..?

John Rennie

Yes. Chlorine exists in nature as $Cl_2$ molecules. These might form coordinate bonds with $ICl_2$ ligands.

user228700

OK, what is delocalized bonding? (Given in that same sentence)

John Rennie

I think that just means a standard bond where there are electrons in a molecular orbital that includes two (or more) atoms.

The point is whether $I_2Cl_6$ is a true molecule or just a complex.

user228700

5:44 AM

OK, I still don't get it :/

user228700

(On the whole)

user228700

I'll think about it some more...

user228700

And also, are u familiar with Bent's rule?

John Rennie

Well in steam the water molecules can form dimers $(H_2O)_2$, but these aren't real molecules, just two separate molecules held together by hydrogen bonds.

user228700

OK..?

John Rennie

5:47 AM

The question is whether $I_2Cl_6$ is a real molecule or just two $ICl_3$ molecules held together by something like a hydrogen bond.

user228700

Ah, OK..?

John Rennie

In the water dimer the hydrogen bonds are longer and weaker than standard chemical bonds, so it's obvious this is just a dimer.

So the question is are the $ICl$ bonds in $I_2Cl_6$ unusually long and weak.

user228700

But my textbook has said the bonds are longer!

user228700

Which means they shouldn't be delocalized bonds, no?

John Rennie

Yes, but the bonds aren't that long. The book is saying they are short enough that they could be real bonds.

i.e. they are short enough that it looks as if there is real overlap between the atoms involved.

Well, that's my interpretation anyway.

user228700

5:50 AM

But, but... appreciably longer?

John Rennie

Well, OK, how much longer would they have to be before we'd say they aren't chemical bonds? And I don't know.

I think the book is making the point that the answer isn't clear cut.

They're longer than you'd expect an ICl bond to be but shorter than a coordination bond.

user228700

I don't know either. Ur interpretation is remarkably clear...I'm just wondering why the word appreciably was used there so I can make a note of exactly what they mean when they use this word.

user228700

@JohnRennie OK...

John Rennie

Hmm, yes, appreciably isn't a precisely defined scientific term :-)

user228700

Anyway, OK, I'll ask what Orthocresol thinks about this particular word being used when he(/she) comes around.

user228700

5:54 AM

Are u familiar with Bent's rule?

John Rennie

Never heard of it, but I might know it by a different name. let me have a quick Google.

user228700

Oh! OK...

John Rennie

Hmm, OK. This all sounds suspiciously like waffle to me. Hybridisation doesn't really exist - it's just a useful simplification - and Bent's rule seems like it's pushing the idea of hybridisation a bit far.

user228700

:-P OK. Never mind then. Orthocresol seems to have an opinion regarding this so I'll ask him(/her) when h/she comes around...I've already asked, actually :-P

user228700

Thanks so much :-)

John Rennie

5:58 AM

Remember that $sp^3$ hybridisation is a way we can describe the orbitals ina tetrahedral symmetry.

So if the symmetry is a distorted tetrahedron the $sp^3$ idea won't apply. We'll end up with some distorted version of it.

If i understand Wikipedia correctly Bent's rule is giving us guidelines as to how the hybridisation is changed by the distorted symmetry.

user228700

Something like that...for when substitutions are made with more electronegative/electro positive atoms...

John Rennie

Yes. remember that this idea is based on constructing molecular orbitals from combinations of atomic orbitals.

So, take a bit of $s$ and a bit of $p$, add them together and you get something that looks like a molecular orbital.

But this is only an approximation. Molecular orbitals aren't combinations of atomic orbitals.

user228700

Yes, OK...

user228700

6:15 AM

Thank you :-)

user116211

@JohnRennie Holy hell!

user116211

That's insanely the greatest truth.

user116211

LCAO is just a method to approximate MO.

user116211

Most introductory books forget to mention this very truth.

user116211

We need more orbitals than AOs to accurately predict MOs.

user228700

6:55 AM

@MAFIA36790 Yes, most do, including NCERT :/

user228700

The hell is Holy Hell BTW? (:-P)

John Rennie

A hell that is full of holes?

user228700

Haha, this reminds of Fred's (or possibly George's) joke after his ear got cut off while flying to The Burrow from the Dursley's place at the beginning of the Deathly Hallows; he says "Ha. I feel saint-like 'cause now I'm holey"

David Z

7:32 AM

@DanielSank because it's about operators and Hilbert spaces that occur in the context of quantum theory specifically. And as I said, this is one of those cases where the line between advanced theoretical physics and math becomes blurry. I probably would not make the analogous argument for a question about, say, Newtonian mechanics.

user228700

7:44 AM

@JohnRennie: Quick question. Back bonding is definitely not the same as a dative $\pi$ bond, correct?

John Rennie

7:56 AM

@Kaumudi I'm not sure what back bonding is ...

Gosh I'd forgotten what a vague subject chemistry is. Assuming I understand Wikipedia correctly then no back bonding and dative bonding are not the same.

Snigger :-)

iflscience.com/space/…

user228700

8:56 AM

OK, thank you :-)

user228700

Don't you worry sir! I will keep on reminding u how irritating Chemistry can get and you are welcome! (:-P I'm just kidding)

Secret

9:12 AM

Backbonding is when electrons fill in the $\pi*$ antibonding orbital of a group known as a pi acceptor. This bonding is usually found in metal complexes as the metal d electrons fill in the vacent pi* antiorbitals of the ligands and is responsible for stablisation of the metal-ligand bonds of these complexes by spreading out the negative charge density from the metal centre of low oxidation state metal ions

Emilio Pisanty

9:30 AM

@heather @MAFIA36790 please have a second look here.

THE LONE WOLF.

9:51 AM

hey guys .its Vidyanshu Mishra here.I just want to all of you look at my question physics.stackexchange.com/questions/286964/… as i have edited it.I want this question to be reopened for the following reasons:

Physics Meta

0

Q: How are these two questions duplicates?

I recently saw that this question has been closed as a duplicate, and I couldn't disagree more, but for some reason it has passed review with a Leave Closed outcome. The currently-closed question asks, in essence, Could someone [...] explain why the reflections [between two opposite mirrors...

discussion closed-questions specific-question

THE LONE WOLF.

1.i have edited it properly and have differentiated it from other question

2.How many time we see a question on physics SE which get around 50 upvotes and 1.5k views in 2 hours.

3.The knowledge associated with this question will not spread ( as after seeing a question as duplicate,people ignore it).

John Rennie

@THELONEWOLF. The duplicate question specifically asks: Would it be improper to drop the units of measurement of 0m like this in an academic paper? and this seems exactly the same question you are asking.

David Z

@THELONEWOLF. your question was already reviewed for reopening, and the reviewers concluded that the edit did not differentiate it from the other question

THE LONE WOLF.

still working,do not want to let such a great question RIP.

the question of which my question is said of being duplicate ,deals completely with dimensions as it ask about dimensions of 0m and ask o consider it in meaning of 5m+3kg

user116211

10:08 AM

Done!

user116211

0

A: Heisenberg's Uncertainty Principle

Lagrange and Poisson brackets: A step back to the Classical realms. Consider two variables $q_i, p_i$ given as a function of two parameters $u,v\,.$ The the Lagrange bracket is given by $$[~u,\, v~] ~= ~ \sum_{i~=~1}^n \left(\frac{\partial q_i}{\partial u}\frac{\partial p_i}{\partial v}- \frac{\...

THE LONE WOLF.

@MAFIA36790 was it for me??

Secret

$$[~u,\, v~] ~= ~ \sum_{i~=~1}^n \left(\frac{\partial q_i}{\partial u}\frac{\partial p_i}{\partial v}- \frac{\partial q_i}{\partial v}\frac{\partial p_i}{\partial u}\right)\,.\tag{I}$$

u,v should be q,p? inside the [,]?

nvm, misread

user116211

@Secret no?

user116211

WTH!

user116211

10:12 AM

The answer which should be upvoted is downvoted.

Secret

@MAFIA36790 I misread the lagrange bracket's convention, since I am used to see stuff being operated on to appear inside the operator

user116211

And the answer which is presumably not giving the correct picture is having +5.

user116211

flippiefanus is correct.

user116211

Don't know why he was downvoted.

user116211

Uncertainty principle is due to commutativity of operators.

user116211

10:15 AM

And not just fourier conjugacy of position and momentum.

user116211

It's a general result.

user116211

I need some coffee after spending an hour on writing it.

David Z

@MAFIA36790 That's a perfectly fine answer

Or, well, I think it's not the clearest answer, but the approach is correct, at least. You can certainly justify HUP using waves.

user116211

@DavidZ yes.

user116211

But as I said, it is not giving the whole picture.

David Z

10:23 AM

Yeah, and that's okay.

Answers do not, in general, get upvoted for giving the whole picture, nor downvoted for failing to do so.

Now, I have to imagine the question is a duplicate of something, but that's another matter entirely.

user116211

@DavidZ Well, there are many posts on Uncertainty Principle here. You need to dig deep ;P

ACuriousMind

@MAFIA36790 Your answer could use a bit more motivation for what you're doing between the definitions. After the first screen width, the average reader will be like "Huh, the question asks about the uncertainty principle, why is this answer telling me about Lagrange brackets?", and to be honest I don't see why you went into so much detail there (I've seen this odd "Lagrange bracket" like...once before, I think, and I don't get what it adds to the answer).

@DavidZ searching for it, but only coming up with related posts, no clear duplicate - the HUP is so confusing to many people that they tend to ask their questions in very different manners, even if the answers may be the same ;)

user116211

I thought it would act as a fine start to Poisson bracket which is more relevant here. But I agree with you.

user116211

Anyways, back to my studies after a warm shower T__T

David Z

Yeah... I wish there were a canonical question to address this

John Rennie

10:29 AM

0

Q: What should be considered Smallest charge

Today my friend asked me that what is the smallest magnitude of charge a body can have. I thought it in two different ways: 1.since electron is a fundamental particle and it is said to have a charge of magnitude 1.6 x 10^-19 which is greater than 0. 2.If we consider a neutral body such as a cup o...

charge

@THELONEWOLF. presumably this is related to your previous question, since the answer comes down to whether 0C (C for Coulombs) is a dimensionful quantity.

user228700

Here is a sort of mathematical poem that I'm unable to make sense of:

ACuriousMind

1

A: Heisenberg's Uncertainty Principle

Yes, it is due to the wave nature as mentioned "that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects" on https://en.wikipedia.org/wiki/Uncertainty_principle. It makes sense that for us, the observers, uncertainty is natural at quantum levels. However, s...

Is that an answer?

user228700

ACuriousMind

It just quotes a half-sentence from the Wiki article and then goes off the rails to say the author doesn't believe in the uncertainty principle

DHMO

Descartes' theorem

In geometry, Descartes' theorem states that for every four kissing, or mutually tangent, circles, the radii of the circles satisfy a certain quadratic equation. By solving this equation, one can construct a fourth circle tangent to three given, mutually tangent circles. The theorem is named after René Descartes, who stated it in 1643. == History == Geometrical problems involving tangent circles have been pondered for millennia. In ancient Greece of the third century BC, Apollonius of Perga devoted an entire book to the topic. René Descartes discussed the problem briefly in 1643, in a letter to...

user228700

10:44 AM

Ik, Ik, I checked that.

DHMO

"kissing" refers to, you know, mutually tangency

"bend" revers to curvature

user228700

Curvature?

DHMO

yes, which is 1/radius

so zero curvature = straight line

user228700

OK. I wasn't able to understand what "bend"=curvature is. Thanks!

DHMO

"The bend is just the inverse of/The distance from the centre."

THE LONE WOLF.

10:50 AM

john,exactly,when i told my friend about success of the question "should zero be followed by units and told him the answer,he asked me that question and specificaly he was asking that is 0c possible.I got u.

David Z

@ACuriousMind It counts as an answer for purposes of deciding whether to delete it, or (to a non-mod) for purposes of deciding whether to flag it as NAA.

i.e. I'd say the proper response is to downvote, but it's not deserving of moderator action.

Physics Meta

0

Q: Is there a safe way to contact other users privately?

A short while ago I've received a request from another user to connect via facebook. Well, apart from the fact that I'm not very active on facebook, I'd definitely not like to provide any contact details on this site. That said, I was wondering whether there is some mechanism on this site whereby...

feature-request

ACuriousMind

It will need 8 more downvotes to nullify the rep that the user already got from it. Imagine the second paragraph wasn't there (it doesn't actually address the question). Is "Yes, it's due to the wave nature, see Wikipedia" really more than a link-only answer, which we would delete?

David Z

Yes, that is more than a link-only answer.

ACuriousMind

I am beginning to see why most of my declined flags always were the NAA/VLQ flags ;)

David Z

11:00 AM

Indeed ;-)

If this answer were literally just "Yes, it's due to the wave nature, see Wikipedia", it's pretty much the least helpful answer you could write (which makes it a prime candidate for downvoting), but it is still an answer.

ACuriousMind

Sigh, I guess you're technically right

David Z

That's one of those things where, if 20k users were to vote to delete it, I wouldn't complain, but it's not really a case for using mod powers.

← 12 messages moved from Physics moderation

DHMO

11:23 AM

@ACuriousMind @rob Congratulations.

ACuriousMind

@DHMO A bit late, but thanks anyway ;)

Slereah

Hello

user116211

Hallo!

user116211

What happened to your Devil's Poison?

Slereah

Still fermenting

I'll take it out at the end of the month

See how it got

probably not well

I'll do a second batch soon, for better results

user116211

11:33 AM

Good! Just need to open a brewery now.

user116211

And you will be millionaire.

Slereah

I'm not sure selling it would be very profitable

user116211

Then you can buy an island and do GR peacefully.

Slereah

I buy the apples at the supermarket, that's already a pretty hefty price

user116211

;/

ACuriousMind

11:34 AM

Clearly, you also need to become an apple farmer

Slereah

I have no orchard!

heather

Hmm...yeah, that might be a problem

user116211

11:58 AM

So, that kpv's answer got +1 and all other including mine got downvote. hmm.

heather

@MAFIA36790, what question?

(was your answer on)

user116211

4

Q: Heisenberg's Uncertainty Principle

I have heard various definitions of the uncertainty principle. Yet I cannot quite comprehend how it is true. Nevertheless, something tells me, it is a consequence of the wave nature of light/electron which gives the intrinsic nature of uncertainty even if we don't measure it. Is it true that th...

quantum-mechanics uncertainty-principle wave-particle-duality

heather

Wow, nice answer! Who would've downvoted that?!

user116211

Hmm.

user116211

Might be my post is too long.

user116211

12:00 PM

hi @balarka.

Balarka Sen

hi

user116211

@heather But the thing that kpv's answer got an upvote stings me more than the downvote.

heather

@MAFIA36790, why? (And, as an aside, yours isn't much longer than the highest upvoted answer.)

user116211

Because apart from the first statement, everything is his own speculations.

heather

Interestingly, kpv's is also the shortest and least in depth answer.

Probably one person just came through and did it, though.

user116211

12:06 PM

Maybe ;/

user116211

12:42 PM

Okay, someone is downvoting me spontaneously :(

user116211

WTH is this downvoted:

user116211

-1

A: What is induced emf?

It is noteworthy to mention an excerpt from Purcell: The electromotive force was earlier defined as the work per unit charge involved in moving a charge around a circuit containing a voltaic cell. We now broaden the definition of emf to include any influence that causes charge to circulate a...

user116211

;/

user116211

WoW!

user116211

And now this:

user116211

12:44 PM

3

A: Are electric fields produced by static electric charges different from those produced by time-varying magnetic fields?

$$\begin{array}{|c|c|} \hline\textrm{True in Statics} &\textrm{True in General}\\ \hline \mathbf F ~= \dfrac1{4\pi\varepsilon_o}~ \dfrac{q_1q_2}{r^2}~\mathbf{\hat r} & \mathbf F= q(\mathbf E+ \mathbf v\times \mathbf B)\\ \hline \nabla \cdot \mathbf E = \dfrac{\rho}{\varepsilon_0} & \nabla \cdot...

0celo7

@Slereah Are you man enough to drink that stuff?

You might get like

user116211

@0celo7 He will.

0celo7

some disease

user116211

Also, what proof?

0celo7

@MAFIA36790 because he's an addict

@MAFIA36790 Hmm?