3:56 AM
@DevanshMittal OK, speak to you tomorrow :-)

3 hours later…
7:07 AM
I would appreciate it if someone could explain-to-OP/mediate/step-in/vote-to-reopen/vote-to-close here.

7:31 AM

Well, the new Closed Post notices do encourage OPs to post a new question. There have been some complaints about that...
I'm concerned that the Delete Question button will encourage OPs to delete and repost the question, rather than repair the question so that it becomes eligible for reopening. Also, many people, especially new members, believe that deleting poorly received questions is good for their reputation, and do not realise that such deleted questions can contribute to a question ban. — PM 2Ring Jun 17 at 9:13
56

Some user posted a question on IPS twice the past 2 days, and then was rude about the closure of both and the request to edit instead of reposting a slightly recomposed version with the same unaddressed problems as their first one. (from here on, this is just called 'reposting', though I don't me...

Also see meta.stackexchange.com/a/337015/334566 which is linked on Tinkeringbell's question, and note that it's :(
So it seems that the new network policy is to encourage OPs to post an improved question, and to not bother repairing their closed question because it's unlikely to be reopened.

2 hours later…
9:45 AM
0

1º SORRY FOR my ignorance about physics and about the laws of this land (physics.stackexchange) 2º MorePleasure? -> i will be improving on the question... as i keep researching on topic. How do formulate a question about the relationship between the wave collapse and special relativity, as regar...

1 hour later…
10:58 AM
0

I recently discovered PO. I read some posts here on meta but none of them answered my questions. I understand that it is not a SE site, but isn't it the same for MathOverflow? Why can I find MO on the list of sites of SE, but not PO? They do seem to have an "import from PSE" sections. Is it enco...

11:39 AM
Anyone able to help with my question?
1

I am confused with relating commutators to operator product expansions. I am following Kac's Vertex Algebras for Beginners book and trying to relate the notions to usual CFT notions. Take $T(z) = \sum_{n} L_n z^{-n-2}$ and let us define also a comformal primary $\phi(z) = \sum_n \phi_n z^{-n-2}$ ...

It is frustrating me to no end, and I'm sure I'm just making a trivial error/assumption.

3 hours later…
2:22 PM
To what extent is the covariant derivative "just another tensor"? In other words if I take the covariant derivative of some vector, is it strictly correct to write $$\nabla\otimes\vec V$$ or is the tensor product there incorrect because the covariant derivative is also an operator in some sense and has to be treated differently? Or is it just some tensor $\nabla\in V^*$?
I guess by analogy I would argue that a differential operator: $$\hat D_\lambda = \frac{\text d}{\text d\lambda}$$ can't be treated like a real number because the operation of multiplication and addition aren't defined for it, it has to act on something, which is the source of my concern that the covariant derivative, similarly, has to be treated as special in some way

2:45 PM
@JohnRennie: Are you there, Sir? If you are free then I would love to learn from you.

@DevanshMittal hi :-)

@JohnRennie: Thank God and Thanks to you, you are there.
I am thinking to break my whole question into smaller chunks and then ask you. Would that be fine, Sir?

1. Is it true that electric field inside an ideal wire "HAS TO BE" zero?

Let's be precise here:

2:48 PM
Yes Sir, kindly explain.

- if the current through a an ideal wire is constant, i.e. not changing with time, then the field inside the ideal wire has to be zero.
- if the current is changing with time, i.e. if electrons are accelerating in the wire, then the field will not be zero even in an ideal wire.
There is a simple way to see this.
The potential across a conductor, ideal or non-ideal, is defined as the work done per unit charge moved from one end of the wire to the other. OK so far?

Ohk Sir. So, it means the statement above is conditional that the current should be constant.

But, David J Griffiths says that electric field in an ideal wire will be zero as "E = (Rho) J" and Rho for an ideal wire is zero.

He says that unconditionally, and this is why I have this doubt.
Though, I agree that what you say is more convincing that Griffiths.
*than Griffiths.

In the vast majority of cases in circuits we are dealing with situations where the current is constant, or it is changing slowly enough to be treated as approximately constant.

Yes Sir. It is not a fundament Truth, but a general case.
OK Sir. The following is my second question.

2:55 PM
2. We know that inside the battery there conservative and non-conservative electric fields are in balance, when it is not connected to any circuit. But, when we connect it to a circuit it starts driving the current in the circuit, so when the electric fields were in balance already, so what makes the battery to drive the current in the circuit?

Inside a battery a chemical reaction creates free electrons at the anode and consumes free electrons at the cathode. Electrons do not actually flow through a battery - they are created at one end and consumed at the other.

Yes Sir. The chemical force itself I called as Non-Conservative electric field.
And at a certain condition/time this transfer of electrons stops as the opposite conservative electric field resists that.
Is it right, Sir?

When a battery is disconnected the reaction creates free electrons at the anode, but if the battery is disconnected those electrons have nowhere to go so they build up at the anode. Likewise electrons are consumed at the cathode and cannot be replaced when the battery is disconneted.
So a disconnected battery has an excess of negative charge at the anode and an excess of positive charge at the cathode.

Yes Sir.

@DevanshMittal Yes. The charge separation in a disconnected battery creates an electric field that stops the reaction.

3:00 PM
And that charge accumulation generates a conservative electric field in the opposite direction of its action and that resists further accumulation. Is it right?
OK Sir.

@DevanshMittal Yes.

We know that inside the battery there conservative and non-conservative electric fields are in balance, when it is not connected to any circuit. But, when we connect it to a circuit it starts driving the current in the circuit, so when the electric fields were in balance already, so what makes the battery to drive the current in the circuit?

When the battery is disconnected the electrons that the reaction creates pile up at the anode and stop the reaction. When you connect the battery those electrons can now flow off the battery and round the circuit then back into the battery at the cathode. As soon as the electrons flow off the battery the reaction inside the battery reestarts and creates more free electrons.
So the battery drives the current round the circuit because it continually creates an excess of electrons at it anode, and that pushes electrons out of the anode and round the circuit.

Yes Sir. But what motivates the electrons to flow off into the circuit from the cathode, when they were in equilibrium already?

I think I said before that the conduction electrons are like a fluid that flows through the wires. The battery increases the density of the electrons at the anode i.e. there are more electrons per unit volume at the anode than at the cathode, and this increases the pressure.
So in effect the electrons are flowing froma high pressure region at the anode to a low pressure region at the cathode. Just like any fluid will flow from high to low pressure.

3:10 PM
Sir, I still have doubt there. When the fluid was already stationary in a pipe at a place, now even if we extend the pipe length, the fluid should not flow.
When the electrons were in equilibrium in the battery, then even if we connect the battery to wires, the current should not follow.
*flow

I need to go now I'm afraid. My lunch is ready. I will be back in an hour or so, or we chat again tomorrow.

Sure Sir. Thanks a lot.

@Charlie It's not a tensor

What kind of object is it? Is it just an operator then?

@Charlie It's a connection

3:15 PM
so that tensor product equation wrote above is undefined

Yeah, that's not how you write it. Just write $\nabla V$

Is it's exact nature something that will be a bit mysterious until I look at more advanced treatments of GR that go into detail about bundles and stuff?

I mean, you could define the derivative to be written as $\nabla \otimes V$, since we also don't have a problem writing $\nabla \times V$ for the curl in ordinary vector calculus, but neither of those makes $\nabla$ into a vector or tensor

ah yeah I can see that
I don't mind not fully understanding it if it's something more advanced that I'll eventually get to, but knowing that it's definitely not just a tensor is helpful for now

@Charlie Yeah, I would say that understanding what "connection" or "parallel transport" really means requires at least a bit of the bundle machinery to fully appreciate

3:20 PM
just out of interest will more advanced GR texts generally go into detail about that? Or would that be something I'd need to find in a diff geo textbook
I'm just using Schutz atm since it's what was recommended for my course I just took

@Charlie Schutz is an easy read, but he is very light on rigour.

lol

AI AI AI AI AI AI AI
It's wednesday my dudes, so I'm back

@JohnRennie Yeah I get that feeling, rigour in GR seems particularly daunting

3:22 PM
@JingleBells And yet you are still kindly asked to not spam the chat with gibberish.

I will get back to it for now, ty

@Charlie The more mathematically inclined probably will, but "advanced" does not always equal "more mathematical" in physics :P

To a chemistry student like me those phrases are basically synonyms :|

@ACuriousMind but... it's wednesday :(
it's a meme

Ah, if it's a meme I take everything back. Clearly you're allowed to do anything if you say it's a meme!

3:26 PM
@ACuriousMind yay thanks

@JingleBells most of us here are not teenagers. Memes do not dominate our lives as they dominate yours.

Anything is a lot
Chill guys, I'm joking. I'm sorry lol
If to be a grown up means to be serious and grumpy I wanna be peter pan

There is a space on the humor spectrum between "grumpy" and "infantile" :P

You go right ahead. See how far it gets you.

:(
Well, I hope at least I made the "infantiles" here laugh
@JohnRennie What does that mean? I thought to behave like a child all the time makes people take you seriously
I'll be releasing my product soon... on eBay... Amazon wants stupid UPCs which cost a lot of money
I don't even know if the product will sell, so I'll try validating first, selling a few and then moving to amazon

3:33 PM
I was watching the start of a lecture series by Frederic Schuller on QM where he builds from a mathematical perspective, he basically starts by saying that the usual "quantum states are rays in Hilbert space" is not strictly correct, is this just mathematical pedantry? I can link the video
fwiw Frederic Schuller is an absolutely baller lecturer

@Charlie As so often, it depends on what you mean by "state" :P

The first axiom of QM being that quantum states are in fact "positive, trace-class, linear maps" $$\rho: \mathcal H\rightarrow \mathcal H$$

Ah, see, that includes stuff like density matrices, i.e. "mixed states". The "ray in Hilbert space" is a "pure state"
People often mean "pure state" when they say "quantum state", but not always

ah so it's not wrong, it's just a special case
in a way
Shankar did a small section on the density matrix, when we say pure state we're talking about an ensemble of identically prepared systems all in the same state right

That's one way of looking at it
We're getting dangerously close to interpretation again ;P

3:38 PM
oh no D:
In that case I'll leave it, I was just temporarily saddened that the nice new QM formalism I'd learned was actually wrong

Grr!
I'm trying to work through the review queues here!

grats on breaking 300k rep on physics SE btw John

@Charlie Thanks :-)
Though I kind of limped across the line.

4:07 PM
Could someone take a look at this answer and tell me if it makes sense? It feels very off to me, but I really can't be sure.
Apparently not...

@Philip If you want the snazzy one-box you have to post the link without any other text in the message

@ACuriousMind Aha!
-1

The question asks for a constructive procedure: any vector $\mathbf{v}$ can be written in terms of two scalar functions multiplied by the two allowed vectors $\mathbf{x}$, and the gradient, respectively. These scalar functions depend on $\mathbf{x\cdot\nabla}$, $\mathbf{x}^2$ and $\nabla^2$. For...

:cheers:

@Philip I find both the question and the answer vague enough that I don't want to call the answer wrong :P
Personally I wouldn't have answered the question unless it defines what "any vector $v$ constructed from $x$ and $\nabla$" actually is supposed to mean mathematically
Because this way I don't know whether the claims in the answer are correct since it's not clear what entities we're talking about to begin with

@ACuriousMind Hehe. I just find it strange that the user seems to claim that any vector operator can be written as a $$f(x^2 , \nabla^2, x \cdot \nabla) \mathbf{x} + g(x^2 , \nabla^2, x \cdot \nabla) \mathbf{\nabla}$$
When I asked how $\mathbf{L}$ could be written like this, I didn't quite understand the response. And seeing as I can't think of any other vector operators, I can't quite prove him wrong. But it seems like the fact that I can't think of any others is more a classical hangover than anything else...

The claim is certainly wrong if you consider the spin angular momentum operators. But that's the thing - the "any vector" in both question and answer is vague enough that these may or may not be included here

4:17 PM
I tried to answer it the way I understood, sort of pointing out what you did, that I can't quite define what a vector operator is unless I begin somewhere... Good to know it's not just me. Yeah, I spoke of spin too. But if we restrict ourselves to only external degrees of freedom, can you think of any other vector operators?
I doubt it, but it's been really nagging at me for a while.

Honestly, this seems like a typical case where Weinberg has expressed a very straightforward thing very inscrutably :P

HA! Yes!
:P

@Philip E.g. the electric and magnetic fields in quantum optics, or a current operator in many-body physics are all vector operators

Ah good, thanks!

@BioPhysicist it only occurred to me yesterday who you are with a new name :P

5:08 PM
@Charlie Yeah, I'm sneaky like that :p

which I assume you don't want to be referred to by anymore

6:00 PM
This might be interesting for anyone working in solid-state or condensed matter / materials physics. It's a Hot Network Question with answers that have shown some really unexpected ways around the problem of not having enough compute power to go beyond a 2x2x2 cell:
9

I could not find a satisfactory answer to this question. Is 2x2x2 enough, or is higher needed? I know the convergence test must be carried out, but increasing supercell size enormously increases computational time, and I am using the Python program Phonopy.

6:48 PM
@JohnRennie Oh yeah, I see that pretty regularly but mine says "You've been kicked for 58 seconds" I don't know why..

4 hours later…
10:28 PM
Is it possible to go ahead of time and do the theoretical physics math for if CPT-symmetry is violated? Or would that math depend on the experiment we used to hypothetically violate it?

@0xFFF1 what's "the theoretical physics math"?
If CPT symmetry is violated, then the theory that describes reality does not fulfill one of the assumptions of the CPT theorem, in particular it cannot be a relativistic QFT like the standard model. So in order to do any math you need to propose an alternative theory that agrees with all our experimental results but is not the Standard Model, and has some way to evade the CPT theorem.
And a single experiment with CPT violation won't really tell you whether that theory is correct either, it just tells you the Standard Model would be wrong.

10:58 PM
I saw a user ask a question on the main site that I almost answered but wasn't totally sure about so I didn't. I would argue that someone sat stationary on the surface of the Earth is not an inertial observer (even if we ignore the spinning of the Earth), because if I drop something in mid air it falls downwards, so even locally spacetime doesn't look flat to me
I think what I'm describing is basically demonstrated by the Pound-Rebka experiment