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user351417
12:47 AM
@EmilioPisanty Sure. I've always flagged homework with NotAnAnswer. I'll raise mod flags next time.
 
user351417
(I'm assuming that it was a homework answer because I think I remember something of the sort on that question and there isn't anything particularly remarkable over there right now.)
 
2:04 AM
@Chair Oh, yeah, if the reason you're flagging is that the post is an answer to the underlying problem in a homework-like question, use a mod flag for that, like Emilio said. The NAA flag generally doesn't apply to those.
 
 
2 hours later…
3:57 AM
@Qmechanic man... do you ever sleep? physics.stackexchange.com/questions/430687/…
 
 
3 hours later…
user351417
6:35 AM
@DavidZ @EmilioPisanty I don't object or anything, but aren't these discussions more on-topic on the Physics Meta chat?
 
Do physists and engineers really understand mathematics and then apply it or do they just apply it?Iam not sure whether this is the right chat to ask it.If not please direct me.
 
@Chair No. The other room is named Physics Meta only because it's associated with our meta site and nobody ever bothered to change the default name. (Which we should probably do...) It's not meant to imply that meta discussion should be held in that room. The scope of both rooms is the same; the other one is just a backup for when this one is busy.
Though that could certainly change if someone makes a proposal on Physics Meta that we change the scopes and the community approves of that proposal.
 
@Mohan I guess it depends what you mean by understand mathematics
 
6:53 AM
Nobody understands mathematics
 
I tried to understand the operators like curl, divergence,etc..(but still now I couldnt understand them clearly).I think, these operators are widely used in physics and engineering.Without understanding them how can we apply them?I did undergrad in engineering.But my teachers didnt realy insist on understandin math, they just taught me to apply it.(by understanding I meant to learn how and why these operators give the results they do.)
 
7:33 AM
My doubt is whether this is the right way to learn or should I try and understand these things?
 
@Mohan what areas of physics are you studying. Basic stuff or the more advanced areas like QFT or general relativity?
 
Divergence isn't too hard to understand
It's the infinitesimal flux
Pick an infinitesimally small volume, the divergence is the flux of your vector field in that volume
 
Just the basic stuff.From your answers it seems, understanding things is the right way to learn.Ill try to understand it.Thanks for helping me guys.
 
 
4 hours later…
Anonymous
11:19 AM
@Mohan Well, if you're not understanding what you're doing then that's definitely not the right way. But, perhaps we should not forget that engineers and physicists have different priorities compared to mathematicians. If they spent all their time trying to understand the math better, they'd not have much time for anything else :P Anyway, as far as the basic differential operators are concerned, Khan Academy covers them really well in an intutive fashion.
 
1:04 PM
@Slereah they just get used to it?
 
1:32 PM
There's now this pesky sidebar on PSE. "Home", "Questions", etc. It takes up a lot of space of the page. Is there a way to hide it?
Why was this changed anyway? The bar was perfect when it was at the top of the page.
 
Sep 19 at 12:03, by John Rennie
@bolbteppa go into your profile, then the Edit Profile & Settings tab, then click Preferences on the left then tick the option Hide Left Navigation. That removes the wasted space on the left. You can still get the left menu by clicking on the hamburger icon.
 
@user2646 Thank you.
 
np
 
I have another question. How do I see the "new" questions.
I can't see that option anymore on the main page.
 
1:43 PM
@user2646 Thank you.
 
:-)
 
But why don't I see that option on the main page? physics.stackexchange.com
 
It is under "Home" on the left panel.
 
Aha, but I removed the left panel.
 
:-/
 
1:45 PM
Guess I'll have to stick with the panel
I don't like the new layout.
Thank you for the help
 
you're very welcome
 
2:30 PM
@Avantgarde Hide the left panel (you know it makes sense). Click on the hamburger at the top left:
This opens a menu. Click on Questions to see new questions.
 
3:02 PM
@JohnRennie Done! Thanks!
 
@Avantgarde I wasn't sure about the new layout at first, but now I've got used to it I really like it.
It's mildly annoying that it now takes two clicks not one to get the new questions page, but only mildly annoying.
 
Anonymous
3:16 PM
@JohnRennie I never really understood why the "Active" page is set as default (that's the page that is linked when you click on the site logo) rather than the "Questions" page, given that most of the active users care more about the new questions (which they can answer). Also, there are just 5 options on the hamburger menu which could easily have fitted on the top of the page too!
 
Anonymous
To avoid the annoying double-clicks though I saved the "Newest Questions" tab as a bookmark on the bookmarks toolbars :P
 
@Blue I agree. But then the design is really targetted to Stack Overflow, which is where the SE make all their money.
And the new questions page is next to useless on SO because there are so many new questions per second (or whatever unit time)
@Blue I've done the same :-)
 
Anonymous
@JohnRennie The active questions page would be equally useless :P
 
Anonymous
Anyway, whatever works for them I guess
 
Anonymous
I like that they're doing some work on "being able to easily find answerable questions" front
 
3:19 PM
I'm disinclined to make a fuss. After all the PSE has been a great benefit for me and ... it's free! :-)
 
Anonymous
Hehe, that's true of course :)
 
Anonymous
The SE sites are really unique
2
 
3:40 PM
@Blue "Most of the active users" != "Most of the users looking at the site"
We shouldn't forget that we active users are a tiny fraction of the total traffic
Also, I always looked at the active tab, even when it wasn't the default. But maybe I'm weird :P
 
@ACuriousMind I will at least glance at every question posted on the PSE, and for that I need the questions in time order.
 
Anonymous
@ACuriousMind Don't we already have a "Top Questions" page for external visitors? That is, the page you get when you visit https://physics.stackexchange.com/
 
Anonymous
I'm not sure what purpose "modified 2 mins", etc ago would serve for the visitors
 
@JohnRennie Ah, well. Especially since I've become a moderator, I'm at least equally interested in every other activity that goes on, too.
@Blue Yeah, good point. Maybe it's just the personal preference of the dev who coded it :P
 
Anonymous
Another thing I just noticed: Even the Top Questions page contains downvoted questions....that's weird
 
Anonymous
3:46 PM
Do we really want to display the (objectively) bad questions to the visitors?
 
Anonymous
4:02 PM
23
Q: A Jim for all seasons

JimNote: This contest is currently inactive. It may be resumed again in the future. So there has been a recent influx of new Jims (meaning maybe 2 in a while). Now, we all know that I am the one true Jim, lord of all other Jims, but nonetheless it gets confusing when someone pings Jim and Jim gets ...

 
Anonymous
LOL
 
Anonymous
I never saw this
 
Ah, good times
It was a jimteresting idea
 
"Don't ask about asking, just ask"
Alright.
Here we go.
The site is populated with multiple questions regarding the premise: a free electron cannot absorb a photon, i.e. we can only have Compton scattering and not complete absorption.
The same can be demonstrated via conservation laws, most notably in this answer, quite explicitly.
So, for free electrons, it is always Compton scattering, never a complete absorption event.
But, there are also several questions on the site, e.g. this one
... which concern Photoelectric effect instead.
In PE, the electrons are bound, not free.
 
3
A: How can the Alcubierre drive itself move faster than light?

enumarisYou're making a mistake that is very easy to make when transitioning from understanding special relativity to general relativity - which is, you aren't thinking of things purely locally. When we go to GR from SR we lose a very important part of physics - the existence of global inertial reference...

this safesphere dude just comes in and puts words in my mouth to criticize my answer...uncool...
 
4:12 PM
The only part I do not get is, if we think in terms of conservation laws, how is the aforementioned answer "cured", i.e. how is the inconsistency resolved, by the fact that the electrons aren't free, but bound?
Offhand, the fact that the electron isn't free but bound, would only indicate that we have something else to share the momentum transferred by the incident photon.
 
@TheDarkSide because the electron can exchange momentum with whatever it is bound to.
 
When the electron is bound, it means there's "other stuff" around it. The momentum that couldn't be conserved in a 2-body problem can be conserved by this "other stuff"
 
Yes. Exactly. But ...
 
I think you answered your own question with your second statement.
 
@JohnRennie Yeah it's annoying to find your way through to the new questions.
 
4:15 PM
Perhaps. But let's think of it in terms of Rob's answer, or a simpler version of the same answer where we consider ourselves in the rest frame of the electron originally.
 
In any case, I don't see much of a (good) difference with the new layout.
 
How does this other stuff enter the derivation, and mitigate the inconsistency, is something I haven't been able to resolve.
 
@TheDarkSide It's there, so it can participate in the scattering. Quantum scattering events are not collisions between billard balls.
 
e.g. consider a simple picture where the electron is bound to the nucleus.
@ACuriousMind True. (But returning to the derivation ...)
 
Electrons which are bound share an electro-magnetic interaction with the atoms it's bound to. This electro-magnetic interaction will mean that the atoms the electron is bound to can share in the interaction since if you move the electron around with the photon, you are also moving the atoms around.
@TheDarkSide are you asking for a rigorous mathematical calculation to show how momentum and energy are conserved in a 3-body collision?
 
4:19 PM
@TheDarkSide In the derivation, you just add a term $p_\text{stuff,i}$ to both sides of the momentum conversation equation.
 
@enumaris Yes. qualitatively I understand that, that's why I wrote that sentence.
 
And since there aren't any other constraints on it, it'll be able to absorb all "excess" momentum.
 
@ACuriousMind Yes. I am only trying to construct the argument. So, in energy conservation, I'll have an extra $M_N c^2$ term, and nuclear recoil KE can be neglected due to the large mass, right?
 
Why would you neglect it? The recoil is the very thing that makes the absorption possible!
 
To see that momentum can be conserved is easy - just do what @ACuriousMind said. But to actually calculate the E.O.M. for a given 3-body collision (given some impact parameter) seems quite hard to me. I'm not sure I've seen it done in the past. However, energy and momentum conservation will both go into these equations as constraint equations. In the 2-body case, it was just to show that such constraint equations cannot be simultaneously satisfied if the photon is totally absorbed.
 
4:22 PM
@ACuriousMind True, but for whatever momentum is exchanged, aren't we dividing by mass of a nucleus which is substantially smaller, so the terms shrinks considerably.
@enumaris Right.
@enumaris But maybe just showing it can be satisfied in this case would be sufficient consolation ...
 
@TheDarkSide just follow @ACuriousMind's advise then :)
 
Alright. I am onto the maths now. Will return in a few minutes.
 
Yes.
Just one point.
@ACuriousMind So, basically, aren't we claiming that it can eat some appreciable amount of momentum, but eats up very little of energy owing to large denominator?
Is that correct?
 
 
4 hours later…
vzn
8:17 PM
@TheDarkSide think you have some deep questions & think they point toward a new idea/ understanding. there are some new ideas about energy dissipation in atoms leading to quanta... can discuss this more at length with the a priori understanding that its a new research area.
 
8:54 PM
@Semiclassical So it's Griffith's 4.18
 
On linear dielectrics in a capacitor
@Semiclassical Here is both question and solution: lief.if.ufrgs.br/~ambusher/griffiths/Pace_EM07.pdf
 
ah, yeah
 
So the Gaussian cylinder has its symmetry axis in the $z$-direction I guess
And if you look at the answer to a), they claim the displacement due to the positive plate is related by $\vec{D} = \dfrac{\sigma}{2} \hat{z}$
 
Sounds right.
 
9:00 PM
Why?
 
First off, I know there's a statement of Griffiths as to when one can use Gauss's law on the electric displacement
And the main point is that it'll be true in this case
 
Homogenous linear dielectric?
 
As such, you can compute the electric displacement from $\nabla \cdot \vec{D}=\rho_{f}$ just like you would for a conductor
It's more specific than that, I think.
 
Throughout all space - but that's not necessary if the field is zero outside the dielectric anyway?
 
right, page 182
 
9:03 PM
Mkay
Yeah sure
 
the key point is that, in this problem, the system has rotational and translational symmetry in the xy plane
 
Agreed
 
as such, the electric field/polarization/displacement can only have vertical components and be functions of z alone
as such, you can't end up in a scenario like figure 4.21
the simpler statement, though, is probably his "conclusion" (and the parenthetical) on the next page
 
The one I paraphrased?
 
right
Note that it'd fail if you had any net charge in the system
in which case, uh, good luck
(it's probably not so bad, but it's definitely a step up in difficulty)
 
9:08 PM
I think he mentions something like that?
 
Probably
 
Page 186
 
yeah
ok, anyways. is it clear why this enables him to deduce $\vec{D}$ here?
 
Not really :P
I am assuming the electric field outside the capacitors has to be set to zero, ok?
Ideal case of infinite parallel plates
 
@Lozansky yeah---it's the same charge distribution
Only difference is that you've got $\nabla \cdot \vec{D}=\rho_f$ instead of $\nabla \cdot \vec{E}=\rho/\epsilon_0$
 
9:13 PM
So that we only need to worry about the displacement field inside the capacitor. Then $\mathbf{D}$ points in the $z$ direction and the surface integral yields $DA$ then, no??
 
Right.
 
So where does this factor $1/2$ come from? Clearly the charge is related by $Q_{\text{enc}} = \sigma A$??
 
There shouldn't be a half when you're doing a pair of parallel plates
In a single charged plate, the half comes from the fact that the electric flux is in both directions (up and down)
 
Yeah but I am only enclosing surface of the $+\sigma$ plate?
 
sure. but you're still accounting for the action of the $-\sigma$ plate, since if it wasn't there you'd be getting field above the positive plate
actually, I should be careful and check which plate is +/-
okay, he's got positive charge on the z=0 plate and negative charge on the z=2a plate
 
9:18 PM
Yes
 
point remains: The fact that you have zero field outside signifies that said field is due to both plates, not just the positive one
As such, the field you get in between will be the combination from both plates individually. but each would give D=sigma/2, so you shouldn't be shocked to get D=sigma in total
This is no different than how Gauss's law worked back when you were doing a parallel plate capacitor in the absence of dielectric
 
OK so if you consider $\oint \mathbf{D} \cdot d\mathbf{a} = Q_{\text{enc}}$, where we let the cylinder go from $z=-\epsilon$ to say $z=a-\epsilon$, what would your surface integral equal and what would the enclosed charge equal?
 
should be $D(a-\epsilon)A-D(\epsilon)A = D(a-\epsilon)A = \sigma A$
i.e. $D=\sigma$ for any $z\in (0,a)$
 
Holy crap I almost emailed my prof by his first name
I am so tired
 
The reason I do not go above $z=a$ is because it is not clear $D$ should be constant going from slab $1$ to slab $2$
 
9:24 PM
actually, that's a good point and something I'm forgetting
I think Griffiths' passing comment in his earlier "conclusion" assumes that all space is filled with the same homogeneous linear dielectric
which is plainly false here
That said, my remark regarding his Figure 4.21 holds: The polarizations, like the rest of the fields in this problem, are vertical. As such, it's not hard to deduce that the integral he's drawing in 4.21 will vanish even if the boundary is dielectric1-dielectric2 rather than dielectric-vacuum
So Gauss's law still works just fine and there's no need to distinguish $z$ in slab 1 versus $z$ in slab 2 when it comes to the displacement field
to the extent that there's charge on the slab-slab boundary, it'll be bound charge not free charge
 
@Semiclassical Do you mean $D(-\epsilon)$? Also, how do you justify dropping the $(a-\epsilon)$ factor?
 
I did mean $D(-\epsilon)$ (which is zero), yeah
But that's not an $(a-\epsilon)$ factor. It's $D(z)$ at $z=a-\epsilon$
And the point is that you get $D(z=a-\epsilon)=\sigma$, which doesn't depend on $\epsilon$
so therefore it's enough to say that $D(z)=\sigma$ for any $0<z<a$, i.e. $D=\sigma$ in slab 1
(my remarks above further imply that $D=\sigma$ works for slab 2 as well)
 
OK good. But then I don't see the point in first considering the contributions from the plates separately?
 
What?
I haven't been considering the contributions from the plates separately.
 
Not you
The wesbite I linked
 
9:34 PM
I'm considering both at once. If I weren't, then the field wouldn'nt be zero outside
ohh
Yeah, I agree
It's not necessarily wrong, but I do consider it superfluous
 
I don't see how it is not wrong
 
How so?
 
If only one plate is considered, then the direction of the field is not along the symmetry line of the cylinder
 
Sure it is. One plate by itself is still going to have rotational and translational symmetry
as such, the displacement field of a plate would still have to be perpendicular to the plate
It wouldn't be zero below and nonzero above, but the field itself still points in the same direction
 
Yes but then the field has a discontuity at the boundary
 
9:40 PM
Not at the slab-slab boundary it doesn't. For the top plate you'll have $D=\sigma/2$ for $z>0$ and $D=-\sigma/2$ for $z<0$. for the top plate, it'd be $D=-\sigma/2$ for $z>2a$ and $D=-\sigma/2$ for $z<2a$
Neither of those is discontinuous at the slab boundary, so when you combine them to get the total $D$ you don't get any discontinuity there
From the perspective of the free charge, the slab-slab boundary isn't there
Bah, $D=+\sigma/2$ for $z<2a$
I guess I should've subscripted those as $D_+$ and $D_-$ respectively, with $D=D_+ + D_-$ total
(also, these are really $D_z$)
 
But then you have two contributions to the displacement field; one for $z<0$ and one for $z>0$. Together this should give $D = \sigma$, not $D=\sigma/2$.
 
Yes, and they do
That's what I was saying
oh, wait
What?
For any given $z$, you'd get a contribution from the positive plate and a contribution from the minus plate
 
I am only considering the positive plate...
Per the website's solution
 
In what sense do you get two contributions? From the perspective of the free charge, all you have is a single positively-charged plate
At best you might think of said plate as being a piece of metal with some thickness. But in that case the charge is distributed on both of its surfaces
 
Yes, but you have a displacement field at both ends of the cylinder
 
9:48 PM
From where to where is your Gaussian cylinder extending?
 
As before
$z: -\epsilon \to a-\epsilon$
 
Okay. But if you're now doing the positively charged plate by itself, then the field below is not 0
The only reason why we were able to claim that before was because we were considering both plates at once
 
Yes, that's why you get two contributions
 
You get two contributions to the surface integral, sure
But that means you've now got $(D(a-\epsilon)-D(-\epsilon))A = \sigma A$, and if you assume that $D(-\epsilon)=-D(a-\epsilon)$ then you get $2D(a-\epsilon) A =\sigma A\implies D(a-\epsilon)=\sigma/2$
The flux hasn't changed at all, since that's just $Q_{enc}$ and that's the same regardless of whether we do just the positive plate or both plates at once
 
Yeah true
 
9:56 PM
What changes is what area that flux is spread out over, and therefore the magnitude of the field over such
When we have both plates, we only have flux in one direction. When we have one plate, we get flux in both directions and thus half the field as before. But then combining both plates gives us the same total field.
this is all forced upon us by superposition and symmetry
 
Hmm...
How does the field due to the free charges differ from the total field in this case?
 
Well, keep in mind that 'Gauss's law for free charge' is $\nabla\cdot \vec{D}=\rho_f$ whereas Gauss's law for the total charge is $\nabla\cdot \vec{E}=\rho/\epsilon_0$
You can ask about what electric field the free charges would generate, and that'll be just $\vec{E}_f = \vec{D}/\epsilon_0$
The total field will be $\vec{E}=\vec{E}_f+\vec{E}_b$ where the latter is the field due to bound charge
 
So is the field from polarization contained in $\mathbf{E}_b$?
 
right
The point of a linear dielectric is that the electric field of the bound charges, which are generated in response to the electric field of the free charges, is given by $\vec{E}_b = -\chi \vec{E}_f$
hmm, I feel like I'm saying that wrong
 
OK now say we distribute dipoles $\mathbf{p}$ in the dielectric, arranged such that $\int \mathbf{p} dV = - \mathbf{P}$. Does this mean $\mathbf{E}_f = 0$ or $\mathbf{E}_b =0$?
 
10:10 PM
If you're thinking of a polarized material, then that should certainly be $\vec{E}_f=0$
if there's no free charge in the problem, then $\vec{E}_f=0$
 
Alright cool
 
right, my claim earlier that $\vec{E}_b=-\chi \vec{E}_f$ is definitely wrong. The definition of $\chi$ comes from $\vec{P}=\epsilon_0 \chi \vec{E}$
 
That's right
 
that does leads to $\vec{E}_b\propto - \vec{E}_f$, but not with $\chi$ as the proportionality constant
$\vec{E}_f = \vec{D}/\epsilon_0 = \vec{E}+\vec{P}/\epsilon_0 = (1+\chi)\vec{E}$
So $\vec{E}_b = \vec{E}-\vec{E}_f = -\chi \vec{E} = -\frac{\chi}{1+\chi}\vec{E}_f$
there we go
Which reduces to $\vec{E}_b=0$ if $\chi=0$ (no dielectric response) and $\vec{E}_b = -\vec{E}_f$ if $\chi\to \infty$ (a strong dielectric acts like a conductor and cancels internal fields)
 
10:26 PM
Nice
 
It’s a good way to check your answers when working linear dielectrics
Though in the present case you’ll have two such susceptibilities
 
It's late, I'm off now
Thanks as always Semi
 
Night
 
0
A: Consequences of infinite redshift in the event horizon of a Schwarzschild black hole

enumarisThe error in your premise is that you have an observer who is hovering at the event horizon which is impossible. No observer can hover at the event horizon, once they touch the event horizon they will invariably fall into the black hole and eventually meet the singularity. For such in-falling obs...

I think my answer's comment section has gotten too long
(Partially my fault, because I felt the need to defend my answer)
maybe a mod should move the comments to a chat?
 
It gives you an option to move it yourself once it gets to a certain point. I suppose you're not at that point yet though
 
10:42 PM
ah...
oh well
 
Also safesphere refuses to talk in the chats even though he tends to have long comment convos
Or at least that's what happened with one of my answers with comments moved to chat
 
hmmm
I feel like he doesn't have a strong grasp of GR...and yet makes strong statements about GR...
 
11:01 PM
[Insert Dunning-Kruger reference here]
 
"Know more by knowing less"
 
indeed
 
11:56 PM
mmmhm, was presented with a cool little tool for ML today
DataRobot basically automates and streamlines finding predictive models
it's mostly built on open source libraries though so I wonder how expensive it is to just streamline that process...
 

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