3:49 AM
@vzn Actually no I was not aware of the headline at the time of writing. I have some students doing a project on the restricted 3-body problem...
and they had never heard of 2010 or the reference to the Lagrange points there... maybe there is none and I’m just mistaken.

Hi GooD MoRninG..
Two Capacitors $C_1 = 2\mu F$ and $C_2 = 6\mu F$ are charged separately to the same potential of 120v.
(a) Connect the negative plate of $C_1$ to the positive plate of $C_2$ (like open circuit).
(b) Connect the negative plate of $C_1$ to the positive plate of $C_2$ and connect the positive plate of $C_1$ to the negative plate of $C_2$ (like closed circuit).
Find the charge on each capacitor in (a) and (b)
Pls help me out how to solve this situation????????

1 hour later…
5:23 AM
Hi! So I am learning multivariable calculus, and I have a doubt. In one of my assignments, I was asked to prove that if a multivariable function had continuous first partial derivatives on a domain $D$, then that function is continuous as well.
Now, when I tried working out the problem, I noticed that continuity of first partial derivatives wasn't a necessary condition. With only the condition that the partial derivatives are defined everywhere on the domain $D$, I could prove that function was continuous on the domain $D$.

Hi @JohnRennie Sir Pls look at my question. How to solve it. I have two answers $120\mu C$ & $360\mu C$

5:44 AM
@FakeMod Solved!

6:09 AM
@123 if you connect the capacitors like this then the charges cannot change:
The topmost charge of +240 μC cannot change because the charge has nowhere to go, and likewise the bottommost charge of -720 μC. And the charges on the two plates of a capacitor have to be equal and opposite.
@123 if you connect them in a loop, as in (b), then you'll get this:

1 hour later…
7:26 AM
Thanks a lot @JohnRennie .
You rock... I made a mistake in loop i am connecting it as series not parallel. That's why i got wrong answer.

7:40 AM
@JohnRennie Does capacitance of capacitors change or always same whether it is connected in series, parallel or separate. Pls explain.

7:53 AM
Hey I was wondering if I could ask something about condensed matter theory and theoretical chemistry here.
Theories in chemistry and Condensed matter physics both are about the physics at length scales of atoms.I think that they are interested in different kinds of phenomena, but the underlying motivation is similar in the sense that they try to explain emergent phenomena using some effective models.
Is this the right way to think about these two fields? Could anyone throw some light on this?

8:24 AM
@Charlie in principle there can be other changes that leave the stationary points invariant (e.g. if you just add the Lagrangian to itself the stationary points obviously also don't change), but the total time derivatives are the only ones that leave the stationary points invariant regardless of the specific form of the Lagrangian

9:01 AM
Who is gnat? They seem to be a profilic voter in several communities, but when I open their profile, it says they're only a member of the community I'm looking from

@NiharKarve Users can choose to hide communities they're part of in that overview, this user has probably hidden all their communities so you only see the one you're currently on

Incredible, they've cast 14,000 votes - that's more than Qmechanic!

9:30 AM
@NiharKarve gnat and Qmechanic are the fourth and fifth most prolific voters on physics.SE, see physics.stackexchange.com/users?tab=Voters&filter=all

@ACuriousMind Yep, that's where I saw it

10:07 AM
is Diff(M) always infinite-dimensional (in physically relevant cases)?
I'd like to learn a bit about gravity as a gauge theory, but I don't really want to learn about infinite-dimensional manifolds/groups first

It's a linear combination of $\partial_{\mu}, \partial_{\mu} \partial_{\nu},...$ derivatives so of course it's infinite dimensional

oh yes, of course
thank you

the gauge group of GR is not Diff(M)

I don't see any notes framing GR in terms of diff really
@NiharKarve these are good, still thinking about the GR sections every now and then though

see physics.stackexchange.com/q/346793/50583 and its linked questions for discussions on what gets "gauged" in the context of GR

10:09 AM
By the guy who first thought this stuff up

I was inspired by physics.stackexchange.com/questions/4359 and the like
I'd like to understand the shortcomings of that formulation

24

Currently, (classical) gravity (General Relativity) is NOT a gauge theory (at least in the sense of a Yang-Mills theory). Why should "classical" gravity be some (non-trivial or "special" or extended) gauge theory? Should quantum gravity be a gauge theory? Remark: There are some contradictory c...

thanks all

10:41 AM
3

Main idea to introduce gauge field for every generator, is to provide invariance under some group of transformation, in your case under group of diffeomorphism and local Lorentz transformations (local version of global Poincaré group). This logic is very similar to gauge invariance, where we intr...

That's pretty good
4

In 2+1 dimensions general relativity with Einstein–Hilbert action with or without cosmological constant is equivalent to a gauge theory with a gauge group one of $\mathrm{ISO}(2,1)$, $\mathrm{SO}(3,1)$ or $\mathrm{SO}(2,2)$ (depending on the presence of cosmological constant and its sign) and a ...

makes the same point about the action being linear vs quadratic as the paper I linked to above, this seems to be a really big issue

11:17 AM
@ACuriousMind ah ok ty

2 hours later…
12:53 PM
99

No waffling, right to the point: What? When? Where? Table support 2020-11-23 Meta Stack Exchange & DBA Meta More table support week of 2020-11-30 DBA Stack Exchange Even more table support week of 2020-12-07 Network-wide launch (if no major issues found) That's right. It's finally...

... since it seems this won't be featured on MSE

huh, neat

2 hours later…
2:32 PM
How do you define a one dimensional Minkowski space? and what is the metric? $(\Bbb R^{1,0},g)$ where $g=dx$?

2:43 PM
@geocalc33 Why would you try to define a one-dimensional "Minkowski space"?
by definition it should be a n-dimensional space(time) with signature (n-1, 1), that doesn't make a lot of sense for n=1

wouldn't that just be a line lol
Minkowski line

a real number line :P

it would be a spacetime with either no space or no time

:-)
actually, I think you need the second dimension to get the irrational numbers

2:59 PM
@user6232128 ?

@NiharKarve how else are you going to construct the square root of 2?

I see, you mean geometrically?

yup

but the rationals do not occupy a special place on the real number line any more than sqrt(2) does

true

3:05 PM
-1

The blue circles created around the lines of force are completely false. This is why this is a computer rendering, no basis in reality. There is no magnetic field inside a condenser other than the created by the plates itself, which should be zero at the point in the center and only exist at the...

If I understand the low quality posts deletion policy, this would stay?
because it's still an attempt to answer?

not all irrational numbers are constructible geometrically (with compass and straightedge) anyway, cf. "squaring the circle" or "trisecting the angle" impossible constructions

again true

@Yashas The line between "nonsense" and "wrong attempt at an answer" is sometimes blurry

lol

I can see arguments either way in this case

3:10 PM
@ACuriousMind have you dabbled in Galois theory?

I took algebra courses where we did it, but I haven't retained much active knowledge about it
i.e. I can understand people talking about some stuff involved, but I couldn't tell you much about it just on my own
I remember field extensions and Galois groups but not how the latter actually tell me about stuff like constructibility :P

I was actually compelled to learn a bit about it after seeing the first page or so of Wiles' proof in Simon Singh's book, "Fermat's Last Theorem"
back when I was a kid it was my favourite book
little did I know they were talking about the absolute Galois group...

Good book.
Wiles' introductory book to number theory is known to be notoriously difficult.

3:26 PM
Yo...

but, they are all still in shock after Yitang Zhang's contribution
Yo

Does capacitance of capacitors always same or change when they are separate, parallel, series, parallel series mix???

"shock" in the sense of his humble beginnings

This website talks about adding capacitance together in a circuit @123, electronics-tutorials.ws/capacitor/cap_6.html

@Charlie thanks, I know total capacitance is added. What about individual capacitor capacitance when in parallel, series and separate. It change or same

3:56 PM
I assume given the way they add that in series each capacitor just has the capacitance it has on its own and in parallel they each have $1/C$ where $C$ is the value they have on their own @123

Greetings to everyone. Does anyone know any good books on Perturbation Theory? Books like Shankar's make nice arguments but I would like to see something more rigorous.

@AndreasMastronikolis what sort of rigor are you looking for? Usually, perturbative expansions are pretty much just Taylor series where you stop after the first few terms and the only rigor I'd associate with that is just Taylor's theorem

4:22 PM
@ACuriousMind Why define a 1-dimensional Minkowski space? Because I want to form the product manifold: $\Bbb M^{3,1}=\Bbb M^{1,1} \times \Bbb M^{1,0} \times \Bbb M^{1,0}$

@ACuriousMind I just found Kato's "Perturbation Theory for Linear Operators" and I think that's the style I want.

user486313
4:49 PM
@JohnRennie yuck, truly boring stuffs ...

5:03 PM
@Charlie Sorry i was busy. Now i am free. It means capacitance of capacitor is not fixed. It is different in different configurations.

@123 the capacitance of a single capacitor is determined by its design, and it is fixed. It does not depend on how the capacitor is connected in a circuit.
The capacitance is given by C = εA/d, where A is the plate area and d is the plate spacing.

@JohnRennie Sir you showed me pictures in the morning. But i calculation does not give me that answer.
Yes i have analyze this formula in q, V, C sense.
It means in series or parallel or separate capacitance is always same?

the capacitance of a single capacitor does not depend on how it is connected to a circuit, and I don't understand why you think it might

5:19 PM
Hi @ACuriousMind . My English is not very good. Pls clear me. Capacitance remain the same in separate, parallel or separate. Is it correct?
Because charlie said capacitance depend on configuration.

The total capacitance of the circuit depends on how its components are connected
the capacitance of a single capacitor doesn't change

@ACuriousMind Thanks. Is it possible for you to solve this problem. I did not found correct answer.
Two capacitors $C_1 = 2\mu F$ and $C_2 = 6\mu F$ charged separately to the same voltage of $V_1 = V_2 = 120volts$. Then negative plate of $C_1$ connect to the positive plate of $C_2$ and negative plate of $C_2$ connect to the positive plate of $C_1$. Find the charge on each capacitors.

I'm not really interested in solving that exercise, sorry

Ans : $120\mu C$ , $360\mu C$
@ACuriousMind Pls give hint than to solve it.
I got the answer of $360\mu C$ not found way of solving $120\mu C$

@123 that's the answer I showed. Are you saying you're not sure how to derive it?

5:27 PM
@JohnRennie yes .. I can not find that answer. I tried but did not get the correct answer.

Like what happened to the voltages. We need to add them or not in loop. $V = V_1 + V_2$ or any other thing.

When you charge the capacitors separately to 120V the charges on them are given by Q = CV, so the charges are 240μC on C1 and 720μC on C2. OK so far?

right..

And you are connecting them together but with the positive of one capacitor connected to the negative of the other capacitor. So you are connecting them as shown on the left in my diagram. Yes?

5:30 PM
Then i tried in parallel $C = C_1 + C_2$ also in series $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$.
Yes..
In series $C = 1.5\mu C$ , $Q = CV = (1.5\times 10^-6) \times (240) = 360\mu C$ that way.

Now the two top plates of the capacitors are connected together so charge can flow between them. If you calculate the total charge on the two top plates it is +240 - 720 = -480μC. OK so far?

Ooohhh.. I see. OKay..

And likewise there is +480 μC on the combined bottom plates.

Yes..

Now this charge of 480μC is disrtibuted between C1 and C2, and we have to work out how much of the charge is on C1 and how much is on C2.
The way we do this is to note that C1 and C2 are in parallel so the voltage across them must be the same. Yes?

5:39 PM
Yes.. Sir.
Roger That.. You rock.

The voltage across C1 is given by V = Q1/C1, and the voltage across C2 is likewise V = Q2/C2, and we've just agreed they are the same, so Q1/C1 = Q2/C2. Yes?
No, the voltage changed when we connected them together. We'll find out what the new voltage is when we do the calculation. For now just call it V.

$Q_1 = Q_2 = 480\mu C$?

No, Q1 + Q2 = 480μC

Aaah Ookay...

i.e. 480μC is the total charge for both capacitors

5:43 PM
$Q_1 = Q_2 = 240\mu C$?

Now take our equation Q1/C1 = Q2/C2 and rearrange it to Q2 = Q1 x C2/C1. OK so far?

Yes..

And C2 = 6μF while C1 = 2μF so C2/C1 = 3 i.e. Q2 = 3Q1. Yes?

Okay Okay. I got it. it means we don't know individual $Q_1 , Q_2$ . We $Q_1 + Q_2 = 480\mu C$?
Yes...

Yes. We get a pair of simultaneous equations:
Q2 = 3Q1
Q1 + Q2 = 480μC
And we need to solve them to get Q1 and Q2.

5:47 PM
Yes....... Sir thank you soooooooooo much @JohnRennie . Without your help it was impossible for me to solve it...........

:-)

6:00 PM
Thanks a lot @JohnRennie Sir for the help and time as always.

7:00 PM
Where roughly in qed do we actually impose some restriction on the energies that photons can occupy? Because given the creation/annihilation operators $a^\dagger(k)$ it seems like we can just create states of arbitrary momentum and by extension arbitrary energy. There's no $n\hbar\omega$ restriction like there is in classical ED
surely we don't just put this in by hand

@Charlie If you create a photon with momentum $p$, that fixes its energy as $E = pc$, and hence it has angular frequency $\omega = E/\hbar$
what "restriction" do you want there to be?

but these energies aren't discrete here are they?

why would they be?

if $p$ in $a^\dagger(p)$ was confined to some discrete values then $E$ by extension would be in $E=pc$

What's discrete about photons is that if you fix $\omega$, then you only can get energies of $n\hbar \omega$
but the $a^\dagger(k)$ describes photons of all frequencies, and hence also all energies

7:06 PM
um
I don't see why you can fix $\omega$ and get photons of varying energy
wait, can two photons have the same frequency but different energies?

oh man now I'm really confused, have I been living a lie
ok good
lol

0

A few days ago, I asked a question related to the addition of magnetic flux densities when there are multiple magnets in a specific piece of equipment. For some reason, it was considered off-topic, even though it is about the physics involved and has nothing to do with engineering. How do I repor...

the $n$ is the number of photons

ohhh

7:09 PM
...you wrote that down first, what did you think it was? :P

I was thinking if I had one single photon, it can only occupy discrete energy levels

that's...not how photons work

wait seriously?
wtf

you're thinking of the harmonic oscillator :P
a photon of frequency $\omega$ always has energy $\hbar \omega$
it's not bound or anything, just a free photon zipping around, there are no "energy levels" for it

oh maybe I shouldn't have used the word occupy
maybe better wording is "photons cannot have arbitrary non-zero energies"
but their frequencies exist on a continuous spectrum
wait maybe they don't
uh oh

7:17 PM
I'm not sure what your problem is here. A photon can have any frequency $\omega\neq 0$, there is no restriction on that. A photon that has frequeny $\omega$ has energy $\hbar \omega$. There's nothing more going on here

omg it actually doesn't
well
this has been troubling
I guess this comes from chemistry, where electrons are confined to bound states of particular energies and photons can only be released if they have as much energy as the difference between the initial and final electron state
I just assumed that therefore photons have discrete energies

7:55 PM
Free photons aren't bound, though
Free particles in general in QM don't have discrete energies

The use of the word "quantisation" in explanations of blackbody radiation completely threw me the wrong way. The interesting point was that the bound states of the electrons had discrete energy levels, not that photons can only ever exist in a discrete spectrum of energies

The point of quantization for blackbody radiation is that EM waves come in discrete packets
but those packets can be of any energy
The thing that doesn't exist in classical EM is the relation between the wave's frequency and energy

I always interpreted "EM waves come in discrete packets" to mean two photons can only differ in energy by integer multiple of some number
I am also just now learning (within the last 30 minutes) that there is no "minimum energy" that photons can have, at least in principle

indeed not
plus, energy of a photon is frame dependent
due to the blue/redshifting

yeah
so many things now make sense
In less enlightened times (aka <1 hour ago) I thought that $\hbar\omega$ was just a number, and that if you examined some random free photon it had to have energy $n\hbar\omega$. It then seemed reasonable that $\hbar\omega$ was just the lowest energy a single photon could have

user486313
8:13 PM
sup

The Ehrenfest theorem, named after Paul Ehrenfest, an Austrian theoretical physicist at Leiden University, relates the time derivative of the expectation values of the position and momentum operators x and p to the expectation value of the force F = − V ′ ( x ) {\displaystyle F=-V'(x)} on a massive particle moving in a scalar potential V ( x ) {\displaystyle V(x)} , Although, at first glance, it might appear that...
Although, at first glance, it might appear that the Ehrenfest theorem is saying that the quantum mechanical expectation values obey Newton’s classical equations of motion, this is not actually the case.[2] If the pair ${\displaystyle (\langle x\rangle ,\langle p\rangle )}{\displaystyle (\langle x\rangle ,\langle p\rangle )}$ were to satisfy Newton's second law, the right-hand side of the second equation would have to be

$${\displaystyle -V'\left(\left\langle x\right\rangle \right),}$$

which is typically not the same as

user486313
@Yashas nice

Why should it be $-V'(\langle x \rangle)$ if it were to obey Newton's laws ?
It's not obvious for me whether expectation of potential derivative should be taken or evaluate the potential at expected value. In classical mechanics, there is no notion of expected values. I don't get how extending it as an analogue requires the expectation of position to be used instead of the potential.

@Yashas ...because that's $F(\langle x\rangle)$, and Newton's law is $F(x,\dot{x},t) = m\ddot{x}$?

still not obvious or I need to sleep

8:20 PM
@Yashas The article isn't making any deep claim about expectation values, it's just saying that the pair of functions $\langle x\rangle(t), \langle p\rangle(t)$ (forget that they're expectation values, they're just numbers now) is not automatically a solution to the classical equations of motion

I understand $F = -\frac{\partial V(x)}{\partial x}$ but I don't get how it was decided that having the expectation of x instead of expectation of V is the correct analogue in classical mechanics.
this whole expectation values and classical limit business is breaking my head

again, the article isn't saying anything about "correct analogues"
it says that:
> the quantum mechanical expectation values obey Newton’s classical equations of motion
is false

but the next line in the article suggests what was expected if it had to obey Newton's laws?

@Yashas Yeah, but that's not about any expectation about "analogues", it just writes down what the statement "the expectation values obey classical equations of motion" would technically mean
the classical equations of motion are differential equations for a pair of functions $(x(t), p(t))$
the pair of functions $\langle x\rangle(t), \langle p\rangle(t)$ does not fulfill these equations
and that's it
you can still say that Ehrenfest's theorem is the analogue of the classical equations of motion in QM, that's not the claim under debate there

Do they obey the classical eom in some limit?

8:33 PM
the article also talks about that :P (they do in the case where the wavefunction is highly localized)

Is this like saying a slightly wavy line looks like a straight line if you zoom far enough out, but the wavy line doesn't actually satisfy the equation of a mathematically straightline
ah

9:06 PM
Happy Thanksgiving.
6