Suggestion for the site designer is when being asked to review a First Answer (say), both the question and answer should be displayed by default, instead of the answer only. Most answers don't make much sense without the context given by the question, and the reviewer has to dig for it.
@RonaldVilliers The key point is that a lot of the dark matter is outside the visible matter i.e. the visible matter is embedded inside a much larger disk of dark matter.
@ACuriousMind Could it also be that condensed matter groups are typically larger, and condensed matter physicists just ask questions within their group
@B.Brekke I don't think most questions here get asked by people who are "in a group"
even most of the hep-th questions are the typical questions students taking a course or reading a textbook have, not the questions hep-th researchers have
if your explanation was correct, we should see proportionate amounts of hep-th and cond-mat questions at "low" levels, and then a drop-off as the cond-mat people join a group and stop asking while the hep-th people continue :P
Why is the sum of the voltages of the capacitors in series equal to the voltage of the battery? My background knowledge is only up to electric potential and capacitors. I know nothing about Kirchhoff or above.
The one thing I only know is charge quantity is same on all the capacitors in series.
@MethNoob suppose you start at the minus terminal of a battery. For convenience, call that 0 volts. Then the potential at the positive terminal will be the battery voltage.
As you move around the loop, you’ll pass across each capacitor in series. For each, there willl be a voltage drop
And once you reach the end, you’ll be back at the minus terminal of the battery. But then you know what the potential is, ie, it’s just zero
So the voltage rise from the battery must equal the voltage drops from the capacitors
@ACuriousMind sanity check: suppose a phase space function is not classically conserved w/r/t the Hamiltonian. If we quantize this, then we know at least that this observable won’t commute with the Hamiltonian, regardless of the quantization?
You can look up details on how batteries work (huzzah chemistry)
Well, you can certainly say this: if you move a positive test charge from the positive terminal to the negative terminal, it’s potential energy will have decreased
The details behind why the PE is higher at one terminal than another is a matter of chemistry
For the purpose of the loop rule we just take it as a given that there exist such a thing as batteries which generate gains in electric potential energy
@ ACM or to put it differently: if an observable commutes with the Hamiltonian in QM, then it must be conserved classically
Potential is 0 on negative and N V on positive terminal. Potential difference is NV. "So the voltage rise from the battery must equal the voltage drops from the capacitors" is little bit confusing.
Then the change in voltage from the positive terminal to the negative terminal can be found in two ways: either you go back across the battery, or you go across the capacitor
Both ways should give the same answer, b/c electric potential only cares about where you are not how you get there
This discussion only makes sense once the capacitor is in equilibrium with the battery, tho
Ie once the capacitor has stopped charging
Before that, we have to include the resistance of the wire if the story is to hang together
@Semiclassical No :) You could theoretically get unlucky that the exact value of $\hbar$ is such that the quantum corrections exactly cancel the classical PB
but that's vanishingly unlikely, so I'd take it as true for all practical purposes (and in particular it's true if you view $\hbar$ as some sort of formal variable instead of a mere constant)
@ACuriousMind lol, true. But no non-conspiratorial way of doing it :P
I say conspiratorial, but I could see a scenario where you handle some things classically and not others, and thereby end up with hbar-dependent parameters in the classical Hamiltonian
In which case that scenario is a bit more plausible
I think one scenario in which you get this is if you have an "anomaly" where some classical symmetry gets naively quantized to something with non-vanishing commutator and then you try to "correct" the symmetry so it becomes a quantum symmetry again
i.e. you might have some $f$ with vanishing Poisson bracket but non-vanishing Moyal bracket with the Hamiltonian and then you try to find a "simple" correction $g$ for $f + g(\hbar,p,q)$ such that the commutator/Moyal bracket vanishes
Suggestion for the site designer is when being asked to review a First Answer (say), both the question and answer should be displayed by default, instead of the answer only. Most answers don't make much sense without the context given by the question, and the reviewer has to dig for it.
