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Alfred Centauri

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Alfred Centauri
Mar 15, 2022 00:47
Thoughts? twitter.com/LKrauss1/status/1502771393615773698/photo/1
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Alfred Centauri
Aug 12, 2021 01:32
@ACuriousMind I won't disagree that that is true in a reasonably broad sense. Still, I'm not satisfied with that bromide. There's more there than meets the eye. What is one's job?
Alfred Centauri
May 22, 2020 21:18
Crap, I just discovered that Ben Crowell had his account deleted. What the heck is going on? I'll miss his grumpy comments that kept me on my toes.
Alfred Centauri
May 1, 2019 02:33
@ACuriousMind ACM wrote "I'd argue that moderators should have an understanding of what this community values". True enough but, I'm afraid, essentially vacuous. It's precisely (IMHO) the values of this community that aren't quite clear and are, in fact, 'in play'. Maybe I'm wrong about this. I don't claim to have the standing to speak for the community. But I'm skeptical that a moderator can, in general, have such an understanding.
 

 Discussion on question by catmousedog…

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Alfred Centauri
Dec 13, 2020 21:47
I'm not sure I follow. On the one hand, you understand that a quantum object cannot have both a definite location and a definite momentum (there is no state that is an eigenstate of both the position and momentum observables). So, what would it mean to simultaneously measure both the position and momentum of a quantum particle? Recall that, according to QM, the particle must be in an eigenstate of the observable just after the measurement of the observable. Does such an eigenstate exist for this simultaneous measurement you have in mind?
 

 Discussion on question by cccube: An …

Imported from a comment discussion on electronics.stackexchang...
Alfred Centauri
Jun 9, 2020 18:45
cccube, done. I can see you're still having some difficulty with this but you'll get the hang of it.
Alfred Centauri
Jun 9, 2020 18:45
@pipe, there was a reason I came here, edited, and posted an answer. And that's all I have to say about that.
Alfred Centauri
Jun 9, 2020 18:45
@pipe, yes, I did. Did you look at the link in my comment?
Alfred Centauri
Jun 9, 2020 18:45
cccube, I edited the schematic to better match the hand drawn schematic here. I will revert it if you're unhappy with it.
 

 Discussion between Alfred Centauri an…

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Alfred Centauri
Jun 3, 2020 16:08
Great!
Alfred Centauri
Jun 3, 2020 15:59
If you haven't (or don't remember), take a look at this and let me know if it helps: math24.net/…
Alfred Centauri
Jun 3, 2020 15:58
OK, I'm clear on that now (just wanted to make sure we're on the same page). So, as outlined earlier, this solution is found by inserting the ansatz psi(x)=exp(rx) into the ODE. When you do this, you get a quadratic equation in $r$. This equation is know as the characteristic equation. It has two solutions since it is a quadratic. You've seen this before, correct?
Alfred Centauri
Jun 3, 2020 15:51
I'm sorry, I don't understand what you're looking for then. By "general solution of the TISE", what do you mean? Do you mean the general solution to the TISE for a specified potential? Or are you thinking of some kind of meta solution for the TISE with unspecified potential?
Alfred Centauri
Jun 3, 2020 15:28
OK, the TISE with fixed potential is just a linear, second order ODE. So, as usual, you try the solution psi(x) = e^rx and find that there are two values of r that satisfy the ODE (which is what you expect - a 2nd order ODE like this should have two independent solutions). The general solution is just an arbitrary linear combination of the independent solutions. Does this sound familiar to you?
Alfred Centauri
Jun 3, 2020 15:12
Have you had a course in differential equations yet?
Alfred Centauri
Jun 3, 2020 15:12
@ThePointer, no, that's not how it works. Equation 3 is general (holds for any $\psi$). Equation 1 holds only those $\psi$ that are eignenfunctions of the energy operator, i.e., that 'pass through' the operator with scaling factor $E$. The subscript is just a label to help keep everything straight.
Alfred Centauri
Jun 3, 2020 15:12
@ThePointer, note that it's more or less customary to label state vectors (whether an abstract ket or a position basis wave function) with their eigenvalue if they are eigenkets of an observable. For example, using abstract kets, you'll see something like $N|n\rangle=n|n\rangle$ or $X|x\rangle=x|x\rangle$. A general ket, on the other hand, is often denoted by $|\psi\rangle$
Alfred Centauri
Jun 3, 2020 15:12
@ThePointer, that's just the result of combining the 1st and 3rd equations, i.e., $H\psi_E(t)=E\psi_E(t)=i\hbar\frac{\partial}{\partial t}\psi_E(t)$
Alfred Centauri
Jun 3, 2020 15:12
@ThePointer, yes, I thought you knew this (from reading your question). The classical Hamiltonian (total energy) for a particle in potential is $H = \frac{p^2}{2m} + V(x)$. The QM Hamiltonian operator is $H=\frac{P^2}{2m} + V(X)$ where $P$ is the momentum operator and $X$ is the position operator. On the position (wavefunction) basis, the Hamiltonian operator takes the form in your question. TDSE is a non-trivial statement about the Hamiltonian operator: the Hamiltonian is the generator of time evolution
 

 Discussion on answer by BioPhysicist:…

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Alfred Centauri
May 21, 2020 21:26
@BioPhysicist it's a temporary deletion: (1) I wanted to let the comment thread 'cool down', and (2) I wanted to take the time to incorporate the comment thread into the answer (which was more of an extended comment in its original form)
Alfred Centauri
May 21, 2020 20:10
@DataXplorer, ditch the Earth and the twins. Just think about two (ideal) clocks (with accelerometers) that are co-located at one event, separate, and are co-located again later. Stipulate that the accelerometer of one of the clocks always reads zero. That clock will show a later time at the second event than the other clock shows regardless.
Alfred Centauri
May 21, 2020 20:10
(4) That special world line is precisely the one for which an accelerometer on the clock reads zero always, i.e., it is the world line of a clock that is inertial. (5) The reading of an accelerometer on a clock on any other world line through the events must read non-zero acceleration along a portion of the world line. (6) All observers agree on the readings of the accelerometers and the readings on the clocks.
Alfred Centauri
May 21, 2020 20:10
@DataXplorer, once you 'get the hang' of SR, the twin 'paradox' seems trivial. (1) There are two events - the moment the twins separate and the moment the twins reunite. (2) There are many (an infinity) of world lines through these two events but one is special - it is the world line with the maximum elapsed time between the events (according to a clock on that world line). (3) That is, the elapsed time along any other world line (according to a clock on that world line) is less. (cont.)
Alfred Centauri
May 21, 2020 20:10
I've made the point I wanted to make and that's all I have to say about that.
Alfred Centauri
May 21, 2020 20:10
As you know, in SR, there is coordinate acceleration and there is proper acceleration (that measured by an accelerometer). An accelerometer attached to an object at rest in the coordinate system of the Sun reads non-zero acceleration (even in the Newtonian context). Yes, I know that these SR problems ignore gravity but many tend to forget that, and they go on to think that a statement made in that context is generally true. I would at least add an explicit disclaimer about that.
Alfred Centauri
May 21, 2020 20:10
An object at rest relative to the Sun cannot be in free fall so how can it be inertial?
 

 Discussion on question by OverLordGol…

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Alfred Centauri
Apr 19, 2020 01:59
OverLordGoldDragon, are you picturing photons as point particles with definite locations?
Alfred Centauri
Apr 19, 2020 01:59
"Eventually, r is large enough that mutually closest photons are light years apart" ???
 

 Discussion on question by Eric Sumarn…

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Alfred Centauri
Sep 25, 2019 22:39
Just a gentle reminder folks, the comment section is not intended as a 'short answer' section.
Alfred Centauri
Sep 25, 2019 22:39
Eric, to be clear, are you unsure if the phrase "meter per second squared" means $(m/s)^2$ or $m/s^2$?
 

 Discussion on question by Sebastiano:…

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Alfred Centauri
Aug 13, 2019 23:23
Sebastiano, don't take anonymous downvotes too seriously. Your question is interesting but not completely 'fleshed out'. Have you done some research on this question (here and elsewhere)? If so, add some additional context to your question by referring to that research and explicitly state what conceptual issue you're still grappling with.
Alfred Centauri
Aug 13, 2019 23:23
Sebastiano, note that the uncertainty principle doesn't so much imply an inability to precisely know the trajectory of a quantum particle, it implies that quantum particles do not have definite trajectories to speak off (in the orthodox interpretation of quantum mechanics).
Alfred Centauri
Aug 13, 2019 23:23
Are you familiar with the Ultraviolet catastrophe?
 

 Discussion on question by Katz389: Wh…

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Alfred Centauri
Aug 6, 2019 23:01
@Katz389 Honestly, I don't buy that. I don't think you're being completely honest with me. Time dilation due to uniform relative motion is a straightforward result from the Lorentz transformations so I think you know where the idea came from. I think you also know that the thought experiment is not an example of uniform relative motion since clock A changes reference frames. I think you probably know that this must have an effect that should be taken into account.
Alfred Centauri
Aug 6, 2019 05:57
@Wolphramjonny, only the OP has the standing to answer my question, and I do welcome a response from Katz389.
Alfred Centauri
Aug 6, 2019 05:57
Welcome New contributor Katz389! I've flagged your question for moderator attention. Something doesn't seem quite right here. (1) Symmetric time dilation due to uniform relative motion is an unambiguous prediction of SR (2) You seem to be suggesting that this isn't logical, i.e., you seem to be suggesting that SR is logically inconsistent (3) SR isn't logically inconsistent. I'm concerned that you're looking for a debate (or argument) rather than an answer. Are you trolling PSE?
 

 Discussion on answer by Bob D: Intuit…

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Alfred Centauri
Jun 23, 2019 13:10
@svavil wrote ". You might be thinking that they get charges of Q, -Q, Q, -Q; this is not what happens" - I'm afraid that it is just plain wrong (despite 8 up-votes!). It's true that the inner plates have the same potential but it's not true that the have the same charge. From an electrostatics perspective, the charge distribution that minimizes the energy is not that the inner plates have the same charge but, rather, that the inner plates have the equal and opposite charge of their respective outer plates. There are plenty of references online that show this. (...)
Alfred Centauri
Jun 23, 2019 13:10
@svavil, (...) Indeed, if it were the case that the inner plates have the same charge, the formula for the equivalent capacitance of series connected capacitors would be wrong since it is derived under the assumption that "they get charges of Q, -Q, Q, -Q", i.e., that the electric field due to the charge distribution is confined to the region between the plates of each individual capacitor.
Alfred Centauri
Jun 23, 2019 13:10
Bob D, I agree that, given identical initial conditions, two identical series connected capacitors will have equal voltage across. But I'm not sure what you mean by "total charge stored". A capacitor doesn't store electric charge (a capacitor stores energy) since each plate holds equal and opposite charge. So, in that sense, it's true that in both cases, the total stored charge is zero. However, the amount of charge that flows through the battery to charge the capacitor(s) is not the same in both cases which implies that the total stored energy is not the same.
Alfred Centauri
Jun 23, 2019 13:10
@Clonkex, the canonical explanation is to say that there is a displacement current (related to the changing electric field) between the plates of a capacitor that is charging / discharging.
Alfred Centauri
Jun 23, 2019 13:10
"When a second equal capacitor is connected in series with the first, the total battery voltage splits across each capacitor and is V/2 ... In other words, in contrast with putting the second capacitor in parallel with the first, the battery does not "charge" the second capacitor. The existing charge is simply redistributed." - this isn't correct is it? First, let me summarize what I think you're describing: (...)
Alfred Centauri
Jun 23, 2019 13:10
(1) Initially there is just one capacitor with capacitance $C$ across the battery with voltage $V$ across and zero current through (DC steady state). This capacitor has charge $Q = CV$ on its 'top' plate and $-Q$ on its 'bottom' plate. (2) a second, identical (uncharged) capacitor is 'instantaneously' placed in series with the first. Now, it seems to me, there is no transient. The circuit is in a valid DC steady state solution with voltage $V$ across the first capacitor and zero voltage across the second capacitor and so no redistribution of charge takes place. (...)
Alfred Centauri
Jun 23, 2019 13:10
I believe it's impossible for this circuit to 'end up' with voltage $V/2$ across each capacitor for this scenario. For there to be voltage $V/2$ across the second capacitor (in steady state), there must be a transient charging current 'down' through that capacitor. Since the capacitors are in series, they have identical current through. So, a current that charges the second capacitor to $V/2$ would also charge the first capacitor to voltage $3V/2$ which is inconsistent with KVL.
Alfred Centauri
Jun 23, 2019 13:10
@Clonkex, wrote "Because current doesn't flow any further than the first cap in the series?" - consider the following series connected components: ammeter - capacitor - ammeter - capacitor - ammeter. Next, connect this series combination to some external circuit, e.g., a voltage source in series with a resistor. It seems that you're claiming that only one of the ammeters will read a non-zero current but, in fact, all three ammeters will read the same non-zero current. Despite the fact that no charge flows from one plate to the other of a capacitor, there is nonetheless a current through.
Alfred Centauri
Jun 23, 2019 13:10
@Luaan wrote the "negative" plate of one capacitor is connected with a conductor to the "positive" plate of the succeeding capacitor, so their charges are the same. - I don't believe that is correct. It's true that the two connected 'inner' plates have the same potential but that doesn't imply the plates have the same charge.
 

 Discussion on question by zane scheep…

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Alfred Centauri
Apr 5, 2019 16:44
"At the same moment in time." - what does that mean? Even in flat spacetime, events that are simultaneous in one IRF are not in relatively moving IRFs but at least IRFs are global. In curved spacetime, there are only local IRFs.
 

 Discussion between Alfred Centauri an…

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Alfred Centauri
Nov 9, 2018 16:59
Correct, there is no magnetic field, time varying or otherwise and thus, the electric field is conservative. This is just the way it is. The chemical forces that produce the emf are not identical to the electric field which, again, is conservative. This is not hard to see and I don't understand why you insist otherwise.
Alfred Centauri
Nov 9, 2018 01:54
Sorry, should have been "... forces does not imply that the electric field is non-conservative..."
Alfred Centauri
Nov 9, 2018 01:39
In summary, you seem to saying that a non-zero emf demands a non-conservative electric field despite Ben's answer and the Wiki article.
Alfred Centauri
Nov 9, 2018 01:39
Once again, you're avoided answering the simple question I've asked. Ben's answer is clear as is the Wikipedia article. The emf due to chemical and thermal forces does imply that the electric field is non-conservative (look closely at the three integrals for the emf at the Wiki article). You've stated that the electric field of the battery is non-conservative but, if that were so, there must be a time dependent magnetic field.