When summing over Feynman diagrams, I sometimes find the dimensions strange, because the larger diagrams have more internal variables to integrate over. So I am adding diagrams D1 dx1 and D2 dx1dx2dx3 and it looks a bit strange to me
@FakeMod Full answers to HW questions should be flagged for deletion via custom mod flag. How you vote on them is - as with all posts - your own choice.
@FakeMod I'm not comfortable with downvoting answers that are technically correct because that may confuse readers into thinking the answer is incorrect.
@ZeroTheHero Could you tell me if the first answer is true? https://physics.stackexchange.com/questions/540489/physical-interpretation-of-triplet-state
I've uploaded files on GitHub that I want no one to be able to access except me. I've created a private repository with no collaborators added. Is that the option to make it completely private? (e.g I don't want anyone to steal or look at my code)
@ACuriousMind FWIW, in the Python room, we do not allow people to ask for answers to their fresh questions. That's primarily to avoid having parallel discussions on the main site & chat. We tell them they need to wait 2 days, when their question becomes eligible for a bounty. However, we do allow people to ask for "meta" assistance with their question, to help them ensure that their question is up to community expectations.
@NovaliumCompany That sounds correct to me, but I've never used that feature so I'm not certain. (I mostly just post Gists, and most of those are public).
I need help with a History question: The Japanese foreign market is mainly aimed at America, the second most important market is: Bulgaria China Korea SSR
Pretty sure it's not the first one
I tried Googling.
plez
I suppose the question is concerned with 1950s (post-WWII) stuff cuz that's what we're studying
The question is so general that it kills my brain cells
This question suspiciously sounds like a reading comprehension question - one to which the answer is trivial if you've read/understood the reading material it presupposes...
Why you think random physicists on the internet know Japanese post-WWII economics is beyond me, though :P
@JohnRennie Welp, I guess they should teach us to Google properly, cuz in 21st century it proves much more useful than remembering random seemingly useless facts.
They should teach us to think, not to memorize and apply formulas. My school has no idea what creativity is, what learning from failure is... it's such a disgusting system, glad it's over soon (for me)
The world moves forward but schools stay in 20th century
A little while ago I read a touching novel where the main character has Alzheimers but manages to keep going by continually Googling for all the stuff she has forgotten due to the disease. How well this would work in real life I'm not sure.
@JohnRennie This would need the disease to only make you forget technical knowledge. You can't google that the face in front of you is your child, you can't google if you think it's the '50s and you don't know what the shiny rectangle in your hand is...
@NovaliumCompany Playing devil's advocate: It's not about the facts themselves. It's about the skill of reading a text, extracting and retaining information from it and being able to answer such reading comprehension questions.
This skill doesn't really change with the advent of Google - sure, you can Google plenty of facts. You likely could also just have read the corresponding section in your text book. In both cases what's being tested is not your ability to memorize the fact but your ability to extract it from a larger source of information.
@ACuriousMind I can still read a text, extract and retain information and answer comprehension questions about something that I actually care about. Something that I'm curious about and I'm eager to learn. In my opinion, learning should be a choice. It should be a journey each person travels through, in their own way, in their own time. I understand that that would be catastrophic for the world and that most people are not particularly curious or eager to learn but still.
I agree that the educational system works for the whole and is, for now, the most optimal system the world has for educating the world but for me personally, it's incredibly dull. What am I even talking about, we'll probably be downloading info with a cable to our brains in a few decades
The author of this answer just edited it. In the table at the end of the answer there are blue rectangles after each planet name. At least, that's what they look like in the Samsung browser (which I think is a fork of the Chrome Android browser). Those rectangles are links to Wolfram Alpha. What are those rectangles supposed to look like? I had a brief look at the raw source of the post but it uses a lot of LaTeX that I'm not familiar with.
trying to decide if the following is in the spirit of "adiabatic invariance"
Suppose you have a rotating body, with initial angular velocity $\omega_0$, subject to an angular acceleration $\alpha<0$. I track a point on that body which is at a distance $R$ from the axis of rotation.
If the angular acceleration is zero, then this is uniform circular motion and the total acceleration is just the centripetal acceleration: $|\vec{a}|=\omega^2 R$
This no longer holds if the angular acceleration is nonzero. In that case, we have tangential acceleration $\alpha R$ and thus $|\vec{a}|=\sqrt{(\omega^2 R)+(\alpha R)^2}$ where $\omega=\omega_0+\alpha t$
However, if $\alpha$ is small, then the tangential acceleration may be neglected and thus $|\vec{a}|\approx \omega^2 R$ once more. In that case, one has $a\approx v^2/r$ despite the lack of uniform circular motion (i.e., despite the fact that $\omega$ is not constant).
That vaguely reminds me of the concept of an adiabatic invariant but I can't tell if I'm just making it more complicated than it needs be.
(should have been $|\vec{a}|=\sqrt{(\omega^2 R)^2+(\alpha R)^2}$, oops)
user434058
1:12 PM
Should I write something like "Hope this helps!" at the end of my answer? (Bob D always does this, so I am wondering whether it might be a good practice or a useless one)
@Semiclassical Could you comment about whether the first answer to this question is true? https://physics.stackexchange.com/questions/540489/physical-interpretation-of-triplet-state/540566#540566
It would at least be better than the flood of strangely intrusive remarks about other people's health and mental well-being that are currently en vogue in the emails I'm getting :P
@Slereah That's nothing. One of my colleagues always ends his presentations with a slide that says (translated) "If it's all getting too much for you, just work faster."
I haven't yet figured out whether this is meant to be humorous or not :P
However, that only works for the z-axis. You needn't have $S_{1x}=S_{2x}$, for instance
In other words: Each of the two M=0 states can be said to have opposite z-components. But, for the singlet state (L=0, M=0), this is also true for the x- and y-components. As such, you can say "spins are opposite" without any real issue.
With the L=1, M=0 state, by contrast, you can't make this further statement. So i that case you can only say that the two particles have opposite z-components of spin.
@Semiclassical That strikes me as a dubious ontological claim. You can say that the two particles have opposite z-components of spin after their z-spin has been measured. Prior to measurement, this state has no definite z-components for the individual particles!
"As more distant stars are revealed in this animation depicting an infinite, homogeneous and static universe, they fill the gaps between closer stars. Olbers's paradox argues that as the night sky is dark, at least one of these three assumptions about the nature of the universe must be false."
and therefore $L_{1z}|L=1,M=0\rangle = - L_{2z}|L=1,M=0\rangle$
I guess the less objectionable phrasing is that, if you measure the total z-component, then this quantity has a definite value for the L=1, M=0 state. From that, you infer that the individual z-components will come out as opposite once measured.
@Semiclassical Sure, but since this is an entangled state when tracing out the other particle you get a mixed state, not an eigenstate of $L_{iz}$. I think this is different from the situation where you do have definite pure states for the component particles, and the way we talk about the state should reflect that
Yes, this makes for awkward phrasing, but the situation is simply rather unclassical and awkward.
Fair enough. This point of phrasing is something I've gotten into trouble with in the past
in particular, there's a footnote to the long philosophy preprint I contributed to:
"This response to Wayne Myrvold’s request for clarification (see the preceding note) was inspired by a footnote in _Wahrscheinlichkeitstheoretischer Aufbau_ (von Neumann, 1927b, p. 249, note 9). In this footnote, von Neumann points out that the Hamiltonian for the harmonic oscillator $\hat{H}$ has a discrete spectrum even though it is the sum of two terms, $\hat{p}^2/2m$ and $\alpha \hat{q}^2$, that both have a continuous spectrum. The value of $\hat{H}$ will not be the sum of the values of $\hat{p}^2/2m$ and $\alpha \hat{q}^2$ even though the expectation value $\hat{H}$ will be the sum of …
Doing Olber's paradox with specific systems is simple enough. Take stars on a grid of length $\ell$, then the total luminosity is just
\begin{equation}
L = \sum_{i,j,k} \frac{L_0}{4\pi\ell^2(i^2+j^2+k^2)}
\end{equation}
which is enough to show divergent. I have attempted to do it for a more rea...
That's perhaps not quite the same issue, but it's a reminder of how the sum of two operators can have a quite different status (for a given quantum state) than either operator by itself.
@ACuriousMind Of course, by that token, the phrase "the spins are opposite" is questionable even for a singlet state
simply by virtue of any spin components only being well-defined after the measurement
the issue is that this language of “opposite spin” is based on simple geometrical addition of spins, and it comes with its difficulties.
In $\uparrow\downarrow$ the spins are opposite...
but it’s not an eigenstate of $L^2$ so... not sure what it would mean to say spins are opposite in this state. It is true that the projection is 0 so in this sense the spins are opposite...
but $\downarrow\uparrow$ has the same property.
the difficulty is that $\uparrow\downarrow+ \downarrow\uparrow$ has $0$ projections, is made of “opposite spins”, but does not have $S=0$.
when you couple 2 spin-1/2 particles you get either $S=1$ or $S=0$ so there’s nothing especially important about $S=1$. In fact methinks the $S=0$ states are probably more useful precisely because they are independent of the direction of the quantization axis.
Of course there's no such thing as an infinite convolution - if your question is to make any sense we have to recast it as a question about limits of finite convolutions. And then we have to ask in what sense the limit exists.
We never get anything interesting as a limit in $L^1$ norm:
...
Hi guys today I was on my terrace with bottle filled with water, and bottle had holes in its bottom, and earlier the bottle was sealed pack but some amount of water was coming out, and I open the bottle Cap and water start coming out from the holes at very fast rate. And suddenly the bottle had felt down.
I am studying a propagator in real space/real time, but I have to discretize time. It is the Hubbard model, hence nearest neighbor hopping. Is it correct that the particle can only jump one time per time step?
It feels like my propagator would then be hugely dependent on how small my time steps are
@Semiclassical So, it's alright to say for the singlet state has opposite spins, because, $S^2$ operator takes the same value that of $S_z$. For the triplet state, the case is different. All what could be said about the latter is the projection is opposite, not the "magnitude". And the point where I stuck is, the magnitude confused with the projection.
@Student404Mus it's alright compared with the triplet state. As ACM and ZTH both emphasized above, the language of "opposite spins" is problematic for other reasons.
with the more sensible statements being... For any M=0 state (singlet or triplet), the results of measuring the z-component of spin for the two electrons are guaranteed to be opposite. For the L=0,M=0 state, this furthermore holds regardless of the axis you measure along. For instance, the results of measuring the x-components of spin for two electrons will also be opposite.
For the L=1,M=0 state, there is no such guarantee: the results only have to be opposite if you measure along the z-axis, and don't have to be opposite if you measure both particles along another axis (e.g. the x-axis).
In the case of the covariance principle and the covariant formulation of EM is there a reason to prefer covarience over contravariance? Or is this just a convention
Surely you can just construct everything contravariently and you just end up with the same equations that transform oppositely
if you want a statement in terms of probability amplitudes, I'd say that $\langle \uparrow_x \uparrow_x|L=1,M_z=0\rangle \neq 0$
huh. if i do the computation in mathematica, it looks like the probability of opposite results is zero if you measure in the x-direction for the L=1, M_z=0 state
here's a quick computation
consider the state L=1, M_x=0. Up to a factor of 1/sqrt(2), that's $\uparrow_x \downarrow_x+\downarrow_x \uparrow_x$
but in the z-basis, we have $\uparrow_x=(\uparrow_z+\downarrow_z)/\sqrt{2}$ and $\downarrow_x=(\uparrow_z-\downarrow_z)/\sqrt{2}$
so therefore we get $$\uparrow_x \downarrow_x+\downarrow_x \uparrow_x=\frac12(\uparrow_z+\downarrow_z)(\uparrow_z-\downarrow_z)+\frac12(\uparrow_z-\downarrow_z)(\uparrow_z+\downarrow_z)$$
which, upon distributing and cancelling, is just $$\frac12\left( \uparrow_z \uparrow_z-\downarrow_z \downarrow_z\right)$$
Peter Shor just posted an answer to that triplet state question, but he messed up the Mathjax. physics.stackexchange.com/a/540618/123208 I'd fix it, except I'm on my phone, and I'm not sure whether he intended to include those escaped square brackets or if he thought they'd function like double dollars.
so to sum up: If M_z=0, you definitely see opposite results for the z-components. If L=0, M_z=0, then you see opposite results for the x-components as well. But if L=1, M_z=0, then you instead see identical results for the x-components
(and presumably the same is true for the y-components)
In my opinion, since |L=0, M_z=0> and |L=0, M_x=0> are the same up to a fixed rotation they sufficiently need to be defined in one frame to be distinguished. Whereas, |L=1, M_z=0> and |L=1, M_x=0> are identical need to be defined in different distinguishable spaces.
So basically I and a friend observed that the field of dipoles effectively partition space into two halves, one where charges will move towards the dipole and one where charges move away from the dipole
and then for quadrupoles, it does similar things to a point charge but now it is 4 qudrants
Now looking at gauss law for forces, which is $r^{-m+n+1}$ for a n dimensional charge density in m dimensional ambient space
it seems natural to interpret a dipole is basically a -1D dimensional charge density
Likewise, a similar curious phenomena is observed in relativity, and its 2 time extension in the scifi orthogonal, where spacetime is partitioned into 2 (light cone, elsewhere) and 4 regions (light cone, elsewhere with 2 constrained degrees of freedom) respectively
Thus it seems for a nD, object, n counts the number of constraints or degrees of freedom depending on whether n is positive or negative
so for a -1D dimensional object, it has one constraint, e.g. a directionality in the ambient space, or some other things that can only move in one direction
Using these, it is then straightforward to extend the interpretation to iD by using $i^2=-1$. Coincidently, rotating the lorentz space in a certain way can interchange space and time axes, so it seems to checks out
For n fractional, it is just the usual hausedoff dimensions
however, I have not figure out how to interpret $e^{i\frac{\pi}{4}}$ yet
@PM2Ring Well, I was actually shocked at that level of reaction from Acuriousmind. Clearly this is way way out of the league compared to my usual weird ideas
I cannot tell if I am overstepping the line from philosophy into crank territory here
Honestly, I don't usually read what you post here very closely. But I was just thinking that if the first thing I'd seen if I'd stepped into this chat room was someone going on about dipoles in negative dimensions - something which makes no sense at all in mainstream physics I should stress - I wouldn't have stepped by a second time.
This isn't "philosophy", you're just saying things that don't make any sense physically
@Secret :) FWIW, I kinda like your idea of using negative dimensions to denote constraints, in complement of positive dimensions denoting freedoms. But it doesn't suit this room's predominantly mainstream physics culture.
right I should keep an eye not to disrupt the mainstreamness. I think worldbuilding will indeed be a better place for that kind of wild idea
I just did not realise I stepped over the line this time since in my past posts, most users are still ok and find the mainstream common ground in it to discuss weird ideas in mainstream context
@Secret I suppose that when it's clearly obvious that you're talking about weird stuff (eg your recent posts that talked about Brahman & simulation) then people who aren't interested in that stuff will just ignore those posts. But if it's not clear & obvious that you are talking non-mainstream stuff, then the mods & ROs are (rightly) concerned.
yeah I clearly overstepped the line into the non mainstream this time, sorry about that
I think most of my recent philosophical posts, some of which Slereah, Semiclassical and others have give some interesting comments, do not touch the mainstream physics, so those are legitimately philosophy
but I think I did not realise this negative dimension thing gone too far into non mainstream, which indeed I agree they are rightfully concerned now I am made aware of that
Maybe you should find a more suitable venue for your musings. You have an interesting imagination, but I feel that the regulars here aren't really a great audience for your ideas.
Just throw the witch from a cliff if she flies away then....you know she indeed is one. If not then check for someone else. Try and error is the best way to do it.
The irony is that at least in England in the 15-16th centuries people would claim to be witches because it offered them a degree of status and protection.
I mostly remember it because I got assigned to read all the parts of one of the main characters, so like it or not I was quite involved in the class read.
> Double, double toil and trouble; Fire burn and caldron bubble. Fillet of a fenny snake, In the caldron boil and bake; Eye of newt and toe of frog, Wool of bat and tongue of dog, Adder's fork and blind-worm's sting, Lizard's leg and howlet's wing, For a charm of powerful trouble, Like a hell-broth boil and bubble.
Double, double toil and trouble; Fire burn and caldron bubble. Cool it with a baboon's blood, Then the charm is firm and good.
No. I didn't even really realize it until right now. I'm not sure how easy it would be to figure out who designed a syllabus ~10 years ago. Though it might not be too hard to find out if those books were required reading by the province or something.
Googling comes up with a range of possibilities including the option that Shakespeare just thought the word sounded good and it means absolutely nothing :-)
"Mind you, the Elizabethans had so many words for the female genitals that it is quite hard to speak a sentence of modern English without inadvertently mentioning at least three of them." — Terry Pratchett, in alt.fan.pratchett
@PM2Ring I seem to recall the sort-of inverse story where Puritan doctors had trouble treating their patients because they'd describe anything below the waist as their "ankle", but I can't remember where I read that
first, i talked about the warm-up questions they submitted ahead of time. (the point was to derive the expression for the centripetal acceleration)
then I put them into smaller groups to talk about the lab video that we provided them. this didn't go as energetically as I'd hoped but that wasn't a huge shock
once I felt like they'd come to grips with the video, I brought them back together and walked them through the data analysis (which in this case was basically just "show them Google Sheets formulas")
and gave some suggestions as far as what their next lab report would look like
Ok, this is a huge long shot, and I also apologise if this is a poor question (I'm a newbie here...). This doesn't seem to me a strictly perfect fit for Chemistry.SE, Biology.SE, or Photography.SE either, so hopefully it's acceptable here, given that there is some level of Physics involved...(ple...
@AaronStevens I understand where that's coming from but I think there isn't really a formulation that fits into the word limit of the close banner that captures what we mean by "engineering". See physics.meta.stackexchange.com/a/6136/50583
Doing Olber's paradox with specific systems is simple enough. Take stars on a grid of length $\ell$, then the total luminosity is just
\begin{equation}
L = \sum_{i,j,k} \frac{L_0}{4\pi\ell^2(i^2+j^2+k^2)}
\end{equation}
which is enough to show divergent. I have attempted to do it for a more rea...
Of course there's no such thing as an infinite convolution - if your question is to make any sense we have to recast it as a question about limits of finite convolutions. And then we have to ask in what sense the limit exists.
We never get anything interesting as a limit in $L^1$ norm:
...
