I feel like sometimes I put a lot of effort into an answer and it doesn't get voted on, not because it's not good (though I can't know for sure) but because there are too many other answers for the question that have received 1 or 2 votes
Aside from putting a bounty on an old question and hoping people scroll down to look for new answers, is there another way I can draw attention to a new answer?
@NauticalMile And here's JohnRennie good-naturedly ranting about getting too many upvotes for the wrong kind of answers. Rep is a nice gratification, but you really shouldn't be writing excellent answers to get rep - for that, you should be writing many easy answers.
Hypothesis: If a model predicted a solution that is never realised, it means the model is either incomplete or we are missing some information that prevent those solutions from being realised
I wonder what prevents the spontaneously moving solution from being realised. Is it because the action that corresponds to it is not mimimised? Or is it because of thermodynamic reasons?
@Secret Here's a simpler solution: The case that a point particle rests perfectly still at the exact top of a perfectly flat dome of that exact shape just never occurs in real life.
In reality, there are so many other factors influencing what happens - the pathological solution for Norton's dome just doesn't correspond to a prediction about reality, since its pathological nature relies on all the idealizations we make.
I found it amazing that you guys always knew the simple solutions while I am only aware of complicated ones. The same thing happens in real life where I am best known among my friends to overthink a problem but that the solution is actually obvious
@ACuriousMind I'm trying to prove: $M\ge0$ is the lcm of $a_1,...,a_n$ iff it is a common multiple of $a_1,...,a_n$ which divides every other common multiple.
Working on $\Rightarrow$ first.
So I need to show that with $K$ another cm, $M\mid K$.
This in turn means that we must have $K=cM$ for some integer $c$.
I don't know how to show this.
@ACuriousMind ($\Leftarrow$) If $M\mid K$ and $M,K\ge 0$, $M\le K$. But since this holds for all cms, it holds for $K=$lcm. Thus $M$ is the lcm.
@0celo7 Look, I believe the point of such exercises is that you try things, and fail until you (feel like) you got it right. Me validating each of your steps defeats the whole purpose of you training how to prove things.
If you really want the answer, just ask it as a question on math.SE, they're fine with such questions.
No, I just think a hint from the teacher would have been in order.
I can't do the induction step.
And the book ranks this as an "easy" problem
@ACuriousMind Meh, I wrote a little paragraph explaining why it has to be a multiple. I've no desire to spend another hour figuring out the exact inductive step.
@vzn That feels like the kind of wild speculating that people engage in every time there is a surprise announcement. See all the crazy preprints that appeared right after OPERA's initial superluminal neutrinos announcement.
@ACuriousMind Ok, let $p_1,...,p_k$ be the set of primes which can factor all $a_i$s. Then $M=p_1^{e_1}\cdots p_k^{e_k}$ and $e_i>0$. (Prove the inequality by contradiction.) Let $K$ have the expansion $p_1^{f_1}\cdots p_k^{f_k}p_{k+1}^{f_{k+1}}\cdots$. It is clear that $f_i>0$ for $i\in\{1,...,k\}$. [cont. in next post]
Now, since $M$ is the least common multiple, it is clear that the $e_i$ cannot be smaller and still be a common multiple. It then follows that $e_i\le f_i$ since $K$ is a common multiple.
Then $K=M\cdot p_1^{f_1-e_1}\cdots p_k^{f_k-e_k}p_{k+1}^{f_{k+1}}\cdots$$\implies $ $M\mid K$.
@dmckee agreed there is some wild theorizing right now, but its not entirely; do you have any of your own idea behind the gamma ray burst? its anomalous wrt black hole merger. sometimes there is a subtle difference between wild speculation and educated hypothesizing. the star collapse idea was put forward by a reputable scientist. lets not forget how outlandish the original idea of black holes was! now, collapsing pairs verified! possibly also with accompanying massive gamma ray burst!
@DanielSank The mod tools don't show anything of interest, but I'll keep it in mind.
1 hour later…
user116211
7:07 AM
@DavidZ Seeing Daniel's concern, I also want to air my concern over some simultaneous downvotes on my posts. For two days, I've been getting downvotes simultaneously at most of the SE I'm active. However, I'm not sure. Just by the by.
Block A is connected to a rope, which is connected at the other end to Block B. There is a force pulling block A up. The net acceleration of the entire system is 3 m/s^2. I want the tension at the top of the rope. The rope weighs 4 kg and block B weighs 6 kg. Is it correct to say that the tension at the top of the rope is (4 + 6)* 3 = 30 N?
>CuriousOne| So why do we see all this talk about "particles" in quantum mechanics? Because (beginners level) quantum mechanics is really a non-relativistic single quantum theory. It avoids all the mathematical problems with relativistic quantum field theory and can make reasonable statements about system with low energy bound states and low energy scattering. We don't have to worry about ever seeing more or less quanta than >CuriousOne: we have put in and it pretends to have a simple interpretation in terms of "particles".
In a comment to my question someone stated the following:
"photons do not travel at some definite number of oscillations per
second. In fact, they do not "travel" at all, no more than electrons
or other quanta do, as by the Uncertainty Principle they don't have a
definite speed and/or ...
This OP's question also makes me wonder: does QM allow a particle to disappear from a point and reappear in another point that is not continuous to it? if so, what is the explanation?
I have no problem picture this, but is this view also techincally speaking, still classical in a sense?
So basically the only thing that can in principle be detected is a distribution of observable after repeated measurements and that it follows some underlying rule parametrised by time, and you cannot observe things like "something blink in and then out of existence"?
Is this the quantum (classical free) way to interpret quantum?
I am actually more curious on how people can bring two quantum states into coherence (let alone entangled)
Given that we cannot detect the phase of the amplitude directly (unlike an oscillating EM field) what do people use to bring the two states to coherance so they can start interfering (because they cannot see whether the phases are correlated until they measure the system, but that will end up changing it?
"I am an 13 year old boy who is very interested in science and I am totally embarrassed after reading the equation from Harold white 's warp field mechanics."
@yuggib His TA advised against even showing him, since he "probably wouldn't like it". I'm still considering sending an e-mail, at least, with the files so he can look at them.
(elaboration of previous question) Given that we cannot detect the phase of the amplitude directly (unlike an oscillating EM field) what do people use to bring the two states to coherance so they can start interfering (because they cannot see whether the phases are correlated until they measure the system, but that will end up changing the system hence the states?
Mathematically, two states can interfere if they are coherent and lies within the same hilbert space. What is the physical process(es) that allows one to bring one of the states into the same hilbert space and correlate their phases thus allowing them to produce an interference pattern?
well, I don't necessary need them to be entangled, I just want to find out how to experimentally make two states interfere
and some wikipedia and PSE search suggest I need to first have the states under coherence
which then brings in to the question on how we can do that, because the phase is not directly measurable, hence how to exert some control on align their phases?
but mathematically it is pretty straightforward to prove that not all the vectors on $\mathscr{H}_1\otimes \mathscr{H}_2$ (or a proper symmetrization if $\mathscr{H}_1=\mathscr{H}_2$) can be written in the form $\varphi_1\otimes\varphi_2$, where $\varphi_j\in\mathscr{H}_j$ ($j=1,2$)
the ones that cannot be written that way, they are "entangled"
it is sufficient to turn on (for a very small time) essentially any interaction between the two systems to introduce correlation even if you start from an uncorrelated system
however I don't know how they do it in practice
there are surely lots of papers on that, for I think those experiments where worth Nobel prize(s)
the experimentalists would know much better than I do
I am a mathematician, for me existence is sufficient :-D
@yuggib: Just stumbled over your comment discussion with CuriousOne. Physicists indeed are not consistent with using the word quantum, you can't really blame him for that. It's completely standard to talk about particles as "field quanta", and thinking about a photon as a "quantum of energy" - which Planck called the "quantum hypothesis" - is indeed the origin of the word.
Maybe you are right, but it is only a source of confusion in my opinion. Historically quanta were really discrete amounts of some measurable quantity and not "excitations of a field" or whatever.
@Slereah That's about the quantum operator associated to measuring the classical phase of an electromagnetic field, not about measure a relative phase between quantum states.
@yuggib I agree that the word "quantum" or "quantized" is overused and that is confusing, but, well, it's kind of arbitrary which part of what it describes you want to keep calling quantum and for which part of it you have to invent a new word (not that you'd get anyone to follow your choice, anyway...)
@Slereah yuggib's point is that the $\lvert p\rangle$ we so like to write down doesn't lie inside the Hilbert space.
@yuggib Yeah, it's misleading terminology, but that doesn't change that it is also standard terminology. You can't blame a physicist for using physics terminology ;)
@Danu : are you stalking me? Only comments like this appear to be deliberately derogatory. And since you have not replied to my correction, dishonest to boot.
@JohnDuffield The query Qmechanic linked catches all questions where "gravitational wave" is written in either the question or the answer - and one of the answers (and one comment) on that question contains it.
QUOTE: So could an analogy be made for gravitational waves where the sun disappeared and re-appeared about 250 times per second. Eight minutes later would we feel those gravitational pulses. In short are gravitational waves like gravity turned on and off? ========== Hmm, other than "the sun disappear and reappear 250 times a second", I don't see why this question falls outside the context of mainstream physics enquiry...
Mathematically such scenario should be easy to be plug into existing GR models, and I never knew of any "unphysical" condition is sufficient to be considered to have the question fall outside mainstream physics
Most non mainstream physics questions I have read so far, one can clearly see some alternate models at play, but here it is only a (possibly not realisable) physical condition, I don't see how that is off topic due to non mainstream
plus what Slereah just mentioned. I don't think this is off topic
@Qmechanic Eh, kinda hard because while I'm not sure "non-mainstream" was the right reason for closure (it seems kind of "letter but not spirit of the law" to me), I also don't really get what the question is asking.
@JohnDuffield As I said before, I'm not interested in any further discussions with you. And of course you can tell yourself that I'm stalking you, if you like that idea :)
@ACuriousMind The OP is basically asking what we would felt for an oscillating gravitational field caused by the sun turning on and off its gravity at 250 Hz
@Qmechanic I think that the question is now, after heavy editing, pretty decent.
The edit makes it very explicit that the scenario is just a poor way of describing a maybe-not-so-terrible idea because the OP isn't familiar enough with the material to cook up more realistic scenarios.
So if I understood correctly, in the span of 8 minutes, how will the gravity changes as perceived at earth when the sun turn on and off its gravity at 250 Hz ----- ok.
I was unable to generate it purely by my own hand---I had to mostly steal it from a TeX - LaTeX answer.
@Secret Only tex
@ACuriousMind The torus shape in that one is kind of ugly. But if you think the shape of the knot is weird: It's not, it is exactly what that thing really looks like (...if only I could show which parts are in front of and behind the torus...... :( )
yup, gotta learn some tex to produce the rigorous versions of yours
@Slereah We the chemistry community requires all authors to produce original graphics. Graphics of others can only be used under permission of the respective authors
It's a velocity map imaging device. Basically you send a molecular beam which is then photolysed by a laser. Another laser then ionise the fragment which then hit the detector
We then use the distribution of the ions to work out the dynamics of the photolysis
In the section 'The free particle' in 'Introduction to quantum mechanics' by Griffiths page 65. He has the wave equation as $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\phi(k)e^{i(kx-\omega t)}dk$$ He then makes statements about if $\phi(k)$ has a large spread then the momentum is ill-defined. Does $\phi(k)$ have such a strong influence on momentum because of the form of the equation of the expectation value of momentum?
@JohnDoe you'd want to look at the one for the variance of momentum, actually, which is $\langle \phi^2\rangle - \langle \phi\rangle^2$ (or maybe the other way around? I mix them up too often). Ill-defined momentum means the variance is large, more or less.
@Danu and @ACuriousMind : The reason I closed it as non-mainstream was because of the sentence the Sun disappeared and re-appeared about 250 times per second, which violates locality/causality. If OP had written e.g. the Sun performs an oscillatory/orbital motion about 250 times per second, it would be better.
@Qmechanic Yes, I'm not saying you made a bad decision in closing it. I think the huge edit made a big difference, and would've voted to close as well before it was implemented.
@DavidZ Why would the variance of momentum be $\langle \phi^2\rangle - \langle \phi\rangle^2$, should it not be $\langle p^2\rangle - \langle p\rangle^2$?
@DavidZ Are you quoting directly from Griffiths book or is this something that may have been omitted?
John, let me try to explain it from a different angle: If $\phi(k)$ is a delta-function then you're dealing with an eigenstate of $p$, right.
If $\phi(k)$ is nonzero for not just one value of $k$ but (in)finitely many, you get a superposition of momentum eigenstates.
This means that for every $k$ such that $\phi(k)\neq 0$ there is a chance of you measuring the momentum to be $k$ (recall that measurement is essentially projecting onto any of the eigenstates with probability for each given by norm-squared)
Hence if $\phi(k)\neq 0$ ("has support") for a large range of $k$ it means that an experiment can result in many different momenta being measured.
In a comment to my question someone stated the following:
"photons do not travel at some definite number of oscillations per
second. In fact, they do not "travel" at all, no more than electrons
or other quanta do, as by the Uncertainty Principle they don't have a
definite speed and/or ...
Why doesn't a pencil write if its tip is heated in a candle flame?
