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02:11
Does a magnetic moment precess in a magnetic field, classically?
 
2 hours later…
03:44
@SirCumference Ho, what's up this Hanukkah? Tell us, what are you doing these holidays?
@NovaliumCompany Regression is a tensor based method (you may use vectors or matrices in case you need, I don't know it changes everytime depending on your aim, general formulations are very hard and I don't understand it.) So, there is no need of graph. In case you are a bit confused with Chaos, it's something very more than mathematics. I had a great stuff for it. Lemme search it, you may like it.
@NovaliumCompany Chaos theory is a very general thing, you may replace it with what you like.
@NovaliumCompany I think that you want to know about Chess AI, that's simple with deep nets/neural nets, lemme explain the mechanism.
____________________________________________________________________
Take one neural database but 2 nets using same database of nodes and weights, then teach the board to accept the moves which are legal in chess, leave your net playin continuously for a week. You'll get a very very advanced AI, that can even defeat you in chess.
04:21
Now, this guy knows some serious physics. Physicists here can learn a lot from him.
-5
Q: Does this explain Dark Matter and a Unifying Theory?

yearzeroIt would be super cool if you would read the whole thing before down-voting Speed of light: 299,792,458 meters per second I'm convinced it has to be this speed to allow a quantum/classical boundary. A Femtosecond holds the key of 0.3 micrometers. An object with this width is going to be auto-ob...

04:51
@AbhasKumarSinha Pretty great, just relaxing at home. How bout you?
@SirCumference same, watching Solar Eclipse on Computer.
there's a solar eclipse?
@SirCumference Yes, in India. :)
neat, a total one?
@SirCumference Yes a total one, ring of fire
04:54
dang that's awesome, never gotten to see one before
@SirCumference Decade's last one.
@SirCumference Where you live?
@AbhasKumarSinha that is weird to think about, we're entering the 20s
@AbhasKumarSinha the US
@SirCumference Yes, we'll now see memes like Only 00(s) kids will understand :P
@SirCumference well it's too north.
@SirCumference Well that's not a good, thing, it's very intense and can blind anyone in minutes if seen directly.
It's not even advisible to watch it's reflection on water indirectly.
they make glasses for it though don't they?
@SirCumference Yes, only Government approved ones.
@SirCumference Astrologers here now are fully pumped on TV. XD
04:58
preaching to the choir on that one :P
the weirdest thing is to be grouped up with them
Their Prediction: A Tsunami, Earthquake or heavy showfall in 15 days.
@SirCumference yes
i tell family i'm a prospective astronomer and they ask me to interpret the stars
@SirCumference no, astrologers and astronomers are different, if they are confused.
lol
@SirCumference lol,
Well that's a good Introduction to Einstein's Field Equations.
05:02
You planning to learn GR?
@SirCumference Yep/
@SirCumference You? You know it?
The basics, still only an undergrad
GR's a grad class here
@SirCumference Well it's highschool here :P
@SirCumference Do you understand Covarient Derivatives?
not really :P
@SirCumference ah okay :)
05:06
I guess my advice is starting with getting a strong intuition in classical mechanics. as a high schooler i kind of jumped to trying to learn QM and GR before i really knew the basics
all higher level physics is intended to reduce to classical mechanics under everyday conditions.
the so-called correspondence principle
@SirCumference Well I think I've enough of Classical mechanics to grab QT, GR. I was a State Olympiad Topper in my High School for a year.
oh well that probably works lol
hamiltonian mechanics is pretty helpful for QM, but other than that you're probably set
@SirCumference Yes, I've completed whole L&L First part (left some unnecessary stuff, they were a head ache).
you want to do astro or just interested in GR?
@SirCumference I'm basically into QM, trying to Grab QFT and string theory, so, I've moved from QM to GR now.
@SirCumference Only thing I'm interested in elementary mathematics :P
05:11
"elementary"? :P
fairly certain covariant derivatives are from diff geo
@SirCumference Yes Geometry, integral, those small things and puzzles are really cool! :)
@SirCumference Yes, tensor analysis.
@dmckee Good Morning, sir, i've a doubt with a proof...
This full ones come in every 144 years and this exact one will come after 500 years.
@dmckee The author here converts a indefinite integral to definite one on LHS and doesn't changes it on right. Is that legal?
For example - $$ \int \dfrac{d}{dq} \left (\dots \right) dq = (\dots) \int \psi^* \psi dq \Rightarrow \int \limits_{-\infty}^{+\infty} d \left( \dots \right) = (\dots) \int \psi^* \psi dq $$
Is that transformation legal?
@AbhasKumarSinha It's common in QM for writers to drop the ∞ bounds when there's no ambiguity
bad notation but it is what it is
@AbhasKumarSinha Neat, though I can't say I understand the hindi :P
05:27
@SirCumference oh okay, then definition of orthogonal vector is : $$ \int \limits_{-\infty}^{+\infty} \psi^* \psi dq = 0 $$?
@SirCumference ah okay :)
@AbhasKumarSinha well in the context of wave mechanics, then yes...iirc
@SirCumference oh okay, thanks :)
but more generally two vectors are orthogonal provided their inner product is zero
@SirCumference Inner product is defined with an integral (unbounded one)
well in this case yep
but for example combining a bra vector with a ket vector is also an inner product
05:30
@SirCumference so, they convert unbounded inner product to the bounded one in QM?
the most general definition is usually found in linear algebra books, but in this case the inner product is indeed an integral
@SirCumference oh okay :)
Thanks :)
@SirCumference see, PM replied a troll on twitter - twitter.com/narendramodi/status/1210070312760557570
Luckiest guy on earth :O
@ayc hi :-)
@JohnRennie Hello, I've problems understanding Metric Tensor...
How is that defined?
05:46
@AbhasKumarSinha it depends on what you mean by "defined".
The metric tensor is the solution to the Einstein field equations.
@JohnRennie Well, that's the problem, I can't follow those on internet.
@JohnRennie oh okay, lik psi function on Sch Eqn?
This isn't obvious at a first glance because the equation is written in terms of two derived objects, the Ricci tensor and the Ricci scalar.
@JohnRennie Yes.
@JohnRennie I don't understand the part $ \eta _{uv} \rightarrow g_{uv} $'
The thing is that never (or almost never) write down the Einstein equation then solve it. It's just to complicated for that to be practical. Instead we usually use guesswork to find exact solutions.
Metric Tensor becomes minkowski metric?
is that equivalance principle?^
@JohnRennie So, pertubation like theory exists for it?
05:51
The Minkowski metric is the solution to the Einstein equations in a vacuum.
Actually it's just one of many vacuum solutions, but it's the lowest energy one.
@JohnRennie oh, okay, then How do you define Ricci Tensor?
When you start looking at exactly what these object are it starts getting very complicated very quickly.
@JohnRennie oh, okay, then what it does?
05:53
The Ricci tensor is a contraction of the Riemann tensor, but it can be written in terms of the metric.
@JohnRennie Ah, okay, got the first part...
3
Q: Ricci tensor given through the metric

JonThis is a request for references, mainly for educational aims. In textbooks about general relativity, it is common to present the Riemann and Ricci tensors using the Christoffel symbols. This is easy to understand because it is a straightforward way to perform practical computations and the formu...

@JohnRennie Riemann-Christoffel Curvature Tensor and Ricci Tensor, do they have any relation?
All the tensors can be written in terms of each other. I suppose the Riemann tensor is the most fundamental object, from which you can easily get all the others.
$$ \begin{equation}
\begin{aligned}
R^{\rho}{}_{\sigma \mu \nu}& = \partial_{\mu}{\Gamma_{\nu \sigma}{}^{\rho}}- \partial_{\nu}{\Gamma_{\mu \sigma}{}^{\rho}}
+\Gamma_{\mu \kappa}{}^{\rho}\Gamma_{\nu \sigma}{}^{\kappa}-\Gamma_{\nu \kappa}{}^{\rho}\Gamma_{\mu \sigma}{}^{\kappa},\\
\Gamma_{\mu \nu}{}^{\rho} &= \frac{1}{2}g^{\rho \sigma}\big(\partial_{\mu}{g_{\sigma \nu}}
+\partial_{\nu}{g_{\sigma \mu}}
-\partial_{\sigma}{g_{\mu \nu}}\big).
\end{aligned}
\end{equation} $$
@JohnRennie This one, I don't understand the intitution.
05:56
I did say it was complicated :-)
@JohnRennie lol...
I don't think the best way to learn GR is to get into the details of precisely what all the tensors are.
I doubt any GR book would work that way.
@JohnRennie Yes, I too think...
The intuition about exactly what all the tensors are and how to manipulate them will come as you learn.
 
1 hour later…
07:02
@AbhasKumarSinha Thank you for trying to explain to me but I still find no correlation between the answer and my question :p
@JohnRennie yo happy holidays
@NovaliumCompany hi, thanks :-) And the same to you!
07:38
Happy holidays all :-)
@JohnRennie Sir many people are saying that mass will not increase with velocity, but as far as my readings are concerned and the Feynman Lectures on Physics it’s the case that mass increases with velocity. What’s your view?
@NovaliumCompany Hahahahaha
@dmckee Sir are you here?
@Knight mass is not a useful concept in relativity. Rest mass is useful since it is an invariant.
76
A: Why is there a controversy on whether mass increases with speed?

Ben CrowellThere is no controversy or ambiguity. It is possible to define mass in two different ways, but: (1) the choice of definition doesn't change anything about predictions of the results of experiment, and (2) the definition has been standardized for about 50 years. All relativists today use invariant...

probably relevant
So saying mass increases is kind of meaningless as it depends on how you define mass.
By contrast rest mass is the norm of the four-momentum so it is clearly and unambiguously defined in a manner that does not depend on your choice of coordinates.
@JohnRennie Okay. So when momentum increases with velocity, can we say that mass has it’s contribution too in increasing the momentum (when velocity is too much)
07:46
When you say momentum you are talking about the three momentum $(p_x, p_y, p_z)$ and that is a coordinate dependent object. So its value will depend on the observer. That makes it useful only if you are always sticking to the same single inertia; frame.
If you're only ever going to be working in this frame then you could define a relativistic mass that increases with speed, but by doing this you're painting yourself into a corner because then you can't work in other frames or in a frame independent way,
It's one of the key things to learn about relativity that you try to use a frame independent approach if at all possible.
That means only ever working with four-vectors and scalar invariants.
@JohnRennie Then what does this equation really tell us $$ m = \frac{m_0}{\sqrt{1-v^2/c^2}}$$
@Knight Basically nothing, or at least nothing useful.
@JohnRennie hahahahaha, I really talking to you. Okay I think i’m getting you, the mass* doesn’t increase with speed but the momentum changes as $$ p = \frac{m_0 v}{\sqrt{1-v^2/c^2}}$$ and not as $$ p= m_0 v$$
Am I right?
I forgot like, the sentence I tried to write was I really like talking to you
@Knight correct! :-)
@Knight you can edit your messages, by the way
07:59
@DavidZ thank you very much for that link. I really like people who always strives to help no matter if you know them or not. You are just a very nice guy.
5
@JohnRennie Thank you sir.
@Knight you're welcome.
Relativity is kind of like learning to ride a bicycle. It seems impossibly hard at first then suffenly you get the knack and it all seems easy.
@Knight thanks, very kind of you to say :)
@JohnRennie Sir I hope one day I will meet you in person for discussing GR .
@Knight maybe you'll get to the UK some day ...
@JohnRennie Yeah. I hope
08:28
Ooh, ooh, there's going to be [a near total solar eclipse in Chester])timeanddate.com/eclipse/in/uk/chester?iso=20260812). I have a few years to wait though.
09:14
Motion experiment:
consider an oscillating bead at the x axis that oscillates at some frequency f
now consider a pin head that touches the floor at some frequency F to the left of the midpoint of the oscillation of the bead
If f >> F, then the pin head will touch the bead always, thus to it, the oscillating bead behave like an extended object
but if f is not high enough, then there is nonzero chance the huge force exerted by the colliding bead to the side of the pinhead will knock the pin out
And hence, metaphysically speaking, the solidity of a classical rigid body can be emergent from the regular motion of some smaller object
A similar principle already showed up in resonance phenomena, where a mismatch in the natural oscillation frequency of an object vs the frequency of its applied electromagnetic field mean that the object either felt the mean field, or get dragged along by the field to have any observable effects
Under this interpretation, it is thus easy to rationalise why classical objects have well defined positions while their quantum counterpart do not:
Classical objects evolved much quicker than a given test object so whenever a contact occurs, the test object always end up bumping into the classical object at that moment, while at any time when there is no contact, nothing can be said about the point in of the classical object
meanwhile, quantum objects evolves comparably to the the test object, thus there is some chance the test object will miss it and hence no contact occurs, resulting in a probablistic outcome
However, this interpretation may not be correct as there remains to be explained on how interference can occur under such an interpretation
The uncertainty principle can be reproduced by this interpretation however:
1. If the intrinsic motion of an object is very high, than it has a very high variance to its momentum by definition of d/dx. Then given a test object that contact with it, the probability for there be a contact approaches certainty
2. Likewise, if the intrinsic motion is very low, then it has a low variance to its momentum. Then the object will miss the test object will approach certainty. Since we rely on the contact with the test object to determine the position of the target object, this translates to high uncertainty in its position
 
4 hours later…
14:01
0
Q: My quarter reputation is more than my total reputation!

FakeMod What is this? Is this some sort of wizardry? Or is the server bad at counting? Jokes apart! Why is my quarter rep more than my total rep?

14:24
ok so after a more detailed analysis, in order to describe superposition and interference, what you end up with is basically a vector sum of the velocity fields that govern the object's momentum. This means the whole interpretation falls back into Bohm mechanics so there is nothing new here
although, it can still be argued that it enriched the ontology of Bohm mechanics somewhat. Here every bohmian velocity is interpreted as a complicated but periodic motion where the object's initial condition is uncertain (similar to ordinary Bohmian). Then the mismatch of the periodic motion of these objects with the test object (the measuring apparatus) give rise to nonzero probabilities
The classical limit is then recovered when the object's periodicity is much faster than the test object, thus causing the test object to interact as if the object's periodic trajectory is a solid object
The wavefunction then via the Bohm equation, prescribes the velocity field that the objects to be followed
I think I will need to check if I can derive Born's rule using this interpretation, starting with the simple case of a harmonic oscillator contacting with another test object which lowers to the oscillator periodically
We can set up the toy model as follows:
Let the test object to be at position $x_0$ when $t = Tn$ for some real period $T$ and $n \in \Bbb{N}$
Now the classical equation of motion of the harmonic oscillator is given by:
$x_h(t) = \frac{1}{2}x_0\cos (\omega t)$
Then, contact occurs whenever $x_h(t) = x_0(t)$
actually correction, the test object's equation of motion is given by $x_t(t) = x_0 \delta_{t,Tn}$
so contact occurs when $x_h(t)=x_t(t)$
Solving this, we have:
$\frac{x_0}{2} \cos (\omega t) = x_0 \delta_{t,Tn}$
$\frac{1}{2} \cos (\omega t) = \delta_{t,Tn}$
$\frac{1}{2} \cos (\omega t) = 0, t \neq Tn$ or $\frac{1}{2} \cos (\omega Tn) = 1$
$t = \frac{\pi}{\omega}(\frac{1}{2} + n)$ or $\cos (\omega Tn) = 2$
uh...
looks like this actually is sensitive to initial conditions, thus we are no longer dealing with Bohm here
So that means, this is not an interpretation of quantum mechanics nor Bohmian mechanics
hmm...
(Btw, I am assuming the motion of the test object and the object are orthogonal so I can simply its equation of motion using only deltas, since contact is only relevant in the x axis)
So if I want to check for all possible initial $x$ I should set $x_h(t)$ as:
$x_h(t) = x_0\cos (\omega t + t_0)$
Solving this for $t = Tn$ we have:
$x_h(Tn) = x_0 \cos (\omega Tn + t_0) = x_0$
$\omega Tn + t_0 = (\frac{1}{2}+n)\pi$
And solving for $t\neq Tn$ we have:
$x_n(t\neq Tn) = x_0 \cos (\omega Tn + t_0) = 0$
if $x_0 = 0$, then contact is certain, if $x_0 \neq 0$ we have:
$\omega Tn + t_0 = n\pi$
Now let $f = \frac{1}{T}$ we thus have the equations:
$\frac{\omega}{f} + t_0 = (\frac{1}{2}+n)\pi$ or $\frac{\omega}{f} + t_0 = n\pi$
if the object periodic motion is faster than the test object, then we have $\omega >> f$, this gives:
$\frac{\omega}{f} \sim \omega$ and so we have:
$\omega + t_0$ on the LHS
so contact is entirely dependent on the motion of the object itself as expected intuitively
however, this contradicts the classical case because the fact that it need to satisfy those equations means contact is not guarantee for all initial conditions even as the frequency of oscillation of the object dominates the probing rate of the test object
15:08
The f*cking wikipedia page of Palpatine spoiled me The rise of skywalker
They have literally put a spoiler there without any warning
so this model does not reproduce classical mechanics, let alone quantum mechanics
We can further support this by doing the same analysis using $f >> \omega$ instead and the result is that the initial condition determines whether contact will occur
Thus compared to Bohmian mechanics, this "periodic momentum model" predicts an extra observation of frequency dependent measurement that depends on the ration between the periodic momentum of the measuring apparatus and the system, which is never observed in experiment, hence such a model is falsified as a interpretation of quantum mechanics and classical mechanics
@NovaliumCompany Wikipedia is always full of spoilers. They don't have a concept of warning for it because it would have to be on every page :P
lol
they should wait a bit for the movie to cooldown a d then update the page
Do my analysis make sense, or is incomprehensible and you have no idea what I am doing?
@NovaliumCompany Eh, it's not like the spoiler really ruins anything. The franchise did a horrible job at setting everything up, so really the spoiler could have been anything and everyone would just be like "oh, ok".
15:19
@AaronStevens ye, I just came back from watching it. It works emotionally tho
@NovaliumCompany Wikipedia's mission is not entertainment, but accurate information. If you don't want spoilers, don't read about the topic you don't want to get spoiled about there!
And "Works emotionally, but makes no sense in hindsight" is something of a running theme with Abrams :P
@NovaliumCompany I guess? Maybe I'm just not enough of a Star Wars fan. This final movie could have handled it better by making it more of a focus rather than shoving in all of the extra random stuff
I agree that the movie is one hell of a mess but it made me feel something and that's worth my 10 Leva.
@ACuriousMind My opinion is that they should simply wait a month after the movie's release, not a big deal.
@AaronStevens happy new year! And be lated merry Christmas!
Ok this makes sense
Unless the frequency $\omega \to \infty$ there are sampling frequency of the test object to miss the system
an infinite oscillation is unphysical, hence this model does not reproduce any experiment
I just hope people will discuss with me when I am actually doing physics modelling like a physicists will do here
Clearly having been inactive for such a long time from SE have massively slashed the response probability
particularly for h bar where I don't see many regulars around except maybe 3 highly inactive ones
16:20
0
Q: How to prove the roots of cosine is dense or sparse when a tend to infinity?

SecretSo I am a bit surprised that there isn't a question on this. Recall the first time when you came across the cosine function $\cos (ax)$ as $a \to \infty$ graphically, it intersect the x axis more and more frequently, result in the limit to diverge Now, more familiar with the different kinds of i...

when you knew the intuition fail you because of dense and sparse sets at infinity
Would not thought of such question had the above analysis is not carried out
 
1 hour later…
rob
rob
17:30
0
Q: What specific chemical/ physical properties of rubies enable them to be used as the gain medium in a laser?

Tariya PosharIt would be really helpful if you included the chemical elements involved (Chromium and Corundum) and their contributing properties that answer the question. Thank you!

I'm working hatd to resist the temptation to edit this question, so that it would ask why rubies make such excellent magical slippers. Deep breaths, Rob.
17:57
The bizarre speculative physics of static motion:
Consider some object X which does not change in spatial position x
Now consider some object Y that is placed at some distance d away from X
Then X is in static motion when $\frac{dx}{dt} = 0$ but $\frac{dd}{dt} \neq 0$
This can be modelled by that the objects stay where they are in a given coordinate system, except the metric is time dependent
Strange things can happen for static motion, such as X and Y can collide with each other despite visually speaking they stay separated
The dual to static motion is moving stillness, which is when $\frac{dx}{dt} \neq 0$ but $\frac{dd}{dt} = 0$
motionlessness thus gain a new meaning: it is when both derivatives are zero
0
Q: How can I re-flag a question?

FakeModThere are some posts which I have flagged and some action is genuinely required for those posts. Here are some examples :- Is this an eigenfunction of a ladder operator? Eigenvalue of operator? Weightlessness on a vibrating plate These are the ones which I am quite confident about. I would sa...

 
2 hours later…
19:55
@Qmechanic Hi and best wishes. Please can you edit my last question adding the tags and can you tell me if my question is clear? Thank you very much.

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