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12:00 AM
@0celo7 I think you already said that unless I'm experiencing a CTC
 
@Obliv no I actually lied earlier
 
I thought you were being sarcastic and searched it up myself
@0celo7 what is $X \approx Y$
is that homeomorphic
dude @0celo7 how do I go about proving this result in the last box hyperphysics.phy-astr.gsu.edu/HBASe/waves/wavstd.html
from what I am reading: $\hat{f(\xi)} = \int_{-\infty}^{\infty} f(x)e^{-2\pi x i \xi}~dx$ so am I supposed to integrate the wave equation w.r.t. one of the variables?
no wait I have to use a fourier series.
 
12:22 AM
@AlfredCentauri I voted for you
@AlfredCentauri You'll always be the moderator of my heart
 
:wipes tear from eye:
 
This will make you feel better
 
@BernardMeurer you're right, the tear is gone ;<)
 
It's saxophone metal, it's awesome
 
user218912
@Obliv why are you learning from there?
 
12:29 AM
@bl00 It was one of the few first search results. I'm not going to cite it or use its information
 
user218912
@Obliv read tong's notes. :D
 
lol..
 
user218912
the ones on electromagnetism have a lot on waves and fourier stuff.
 
I am more interested in mechanical waves though
 
@Obliv what
fourier's theorem is very hard
just assume it
 
12:34 AM
I wanted to have a bit more rigor in the treatment of the theory for this lab instead of : a standing wave is : $A\cos(kx + \omega t + \phi) - A\cos(-kx + \omega t + \phi)$
 
@0celo7 Get Reb, I want to play flute
 
user218912
@Obliv tbh I never learned about mechanical waves o.O
 
user218912
it's not in tong's notes or poisson's notes.
 
Yeah I'll probably just end up going with the standard stuff in the textbook then. Oh well, maybe next lab will be more interesting
gtg for now
 
user218912
wow searching for lecture notes I'm finding so many good things online.
 
user218912
12:46 AM
when will I have time to read them though :(
 
@bl00 @Obliv One of you need to change avatars, its really confusing
 
user218912
i'll change.
 
user218912
better?
 
Yeah, that actually does help, thanks :)
 
no
 
user218912
12:55 AM
blue block is boring.
 
@bl00 I like that shade of blue
#007DC3
 
user218912
I specially formulated it.
 
user218912
in my blue lab.
 
1:38 AM
anyone around?
 
@0celo7, I'm around.
Do you know Mathematica?
(Wolfram Mathematica, I mean)
 
1:56 AM
no
I wish I did
 
oh, man.
Thanks anyway!
 
2:19 AM
Nov 14 '15 at 17:57, by 0celo7
Oh the pun man who cant do calculus
Ruthless :/
No reason for me posting that now, just strange I've been here for almost a year
 
user228700
2:52 AM
Hi everyone :-)
 
Hello.
 
3:15 AM
Why does light obey the Fermat's principle of least action?
 
@SwapnilDas Fermat's principle is the principle of least time, not the principle of least action, but the reason why light obeys Fermat's principle is due to the way photons behave in quantum mechanics.
@SwapnilDas Basically light has a probability to on every possible path between two points, but the paths that do not obey Fermat's principle tend to destructively interfere with one another, whereas the paths that come close to obeying Fermat's principle add constructively, so the probability that light travels on a path that comes close to obeying Fermat's principle is very high.
@SwapnilDas If you want more detail, you can read the book QED: The Strange Theory of Light and Matter by Richard Feynman.
Can someone take a look at my question here?
2
Q: How do we prove that the angular velocity vector is equal to a limit?

Keshav SrinivasanThe angular velocity vector of a rigid body is defined as $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$. But that's not how most people intuitively think about angular velocity. Euler's theorem of rotation states that any rigid body motion with one point fixed is equivalent to a rotat...

 
3:46 AM
@ACuriousMind THE LEGION IS SAVAGE
 
Oops, I often get confused. Thanks! @KeshavSrinivasan
 
4:03 AM
@SwapnilDas You're welcome.
@SwapnilDas By the way, are you a Hindu? If so, you may be interested in the site Hinduism Stackexchange:
@SwapnilDas I'm the moderator of that site.
 
4:18 AM
@KeshavSrinivasan Uh, I would say that the fact that light travels along the path of least time comes from Maxwell's equations, which have nothing to do with quantum mechanics.
 
@DanielSank Well, yeah, classically it comes Huygens' principle, which comes from Maxwell's principle, but quantum mechanically it comes from the phase of the amplitude of a photon to go from one point to another.
 
4:55 AM
@KeshavSrinivasan Are you physics Grad?
Yes, I'm a Hindu. But I believe more on in Physics much more than Hinduism.
@ManishEarth Hi, are you from IITB?
 
@SwapnilDas Yes, I have a Bachelors double majoring in Physics and Math and a Masters in Math.
 
user228700
5:57 AM
@JohnRennie: Hi sir, morning :-)
 
Morning Kaumudi :-)
 
user228700
I'd pinged you at the CSE chat...
 
user228700
They're having a discussion about the elections (sort of), so never mind. I'll ask again, later, after thinking about it some more..
 
@Kaumudi the text you quoted (the text not the image) seems rather bizarre ...
By resonance do you mean this:
In chemistry, resonance or mesomerism is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by one single Lewis structure. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical structures). Each contributing structure can be represented by a Lewis structure, with only an integer number of covalent bonds between each pair of atoms within the structure. Several Lewis structures are used collectively to describe the actual molecul...
 
user228700
@JohnRennie I quoted two texts (not image)...which one are u talking about?
 
user228700
6:10 AM
@JohnRennie Yep.
 
in The Periodic Table, 2 hours ago, by Kaumudi
"Now, if the charge or free electron is moved from place to place (resonance takes place), any reacting species finds it harder to get to the charge or radical. Also if it is moved rapidly, as is the case in resonance, critical charge densities or electron densities for reaction and subsequently high energies do not build up for any periods of time but are spread over different locations. Hence we get a lower energy."
That seems to be largely meaningless
 
user228700
@JohnRennie Yeah, I know :/ I was very suspicious about this.
 
user228700
The whole text can be found here:
 
user228700
 
Quantum objects like electrons don't have a position, just a probability distribution. When we have resonance ina molecule it just means the probability distribution for the electron(s) is spread out over large parts of the molecule.
 
user228700
6:13 AM
@JohnRennie Yeah, OK...
 
It doesn't mean the molecule oscillates between the two forms.
The stuff you quoted seems to be making things far more complicated than they are.
 
user228700
@JohnRennie Yes, I know this. I was wondering why people keep saying "resonance reduces energy".
 
That's a silly statement
 
user228700
And yes, it seems to have an overall stabilizing effect on the molecule but I was wondering why.
 
If you mix up a bunch of nuclei and electrons they will arrange themselves into the lowest energy configuration.
 
user228700
6:15 AM
@JohnRennie Really? Because I hear it all the time! :/
 
And that may be one in which electrons are delocalised over the whole molecule.
 
user228700
@JohnRennie Hm. Does this delocalization of electrons always produce a stabilizing effect?
 
I suppose it's a general rule that the more delocalised a particle is the lower its energy is.
 
user228700
Right, OK. And this is a consequence of...what?
 
That comes from the uncertainty principle. The more tightly you constrain a particle the more you put up its ground state energy.
 
user228700
6:17 AM
@JohnRennie Ah, OK...
 
user228700
This is the most useful explanation I've received. Thanks so much! :-)
 
But so resonance decreases energy is misleading because it sounds like we have some causal process i.e. we start with a molecule then introduce resonance and that decreases the energy.
 
user228700
(As opposed to "We're talking about Quantum Chem.; it's not supposed to make sense")
 
user228700
@JohnRennie Yeah, yeah!
 
@Kaumudi The funny thing about intuition is that it changes as you learn more. After xxx years quantum mechanics and relativity now make perfect sense to me and seem actually rather obvious.
But for the newcomer I concede that they can seem baffling
 
user116211
6:20 AM
Are you guys studying resonance?
 
user228700
@JohnRennie Wow, that's rather astounding, to say the least :-)
 
@MAFIA36790 in molecules - that isn't what we physicists normally mean by the term ...
 
user116211
It's just quantum superposition of base states.
 
@Kaumudi it just happens. It'll happen to you too, wait and see :-)
 
user116211
Feynman explained it in his book.
 
user228700
6:22 AM
@JohnRennie :-) @MAFIA36790: Nah, these are all the details I need, thanks :-)
 
user116211
I have some posts related to that.
 
user116211
9
Q: Why is the original Pauling's theory of resonance that uses superposition of wavefunctions not used today?

MAFIA36790I have recently studied superposition of states (also the famous Schrodinger's cat), measurement problem, decoherence & so on. I then read the resonance concept from Atkins' Physical Chemistry where it was written as: It is the superposition of wavefunctions representing different electron di...

 
user116211
5
Q: Why is the resonance concept not required in molecular orbital theory?

MAFIA36790In valence bond theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why is it told that "MO theory provides a global, delocalized perspective on chemical bonding"? (This is a follow up to What is actually the difference between valence bond theory and molecular o...

 
user116211
7
Q: Trying to understand the statement: 'Resonance is not a flickering between the contributing states. '

MAFIA36790So, the famous resonance definition: it is the weighted average of different Lewis structures.... Well, this is purely wrong. Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'-Peter Atkins. For-ins...

 
user116211
Related:
 
user116211
6:26 AM
9
A: What does the circle in a drawing of benzene ring represent?

MAFIA36790This is a typical example of the downfall of the Lewis structures. $\ce{C_6H_6}$ can be represented by more than one Lewis structure: Dewar structures: Kekule structures: Does anyone of them actually represent benzene? No! If it were so, then some of the $\textrm{C-C}$ bonds would be sm...

 
user116211
Damn internet; I was writing a post; but couldn't post it yesterday; and now seeing somewhat related post has already been posted T__T
 
user116211
That's the most depressing thing ;/
 
user228700
@MAFIA36790: Whoa, that's a loot of links. Thanks, I'll check 'em out when I have the time :-) For the time being, I'm more than satisfied with JR's explanation
 
user116211
@Kaumudi, No, I didn't tell you to read it immediately or later; I just showed that I have many posts on resonance.
 
user228700
@MAFIA36790 Uh, OK :-P
 
user228700
6:37 AM
@JohnRennie: Since I have a shitty textbook, I do this a lot more than u realize:
 
user228700
 
user116211
@Kaumudi It's looking right.
 
user228700
These are the notes that I will revise, later on and these are essentially an amalgamated version of bits and pieces from your explanation :-D
 
user228700
@MAFIA36790 No, I wasn't asking for a review; I was merely pointing out the fact that these notes consist of @JohnRennie's explanations :-D
 
user116211
okay.
 
user228700
6:40 AM
@JohnRennie: No comments? :-P
 
user228700
OK, guess you're busy at work.
 
@Kaumudi That looks fine. Note that invoking the uncertainty principle is a bit armwaving. It's true, but the reality is a bit more complicated.
 
user116211
@JohnRennie There is a precise calculation on it in Feynman.
 
user116211
You can check it online in CalTech site.
 
user228700
@JohnRennie You also missed my point! I was showing you how my notes consist of ur explanations a lot!
 
6:49 AM
@Kaumudi Ah, OK :-)
 
user228700
(And it's alright if the reality is a bit more complicated; I only need the essence of it at this stage)
 
user116211
0
Q: What charecteristics define a wave for physicists?

SRSAny superposition $y(x,t)=f_1+f_2$, of two arbitrary function $f_1(x-vt)$ and $f_2(x+vt)$ of $x\pm vt$ satisfies the wave equation in one-dimension. Will it be called a wave if the function $y(x,t)$ doesn't have any periodicity? For example, consider the aperiodic functions $y(x,t)=A\exp[-\frac{(...

 
user116211
WTH is that.
 
user116211
Do physicists consider the function different than math people?
 
user116211
Anyways, @JohnRennie, could you help me in Real analysis theorem?
 
user116211
6:51 AM
There is no 0celo :(
 
user228700
@MAFIA36790 You say that like @0celo7 is a thing, not a person, lol :-P
 
user116211
;P
 
Why are pure math guys harsh, while applied ones cool?
Just asking from personal experience :P
 
7:32 AM
@MAFIA36790 I doubt I can help with a maths problem as my knowledge of maths is rather sketchy. 0celo7 is presumably asleep at the moment, but will be around later.
 
user116211
okay, sure; I will try asking Balarka after re-reading the content again.
 
[Division by zero]
Zero power rule: Given $0 \in \textrm{WIP}, n\in \mathbb{Z}$, the following can be derived:
\begin{align}
0^n & =0^n + 0^{n-1}\\
0^0 & =1\\
\pm 0^2 & \neq 0
\end{align}
WIP=work in progress, constructed by breaking the left distributive law, left additive identity and right additive inverses
If we are are lucky, we might be able to localise the violations of these real number axioms to only terms involving $0,1 \textrm{ and } q$, thus allowing division by zero to be appended into the real number line
Here, we do have a zero term, 0 has only a unique left multiplicative inverse (due to violating left distributive law to stop $x0+x=x+x0=0$, followed by violating left additive identity to stop $0^2=0$. Therefore 0 is a right additive identity with a left multiplicative inverse
Multiplicative associative law remains to be checked. The additive inverse of $0^2$ and all powers of 0 remains to be solved in order to not let any zero powers to become 0
Once we ran through all the associative laws, and hence having an associative cayley table, we will check this with the MSE guys
 
7:49 AM
@Secret: $0^0$ is normally treated as undefined.
 
It is in the complex numbers, because of the sharp discontinuity of the function $z=x^y$ at $(x,y)=(0,0)$
But even then, in most applications such as binomial theorem and factorials, $0^0$ is often conventionally defined to be 1 because it makes the formulae easier
217
Q: Zero to the zero power - is $0^0=1$?

StasCould someone provide me with good explanation of why $0^0 = 1$? My train of thought: $x > 0$ $0^x = 0^{x-0} = 0^x/0^0$, so $0^0 = 0^x/0^x = ?$ Possible answers: $0^0 \cdot 0^x = 1 \cdot 0^x$, so $0^0 = 1$ $0^0 = 0^x/0^x = 0/0 = \text{undefined}$ PS. I've read the explanation on mathfor...

(Below are my own investigation on the topic a year ago)
9
Q: 0 to the power of 0, what does the essential discontinuity actually look like?

SecretSo having watch this clip by Numberphile which explains why $0^0$ is undefined https://www.youtube.com/watch?v=BRRolKTlF6Q And also this http://mathforum.org/dr.math/faq/faq.0.to.0.power.html And also this Zero to the zero power - Is $0^0=1$? I understand how when given a function of the for...

 
WIP however is not related to this because we are strictly speaking, not working in the complex numbers. Suppose we are lucky enough (which might be quite slim, considered I have been working on this division by zero stuff for perhaps almost 8 months now (in my free time)) to have it to be self consistent, it is not clear how calculus on this number system with be affected
 
user116211
8:19 AM
@JohnRennie, I wrote a post on the zero vector post yesterday; but due to poor connectivity, couldn't post it. Now, today I'm seeing another user has answered it with the same intention as that of my post. Would I post it still? I wanted to cover the rigorous definition of vector space as a unitary K module with a division ring.
 
If I find myself in this situation I decide whether my post adds anything new. If it does I would answer but say something like: I'd like to add the following comments to AUser's answer ...
 
user116211
@JohnRennie okay.
 
That way you're acknowledging that an answer already exists and it makes it clear you're not just plagiarising the existing answer.
 
user116211
The nutshell was that the zero vector is the identity of the abelian group associated with the vector space. So, it's inherent in the structure of Vector space.
 
user116211
He seems to have aired it well by mentioning it in the axioms.
 
user116211
8:24 AM
So, I'm not seeing what my post would add more than him other than the rigorous definition of vector space in abstract algebra.
 
user116211
Hmm, I BLAME my damn connectivity for that ;/
 
user116211
Anyways, I might complete it in off-time and post it.
 
user116211
What might happen in worse?
 
user116211
Downvotes?
 
user116211
Nay, don't care.
 
user116211
8:35 AM
Not sure, though.
 
user116211
The main thing is the excitement I get writing things I've recently learnt and still learning ;) As writing it makes it crystal clear.
 
1
Q: How to contact another user?

VickHow can i contact another user for comments or to ask them to answer particular questions? Is there a way to have instant messaging on the SE?

 
user116211
@PhysicsMeta I don't think it's possible.
 
user116211
Unless the concerned user is active in chatroom.
 
9:19 AM
in Mathematics, 21 mins ago, by Secret
Some abstract algebra "findings"(?):
Too lazy to type it again
 
Btw, an advice that might help on 0celo7 to better his linear algebra. Tell him to code something in MatLab and I can assure you that after that he will be thinking with matrices
O wow, I got something that is $(aa)a\neq a(aa)$...
now that is very weird...
 
user116211
@Secret Every magma preserves associativity.
 
user116211
Most of the structures we work with are magma.
 
user116211
9:34 AM
Hello @Qmech.
 
user116211
Sometimes Royden is vague.
 
In that case, whatever WIP is, it is not even a magma
 
Hello @MAFIA36790.
 
user116211
I've spent literally hours to comprehend the theorem.
 
If it is self consistent, then it is something that is not well studied, and perhaps not very useful
 
user116211
9:36 AM
@Secret So, what structure are you working with?
 
You knew the answer, it is always that
Even that Rock paper scissor like + structure I derived some days ago (which is confirmed by my maths peers to have no zero elements) is much nicer than this thing
 
Except for the h bar and MSE chat room, most chat rooms are dead.
 
From my computations done so far, WIP has a two sided multiplicative identity, obeys right distributive law, is closed, left inverses always exists while right inverses don't exist for 0 by construction, is noncommutative in + and $\times$ and nonassociative in $\times$ and (possibly) $+$, 0 is a right additive identity
The few interesting thing I have so far is that there is a pair of zero power theorems, where for $n\in \mathbb{Z}$
\begin{align}
0^n & = 0^n + 0^{n-1}\\
0^{n-1}+0^n & = 0^{n-1}
\end{align}
which means one can impose some kind of noncommutative partial order to the powers of zeros such that $0^n > 0^{n-1}$ on the left, while $0^{n-1} > 0^n$ on the right
 
9:56 AM
@357 : ::pondering if physicists call sand cake for yellowcake:: :)
 
update: It seems for the 9 blocks in the $\times$ associative table, 5 of them are actually associative. Preliminary observation suggests it is piled with 1 s, thus that might suggest the real axiom violations might be localised to 0 and its inverse after all
There are 8 nonassociative $\times$ cases:
 
user228700
@Secret: Hi :-) Are u familiar with the VSEPR theory?
 
@Kaumudi yup, what's the question?
 
user228700
The formula $AB_3E_2$ corresponds to a trigonal planar shape, yes? ($E$ is the number of lone pairs on the central atom, $A$)
 
user228700
I imagine that the effect of the two lone pairs on either side cancels? Sort of like this:
 
10:07 AM
Yes because lone pair repulsion > bond pair repulsion, thus by VSEPR, the two lone pairs will go to the axial position of the trigonal bipyramid thus the molecule (ignoring lone pairs) will be trigonal planar
 
user228700
 
user228700
So, this is correct. Alright! What about $AB_3E_3$?
 
O wait sorry mistake:
Since lone pairs > bond pairs, to minimise repulsion, lone pairs need to be in the equatorial position (so it has one neighbour at 120 deg, and two neight at 90 deg, compared to 3 neightburs at 90 deg each, thus the molecule is (generally) T shaped
 
user228700
@Secret Ohh. What are the guidelines to figuring this out? Discern which is which..?
 
It is difficult in the most general case, but basic guidlines is you want lone pair-lone pair to be furthest away first, followed by lone pair-bond pair, and then finally bond pair-bond pair
whcih means for each case you want to minimise number of neighbours for each interaction, and also to maximise the angle in that order
 
user228700
10:14 AM
@Secret Right. But in the case of trigonal planar the lone pairs are the farthest they can be from each other! 180°.
 
user228700
Hang on, in T-shaped, what's the angle b/w the lone pairs?
 
you want lone pair-lone pair (lp-lp) and bond pair-lone pair (bp-lp) to be having the fewest neightbours and furthest away possible. To compare between T shape and Trigonal planar, let's make a table:

T shaped:
lp-lp 120 deg x 1
bp-lp (120 deg x 1 & 90 deg x 2) x 2 = 120 deg x 2 & 90 deg x 4
bp-bp 90 deg x 2 & 90 deg x 1 & 90 deg x 1 = 90 deg x 5

Trigonal planar
lp-lp 180 deg x 1
bp-lp (90 deg x 3)x2 = 90 deg x 6
bp-bp (120 deg x 2)x3 = 120 deg x 6
In trigonal planar, yes you minimise lp-lp, at the cost of a big lp-bp. Thus molecules are geneally T shaped for AX_3E_2

The angle for lp in T shaped is 120 deg
bp-lp and lp-lp can compete, but bp-bp is generally much smaller than the other two categories
 
user228700
@Secret Hang on, isn't this T-shaped:
 
yes AX$_3$E$_2$ is generally T shaped as otherwise you have a large bp-lp contribution
 
user228700
 
10:27 AM
yup, that's T shaped
 
user228700
So, in this, bp-bp are 2 (90°) and 1 (180°), right?
 
For the F sticking out on the left, 2 (90°), For the two axial ones, 1 (90°) & 1 (180°) each
 
user228700
Alright, u know what, never mind. I'll just accept this and move on.
 
user228700
Thank you :-)
 
@EmilioPisanty Hahaha nice
Congrats! May the gods (or rather the research group leaders) smile upon you
 
10:31 AM
@Kaumudi This will be T shape no matter what, since all 6 positions have the same number of neightbours and angles (a octahedron)
 
user228700
@Secret This will be T-shaped?! I thought the max. number of positions in T-shaped is 5!
 
I once thought of a question, to which I came up with an answer, but I would still like to hear what you guys think:
according to N3L, every force has a reaction
 
user228700
@DHMO Ah, welcome to The h Bar! :-)
 
but does the "reaction" come with another reaction? like, the reaction of the reaction?
 
@Qmechanic I've seen you around chat a lot (more than usual) lately!
 
user228700
10:34 AM
Everyone wait!
 
@Kaumudi thanks
 
user228700
I (an amateur) am going to attempt to answer @DHMO's question!
 
$AX_3E_3$ (you are not asking for $AX_2E_3$ for that particular message) has 6 electron densities, thus initially you want to arrange them all such that they are all furthest away from each other possible (which will be an octahedron). You then apply the bp-bp < lp-bp < lp-lp ordering. However, no matter where in the octehedron you put the lp, two of them has to end up opposite to minimsie a lp repulsion (the bp-lp repulsion wil be the same regardless because it is 90 away ffrom other electron densities),
 
@DHMO because I am not very satisfied with my answer
@Kaumudi I think we use X instead of B
 
user228700
@DHMO Firstly, I'd like to clear up something; there is no action and RE-action, which is to say that there is no cause-effect relation involved. If I push a table with a force of magnitude $F$, it is incorrect (and often misleading) to think about this force as the action.
 
10:37 AM
@Kaumudi bingo
 
user228700
@DHMO OK, so are u clear about this?
 
@Kaumudi well, I said I have an answer but am not satisfied with my answer
because my answer involved causality
 
user228700
@DHMO What was ur answer?
 
@Kaumudi essentially the same as yours
 
user228700
@DHMO Huh? But that wasn't even an answer :-P
 
10:39 AM
@Kaumudi eh... alright
 
user228700
@DHMO So ur question has been answered..?
 
@Kaumudi well, essentially your answer invoked causality as well
 
user228700
@DHMO That wasn't an answer! I was trying to clear up the misconception (that u may have had) regarding causality! I wasn't invoking it!
 
@Kaumudi alright
you may continue your answer
sorry for the interruption
 
user228700
NP :-)
 
user228700
10:42 AM
Essentially, what u call "the reaction of the reaction" is the action itself. We are talking about only two forces.
 
I see
 
user228700
If I push a table with a force of magnitude $F$, my hand experiences a force (of the same magnitude, $F$), that the table applies on it.
 
is there an explanation of N3L from a quantum level?
 
user228700
@DHMO Ah, OK, if that's what u seek, I'm useless. Although, I doubt if anybody will be able to give u a satisfying explanation.
 
@Kaumudi why?
 
user228700
10:48 AM
@DHMO Never mind that; I may be completely wrong. I just don't feel like an explanation in that sense would be useful at all.
 
@Kaumudi alright
 
user228700
Anyway, @JohnRennie: Can you help this person?! ^
 
3
Q: Quantum explanation of Newton's Third Law of Motion

Mohammad Al-TurkistanyNewton's law states that for every action there is equal and opposite reaction. This law explains how rockets fly in space and accounts for the the majority of the lift action generated by a airplane's wings. Is there a fundamental quantum explanation of the third law of motion?

 
user228700
@DHMO: JR is the King of explanations around here.
 
@Secret thanks
 
10:49 AM
Newton's 3rd law is the consequence of the conservation of momentum
however, Newton's 3rd law is not compatible with relativity without generalising to 4-momentum
The law works fine in the quantum level
@Kaumudi No, ACM is
though JR is king for anything not QFT or maths related
 
user116211
@DHMO Causal relation is implied in the definition of force.
 
@MAFIA36790 why?
 
user228700
@Secret This is what I meant.
 
user228700
@MAFIA36790 :-O EXPLAIN! (Please)
 
user228700
@Secret Uhh. I don't understand this at all :/ Can you please 🙏 show a diagram of this?
 
10:57 AM
Drawing... please wait
 
user116211
There is a paper on this by Jefimenko; couldn't remember the doi though.
 
user228700
@MAFIA36790 Can u please explain how a causal relation is implied in the definition of force..?
 
user116211
@Kaumudi Force is defined to be the cause of acceleration.
 
user228700
@MAFIA36790 Dude, no. I meant this:
 
user228700
25 mins ago, by Kaumudi
@DHMO Firstly, I'd like to clear up something; there is no action and RE-action, which is to say that there is no cause-effect relation involved. If I push a table with a force of magnitude $F$, it is incorrect (and often misleading) to think about this force as the action.
 
user116211
11:03 AM
And thus $\mathbf F= m\mathbf a$ is a causal relation by definition.
 
user228700
^That's correct, yes?
 
user116211
@Kaumudi I agree with this; but this not causality.
 
user116211
Action-reaction is a misnomer.
 
user228700
@MAFIA36790 OK, I may be a little confused regarding the terms.
 
user228700
@MAFIA36790 Yeah, this is all I was looking to clear up.
 
user116211
11:04 AM
It's solely the conservation of linear momentum of an isolated system.
 
user228700
@MAFIA36790 Yes, alright, thanks.
 
user228700
@Secret: Waiting... (Thanks so much! :-})
 
@MAFIA36790 what is causality in physics?
 
user116211
$$\int_{t_1}^{t_2} \mathbf F_{21}\cdot \mathrm dt=- \int_{t_1}^{t_2} \mathbf F_{12}\cdot \mathrm dt$$
 
user116211
It is the conservation law of linear momentum of a system composed of two bodies.
 
user116211
11:09 AM
> In principle, $\mathbf F_{21}$ and $\mathbf F_{12}$ could have quite unrelated values at any particular instant, as long as the above integrals are equal. However, failing any evidence to the contrary, we assume that they are equal and opposite at every instant.
 
@MAFIA36790 I don't understand the second sentence. What does "failing any evidence to the contrary" mean?
The mighty @JohnRennie is here
 
Hi DHMO, what are you asking about?
 
@JohnRennie about N3L
 
Seems less clear cut, as right structure has less bp-bp and lp-lp but more bp-lp
 
11:11 AM
N3L?
 
@JohnRennie Newton's third law
 
Aha! What about it?
 
@JohnRennie How do you explain it in quantum?
I've heard of something called virtual photon
 
Quantum mechanics doesn't explain Newton's third law. The third law is a statement of conservation of momentum, and conservation of momentum is due to a symmetry called shift symmetry.
 
@JohnRennie what about virtual photons?
 
user116211
11:16 AM
> In situations, in which objects influence one object at a distance, Newton's Third Law may cease to apply. For no interaction is transmitted instantaneously, and if the propagation time cannot be ignored in comparison with the time scale of the motion, the concept of instantaneous action and reaction can no longer be used.
 
@DHMO Virtual particles are a mathematical device used when calculating scattering probabilities in quantum field theory. Virtual particles don't really exist - they are just a computational device.
 
@JohnRennie so Hawking radiation is completely fake?
 
@JohnRennie, I thought virtual particles had been proven to exist - their effects are measurable, or something.
 
@DHMO Absolutely not. However the usual explanation that Hawking radiation is due to pairs of virtual particles is just a metaphor and not actually what happens. For more on this see:
12
A: An explanation of Hawking Radiation

John RennieTo answer this we need to talk a bit about how particles are described in quantum field theory. For every type of particle there is an associated quantum field. So for the electron there is an electron field, for the photon there is a photon field, and so on. These quantum fields occupy all of s...

@heather You're probably thinking of the Casimir effect, which has been experimentally confirmed. However it is misleading at best to say this is due to virtual particles. It is due to the way quantum fields behave.
The calculation of the Casimir effect involves virtual particles, but I must emphasise that they are just a computational device.
 
@JohnRennie, ah, so they really are only a calculational device. Huh. Then I guess any further questions along this line would be in the realm of philosophy ("do calculations exist" type questions). Thanks!
 
11:21 AM
@heather read the article by Matt Strassler that I linked above.
@heather in quantum field theory particles are surprisingly elusive objects. In an interacting quantum field theory like the Standard Model they aren't unambiguously defined i.e. you cannot say exactly what particles are present at any moment.
 
user228700
@Secret Ohhh. OK, I think I understand this now. Thanks so much! :-)
 
@heather Which is related to how new particles get created in colliders like the LHC.
 
@JohnRennie, the article is incredibly enlightening - I'm still reading it, but it already has made clearer a whole lot of things. Thanks for clearing up the confusion!
 
user228700
@MAFIA36790 Interesting! Thanks for this! (Although, I don't see how there will be evidence to the contrary at any point).
 
@heather Matt Strassler is a terrific writer and something of a role model to me! There are a lot of really good articles on his blog.
 
11:27 AM
@JohnRennie Could you explain this?
 
@JohnRennie, okay, I'll look around on his blog.
 
@DHMO Have a look at:
11
A: What is the cause for the validity of Newton's third law?

John RennieWe start by noting that force is the rate of change of momentum. Let's suppose you and I are floating in space (so we are the only two interacting bodies) and you're pushing me so I feel a force $F_{me}$, then: $$ F_{me} = \frac{dp_{me}}{dt} $$ where $p_{me}$ is my momentum. But we know that m...

 
@JohnRennie thanks, but why is momentum conserved?
 
@DHMO See:
3
Q: Conservation of Momentum

N.S.JOHNConservation of momentum can be derived during collisions by using Newton's laws of motion. But in other cases, do we simply take it like an axiom ?

 
@JohnRennie thanks, but why is noether's theorem true?
 
11:38 AM
Noether's theorem can be derived from the principle of least action, so Noether's theorem is true because the principle of least action is true.
The next question is why is the principle of least action true?
And the answer is that there is no fundamental reason why the principle of least action must be true,
but when we do experiments we always find that is is true.
So the principle of least action is established by observation.
 
user228700
@Secret: Are u still around..?
 
I have to go now. Back in a few hours ...
 
@JohnRennie I see, thanks
 
Well, there's some kind of "explanation" in QFT in that the only paths with high amplitudes are those where the action is stationary, as otherwise the phase vary too rapidly and average out in destructive interference (As usual, ACM can flesh out the details)
@Kaumudi i am still here
 
user228700
@Secret OK. I have a small question about VSEPR again, do you mind..? (I won't ask u to draw diagrams this times, I promise! :-P)
 
11:42 AM
@DHMO Your profile doesn't say whereabouts you are in the educational system. The principle of least action and Lagrangian mechanics is something you wouldn't learn until the second or third year of a physcs degree.
So don't worry if it all seems a bit confusing :-)
 
@Kaumudi Shoot (ask the question)
 
user228700
@Secret One of the postulates of VSEPR states "A multiple bond is treated as a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair"
 
user228700
What does this mean..?
 
It basically means that assuming the electron density of a multiple bond is localised in the same way as a single bond, thus a bond pair in a multiple bond is treated like a single bond pair
 
user228700
"A single super (bond) pair"?
 
11:47 AM
that means, while there are more than one electron pair that consists it, it acts like a single bond pair
 
user228700
@Secret 👍 Thanks!
 
MAFIA: The + associative results are computed: It seems it is trivially associative. I.e. given a column of elements (a,b,c) in WIP, it will always end up as (a,b,c) no matter what happens for the other two terms and where you put the brackets
i.e. the 9 blocks of the + associative table is literally the column (a,b,c) copied 9 times
Now, given this "absorbing like" property, I am not sure if right distributive law will hold. If it does, then I believe WIP will pass all the test for self consistency
 
@JohnRennie I reveal next to nothing about myself, for obvious reasons.
 
@DHMO you do same always
 

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