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1:30 AM
@skullpatrol Idk I haven't talked to him in a while. He's at princeton though so I bet he's busy.
@SirCumference Hey man hows everything did you graduate?
@ACuriousMind Not surprising what he's done amidst the pandemic considering his massive ego. Maybe STEM people should take more philosophy classes or just not have opinions on matters that are larger than their sphere of knowledge.
It pains me to read the blind optimism some have over technological development. I will admit I was once that naive but it's really more complicated than simply making more technology to do things for us.
How we live our lives shapes the planets future and if we want to live in balance we can't be slaves to our instincts :\
Awareness and empathy is crucial at this point.
 
 
1 hour later…
2:48 AM
2
Q: Determine if a coincident point in a pair of rotated hexagonal lattices is closest to the origin

uhohA pair of hexagonal lattices with one scaled by the square root of a rational number $r = \sqrt{\frac{m}{n}}$ and then rotated will produce a variety of different coincident points. For the first lattice let $$x, y = i+\frac{1}{2}j, \ \frac{\sqrt{3}}{2}j$$ and for the second $$x, y = r\lef...

any crystallographers here? I've just added a bounty
 
3:22 AM
If anyone could take a look at my question, that would be great
0
Q: Galilean Transformations Derivation

DarkRunnerI'm trying to understand the Galilean Transformations, as shown in my book. Here's the situation, first and foremost: Two observers, R (uses Roman Coordinates for its Reference Frame) & G (uses Greek Coordinates for its Reference Frame) are traveling towards each other at a constant velocity $v<<...

@uhoh Any ideas?
@jmac It would be greatly appreciated if anyone could help
 
4:16 AM
@DarkRunner I know practically nothing about Galilean Transformations, and you really shouldn't ping people in the chat asking for help unless you know they will be interested in helping. (see the same guidelines I linked earlier)
 
Yes, I understand. Unfortunately, I've waited patiently the whole day hoping for an answer and I see none. @TheCuriousMind's prediction was wrong, after all.
 
4:55 AM
Why connecting batteries in parallel make the voltage stay constant but the amps add? I understand that in some way the current adds up, I can see that, but I'm a bit confused why the electromotive forces don't add up on the end
Sure, if those were rivers, in the end, their current would add, but why the electromotive voltage force doesn't add as well? Each battery should contribute to pushing electrons?
 
5:28 AM
@JingleBells the positive terminals of the batteries are all connected together so their potentials must all be the same. Likewise the negative temrinals.
 
@JingleBells The analog for a river would be that the pressure at the top and bottom of each stream is the same, and the resistance determines how much water flows.
 
 
1 hour later…
6:59 AM
@JohnRennie sorry I still don't understand
If we imagine that there's a group of people pushing electrons on all positive terminals, then in the "pushing force" should add?
 
@Jim I had already commented. I downvoted a day or so later. In response, I got this comment:
@PM 2 RING This is why the Australian forests were in fire a year ago, because you do not find helpful to think of electrons as little spinning balls. — user36636 3 hours ago
 
7:54 AM
@JingleBells Maybe it will help to think in terms of waterfalls. Let's say that a single battery is like a waterfall that's 1 m wide with a drop of 10 m. Four batteries in series is like a waterfall that's 1 m wide with a drop of 40 m, so the water hits the bottom with more force. But four batteries in parallel is like a waterfall that's 4 m wide with a drop of 10 m, so you get 4 times the volume of water, but the force at the bottom is the same as the original waterfall.
 
@PM2Ring You. are. a. genius.
This was the clearest explanation I've ever gotten about anything.
2
Such a great analogy.
Thank you!
 
No worries. :)
 
 
1 hour later…
9:29 AM
In Peskin's introduction to phi^4 theory the symmetry factor of diagrams is discussed. This is not the case in Fetter & Walecka introduction to diagrammatics for fermionic theories. Is the symmetry factor not important for fermionic theories?
 
9:46 AM
@DarkRunner Our site works asynchronously, there is no promise that a day is enough t get an answer, and especially not on the weekend (many people usually browse the site during work :P). You are not entitled to an answer, and claiming you've "patiently waited the whole day" when you posted the question for the second time here 9 hours after it was posted is an impressive stretch of the notion of a whole day.
If you feel your question is not getting enough attention, offer a bounty on it (you can do that two whole days after the question was asked, which is unequal to 18 hours) instead of bothering random people in chat to look at it.
@B.Brekke The arrows on fermionic lines make these diagrams not have a symmetry factor
 
@ACuriousMind Sounds like I got lucky then. One less thing to calculate
 
At least I think that's the case. It's been a while since I actually calculated any diagrams :P
 
10:38 AM
@PM2Ring I have to say, "Quantum mechanics did wildfires" was not on my bingo card
 
I, personally, am not impressed by that stretch of the notion of a whole day being 9 hours; given an 8 hour work day. But on the other hand, for example , working the whole day, without a lunchtime break would be impressive :P
 
@ACuriousMind Do you believe in determinism?
 
@FakeMod What exactly do you mean by that?
 
::makes popcorn::
 
In the history of science, Laplace's demon was the first published articulation of causal or scientific determinism, by Pierre-Simon Laplace in 1814. According to determinism, if someone (the demon) knows the precise location and momentum of every atom in the universe, their past and future values for any given time are entailed; they can be calculated from the laws of classical mechanics.A desire to confirm or refute Laplace's demon played a vital motivating role in the subsequent development of statistical thermodynamics, the first of several repudiations developed by later generations of...
@skullpatrol ::grabs armor::
 
10:44 AM
@FakeMod Of course I do not think Laplace's demon works - today we know quantum mechanics, and that therefore the proposition to know every position and momentum at once is inconsistent to begin with
But, on the other hand, it also obviously works in situations where the classical approximation is good enough
 
@ACuriousMind Yeah, but what about determinism at smaller time scales. At the time scales where the evolution of the system won't be affected by quantum uncertainty?
@ACuriousMind Yeah, I am only talking about macroscopic fat human like scales.
 
The applicability of quantum mechanics is not about scale as such. The sun - a very macroscopic object - burns only because fusion is allowed by the laws of quantum mechanics - a classical account of fusion knows not about quantum tunneling and could not sustain the sun.
So Laplace's demon cannot predict the sun.
And also Schrödinger's cat is designed to transfer the indeterminacy of the "small" (=radioactive atom) to the "large" (=the cat and poison vial)
Now you can try and save determinism by saying "we know the quantum state of everything" instead of talking about classical position and momentum, but then you run into the measurement problem
E.g. Many Worlds is "determinist" in the sense that the overall wavefunction of its multiverse evolves deterministically, but it cannot remove the indeterminism experienced by observers in one of the branches
 
@ACuriousMind Hmm... That's why I was thinking about small time scales where manifestation of quantum mechanics in the macroscopic world in negligible. The final point which I had to make was how could such a deterministic system be imagined. I can't wrap my head around the idea that knowing everything about the current state will let us uniquely determine the future.
 
hidden variable theories are determinist but necessarily non-local, meaning in the quest to save the notion of "one cause always has the same effect" we have given up the notion that "causes" and "effects" must somehow be spatially related at all.
@FakeMod It depends on what you mean by "the future" :P
 
@ACuriousMind Before I knew anything (except the name) about the Many Worlds interpretation, I thought it was just too fantasized and thus BS. But now it does make some sense.
@ACuriousMind Like, what am I gonna eat today? Or more technically, which particles are gonna enter my mouth today?
 
10:59 AM
@FakeMod short answer: "no."
:-)
 
@FakeMod We can easily imagine Schrödinger's meal dispenser: It gives you one meal if the atom has decayed, and another if the atom has not decayed.
 
Correct me if I am wrong (very likely), but I understand that in QM, when we say something is in the state of superposition of multiple different states, we are just trying to (proportionately) express what the specific state of that "something" could be. But once we measure its state, we get to know the exact state which it is in. So in ky understanding, superposition is just a mathematical tool (equivalent to probability). So when we say I am a in a superposition of two different states[contd]
[contd] we don't really mean that I am currently in both of those states. It's just that we don't know yet what the specific state I am in.
How much mainstream does that sound? :-)
 
That is the notion of classical probability.
 
@ACuriousMind Really, why didn't Schrodinger think of this. Killing cats was so cruel. ;-)
 
I thought that was just the distinction between the various interpretations, is a system actually in multiple states before measurement or are there hidden variables?
 
11:02 AM
The content of Bell's theorem is precisely that such a local realist account of probability cannot explain quantum mechanics
If you want to believe that you really do have a specific state of the superposition and the quantum state just reflects not knowing, you have to be non-local, i.e. believe in spooky action at a distance where measurements on one particle instantaneously change the state of another across arbitrary distances
 
Ah yes
 
It turns out most people like locality more than they like realism, so the party line is that quantum mechanics says the superposition is fundamental in the sense that it does not reflect a lack of knowledge but the actual state of the world
 
@ACuriousMind Ah, yes. BTW, why don't we notice spooky action at distance in classical mechanics. Because the conservation laws are also present in classical mechanics, I do expect any kind of spooky action at a distance to happen.
@ACuriousMind That makes me uncomfortable :-)
 
(there's also plenty of interpretations in between where e.g. quantum states are relative to an observer and the notion of "actual state" doesn't make sense at all)
@FakeMod Why would conservation laws lead to action at a distance?
 
Does asking "Which of the interpretations is correct?" make sense? If yes, then what is its answer.
 
11:07 AM
A conservation law is not something that is imposed from the outside in the sense of "If you take charge way, the universe will magically make it appear somewhere else"
 
I feel like the phrase "which of the interpretations is correct" is almost a self-contradictory question by definition
 
A conservation law is an observation about how the (local) dynamics of a mechanical system work - it is deduced from the dynamics via Noether's theorem, not an additional fact!
@FakeMod What do you mean by an interpretation being "correct"?
There's no 'right' answer to that question, because it depends on your philosophy of knowledge: What do you think knowledge is? What makes a belief "true"? Does it make sense to apply that notion to an interpretation?
 
27 mins ago, by ACuriousMind
@FakeMod What exactly do you mean by that?
 
@ACuriousMind Let's say a stationary bomb explodes in exactly two pieces. I measure the momentum of one of the pieces, then I automatically get to know the momentum of the other piece as well. Why isn't this call spooky action at a distance? Before any measurement, the momenta were in a superposition (were they?).
@ACuriousMind Correct interpretation will give us correct physical results. So in that way, are all of those valid?
 
@FakeMod That's an entangled state. Why would you gaining knowledge that the other piece has a specific momentum be action at a distance? See also physics.stackexchange.com/q/3158/50583
@FakeMod They are physically indistinguishable. Whether that means they are all equally "valid" or "correct" depends, again, on your idea of what that means.
I personally think they are all equally useless.
 
11:15 AM
@ACuriousMind Then, why are people coming up with different interpretations if all the existent ones are good enough to explain stuff?
 
Because they want to figure out how the world really is. They began doing physics because it seemed to explain the nature of reality, and when it handed them a picture of reality they didn't like they started adding bells and whistles to it that made it more palatable.
Humans want stories, but "put these numbers into a mathematical machinery of Hilbert spaces and states with no fixed underlying ontology and you get predictions of probabilities out the other end" simply isn't a good story.
 
@ACuriousMind Lubos' answer made complete sense to me and it goes really well with my intuition as well, so thanks!
 
I have to stress that this is my personal opinion. People with different philosophical ideas about knowledge and reality will tell you very different things.
 
@ACuriousMind I do like the idea of understanding stuff instead of just mathematically modelling it, but understanding it easily or intuitively (is nice but) isn't really necessary thing, IMO.
@ACuriousMind Anyways, thanks for your time. Gotta learn something new to refresh my mind from the same old stuff going on for 2 years :-)
 
11:47 AM
When study for the GR exam I sat, I learned I'm really weak at coordinates, coord transformations, line elements, the basic coordinate stuff needed, what area of math would specifically coordinates fall under so I can get more familiar? Also are there any particularly highly regarded books/lectures on that topic I should know about?#
Fairly broad question I know, but I wasn't able to find an overwhelming amount of resources specifically targeting coordinates in preparation for gr
 
"specifically coordinates" is a bit narrow. Try intro texts to differential geometry.
 
1
Q: Book on coordinate transformations

AjoyI am looking for a book that covers various coordinate systems in 3 dimensions, various methods of representing rotations and other transformations like rotation matrices and quarternions, including algorithms for conversions between various coordinate systems and representations of transformati...

@Charlie might help 👆
 
That answer recommends looking into engineering text which I hadn't thought about doing, I might look into that thanks
I'm working my way through some diff geometry notes atm, hopefully if I see enough of them I can work on it slowly
thanks guys
 
@skullpatrol lol :)
 
:-)
@Charlie that sounds like a review of basic applications may be called for
 
11:59 AM
basic applications?
 
engineering
 
ah I see, yeah I will definitely get to that
I am missing a lot of undergraduate training in physics, I used to practice the dark arts (chemistry) so my mathematical knowledge is filled with some pretty substantial holes
 
@Charlie It's the other way around with me. :-)
Though it's not undergrad but HS level Chem issues with me.
 
In a lot of ways chemistry is actually harder to learn than physics when you're below postgraduate level, there is no mathematical rigour, a lot of inspired leaps of logic and taking things to be true for granted
 
@Charlie haha true!
 
12:10 PM
Obviously the maths is a lot simpler, calculus is about as far as you go in undergad chemistry but still, there is very little maths training in undergrad chemistry, I had one module in 3 years and it was in first year and we basically did HS level calculus and some error analysis (which no one learned properly because they never knocked off marks for error analysis in our lab reports)
If the satisfaction of being able to derive things from relatively fundamental concepts in physics is something you appreciate you will have a daily aneurysm studying chemistry, fig.1: me
 
Yup, even in physical chemistry derivations are too time consuming.
While biochemistry is a pure test of memory.
 
I loved the biochemistry modules we did in my degree, the module was a joint module between us and the biologists
so the biology part had to be easy enough that we could do it, and the chemistry part had to be easy enough that they could do it, so overall the module was much easier than all the other ones we did
 
physics and engineering are sorta related like that
except engineering physics
they take away the easy stuff
@Charlie filling in those holes is going to be time consuming, pal
 
12:27 PM
oh for sure
I sat a GR exam 2 days ago but I can't do a lot of basic electrodynamics
everything is mixed up and in the wrong order for me, but I'm happy to go back and learn it all
fortunately the availability of resources in physics is so, so much better than they are for chemistry
@skullpatrol are you a maths major or something?
 
yeah, I found chemistry has a "I have a secret" approach to resources and it is your job, as a student , to find it
and that^ is only part 1, next you gotta memorize the important results for the exam
 
1:29 PM
Hello, can I solved a question but I'm not sure about my result. Can someone check please?
Question: Write the most common $x(t)$ solution for a driven harmonic oscillator given by $\ddot{x}+ax=\sin(bt)$ motion equation, $a,b\in R^+$
My result: $x(t)=\frac{1}{a-b^2}\sin(bt)$
I found the solution by just rewriting the general $x(t)$ equation for driven harmonic oscillators.
General motion equation for driven harmonic oscillators given by $\ddot{x}+\frac{1}{\tau}\dot{x}+\omega_0^2x=A\sin(\omega t)$
So, $\omega=b, \omega_0^2=a, A=1=\frac{F_0}{M}, \frac{1}{\tau}=0$
And the solution of $x(t)$ for $\ddot{x}+\frac{1}{\tau}\dot{x}+\omega_0^2x=A\sin(\omega t)$ is $x(t)\frac{A}{\sqrt{(\omega_0^2-\omega^2)^2+(\frac{\omega}{\tau})^2}}\sin[\omega{t}-\arctan(\frac{\omega/\tau}{\omega_0^2-\omega^2})]$
When I put the given variables into the general solution of $x(t)$, I find my result.
Am I correct?
 
1:57 PM
@ICCQBE If you just insert your solution to the diff. equation you can check your result. Seems like you have a correct solution, but lacking a integration constant?
 
Let me check please. Thanks for the correction.
 
2:27 PM
I tried to solve the equation manually but there were at least 10 terms, I couldn't handle it but I checked the solution using symbolab and it show that the solution is right.
Thanks for the adive @B.Brekke
 
 
1 hour later…
3:31 PM
Hey @ACuriousMind I thought about the stuff we discussed. And I subsequently read the first few pages of Griffiths QM. He talks about the orthodox position which highly relies on the Copenhagen interpretation. This class of people believe that the object was not localized at a specific place before they took the measurement. It was only while taking the measurement that the wavefunction collapsed. [Contd.]
[contd.] One peculiar/unintuitive result of this interpretation is the famous, "Does the moon disappear when I am not looking at it?". Now, my doubt is that when we look at the moon, we are looking at the light which was emitted a few milliseconds ago. So when we measure something visually, we measure the state of that thing before the particular instant of measuring. So this means that if the moon really disappears we should get a few milliseconds of black sky in our observations [contd.]
[contd.] before the moon returns. Also, since information cannot travel FTL, thus the collapse of the moon's wavefunction should also happen a few milliseconds after our start of measurements. But in practice, AFAIK, we don't see a momentary black siy before seeing the moon, so this orthodox approach must be wrong (very less likely) or I am misinteroreting something (very likely). Please find the fallacy in my arguments.
Thanks!
 
As a sidecomment, the fact that the wavefunction of the moon hasn't collapsed before you look at it doesn't mean the moon disappears out of existence
 
@FakeMod The Copenhagen interpretation does not predict that things "disappear" when you are not measuring them
It asserts that the wavefunction spreads undisturbed when you're not measuring things, and then suddenly collapses when you do
 
@ACuriousMind But it does predict that stuff is spread out and probably fuzzy like orbitals when you're not measuring it. Doesn't it?
 
So when no one is looking at the moon, the moon's wavefunction spreads (with finite speed, in accordance with its Schrödinger equation).
 
@ACuriousMind yeah, exactly. So shouldn't we initially see a blurry moon all acroos the nightsky?
 
3:41 PM
All the while, it's still sending out photons. So you get a giant entangled state of the schematic form $\lvert m_1\rangle \lvert p_1\rangle + \lvert m_2\rangle \lvert p_2\rangle + \dots$, where $m_i$ are the possible positions of the moon and $p_i$ are the possible photons emitted from that position
when you look at the moon, the whole entangled system "moon + emitted photons" collapses into one of the $\lvert m_i\rangle \lvert p_i\rangle$, i.e. one particular position of the moon with the particular set of photons consistent with that position
it's the same as the two entangled particles from before, just on a larger scale.
 
@ACuriousMind But how can that happen instantaneously?
 
@FakeMod Again, it's the same kind of not-really-instantaneous-event as in the case of the two entangled particles with spin, and when you measure one of them, the whole entangled state of both of them collapses.
"How" this happens is the measurement problem.
 
@ACuriousMind So, I suppose it boils down to spooky action at a distance?
 
You can just say "action at a distance", no need to say 'spooky' everytime :P
 
@ACuriousMind :D
 
3:48 PM
And whether or not it is 'action at a distance' depends on whether or not you think the wave function is a "real thing that exists", whatever that means
"The Copenhagen interpretation" is not a formally defined interpretation, it's more like folklore, so it doesn't have a well-defined stance on this
What is true regardless of interpretation is that you cannot use this to send information
In what meaningful way an "action at a distance" that cannot transmit information to anyone being at that distance is still an action is left for the reader to decide.
 
@ACuriousMind yeah, I read something like that on Bell's theorem Wikipedia page...
 
4:07 PM
@ACuriousMind Do you know about the Galilean Transforms? If so, can you help me?
 
@DarkRunner You've been told at least twice already to not ping random people with questions.
I'm not interested in helping you under these circumstances.
 
 
2 hours later…
6:19 PM
ouch
 
Yeah, ACM was quite cruel... :P
 
what does the spacetime manifold which presents a galaxy converting its whole mass into gravitational radiation in one second look like?
 
@FakeMod Was I? What is cruel about not obliging someone who repeatedly ignores the stuff you tell them?
@CaptainBohemian ...why would any galaxy do that?
There's no magic button you can press to "convert mass into gravitational radiation", it happens under very specific circumstances (like merge events of stars or black holes)
 
@ACuriousMind He/She was a newcomer desperate for help. You are not at all obliged to help them however tellng them off was a bit rude, IMO. But, yeah, let's end it here. We ain't gonna get anywhere by discussing this.
@DarkRunner You could try posting your question in the Problem Solving Strategies. You might find your answer there.
 
yes, i was about to suggest the same thing
 
6:31 PM
@FakeMod I'm asking because I genuinely don't want to be cruel. Firm, yes. Cruel, no.
 
"cruelty" would be kicking him off the chat
 
@ACuriousMind I don't know if any galaxy can actually do that, but I read from an abstract that they have developed a code which can stimulate that kind of spacetime.
 
0
Q: Answer posted with extremely little potential for acceptance

EdouardDoes an accepted edit to a question allow the subsequent withdrawal of an earlier answer (which had been posted to it) to be accomplished without any resulting potential for limiting or reducing the answerer's ability to post further answers? (If I would've read the biographical info offered by t...

 
@ACuriousMind It felt a bit weird since I just pinged you a few minutes befor DarkRunner did so. Yes, I know you and I was continuing an old discussion. However, it might strike someone as unfriendly that the regular folks can ping whomever they want but the unknown folks can't do so without being reprimanded.
Hey @BioPhysicist In your recent answer (which you deleted), the case of an equipotential plane parallel to two oppositely charged infinite planes is a valid example of an equipotential surface across which the field is zero everywhere
Hey people, I really think that people think that there's something wrong with this answer of mine. Could anyone please be kind enough to tell me what exactly is wrong about my answer?
 
6:47 PM
@FakeMod When you pinged me the first time, I was startled by the sound produced by your ping. At that time, I focused on doing something else on computer while just leaving this chat window open.
 
@FakeMod Yep, I deleted my answer for a reason. If I wanted feedback on it I would have kept it up :)
 
@BioPhysicist alright, no worries :-)
Do you have a reference for this claim? E.g. two straight planes intersecting is not conventionally called a "surface" in most contexts. See Emilio's answer to a similar question for how to think about intersecting surfaces formally. — ACuriousMind ♦ 15 mins ago
@ACuriousMind FWIW, I never talked about intersecting straight planes in my answer...
 
7:04 PM
@FakeMod Note the "e.g.", I'm just giving an example of two intersecting surfaces. You're claiming that the two intersecting surfaces are considered "one equipotential surface". I'm saying that's not what "surface" means.
 
@ACuriousMind Oh, I get it now. Anyways, I agree with the dupe that Biophysicist linked to, so should I delete my answer?
 
To wit, "surface" usually means a (sufficiently regular) 2d submanifold. Two surfaces intersecting is not a surface (because it is not a manifold at the points of intersection).
 
@ACuriousMind yeah, it isn't...
 
@ACuriousMind What would be a better word then that could be used more generally?
 
youtube.com/… how is this so good
 
7:11 PM
Hey, I got some MathJax rendering issues. Can anyone help?
This is how it looks while editing.
This is how it looks to the reader ^
What should I do?
 
@BioPhysicist I don't think there's any more specific word than "subspace" once you leave the realm of manifolds. Usually you want manifolds in physics because you want to do some sort of differential geometry on your surfaces/volumes/whatever
You can model spaces like two surfaces intersecting as stratified spaces but I haven't really seen that in use outside of physics specifically caring about singularities (and even then they usually prefer talking about things like orbifolds)
@FakeMod Restart your browser (and link the answer next time :P) I had a look at physics.stackexchange.com/a/557770/50583 and it looks like the first version in both cases to me.
 
@ACuriousMind That's nice :-) I am on a tab, so I expect such glitches to occur, but as long as it looks fine to most of the people, I suppose thats good enough.
Yo! Restarting the browser worked. Danke ACM!
Wow, "danke ACM" could be a really nice silly pun :D
 
7:31 PM
...where's the pun?
 
I feel like I'm getting deja vu
 
Yeah, not the first time I was confused about a "pun". I'm beginning to suspect @FakeMod uses 'pun' synonymously with 'joke' :P
 
:P
 
@FakeMod Thanks for the help.
 
7:52 PM
@ACuriousMind "dank ACM" and "danke ACM" 😶
 
8:45 PM
Hello
I have question that is related to force, linear and angular momentum
(Sorry for the bad paint skills, I would like to learn how to draw scientific models or figures)
Here, as you can see there are a rod and ball.
The rod with mass $M$ and length $L$ lies on a frictionless horizontal plane. A ball with mass $M$ and speed $v_0$ approaching to the rod directly.
The question is, what point of the rod that is above the center of mass of the rod must ball hit so both ball and the center of mass of the rod would have speed $v$?
I made 2 solutions for that question. First one, using linear and angular momentum. Second one, using torque, angular momentum and impulse.
My first solution: Since there is no external forces acting on system, the linear momentum conserves. So we can use conservation of linear momentum.
$\vec{p}_i=\vec{p}_f$
$Mv_0(\hat{x})=MV(\hat{x})+MV(\hat{x})$
$v_0=2v \Rightarrow v=\frac{v_0}{2}$
So I found the speeds of both ball and the CM of the rod.
Then now, I can use conservation of angular momentum.
$\vec{J}_i=\vec{J}_f$
Lets say that the vertical distance between ball and the center of mass of the rod is $d$.
$dMv_0(-\hat{z})=dMv(-\hat{z})+I\vec{\omega}$
$\vec{\omega}=\omega(-\hat{z})$
$dMv_0=dMv+I\omega$
$v=\frac{v_o}{2}, I=\frac{1}{12}ML^2$
$dMv_0=\frac{1}{2}dMv_o+\frac{1}{12}ML^2\omega$
$\frac{1}{2}dMv_0=\frac{ML^2\omega}{12}$
So, here I found $h=\frac{L^2\omega}{6v_0}$
 
9:06 PM
@ICCQBE Hi, it seems you know some Special Relativity. Can you helpl me with my question please?
 
Yeah, sure but I'm not really advanced in that topic, but I can try to help.
 
Thanks. Here it is:
Let me edit this picture
 
Addition to my question: I found the $\omega=\frac{6hv_0}{L^2}$, when I put this into $d$, it gives me $d$ so I can not go any further than my result for $d$. If you can check my solution, I would be very happy.
@DarkRunner Alright, I'm very excited to see your question :P
 
I've circled in red what I don't understand
Whoops wrong image
Finally
So here's the idea:
Two observers, R (uses Roman Coordinates for its Reference Frame) & G (uses Greek Coordinates) are traveling towards each other at a constant velocity 𝑣<<1.
@ICCQBE They synchronise their clocks so they pass each other at Event O and 𝜏=𝑡=0
The graph above shows their worldlines
The problem is that I don't understand why those parallel lines (circled in red) are there
@ICCQBE If you could help, that would be great
 
Hmm, I'm looking at the diagrams and trying to understand.
 
9:15 PM
Maybe I can explain a bit:
 
I have never seen that Greek and Roman frames of reference.
 
The worldline of R in the Reference Frame of G is linear because it's moving at a constant velocity. The worldline of G in the Reference Frame of G is $z=0$, because you're not moving in your own reference frame
@ICCQBE Yeah, it looks surreal
 
Hmm, i'm trying to understand but this question looks really out of my level.
 
@ICCQBE I've already asked on Physics SE, but nobody's responded. Any ideas?
 
It seems like it contains time as a paramater, i mean it contains space-time, i'm not sure.
Can I learn the name of the book? I want to take a look at it If i can find online copy of it.
 
9:25 PM
Yes, here
But I don't think there's an online copy;
 
Yeah I couldnt find an online copy.
I wish I could help you but I don't have any idea about the content too :(
 
Thanks for trying
 

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