« first day (1985 days earlier)      last day (3242 days later) » 

15:00
Exponential
or sine and cosine
since it's an image from wikipedia I assume the solution is right underneath
Analytically, plug in the ansat exp($\lambda$t) into the ODE. Then cancel the exp terms, this left you with a quadratic equation (the characteristic equation). Then solve for $\lambda$ to get the general solution (+ or - depend on which one won't cause the terms to to blwo up to infinity)

This equation is however a standard undamped harmonic motion, thus most of us will just memorise the solution anyway
some interesting stuff about moment of inertia
https://socratic.org/questions/can-moment-of-inertia-be-negative
hehe
anyone knows about universe expansion phenomena? (because i got a question)
15:08
just ask i'm sure someone will answer
Okay..
when universe is expanding light gets redshifted, right?
but why does it get redshifted i mean, universe is expanding it means some extra part is getting added to our universe or our universe is like a wrinkled sheet which we stretch and the area increases?, does my question make sense?
Doesn't change the fact that objects are moving away from each other
pardon Obliv, but i didn't get you!
Well, if your question is why does redshift occur in the expansion of the universe, the simple answer is that objects are moving away from each other.
@ramsay The standard analogy is the surface of a balloon that's being blown up.
15:19
hey @barrycarter
@Obliv Word!
@barrycarter that analogy isn't that good though, is it? Since there's no exact 'center' to the expansion and if you deflated a balloon you would get to a center.
@barrycarter yup, that answers my question, but if i consider it to be balloon it means distance between sun and the earth increases, right? but no such thing happens, why?
@Obliv @ramsay OK, consider the surface of a sphere whose radius is increasing. Probably better than the balloon analogy anyway.
@ramsay The rate of expansion is very slow, so we don't notice even for nearby stars, much less for the Sun (which I suppose is technically a nearby star too)
@barrycarter if this is the analogy it means light should not redshift
15:23
@ramsay i think the rate of expansion of the universe is around 70 kilometers per second per megaparsec
@ramsay Well, no, if light had to travel on the surface of the sphere, it would redshift.
so since the earth is only like 8 light minutes away from earth it shouldn't really be felt locally
@Obliv Your units are hertz, wow :P
@ramsay Are you studying physics, or concerned about its accuracy as relates to other theories of the universe?
@barrycarter i don't think light will redshift if sphere increases it's radius,here sphere means like we keep on coating a sphere to increase size but not like balloon
@barrycarter physics+astro
@ramsay Well, pretend the sphere is being inflated. As it's inflated, the distance between any two points increases, so light, traveling on the sphere's surface, must be redshifted.
@ramsay OK. If you have epistemological concerns about physics, there's another way to go with it.
15:28
@barrycarter hmm, but it implies distance between galaxies should increase, but no such thing happens
@ramsay I think it does, actually, Can you source that it doesn't?
@ramsay what do you mean by coating? If a sphere is increasing in radius, surely the points on the surface are getting further from each other?
@barrycarter but i want to specialize in GR+QM
@ramsay Oh good, two subjects I know nothing about.
haha, but i think you were chatting with someone related to spacetime, right?
and this means you are know GR
15:32
@ramsay Damn chat logs :P Yes, the great @JohnRennie was explaining the metric of spacetime to me.
My interest, however, was in solving a constant acceleration SR problem.
As it turns out, the invariance of the line interval wasn't helpful to me, although I'm sure it could be if I knew how to use it. I instead turned to a discrete approach.
well, i have not started relativity or QM so far only the things i know is very very basic things
You probably still know more than me. I still believe God does not play dice with the universe, and that General Relativity is a guy who works for NASA.
i love "God does not play dice with the universe."
15:36
I believe QM is a result of bad methodology.
My views are not mainstream.
but atleast you know spacetime, i don't even know that
I'm sure you have some understanding of spacetime. There's a good chance you live in it.
but that understanding i very very $\to \infty$ basic
@ramsay Well, if you're just starting your studies, nothing wrong with that.
is your age > 18 or 17 ?
15:42
@ramsay What did you have in mind? ;)
$I = \frac{\tau}{\alpha} = \frac{{N}~{m}}{\frac{rad}{s^2}} = \frac{kg~m^2}{rad}$ Why is moment of inertia divided by radians? I get that it's a dimensionless unit but does it mean anything?
@Obliv It just means that you use radians to measure the angular acceleration.
I think you lost a time unit there
No wait, never mind
So your moment of inertia is constant when you're accelerating?
oh okay
@Obliv If you use, for example, degrees, you get a different numerical answer. Of course, if you consistently use degrees in all formulas, it would also work out fine in principle (but I don't recommend this).
15:49
All hail @JohnRennie
@barrycarter that's what Einstein thought too, and it consigned him to decades of irrelevance to modern physics.
@JohnRennie Yeah, but I'm smarter than that guy.
okay since $\theta$ is a ratio of lengths. Differentiating the lengths with respect to time and getting acceleration doesn't change the fact that it's a ratio
@barrycarter wow, that's two of us
@JohnRennie But seriously... I believe that the detectors of an event themselves influence where the particle goes. I know very little about this, and one day hope to casually learn more, but, for the moment, that's my "theory".
@Obliv That may have been in reply to me, but are you coming up with constant inertia for something that spins faster and faster?
15:53
no that was in reply to @MarkMitchison . No I'm just coming up with moment of inertia with relation to force and acceleration
doesn't have to be constant force or acceleration
(but the moment of inertia is constant for an object with angular acceleration)
@barrycarter that is the discovery of QM hot water...of course the detectors influence the subsequent behavior of a particle
@Obliv OK, but your rate of spin can increase, right? radians/second/second ?
yea
@yuggib But I believe they influence the prior behavior. But I want to learn more about that record-and-erase thing.
@Obliv I'd think the inertia would increase with time in that case.
@MarkMitchison It is?
15:54
this is for a rigid body so inertia can't change
@barrycarter you should look up the definitions en.wikipedia.org/wiki/Moment_of_inertia
OK, I'm probably wrong
it's a bad name, admittedly
it doesn't mean inertia
(in the modern sense)
wait what @MarkMitchison what if the objects move closer or further from the axis of rotation? Surely it changes regardless of angular acceleration?
@MarkMitchison Oh, duh. For an object to maintain constant angular velocity, you don't need a force.
15:55
it's the same way that the mass of an accelerating object doesn't increase
@barrycarter ... what about that force we talked about on friday ._.
@Obliv I was just talking about rigid bodies
oh okay
@Obliv I have no memory beyond the last few moments :) Remind me?
I always forget that force is a timeless unit.
the force required to keep the object from flying off tangentially. If you assume no tangential velocity then okay I get that
15:57
@Obliv Oh crap, I probably just confused you with that. That's just the electrical force that holds solid objects together.
but that would mean being on the axis of rotation
yes but that's still a force
If you hold a pencil from one end, the other end doesn't fall to the ground due to gravity, because the electrical force holding the pencil together is stronger.
doesn't each person in a spinning amusement park ride have an angular velocity? or is it just the ride as a whole that has one?
Technically, rotating an object creates stress or strain (ask engineer @0celo7) that weakens the material.
@Obliv They do, but they're probably holding on to the handlebars, or there's sufficient friction that they don't slide off.
Oh, I thought you said merry-go-round.
If the centrifugal force is strong enough it can pin them to the side of the ride... if it wasn't for the normal force, they'd go flying off across the park (which would be kind of cool)
perhaps there is no force related to the constant angular velocity that they feel, but in order for them to even feel a constant angular velocity they must not fly off and that requires something to hold them on, right?
16:04
@Obliv Correct.
They feel a "centrifugal" force because they "want" to go in a straight line
All hail @JohnDuffield
@JohnRennie : or maybe it consigned modern physics to decades or irrelevance.
In the record-then-erase experiment, how long can you wait between recording and erasing?
@barrycarter the moment of inertia is different for different points of an object, right?
@Obliv Do you mean the actual inertia? As someone noted earlier, the moment of inertia is for the entire object.
@barrycarter I've not seen a mathematical proof of that
So I'm not convinced.
I require rigor
@JohnRennie You around
16:17
@0celo7 You're an engineer. You require a calculator and some note paper.
I learned a very interesting fact about geodesics yesterday
@barrycarter but isn't there a different torque being applied based on where the object is?
@Obliv Correction: do you mean the angular velocity?
@0celo7 well, I'm kind of long and thin, but yes I'm here
SHIT I left my book in the office
@JohnRennie So
16:18
(muttering) square candies that look round
@Obliv The torque is normally applied to the outer edge of the disc, correct?
yes or sphere or box etc
Let $M$ be a Riemannian manifold and $\gamma$ a closed geodesic on $M$. Then there is a curve homtopic to $\gamma$ with a smaller length than $\gamma$.
@Obliv So when you say torque being applied to some other point, what do you mean?
like isn't torque $r \times F$ so if the force is the same to each point but the radius of the rotation differs, shouldn't there be a different moment of inertia?
@0celo7 homotopic means the two curves can be continuously transformed into each other?
16:21
@JohnRennie Yes. (Actually you can show the transformation is smooth if it's small.)
@0celo7 Isn't that just an issue of scaling? Find one homotopic curve and multiply be epsilon?
I guess that's obvious if M looks like R^4 in the region containing the curve, or am I missing something
@JohnRennie Obvious is not a proof!
@Obliv You mean the implicit torque generated by the surrounding molecules so the object doesn't fall to pieces?
@barrycarter ...what?
16:22
Can't all closed curves be shrunk to a point in R^4?
yes
@0celo7 The length of your homotopic curve can be scaled down by multiplication, no?
@JohnRennie Yes, but this works for any manifold with positive curvature.
Oh, I forgot that condition.
It's in general not true for manifolds with negative curvature!
Actually, do manifolds with negative curvature even admit closed geodesics?
What do you mean by curvature? Scalar curvature?
@barrycarter What is "multiplication"
16:24
@Obliv I feel really bad that I got you started on this. It's something most people take for granted: solid objects remain solid under reasonable conditions. But yes, in Newtonian mechanics, everything moves in a straight line unless a force is applied. So, if something is NOT moving in a straight line, there is a force present. In general relativity, they just make up stuff about space being curved :)
for ex: @barrycarter if you try to rotate a cylinder about an axis of rotation at one end of the cylinder, its moment of inertia would be higher at the edges, right? since $I = \frac{\tau}{\alpha} = \frac{r \times F}{\alpha}$
@0celo7 Your homotopic function maps to R^n yes?
@JohnRennie I think it's that the sectional curvatures are all positive.
I haven't worked out the full proof.
assuming the same force is used
@barrycarter What?
16:26
@Obliv Well, yes, but the molecules work that out by themselves... give it torque at one point and the torque "spreads out" as needed.
wait.. the angular acceleration would not be the same would it?
@0celo7 I thought you were mapping to R^n, there's a good chance I have no idea what I'm talking about.
@Obliv Well, yes, because the angular speed must remain the same for the entire object.
Not the physical/distance speed, just the angular speed.
@barrycarter maybe...
@JohnRennie If you've never heard of sectional curvature
@0celo7 And if you have a curve in ... did you mean about my not knowing what I'm talking about?
that's understandable
16:27
that's strange. I don't know how the torque could be the same for each point. I need to see some proof of that
@barrycarter Yes.
I have a curve on a manifold
@Obliv Sigh. Are you just starting in physics? I'm worried that I've broken you.
It's an embedded circle where the image is a geodesic
And there is a continuous deformation of this curve that is guaranteed to have a smaller length
@0celo7 And can you repeat that process with the new curve?
@barrycarter No, it won't be a geodesic (in general)
16:30
I still have no idea what I'm talking about, I'm just inserting random sentences that attempt to sound intelligent.
This is interesting on manifolds which are not simply connected.
@0celo7 sexual curvature? Sounds fun :-)
I wonder what a good positive-curvature non-simply connected manifold is.
OK I'll go away and Google sectional curvature ...
@JohnRennie No, sectional, you dirty old man.
16:31
In Riemannian geometry, the sectional curvature is one of the ways to describe the curvature of Riemannian manifolds. The sectional curvature K(σp) depends on a two-dimensional plane σp in the tangent space at p. It is the Gaussian curvature of the surface which has the plane σp as a tangent plane at p, obtained from geodesics which start at p in the directions of σp (in other words, the image of σp under the exponential map at p). The sectional curvature is a smooth real-valued function on the 2-Grassmannian bundle over the manifold. The sectional curvature determines the curvature tensor completely...
I think sexual curvature sounds more entertaining though ...
@ramsay asked me if I was over 18 earlier... but never followed up. The heartbreaker!
@JohnRennie Have you been reading about breasts again?
I want a hands-on experience with those
There are questions in mathematica.stackexchange.com re how to model a human body.
Screw reading about those
16:35
@0celo7 Toothless girl not giving you any?
@barrycarter Nope
@0celo7 Pro tip: if toothless girl also has an Adam's apple, might be time to seek out another.
@0celo7 Now, where did you get that joke from...?
@ACuriousMind Don't know, is it funny?
Mar 5 at 11:48, by ACuriousMind
@HariPrasad Most students will prefer to have more...hands-on experience with the subject
16:36
Now comes the "oral exam" jokes...
@barrycarter I can guarantee that's not the case
@0celo7 Oh, I forgot. At her age, she may not have breasts.
Well, I'm sure she does? Don't all people have one?
And to think that we used to discuss physics in the chat room
@barrycarter Sheeeeeiiiiiiit
#rekt
16:37
@JohnRennie Liar! That only happened like once.
@ACuriousMind I can't remember to go on a date with my gf and you expect me to remember a shitty joke you made?
I have a serious brain issue
@0celo7 In order to have a brain issue, you must ... nah, never mind.
Can't you and 0celo7 get back to discussing manifolds. I learned something by following that conversation.
I like the manifolds of a BBW?
Open balls, hur, hur
16:38
@0celo7 ...yeah, I do :P
@JohnRennie Your method was not useful in helping me find a formula for constant acceleration, so I'm using a discrete approach instead.
@JohnRennie Geodesic balls are sexy.
"I'd like to open her sets"
I told you the invariance of ds^2 wasn't useful for describing accelerated motion, and I told you where to find a beautifully clear derivation of it. What more do you ask of me?
"If I complement your closed sets, will you open them for me?"
16:39
@barrycarter Other way around.
@JohnRennie Then what the hell was ds^2 all about? You said that would lead me to infinite understanding.
@0celo7 You're right, fixed.
That's the basic symmetry that underlies relativity, but as a principle it's too basic to be useful in many circumstances.
okay @barrycarter if what you're saying is true, that moment of inertia is constant for a rigid body rotating around an axis, then if you applied the same torque to each point in the body you would get the same angular acceleration. This is true. What I was confused about was that if you applied the same force at each point in the object you would get a different angular acceleration (since radius changes and $\tau = r \times F$) and consequently a different moment of inertia.
If you were trying to prove Fermat's last theorem would you start from Peano arithmatic?
@barrycarter wait that last part..
16:43
@JohnRennie Well thanks a lot Mr "understand ds and you understand everything"
but obviously for a rigid object the particles can't be rotating at different angular velocities so the torque has to even out somehow, that part I guess I won't be able to know.
@barrycarter haha
@Obliv I think what you said is correct. Essentially, consider a bunch of M&Ms in a disk-like circle. If you push on one of the M&Ms, what happens is different from what happens with a solid disk.
@ramsay OK, so what was all that about?
@JohnRennie Personally, I probably would. But, in this case, I thought you were giving me the magic key to relativity.
@barrycarter do you have insight on what happens to a solid disk? Is it newton's 3rd law on steroids?
@Obliv Well, the question to ask is: how is a solid disk different from a bunch of particles that just happen to be close to each other?
16:46
oh the EM forces between them that keep them connected. okay I understand what you meant earlier now
@Obliv Remember, when you apply "torque" to a liquid, different parts can spin at different rates (sort of like a whirlpool).
if you are over 18 and know GR then it is normal, but if you are less than 18 and know GR it means you are super nerd!
theory by our school students and a total waste :P
@Obliv Yes. The force that holds solids together insures enough "torque" is applied to each particle to keep it in angular rotation.
@ramsay Oh, and here I thought tit involved breasts.
@ramsay No, I'm old. 55.
@JohnRennie Of course.
I start GR from ZFC.
It's the only way.
Otherwise you're not guaranteed to be correct.
@0celo7 But you can't, physics involves the real world, ZFC doesn't.
16:48
Guys
haha! but i don't think any of us is a girl!
and you are super nerd
I'm feeling a bit light regarding QFT books
@barrycarter The mathematical theory of GR, for fuck's sake
You need additional postulates to model the real world.
@barrycarter Oh God, here we go again
16:48
I only have Peskin and Weinberg for basic QFT
Should I get Zee
Zee is terrible @Slereah
@ramsay Absolutely. Why are there no girls in here? I demand the mods do something about this.
could someone suggest a strategy for this? ctrlv.in/736940
How would you know
@0celo7 Yes, but you can't go from ZFC to GR without some observational postulates.
16:49
You don't even believe in QFT
@Slereah I own it.
@barrycarter but sometimes my sis uses my account =, so i will be girl sometimes
@barrycarter How do you know there aren't?
@barrycarter GR is Lorentzian geometry + a PDE.
That's it.
@ramsay How old is she? Although I feel creepy asking. And does she ever come to this room?
@ACuriousMind Good point. @Slereah's av sort of looks female.
16:50
@JoeStavitsky the charge on all the capacitors has to be the same, so the voltage across each capacitor is inversely proportional to C.
@0celo7 Yes, but you still need the postulates of relativity (light moves at a constant speed, etc). ZFC does not have those.
@barrycarter I've had men, women and skulls as avatars, that picture tells you nothing
@JohnRennie, whch c? C_t or C_n?
@ACuriousMind You don't mean... people LIE on the Internet? :O
@barrycarter Light moves at a constant speed is physics, it does not matter for the math.
16:51
@0celo7 But you need that fact to derive a mathematical theory.
@barrycarter yeah she comes to this room! if someone asks question about waves with my account she is my sis, and she is 14
@ramsay Way to kill the mood, dude.
@ramsay How old are YOU anyway?
@barrycarter I postulate the mathematical theory
@barrycarter me is 15!
16:52
@0celo7 OK, I see what you're saying. You're saying of all the possible mathematical theories than can exist, one is guaranteed to model GR.
@barrycarter What?
@ramsay Wow. Really not a great day for flirting with people.
@ACuriousMind Women don't do physics
@barrycarter haha!
@0celo7 Just what I said. You're saying: of all the theories derivable from ZFC, GR is one of them.
16:53
Well yes
@JoeStavitsky what are C_t and C_n? If the charge is Q then the voltage across the 10uF capacitor is V_1 = Q/10e-6, the voltage across the 25uF capacitor is V_2 = Q/25e-6 and so on. Add them all up and equate them to the total voltage.
It is
@0celo7 are you a female?
@0celo7 In other words, all reality can be described by some mathematical theory that starts from ZFC, and possible even ZF.
@ramsay We're not sure what he is.
Obviously not since ZFC isn't complete
16:54
@ramsay No.
Wait, define "female"
@Slereah But it doesn't have to be to describe the real world.
With the PC police around I'm not sure what that means
Sexist!
Are you sure
The member of the species that has the larger zygote.
If you have a countable number of statements, then you can axiomatize them
Per Craig's theorem
16:55
Female (♀) is the sex of an organism, or a part of an organism, that produces non-mobile ova (egg cells). Most female mammals, including female humans, have two X chromosomes. == Defining characteristics == The ova are defined as the larger gametes in a heterogamous reproduction system, while the smaller, usually motile gamete, the spermatozoon, is produced by the male. A female individual cannot reproduce sexually without access to the gametes of a male (an exception is parthenogenesis). Some organisms can reproduce both sexually and asexually. There is no single genetic mechanism behind...
@0celo7 needs rigor. We need to prove mathematically what a "female" is
But that's only for a countable amount
@JohnRennie so the voltages add algebraically?
not a standard definition on wiki
@Slereah A lot of the Axiom of Choice stuff applies to theoretical things that don't occur in real life... probably.
16:55
Well yes, PROBABLY
@Slereah Why are you looking for QFT books?
I'm doing QFT, haven't you heard
I'm doing all of Peskin!
@Slereah Yes, but the Axiom Schema of Reduction is technically not a single axiom.
It's an axiom scheme
@JoeStavitsky yes if a DC voltage is applied. In fact I think they'd still just add up with an AC voltage, though I'd want some time to check that.
16:56
Basically the same thing
As far as this is concerned
Syntactically it functions the same way
@Slereah Uh, I'm not sure that's true. The schema can apply to ANY set and condition and create a new axiom.
@Slereah However, since the number of proofs you can write down in ANY axiom system is countable, you're sort of right.
It's just your typical pick-and-replace syntactic rule
@Slereah just ask ACM for his lecture notes
@Slereah Yes, but the theoretical number of axioms you can create from it is larger than any transfinite number.
But they are all recursively enumerable, though
Nothing dazzling
16:58
@Slereah I agree with you, though. But I also believe the universe is not uncountable.
Well that's more of a philosophical position
Although, to be fair
@Slereah Are they? You're saying the entire category of sets has recursive enumerability?
I think that, if it's not the case
We won't know it
@JoeStavitsky actually the simple way is to just calculate the total capacitance, C_t, in the usual way then Q = C_t V. The capacitor with the biggest voltage drop is the 5uF capacitor where V = Q/5e-6. Set this voltage to 30 and you can calculate the total voltage.
Probably
I don't think you can make an uncountable amount of experiments
16:59
@Slereah I don't think all sets are in the sequence: R, P(R), P(P(R)), ie, the aleph sub ns, if that's what you're saying.
So in the end it will be the same
I dunno
@JohnRennie ty so much. I may have a couple more. WOuld you mind if I asked you in a few min?
@ACuriousMind : So what would you recommend

« first day (1985 days earlier)      last day (3242 days later) »