The h Bar

@Doraemonドラえもん Hint. Suppose you use a frame moving to the right at the same velocity $u$ as the end of the spring. In this frame the end of the spring is stationary and at time zero the mass is moving to the left at speed $-u$.

Slereah

What is the difference between an analytic function and a regular function

user246160

@JohnRennie Ok. No. In that case definitely it won't be in SHM as the block will be standing still. But still, how do we explain this discrepancy in the two reference frames ? Suppose we do it the other way round. We hold the point where u velocity is applied (as shown in the diagram) still and give a velocity u to the block towards left. In that case SHM does occur.

9:37 AM

@Doraemonドラえもん In the moving frame the motion is SHM because it's just a mass on a spring. In the lab frame the motion is SHM + a constant linear velocity.

user246160

@JohnRennie "In the moving frame the motion is SHM because it's just a mass on a spring." Really ? That is so difficult to visualize :/

user246160

Though theoretically it seems correct

hard to visualise? It seems really straightforward to me. Maybe I've spent too many years doing physics :-)

user246160

@JohnRennie You're an expert :P And I have just started with Classical Mechanics :P Anyway, do you know any book or source which explains this concept of SHM from moving frames with diagrams ? I think i need to "think" a lot about this before i can visualize it :)

@Doraemonドラえもん Suppose I have a mass on a spring on my (frictionless) desk and at time zero I give the mass a velocity $v$ to the right. Are we both agreed this is a simple SHM setup? That is, if you're standing behind me looking over my shoulder we'd both agree it's SHM?

user246160

9:46 AM

@JohnRennie "Mass on a spring"...by that do you mean the spring is horizontal and attached to the mass ? (like the above diagram )

@Doraemonドラえもん Yes

user246160

@JohnRennie But is the spring attached to a wall or something ? Because otherwise the spring will too start moving along with the block with a linear velocity

@Doraemonドラえもん The end of the spring is fixed to a nail I've hammered into my desk. Honestly, you're making this far more complicated that it need be!

user246160

@JohnRennie Okay I get it now. So yes, when we give the mass a velocity v towards right, then indeed it is a simple shm setup !

Great :-)

Pissed off layman

9:50 AM

:-)

And now you jump in a car and drive past me from left to right (the same direction that the mass is moving) at a speed $v$. That is, when you pass me both you and the spring have the same speed $v$ to the right.

user246160

@JohnRennie Okay..i got it till there

Obviously we haven't changed my the mass, spring and my desk so the mass is still happily bouncing around performing SHM. Yes?

user246160

@JohnRennie yes..right

But when you look out of the car window you see the mass motionless (because it's moving at the same speed you are) and you see me, my desk, the nail and the end of the spring moving to the left at $-v$. Yes?

9:55 AM

@JohnRennie will there no energy stored in spring

1/2 k x$^2$

user246160

@JohnRennie Here is where I have a problem. The velocity of the block in SHM will be $v$ only at the mean position of SHM and not at all positions. So only at mean position I will see it motionless.

user246160

@koolman Could you please wait a bit...JR is explaining that only :)

Okay

Yes, the point is that you see the position of the block as something like $$x = A\sin(\omega t) - vt$$

While I see the position of the block as: $$x = A\sin(\omega t)$$

user246160

Yes, exactly. So in my frame it won't be SHM, right ? It will be some weird oscillation!

user246160

9:58 AM

@JohnRennie

@Doraemonドラえもん Well I guess it's a matter of terminology. In your frame the motion is a combination of SHM and linear motion, but there is still an SHM component. Anyhow the point I was getting at is that this is the way to understand the problem you were given.

user246160

@JohnRennie Well. Yes. I can sort of visualize it now. But still I have one more confusion remaining. In the above problem i.stack.imgur.com/O5Ai6.png when the block and spring system are moving together with a velocity u the block will continue showing SHM, right? In that case is some energy stored as spring energy ? Will the spring keep expanding and contracting as it moves with velocity u even in absence of friction ?

user246160

Consider the length of the spring i.e. the distance between the position of the mass and the position of the end of the spring.

The length if the spring is just some distance, and that distance doesn't change depending on who is looking at the spring. If I measure the length to be 10cm then so will you driving past in your car at some speed $v$.

And in my frame the length of the spring will be given by $$\ell = \ell_0 + A\sin(\omega t)$$

So that equation must also be true for you driving past in your car.

We may disagree about the velocities of the mass and the end of the spring, but we will both agree on how the length of the spring changes with time.

user246160

10:13 AM

@JohnRennie Okay, so the length of the spring keep changing according to the SHM equation $x =l_o+ A\sin(\omega t)$ in the above diagram ? i.stack.imgur.com/O5Ai6.png And so the energy of spring + block system will not always be 1/2 mu^2...right ?

Kinetic energy isn't frame invarient. That is if you're moving at a different speed to me then the kinetic energies you measure are different to the ones I measure.

user246160

@JohnRennie Right. I agree KE is same in any frame. But what I am asking is that will the potential energy in spring vary with time ?

This is a source of much confusion. It happens because we will agree on the fiorce the spring exerts on the block, but we disagree about the distance the block moves in response to that force. So when we calculate the work done $F\Delta\ell$ we get dfifferent numbers.

@Doraemonドラえもん The potential energy in the spring is always $\tfrac{1}{2}k(\ell - \ell_0)^2$ for all observers.

user246160

@JohnRennie Yes yes...that is what I was asking. Thanks. $PE=\tfrac{1}{2}k(\ell - \ell_0)^2$ is for all observers. Now $l-l_o=A\sin(\omega t)$, right ? But here, how do we determine the amplitude $A$ in this diagram ? i.stack.imgur.com/O5Ai6.png

user246160

We don't know what will be the amplitude of oscillation when the spring is given $u$ velocity....

10:21 AM

If $\ell_m$ is the maximum length of the spring then the amplitude is $\ell_m - \ell_0$. Yes?

user246160

@JohnRennie Umm, but even $l_m$ is unknown, isn't it ?

@Doraemonドラえもん Bear with me :-)

user246160

Sure...go on :)

user246160

(I should have said that actually :P)

We're agreed that all observers agree on the length of the spring. Yes? Therefore all observers agree what $\ell_m$ is, and therefore all observers agree on the amplitude. So far so good?

user246160

10:25 AM

@JohnRennie Well, yes. Only if we practically know what $l_m$ is. In this i.stack.imgur.com/O5Ai6.png block-spring diagram we have no idea about what $l_m$ will be !

My point is if all observers agree on $\ell_m$ then any observer can calculate it and get the required result. Yes?

user246160

@JohnRennie Yes. I agree !

user246160

But we cannot theoretically calculate $l_m$ here i.stack.imgur.com/O5Ai6.png?

Remember my analogy of the spring on my desk and you driving past? Well for me it's really easy to calculate $\ell_m$ by equating KE and PE $$\tfrac{1}{2}mv^2 = \tfrac{1}{2}k(\ell_m - \ell_0)^2$$ Yes?

user246160

@JohnRennie Yes, there it is easy to calculate it. Got it! But how do we calculate $l_m$ when the spring is not attached to any nail and moves with the block ?

10:32 AM

You mean as in the problem you posted? If so the spring is attached to something (whether it's a nail or not) and that something is moving at a speed $u$. That's the point of the question. The end of the spring has to be fixed to something or it wouldn't move with a constant speed $u$.

user246160

Ok, so I suppose that free end of the spring is attached with a wall which moves with a velocity $u$. Now in reference frame of wall we can write, $(1/2)mu^2=(1/2)k(A)^2$ where $A$ is the maximum amplitude, right ? That means at maximum compression or elongation a potential energy of $(1/2)mu^2$ is stored in the spring. Right ?

user246160

@JohnRennie

Yes

And that's the point. The amplitude $A$ is the same for all observers so we calculate it in the rest frame of the wall where the calculation is easiest.

user246160

@JohnRennie Umm, so maximum energy of the spring + block system in subsequent motion should be $(1/2) mu^2 + (1/2) mu^2=mu^2$ (KE+PE), right ?

DenisKa

isn't this the script for mathJax?

javascript:(function(){if(window.MathJax===undefined){var script = document.createElement("script");script.type = "text/javascript";script.src = "http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML";var config%

10:45 AM

@Doraemonドラえもん be careful with the kinetic energy since it does vary for different observers.

user246160

@JohnRennie I know that KE is different for different observers. So suppose I am trying to calculate maximum energy possible for block+spring in ground frame, what should it be ?

user246160

$(1/2)mu^2$ must be the KE

user246160

But what about maximum spring potential energy ? As we know that in this case the motion will be a superposition of linear motion and SHM

Emilio Pisanty

@JohnRennie Yeah. To be honest, though, I was quite impressed by the case that Philip Gibbs makes for viXra over on quora, quora.com/Are-there-any-serious-papers-on-viXra

@Doraemonドラえもん At the extreme of the motion, i.e. when $\ell = \ell_m$, the mass and the spring are momentarily stationary with respect to each other. Yes?

user246160

10:50 AM

@JohnRennie Yes, right

@vzn cqc2t.org/biography/102

So at that moment the mass and the spring are both moving at velocity $u$. The total energy is then KE + PE, which is $\tfrac{1}{2}mu^2 + \tfrac{1}{2}kA^2$ and ...

aha, I now understand what you did.

We've already shown that $\tfrac{1}{2}mu^2 = \tfrac{1}{2}kA^2$

So yes, I agree. Total energy is $mu^2$.

Sorry about that, I didn't follow your working at first :-)

user246160

@JohnRennie Yeah! So the maximum possible total energy should be $mu^2$. But I have no idea why my textbook says that the maximum energy should be $2mu^2$ here i.stack.imgur.com/BS2Ag.png . They state option (C) as the correct option. Are we missing something ? :P Or is $mu^2$ the final and correct answer ?

user246160

(My textbook might be wrong sometimes)

10:57 AM

Hmm, textbooks can be wrong, but it doesn't happen often. It seems more likely I've missed something in the working ...

user246160

Maybe we made a mistake in PE calculation somehow. KE is definitely 1/2 mu^2....

I have to get back to work for a bit. I'll have a think about it.

user246160

Ok sure..no probs :)

user246160

bye

user246160

@JohnRennie

Swapnil Das

11:20 AM

@Doraemonドラえもん Which book are you using for physics?

user246160

@SwapnilDas I not using any book as such...just institute study material

Swapnil Das

Oh great.

I use Physics Galaxy :)

@Doraemonドラえもん In which class are you?

user246160

12

Swapnil Das

Oh. So Doraemon is 7 years older than Nobita :P

11:36 AM

@Doraemonドラえもん I've realised the problem is far simpler than I thought.

user246160

@JohnRennie Hurray...even I solved it just some time ago you know.! the maximum energy will be when the block is returning with u velocity making relative velocity 2u w.r.t ground frame. So maximum energy will be (1/2)m(2u)^2

Yes, exactly :-)

user246160

@JohnRennie Thanks a lot! I wouldn't be able to do it without your help ! Thanks a ton :) It is a lot clearer now :)

Which makes me wonder where the original calculation went wrong. I'll have to have a think about it.

user246160

I found it. I am writing the steps. Wait a min

user246160

11:44 AM

Velocity of the block as a function of time is $-u\cos(\omega t)+u$

user228700

@JohnRennie: Are u sure I shouldn't give TNT a call?

user246160

And extension of the spring as a function of time is $\frac{u}{\omega}\sin(\omega)t$

user246160

Where $A=\frac{u}{\omega}$

@Kaumudi.H I guess it won't do any harm, but all they can tell you is that it's in the hands of the customs office. You could trying ring the customs office, but I'd guess they will just tell you wait and they'll get round to it.

user246160

So total energy as a function of time is $\frac{1}{2} m [ (u-u\cos(\omega t))^2 + (u\sin(\omega t))^2 ] $

user246160

11:47 AM

And maximum value of this function is $2mu^2$

user228700

@JohnRennie Sigh. This is too much. How much time are they going to take?!

@Kaumudi.H I know it's frustrating, but I honestly don't think you can do anything but wait. I'd be inclined to leave it until Monday and ring then if there is still no progress.

user228700

Monday?! I feel like dying, ugh. (I also have a terrible headache but that's besides the point :-P)

user246160

@JohnRennie Our mistake was that we though maximum energy will be at maximum compression/elongation but it will be rather at the mean position.

@Doraemonドラえもん Aaaah

user246160

11:49 AM

When the velocity of the block is $2u$

user246160

w.r.t ground

user246160

:)

Oh well, you got there in the end :-)

user246160

@JohnRennie Well it was because of you and the shm equations you showed me :)

@Kaumudi.H You have a terrible headache? You mean besides the customs office?:-)

user246160

11:52 AM

@Kaumudi.H Amritanjan laga le :P

user246160

^ I ain't joking...it is seriously good for headaches ! amrutanjan.com You can buy it from medical stores :)

@Doraemonドラえもん yes, but is it any use for the pain caused by customs offices? :-)

user246160

@JohnRennie Haha....no XD I thought she has migraine :P

user246160

@JohnRennie How much is the custom duty on the laptop? :P

@Doraemonドラえもん I can't remember but it wasn't much. The problem is that the customs office insist on inspecting the laptop to make sure there isn't anything illegal on it, though what they are looking for I have no idea. They won't release the laptop for delivery until they've inspected it.

user228700

11:59 AM

@JohnRennie :-P I don't speak with my friends at all. Today, I spoke with my best friend for about 3 hours straight after months, it being his birthday and all and now I have a headache and I'm exhausted af. I dunno how I'm going to survive college, at this rate.

user228700

@Doraemonドラえもん What are u, their sales representative? :-P

The thing about college is that you're with a large group of people, all about the same age as you and many with similar interests to you. So there are lots of people around to chat to, lean on, provide mutual support and encouragement for, etc.

That makes things a lot easier even though it is hard work.

I've read you solving the problem earlier @JohnRennie @Doraemonドラえもん but I still do not understand why $\frac{1}{2} m u^2$ get stored as potential energy in the spring - if you don't mind, could you give me a hint? =) I do understand that if that happens, the maximum energy in the system is $2 m u^2$ but I don't understand that step and it's bugging me :|

user228700

@JohnRennie Sure sure, I get that. But you see, I've been in social isolation for about 5 months now and I feel like I'm beginning to become outright antisocial. "Many people your own age" is anxiety-inducing.

user246160

@JohnRennie Oh! I don't think that is something to worry about. Unless there are things like explosives or say ISIS stuff :P they won't block it. My cousins overseas have sent me several electronic devices by postal service over the years and they were never blocked :)..anyway good luck @Kaumudi.H :P ....i hope another cyclone doesn't carry it away :D

user228700

12:04 PM

:-) Thanks.

@Doraemonドラえもん how long did they take to get through customs? Kaumudi has been waiting since Monday.

user246160

@Kaumudi.H Not yet, but they should seriously give me money for I am promoting their brand :D

user228700

Sigh, yes I have.

user228700

@Doraemonドラえもん x'D Why don't u contact them, see if they're interested?

user246160

@JohnRennie The last one I got was 2 years back....It took about 1 week after reaching india

user228700

12:06 PM

Oh God :'-( Sigh, 4 days to go, then...hopefully.

user246160

@Kaumudi.H The journey is more important than the destination XD When it reaches you, after few days you will lose interest :D...so be happy and eat biryani =D

user246160

Yahoo...i will have biryani today..i am so happy :D

user246160

balle balle

user228700

Wut? -___- No. The excitement is wearing off, not building up.

user228700

@Doraemonドラえもん Ah, one of the simplest pleasures of life; food 😍

user246160

12:09 PM

@Kaumudi.H FOOD IS LIFE! :D

user246160

eat and be happy !

user228700

x'D Sure thing. I'm gonna go see what I can do about my effing headache. Enjoy your biriyani, bye :-)

user246160

bbye...tata...alvida!

user246160

@Sanya (1/2) mu^2 is not the PE...see I wrote the expression for PE above

user246160

25 mins ago, by Doraemon ドラえもん

So total energy as a function of time is $\frac{1}{2} m [ (u-u\cos(\omega t))^2 + (u\sin(\omega t))^2 ] $

user246160

12:12 PM

The second term is the PE as a function of time

user246160

$x=A\sin(\omega t)$. So $(1/2)kx^2=(1/2)k(A\sin(\omega t))^2$ but $A\omega=u$ and $k=m(\omega)^2$

hello

@Ramanujan, I'm not sure what you mean by "wall", I'd be glad to explain, but could you elaborate?

@Doraemonドラえもん because we just assume that the mass will go into a harmonic motion?

@heather hey heather

10 hours ago, by DanielSank

$$D^2 f = - C f$$

@G.Bergeron yeah, it seems I am still bad at writing my constructions in a formal mathematical language. I suspect one strange thing is probably cartesian products cannot be used this way to single out rows of a nxn table. I have to check with the maths chat to see what in the language I have screwed up

As for the construction itself, it is trying to rewrite the associative law in terms of rows, and found that this basically put the two tables of the Light's associativity test on top of each other thus allowing one to check the associative law as one fill in entries in the cayley table so that the resulting algebra is guareenteed to be associative

12:27 PM

c sin -> c cos -> c -sin

@DanielSank, $f(x) = c \sin(x)$?

Then what is the prob @heather

Now I am going to edit them

user246160

@Sanya We don't assume that it goes into SHM. It exhibits SHM because there is a spring and not a string. See my and JR's 3 hour long discussion on this :P When we pull the spring the block doesn't immediately start moving.

@koolman, I'll be talking with Mew later. Follow what I said earlier and then we'll talk about where we stand. As in: fix the MathJax. I saw what Mew said about images. As I said, I'll be talking with him. Please, just do it.

This is not up for argument.

@heather I have done what you want

My new ques is also like that you said

user246160

@koolman That question is already solved! See my and JR's chat

12:37 PM

@Doraemonドラえもん yeah ,I know but I have posted that before that

You got the ans

@Doraemonドラえもん hmm, what John said doesn't convince me - but I guess I'd have to take that up with him :D (I've read through your conversation from 3h ago :D )

user246160

@koolman I think heather is talking about using MathJax..you did not use MathJax for writing the equations

Where

user246160

@Sanya Wait...which part you don't understand? Did you get that the block does not start moving immediately when the spring starts end moving ? That is the sole cause for inducing SHM....In case of string the block will start motion immediately and there will be no shm

user246160

anyway you can take it up with JR

user246160

12:39 PM

I gotta go now

user246160

good luck with that :)

user246160

@koolman on that website

user246160

physics.qandaexchange.com/?qa=562/…

@koolman It's better, but you should still format the math, which you didn't

@Doraemonドラえもん I think this is the crucial point where I'm kinda at odds with the idealisation/assumption ... but yeah, maybe I'll ask John if he comes back

see you around and thank you

12:40 PM

@Doraemonドラえもん i have already done there

@heather which one

of the electric field one; you haven't even started to edit the nuclear one.

@heather as mew said we can post image of equation

@koolman, did you even read what I said?

15 mins ago, by heather

@koolman, I'll be talking with Mew later. Follow what I said earlier and then we'll talk about where we stand. As in: fix the MathJax. I saw what Mew said about images. As I said, I'll be talking with him. Please, just do it.

Ohk

But @heather do it as soon as possible

Please

koolman, i'm not reopening those questions until you do the mathjax.

period.

1:08 PM

have a good day everyone

(::heads to bus stop::)

@heather have a good day at school

Ramanujan

1:29 PM

This wall @heather

Escape The Kugelblitz Challenge | Space Time | PBS Digital Studios

►

1:51 PM

@G.Bergeron A top-dimensional differential form (which is what you ahve to act on to get the determinant) is nothing but a volume form, so our two statements are secretly the same.

@Danu dear lord that paper is written terribly

3:38 PM

Why is it that $$\lvert m=0, s=0\rangle = \frac{1}{\sqrt 2}(\lvert\uparrow\rangle \lvert\downarrow\rangle - \lvert\downarrow\rangle \lvert \uparrow\rangle)$$

@Argon ...because if you evaluate the operators for $m$ and $s$ on it you see they give zero?

What, exactly, is the question?

How does one come up with this?

It's called Clebsch-Gordan decomposition

@Argon If you want to understand it in this specific example, then if you agree that $\lvert \uparrow\uparrow\rangle$ is the m=1,s=1 state, you can just apply the spin lowering operator twice to find the components of the s=1-triplet

But the tensor product space of the two two-dimensional 1/2-spin systems is fourdimensional, so there's a fourth vector which is orthogonal, and necessarily must form a singlet, i.e. m=0,s=0

@ACuriousMind Why twice?

@Argon Because you know that the s=1 triplet has the states m=1,m=0,m=-1 in it, so you apply it once to find the m=0,s=1 state, and twice to find the m=-1,s=1 state.

So then you've got the basis vectors of the triplet in the product space, and there's only a single other orthogonal direction left - and the only rep with only one vector is the s=0 singlet, so it must be a m=0,s=0 vector.

vzn

4:02 PM

@Secret ok thx interesting, but none of his papers seem to assert "wavefn is real"...

When my book tries to determine total spin of a two particle system, it only looks at $S_z$, since the four possible states are eigenstates of this operator. What about the other operators?

@vzn I think it is one of his collaborators, and I suspect that paper was recently in thus had not gone online yet

Stef Simmons's profile does not have any relevant paper either, except a 2006 one

vzn

4:19 PM

@Secret ok here is another ref that came up, Lundeen, "direct measurement of the wavefn" physicsworld.com/cws/article/news/2011/jun/15/…

Q: Rainbows at the surface of calm cold water : mechanism?

0

Pictures Bigger versions : First | Second On this one we see the other side of the "main" arc and the white surface of water (see below). Observations So I observed these rainbows in a calm river. Like classical rainbows, the phenomenon requires the sun and creates a circle of colors at ...

Strange...

It is like a rainbow that can only bee seen under reflection

Q: Define ever renewing "now" . Define living being in "now"

0

we always live in present only , physically. please elaborate the "now" and how we are happened to be locked in present ? Also define consciousness , awareness, intelligence in the view point of physics.

cosmology

"Also define consciousness , awareness, intelligence in the view point of physics."

"Also" :o

@Argon I'm not exactly sure what you're asking. There are more than "four possible states", but since $S_z$ is self-adjoint, it has an eigenbasis consisting of four vectors since the space is four-dimensional.

4:37 PM

@ACuriousMind Griffiths outlines only the states up up, down down, up down and down up.

Well, you can take linear combinations of these too

Exactly. They form a basis, and all linear combinations of them are states, too

But still, he asks "what is the total angular momentum of the atom?" and goes on looking only at $S_z$

Well, you should look at $S^2$ for that, but since $S^2$ and $S_z$ commute, you can find an eigenbasis of $S_z$ that's also one of $S^2$, so the two approaches are equivalent and looking at $S_z$ is easier.

@ACuriousMind That makes sense, thanks.

A: Is this statement about the impact of the fine-structure constant accurate?

@heather your solution is correct, but note that $f(t)=A\sin(t)$ works for any $A$.

... In the case where $C=1$, i.e. $D^2f = -f$.

Emilio Pisanty

4:47 PM

booya

0

In the form you have phrased it, the statement is false or, at best, incomplete. In particular, the specific claim that In a universe where the constant is a fraction higher, electrons would orbit closer to the atomic nucleus. is essentially meaningless. The problem is the assertion that, ...

I've been wanting to write that for ages

5:39 PM

@EmilioPisanty It's a good answer.

arse licker

I prefer a French liquor myself.

2

6:10 PM

Thank you! Can you please tell me a little bit about the difference of the MEANING of the Christoffel symbols, and the Riemann tensor? — Árpád Szendrei 2 mins ago

Well, he's not asking much. And in a comment!!!

I have a truly remarkable explanation which this comment is too small to contain :-)

2

tell him that the Riemann tensor measures the atmosphere and the Christoffel symbols are used to find genomic patterns

The Christoffel symbols are gauge fields, the Riemann tensor simply their field strength. Now you have reduced the problem to gauge theory, which is much simpler to understand ;P

or something like that

@AccidentalFourierTransform you're in a good mood this afternoon :-)

@JohnRennie why do you say so ;-)

ah, I wish... so much to do today

and ive already wasted all morning and afternoon

yet, here I am

6:15 PM

I'm glad to see we have (at the time of writing) two connoisseurs of ancient jokes :-)

Alex

6:27 PM

I want to show that $\int \frac{\partial \vec{B}}{\partial t} \cdot d \vec{a} =\frac{\partial}{\partial t}\int\vec{B} \cdot d \vec{a} $ for a magnetic field, using Leibnitz rule. Does anyone know if it is true that the components of the first derivative of the magnetic field is continuous? (as this is required by Leibnitz rule).

Q: Downvote to remove a comment

@Alex physical fields are usually assumed to be smooth unless you have a good reason to say otherwise

Alex

@ACuriousMind Oh okay, great thanks.

Emilio Pisanty

6:47 PM

@DanielSank I saw you were in chat and I was about to ask whether you'd seen yesterday's SMBC

until, of course, the star board made the comment moot

;-)

I also now realize

the star board is to starboard (in a reasonable reference frame). coincidence?

7

Physics Meta

6:58 PM

0

I think there should be a mechanism to remove a bad comment if it has been downvoted many times . Since such comments degrade the quality of discussion.

feature-request comments voting

@EmilioPisanty haha

rob

7:51 PM

@EmilioPisanty That realization is just so ... right.

Pissed off layman

the star board is to star board as a circular argument is to a circle :P

Pissed off layman

8:07 PM

almost halfway @0celo7

We miss you.

Bernard Meurer

9:07 PM

@ACuriousMind Are you there?

or @DanielSank

9:25 PM

hello everyone

@Ramanujan, that feature has been removed for all users.

It was meant to be a place where users could comment to other users, but it was removed as it didn't serve any purpose, really.

9:41 PM

We have one question where a potential $U = U_0 e^{-|x|/x_0}$ and $U_0 > 0$ is given. How can I go about determining if "the potential supports discrete states, continuous states or both?"

@BernardMeurer yo

@DanielSank, hi

Bernard Meurer

@DanielSank Check hangouts