4:40 AM
@JohnRennie Hello. I still have some problems about the Kruskal–Szekeres diagram. Can you help me clarify them?

@CaptainBohemian hi :-)

@JohnRennie Do all constant $r\neq r_s$ curves approach the light cone $T=\pm X$ when t approaches $\pm \infty$?

On the KS diagram lines of constant $r$ are hyperbolae with the $X=\pm T$ lines as their asymptotes.
I'm not sure what you mean by approach the light cone

isn't $T=\pm X$ the light cone of the KS diagram?

Those lines are the event horizon.

5:06 AM
@JohnRennie But the even horizon is the black hole's boundary where no causal curve can escape towards infinity and that's just the light cone, isn't it?

At any point the light cone at that point is the cone traced out by light rays passing through that point.
So on the KS diagram it is the two lines of gradient 45° passing through that point.
So the lines $X = \pm T$ are the light cone for the central point on the diagram.

5:25 AM
@Charlie Hello
@FakeMod I can only read u sory :/

5:40 AM
@JohnRennie OK. You are right that the hypersurface $T=\pm X$ is just the light cone of the origin of the KS diagram, not the light cone of the whole KS diagram while the event horizon is the hypersurface $T=\pm X$. So what does that all curves $r\neq r_s$ approach the event horizon $T=\pm X$ when t approaches $\pm \infty$ mean? All test objects stationary on the KS diagram would approach the event horizon when t approaches $\pm$? Why is so?

They don't really approach the event horizon of course, it's just the result of the non-linear mapping between Schwarzschild and KS coordinates that makes it look as if the lines of constant $r$ are approaching the horizon.

@JohnRennie so the KS coordinate system is not ideal in that it may lead to the misunderstanding that all test particles stationary on the KS diagram will approach the event horizon when t approaches $\pm \infty$?

All diagrams have their uses. The KS diagram is very useful for understanding the behaviour of objects crossing the event horizon. It's not so useful for distances large compared to the event horizon radius.

5:58 AM
@JohnRennie OK. In addition, I also find another strange place on the KS diagram: the curve $r=r_s$ is at the origin at whatever finite $t$ but runs across $T=\pm X$ when $t=\pm \infty$.

Hi there! Is $n$th order degenerate perturbation theory contained as a limit in $n+1$th order non-degenerate PT? Is there a concise way to form this limit?
I imagine I have some symmetry breaking parameter $\delta$ which allows me to control $E_1 - E_2$ such that I can bring the gap the zero in a controlled way.

@JohnRennie does what I said in my last message also the result of the inaccurate mapping from the Schwarzschild coordinates to the KS coordinates?

I'm working now for a bit I'm afraid ...

when I form the limit of $E^{(2)}$ for $\delta\to 0$ I would hope to obtain the coefficient

@JohnRennie then I may ask you later. I am going to eat lunch now.

6:06 AM
OK :-)

@ZeroTheHero Well, yes I understand the first part if your message. For the second part, IMO, there's no reason to not "invite" any user who wants to contribute in keeping the site clean. And I don't know why that room should be even made a gallery room. I can cast close votes, so I could be of some help to the community. I don't see what's the point in stopping me from doing that.
@ZeroTheHero And if you are under the impression that the chatroom (that I proposed) is a "cabal" or, in any way, secretive, then you are misunderstanding my proposal. I only wanted an open and visible to all chatroom. It is in no way any special group of sword-yielding elite users :-)

2 hours later…
7:46 AM
$v(t) = t^2$ <-- velocity-time graph where $\frac{dv}{dt} = 2t$ therefore at exactly 3 seconds after I have started running, my instantaneous velocity (my velocity at exactly the 3rd second) would be $6m/s$ right?

8:21 AM
@JingleBells No! Just find $v(3)$, in this case $3^2=9\:\rm m/s$. What you've found by putting $t=3 \: \rm sec$ in $\mathrm d v/\mathrm dt$ is the instantaneous acceleration at $t=3 \:\rm sec$, which is $6 \:\rm m/s^2$.

@FakeMod Oh
just as I thought I was finally getting it
I think it's clear now, i'd have to think about it a bit
$y = -\frac{2}{3}x + 4$ why is the slope of this function $-\frac{2}{3}$ and not $\frac{2}{3}$? The y-intercept is 4 and the x-intercept is 6.
I understand that y decreases 2 units for every 3 units of x but do we determine the sign based on whether the slope is upwards or downwards?

4 hours later…
12:02 PM
pff the chatroom is so dead :P

It's filled with people

hello

12:22 PM
@JingleBells Slope of a line is -(y-intercept/x-intecept). However straightway do dy/dx, it's worth lesser effort.

12:50 PM
1

So in this example I want to find the extension produced in the spring. The two methods are given in the image.( First one is by force method and the other one is energy method). The two yield different results. Where did I commit the mistake? In the second method I have assumed that kinetic e...

We meet again
Is there another question that is asked as often as this one is?

1 hour later…
2:14 PM
@FakeMod What you're proposing would primarily affect new users, and not every user can access chat (even if the rep threshold is low). Moreover, the chat feature is not used much expect by a small subset of users, so it would de facto have a selective reach. Given the controversy around this CURED room, you should at least stand up and make the suggestion on Meta about this: I hope to be the first to downvote this foolish idea.
However frustrating it is to run out of close votes, the reality is that concerted action to work around that limitation would IMO be harmful.

2:26 PM
@ZeroTheHero Yeah, I was thinking of making a Meta post, but seeing lesser support than I expected, I suppose making a post on Meta would not make any difference as to the creation of the room. I have satisfactorily convinced myself that the room should not be created, and that's why I do not intend to push this further :-)
@BioPhysicist I was quite sure that this would have beenasked before (since it's really common), but I couldn't find any dupes. So after 10 minutes of searching, I decided it to answer it myself :-)

@JingleBells Yes, that's exactly what the sign of the slope tells you

2:47 PM
@FakeMod Interesting. I would say it is one of the most frequently asked mechanics questions I see on the site.

I have to confess I mostly just don't read any questions that involve springs :P

@BioPhysicist StackExchange's search sucks, so I did a Google search with site:physics.stackexchange.com. Still nothing similar showed up...

@ACuriousMind Because they involve a lot of potential problems? :P

Heh.
It's more that I tend to find Newtonian mechanics questions boring, and springs in particular extra boring...

@ACuriousMind AFAIK, I'll never get bored of classical mechanics... Newtonian mechanics may seem boring...

2:55 PM
Classical mechanics is fine as long as it involves words like "gauge theory" or "symplectic structure" :P

@ACuriousMind From more of a teaching / tutoring perspective I enjoy Newtonian mechanics in the sense that it is (should be?) used to teach students how to start thinking more like a physicist.

Yeah, I'm not trying to say it's unimportant or anything, it just doesn't excite me

Although many physics teachers fail in this regard
Don't let Newton hear you say that

oh, he heard me

dang it

2:57 PM
when the zombie apocalypse starts, he'll come running straight for me

"Is this not exciting enough for you?!" as he projects apples at your head

"You have to evade Zombie Newton throwing apples at you" sounds as if it could've been a platformer game in the 90s

At least ACM doesn't have to worry about zombie Feynman. ;)
4

@ACuriousMind Do Newtonian mechanics apply to 8-bit universes?

Yes, but you first need to compress them down to 4

3:10 PM
Yeah my physics teachers definitely skipped over that part of the text book

3:25 PM
@GuruVishnu, I saw a rocket land in a way it never would

Falcon 9? You'll see another tomorrow with the DM-2.

It seems it didn't turn off its engine after touchdown. May be just hovering? :P

Yeah, one would feel it was just about to lift off
bye, I need to go for dinner.

0

I just asked a question about how to figure out how fast a room will cool, and what the name for this type of problem is, so I can do more research, and potentially find the answer to it in the responses, or if someone tells me what type of problem it is, by searching Google for the right phrases...

4:06 PM
What kind of physical interpretation do you propose for a reversible thermodynamic process that has a -ve value of dq/dT?

4 hours later…
7:50 PM
45 minutes, and the USA sends two men in the space, after nearly a decade!

3 hours later…
10:25 PM
@peterh-ReinstateMonica oops

@BioPhysicist Sad :-( Space SE folks say, had they 10 minutes more, they chould have launched. Next try about 2 days later ($\approx$ 22:30 saturday in CET)

Yeah I heard them say that on what I was watching