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12:01 AM
@ChrisWhite yeah visiting places in person made a HUGE difference
I absolutely fell in love with Berkeley
you said you'd be starting at Berkeley in September right?
 
@FenderLesPaul Berkeley is nice, and bustling. For the place alone I would prefer Santa Barbara, but Berkeley has a big advantage on the "things happening all the time" front. If you care for that kind of thing. And more choices in exotic food, too.
 
12:21 AM
not to mention you'll be nearer to the capital of raider nation :P
 
12:33 AM
Oh, he's gone? That was a good example of a "drive by" comment :D
 
user54412
@dmckee Yes. Evil.
 
user54412
@BernardMeurer It makes beautifully fast code. But we do all our development on gcc. icc is slower to work in new standards (do they even fully support c++11 yet?), and has more subtle optimization bugs. It also lets you get away with more that it shouldn't -- "You called sqrt(), so you must surely have meant to include cmath, let me do that for you silently."
 
user54412
I still half suspect intel chips have a secret flag that slows down code not compiled with icc.
 
user54412
Except there are now other architectures where icc makes better code than the compiler made for that architecture.
 
user54412
@FenderLesPaul Yes, September 1 if all goes well.
 
12:54 AM
Did you @dmckee watch the video I linked?
 
@ChrisWhite Have you seen what icc does with bad branch prediction? It's bloody magical
 
user54412
@BernardMeurer ?
 
Read the update in the end
 
user54412
1:09 AM
ha
 
user54412
that's only icc 11
 
user54412
icc 16 probably solves the halting problem
 
LMAO
HA
@Obliv IT WORKS
My Euler calculator works :3
with arbitrary precision
 
user54412
SO: where you can get +1745 for answering an optimization question with unoptimized code, drawing all the wrong lessons from it
 
@ChrisWhite Why is x&1 faster than x%2 for testing if something is even?
 
1:14 AM
mit ocw has a new video lecture course on graduate level string theory - link
 
user54412
@BernardMeurer It's not. They'll generate the same machine code. If they don't, either your compiler sucks or x is some weird type where those two expressions are different in corner cases.
 
@ChrisWhite Oh, nice
Also, earlier you were saying you are against boost and things like that. How would you deal with arbitrary precision using only the stdlib?
 
user54412
As a general rule, the first 1000 times you want to do bit twiddling for performance, the compiler has already thought of that.
2
 
@ChrisWhite Makes sense, smart people wrote it after all
 
user54412
@BernardMeurer I... wouldn't :p But if forced to I would make a class that acted like numbers (define all the arithmetic operators and such) but internally had an array with on-the-fly memory allocation and such, and defined all operations on that data structure.
 
user54412
1:19 AM
Maybe bundling allocations into large chunks, creating a virtual heap that my own code micromanages for its purposes, if performance is an issue.
 
:O
You're a c++ monstah though, I'm a little far from that point
is the -Ofast flag any good? is it pure magic?
 
user54412
what compiler?
 
GCC
No money for Intel CPU ;-;
 
user54412
oh, that uses fast math -- things like the fast-inverse-square-root algorithm used in graphics
 
user54412
it's... a little dangerous
 
user54412
1:22 AM
your results will no longer be IEEE compliant, and in general will be wrong in the last few bits
 
user54412
on the one hand, if your floating point algorithm blows up because of inaccuracies, it's the algorithm's fault
 
fast-inverse-square-root? That magic number crap?
 
user54412
on the other hand, in most real applications it's only a matter of time before things become too inaccurate, and fast math hastens your program's demise
 
user54412
to be honest, a lot of optimization lore is built around hardware traits that simply aren't true anymore
 
How do you benchmark your code? Like I have my code for the euler function and someone else's, how can I accurately check which one is faster?
Also, would you have any recommendations for writing testcases for C++ code?
I feel like I'm question-avalanching you; I'm sorry
 
user54412
1:27 AM
vectorization gets you 2-8x speedup, pipelining is a factor of 5-50, and fundamentally better hardware is some additional factor -- the result is compared to 20 years ago, floating point operations are often free
 
user54412
what hasn't improved over the years is memory access speed -- they invented cache hierarchies, and that's about it
 
user54412
@BernardMeurer run them? :p
 
@ChrisWhite Pff, but how do I chronometer it?
I know it in Python, but dunno in C/C++
 
user54412
@BernardMeurer google makes a decent unit testing framework I've used before
 
user54412
@BernardMeurer well if your code takes long enough you can just time it externally from the command line
 
1:31 AM
@ChrisWhite It's blazingly fast, I just want to know if it's fast enough for me to win this bet :p
 
user54412
27
Q: Timing algorithm: clock() vs time() in C++

blazeFor timing an algorithm (approximately in ms), which of these two approaches is better: clock_t start = clock(); algorithm(); clock_t end = clock(); double time = (double) (end-start) / CLOCKS_PER_SEC * 1000.0; Or, time_t start = time(0); algorithm(); time_t end = time(0); double time = diff...

 
@ChrisWhite If we go to DanielSank's wedding will you be my date?
<3
 
user54412
You've offered your love to half this chat room ;P
 
Somewhere I have a link to a (now rather old) presentation about how compilers bit-twiddle. Money line "gcc is smarter than the video codex programmer on all platforms".
 
@ChrisWhite Tsc, don't slutshame me
 
user54412
1:38 AM
On the off chance someone can help with html: I'm making a table in a github wiki, and it automatically alternates row colors. How can I turn this off given that I can't control the style sheet?
 
I wish, I really do
 
Link for above: linux-kongress.org/2009/slides/… It's from 2009, so it is quite ancient these days, but i keep it in mind for humility.
2
 
"Warning: contains assembly language code." ::closes webpage::
 
indeed, assembly language keeps me humble
 
1:55 AM
Another good lines in there "very few humans know this" about something a compiler does because it's faster on one manufacturer's x86_64 architecture and the same speed on others.
 
@ChrisWhite Alright, it's blazingly fast
to my eyes at least
 
user54412
2:12 AM
@dmckee "Fancy data structures help on paper, but rarely in reality" and "If the data structure can’t be explained on a beer coaster, it’s too complex." These should be required learning for CS-students-turned-programmers.
 
Holy crap C++ is just silly fast
Still reading that pdf, can't understand any of the assembly though
 
user54412
There's some pretty excellent stuff in there. Wish I could hear the talk that went with it.
 
yup
@dmckee thanks for sharing :-)
 
 
2 hours later…
 
2 hours later…
5:33 AM
@BernardMeurer "From the Standard model of Particle Physics and Beyond!" :)
M1-3 Neutrinos
Hagen boson
Pomeron
 
6:23 AM
23
A: Have I embarassed my supervisors by solving a problem that a PhD student in my group was working on without success?

Anonymous MathematicianOf course I can't say anything definitive based on the limited information in your question, but one possibility is that you may have inadvertently committed a faux pas. Your advisor may be upset that he/she didn't anticipate this possibility and head it off. One of the basic rules of the mathe...

^ ever seen an answer sitting at 23 = +43 -20 !!!
It is good for communities when people voice their disagreements openly, even when the author has 97k reputation.
Downvoted because the idea that an undergraduate can take any blame for taking an assignment, and performing it too well for diplomatic niceties, is beyond my ability to accept. — Daniel R. Collins Apr 13 at 4:46
when you started trying to prove the theorem, what did you think was going to happen to this graduate student if you succeeded? is holding the OP responsible for not thinking carefully. I strongly disagree with this sentence. What is the OP supposed to think? Why should he think what is going to happen? He just took the topic from his advisor and worked on it. What did he do wrong? What else can an undergrad do when he is given a thesis topic to work on? — scaaahu Apr 13 at 9:16
The posting explicitly says "the supervisor offered me a thesis on the topic". That being the case, it seems absurd to suggest a rule was violated. "you don't compete with graduate students by working on their thesis problems" seems like a rule for faculty. A student who is "offered" "a thesis on the topic" cannot be considered to be violating such a rule because he succeeds. — Michael Hardy Apr 13 at 5:11
(As I post that, I realize that 97k rep is the reason why this collected +43, considering that the premise is plain wrong.)
But still, one data point where not having a silent-downvote culture helps the community. Overall, this would be a thread that this community would be really proud of.
("Thinking out loud") :P
 
user54412
7:03 AM
@TheDarkSide Uh, why do you say that? That answer is completely right. It doesn't say the person was in the wrong, but that a faux pas was committed.
 
user54412
None of those commentators seem to know what they're talking about.
 
@BernardMeurer That's cute. Get your girlfriend one.
 
@ChrisWhite See, it could be that we disagree here. With all due respect to the OP, this answer is incoherent at best. Contrast "I don't mean to suggest that you are primarily to blame", with (later in the para), "However, when you started trying to prove the theorem, what did you think was going to happen to this graduate student if you succeeded?".
And then, there's:
 
@DanielSank She's not my girlfriend!
 
@BernardMeurer Ok.
 
7:15 AM
:p
 
"It comes across as callous to be more concerned with getting credit yourself than the repercussions for this graduate student."
 
I always wanted to say that
 
Why?
 
idk, sounds like a frustrated teenager thing, never had the opportunity to use it
 
(contd.) Does one really expect the undergrad to question the adviser - why are you assigning me a problem that your grad student is already working on? It could be case of contrasting cultures, but I find that a little difficult to visualize, given the kind of power structures I'm familiar with.
So, it is really not the undergrad's fault that he's in this uncomfortable situation, and realistically, there's nothing he could've to prevent getting into this. Sure there's a faux pas, and blame predominantly goes to the adviser, but the discussion starting from "However, when you started trying to prove the theorem, what did you think was going to happen to this graduate student if you succeeded", seems to suggest that OP is accusing the undergrad to be eating off the grad student's pie.
I don't quite agree with that.
@ChrisWhite - ^ ( when you see this ) :)
@DanielSank - Congratulations on the upcoming wedding. :)
Bye guys. :)
 
7:23 AM
@TheDarkSide ...?
 
@DanielSank From the chat logs:
2 days ago, by DanielSank
@BernardMeurer @ChrisWhite I'm getting married in July. If you want to come by we could probably split the cost of stuff.
Hence.
 
@TheDarkSide Oh right. I think @BernardMeurer didn't understand what I meant by that.
I meant that if he and @ChrisWhite wanted to get married we could split the cost of the venue, etc. ;)
 
WHAT!
I should've read the logs properly. I just saw a bi-starred message right below it, and jumped the gun. Sorry. Enjoy single-hood.
:D
 
I am getting married.
 
@DanielSank You're right I hadn't
 
7:33 AM
I'm just pointing out that @BernardMeurer misunderstood my joke.
 
As usual :p
 
Well, it's probably because I don't deliver them well.
 
I don't get jokes in english very well
it't not a funny language :p
 
7:49 AM
@BernardMeurer It's pretty good for humor.
Not as good as Russian.
 
 
1 hour later…
user147690
9:21 AM
Congratulations @ChrisWhite @JohnRennie
 
10:47 AM
Guys
Why are Hilbert spaces separable in QM
Despite the position and momentum basis being not countable
Oh it's because position basis isn't part of the Hilbert space
nvm
What's an example of a countable basis for a free particle in basic QM?
 
11:54 AM
@Slereah $\mathrm{sin}(nx)$ ?
 
I s'ppose
What is it an eigenvector of?
 
@Slereah The differential operator $\partial_x^2$, i.e. $p^2$
 
true dat
 
Why is this room called "The h Bar"? A reference to Planck's constant, presumably. But why "Bar"?
 
H-bar or h-bar can refer to: H with stroke, a Latin letter H with a doubled horizontal stroke (Ħ ħ) reduced Planck constant, which the above symbol represents as a mathematical symbol, ħ = h/(2π) Antihydrogen, an antimatter element represented by the symbol H Steyr AUG H-bar, the light machine gun version of the AUG assault rifle...
 
12:09 PM
@Loong Ah, right. The scaled version. Forgot about that. Thanks.
 
A bar is a place where people congregate
 
@FaheemMitha : and talk.
 
12:31 PM
Yes, I got the play on words. You forgot the most important verb. Drink.
 
We don't actually drink here, though
Well we can, but independantly of being there
 
@Slereah Yes, I meant an actual bar.
Or pub, or tavern, or whatever other name one might choose.
 
Feel free to drink while here, though
Might make for more fun conversations
 
12:55 PM
after you sleep it off you may find yourself banned
 
1:10 PM
@Slereah I don't really drink, personally. I might if someone offered me good red wine for free. But that rarely happens.
 
1:47 PM
@Bass That's not a wavefunction of a free particle, as it is not square integrable.
@Slereah The countable bases for free particles are not different from countable bases for bound particles, as they share their Hilbert spaces.
 
ah yes right
 
2:04 PM
@DavidZ may I nominate ACuriousMind to be added to the list of this room's owners?
 
Oh @ACuriousMind
yesterday, by Slereah
Is the argument that perturbation theory doesn't work because 1) it's basically a polynomial expansion in $\lambda$, so it assumes that the propagator is analytic 2) the theory isn't defined for $\lambda < 0$ 3) same argument as compact support functions not being analytic
 
@skillpatrol No.
@Slereah 2)
If the perturbation series converged, it would have to have a radius of convergence
 
Ah, and you can't because it is on $[0\infty)?$
Closed around 0
 
But any radius of convergence around zero coupling would imply that the series converges for a negative coupling.
Which...could happen, but contradicts all physical expectations.
It's really just a heuristic argument
 
Why can't we have a non-shitty interacting QFT
I should look into lattice QFT
 
2:10 PM
@Slereah Hahahahahaha...good luck with fermion doubling :P
Lattice methods get rid of all those annoying divergences, but they have incredible difficulties getting fermions right
 
Oh well
I'm fine with just doing bosons :p
 
Also, you never know how large the lattice needs to be
 
How does it handle gauges?
 
@Slereah Fine. Lattice gauge theory works quite well
 
Relativistically speaking, if you are accelerating at a constant rate, can you measure your distance from a starting point using $\int v(t) \, dt$ or is this formula suspect because distance changes after you've traversed it? In other words, is your "odometer" an accurate measure of current distance from a starting point?
 
2:12 PM
I would quite like to be able to do non-scattering states in interacting QFT
So it might b cool for that
 
I rouse the beast @JohnRennie
 
@Slereah Oh, you don't do things with "states" in lattice QFT
 
I'm here but eating my lunch ...
 
You just compute the expectation values of observables
 
Same thing
 
2:13 PM
in the vacuum state, usually
 
Are there any more complicated ones
 
They say that the great @JohnRennie can multitask. Surely my miniscule problem is but a bite of lunch?
 
That aren't the usual
vacuum, thermal state and scattering
 
It's typing with greasy fingers that's the problem
 
Boy if I had a nickel...
 
2:14 PM
Anyway, what's up?
 
Oh, the question I asked earlier above. I can re-type it if you'd like.
 
hang on, let me go wash my hands ...
 
@barrycarter @JohnRennie That one.
:29072769
7 mins ago, by barrycarter
Relativistically speaking, if you are accelerating at a constant rate, can you measure your distance from a starting point using $\int v(t) \, dt$ or is this formula suspect because distance changes after you've traversed it? In other words, is your "odometer" an accurate measure of current distance from a starting point?
Ah, of course.
 
Distance is always $\int v(t)dt$, where $v$ is the coordinate velocity and $t$ is the coordinate time. However this is frequently not useful.
Suppose you are hovering a fixed distance from a black hole. This is an accelerating frame because in this frame your proper acceleration is non-zero. If you drop an object then that object is in a non-accelerating frame.
You can measure the velocity of the object and use $\int v(t)dt$ to get its distance, but all you are doing is calculating the coordinate distance, which is not much use.
 
I'm not sure I follow your example. Do we need something as extreme as a black hole here?
 
2:23 PM
An accelerating frame contains an event horizon that is very similar to a black hole event horizon. The two cases are more similar than you might at first think.
 
I ran into this earlier, but I thought that only applied if you were accelerating constantly as viewed from an inertial frame?
 
See this answer of mine for more on this.
 
OK, that seems really complicated for what I'm asking, but thanks.
 
@skillpatrol I think my (newly acquired) God like powers including making other users ownsers of this room. However even if ACuriousMind wanted to be a room owner, I suspect exercising those powers with the OK from the mods would quickly return me to mere mortal status.
@barrycarter You can't really understand acceleration in relativity (either sort) without understanding the Rindler coordinates, and these are more subtle than they appear at first sight.
 
@JohnRennie I'm trying to follow the treatment from the "relativistic rocket" pages, but derive the equations from first principles. I'm pretty close, but I'm running into a problem with distance.
 
2:32 PM
@JohnRennie he has already refused
 
@barrycarter we would normally do the calculation in the earth frame, where everything hould be straightforward and $s=\int v(t)dt$ has the usual interpretation.
 
@JohnRennie Correct. But from an accelerating frame, this won't work, right?
 
It's only when you're trying to work in the rest frame of the accelerating observer that things get screwy.
@barrycarter Right
 
OK, that's what I thought. My "odometer" may read 1 light year, but I may look back and see my starting point only 0.5 light years away?
 
If you're working in the accelerating frame things are a lot more complicated.
@barrycarter you have a major conceptual problem ...
 
2:36 PM
Yes?
 
In the accelerating frame the rocket is at rest so the distance it moves is zero.
In the accelerating frame you're calculating the motion of the earth.
That's a surprisingly hard calculation to do.
 
OK, let me rephrase that. Suppose I look back at the Earth frequently, note down its velocity, and integrate/add up those velocities.
Integrating Earth's velocity curve wont' give me distance, correct?
 
@barrycarter that will give you the distance of the Earth from the rocket in the accelerating frame (Rindler) coordinates.
Which is not the same as the distance in the Earth coordinates
 
@JohnRennie OK, but not in "real" coordinates, right? If I "look back".. but is it the same as in rocket coordinates?
 
@barrycarter you've lost me.
 
2:41 PM
@JohnRennie OK, let me try again...
 
Give me a moment, just going to wash my plate ...
 
@JohnRennie Suppose the accelerating frame drops a pebble every time it sees the Earth move one more light second away, and the pebble instantly decelerates into Earth's frame (just so we won't have moving pebbles). Is the distance to Earth equal to the number pebbles at any given time?
 
You can't drop pebbles and have them decelerate. Objects keep their velocity unless forces act on them
 
@ACuriousMind They're magic pebbles.
 
I find that "magic" objects make for poor thought experiments in relativity.
 
2:46 PM
Are you OK with instant acceleration in relativity, even though it's not realistic?
 
Because the assumption of their very existence frequently violates relativity itself, for example "rigid" objects.
 
I tend to agree with you, but I find it's difficult to derive formulas without some fictitious thingies.
Like instant acceleration. Or are you saying that's a no no?
 
"instant" acceleration is pretty much a no-no. For example, the twin "paradox" is only a paradox because the acceleration phases are neglected.
 
@ACuriousMind OK, can I have the pebbles remember their distance to Earth when they're dropped and quickly (but not instantly) anchor themselves to that position?
 
@barrycarter: as soon as you move away from the Minkowski metric there are no simple transformations between frames like you get with the Lorentz transformations.
 
2:50 PM
@JohnRennie So you're saying I can't derive constant acceleration formulas using SR?
 
@barrycarter How? "Distance" is not a frame invariant.
 
@ACuriousMind Well, OK, if we regard the Earth as fixed, the pebble has coordinates in the Earth frame when it's dropped, yes?
 
What you can do is calculate the spacetime trajectory of the accelerating observer, including calculating the proper length of that trajectory. but you can't simply invert this and calculate the Earth's spacetime trajectory in the accelerating frame.
 
I won't regard the Earth as "fixed" because relativity is all about not choosing a preferred reference frame, but yes, a pebble has coordinates in the Earth's frame when it is dropped
 
For that you need to solve the geodesic equation using the Rindler metric. And that's hard.
 
2:52 PM
@JohnRennie The question I want to ask: for Earth observer, how far is accelerating observer and vice versa (not the same because different frames). Are you saying that question is not possible to answer using SR?
 
Ah, here is the message from @ChrisWhite that I was searching for:
May 4 '15 at 15:13, by Chris White
Seriously, it's not like it's considered acceptable to teach biology in terms of riddles, or topology in terms of brainteasers, or mechanical engineering in terms of limericks. Why are we still teaching relativity in a maximally confusing way?
 
@ACuriousMind OK, OK, regard the Earth as "inertial". That's what my magic pebble does.. it changes velocity until its Earth coordinate is fixed.
@ACuriousMind Fair enough, but the physics FAQ suggests you CAN use SR to derive constant acceleration formulas.
 
@barrycarter How does it know how to do that?
 
@barrycarter I'm not sure the question makes sense, but maybe I've misunderstood you.
 
In particular, how can you determine when it is moving at earth's speed.
 
2:55 PM
@ACuriousMind It's a super-intelligent pebble-shaped life form, ok? I mean, you agree it can be done in theory, right?
@JohnRennie OK. Bob accelerates away from Carol uniformly at $a$ in his reference frame. I've glossed a bit, but is that setup OK?
 
@JohnRennie If Bob looks at his clock at time $t$ he sees Carol moving away at Tanh[a t] (it's fairly easy to establish this). Still OK?
 
@barrycarter No, I don't. I've learned not to presuppose that anything I intuitively think should be possible to be possible. Unless you give me a procedure for doing what you want, I'll not accept it as possible.
 
@ACuriousMind OK, but you agree that the Earth x coordinate exists at the time of drop, right? And that if the pebble is mobile, it can fly to that point right?
@JohnRennie OK, that died sooner than I thought it would. What's wrong with what I said?
 
user116211
2:58 PM
> Not only God knows, I know, and by the end of the semester, you will know.
 
user116211
Legendary.
 
@barrycarter I don't know what it means for a coordinate to exist, and I'm not sure what you mean by "fly to that point". What point?
 
@ACuriousMind The same point (in Earth's reference frame) where it was dropped.
 
@barrycarter How is it going to determine that point?
 
In Carol's frame (the non-acceleraing frame) Bob's velocity is given by $v = tanh(a\tau)$ where $\tau$ is Bob's time. But at time $\tau$ in Bob's frame Carol's velocity is not the same $v$.
 
2:59 PM
@ACuriousMind OK, it can look at Earth, determine its relative velocity and position and adjust until it has correct position and 0 velocity wrt Earth.
 
Sorry about the typos, I'm eating a custard tart
6
Who starred that? :-)
 
@JohnRennie you can unstar it now. ;-)
 
@Loong Oh yes! The POWER!!!
 
@barrycarter Okay, you're dropping tiny rockets that will match their velocity to Earth's. How are they going to find the "point where they were dropped"?
 
@JohnRennie This is good stuff, reading.
@JohnRennie OK, I thought velocity was an invariant of SR. If X is traveling at $v$ with respect to Y, then Y is traveling at $v$ (or negative v if you prefer) with respect to X. Are you saying that is NOT true?
@ACuriousMind They can record that the instant they drop, no?
 
3:04 PM
@barrycarter It is not true. That is true only of inertial motion in flat spacetime.
 
@JohnRennie So if Bob stops accelerating, it will become true?
 
@barrycarter yes
 
@barrycarter Well...*what* do they record? They record time and position in the accelerating frame, if you say that they are able to compute position and time in the earth's frame, then you are presupposing they are able to do the very calculation you want to use them for
 
@JohnRennie OK, that's the part I wasn't grasping. I'm still having difficulty accepting it.
 
@barrycarter I can easily show it.
 
3:06 PM
@ACuriousMind Well, the instant they're dropped they're in an inertial frame, not an accelerating frame. So they should be able to handle it.
@JohnRennie OK, please do, unless it involves black holes.
 
@barrycarter Hm. There's actually a deeper question there: How does "dropping" things actually work in relativity? How do they switch from an accelerating to a non-accelerating frame in an instant?
 
@JohnRennie The velocity with which two observers view each other is non-symmetric and the magnitude of non-symmetry depends on the acceleration of the accelerating observer?
 
In Carol's (non-accelerating) frame Bob starts at rest, then his velocity increases with time and approaches $c$ asymptotically. Yes?
 
@JohnRennie Well, yes, he can't pass $c$, so yes.
 
@JohnRennie assuming you meant "without the OK"... and it would probably get you a "hey, don't do that". I mean, nothing you can do as room owner is irreversible, so if you mess up, it's not that big a deal.
 
3:08 PM
@DavidZ trust me, I shall be treading very carefully at first :-)
 
@ACuriousMind If I drop something from an accelerating craft, it won't continue accelerating, right?
 
@DavidZ can I delete a post (that isn't my own)?
 
@barrycarter Yes, I agree. I'm just not sure what happens in the perception of the dropped thing, since we say that an accelerating frame is different from a non-accelerating frame.
 
@JohnRennie I don't think so? But maybe. You can try it.
 
@JohnRennie Try it with this one.
 
3:10 PM
@ACuriousMind I can't see any way to delete the message. Clicking on the left triangle doesn't give me a delete option.
 
@ACuriousMind Hmmm... I see its velocity curve (or the inverse curve) as flattening out. That's still continuous, although you've got a point re derivatives.
 
@JohnRennie I guess your powers are only demigod-like, then ;)
 
With great power comes custard tart.
 
@DavidZ I saw a comment in the mother meta that room owners could move posts, including moving them to a "bin", but I can't find any way to do that.
 
They love moving me to the trash in other rooms
 
3:11 PM
@barrycarter I've finished the custard tarts now - very nice too. I do like a good custard tart.
 
@JohnRennie Go to "room > move messages" from the link at the top-ish right
at least that's how I do it
 
@JohnRennie OK, please continue with your illuminating example.
 
1 message moved to Trash
It works :-)
 
Ta-daa!
 
What a strange place.
 
3:13 PM
Yeah. So you can do that if there are ever messages that need to be removed from this room.
 
@barrycarter OK in Bob's (accelerating) frame Carol starts at zero speed, her speed rises with time, but then Carol's speed slows again as Carol approaches the event horizon in Bob's frame. Carol asymptotically approaches zero speed.
So we have two completely different asymptotic behaviours in the two frames
 
@JohnRennie Wait a minute... doesn't Carol's speed approach $c$ with respect to Bob? I don't see why Carol's speed decreases? At what point does that happen?
 
@barrycarter This is what I keep telling you. Bob's spacetime is radically different to flat spacetime. It contains an event horizon very similar to a black hole event horizon.
 
@JohnRennie OK, but, at any time, Bob can stop accelerating, right? And then things return to "normal"?
 
Yes
 
3:16 PM
@JohnRennie So Carol's speed would suddenly increase drastically as Bob moved back into an inertial frame?
 
Yes, and at the same time her distance from Bob would increase drastically.
 
0
Q: Is showing thanks like this acceptable?

Anubhav Goel Thanks for your answer. Because i gave the bounty to other answer i will upvote other answers of yours in this site. can we do something like this?

 
@JohnRennie And this applies no matter how slowly Bob is accelerating?
 
The more slowly Bob accelerates the farther away the horizon is, but at any acceleration > 0 the horizon is there
 
@JohnRennie Right, right. But you're saying Bob at .999c and accelerating at .01 m/s^2 is dramatically different than Bob at .999c in an inertial frame?
 
3:19 PM
The horizon distance is $c^2/a$
 
@JohnRennie You seem to be saying non-zero acceleration has a discontinuous jump.
 
@barrycarter why am I saying that?
If you change discontinuously from inetrial to accelerated motion then the metric changes discontinuously, but that isn't physically reasonable.
 
@barrycarter For infinitesimal acceleration, the horizon is infinitely far away, where is the discontinuity?
 
@ACuriousMind : I know the answer to this. .
 
For the sake of argument, just suppose the horizon's distance is 1/a for a the acceleration. That's not discontinuous, but it does exactly what John says how the horizon behaves.
 
3:21 PM
Incidentally I'm about to start on some cranberry, orange and oat slices
2
 
@JohnRennie I really really REALLY wish I had more time, because I'm pretty sure I can break you again :) But I don't. The problem I have is you say Carol's velocity goes to 0 in Bob's frame, but when Bob stops accelerating, it jumps back to .999c or whatever. However, I will have to defer until later.
 
@JohnRennie This is a chat, not a text version of instagram
 
I am saying that, and it's true, but these are just coordinate velocities and distances.
 
I also don't know what an "oat slice" is
 
@ACuriousMind blame Barry, he's the one who interrupted my lunch :-)
 
3:26 PM
"Deliciously moist golden cakes topped with honey coated toasted oat flakes"
That makes me almost hungry
 
Oh well, back to reading about axions and wormholes again. I do like reading science fiction ;-)
 
Aw come on, axions aren't that fanciful!
 
Axions as a component of the stress-energy tensor that behaves like exotic matter?
 
Who knows
I don't know much about axions
 
@JohnRennie : there's a lot of science fiction around. I used to write science fiction.
 
4:18 PM
wikipedia is the best place to learn math.
 
Eh
Depends for what
 
generally.
 
user116211
4:38 PM
Congratultions @kyle Kanos (hmmm... you're not active recently) for completing 10000 CV reviews :)
 
vzn
4:56 PM
@kevinTahN. if you have any interest in (math) research plz let me know. what is quantconnect?
 
5:17 PM
@vzn Quant connect is for people who want to predict (or chart or otherwise know about) "stock prices" and the link. Similar to quant.stackexchange.com
@kevinTahN. If you like M.U.L.E. (the game), email me at carter.barry@gmail.com and/or read github.com/barrycarter/bcapps/tree/master/YAMC
 
@barrycarter Pro tip: Writing [name-of-se.SE] creates a direct link to the SE site you mean. E.g. [quant.SE] gives Quantitative Finance.
 
@ACuriousMind Nice, thanks! I love Astronomy and Mathematics and Mathematica
Coolness
And of course Quantitative Finance
 
Also works in comments on the site, not only in chat
 
@ACuriousMind Also awesome. Area 51
Now I just feel I have to use it even when unnecessary.
 
user116211
@barrycarter Ha!
 
user116211
5:22 PM
Great conclusion ;P
 
As an Astronomy-er, I sometimes wonder about the Mathematics of the stars, and sometimes even use Mathematica to compute formulas, not just for stars, but for Quantitative Finance too, since I want to maximize my Personal Finance & Money
 
...why do we have a tag ? What has that got to do with physics?
 
@ACuriousMind good question
We could probably merge it into or something
A tag merge request on meta might be appropriate, just to collect comments on the idea
 
5:56 PM
What's wrong with having general and specific tags?
Or is the point that this is engineering?
 
@DanielSank My main point is that it's engineering, and that the physics part of it is too specific to be useful - I cannot conceive of a physicist thinking "Why, yes, I want to read exclusively questions about the measurement of energy efficiency!"
 
@3075 I always found Wikipedia's math articles mediocre at best. They try to be too formal.
 
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