@bolbteppa think its a much bigger deal than ppl realize, making my neurons trigger/ buzz, found related material, think its compatible/ consistent with SLC theories, makes me wanna write paper on it! alas only a few ppl in the world are tracking similar areas directly/ seriously, although some are quite eminent authorities :) eg arxiv.org/abs/1701.08300 nytimes.com/2020/06/25/magazine/…

vzn1.wordpress.com/2021/01/24/subquantum-revolution-begins-now

@vzn you know one need's to be careful basing any conclusions on a single experiment

Remember the speed of light fiasco a few years ago etc

@vzn would you not try to read like the first chapter of an established QM textbook to try to get a sense of what the main thing is claiming

hmm thanks @bolbteppa that's what i was thinking

sorry for removing, i figured there's no point badgering people in here with my study issues

but thanks for your comment :)

oh i missed that what did you say?

ahh gotcha

yes good point

i'm already in the physics department ~50% of my lab days anyway due to research project, and know people in the department

@bolbteppa QM textbooks are great + so are physics history books. there are already replication(s), see the blog post. agreed it is early to make any definite conclusions. however, conversely, sometimes it takes scientists many, many years to recognize genuinely new phenomena, and to recognize when theories are incomplete or need (major) revision or even near-replacement, there are many cases of that in the history books. think almost nobody is looking at the experiment extremely carefully...!

v.a should be 0 which would imply that d²θ/dt is 0, how can we justify that

I know that d²θ/dt is 0 for uniform circular motion but here we are dealing with any general motion, it would be great if anyone can provide general explanation for this

Not clear what you're asking, if $\mathbf{v}$ has constant speed then $||\mathbf{v}||^2 = \mathbf{v} \cdot \mathbf{v}$ is constant so it's time derivative is zero but it's time derivative is also $2 \mathbf{a} \cdot \mathbf{v} $ so acceleration is perpendicular to velocity

So, it's my birthday today!

Let's make HBar great again!

I wished Guido van Rossum (creator of python) on 31st Jan and got his wishes today on 1st feb!

:-)

@RewCie Happy Birthday, AKS :)

@RewCie HBD AKS!

physics.stackexchange.com/review/suggested-edits/328733 <--- is this edit correct?

@Yashas yes, general relativists almost always write "tensor" instead of "tensor field"

which is why the edit is kinda superfluous

although it is technically correct

@NiharKarve Thanks buddy :-)

@satan29 Thanks man :-)

@bolbteppa yeah I get, thanks

happy birhday @RewCie

@bolbteppa It's just a very explicit demonstration of weak measurements and open quantum systems, see physics.stackexchange.com/a/610331/50583. Note that the authors themselves do not claim that anything here is unexpected or inconsistent with standard quantum mechanics, they're just following an unfortunate precedent in their field to connect these kinds of observations to Bohr's "quantum jumps"

Cool will try and make sense of it

@vzn ^

Hi all, sorry to bother, but I need a feedback to be sure. I am trying to understand why being inclusive in the final state is important for infrared safety in QCD. For what I have seen in e+e- -> hadrons, to avoid singularities we have to consider all the Feynman graphs, degnerate in the IR limit, order by order. In this way we avoid problems due to emission of soft gluons and collinear soft partons.

But this isn't enough because divergencies arise even in the case of non soft collinear partons. And this proccesses are present because we are being fully inclusive respect the final state. Without being inclusive I would miss these contributions to the cross section, and there wouldn't be a finite result. Is this right?

@antimony Thanks Antimony! :-)

@JohnRennie that seems to put you waaaay ahead of the field

Leaderboard for total gold/silver/bronze badge numbers

though @Qmechanic still beats you on total bronze and total badges

@EmilioPisanty he has a shedload of Announcer badges because he has posted so many "and links therein" comments.

@JohnRennie ah, that's what it is

I don't think his/her Announcer badges would fit in a shed, though

Hello Guys...

Hi @JohnRennie Sir.

@123 hi :-)

BTW @mods @Qmechanic @ACuriousMind @etc, is it me, or is it odd that the fourth user with the most votes on this site has voted 15k times, but only has 101 rep?

Or is there e.g. some well-known and above-board bot mechanism that's partially responsible?

Hi all. Is there a reason why Kruskal (and Penrose) diagrams are drawn with rotated axes? I know that the axes still form a 90° angle, but wouldn't be visually easier to rotate the whole picture by -45°? Is there a reason behind this choice of how to show these diagrams?

this picture VS the following

@newUser the first example on Wikipedia looks like this

@EmilioPisanty I am studying Poisson's and Wald's books, I didn't check on Wikipedia

... but it's still possible that the overwhelming bulk of the professional literature uses the axes in your first picture. It's not my domain, so I can't say much more.

Maybe @JohnRennie can help.

I know that there is no difference by just rotating the axis. I was just wondering whether there were some other reasons why Poisson and Wald show the diagrams with this orientation of the axes. To avoid confusion with Cartesian coordinates maybe?

@EmilioPisanty It's a bit odd, but that particular user just seems to spend a lot of time on SE reading and upvoting things :P (you'll find them in the top voters of many sites)

@ACuriousMind huh. Fair enough.

I don't see any particular evidence this voting is automated, and even if it were, it is unclear whether that's against SE policies to begin with

@ACuriousMind yeah, fair enough. If nothing else, they have a Fanatic badge, which points to on-site presence.

@newUser the angled lines are not the axes, they show the location of the event horizon. The $u$ axis is horizontal and the $v$ axis is vertical, but we usually don't draw the axes as they don't give us any useful information.

seriously, the things some new users do...

To be fair, that's kind of how mathjax works

Mathjax just converts everything to that weird HTML format

$M_1$

@Slereah you mean MathML?

The true face of Mathjax

@Slereah just because it's also XML doesn't mean it's HTML ;-)

all is HTML

or LISP I guess

@Slereah eh

if you'd said All Is XML then yeah, I'd kinda agree

I mean XML isn't even the first markup language

I forget the name of the original ML

particularly after tinkering around with exporting to x3d for a long time

> SGML, the first standard descriptive markup language ?

Maybe

Or this I guess

eh

@JohnRennie thanks. What are the definitions of $u$ and $v$ axes then? not really sure what they are. I only have coordinates $U$ and $V$

@bolbteppa saw all that when 1st posted, while helpful/ insightful dont think its comprehensive/ definitive. there are some very subtle/ complex/ complicated issues involved. it seems possible even the paper authors are not fully aware of the whole situation/ context/ background etc., think its very worthwhile to discuss further but very few people have the bkg/ time/ patience to do so. think its a small miracle its gotten the attn it has so far...

@ACuriousMind Is a diffeological space just a manifold sans the topology condition?

@Slereah The nitpicker in me wants to say he doesn't know what you mean, but the rest of me says that's essentially it :P

What is the nit

I don't know what "the topology condition" is supposed to mean, or, if you mean the "locally diffeomorphic to $\mathbb{R}^n$", what remains of a manifold definition if you remove it

The map from $R^n$ to $X$ doesn't have to be a continuous map, I guess?

Only the overlap conditions and smoothness have to hold

the usual definition of a diffeological space does not suppose a topology on $X$, so "continuous" isn't an available notion

yeah