12:44 AM
Google is being evil again. Who'd 've thunk it?
250

I often have my browser's devtools open since I'm often debugging a snippet. I noticed this message often recently: Why is Stack Overflow trying to start audio? Update. I see it's from an ad? https://static.adsafeprotected.com/sca.17.4.95.js Update 2 It happens when this ad appears, From...

3 hours later…
3:51 AM
Are people having vacation or something? SE chats are awefully quiet in June

4:36 AM
@Secret everyone has gone to Glastonbury

1 hour later…
5:37 AM
@JohnRennie Bunch of hippies! :) (Says the guy listening to Hendrix and Pentangle).

@PM2Ring I grew up in a village only ten miles from Glastonbury

5:55 AM
@JohnRennie Ah! Back then, you had real hippies at Glastonbury for the summer solstice.

@PM2Ring as I recall they were a dirty and smelly lot. The romance of the hippy lifestyle has been greatly exaggerated.

@JohnRennie I guess so. ;) Most of the hippy types I knew back in the 1970s were Australians, although I did know a few from the UK & the USA. And they may not have been as hard-core as the hippies that "invaded" Glastonbury.

6:13 AM
@PM2Ring we used to call them crusties :-)

6:25 AM
the weather these days is so frustrating, either raining like downpour or over hot with vehement sunshine.

We've had a lot of rain in Sydney. Today is the 1st day in over a week that we've had some sunshine.

you are in the southern hemisphere of Earth? Then I imagine you are in the beginning of winter, the opposite to us.
I am the subtropic zone.

@CaptainBohemian Yes. And we had an unusually warm autumn. So the cold temperatures are a bit of a shock. Of course, the climate in Sydney is rather mild, compared to some places. So I can't really complain.

6:50 AM
I like the places with abundant sunshine. I have been such a place, that is Beijing, where there is sunshine almost every day and if it rains, it only rains for a short period rather a whole day or consecutive several days. I feel this kind of place is so congenial. I can feel energetic in daytime most of time because there is sunshine most of time in daytime.
I mean I have been to.

@CaptainBohemian For 7 years, I lived a bit north of Coffs Harbour. It has a nice climate, but maybe not sunny enough for you. ;) I've been back in Sydney for 2 years, and I don't like the colder winters.

A prolonged period being either rainy or overcast often makes me melancholy. I remember there have once been a month being that way not long ago; that kind of weather is really daunting.

7:10 AM
Whatever happened to Blue?

@PM2Ring It's like a person in my Facebook told me he also lives in New South Wales but I foreget the definite place name he said. He said his place can be very hot in summer, like over 40 C and very cold, like below 0 C in winter.

7:24 AM
@CaptainBohemian Yes, it sometimes gets over 40°C, but it almost never gets below 0°C near the coast, where most of the population live. It only gets that cold inland, and in the mountains. Our mountains aren't very tall, but we have a long chain of mountains parallel to the east coast, and during winter our southern mountains get plenty of snow.

morning

7:45 AM
I see Rob & Arpad are still neck & neck on that beta+ decay question. Let's hope the OP doesn't accept Arpad's answer...
@SirCumference You mean Sanchayan Dutta? Hanging out in QC, I expect. He's been rather scarce in this room since Chair left.

8:23 AM
@RyanUnger The cobordism proof thing implies that the fact that a timelike vector field is nowhere vanishing implies that its flow curves don't have any endpoints, and therefore the curve must end on $S_2$, but is that correct
Couldn't a curve end up on some singularity instead
For Lorentzian manifolds a compact manifold can be geodesically incomplete

8:36 AM

5 hours later…
1:26 PM
@PM2Ring well, thank the stars that it wasn't hnq'd

1:43 PM
@Slereah wot r u talking about
What is a cobordism proof thing?

The Geroch paper I passed along a few days back
"Since this possibility has been excluded by hypothesis, we conclude that every such curve $\gamma$ must have a future endpoint. This endpoint cannot occur in the interior of $M$, for $\xi^\mu$ vanishes nowhere, nor on $S$, since $M$ is isochronous. Thus, the future endpoint of $\gamma$ lies on $S'$"
But if I consider let's say the Clifton-Pohl torus, I get an inextendible curve without an endpoint yet with a finite proper time parametrization
So I am wondering if that is an issue

2:02 PM
@Slereah what is isochronous
also he says there's an endpoint
do you buy that much at least?

Isochronous is old timey speak for time-orientable
Well, the previous part of the proof shows that there is an endpoint because if there isn't, the spacetime contains a CTC

what does this have to do with cobordism

But it assumes that, if $\gamma$ has no future endpoint, then it may be parametrized by a continuous variable $t$ with range zero to infinity
@RyanUnger the proof shows that "reasonable" compact spacetime cobordisms must keep the same topology

uh can you link the paper again

Basically he assumes that a curve without an endpoint in the interpolating spacetime must range over some unbounded interval
But I'm not sure this is true
would be true with a Riemannian metric, but alas

2:07 PM
the fuck is a "compact geometry"

"a compact geometry means the manifold is compact."

@Slereah just to mess you up some more, I'm not convinced that the usual thing with $\chi$ works for a manifold with boundary
you need to use the Poincare-Hopf theorem, I don't actually know how that works for manifolds with boundary

I think he talks about it, or at least some other paper does

actually scratch that
you need to know what the double of a manifold with boundary is

The conditions for a Lorentz metric on a manifold with boundaries

2:09 PM
then you double $M$
the Euler characteristic doubles
so you put the vector field on the double
and then restrict to one copy
so that's fine

fortunately $2\times 0 = 0$

since $2\cdot 0=0$
@Slereah ok so now what is the issue

Basically the statement on the proof of thm. 2
"Suppose first that $\gamma$ has no future endpoint. Then $\gamma$ may be parametrized by a continuous variable $t$ with range zero to infinity"
Hm, although...
I guess this is technically true even if the curve ends at a singularity

well $M$ is compact right

Since you can reparametrized a curve $(0,1)$ to $(0, \infty)$

2:14 PM
yeah right
so this might be trivial

So yeah that's probably fine
I don't think it needs to be the proper time that needs such a wide range

seems strange tho haha

I have no idea what a curve on the Clifton-Pohl spacetime looks like tbh
I don't know what the "singularity" looks like

I don't know what that torus looks like

It's the torus in O'neill regarding geodesically incomplete compact spacetimes
In geometry, the Clifton–Pohl torus is an example of a compact Lorentzian manifold that is not geodesically complete. While every compact Riemannian manifold is also geodesically complete (by the Hopf–Rinow theorem), this space shows that the same implication does not generalize to pseudo-Riemannian manifolds. It is named after Yeaton H. Clifton and William F. Pohl, who described it in 1962 but did not publish their result. == Definition == Consider the manifold M = R 2...

2:19 PM
yeah I have no idea

If a compact manifold has a singularity and you perform the boundary construction of that singularity, what does the resulting space look like???
A mysterious thing
Considering the construction of the Clifton-pohl torus it seems to be very curly geodesics that are incomplete
Not sure what that means for boundary construction

@Slereah if it just winds then there's no endpoint
take an irrational line on the flat torus

yeah
Those ones have infinite proper time, tho
I suppose that there doesn't have to be a "physical" singularity point for geodesically incomplete spaces
Although I should try the Penrose construction for boundaries
I'm curious what the resulting space is
The GKP boundary
though I forget how much of a reasonable spacetime you need for GKP
Might not work for Clifton-Pohl
I suspect it would not
Compact spacetime so probably $\forall p, I^+(p) = I^-(p) = M$
I think you need it to be at least distinguishing
I guess you could work out some topology by considering classes of curves instead but that seems trickier

1 hour later…
4:03 PM
@PM2Ring Welp, I'm asking since I noticed he's also missing a mod diamond now

4:25 PM
looks like experimental control of timespace fluid waves to me :o o_O

4:47 PM

4:58 PM
@JohnRennie Yeah, wonder what happened
Though I guess it isn't my business

@SirCumference probably too busy now he's made it to one of India's top universities!
4

@ACuriousMind Oh wow :/

Wtf
@ACuriousMind am I not allowed to run for mod?

@RyanUnger cf. meta.stackexchange.com/q/274114/263383, so given that we just recently had an election you should be elegible for the next ;P
@Slereah I've never understood some people's aversion to pants.
@EmilioPisanty I'm trying to figure out how "hnq'd" is pronounced..."hunked"?

@ACuriousMind they don't have anything to hide

@ACuriousMind "hanked"?

5:36 PM
@Slereah Yes, but they also don't have any pockets to put their wallet, keys, etc!

"henked"?

@ACuriousMind that's what the fanny pack is for

@EmilioPisanty (ge)henkt is German for "executed (by hanging)", so I guess that fits...

@ACuriousMind does suspend mean chat or main site
Being chat suspended is easy
Although I don’t think I have been since I returned
Very curious
Can you see if I’ve been flagged

@Slereah ugh

6:13 PM
@EmilioPisanty Indeed! I did consider editing $\beta+$ into the title... Just kidding. :)
Speaking of questions with Latex in the title, I see nobody's tackled this metrology question yet.
3

One of the most "embarrassing" failures of modern physics is the current disagreement of measurements of the gravitational constant $G$, well beyond reported uncertainties (and agreeing only to about 1 part in $10^5$). Part of the difficulty is that it's impossible to "screen out" the gravitation...

How would you do a higher precision G measurement? A huge Cavendish balance in a trans-Neptunian orbit? Or maybe an orbit in a plane perpendicular to the ecliptic.

@Qmechanic can I just briefly tell you off for the title edit at physics.stackexchange.com/posts/488519/revisions?
Element symbols work just fine in their own, and unnecessary MathJax just makes them harder to read in situations where MathJax is not enabled
Say, external searches, or the inbox and achievements boxes in other SE sites
@PM2Ring not a particularly good question, I think
Basically, any question that boils down to "is everyone in this particular field, that I don't have any detailed knowledge of, completely incompetent", generally has a simple answer, "no".
Ditto with "I looked but I haven't found any discussion of X in the literature" - the problem is with the "I looked" and not with the literature.

6:35 PM
@EmilioPisanty Fair points. I suppose you need to dig pretty deep in the literature to find all those mundane details. Still, it'd be nice to see an answer that explains why G is so hard to measure.

6:58 PM
@PM2Ring it would
though I suspect that the unconstructive way that this one has been phrased has pretty much guaranteed that no serious metrologist will answer it

Ok, but the OP does finish with "If not, what am I misunderstanding about the physics of the situation?" But I totally agree that the question should be rephrased so that it doesn't imply that the physicists trying to measure G aren't incompetent.
I'll write a comment to that effect.

7:21 PM
@EmilioPisanty : Well, it is nice to have symbols (that can appear in equation mode) to be in math mode to be consistent but searchability is an important point.

7:46 PM
Please don't link to crackpot articles about obtaining free energy from ZPE. — PM 2Ring 5 mins ago

@PM2Ring He seems to have swapped the link out for a seemingly irrelevant wikipedia link. Now all we need is "a way to do low energy positron generation from quantum vacuum"... sounds totally trivial
... and naturally he adds a new wikipedia link to zero point energy to add to the confusion... ZPE doesn't imply anything about generating positrons from a vacuum, does it?
well except I guess in a virtual particle sense maybe... but not in like a practical generation, get energy from it, sense

8:09 PM
@JMac Correct. You can't do work with the vacuum ZPE. Unless the vacuum is actually in a metastable state, and there's actually a lower state which is the true vacuum.

@PM2Ring ::sigh::
there you go, though.

It's the same as trying to make hydroelectricity at the Dead Sea. Sure, the water has gravitational potential energy, but you don't have a lower location for the water to fall to.

@PM2Ring Yeah, makes sense to me. It's called "zero point"; I always see pseudoscience references to it as like a magic energy device; but the term "zero point" (and like 2 seconds of googling) have always made it pretty clear to me that it's just a minimum potential, so you can't do much with it

Greg Egan has a novel, Schild's Ladder, where they prove that the vacuum is metastable. They successfully attempt to induce a vacuum transition to the lower state. And the new vacuum propagates outwards at half the speed of light, destroying the current universe.
@JMac Oh, yeah. The free energy crackpots love ZPE.

On a slightly similar topic, I recently watched season 2 of the German show Dark on Netflix. I've really enjoyed the first two seasons, but the use of pesudoscience terms to explain time travel is hilarious. They discover "dark matter" which then leads to them finding the "god particle" which enables time travel. The show's great IMO; but the abuse of science is so comical it broke some immersion

8:24 PM
Oh dear. It's understandable that they'd use jargon words that are probably familiar to the general audience, it's a sci-fi tradition, especially in movies & TV. But it does make it a bit cringe-worthy for people who know what the terms actually mean.
OTOH, I can read old pulp sci-fi from the 20s and 30s that has totally ridiculous stuff. But when I'm reading those stories I just recalibrate my expectations so it doesn't break the suspension of disbelief unless they're internally inconsistent.

8:43 PM
Yeah, it wasn't too off-putting, because I wasn't expecting real accurate physics, but it would have been nice if they avoided actual science terms that are so often poorly used

3 hours later…

@DanielSank Is this at the right level for physicists-who-know-stuff-just-not-light-stuff?
It definitely seems to be glossy-cover season around here o.o
... though it seems my colaborators had slightly better luck getting on the cover =P
@PM2Ring thanks for the ping =)