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2:49 AM
Is my answer to this one physics.stackexchange.com/questions/299671/… ok? It seems a bit confusing...
2 hours later…
1 hour later…
5:33 AM
@JohnRennie uh what's the issue here
@RyanUnger what are you asking? Why I've drawn attention to safesphere's comment, or what am I objecting to in that comment, or both?
@JohnRennie I can read the last sentence...I don't understand what the disagreement is about
I can't tell if he's defending JD
and I'm not going to try to read what he wrote
He says:
1. the timelike geodesic is still interrupted (read ends) at the horizon
2. Finally, there exist no frame, in which a crossing of the horizon can be observed and thus is unphysical
3. Open your mind!
1 and 2 are wrong and 3 is offensive
I must be missing the context here
what's the timelike geodesic
The trajectory of an observer falling freely into the black hole
5:39 AM
the commenter says it ends?
he seems to be using the rest of the words correctly
I doubt that's really what he means
Correct. Safesphere says the geodesic ends at the horizon. Not the singularity, the horizon.
@RyanUnger that is what he really means
hmm well
He is defending Duffield's belief that the region of a Schwarzschild black hole inside the horizon is unphysical because nothing can ever get there.
I'll leave corecting people about GR on the internet to you
I can't figure out how people come to such conclusions
hard to argue against them
I have no intention of trying to correct him. I can spot a futile waste of time when I see one.
5:42 AM
I don't understand how coordinates can be unphysical
"Because gravity isn't a geometric property of space-time."
oh dear god
he reads everything Einstein wrote and that's the conclusion he draws?
Observers falling freely into a black hole explode before they reach the horizon.
Duffield's firewall theory.
@JohnRennie something I never did understand and as far as I can tell, does not prevent me from writing math papers, is what happens if you hover outside of the horizon and poke it with a stick
You’ll need some kind of jet pack of course
But just sit a foot away and poke it with a long stick
Can someone please tell me if there's some mistake in the proof of the Gram-Schmidt orthogonalization here: chem.libretexts.org/Bookshelves/…?
@RyanUnger I confess I don't know what will happen either, but I'm sure someone will have done the calculation.
5:58 AM
@JohnRennie I actually suspect something strange will happen. Most of our intuition about GR comes from the geodesic equation —point particles
I really have no idea how extended objects really behave
I might ask this on the main site or has it been done already
I definitely know you can’t just pull back on the stick without consequence
So does the end just get sheared off? But that’s weird because we’re told the horizon is not a special place
Am I making a naive mistake?
6:34 AM
British summer. This time last week we were all complaining how hot and sunny it was :-)
6:46 AM
it's 35 degree C here but the body sense temperature is 43 C
7:24 AM
Remove your body
7:39 AM
that "body sense temperature is 43 C" is what I take from our Weather Channel
I don't know what the air temperature 43 C feels like because I have never experienced it.
@JohnRennie open your mind, man!
everything is just good vibrations resonating with each other
today the moon rose at 5: 07 and set at 18:57, and the sun rose at 05:21 and set at 18:38. So the moon almost meet the sun. This is when the moon is at new moon phase.
@Gyromagnetic hey, I'm old enough to remember the hippy era. It was rubbish! :-)
@CaptainBohemian presumably that's a function of the relative humidity i.e. high humidity inhibits the evaporation of sweat so it makes the air feel hotter than it would feel at zero RH.
With my current exploration of how the Lagrangian works
I think it's time to revisit the van Vleck determinant
7:57 AM
@JohnRennie the humanity indicated in the Weather Channel is 58%.
2 hours later…
10:06 AM
@JohnRennie Turns out that that day in July everyone thought was the second hottest on record in the UK was actually the hottest
@Mithrandir24601 I remember the Cambridge weather could be challenging. In summer it gets very hot and in winter very cold.
I once cycled across the fens to a friend's house in Waterbeach on a February evening and it was about -10°C. Despite gloves I'd completely lost all feeling in my hands by the time I arrived.
10:39 AM
@Gyromagnetic Is this the new Minecraft portal update?
11:37 AM
@NovaliumCompany that's just the DMT kicking in
12:09 PM
Q: Do we really need a partition of unity to define integration on manifolds

Phil-WOn wikipedia, there is the following : A partition of unity can be used to define the integral (with respect to a volume form) of a function defined over a manifold: One first defines the integral of a function whose support is contained in a single coordinate patch of the manifold; th...

just what I need
12:20 PM
A: What force enables us to walk? Friction or normal reaction?

AmitFriction force is proportional to Normal force as well as perpendicular to it. (http://hyperphysics.phy-astr.gsu.edu/hbase/frict3.html). So both are related in my opinion. Normal force is not necessarily in the same line as gravity (for example when the surface is inclined). It is the Friction f...

And it is in Lee too
proposition 16.8
It's one of my pet peeves when people post a new answer that contains no new information compared to what has already been answered for the question.
Quite serendipitous
12:31 PM
I think the point of partitions of unity r.e. integration is that integration on manifolds has to reduce to integration on charts and charts are local they don't cover the global manifold so they let you extend local to global and you need certain topological properties for this to be possible, I wonder if that answer about not necessarily 'needing' them to do integrals is right
In mathematics, a partition of unity of a topological space X is a set R of continuous functions from X to the unit interval [0,1] such that for every point, x ∈ X {\displaystyle x\in X} , there is a neighbourhood of x where all but a finite number of the functions of R are 0, and the sum of all the function values at x is 1, i.e., ∑ ρ ∈ R ρ ( x ) = 1 {\displaystyle...
@bolbteppa Well as I said
It's in Lee
Hopefully this will adapt well to a non-Hausdorff manifold
1:24 PM
Riddle me this, Batman
In a non-Hausdorff manifold, is the set of adjacent point of measure 0
Second-countable case, at least
I think I need to prove that the set of connected adjacent points is $n-1$-dimensional
Oh god
my site is now #5 on google for "non-hausdorff manifold"
2:05 PM
"Proposition 2.5. Let $S$ be a set of compatible apparition points in a non-Hausdorff manifold $M$. Then $S$ is nowhere dense in $M$."
2:35 PM
Is my 5mW/10mm^2 laser really as safe as it seems (compared to the sun's 1kW/m^2)?
Sunlight is 2 times more powerful. Unless i messed up my calculations.
The problem with lasers is that they are focussed by the eye to a tiny point on the retina and they destroy the tissue at that point.
This doesn't happen with sunlight because sunlight has an angular divergence of half a degree or so and this prevents it from being so tightly focussed. However staring at the sun will also destroy the cells in the retina eventually. There are a few cases of this everytime people gather to watch an eclipse.
@JohnRennie Plus there's the issue that not all lasers emit in the visible spectrum, whereas light is primarily in the visible spectrum; so just looking into an infrared laser you might not notice that you're damaging your eyes, whereas looking into the sun is obviously very bright and easy to react to.
2:56 PM
@JohnRennie something funny happened to that post
@EmilioPisanty really? :-)
I'm pretty sure that in a non-Hausdorff manifold, the set of adjacent points is gonna be some union of boundaries of the overlapping charts
hence likely to be $n-1$ dimensional
@JohnRennie it used to have a 7500 next to it but it seems to have gone missing
Now to prove it
2:58 PM
@EmilioPisanty good heavens, you're right!
@JohnRennie I don't understand your last comment there
@EmilioPisanty English humour. Just ignore it.
so, anyways, brunch on the 19th?
I should probably be in brum in the afternoon with enough time to have some form of useful discussion
but Chester-Birmingham by train should presumably be quick and painless?
@EmilioPisanty I think you need to change at Crewe, but then to get anywhere from Chester seems to involve a change at Crewe.
yeah, google maps calls for changes at Crewe and Wolverhampton, and then another train from New Street to University
but it still fits under two hours so I'm not fussed
3:05 PM
Crewe is a massive rail hub. To go south from Chester or indeed anywhere in this area involves a change at Crewe.
But Crewe is an easy station to get round. It's such an important hub that it's well organised.
@EmilioPisanty Wow... how many people get year long suspensions for bad answers? It takes some serious commitment to have positive rep while having all your "top tags" show negative score.
I guess the direct trains go from Liverpool but it's not worth going against the overall direction of travel
@EmilioPisanty you can check the times, but I suspect it will be quicker to go from Chester via Crewe than back to Liverpool and take a direct train.
@JMac I'm unsure how many such bans have been handed out, or to how many users. But for this particular case the Wayback Machine will tell you that this is the third time around.
Serious commitment indeed.
3:07 PM
@JohnRennie eh, I'm not fussed about the details, they can be sorted out on the day.
but I'd like to have an overall layout for the day
Have you been to Chester?
Oh, I thought he was banned for how he dealt with the feedback in the past. If that's the third time for this reason... it seems strange to keep giving him 1 year cooldowns
@JMac it's the longest ban our mods can hand out. Longer than that requires action from the SE staff.
goodness, is Chester really one hour from Liverpool? that's ridiculous, I thought they were essentially equivalent.
ah well
3:09 PM
ron maimon got banned for like 10 years right
@RyanUnger At least a century now
@RyanUnger yes, by SE community managers
@EmilioPisanty the Chester Liverpool train stops everywhere along the line so it takes a long time.
Wait, not quite a century, looks like he's back in 2092 (assuming that '92 doesn't mean like 9992 or something)
@EmilioPisanty it's about a 5-10 minute walk from the station into the centre of Chester if you want to meet up at one of the more scenic coffee houses.
3:11 PM
as I understand it, this happened while he was on a one-year PSE mod-imposed ban, and it was imposed network-wide due to interactions elsewhere on SE
@JMac 2292
(search for that number on this chat if you want the details)
@JohnRennie yep, that works
something like early lunch, then?
wow wtf did he do anyway
@RyanUnger who cares?
@EmilioPisanty lunch or just a coffee?
@JohnRennie coffee's OK with me, but if it takes an hour to get to Chester let's get an early lunch =)
(sorry, I need to skip out for a bit)
In your place I'd grab a big breakfast at the hotel, meet early ish for coffee and chat then get off to Brum and have lunch there.
3:13 PM
@EmilioPisanty you don't think it's curious that someone got banned for 275 years
If I have a set $\{ H_i \}$ of $H$-submanifolds covering my manifold
Is the pairwise intersection $H_i \cap H_j$ never empty
@Slereah non-hausdorff manifolds is veering into crank territory
Is it
I guess it might be if I was saying the universe was a non-Hausdorff manifold
Find out more in my new blog
3:17 PM
@JohnRennie sure, that works too
@EmilioPisanty whatever you prefer. I have generally finished the work I have to do by 9 a.m. so I can wander into town whenever suits you. I'll have a think about a nice scenic coffee shop we can meet in.
@JohnRennie how cute
The station is on my side of town so I can meet you there.
@RyanUnger finishing work at 9 a.m.?
@RyanUnger Would you prefer a meeting in a crack house
I don't think Chester has any crack houses. It's far too genteel for that :-)
3:22 PM
Crack mansion, perhaps
@EmilioPisanty or would you be interested in walking round the city walls rather than coffee? It's about a 40 minute stroll, less if you hurry, and it's pretty scenic.
3:44 PM
@JohnRennie that sounds interesting
@EmilioPisanty I quite often take visitors for a walk round the wall if the weather is good.
It's a pleasant stroll and it's quiet enough to chat as we walk.
Plus you get a bird's eye view of some of the best bits of Chester.
@RyanUnger Not particularly. Ron essentially declared war on SE at some point and, from what I could tell, he was on a campaign of taking every effort to get himself banned. I'm not sure what there is to wonder about in his success.
@JohnRennie cool, sounds good to me
@EmilioPisanty OK, so the provisional plan is that we'll meet up at the station and walk round the walls. Allow an hour or so plus any extra time you want to take for a coffee or whatever.
We can sort the details later. Basically just exchange phone numbers I guess.
3:49 PM
@JohnRennie sure
You have mail
Interesting. Didn't think of the sun divergence.
As of IR in lasers, there is an extra threat....
Some visible (green) lasers have a very high IR component.
@Fermiparadox ... if they're improperly shielded, that is
@Fermiparadox if your laser is 5mW I'd guess it's a class 3A, which is the class that everyday lasers like pointers belong to.
green laser pointers are not green lasers -- they're IR lasers that are then frequency-doubled, and they're meant to completely filter out the IR, because it's extremely unsafe (it can reflect off of glass where the visible beam doesn't, so you risk shining an IR laser directly at someone's eye).
4:04 PM
@EmilioPisanty Yes, some manufacturers don't shield it to cuts costs from what I've heard.
If it isn't shielded properly (easy to test for: arstechnica.com/science/2010/08/…) then do not use it.
pretty messed up that those are just out there on the market
@EmilioPisanty Nice, exactly what I was looking for. I tried something similar with a prism and my phone cam, but failed.
4:08 PM
the source paper for that homebrew test (arxiv.org/abs/1008.1452) does stress that you need to be vigilant about safety procedures
@Semiclassical What's even worse is that some very powerful lasers are available. Not sure why they aren't regulated as they should. There should be a <5mW limit in EU/US if i remember correctly.
(not exactly a shocker, since you're testing whether something is unsafe)
Their marketing strategy lies on "this laser can set things on fire". Unsurprisingly 1 month ago a 10y old was blinded while playing with his uncle's laser in Greece.
@Fermiparadox whoa! Good grief!
>Although awareness about the dangers that lasers can pose has increased, experts recommend that parents who use laser pointers for work should keep them away from their children. That means no playing with them like they're light sabers or aiming them in front of cats to get them to chase the little dot. "Fundamentally, lasers should never be considered toys," Lee said.
4:17 PM
If you're not letting your kid play with horribly dangerous things do you really love them
4:43 PM
(guess I should have a trigger warning for "trauma of being an adjunct faculty")
What a dangerous mistake I made, implicitly ignoring the sunlight's divergence. A 1mW laser could be 170 times "more powerful" than the sunlight.
> I hope you still believe it was all worthwhile. You worked so hard (sometimes!) and it hasn’t seemed to lead anywhere.
that parenthetical is all kinds of f'd up
5:03 PM
How can a system of capacitor have two capacitance at the same time?
@user541396 we'll need to see a circuit diagram to make clear what you are asking?
The question you're asking appears subjective and is likely to be closed
what the actual fuck
@JohnRennie >:(
@RyanUnger huh? What did I do?
5:11 PM
how am I supposed to ask this question
@RyanUnger Did you ask something like "what is the best …"?
@user541396 if you have a single isolated sphere then it has a form of capacitance called the self capacitance.
@user541396 It's not saying that there's actually two capacitances in the system. It's saying that, when the spheres are far apart, then you can approximate the capacitance as that of two isolated spheres in series.
@Loong "What happens if you poke a black hole with a stick?"
So two isolated spheres behave like two isolated capacitors.
5:12 PM
the question is not subjective because I am asking for calculations
When you bring the spheres together their interaction changes the capacitance of the whole system.
@JohnRennie and even if they're not strictly isolated, it can still be a good approximation to say that they are
@JohnRennie @Semiclassical I am trying to say that just by changing the perspective of the potential there are two capacitance ie $ \frac{4\pi e}{1/a + 1/b - 2/d} $ and$ \frac{4\pi e}{1/a + 1/b + 2/d} $
Q: What happens if you poke a black hole with a stick?

Ryan UngerSuppose I'm hovering 1 meter away from the horizon of a Schwarzschild black hole and I'm holding a 2 meter long metal rod. Now I thrust the rod towards the horizon? What do I observe? A satisfactory answer to this question needs to include a discussion of the following: When the Schwarzschild...

@JohnRennie pls answer
@user541396 Given that the self capacitance of a sphere is $4\pi\epsilon_0 r$ you could write the equation for the total capacitance as a function of $C_a$ and $C_b$, but it isn't really.
5:23 PM
@RyanUnger Obviously it just explodes, because nothing can enter a black hole because something something Einstein
It's a function of the overall geometry.
@RyanUnger I think that's a really hard question because the stress in the stick cannot propagate faster than the speed of light and near the horizon the coordinate speed of light gets arbitrarily low.
So when I push the stick at my end can the part of the stick near the horizon transmit the stress back up the stick for me to feel?
idk John that's the question innit
I have no idea what the answer is
I have no idea how you'd go about formulating the problem.
I think there's some sort of paradox here
the horizon isn't special but something very strange happens at the horizon
explain that, science
5:28 PM
who gave me a -1
Q: Thought Experiment - Poking a stick across a Black Hole's Event Horizon

Mr. BoyThe classical explanation of a black hole says that if you get to close, you reach a point - the event horizon radius - from which you cannot escape even travelling at the speed of light. Then they normally talk about spaghetti. But here's a thought experiment. What if I have a BH with event hor...

@RyanUnger not me. I think it's an excellent question and I upvoted it.
@Loong Oh I've actually seen that question
I have upvoted it already lmao
if the answer is correct there, I think we should shoot everyone who has said "nothing special happens at the horizon"
or am I misinterpreting what he wrote
Nothing special happens for a freely falling observer. For a shell observer hovering at some distance x from the horizon the proper acceleration goes to infinity as $x \to 0$.
5:32 PM
So special things happen at the horizon, but only in some coordinate systems.
yeah, uh
what, too soon?
i'm not going to look at lego grad student
yeah, pretty much
@JohnRennie so what does this situation look like to someone passing by on a geodesic
Is the guy just not moving
5:55 PM
@Loong you posted the same link as me
Great minds link alike :-)
link think
How about
Solving the Polyakov string action in a Schwarzschild background
Although I don't think a string can break.
Is there a simple model of two bound particles in GR
Point particles with SHO term?
6:14 PM
A hailstorm...
When am I gonna enjoy nice summer weather
7:20 PM
Is there a Jordan Brouwer theorem for manifolds and closed hypersurfaces?
Like not necessarily spheres
strange, I still get "John Duffield" on my @ autocomplete?
Star this^ if you also get the same
7:35 PM
@Slereah yes.
What theorem mayst that best
@skullpetrol no
Yay! Done with another paper!
JD is suspended on chat, as I understand it via a suspension coming from elsewhere, but he can still enter chatrooms and read, and probably be pinged
let's not star that comment.
7:38 PM
@G.Bergeron done reading or done writing?
@Slereah a properly embedded hyper surface in a manifold with H_1 = 0 is separating
@EmilioPisanty Done writing
Let’s add orientability for good measure
Grants application are coming soon... ;)
7:39 PM
@RyanUnger thx
@G.Bergeron congratulations!
Which one is $H_1$?
Onwards to the next
Is that one of thems topolgoy invariants
Abelianization of the fundamental group
7:40 PM
Q: About separation property of hypersurface

user160211Let N be a complete Riemannian manifold and M be a complete hypersurface in N. M is said to have separation property if N\M is disjoint union of 2 connected open sets in N. Under what reasonable assumptions on N (like simply connected or vanishing homology or cohomology)can ensure that M must hav...

there we go
That’s what I said...
I know
I just wanted
the details
You need to learn reduced homology
Fun fun
I'm afraid this is for non-Hausdorff manifolds
Oh then this answer might not be true
Christ why are you doing this to yourself
7:43 PM
Oh no it's fine
Because you see
This is for the $H$-submanifolds
He enjoys it?
I suspect that if you cut up the $H$-submanifolds along the set of adjacent points and then identify them, you end up with a bunch of disconnected, Hausdorff manifolds
Which would be fine to integrate over
The connected adjacent points forming hypersurfaces in the $H$-submanifolds
well the proof uses some properties of tubular neighborhoods
I'm not sure if they remain true in general
My worry is more that the adjacent points may not necessarily form a proper hypersurface
Non-Hausdorff manifolds can get fairly convoluted
I mean it's obviously true for the ones I can think of, but then that's pretty easy
idk what you mean
whats the issue
7:48 PM
$H$-submanifolds are maximal Hausdorff submanifolds of $M$
So that $M = \bigcup_i H_i$
The adjacent points are the set of points such that there exists another point which isn't well separated
The adjacent points form hypersurfaces in each $H_i$
The plan is to use those hypersurfaces to split $H_i$ into disconnected components
but that will depend on the topology of the adjacent points
wtf they form hypersurfaces?
The proof's in Hajicek
I'm surprised
gotta run
8:17 PM
It's not super hard to see that the set of adjacent points forms hypersurfaces, rly
All the common examples are like that
Points for the 1d examples, likes and circles for the 2D ones, etc
Showing that you get disconnectedHausdorff pieces if you remove them is the hard part, alas
Although I guessI don't need them disconnected
Once they're removed the remaining manifold is Hausdorff by definition
9:11 PM
@user541396 looking at their result again, I actually find myself puzzled as well. Suppose we take the case where the second sphere has the same radius as the first, i.e. b=a
then their results simplify to $C=4\pi \epsilon_0/(2/a\pm 2/d)$
But that seems bizarre. If the two spheres are identical, then why should changing which one has the positive/negative charge alter the capacitance?
9:31 PM
I'm also not convinced their result agrees with the exact expansion given here (iue.tuwien.ac.at/phd/wasshuber/node77.html, via a pretty painful calculation)
In particular, the result given in the link would imply that the total capacitance for equal spheres at large distances is $C=4\pi \epsilon_0 a$, which is the capacitance of a -single- isolated sphere (not both together)
So I'm finding myself very dubious of the result given in the linked Example
(there may be some differences in assumptions, of course, between the example and the linked result.)

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