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!
@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
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.
@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.
On 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...
Friction 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...
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...
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.
@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.
@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.
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
@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.
@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.
@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.
@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.
@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.
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).
@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.
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.
>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.
@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.
@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} $
Suppose 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...
@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.
@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?
The 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...
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 $x \to 0$.
Let 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...
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
@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?
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.)