This question was bugging me for many years.  it was argued that it would take an infinite time for somebody (suppose, an astronaut) to fall into the black hole, given that it is not his time, but the time at spatial infinity (suppose it is the time on the space station). I completely agree with this result.  Now to my question. Forget about the astronaut for a moment, let's focus on the process from the point of view of the person on the space station. Imagine him immortal. Then he will live long enough for the black hole to evaporate due to the Hawking effect. But the astronaut wouldn't reach the event horizon by that point (it is taking infinitely long for him to do so). So after all he will be killed by a flash of light/particles before falling inside the black hole.  Because all frames of reference are good to go, this also holds in the frame of the astronaut. He will not experience 'nothing' when falling into the black hole. Instead, he will almost instantly (his clocks are ticking much slower) be evaporated by the Hawking radiation. Right?  There are another likely scenarios of the end of the outside world (like thermal death) which would take finitely long of the space station time. They would also destroy the astronaut before he crosses the event horizon.  It seems like these arguments also prohibit any matter to be sucked inside the event horizon. Black holes then have no horizons, they are just very dense formations of matter with scales very close to the black hole diameter, but yet larger.  What is wrong in my propositions? After all, it is commonly believed that black holes suck in the neighbouring matter.
This question was prompted by . Notoriously, if you calculate the Schwarzschild coordinate time for anything, matter or light, to reach the event horizon the result is infinite. This implies that the universe ages by an infinite time before someone falling into the black hole reaches the event horizon, so could that person see the universe age by an infinite time?  To be more precise, suppose the observer starts falling from rest at time $t = 0$ and some initial distance $r &gt; r_s$. If we wait for some time $T$ then shine a light ray at the falling observer. Will the light ray always reach the falling observer before they cross the event horizon? If not, what is the formula for the longest time $T$ that we can wait and still be sure the ray will catch the observer? If $T$ is not bounded it implies that observer could indeed see the end of the universe.  I can think of a qualitative argument for an upper limit on $T$, but I'm not sure how sound my argument is. The proper time for the observer to fall to the event horizon is finite - call this $\tau$. The proper time for the light ray to reach the horizon is zero, therefore the light ray will reach the observer before they cross the event horizon only if $T &lt; \tau$. Hence $T$ is bounded and the observer won't see the end of the universe.  I think a more rigorous approach would be to determine the equations of motion (in the Schwarzschild coordinates) for the falling observer and the light ray, and then find the condition for the light to reach the falling observer at some distance $\epsilon$ from the event horizon. Then take the limit as $\epsilon \rightarrow 0$. In principle this seems straightforward, but in practice the algebra rapidly defeated me. Even for a light ray the radial distance:time equation isn't closed form (Wolfram claims it needs the $W$ function) and for the falling observer the calculation is even harder.
The difference between a poor or meh question and a stellar question can often simply be someone understanding it and providing it a great answer. I can't begin to count the number of times I've justified re-opening a question as a moderator by saying:     Look at the answer it got, though. This isn't something we want deleted, this is something we want fixed, because it's obviously valuable. Someone with domain knowledge can easily edit that question based on the answer it received.   This doesn't only apply to borderline-poor or poor questions – sometimes a good edit means the difference between 100 and 10,000 people finding something through typing things into Google. Good titles are hard.   We think we've come up with a way to provide incentive for good, opportunistic edits to happen more often, ideally without questions needing to go through closure or collecting a centimeter of dust before they do. What if we simply rewarded folks that answered and edited questions with a special, relatively difficult to earn badge?  Let's say that you answer something, and:   You edit the question 12 hours before or after answering it. This allows you to edit now, answer later &mdash; or answer now and edit later, when you have the time. Your edit isn't rolled back, or outright rejected if it was a suggested edit The question is not closed for any reason, even simply being a duplicate Your answer has a score of 1 or higher   &hellip; then you've done something that we probably want to recognize. You understood something, you provided the knowledge that you have, and then you provided an edit to make sure that more people in need of this knowledge can find it, while raising the overall quality of the site.  Here's the proposed tiers for the badge (within the above stated criteria):   Explainer (Bronze): Answered &amp; edited 10  questions Refiner (Silver): Answered &amp; edited 50 questions Illuminator (Gold): Answered &amp; edited 500 questions   We think this is going to be pretty hard to game. You can't go leaving streaks of terse snark all over the place on questions you know are going to be closed anyway, and junk edits aren't going to fly.  We're talking about rewarding folks that take a tiny bit more time out of their day just to write a more descriptive title so that folks can find the awesome answer that they wrote. Just doing that alone can make a big difference. Sure, it's a tad hard to get at the gold level, but it's supposed to be. If you routinely edit questions that you answer and have done so for some time, you'd probably get it the day we roll it out.  While this is technically something that we came up with on the quality initiative ( | ) &mdash; this badge could be earned on all of our sites.  What do you think? What did we miss? What could be better about the idea? Delete questions/answers in iOS app There is currently no way to delete questions or answers in the iOS app, even with the required Trusted User privilege. We can edit and vote, but not delete.  Can this option please be added to the app? Error when creating data with special characters and numbers My code will not project and store the feature layer in the scratch.gdb. "ERROR 999999 ... The table name is invalid. The table was not found. [17_5330_51961Lake]"  The interpreter doesn't like "-" characters or beginning or ending in a number. Does anyone know of a work around?  import arcpy import os  mxd = arcpy.mapping.MapDocument("Current") mxdpath = mxd.filePath (filepath, filename) = os.path.split(mxdpath) MCP_Path = str(filepath)  srl = os.path.join(MCP_Path, r"CoordinateSystems\NAD 1983 UTM Zone 17N.prj")  WBYLID = "17-5330-51961"  WBYLIDlabel = WBYLID.replace("-","_")  LakeOut = os.path.join(MCP_Path + "\OutputData\Scratch.gdb", WBYLIDlabel + "Lake") arcpy.Project_management(fl, LakeOut, srl) How come everyone easily takes or say the name of "Voldemort" without any fear in Scorpius's world? I need a 'good' way to initialize the pseudo-random number generator in C++. I've found  that states:     In order to generate random-like   numbers, srand is usually initialized   to some distinctive value, like those   related with the execution time. For   example, the value returned by the   function time (declared in header   ctime) is different each second, which   is distinctive enough for most   randoming needs.   Unixtime isn't distinctive enough for my application. What's a better way to initialize this? Bonus points if it's portable, but the code will primarily be running on Linux hosts.  I was thinking of doing some pid/unixtime math to get an int, or possibly reading data from /dev/urandom.  Thanks!  EDIT  Yes, I am actually starting my application multiple times a second and I've run into collisions. Modify TikZ coil decoration Can I modify the existing tikz coil decoration, so that the downstrokes are not drawn in a certain interval, to produce a picture like this?    Edit: Previous attempts, code and MWE has been moved to or superseded by my own answer. Are there any differences between Skyrim & Skyrim: Special Edition? Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition. How to turn off notifications for individual apps? How to find a 4D vector perpendicular to 3 other 4D vectors? In 3 dimensions it is possible to find a vector c (one of infinitely many) perpendicular to two vectors a and b using the cross product. Is there any way of extending this to 4 dimensions, i.e. given three vectors a, b, and c finding a vector d perpendicular to all of them?  I realize that this can be done by solving the equation system:  dot(a, d) = 0 dot(b, d) = 0 dot(c, d) = 0   and imposing an additional constraint, for instance setting the length of d to or one of its components to 1, but is there a way of doing this that does not involve solving an equation system? How to connect two parents to the same child, with parents not in the same tree? I'm trying to draw a specific tree, with two parents node linked to the same child. The problem is that one of the parent is outside the main tree.  Here's the code I have for the moment:  \documentclass{article} \usepackage{tikz-qtree,tikz-qtree-compat} %\usepackage{gb4e} \usepackage{forest}  \begin{document}  %\begin{exe} \begin{forest} qtree edges/.style={for tree= parent anchor=north, child anchor=south}    [A, grow=north        [C, name=C, grow'=north            [D, grow=north ]            [E, name=t5, grow=north ]        ]        [B, grow=north ]    ]    \hspace{0.8cm}    [$F$, name=F, grow=north ]    %\draw (F.north) -- (C.south); \end{forest} %\end{exe}  \end{document}   I'd like the nodes F and C to be connected. I don't know why, but if I delete the dollar signs, node F doesn't even appear on my document.  Another thing is that I want the forest environment to be embedded in an example environment, but this doesn't work either.  Thanks in advance for your help! Input form label alignment top or left? In university last semester I was asked to prove that $\sin1$ (1 radian that is) is irrational,  and ended up simply using the Taylor Series Expansion. This method provides a very quick solution, but I am curious as to whether anyone has a method for proving this without making use of the Taylor Series Expansion. I feel as though doing so must be possible using some number theory, but am low on ideas as to an alternative approach to the question.  Note:  If anyone is interested in my solution using Taylor Series Expansion (although it is not the focus of my question), here it is :      From    $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} -  \dots$$   We see that    $$ \alpha = \sin 1 = 1 - \frac{1^3}{3!} + \frac{1^5}{5!} -  \dots $$   Given integers $a$ and $b$, if $\alpha = \frac{a}{b}$ then it follows that $b!\alpha \in \mathbb{Z}$, and $b!\alpha = C + D$ where $C \in \mathbb{Z}$ and we have :   $$ D = \begin{cases} \pm(\frac{1}{b+1} - \frac{1}{(b+1)(b+2)(b+3)} + \dots) \text{    if $b$ is even}\\ \pm(\frac{1}{(b+1)(b+2)} - \frac{1}{(b+1)\dots(b+4)} + \dots ) \text{ if $b$ is odd}\\ \end{cases} $$   In each case we can see that $0 &lt; D &lt; 1$, giving us a contradiction.    Thus, we have that $\sin 1 $ is irrational.    $$ \blacksquare $$ I have just dropped a half full latté on my company laptop and am not sure what to do.  Is there a way I can make sure that my laptop is usable again?  What steps must I go through to improve it's chances of survival? iPhone screen randomly turns on After installing iOS 10, I found that the screen on my iPhone 6s turned on at seemingly random times, and it slightly annoyed me. I now know that this feature does what it is supposed to (turn the screen on when the phone enters certain positions), but I want to disable it.  Is there a setting can I use to do so? INC instruction vs ADD 1: Does it matter? I know that a stationary metric is one which possesses a timelike Killing Vector (so I have time invariance).  I read that a static metric is one which has a timelike Killing vector (so it is stationary) AND this Killing vector is orthogonal to a family of hypersurface (of dimension n-1 is n is the dimension of my manifold). I don't understand the "physical" meaning behind this definition ? Is there a way to explain this in some more intuitive way ? Thanks Does anybody know of a method to trigger an event in Prototype, as you can with jQuery's trigger function?  I have bound an event listener using the observe method, but I would also like to be able to fire the event programatically.  Thanks in advance Why does a bezier curve twist when rotated? If I create a bezier curve and rotate it in edit mode it does not accordingly rotate the tilt of the handles which creates an undesirable twisting along the curve.  Here is an example using a bezier circle with a square used for a bevel object:   Here it is after being rotated 90° about the y-axis:   How do I prevent/fix this? To prove Cayley-Hamilton theorem, why can't we substitute $A$ for $\lambda$ in $p(\lambda) = \det(\lambda I - A)$? Cayley Hamilton Theorem states that if $A$ is an $n \times n$ matrix over the field $F$ then $p(A) = 0$.  We note that $p(\lambda) = \det(\lambda I - A)$. Hence, why can't we just substitute $\lambda$ with $A$ and directly prove Cayley-Hamilton Theorem by saying that $p(A) = p(\lambda) = \det(\lambda I - A) = \det(AI - A) = 0$?