Should we stop saying that nothing can escape the event horizon of a black hole post LIGO?
How does some of the black hole (BH) mass escape the event horizon (EH) of either BH, or the merged EH of two merged BHs?
Is tipping mandatory at Restaurants and Bars in Germany? Why does a 'sudo -i' login shell break a here-doc command string argument? In the sequence of five commands below, all depend on single-quotes to hand off possible variable substitution to the called bash shell rather than the calling shell. The calling user is xx, but the called shell will be run as user yy. The first command substitutes $HOME with the calling shell's value because the called shell is not a login shell. The second command substitutes the value of $HOME loaded by a login shell, so it is the value belonging to user yy. The third command does not rely on a $HOME value and creates a file in the guessed home directory of user yy.  Why does the fourth command fail? The intention is that it writes the same file, but relying on the $HOME variable belonging to user yy to ensure it actually does end up in her home directory. I don't understand why a login shell breaks the behaviour of a here-doc command passed in as a static single-quoted string. The failure of the fifth command verifies that this problem is not about variable substitution.  xx@host ~ $ sudo -u yy bash -c 'echo HOME=$HOME' HOME=/home/xx xx@host ~ $ sudo -iu yy bash -c 'echo HOME=$HOME' HOME=/home/yy xx@host ~ $ sudo -u yy bash -c 'cat &gt; /home/yy/test.sh &lt;&lt; "EOF" &gt; script-content &gt; EOF &gt; ' xx@host ~ $ sudo -iu yy bash -c 'cat &gt; $HOME/test.sh &lt;&lt; "EOF" &gt; script-content &gt; EOF &gt; ' bash: warning: here-document at line 0 delimited by end-of-file (wanted `EOFscript-contentEOF') xx@host ~ $ sudo -iu yy bash -c 'cat &gt; /home/yy/test.sh &lt;&lt; "EOF" &gt; script-content &gt; EOF &gt; ' bash: warning: here-document at line 0 delimited by end-of-file (wanted `EOFscript-contentEOF')   These commands were issued on a Linux Mint 18.3 Cinnamon 64-bit system, which is based on Ubuntu 16.04 (Xenial Xerus).  Update: The here-doc aspect is just clouding the issue. Here's a simplification of the problem:  $ sudo bash -c 'echo 1 &gt; echo 2' 1 2 $ sudo -i bash -c 'echo 1 &gt; echo 2' 1echo 2   Why does the first of those two commands preserve the linebreak and the second does not? sudo is common to both commands, yet seems to be escaping/filtering/interpolating differently depending on nothing but the "-i" option. Make rigid body collide with hair particles Using have gone to / been to / been in I understand the difference between these constructions: "has gone to", "has been to", "has been in":   "has gone to" &mdash; there or on his way to; "has been to" &mdash; someone has been there but he is on his way back; "has been in" &mdash; someone is still there;   But now I am puzzled by next statements:   Have you ever been to / in Greece? &mdash; Yes, it's a very beautiful country; When did you last go to Canada? &mdash; I've never been to Canada.   If I use "has you ever gone to Greece" construction in the first case or "I have never gone to Canada" in the second statement will it be uncorrect? Or just will sound odd? May I say "I have never been in Canada" in the second sentence. Or in these case all of the variants are possible, the problem is in how people usually say it (in what how it perfectly sounds)? Spambots causing server crashes. How do I block them? Difference between 'struct' and 'typedef struct' in C++? Does one need a visa to transfer between two airports during a transit through the Schengen area? I am an Indian citizen travelling from New York to Delhi with a layover in Paris where I need to transfer between Paris Orly Airport (ORY) and Paris Charles de Gaulle Airport (CDG).  Do I need to have a visa to make this transfer?  I didn't notice that while booking the ticket. How does a credit card company make any money off of me using their card if there are no fees associated with the card (I don't pay any extra) and I pay off my balance on time every month (so I don't pay any fees ever for this either).  Are they just banking on the fact that many people don't do this and end up owing them money? Joining lots of small polygons to form larger polygon using PostGIS? How to migrate a WordPress installation from one site to another, including all images? C# Unsafe/Fixed Code Is time dilation an illusion? It is said that we can verify time dilation by flying a very accurate clock on a fast jet or spaceship and prove that it registers less time than the clocks on earth. However, the clocks on earth would be moving relative to the clock on the spaceship, and since time always dilates and never goes faster regardless of the direction of relative motion,  the clocks on earth should register less time than the clock on the spaceship.  Is this true? Whenever there is a fast-moving object such as a rocket do all clocks on earth really become slow?  If the rocket with the clock landed after moving at relativistic speed, would its clock and the earth's clock again show the same time since during its travel both appeared slow to each other?  Or is all this just an illusion, ie. the clocks just appear to be slow to each other but in actually run at normal speed, and neither is behind when the rocket actually lands? Is posting someone's comment as an answer okay? I came across  and at the time there were no suggested answers. After reading Mark Hubbart's comment (at the time with four upvotes), which was the correct answer, and seeing that it had been over 30 minutes since his comment, I figured I would provide that answer in my own words, which I did. My answer is longer, and I think more rounded and more useful, but the core of the answer came from the comment. I have no idea if I'd have thought up the answer on my own--a definite maybe to that.  Now had I tried to be "honorable" or something, such as instead of answering, suggesting to the commenter that he make his comment an answer, I would only have accomplished giving away easy rep to some other user not as into gratuitous self-denial.  The commenter had merely to enter his comment into an answer box instead of a comment box! It was his choice! Whether he was too lazy or didn't think his answer worthwhile or some other reason is not my responsibility. Must I bend over backward to help him earn points that he so carelessly left lying around for another person to take? It just took a tiny bit of non-laziness on my part to get my text typed into the box that would reward me instead of the box that wouldn't.  If my perspective is incorrect and the community would rather see such answers-as-comments rewarded and supported, I will gladly accept that. In the meantime, I just want to shake off the tiny bit of guilt my overly-nice and overly-concerned-with-propriety self is feeling. :) In summer, should I close curtains during the day? Multi-OS Portable TeX system I am looking for      an example of a ring where unique factorization does not hold. However the ring is supposed to have the property that every irreducible is already a prime.    So unique factorization should fail because one cannot decompose every element into a finite product of irreducibles. In particular the ring must not be Noetherian. What is an example of such a ring? $f$ a real, continuous function, is it measurable? Let $f: \mathbb{R} \to \mathbb{R} $ be a continuous function. I need to show that is a measurable function.   I tried working with the definition:  Let $f: X \to \mathbb{R}$ be a function. If $f^{-1}(O)$ is a measurable set for every open subset $O$ of $\mathbb{R}$, then $f$ is called a measurable function.  Since $f^{-1}(O)$ also lies in $\mathbb{R}$, I think it is sufficient to show that every subset of $\mathbb{R}$ is measurable. But is this possible?   So far I concluded that $\mathbb{R}$ itself is measurable, since $$\mu(A) = \mu(A \cap \mathbb{R}) + \mu(A \cap \mathbb{R}^c) = \mu(A) + \mu(\emptyset) = \mu(A).$$ How do I need to approach? An operator has closed range if and only if the image of some closed subspace of finite codimension is closed. I tried to install wine using these commands: sudo apt-get install -y wine  And: sudo apt-get install wine  But terminal says that: Reading package lists... Done Building dependency tree        Reading state information... Done Package wine is a virtual package provided by:   winehq-staging 2.16.0~zesty   winehq-stable 2.0.2~zesty   winehq-devel 2.16.0~zesty   wine-development 2.0-3ubuntu1 You should explicitly select one to install.  E: Package 'wine' has no installation candidate  So there are 4 different version of wine. What is the difference between these versions. Which one is better? I wanted to install Internet Explorer from it and I installed winehq-stable. But I couldn't see that among applications in Winetricks. Do the other versions of wine have more applications to install? What is the difference? As someone who doesn't play Team Fortress 2, I occasionally get people asking me to trade, but I have no idea what's a "fair" price for these items. I don't really get this whole trading thing (I've seen people ask for "63 buds", I have zero idea what that means), but I don't want to get ripped off.   Some recent  examples would be:        How can I figure out how much an item is worth? What "lingo" do I need to know in order to trade?