Imagine a very heavy (tens of solar masses) star in its final moments before collapsing to form a black hole. The gravitational force exerted by the weight of the star overcomes the neutron degeneracy pressure and the star continues to collapse inward until the escape velocity at some radius from the star exceeds the speed of light. Now pause at the exact moment this happens.  Am I correct in thinking that at this point we have a black hole, since there is a radius at which light could not escape? If the answer is yes, then from what I understand this black hole must have a singularity. Assuming that a singularity can only exist behind an event horizon, this singularity must have come into existence at this exact moment.  My question is twofold:   Are all of the above assumptions / assertions correct? How does a singularity instantly form?   Apologies if the question is naive - I am just a curious innocent rather than an expert!
Excuse me if the question is naïve. I am not a physicist.  Let's imagine formation of a black hole from a homogeneous collapsing star. At certain moment when enough of matter is inside of a small enough spherical area the event horizon starts to exist. When does the singularity start to exist and why?  The matter below the event horizon still continues to collapse towards the centre (e.g. in the reference frame of the falling matter). What are the necessary conditions for the singularity to appear?  I have found two similar questions but the answers did not really explain this problem to me:
How do I tell if ACR will work with my camera and how do I get the correct version for Photoshop? I am a beginner photographer with a Canon 7d and MacBook Pro. I'm not sure if Adobe camera Raw is compatible with either and really need advice on how to go about downloading the correct version. We're running a couple SQL servers in Azure that are set up with an Always On availability group and Windows Failover Clustering. The servers are Windows 2019 and we're running SQL Server 2019. When we set up the cluster, it was set up to use a Distributed Network Name instead of a static cluster IP address. Thanks to this we shouldn't need an internal load balancer according to these notes: .  I'm struggling to understand exactly how this works though. Based on what I read, it seems like our connection strings will point to the DNS name of the cluster (let's call it AgCluster). If I look in DNS, there is an A record for AgCluster pointing to sql1 and another pointing to sql2. When I use AgCluster in my connection string it seems to always connect me to the primary server, even if I have ApplicationIntent=ReadOnly set. When I query @@SERVERNAME I always get the same server.  So with the Distributed Network Name setup, what should I use in my connection strings to make sure read/write queries go to the primary and read only go to a secondary? Any guides on setting this up in general would be helpful. Thanks! Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that      $$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$   I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$  Doing some manipulation we get:           $$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$      How can I prove this limit? I am looking at the 4-vector treatment of special relativity, but I have had no formal training in Tensor algebra and thus am having difficulty understanding some of the concepts which appear.  One such concept is the notion of contra- and co-variant vectors.  describes a contra-variant vector as varying with the co-ordinates, i.e. the inverse of the reference axes. It gives the example of a change of basis matrix $\mathbf{M}$, and a contra-variant vector $\vec{v}$ and states that in the new set of co-ordinates, we have:  $$\vec{v}'=\mathbf{M}\vec{v}$$  It then states that a covariant vector varies with the co-ordinate axes.     Does this mean that for a covariant vector $\vec{u}$ and change of basis matrix $\mathbf{M}$ we have: $$\vec{u}'=\mathbf{M}^{-1}\vec{u}?$$   I am also having problems understanding the concept of raising and lowering indices. For instance, Wikipedia states that for two four-vectors $\vec{A},\vec{B}$ and using Einstein summation we have:  $$\left\langle \vec{A} \middle|\vec{B} \right\rangle=A^{\mu}B_{\mu}=A_{\nu}B^{\nu}$$  And I am having difficulty understanding how this relates to the other definition of the inner product:  $$\left\langle \vec{A} \middle|\vec{B} \right\rangle =\mathbf{A}^{T}\mathbf{\eta}\mathbf{B}$$  If anyone could help give me a more intuitive understanding of these concepts I would be grateful. How can I take screenshots of YouTube videos in full screen? I am trying to take a screenshot of a YouTube video in fullscreen.  When I pause the video and wait a moment, only the bottom UI bar (play/pause/seek) disappears. The top UI bar (Like/Share/More Info) stays there.  So when I press PrintScreen on the keyboard, the screenshot is polluted by this top UI bar. Is there any way to remove it while the video is on pause? Validity of maximising log-likelihood for maximum likelihood estimation For reasons owing to mathematical convenience, when finding MLEs (maximum likelihood estimates), it is often the log-likelihood function---as opposed to the standard likelihood function---which is maximised.  From what I've gathered, this approach is deemed valid as a result of the monotonically increasing nature of the (natural) logarithm function.  My understanding of a monotonically increasing function is: for all $x$ and $y$ (defined on a subset of the reals), if $x \leq y$ then $f(x) \leq f(y)$.  This, however, does not appear to be the case for all log-likelihood functions; for example: for the log-likelihood $\text{Gamma}(3, 5)$ function, if $x = 0.15$ and $y = 0.46$, then $f(x) = -0.59$ and $f(y) = -1.19$.  Clearly, I've misunderstood this concept. Fundamentally, I guess I'm asking can somebody (preferably mathematically) demonstrate why:  $$ \hat{\theta} = \text{argmax} \text{ } L(\theta) = \text{argmax} \text{ } \text{log} \text{ } L(\theta) $$  where $\hat{\theta}$ is the MLE for a given likelihood function. History of changes to a particular line of code in Subversion Is it possible to see the history of changes to a particular line of code in a Subversion repository?  I'd like, for instance, to be able to see when a particular statement was added or when that statement was changed, even if its line number is not the same any more. What is recordSetVar and where is it used? I was looking at some code and in the 'apex:page' tag, an attribute called recordSetVar was mentioned. I'm new to VisualForce so could someone please tell me when and why is it used? How To Make A Solid Object Drop To The Ground Like It Was Liquid Why is echo a shell built in command? $ which echo echo: shell built-in command. $ which ls /bin/ls $ which cat /bin/cat   Why is echo not an independent utility like ls, ps, cat etc? Why is it shell specific? Any good reasons? Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients  -- contain multiplicative inverses?   I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$. Given a joint density function, what is the conditional expectation $E[Y|x]$? The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)= \left\{\begin{matrix}e^{-y}, \mbox{ } 0\leq x \leq y \le \infty \\ 0,  \mbox{   otherwise}\end{matrix}\right.$$ Evaluate the conditional expectation $E[Y|x]$.    I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.  My solution: (My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)  $$E[Y|x] = \int_{-\infty}^{\infty}yf_{Y|X}(y|x)dy.$$  $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$    $$f_{X}(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy$$   $$= \int_{0}^{\infty}e^{-y}dy $$  $$=\left [ -e^{-y} \right]\Big|_0^\infty $$  $$ = -e^{-\infty}-(-e^{0})=1. $$ $$f_{X}(x) = 1, x \geq 0.$$ Thus,   $$E[Y|x] = \int_{0}^{\infty}-ye^{-y}dy$$  $$= -\int_{0}^{\infty}ye^{-y}dy$$ After integration by parts,  $$E[Y|x] = \left [ ye^{-y}+e^{-y} \right]\Big|_0^\infty$$ $$=0-[e^{0} + 0]=-1$$ Therefore, $E[Y|x] = -1.$  Thanks in advance for reading this long problem and responding! Changing back button in iOS 7 disables swipe to navigate back $\Bbb Q$-linear is equivalent to additive Solve $\int_0^{+\infty} \Big(\frac{\sin t}{t}\Big)^2 \ dt$ Following my question , I am wondering if there are strong views for or against the use of standard deviation to detect outliers (e.g. any datapoint that is more than 2 standard deviation is an outlier).  I know this is dependent on the context of the study, for instance a data point, 48kg, will certainly be an outlier in a study of babies' weight but not in a study of adults' weight.  Outliers are the result of a number of factors such as data entry mistakes. In my case, these processes are robust.   I guess the question I am asking is: Is using standard deviation a sound method for detecting outliers? Does taking levels in a prestige class cause XP Penalty? How to change an application icon programmatically in Android? What's in an Eclipse .classpath/.project file? I'm fairly excited for Unity, as it looks like a promising new direction for Ubuntu.  However, I do have a concern - will it be possible to use Unity without the global menu?  I have my window manager set to focus-follows-mouse/sloppy focus, and find the productivity gains to be immense. Sloppy focus is incompatible, however, with global menus, as it is possible for the focus to change while you move from window to menu.  Will Unity support an option to use window menus while still using Unity?