I got into a discussion regarding the light from the Sun that reaches the Earth and whether it maintains the same path from the moment it was emitted to the point it is processed by our eyes. My thoughts were that each photon reaching the surface of the Earth must have undergone some refraction/reflection while passing through the Earths atmosphere, so while the apparent path of any individual photon can be traced on a straight line directly back to the Sun, the chances of this occurring are slim and from  and one of the answers from , it would appear some bending does occur, so it is near impossible to have a photon reach our eyes that has maintained a straight path from the Sun since being emitted.   The question here is: if light travels in a parallel path, what then explains the phenomenon where a break in cloud cover produces two rays of light that seemingly converge at a point (Sun).    My thoughts were that the sheer size of the Sun and the amount of light being emitted would create such an illusion of the divergence in light beams, but then I am unable to reconcile that with the beams not maintaining their actual path. Any explanation or pointers would be appreciated.
Consider this picture of sun beams streaming onto the valley through the clouds.    Given that the valley is only (at a guess) 3km wide, with simple trigonometry and the angles of the beams, this gives the result that the position of the light source is being a few tens of km away at most.   What is wrong with the analysis?
note: This is a question about relativistic effects. I've included some detail about the spacecraft and its orbit for background, but the question is about relativistic effects and their observability.     When the Parker Solar Probe passes within 8.5 solar radii of the Sun, it will be moving really fast.  A perihelion and aphelion of about 6.6 and 109 million kilometers suggests a semi-major axis of 57.8 million kilometers. The standard gravitational parameter for the Sun is:  $$GM_{Sun} \ = 1.327\times 10^{20} m^3/s^2$$  So using the :  $$v^2 \ = \ GM_{Sun}\left(\frac{2}{r}-\frac{1}{a}\right)$$  the orbital velocity at perihelion will be about 195,000 m/s, or about 25 times faster than a satellite in LEO. This is close to one part per thousand of the speed of light! I think this will be a new record for a spacecraft velocity relative to the solar system.  Depth of a gravitational potential well for an object of mass $m$ is approximately given by:  $$u_G \ = \ -GM_{Sun}\frac{1}{r}$$  At 6.6 million kilometers the spacecraft will be 23 times deeper in the Sun's gravitational well than it would be at 1 AU.  Will there be any particularly unique relativistic effects detectable for the Parker Solar Probe during it's close flyby of the Sun?  Just for example, during one close flyby, how much time will its on-board clock gain or lose compared to some convenient standard solar system timescale? I know there are several different choices, "GPS time" might be one but I don't want to arbitrarily choose an inconvenient one.     above: Parker Solar Probe cropped, from .    above: screen shot from NASA's      note: the link seems to be dead now, but there is plenty of related information available at     above: Illustration of the Parker Solar Probe's orbit, from . How to assign a SSL Certificate to IIS7 Site from Command Prompt Can you advise me whether it is possible or not to assign a SSL Certificate to a website in IIS7 using the APPCMD application?  I am familiar with the command to set the HTTPS Binding  appcmd set site /site.name:"A Site" /+bindings.[protocol='https',bindingInformation='*:443:www.mysite.com']   and how to obtain current mappings   %windir%\system32\inetsrv\Appcmd   but can not seem to find any way to map a site to a certificate (say the certificates hash for example) Everytime, I read about blackbody, I always get confused at the point where it is written     Under thermal equilibrium conditions , the blackbody radiation depends only on temperature. ..... . At equilibrium , electromagnetic waves bounce around with the walls of the cavity.   What is this equilibrium all about? Why is it important here?? If there is thermal equilibrium, why does the body radiate?? Please help. Find the smallest disk in complex plane containing the eigenvalues of a matrix (NBHM 2012) Consider the given matrix  $$\left[\begin{matrix}1&amp; -2&amp; 3&amp; -2 \\1&amp; 1&amp; 0&amp; 3\\-1&amp; 1&amp; 1&amp; -1\\0&amp; -3&amp; 1&amp; 1&amp;\end{matrix}\right]$$  Find out the smallest disk like ($|z-1| &lt; r$ ) in the complex plane containing the eigenvalues of the given matrix.  One easiest method is to find out the eigenvalues from the characteristic equation of the matrix. I can do it so no problem with it.   I am looking for a method to solve the problem without the above method that can be evaluated in very short time. This type of method is possible, I believe.  Thank you for your help "The queen of England's crown" or "The queen's of England crown"? Possible Duplicate:          What is the correct way of these two sentences?   The queen of England's crown The queen's of England crown   Strictly linguistically, sentence 2 should be the correct one, since the crown belongs to the queen and not to England but it sounds really awkward to pronounce. Why does the p-value double when using two-tailed test compared to one-tailed one? Since this has been marked as a duplicate, I want to clarify that it's not about critical values of one- vs two-tailed tests, but the calculation of the p-Value in this case.  I get that in a two-tailed test, you look at both sides of the distribution and therefore you split alpha in half and you need a more extreme test statistic to get a significant result (at the same alpha level).   In my understanding, the p-value is the probability to get this or a more extreme test statistics if the H0 is true. This is easily calculated for a one-tailed test: I just calculate the integral right of the empirical test statistic    For a normal distribution, a z-Value of 1.645 gives me a p-Value of ~0.05. But if I do the same procedure with a two-tailed test, SPSS/Excel doubles the p-value. I get that you now look on the other side of the distribution as well, and therefore I think it's the integral of z(test statistic) + z(-test statistic), but why? I did not get the result -1.645, so I don't understand why I should add the 2.5% of the other side. What does "off of" mean? I have read a sentence from “Diary of a Wimpy Kid 4”:     The teachers have really been cracking down on kids copying off of each other this year.   How to understand "off of"? Please parse it. How come if $\ i\ $ not of the following form, then $12i + 5$ must be prime? I know if $\ i\ $  of the following form $\ 3x^2 + (6y-3)x - y\ $ or $\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$,  then $\ 12i + 5\ $ must be composite number, because:  $12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x  + 1)$  How come if $\ i\ $ not of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,  then $12i + 5$ must be prime? $\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$.  For example: $\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ $ ,when $\ x = y = 1\ $, as proved before,$\ 5*12+5\ $ must be composite number;  $\ 0, \ 1. \ 2.\ 3.\ 4\ $ can't of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,  $\ 12*0 + 5 = 5\ $, $\ 5 $ is prime,$\ 12*1 + 5 = 17\ $, $\ 17 $ is prime, and so on, Why would spacetime curvature cause gravity? It is fine to say that for an object flying past a massive object, the spacetime is curved by the massive object, and so the object flying past follows the curved path of the geodesic, so it "appears" to be experiencing gravitational acceleration. Do we also say along with it, that the object flying past in reality exeriences NO attraction force towards the massive object? Is it just following the spacetime geodesic curve while experiencing NO attractive force?  Now come to the other issue: Supposing two objects are at rest relative to each other, ie they are not following any spacetime geodesic. Then why will they experience gravitational attraction towards each other? E.g. why will an apple fall to earth? Why won't it sit there in its original position high above the earth? How does the curvature of spacetime cause it to experience an attraction force towards the earth, and why would we need to exert a force in reverse direction to prevent it from falling? How does the curvature of spacetime cause this?  When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. So we cannot just say that it fell to earth because its geodesic curve passed through the earth. Why did the spacetime curvature cause it to start moving in the first place? Dimensional Tolerance for US Standard NEMA 1-15 Plugs I am trying to find the dimensional tolerance on US electrical plugs. I have tried looking online, but I have only been able to find that they are called , and have found the picture below showing length and height for the male connector (but not width of the male prong), and I haven't been able to find dimensional tolerances anywhere.     I ask here because I'm assuming there's a US code that governs electrical plugs, but I can't seem to find the standard anywhere. In C++, what is the purpose of the scope resolution operator when used without a scope? For instance:  ::foo(); Is a memory leak created if a MemoryStream in .NET is not closed? I have the following code:  MemoryStream foo(){     MemoryStream ms = new MemoryStream();     // write stuff to ms     return ms; }  void bar(){     MemoryStream ms2 = foo();     // do stuff with ms2     return; }   Is there any chance that the MemoryStream that I've allocated will somehow fail to be disposed of later?    I've got a peer review insisting that I manually close this, and I can't find the information to tell if he has a valid point or not. Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? Automatic tagging of releases Complex number operations: telling Mathematica variables are real Scheduling a Batch job at a particular time I have a Apex batch class which I would like to schedule at a particular date time. The date time is stored in a variable and changes dynamically. ex Datatime scduleDateTime = datetime.now() ; I want to schedule the apex batch file to run at this datetime variable value.  How can I do that using system.scheduleBatch()  I started with this but could'nt see how I can specify a datetime variable in here:  datetime d = datetime.now(); sampleapexbatch b = new sampleapexbatch(d); System.scheduleBatch(b,'Sample Job',XXXXXXXX)   Any help is appreciated! Is there a way to determine the device running an application. I want to distinguish between iPhone and iPod Touch, if possible. Enumerations on PHP I know that PHP doesn't yet have native Enumerations. But I have become accustomed to them from the Java world. I would love to use enums as a way to give predefined values which IDEs' auto-completion features could understand. Constants do the trick, but there's the namespace collision problem and (or actually because) they're global. Arrays don't have the namespace problem, but they're too vague, they can be overwritten at runtime and IDEs rarely know how to autofill their keys without additional static analysis annotations or attributes. Are there any solutions/workarounds you commonly use? Does anyone recall whether the PHP guys have had any thoughts or decisions around enumerations? How can I recursively clear the "Hidden" file flag on Windows Vista? A friend's computer was recently infected with a virus.  I hope the virus has been removed successfully. But, it appears that the virus set the "Hidden" flag on most directories and files on the system.   What's the most convenient way to clear this flag from selected directories and their contents? Integral $ \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx $ I am trying to solve this integral $$ \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx $$ A closed form does exist despite the looks of the integrand.  This problem is from some old high school IMO training courses.  I am not sure how to solve it. The only information that may be of help is $$ \sin x=\sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x=\sum_{n=1}^\infty \frac{ x^{2n}}{(2n)!} \quad \forall \ x $$ Thanks, I am looking for a complete solution, not a description as of what to do.