Why is glass really transparent? Glass is an amorphous polymorph of silicon dioxide, melted and quenched so that grain boundaries grow uniformly and are small with respect to visible light. Its an oxide not a metal and has a large bandgap so visible light doesnt excite the electronic structure.  Now i can imagine how some gases and solvents are mostly transparent dielectrics, but im having trouble understanding why glass transmits light coherently.  How do the little crystals in glass manage to all polarize the same way if they are arranged randomly and aren't free to move? I read glass molecules do move like a magma but very slowly. Is it just that the negative charge polarizes like in a leyden jar?  If thats the case why doesnt the motion of a gas or a fluid affect the coherent transmission of light?
Why is glass transparent? Once I asked this question from my teacher and he replied "Because it passes light.". "And why does it pass light?" I asked and he said, "Because it is transparent.".  The same question again, Why glass is transparent? Why does light pass through it, while for opaque objects, it does not?
I've tried to setup an autologin, but that isn't possible, because my home is encrypted. What are the reasons for this?  PS. Here I have described my problem: . Can I argue that:  \begin{align} \left(\int_a^b f(x) \,dx\right) \left(\int_c^d g(y) \,dy\right) &amp;= \int_a^b f(x) \int_c^d g(y) \,dy \,dx \\ &amp;= \int_a^b \int_c^d f(x) g(y) \,dy \,dx \end{align}  since the same would be true for a product of finite sums? If a friend wants to transfer $7000 to me, what is the best way to do it, especially if we don't share the same banks? We could do wire transfer, but that comes with its own fees. Could we also do direct deposit? Paypal doesn't allow such large lump sum transfers.  This would be US-to-US transfer. Can I use javascript to dynamically change a video's source? I am trying to create Thiessen (Voronoi) polygons, based on points within a defined shapefile extent. The output should be inclusive to the feature.  The output I receive now is the rectangular area around the selected points.    Steps I am using: 1. select points of interest (inside and outside spatial extent) 2. set primary display field to FID (aka:ObjectID) 3. Environment… > Set 'Analysis Mask' to shapefile with feature (in this case, a watershed)  Now here is where the problem starts. I have a set of instructions for arc 9.3, but I am now operating on ArcGIS 10 (Spatial Analyst installed &amp; activated). Also, just to clarify, I did scour the inter-webs for some number of hours- Esri support, gis.SE, Professor Google...  ########  9.3.1 version: Spatial Analyst >> Distance >> Allocation - set 'Assign To' as point shapefile of selected data - set cell size  10.0 methods tried: Analysis >> Create Thiessen Polygons -- Many variations of selected features -- File locations, both inside and out of file geodatabase  I am open to any solution, using Arc 10.0 and/or python would be ideal.  The first image is an example.  The output does not need to be exact.  wiki: Is calling a linear-equation a linear-function, misnomer or completely wrong? From my college life, I remember many professors used to call a linear-equation a linear-function, however:  A standard definition of linear function (or linear map) is:  $$f(x+y)=f(x)+f(y),$$ $$f(\alpha x)=\alpha f(x).$$  Where as linear equation is defined as:  $$f(x)=mx+b.$$  So, linear-equation is NOT a linear-function, according to the definitions defined above.  Though, for $b=0$ the linear-equation becomes a linear-function, but it is not true in general.  Question: Is it misnomer to call a linear-equation a linear-function, or it is completely wrong to say that? And linear-equation must be considered strictly as an affine-mapping. AMD GPU Material in "material preview mode" and "rendered mode", is entirely gray as background. Only outline is displayed I don't know why but now, when i go in material preview or rendered (in solid mode it works fine), my object is entirely gray, or invisible idk, because it has the same color of the background. I've been working for days on this project without having this issue, it suddenly showed up. I tried opening other old projects and suddenly they all do the same so i guess it is some blender program setting that applies for all the projects, but idk how it happened cuz i didn't change anything :/, anyone got any tips? Which is correct if I want to request for a pen?        Can I have your pen please?   May I have your pen please? In the sentence:     All the information you need is in these folders.   I would have expected to find:     All the information that/which you need is in these folders.   Why are those omitted? Creating rounded corners using CSS How can I create rounded corners using CSS? How do we escape the cycle of bad college eating? Mouseover text: And yet I never stop thinking, 'sure, these ingredients cost more than a restaurant meal, but think how many meals I'll get out of them! Especially since each one will have leftovers!'    Randall Munroe really hit the nail on the head with this comic and especially the mouseover text. There's this sort of cycle that I seem to follow when I try cooking, and I need to get my way out of it. It goes like this:   Decide I should cook something. Thumb through my recipe books, find one or two recipes I like. Write down the ingredients for the recipes in a list. Go to the grocery store, stumble around for a while and eventually find exactly those ingredients. Go back home; at some point in time soon after, cook each recipe. Whew, that was a lot of work! Spend the next week or two eating junk again, until I decide I should cook something. (repeat)   Very similar to the comic; sometimes I too let the leftovers go bad (usually not). And every time I rack up a huge bill at the supermarket and think to myself it's justified because of the leftovers or the amount of food. Not to mention, that's a lot of work for just a couple of meals, and a long list of ingredients.  I love cooking fresh meals but that's just so much overhead and so time consuming! So I always resort back to Easy Mac, Ramen, frozen dinners and frozen pizza. Yuck.  How can I escape this cycle and cut down on the overhead? How can I focus on more than 2 recipes at a time without buying giant lists of ingredients and spending tons of money? Is it really worth the trouble or should I just chock it up to being a college student and wonder about the years ahead of me where I'll have much more time to spend cooking one meal at a time? Break a long command to multiple lines I got a shell script and inside I got the command as below #!/bin/sh sh ../jboss/bin/standalone.sh --server-config=standalone-full.xml -Djboss.node.name=node1 -b 0.0.0.0 -bmanagement 0.0.0.0  How can I break the line to multiple lines for better understanding and management #!/bin/sh sh ../jboss/bin/standalone.sh    --server-config=standalone-full.xml   -Djboss.node.name=node1   -b 0.0.0.0   -bmanagement 0.0.0.0  Tried with backslash at the end of each line and it doesnt work. Initializing a member array in constructor initializer Variance of product of multiple independent random variables Finding an expression for $\tan (a+ b + y)$ which involves only $\tan a, \tan b,$ and $\tan y$. How to find this ?   I'm stuck at $1$ and how to $1$ to change the tangent associated with a,b or c.  I meant this ,everybody.  $\frac{tan(a)+tan(b)+tan(y)−tan(a)tan(b)tan(y)}{1−tan(a)tan(b)−tan(a)tan(y)−tan(b)tan(y)}$ Calculate Time Remaining $p = x^2 + xy + y^2$ if and only if $p \equiv 1 \text{ mod }3$? Uniform convergence of derivatives, Tao 14.2.7. This is ex. 14.2.7. from Terence Tao's Analysis II book.   Let $I:=[a,b]$ be an interval and $f_n:I \rightarrow \mathbb R$ differentiable functions with $f_n'$ converges uniform to a function $g:I \rightarrow \mathbb R$. Suppose $\exists x_0 \in I: \lim \limits_{n \rightarrow \infty} f_n(x_0) = L \in \mathbb R$. Then the $f_n$ converge uniformly to a differentiable function $f:I \rightarrow \mathbb R$ with $f' = g$.  We are not given that the $f_n'$ are continuous but he gives the hint that $$ d_{\infty}(f_n',f_m') \leq \epsilon \Rightarrow |(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))| \leq \epsilon |x-x_0| $$ This can be shown by the mean value theorem. My question is : How does this help me to prove the theorem ? At 9th level, the Armorer Artificer (from Tasha's Cauldron of Everything) gains the &quot;Armor Modifications&quot; feature :  You learn how to use your artificer infusions to specially modify your Arcane Armor. That armor now counts as separate items for the purposes of your Infuse Items feature: armor (the chest piece), boots, helmet, and the armor’s special weapon. Each of those items can bear one of your infusions, and the infusions transfer over if you change your armor’s model with the Armor Model feature. In addition, the maximum number of items you can infuse at once increases by 2, but those extra items must be part of your Arcane Armor.  This special weapon is called 'Thunder Gauntlets' (a simple melee weapon) for the 'Guardian' Armor Model, and 'Lightning Launcher' (a simple ranged weapon) for the 'Infiltrator' Model. But I am wondering which Artificer Infusions are appropriate for this special weapon ? Is it only the 'Enhanced Weapon' and 'Radiant Weapon' infusions, or are there others ? Showing that the orthogonal projection in a Hilbert space is compact iff the subspace is finite dimensional Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$.   I have to show that $T$ is compact iff $M$ is finite dimensional.  So if we assume that $M$ is finite dimensional then $\overline{T(B(0,1))}$ is a closed bounded set in a finite dim vector normed space and so it is compact. Which gives that $T$ is compact.  But I am unsure how to prove that if $T$ is compact then $M$ is finite dimensional?  Thanks for any help