Are We the One Whom We See in the Reflection of Mirror?

Are we the one whom we see in the reflection of mirror?

Yes we are! Because where I go, there I am. So, if I look in the mirror and see me , the mirror reflected what I see, and I see me. Theirs no getting away from me so i try and do my best to be ok when i see the reflection from the mirror of me.

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What's the difference between scattering and reflection?

Reflection and scattering are two phenomenon observed in a lot of systems. Reflection is the procedure of diversion of a path of a particle or a wave owing to a non-interacting collision. Scattering is a procedure where interaction between the two colliding particles occurs. Both of these phenomena are extremely important in fields such as mechanics, geometrical optics, physical optics, relativity, quantum physics and various other fields. Scattering is the course where waves get deviated due to certain anomalies in the space.The reflection is mainly govern by the law that the angle of event is equal to the angle of reflection at any given point.

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Are the OWS kids a reflection of our Fatherless society?

Where else are they going to learn camping skills?

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Reflection relating two subspaces

Let $E$ be an Euclidian vector space and $S_1,S_2$ two subspaces of same dimension $k$. Assume that $E = S_1 oplus S_2$. Then there exists $f$ an orthogonal symetry such that $f(S_1)=S_2$ and $f(S_2)=S_1$.Proof: Let $P$ be the matrix of the scalar product on $E$, and choose a basis of $S_1$ and a basis of $S_2$ such that $P$ has the form : $$ P = beginbmatrix I_k & C ^tC & I_k endbmatrix.$$ Let $A$ be a $k times k$ invertible matrix and $f$ be an endomorphism of $E$ such that its matrix is : $$ f = beginbmatrix 0 & A^-1 A & 0 endbmatrix.$$ Then $f$ is a symetry such that $f(S_1)=S_2$ and $f(S_2)=S_1$. But $f$ is orthogonal iff $$^t f.P.f = P,$$ that is, $$beginbmatrix ^tA.A & ^tA.^tC.A^-1 ^tA^-1.C.A & ^tA^-1.A^-1 endbmatrix = beginbmatrix I_k & C ^tC & I_k endbmatrix.$$ Which means that $A$ is orthogonal and $CA$ is symetric. So if $A$ is the inverse of the orthognal part of $C$ in the polar decomposition, then $f$ is an orthogonal symetry.In the general case, if $S_1$ and $S_2$ have an intersection. Denote $S = S_1 cap S_2$, and write $S_1 = S oplus^perp T_1$ and $S_2 = S oplus^perp T_2$. Then $E = S oplus^perp (T_1 oplus T_2)$. Now choose $f : E

ightarrow E$ such that $f$ switchs $T_1$ and $T_2$ on $T_1 oplus T_2$ as above and such that $f$ is the identity on $S$.

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The Countdown Reflection

"The Countdown Reflection" is the 24th and final episode of the fifth season of The Big Bang Theory. It first aired on CBS on May 10, 2012. It is the 111th episode overall. In the episode, featuring astronaut Mike Massimino, Howard and Bernadette get married before Howard goes to space. "The Countdown Reflection" received 13.72 million views in the U.S. and garnered mostly positive reviews

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How to write and start a reflection essay?

Start writing the first draft. Describe your reflective essay project. How did it contribute to your self development? Evaluate the quality of your work within your reflective essay. Did you achieve your goals? Tell how you might have made your project better. Revise your first draft. Be complete when telling "what happened" and use descriptive details to describe your actions. Stick to the point. Concentrate on your reflective essay project as a whole and include the details that really affected the final outcome of the project. Tell about your experiences while working on your essay. Answer the question: "What did you learn?" Write in a clear style that your reader can understand. Show your essay to a friend for helpful feedback.

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Reflection of light from a plane surface [closed]

I can not be sure without seeing the phenomenon, but believe that the colors that you are seeing do not originate in the glass window of your mobile phone. I think that they are caused by diffraction from the structures under the glass: the pixels of the display.I will comment that dispersion of white light into colors does occur in a parallel slab. It is difficult to detect because all of the colors exit parallel to each other, but offset by a tiny amount. To the eye it probably appears that no dispersion is taking place

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Quantizing Solutions to the Reflection Algebra
Here are some comments to help you on the way. Your (QRE) is almost correct: on both sides of the equality the second $R$-matrix should in general be $R_21 P , R , P$ with $P$ the permutation. (So, by definition, you don't see this for the case of a symmetric $R$-matrix.)This change can be understood in terms of the usual graphical notation. For definiteness let's think of time as increasing upwards. The $K$-matrix can be depicted as a sort of "K", with the | depicting the reflecting end, which we think of as a wall or mirror on the left, and the rule for $K(mu)$ is that the 'incoming' (from the right) spectral parameter is $-mu$, which is reflected to the 'outgoing' (back to the right) parameter $mu$. The sign makes sense if we would interpret the spectral parameters as slopes, cf the usual interpretation of the Yang--Baxter equation (YBE) as factorised scattering for 2d integrable QFT. The (quantum) reflection equation then says that the two $K$s --- one above the other on the same wall --- can be slid through each other, rather like the YBE graphically is about sliding one line through the crossing of the two other lines. The arguments of the two $R$-matrices in the reflection equation can be understood as well: as for the YBE, the rule is that the $R$-matrix with incoming lines with parameters $lambda$ on the left and $mu$ on the right has arguments $R(lambda-mu)$, and the parameters follow the lines.The resulting equation is also known as the boundary Yang--Baxter equation, and its solutions have been studied to quite some extent. This should at least be enough for you to find various examples, e.g. for the six-vertex model (XXZ spin chain) with either 'diagonal' or generic reflection.• Other QuestionsAlternative notation for exponents, logs and roots?You can use an explicit predicate and some kind of placeholder like $cdot$ to select arguments to hoist out of the expression. let's use the three-place predicate $E$ to represent an exponential fact. This notation is inspired by internally headed relative clauses in some languages such as Navajo, but it's essentially just a more compact special case of set-builder notation.$$ E(x, y, z) stackreltextdefiff x^y z tag101 $$If we want to write $2^3$ , we write it like so (102):$$ E(2, 3, cdot) ;;;textevaluates to;;; 8 tag102 $$If we want to write $ln(7)$, we write it like so (103):$$ E(e, cdot, 7) ;;;textevaluates to;;; ln(7) tag103 $$To express the a cube root of 14 (like the principal root), we write (104):$$ E(cdot, 3, 14) tag104 $$This notation also admits an immediate generalization to extract more than one thing, for instance:$$ E(cdot, cdot, 4) tag105 $$I think the most sensible interpretation for (105) is that it expands into a set of ordered pairs $(x, y)$ such that $x^y 4$ , but you can also make it return an arbitrary pair instead similar to Hilbert's $varepsilon$ operator (called $tau$ in Bourbaki), which is more consistent with the single-cdot behavior.The notation is unambiguous as long as we always interpret it as applying to a single named predicate, so (106) is ill-formed, but (107) is not. I'm using implies bottom instead of $lnot$ because we could reasonably choose to have $lnot$ bind more tightly to an expression than our implicit set-builder notation, and I'm trying to illustrate a point about resolving ambiguity in the notation.$$ textBAD!;;;;; E(cdot, cdot, cdot) to bot tag106 $$$$ F(cdot, cdot, cdot) ;;;textwhere;;; F(a, b, c) stackreltextdefiff E(a, b, c) to bot tag107 $$There's another problem, which is that not every predicate will be able to uniquely determine all its parameters if all but one is missing. In fact, (104) required a convention in order to make the expression single-valued and deterministic. I'm not sure how to resolve this in general.------Writing first person or third person limited for steampunk?Everything you have a problem with is stuff that isn't a problem, or that you needn't do anyway. Regardless of which POV you choose, the symptoms you are experiencing are common novice symptoms and are easily cured. So go cure them and choose either perspective. It really doesn't matter a whole lot.Asking and answering multiple questions in one post is supposed to be off topic here but I'll give you brief answers to illustrate my overall answer, which is: it's all in your head.Moving place to place: you don't need to describe the action of it. Just start a new chapter/paragraph and describe the room in one sentence and move on. Omit the action of entering the room, and describe the first thing the MC does once she's in the room, eg grabbing a beer, or setting fire to the sofa. Same applies in either 1st or 3rd person.Describing the MC: Don't do it. Who cares what the MC looks like? If he looks bad, other characters will cross the street to avoid him, and if he looks great they'll blow him kisses. The interaction is what matters. Even in 3rd person describing the MC isn't necessary.Dialogue: the he said/she said problem isn't a problem. There are very good solutions on this site. Short version, the word 'said' bothers writers more than readers. Again POV is irrelevant.Action: your example is fine. Write more of it, learn to write it as fast and with as few words as possible. Swap 'dicing', 'had frozen', 'thrusted' for slightly more immediate, 'dice', 'froze', 'thrust'. Read some action novels to pick up more incisive vocabulary, especially verbs. If you still don't enjoy it, write 'they fought.' and move on.Intimate scenes: if you don't want to, don't write them. About 10% of the books I have read include them. That's sci-fi, thrillers, fantasy and humorous versions of those. I'm sure these scenes abound on certain shelves but you don't need to go there. And again POV is irrelevant.------Time dilation derivation of special relativityOne can derive the form of the Lorentz transformation from (1) Galileo's relativity principle and assumptions of spacetime (2) flatness, (3) homogeneity and (4) isotropy as well as of (5) continuity of the transformation between inertial frames. You don't need the Einstein postulate of the constancy of the speed of light. The argument runs as follows. If one derives the most general transformation law consistent with (1) through (5) as well as an assumption of absolute time, then Galilean relativity is the unique transformation law possible between frames. If, however, we relax the assumption of absolute time, then we find that a whole family of transformations is consistent with (1) through (5). Each member of the family is a Lorentz group ($O(1,,3)$) parameterized by a universal constant $c$ that has the inertial frame invariance property. Galilean relativity is the $ctoinfty$ limiting member of the family. See my answer here and here for more details and references.This method does not yield a value for $c$; this value must come from experiment. But we know the speed of light is experimentally found to be inertial-frame-invariant, therefore the relativity that applies to our universe must be $O(1,,3)$ with $c$ equal to the speed of light. We can then think of the Michelson Morley experiment as showing (1) that we live in a universe with a finite $c$ parameter, (2) $c$ is the speed of light and thus (3) light is mediated by a massless particle. The first person to propose this approach was Vladimir Ignatowski.In effect, the above approach is the logical converse of what Einstein did. Einstein took an assumption of the constancy of the speed of light further to Galileo's postulate and inferred the Lorentz transformation and thus relative time. The above approach begins with an assumption of possibly relative time together with Galileo's postulate and derives the constancy of light. It shows special relativity to be what we get from Galilean relativity if we relax the assumption of absolute time.------How to deal with a team in which one of the members doesn't accept critique?This is an awful circumstance. Not being receptive to criticism is the ultimate destroyer of any relationship, not just work/business. Considering oneself exempt from receiving constructive feedback is simply not feasible and realistic in any type of work environment. It's like saying 'I am god'.In the course of remodelling my house, which took several years as I did in on my own with only occasional help of a low skilled laborer, I had a very good relationship with that laborer, not to call him a contractor because he was doing odd work that required little skill. Since, over the years, he also had other projects with other clients, I trusted that he acquired skills in finishing drywall and, when I reached that stage in the renovation, I decided to give him an opportunity to do it under less supervision, essentially being the defacto contractor for that project stage. He did mostly immaculate work except for one irregular corner which came out sloppy but could very easily be fixed. When I submitted that feedback to him and requested that he redo it properly, he got offended and walked out from the job and I never heard from him again. Personally, I am not the kind to give in in that kind of situation because I won't tolerate or condone of that kind of attitude. So he essentially fired himself over his stupid, overly sensitive and immature pride.Look, if I, as a software developer were to get offended every time a QA enters a bug each scrum sprint, I would render myself unusable.If I were you, I would start ignoring this person and wait until his intolerably dismissive attitude towards criticism renders him useless in the team environment and demand that it either change or else... Welcoming feedback and constructively responding to it is critical in the course of any continuously improving effort, which is basically any job that's not in a dead end------Why doesn't a typical beam splitter cause a photon to decohere?Nobody is answering this question, so I'll take a stab at it. Consider the mirror. Suppose you started your experiment by (somehow) putting it in a nearly-exact momentum state, meaning there is a large uncertainty in its position. Now, when you send a photon at it, the photon either bounces off or passes through. If the photon bounces off the mirror, it will change the momentum of the mirror. You could theoretically measure the "which-way" information by measuring the momentum of the mirror after you've done the experiment. In this scenario, there wouldn't be any interference. However, you didn't do that. You started the mirror off in a thermal state at room temperature. This state can be considered as a superposition of different momentum states of the mirror1, with a phase associated to each one. If you change the momentum by a small amount, the phase associated to this state in the superposition only changes by a small amount. Now, let $p_gamma$ and $p_m$ be the original momenta of the photon and the mirror, and let $Delta p_gamma$ be the change in the momentum when the photon bounces off the mirror. When you send the photon towards the mirror, the original state $p_m$ (photon passes through) will end up in the same configuration as the original state $p'_m p_m - Delta p_gamma$ (photon bounces off). These two states $p_m$ and $p'_m$ had nearly the same phase before you aimed the photon at the mirror, so they will interfere, and if the phase on these two states are really close, the interference will be nearly perfect. Of course, a change in momentum isn't the only way for the mirror to gain which-way information. However, I think what happens when you consider the other ways is that they behave much like this, only not anywhere near as cleanly, so they're harder to work with.1 Technically, it's a mixed state, i.e., a density matrix, and not a pure state. But the basic idea of the above explanation still holds.
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Quantizing Solutions to the Reflection Algebra
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