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Astonishingly, the solution to the problem is similar to that of calculating the angular sizes. After focusing on the problem I realized that: Center = Real Center / Depth. The real center is the center of the object at distance = 1. The same way: Angular Size = Real Size / Depth The real size is the size of the object at distance = 1. Center2 = ...


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This of course has been solved many years ago in computer graphics. Look up something called perpsective projection. The simplest and most common way to do this is a point to plane projection. All parts of the scene are projected to a single point, usually called the view point or the eye point. You conceptually suspend the image plane at a fixed ...


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My apologies that I can't follow your question enough to answer everything, in particular I can't see how you project your images. If you extended them in what look like straight lines to you, you'll get radial lines expanding from your position, but it seemed like you wanted a Cartesian grid. However I do want to answer the part about strain and its ...


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An orbit in three dimensions is generally specified by giving how three quantities depend on time, or by giving how two coordinates depend on the third one.You have however omittied saying how $\phi$ depends on $\theta$; I will thus simply assume that $\phi$ is a constant, hence $v_\phi = 0$. In this case the total angular momentum $$ \mathbf L = m\mathbf ...


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Using differential geometry one can show that $$ \vec{p}(\theta) = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 a \sin \theta \cos \theta \\ 2 a \sin \theta^2 \\ 0 \end{pmatrix} $$ $$ \vec{v}(\theta,\dot\theta) = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \frac{{\rm d}}{{\rm d}t} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = ...


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Using energy conservation you can find your answer from $$ V = E-\frac{1}{2}mv^2 $$ Express $V$ in terms of $r$ and you should get the form of the potential, assuming that it is a central force. Then you have to show that the assumption is correct. Is the problem such that you can assume $\varphi =0$?


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Using conservation of angular momentum you have (note that this relation implies a monotonic relation between angle and time) $$ mr^2\dot{\varphi}=J\Rightarrow\mathrm{d}t=\frac{mr^2}{J}\mathrm{d}\varphi. $$ The energy is conserved and given by, $$ E=\frac{1}{2}m\dot r^2+\frac{J^2}{2mr^2}+V(r). $$ Changing variables from time to angle ($\dot r=r_\varphi ...


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If you say that the acceleration of gravity is towards the ground and positive, then you must have distance increasing in that direction as well - so top of building is zero, and ground is 50. In that case $$y(t) = y(0) + v_0t + \frac12gt^2\\ 50 = 0 + 4.9 t^2$$ Or you say that the vertical direction is "up is positive"; then the acceleration of gravity is ...


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You want $X_i=-50$. The ground is zero, down is positive, so the top of the building is at $-50$. There's no universal convention. You're stuck figuring it out from scratch each time. Fortunately once you do it several times you'll get the hang of it.


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Planes of simultaneity in special relativity don't really mean much of anything. The real physical structure of spacetime is in the light cones. The takeaway from "relativity of simultaneity" is not that there are "different time orderings for different observers", but rather that there is no meaningful time ordering for spacelike separated events. They ...


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What you're asking about is the existence of surfaces of simultaneity. In SR, surfaces of simultaneity can be defined by measurement procedures such as Einstein synchronization, and they turn out to depend on one's frame of reference. In GR it gets a lot tougher to do this. We don't even have global frames of reference. It turns out that what you need in ...


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It sounds like you're interested in when a spacetime admits a Cauchy surface. The answer is that every spacetime that is globally hyperbolic has this property. This was proved by Geroch in 1970 (article here, see Section 5). This includes most of the textbook relativistic spacetimes --- Schwarzschild, Kerr, FLRW, and many others. But there are some ...


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A manifold should be thought of as a topological space that locally looks like a $\mathbb R^n$. This is made precise by saying that every point has a neighborhood that is homeomorphic to an open subset of $\mathbb R^n$. If you don't require anything else (except maybe Hausdorffness and the existence of a dense countable subset to make everything easier ...


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This is the way the different "pieces" (i.e subsets/neighborhoods/atlases $\lbrace{O}_{\alpha}\rbrace$) can be "stitched" together to provide a smooth/uniform cover of the whole manifold. Also this gives a way to measure/compare a function or vector on one subset (neighborhood/atlas) with a copy of it on a neighboring atlas. A necessary condition to ...


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This axiom is a "smoothness" property of manifolds. You could define a manifold without this property simply as a space which is locally Euclidean. However, then something which looks smooth in one chart may be highly non-smooth in another chart. For example a smooth curve in one chart may be discontinuous in another chart (since the representation in the ...


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First, the definition above is a right definition to include things like (the surface of) a smooth torus – and any similar ... manifold ... with an arbitrary topology, but exclude objects that wouldn't be locally equivalent to smooth space. If you changed something important in the definition, you would get a difference concept that doesn't agree with our ...


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Some key reading if you want to understand this stuff is chapters two to five of the IERS Technical Note 36, the IERS Conventions (2010). It's not just the J2000/FK5 frame (aka the EME2000 frame) that is associated with some epoch date. Every Earth-centered inertial frame has some epoch date. There are two fundamental reasons why this must be the case: ...


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First of all, the first equation for $ds^2$ is only valid if $f$ is nothing else than the azimuthal angle $\phi$. Second, if you are evaluating $X_i X^i$, the squared distance from the origin without any infinitesimals, then it is exactly equal to $-t^2+r^2$ and nothing else. The polar coordinate $r$ is chosen as $\sqrt{x^2+y^2}$ so its square already ...


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The Jeans equations can be a bit tricky. Their simplest form (in cartesian coordinates, with no particular assumptions) is: $$\frac{\partial\nu}{\partial t}+\frac{\partial(\nu\bar{v_i})}{\partial x_i} = 0$$ $$\nu\frac{\partial\bar{v_j}}{\partial t}+\nu\bar{v_i}\frac{\partial\bar{v_j}}{\partial x_i} = -\nu\frac{\partial\Phi}{\partial ...



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