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The causal structure of spacetime is determined by the lightcones at every point of it, basically. From this you can construct such things as the chronological past (all events that influenced this point), the chronological future (all events that will be influenced by this point), the Cauchy horizon (the limits of what you can predict in the future from a ...


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Partial answer: In the context of general relativity, it is conventional to make use of conformal diagrams (a.k.a. Penrose diagrams to enable us to visualize a spacetime. These bring the boundaries of the spacetime (at infinitey) to a finite coordinate value, and keep lightlike worldlines at 45 degrees. As such, they're the generalization of Minkowski ...


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As a geometric quantity, the value of an "angle" can be determined and expressed in a coordinate-free way: Given three pairwise space-like events, "$A$", "$B$" and "$C$", and given the positive real numbers $\frac{s^2[ A C ]}{s^2[ A B ]}$, $\frac{s^2[ A C ]}{s^2[ B C ]}$ and $\frac{s^2[ B C ]}{s^2[ A B ]} = \frac{s^2[ A C ]}{s^2[ A B ]} / \frac{s^2[ A C ...


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If it is a photon, then you know that $x^{a}x_{a}$ is always zero. You have measured the spatial direction of the photon with your mirror apparatus, so you know the values of the $x^{i}(0)$ at some time${}^{1}$, which we will call zero. Furthermore, we can infer the value of $x^{0}$ at this time from the fact that $x^{a}x_{a} = 0$. Then, all we need is to ...


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I'm guessing that when you say: What is now the spatial distance between the two particles? You mean the proper distance. The coordinate distance is of course just $L$. The proper distance is the distance you would measure if you sat at radius $R+L$ and let out a tape measure until it reached radius $R$. To calculate the proper distance start with the ...


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I found from google books, Numerical Flow Simulation II: CNRS-DFG Collaborative Research Programme Results 1998-2000 page 22. ...


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A tetrahedron composed of steel beams is very strong to shear forces, where a cube would easily collapse without other supports. On the other hand, when it comes to compression forces A cube will have four beams supporting each face and a tetrahedron would only have three. Even then, when a compression force is applied to a face, each steel beam in of ...


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A necessary condition for mutual rest is mutual rigidity; and a necessary condition for mutual rigidity is according to Synge (Relativity: The General Theory, p. 115) operationally ("chrono-geometrically") defined as [...] sending photons from one [...] to another, and receiving back the scattered or reflected photons. The criterion for rigidity is that ...


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Thanks for the question. You can derive this equation from the Pythagorean equation which is only for the the case when the two lines a and b are perpendicular to each other. But if they are at an angle then this is the general equation. Now from the picture $$ \begin{align} c^2&=(a+b \cos \alpha)^2 + (b \sin \alpha)^2\\ & = a^2 + 2ab \cos ...


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In relativity, in order for something to be at rest with respect to something else, both of them must be particles with nonzero mass in their respective rest frames, which are represented by time-like four-vectors. Saying that two photons are at rest one w.r.t another does not make sense, because there exists no inertial frame in which a photon can be at ...


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In the diagram you've drawn, any line with a gradient of greater than one is timelike and any line with a gradient of less than one is spacelike. As your diagram shows, the gradient of all radial lines (i.e. outwards from the time axis) on the hyperboloid have a gradient of less than one. So to the observer whose light cone it is, any object following one of ...


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you have three terms in the equation, which together define the scattered wave. the first two define the amplitude at any given time and place along the vector $\hat{r}$, which points the direction you're interested in. $\hat{r}\times\hat{E}\times\hat{r}$ tells you that the wave is going to be perpendicular to $\hat{r}$ and also in the plane defined by two ...


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First of all the hats on $\hat r$ and $\hat E_i$ indicate that these are unit vectors. $$(\vec A \times \vec B) \times \vec C \neq \vec A \times (\vec B \times \vec C)$$ See Vector Triple product Which means that $\vec A \times \vec B \times \vec C $ without parenthesis has no meaning. So I shall assume parenthesis around the first pair ie $(\vec A ...



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