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Any metric is locally equivalent to a symmetric matrix, which is always diagonalizable. Just taking a reference system co-rotating with the black-hole and you will get a diagonal form to the Kerr metric.


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Mathematicians call lots of things maps, but this concept is really no different from a scalar field. Given a point (which is an element of $M$), the scalar function returns a number (an element of $\mathbb R$). A chart maps an open set $U \to \mathbb R^n$--for each point on $U$, there is a corresponding point in $\mathbb R^n$. That's different from $U ...


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So the function defined by $$f(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi, \cos\theta)$$ is not a coordinate chart but rather an embedding $\mathbb{S}^2\hookrightarrow \mathbb{R}^3$. The reason it cannot be a coordinate chart is as follows: A coordinate chart must be a homeomorphism from the open sets $\mathcal{U}_i$ of the manifold $M$ to ...


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An important connection to relativity can be made here. Consider the infinitesimal displacement in the Cartesian coordinates: $$ ds^2=dx^2+dy^2+dz^2=dx^a g_{ab}dx^b $$ where $a,b\in\{1,2,3\}$ and $$ dx^a=\left(\begin{array}{c}dx\\dy\\dz\end{array}\right) $$ and $g_{ab}$ the metric, $$ ...


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When you find the total (squared) value of some vector in an orthogonal basis, such as the Cartesian system $(x,y,z)$ or indeed the spherical system $(r,\theta,\phi)$, what you're doing is simply adding the squared values of each component of the vector. Taking the velocity, let's think about the different components: What is the velocity in the radial ...


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There is an effortless way, if you accept geometrical reasoning. You know, that $T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2$. Furthermore, spherical coordinates are orthogonal, therefore you can just write: $$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$ Geometrically, one easily finds: $v_r = \dot r$, $v_\theta = r \dot ...


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If you want the vector components of acceleration, it may be insightful to write the laws in vector form: $$m\mathbf{a} = -\nabla{V}$$ In spherical polar coordinates, we have the following components of the gradient: $$(\nabla V)_r = \frac{\partial V}{\partial r}$$ $$(\nabla V)_\theta = \frac{1}{r}\frac{\partial V}{\partial \theta}$$ $$(\nabla V)_\phi = ...


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There are different kinds of frames. A common frame to use is a coordinate frame. For that all you need to imagine is each region of spacetime has a coordinate system that you can use in that region to find and label all the events in that region. An advantage to this is that you can practice using arbitrary coordinate systems even while still doing ...


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reading special relativity [...] I pictured a frame of reference being grid Of course there is no definitive requirement for the grid constituents to be rigid with respect to each other, or being in any particular way "regularly spaced" or "regularly moving". Required is (only) for the grid constituents to be distinctive, for any two grid ...


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This is really just a comment on Dargscisyhp's post - please don't upvote this. The equation Dargscisyhp has given you is a vector equation i.e. the $\vec{x}_i$s are vectors. You do the calculation by adding vectors. If you're uncomfortable with this then you can find the $(x_c, y_c, z_c)$ coordinates for the centre of mass by using three separate ...


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$$\vec{X}_{cm} = \frac{1}{M} \sum_i m_i \vec{x}_i$$ where $m_i$ is your mass of particle $i$, $\vec{x}_i$ is the position of particle i, and $M$ is your total mass.


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In principle, yes, $\varphi$ is a coordinate system, in that it tags every member of $\mathcal S$ with an appropriate $n$-tuple of real numbers. If you know a set of coordinates $(c_1,\ldots,c_n)$, you can use $\varphi^{-1}$ to find the corresponding point in $\mathcal S$. In practice, this will not be useful at all, mostly because the amount of real ...


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Just to complement John Rennie's answer, one can always perform a Lorentz transformation to a coordinate system such as the particle is at rest for a given time. It's called instantaneous rest frame (IRF). This frame changes point to point, unless the particle's velocity is constant. In such a frame, we have $ ds^2 = -c^2d\tau^2, $ where $\tau$ is the ...


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In general relativity, it just has to be continuously differentiable. If you walk along a grid line, it can't suddenly turn, and the clocks can't suddenly change speed. Beyond that, pretty much anything goes. You don't even have to cover everything with a single map, but you do have to have extra maps to make sure it all gets covered somewhere. For example, ...



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