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Why do we need coordinate-free description in the first place? In physics, we have coordinate-free description in the first place. Consider your example: When I am walking from school to my house The description of this consist of you and the school (building) departing from each other, you and your house meeting each other, and, filling in ...


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For high-performance graphics, where this thing has to be done all the time, the most common way to do these sorts of rotations is to store quaternions rather than Euler angles. You can convert between them with these techniques and then use them do to spatial rotations, Quaternions are not the simplest way to go but realistically, it's probably not going ...


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In order to 1.) simplify the calculation and 2.) make the results useful for more complicted geometries, the charged ring is uniformly charged. That is, the charge density is not a function of the angle. The density is the same at any value of $\phi$ that you choose.


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Here is one solution to Einstein's field equations: $$ds^2=dt^2-dx^2-dy^2-dz^2 \text{ on } \{(x,y,z,t):x,y,z,t\in\mathbb R\}.$$ It has a global coordinate system, a global reference frame, no closed time like curves, and a famous name, Minkowski space. Our second solution is a different manifold $\mathbb R^3\times \mathbb S$, which can (but doesn't have ...


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This is a great question, with three great answers, and here I am, a bit late to the party. There are two crucial things which the above answers don't seem to address, so I am going to try to give a really simple explanation at those levels. Locality / Manifolds I'm going to come close to giving you one technical definition of a manifold, following ...


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Why do we need coordinate free in the first place? Let me tell you about a related experience I have had teaching students. When I ask them to define the scalar product then the vast majority will write down something along the lines of $$ \vec{v}\cdot \vec{w} = v_x w_x + v_y w_y + v_z w_z \qquad (1) $$ i.e. a coordinate-based description. However, ...


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Let me first tell you that what you read is very vague. In GR the laws of physics are assumed to be independent of the observer. An observer is represented by the a reference frame the observer uses to measure physical phenomena. There are a set of coordinate transformations that relates observables for different observers. Say for example the speed of a ...


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That's a very good question. While it may seem "natural" that the world is ordered like a vector space (it is the order that we are accustomed to!), it's indeed a completely unnatural requirement for physics that is supposed to be built on local laws only. Why should there be a perfect long range order of space, at all? Why would space extend from here to ...


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In general relativity there might not be a general frame of reference that will look the way an inertial frame of reference looks in special relativity. And the fundamental deep down reason is that we didn't assume there had to be, thus it didn't have to happen. Whether a particular solution to Einstein's equation has one or not is up to experiment to ...


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As I said in my comment above, I could not find any book by Blau. In his lecture notes at page 481 he makes the transformation you mention. In detail he starts with the metric in $(t,r)$ in eq 23.1 which has diagonal terms, makes the transformation 23.3 which introduces $T$, then he finds a suitable $\psi$, and finally he rewites the metric in (T and r) ...


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You use the term frame of reference but we need to be careful what we mean by this. In special relativity this phrase generally means an inertial frame i.e. a frame in which Newton's first law applies. In GR we obviously can't have a global inertial frame because objects accelerate (due to gravity) whenever they are near a mass so their behaviour isn't ...


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That's just a very basic concept of how you choose (curved) coordinates. Maybe think first of the example of an Euclidean coordinate system. There your coordinate basis vector $\partial_x$ is also orthogonal to surfaces of constant $x$. The same holds also true for curved coordinates. If you like you can also parametrize your surface $T=const.$ as a ...


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You just need to plug your expression for $dT^2$ back into the original metric, with the substitution $\psi' = -C/A$. This gets you \begin{align*} ds^2 &= -A(r) \left[ dT^2 - {\psi'}^2 dr^2 + 2 \frac{C}{A} dr dt \right] + B(r) dr^2 + 2 C(r) \, dr \, dt + D(r) r^2 \, d\Omega^2 \\ &= -A(r) dT^2 + \left[ B(r) + A(r) \left( \psi'(r) \right)^2 ...


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I think you are mixing up two different concepts, which is muddying the waters. Firstly, relativity (both special and general) is a geometrical theory and the proper time for an observer has a precise definition as the length of a world line along which the observer travels (give or take a factor of $c$). This length is calculated using the metric. As ...


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Proper time of an observer is time as measured by the observer's own clocks. So it's obviously frame-independent because calculating proper time of a given observer requires to use his own frame of reference.


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To decide if two metrics are related by a change of frame and/or coordinate transformation is called the equivalence problem. It can be solved using the Cartan-Karlhede algorithm. Given a metric $g$ expressed in some coordinates $x_i$, the algorithm computes a set of invariantly defined curvature invariants expressed as functions of $x_i$. For example, the ...


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It makes no sense to demand that $dx' = a(t)\ dx$. Suppose that your coordinate transformation was of the form $x' = x'(x)$. Then you would have $dx' = \frac{dx'}{dx}\ dx$, but $\frac{dx'}{dx}$ would have to be a function of $x$ only, and so it couldn't be $a(t)$. Now suppose we tried to fix that by doing a transformation $x' = x'(x,t)$. Now $dx' = ...


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Different observers both agree on the equations of motion. They both see the same physics, governed by the same laws. However, they deal with different initial conditions. One guy sees a stationary object which seems to be moving for another. But equations governing that object are all the same. Take a very simple example: a particle in flat space. I can ...



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