# Tag Info

## New answers tagged coordinate-systems

2

"Contraction-orthogonality" of covariant and contravariant basis Contravariant vectors or just "vectors" are defined as elements of the tangent space at a given point. In practice, they are defined with respect to a coordinate-vector basis $\mathbf{e}_{(i)}$, where $\mathbf{e}_{(i)}$ is the vector tangent to the $i$-th coordinate line. Then they are given, ...

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The metric being a rank $(0,2)$ tensor transforms under general coordinate transformations $x^\mu \to x'^\mu(x)$ as $$g'_{\mu\nu} (x') = \frac{ \partial x^\rho}{ \partial x'^\mu } \frac{ \partial x^\sigma }{ \partial x'^\nu } g_{\rho\sigma} (x)$$ Now set $x'^\mu (x) = x^\mu + \alpha k^\mu(x)$ in the above expression and take a limit of small $\alpha$. ...

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Pulling together what's been said in various comments: 1) General relativity admits models where spacetime is foliated by spacelike leaves, all of which are indexed by a global time coordinate. The simplest of these models is Minkowski space. All of your observations about models with comoving observers apply equally well to Minkowski space, so if you ...

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Suppose two observers, Alice and Bob, are moving relative to each other since the beginning of the universe. While they do it, they construct the chronologies of all the events of the universe, as they record them in their frame of reference. They will construct different chronologies. However, and this is key, each can reconstruct the other's chronology. ...

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A comoving observer and an observer that has been moving at $0.866c$ since Big Bang will disagree on their measured age of the Universe by a factor of 2. While both measurements are correct, we can say that the comoving observer measures a more "natural" age of the Universe. For instance, the comoving observer is the only observer who will measure the ...

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Suppose we start with an orthonormal cartesian coordinate basis with unit vectors $\left( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right)$. Suppose we wish to rotate to a new coordinate system (may or may not be a coordinate basis, see the following answer which has a good discussion on the difference) with unit vectors $\left( ... 1 1) If the transformation itself is smooth, then the Jacobian will be smooth. This is a desirable property because... 2) The coordinate transformation tells us how the coordinates change, but the Jacobian tells us directly how the coordinate basis vectors change. For example, a transformation from Cartesian to polar coordinates would use the Jacobian to ... 2 The functions are "smooth" because we want to be able to speak about derivatives on our manifold$M$, and for that, it is convenient to have a smooth structure on$M$(one could settle for a$C^k$-structure with$k$as needed, but physicists rarely care for such details). And of course we want to be able to take derivative because we might be interested in ... 1 Perhaps you could explain what exactly led you to question your understanding... The physical meaning of coordinate invariance is pretty simple. It's just that the laws of physics cannot depend upon your choice of coordinates as long as the reference frame you're working in is inertial. So if you were to perform a coordinate transformation from one inertial ... 1 The metric measures lengths in various directions, and also angles between various directions. For example if$\vec{e}_{(1)}$is the basis vector in the$x^1$-direction, it will have length given by $$\lVert \vec{e}_{(1)} \rVert^2 = g(\vec{e}_{(1)}, \vec{e}_{(1)}) = g_{11}.$$ If we also have the basis vector$\vec{e}_{(2)}$in the$x^2\$-direction, then the ...

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The metric is an important concept in general relativity. In GR, vectors correspond to weighted directions in spacetime (by "weighted", I mean any scalar multiple of a vector corresponds to the same direction, but weighted differently). The metric tensor can then tell us about the angle between two directions or the magnitude of a given vector, which gives ...

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(8.21) is the inverse of (8.20) up to first order.

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