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In a Riemannian manifold such as the surface of a sphere you could travel a distance $\epsilon$ in every direction and note that you get a different amount of surface area or volume than if you did so in a flat space. For instance if you walked a distance equal to the distance from the north pole and the south pole ($2\pi R$) you get an area of $4\pi R^2,$ ...

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I don't know if I have to use (Lorentz) metric anywhere here. No you don't. This question is wholly about linearity and the definitions of tensors. Choose a basis $\{\hat{e}_j\}_{j=0}^N$ where my index runs from nought to $N$ to be in keeping with standard notation in relativity, but that is the only link: this question is general. Our tensor ...

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I think your last sentence shows you're on the right track. A one form is a linear functional that maps vectors to real numbers. You give it a vector as an input, and, as you say, it returns the rate of change in the implied direction. Let's say we have a path whose tangent at a point is defined by the vector $v^j\,\partial_j$ - the differential operator ...

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Your proof is entirely correct. Relativity doesn't have to be difficult :) To be clear, the steps in your proof are simply the definition of $U'$, the definition of $X'$, the product rule, the constancy of $\Lambda$, and the definition of $U$, respectively. There is nothing wrong with any of these steps. As for why $\tau$ and not $t$: $\tau$ is defined ...

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I've tried to do it the following way, but I don't know if I can use the product rule as such when matrices and vectors are involved. \begin{equation*} U'=\frac{dX'}{d\tau}=\frac{d}{d\tau}\Lambda{X}=\frac{d\Lambda}{d\tau}X+\Lambda\frac{dX}{d\tau}=\Lambda\frac{dX}{d\tau}=\Lambda{U} \end{equation*} since the Lorentz transformation matrix ...

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There are two different questions that are unrelated to each other. The first one is how the 4-velocity is defined. By definition, velocities are tangent flows on a differential manifold, therefore derivatives must be taken with respect to the parameter you are using to describe the flow with. In the context of special relativity such parameter is the ...

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Suppose you have some four-vector $v$, then the norm of $v$ is given by: $$v^2 = g_{\alpha\beta}v^{\alpha}v^{\beta}$$ Since the dimensions of the left and right sides must agree this shows the metric tensor is dimensionless.

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Start by rewriting the scalar product as a covariant-contravariant contraction, like so: $${\bf u}\cdot{\bf v} = g_{ij}u^iv^j = (g_{ij}u^i)v^j = u_jv^j$$ Now transform the components with your $S$ and $T$ matrices, $$u_jv^j = \left( S_j^a {\bar u}_a \right) \left( T^j_b {\bar v}^b \right) = (S_j^a T^j_b) {\bar u}_a {\bar v}^b = \delta^a_b {\bar u}_a {\bar ... 3 The expression A^{\mu}+B^{\nu} is not a tensor. It is not even an allowed operation, since tensors only add to tensors with the same index structure (and same index "names"). To see that X^{\mu\nu}=A^{\mu}+B^{\nu} is not consistent mathematically, perform a contraction of both sides with the metric tensor g_{\mu\nu}. On the left hand side you get a ... 3 A tensor is something that transforms like a tensor. The sum of two vectors transforms as$$ (A^\alpha + B^\beta)^\prime = \Lambda^\alpha_\mu A^\mu + \Lambda^\beta_\nu B^\nu $$A tensor transforms as$$ (T^{\alpha\beta})^\prime = \Lambda^\alpha_\mu \Lambda^\beta_\nu T^{\mu\nu} $$If A^\alpha + B^\beta doesn't transform like a tensor, it isn't a tensor. 0 If I understand correctly, you're asking how to prove that symmetry of a tensor is coordinate independent, but you seem to be having trouble with the definition of a tensor. Well, you're not the first. Let me give you a definition that might help. First, suppose you have some space (it can be 3-space or spacetime or whatever) and you have a set of ... 1 To see how the components of the electromagnetic tensor transform from F to F' under a Lorentz transformation you take the tensor$$F^{\mu \nu}=\begin{pmatrix} 0 & -E_x /c & -E_y /c & - E_z /c\\ E_x /c & 0 & -B_z & B_y\\ E_y /c & B_z & 0 & -B_x\\ E_z /c & -B_y & B_x & 0 \end{pmatrix}$$and apply the ... 1 Our professor defined a rank (k,l) tensor as something that transforms like a tensor Oh dear. All too common, and all too pedagogically faulty. A tensor is nothing more or less than a linear map from (possibly multiple copies of) a vector space (and possibly copies of its dual space) into the scalar field. If I give you components T_{\mu\nu} (16 ... 3 A tensor is not particularly a concept related to relativity (see e.g. stress tensor), but is a more general concept that describes the linear relationships between objects, independent of the choice of coordinate system. This coordinate independence results in the transformation law you give where, \Lambda, is just the transformation between the ... 4 This is a covariant derivative along a world line (if you would not consider a world line the proper time \tau would not make any sense). So you consider a curve in space time parametrized in dependence of the proper time x^\mu(\tau). Then you have:$$\frac{DA^\mu}{d\tau} = \frac{\partial A^{\mu}\big(x(\tau)\big)}{\partial \tau} + ...

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