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1

The mistake you made is this: $\eta^{\mu}_{\nu} \neq \eta_{\mu\nu}$. When you raise index $\mu$ from downstairs to upstairs, the matrix elements change. $\eta^{0}_{0} = 1$, $\eta_{00} = -1$. That is why if you take the trace of $\eta_{\mu\nu}$, you get 2, but if you take the trace of $\eta^{\mu}_{\nu}$ you get 4.

2

If you are asking about notation, the comment of @anderstood is spot on. If $f:\mathbb{R}\rightarrow\mathbb{R}$ is any real function, writing $fx$ instead of $f(x)$ is confusing, because $fx$ could be the product of $f$ and $x$. However, if $f$ is a linear function (linear in a linear algebraic sense, eg. homogenous linear), then $f(x)=ax$, so identifying ...

1

Given a $(1,1)$ tensor as a map $$\tau\colon V\times V^*\to\mathbb{C}$$ such element can be always expressed as $\tau = \tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\otimes \alpha^{\nu}$, with $(e_{\mu}, \alpha^{\nu})$ being a basis of $V, V^*$, respectively. Using your notation, let us act with $\tau$ upon a vector $v\in V$. We have $$\tau(v) = ... 2 We have the formula for the covariant derivative$$\nabla_\mu x^\nu = \partial_\mu x^\nu + \Gamma^\nu{}_{\mu\rho}x^\rho.$$In particular, if x^\mu is a coordinate vector field, then the covariant derivative is precisely the action of the Christoffel symbols on the vector field. Now one should recognize that the covariant derivative comes from the notion ... 2 Christoffel symbols do not describe gravitational fields, nor do they describe accelerations. Rather, they are related to the Levi-Civita connections that can be used to describe geodesics onto manifold with any general metric. In this respect, being general relativity a metric theory, they will eventually show up whilst calculating the equations of motion ... 0 I am finding a difficulty understanding how a tensorial tool (mathematical tool) is the one that describes an idea so physical like the continuous matter. How can those two ideas relate? That seems like asking how any mathematical tool can describe physics, which is unrelated to the stress-energy tensor itself or general relativity. The answer to such ... 0 Perhaps there are two parts of your question. First, why is the stress-energy tensor a useful and natural construct in relativity? And, second, why are mathematical constructs useful and effective in describing physical systems. The latter is largely a philosophical question that I think is too speculative to discuss here.* Both special and general ... 0 One answer is that the delta operates on the metric tensor, changing its (the metric tensor's) index (one of its indices). Another answer is that the metric tensor operates on the delta, lowering one index. Both answers must be correct and thus they are equal. So you get delta with two bottom indices = g with two bottom indices. The insight is that the ... 0 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, ... 1 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 ... 2 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 ... 1 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 ... 2 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 ... 1 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 ... 1 Suppose you have some four-vector v, then the norm of v is given by:$$ v^2 = g_{\alpha\beta}v^{\alpha}v^{\beta} $$Since the dimensions of the left and right sides must agree this shows the metric tensor is dimensionless. 1 Start by rewriting the scalar product as a covariant-contravariant contraction, like so:$$ {\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 ...

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