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disclaimer: this is my first time asking a question. In general relativity, I have been told that everything stems from the coordinates we have and that from these, we can derive the metric and, thus, …

I am playing around with calculating a line element for cylindrical coordinates. So I tried this in two different ways. First, I took the position vector to be $$\vec{r} = (x^2+y^2)^{\frac{1}{2}}\hat{ …

A formal definition of this can be obtained from wikipedia: A tensor is an assignment of a multidimensional array $T^{}_{}[f]$ to each basis $f = (e_1, ... , e_n)$ of an n-dimensional vector space suc …

I am going to summarize what @ghoster and @basics have said in the comments. As Ghoster points out, $$e^l\cdot \Gamma ^k_{nm}e_k = \Gamma ^k_{nm} (e^l \cdot e_k)$$ We know that, by construction, of ba …

I am reading Guidry's Modern General Relativity, and there is a definition for dual basis vectors that are as follows: If we have three coordinates $x(u,v,w)$, $y(u,v,w)$, and $z(u,v,w)$ where we ass …

The metric tensor is a bilinear map that takes in vectors of the tangent space to the manifold. We can expand the metric tensor as $$g(X_i, X_j) = g_{ij}dx^idx^j$$ Now, say the metric is a function of …

You can think of the vectors as a triangle. The x and y components are the base and height of the triangle. From there, you can work out the trigonometry. For instance, in this picture: We can see th …

I am looking at a problem in Guidry's Modern General Relativity, and the solution contains the following two sentences: In the scalar product expression $A\cdot B = g_{\mu \nu}A^{\mu} B^{\nu}$, the l …

If we take a vector $A$, which has three components, my understanding is that we can write this using Einstein notation as $A_{u}$ where this is actually $A_1+A_2+A_3$. We can also write $g^{uv}A_v = …

I am doing problem 5.11 in Guidry, which asks the following: using $$g_{\mu \nu}(x) = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g_{\alpha \beta}(x')$$ …

Follow-up to this question: Why proper time is a measure of space?. The selected answer to me tells us why proper time is an invariant quantity, but I'm still wondering why we equate it to $ds$. Can t …