Tag Info

New answers tagged

1

For Mathematica the best I know is RGTC. I Used it a long time ago (briefly) for a calculation in IIA SUGRA in 10 dimensions. It calculate gravitational tensors, manages differential forms (also Lie algebra valued ones), calculates Hodge dualities, etc. Personal comment: If you are more intrepid (and FLOSS lover), there is a software called SAGE, which ...


2

Recall the integral definition of the gradient: $$\nabla \varphi = \lim_{V \to 0} \frac{1}{V} \oint_{\partial V} \varphi \hat n \, dS$$ This should tell you the gradient's components transform the same way as those of the normal vector $\hat n$, which is known to have covariant components. You can verify that the normal vector has covariant components by ...


4

Gradient is covariant! Why? The components of a vector contravariant because they transform in the inverse (i.e. contra) way of the vector basis. It is customary to denote these components with an upper index. So, if your coordinates are called $q$'s, they are denoted $q^i$. Therefore, the gradient (or a derivative if you prefer) is $$\partial_i = ...


1

A vector $u^a$ is parallel-transported along the integral curve of tangent vector $V^a \equiv \frac{dX^a}{d s}$ if we have $V^a\nabla_a u^b =0$ (a vector parallel-transported with respect to itself defines a geodesic). The angle between two vectors $v^a$ and $u^a$ is given by $\cos\theta = \frac{v^a u_a}{\sqrt{v^av_a u^bu_b}}$, hence we take: $$ V^a \nabla_a ...


1

"Covariant vectors as expressed in a dual basis" is exactly the same thing that "orthogonal projections to achieve covariant components" Choose one generating system $\vec e_1, \vec e_2$, non necessary orthogonal, a vector $\vec v$ may be expressed by $\vec v = v^1 \vec e_1 + v^2 \vec e_2$. The coordinates $v^1, v^2$ are called contravariant coordinates of ...


9

In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ...


1

You may find this question and especially Emilio Pisanty's answer to be enlighting with regards to what co- and contravariant vectors really are. Now, if I may rephrase your original question a bit: *How do parallelograms and lines in $\mathbb{R}^3$ relate to co- and contravariant vectors in $T_p \mathbb{R}^3$ and $T^*_p\mathbb{R}^3$? The first one is ...


1

A covariant vector is commonly a vector whose components are written with ``downstairs" index, like $x_{\mu}$. Now, the gradient is defined as $\partial_\mu := \dfrac{\partial}{\partial x^\mu}$. As you can see the covariant vector $\partial_\mu$ is the derivative with respect to the contravariant vector $x^\mu$. the contravariant form of $\partial_\mu$ is ...


2

The formula given is horribly unenlightening because it does not seem to use the fundamental fact about differential forms that they are alternating and thus adds $r$ equal terms, it also does not provide the connection to index notation that it supposedly tries to. Let us first understand the idea of the interior product. An $r$-form is something with $r$ ...


3

To derive the field equations more quickly, consider that all the terms in the Lagrangian are "squares": a tensor contracted with itself on all indices. The variation of a square is $\delta(F_{ab} F^{ab}) = 2 (\delta F_{ab}) F^{ab}$ in analogy with $d/dx\; f^2 = 2ff'$. Using this, we have for the variation of the Lagrangian $$\delta \mathcal L = - (\delta ...


3

Comment to the question (v2): Apart from the issue of various overall sign conventions found in the literature, note that: On one hand, there is the Levi-Civita symbol with upper (lower) indices, whose entries are only $0$s and $\pm 1$s; it is a contravariant (covariant) pseudotensor density, respectively. On the other hand, there is the Levi-Civita ...


3

Going by a magic 8-ball a brief web search, the most important steps towards the geometrization of electromagnetism (ie its formulation as a classical Yang-Mills theory in terms of principal connections) should be: Maxwell's equations: James Clerk Maxwell, A dynamical theory of the electromagnetic field (1865) differential forms: √Člie Cartan, Sur certaines ...


4

Your question is one about true mathematical duality, you just do not know it. What you are looking for is Hodge duality, which holds in the exterior algebra of any vector space equipped with an inner product and an orientation, and the differential forms one looks at in EM, GR and elsewhere are just elements of the exterior algebra of the tangent space (or, ...



Top 50 recent answers are included