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There is a sort of analog called gravitomagnetism (or gravitoelectromagnetism), but it is not discussed that often because it applies only in a special case. It is an approximation of general relativity (i.e. the Einstein Field Equations) in the case where: The weak field limit applies. The correct reference frame is chosen (it's not entirely clear to me ...


3

There is a gravitational analogue of the magnetic field. See gravitoelectromagnetism and frame dragging on Wikipedia.


2

$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$ You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that: $$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$, so, we have: $$\begin{align} ...


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

"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 ...


1

I believe this has to do with the production of vorticity. If you look at the vorticity transport equation (incompressible, barotropic): $$ \frac{D\vec{\omega}}{Dt} = (\vec{\omega}\cdot\nabla)\vec{v} +\nu\nabla^2\vec{\omega} $$ you see that if $\vec{\omega} = 0$, then $\frac{D\vec{\omega}}{Dt} = 0$, i.e. there is no production term for vorticity in an ...



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