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The square root of the determinant of the metric can be understood as a particular function of the components of the metric $g_{ab}$ $$\sqrt{-g} =f(g_{ab})$$ By the chain rule we of course have $$\nabla_a \sqrt{-g} = \nabla_a f(g_{bc}) = \frac{df}{d g_{bc}} \nabla_a g_{bc}$$ But we know that $\nabla_a g_{bc}=0$ so that of course $\nabla_a \sqrt{-g} =0$. This ...

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Here's a heuristic calculation: Let $\{E_i\}$ be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then $\mu$ is the canonical volume form $\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ iff $\mu(E_1,\dotsc, E_n)=1$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X ... 6 Comments to the post (v2): Note that \sqrt{|g|} transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of \sqrt{|g|} does not necessarily coincide with the partial derivative of \sqrt{|g|}. Here is a heuristic explanation using local coordinates. The Levi-Civita connection is ... 0 A particle following a prescribed path has its velocity vector parameterized as$$ \vec{v} = \vec{e} \,v $$where \vec{e} is the tangent vector and v is the speed at that instant. This is kind of obvious. But you use the above to find the tangent vector if you know that radial vector \vec{r}. Use \vec{v} = \frac{{\rm d}}{{\rm d}t} \vec{r} = \vec{e} ... -3 What we want us not just$$ \int|\Psi (x)|^2\;dx $$For example$$ p=\int Ψ^*(x)(- i \hbar)\; \text{grad} Ψ(x)\;dx $$If the wave function is not differeciable \text{grad} will come into a terrible result 0 A differential like dx in an equation like$$dz = dx + dy + dx^2 + dx\,dy$$is shorthand for a quantity that will eventually go to zero in a limit. If a differential is divided by another differential, then it has a chance of resulting in a finite, non-zero quantity. Let's assume x, y, and z can be parameterized by a variable t. If we divide the ... 2 Your first line is fine, everything else is wrong (except the one time you repeated something from the first line). How wrong? You don't even have the right units, so pretty much as wrong as you can be. It's like if I asked for the surface area of a house and you said 5m or 8s or 20N or 80K. For a differentiable function f the directional derivative in ... 0 In order for the derivative to exist, the function must be well defined on an interval which includes that instant. Therefore the derivative is also defined on an interval arbout that instant. 0 The divergence of a vector field is a scalar, which means that when evaluated in another coordinate system, you obtain the same value. This means that you can evaluate the divergence of the expression in the coordinate system which has its x-axis pointing in the direction of the vector \vec{a}. In that coordinate system, you have:$$\vec{E} = ...

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There is a certain way that vector calculus operators convert to other coordinate systems from Cartesian. The derivations are a little bit involved but it is a common practice to straight away use the formula. See https://en.m.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates for those formula. Alternatively you can do it using tensor(index) ...

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