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3

One need not follow these steps. Indeed let $\gamma : I\subset \mathbb{R}\to \mathbb{R}^3$ be the trajectory of a particle. It's position at time $t$ is $\gamma(t)$, it's velocity is $\gamma'(t)$ and it's acceleration is $\gamma''(t)$. It's easy to see that $$(\gamma'\cdot \gamma')'(t) = 2\gamma'(t)\cdot \gamma''(t),$$ so the work done by the resultant ...


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Formally, we have the standard dot product identity $$\frac{\mathrm{d}}{\mathrm{d}t}\mathbf{A}\cdot\mathbf{B}=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{B}+\mathbf{A}\cdot \frac{\mathrm{d}\mathbf{B}}{\mathrm{d}t}.$$ Inserting $\mathbf{A}=\mathbf{B}=\mathbf{v}$ gives $$\frac{\mathrm{d}\mathbf{v}^2}{\mathrm{d}t}=2\mathbf{v}\cdot ...


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I'm not completely sure what you want, but honestly the entirety of Spivak's Calculus on manifolds is devoted to exactly that. If you want something that feels familiar, you can simply find $\nabla$ in various coordinate systems in Wikipedia, but if you want a less coordinate-centric view then you're probably going to need to step outside of your comfort ...


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I would say that the Wikipedia page on curvilinear coordinates and the article Mathematical Physics Lessons - Gradient, Divergence and Curl in Curvilinear Coordinates by James Foadi are enough to understand what is going on.


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$$ \nabla \epsilon_{...}=\nabla \sqrt{g}[...]=gg^{\mu\nu}\nabla g_{\mu\nu}[...]=0$$ In the last step we use Jacobi's formula for differentiating a determinant and $\nabla g_{\mu\nu}=0$. Here $[...]$ describes permutations and consists of $-1,0,1$ as you see from this article.


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Consider the drift diffusion equation $$\dfrac{\partial}{\partial t}\psi=\mu\dfrac{\partial}{\partial x}\psi+\kappa^2\dfrac{\partial^2}{\partial x^2}\psi.$$ Dimensional analysis tells us that $\mu$ is a characteristic length per time (drift velocity) while $\kappa$ is a characteristic length per square root of time. This small factoid has curious ...


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Actually here $\rho$ and ${\bf v}$ are function of $(t,\vec{x}) \in \mathbb R \times \mathbb R^3$, as it is usual in the so-called Eulerian description of a continuous body. There is no reference to the curves describing the evolutions of the particles of the fluid $\vec{x}_{\vec{x_0}}= \vec{x}_{\vec{x_0}}(t)$ as instead, it is the standard in the so-called ...


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From the chain rule we have, $\frac{d \rho}{dt} = \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}\frac{dx}{dt} + \frac{\partial \rho}{\partial y}\frac{dy}{dt} + \frac{\partial \rho}{\partial z}\frac{dz}{dt}$ $\therefore \frac{d \rho}{dt} = \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x} v_x + \frac{\partial ...


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Hint: $\frac{\partial }{\partial \dot{q}}\dot{F}=\frac{\partial}{\partial \dot{q}}\left(\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial \dot{q}}\ddot{q}+\frac{\partial F}{\partial t}\right)$ What does $F=F(q,t)$ imply about $\frac{\partial}{\partial \dot{q}} F$?


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I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ...


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$$\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{q}_j}=\dfrac{\partial\dot{r}_i^T\dot{r}_i}{\partial\dot{r}_i}\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}=2\dot{r}_i^T\dfrac{\partial\dot{r}_i}{\partial\dot{q}_j}$$ (A simple chain rule, while bearing in mind that $\dot{r}_i^T\dot{r}_i$ is essentially $\dot{r}_i^2$. You just need a transposed vector on the ...



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