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3

It seems to me that you are making confusion between a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative). Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from $\mathbb R^3$ to $\mathbb ...


1

First some notation $$\nabla \times \vec B \rightarrow \epsilon_{ijk}\nabla_j B_k$$ $$\nabla \cdot \vec B \rightarrow \nabla_i B_i$$ $$\nabla B \rightarrow \nabla_i B$$ Now, to your problem, $$\nabla \cdot(\nabla \times \vec V)$$ writing it in index notation $$\nabla_i (\epsilon_{ijk}\nabla_j V_k)$$ Now, simply compute it, (remember the Levi-Civita is ...


3

The Basic Idea Physically the total derivative tells you how a quantity changes when it is subjected to a space and time dependent velocity field. In physics we usually call it the material derivative. An Intuitive Example Suppose $\rho(\mathbb{x},t)$ measures the temperature of a fluid, according to a thermometer immersed at the point $\mathbb{x}$ and ...


1

Conservation laws are expressed in the form, $$ \frac{\partial q}{\partial t}+\nabla\cdot\mathbb T =0\tag{1} $$ for some quantity $q$. Here, the term $\mathbb T$ depends on what the quantity $q$ is; for the Euler hydrodynamic equations, it would be: $$ \mathbb T=\begin{cases}\rho\mathbf u & q=\rho \\ \rho\mathbf{uu}^T+p\mathbb{I} & q=\rho\mathbf u \\ ...


0

Let's consider a function $f$ that depends on time $t$, position $q$ and momentum $p$. This is just an example related to Lagrangian physics regarding notations. One wants to measure the variation of $f(t,q,p)$ with respect to time. We know that $f(t,q,p)$ has an explicit dependence on time but nothing keeps $q$ and $p$ to also depend on time. Then one can ...


2

It's the chain rule. We have $$ r_{ij} = \sqrt{w} $$ where I've defined $w$ as all those terms underneath the square root. So $$ \frac{d r_{ij}}{dt} =\frac{d r_{ij}}{dw} \frac{dw}{dt} = \frac{1}{2 \sqrt{w}} \frac{dw}{dt} = \frac{1}{2 r_{ij}} \frac{dw}{dt} \,, $$ where to go to the second expression I've used the chain rule (the rest is just ...


1

The conservative answer is: It depends on context. Different authors mean different things. See also this Phys.SE post.


3

It is definitely an ambiguous notation, but one that is quite conventional. You should interpret it as: $(\partial_a X_\mu)(\partial^a X^\mu)$. For instance, often the Klein-Gordon Lagrangian is written as: $$\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \cdots $$ which should be interpreted as: $$\mathcal{L} = \frac{1}{2} (\partial_\mu ...


1

In my experience, the correct interpretation is $$ \partial_\mu X \partial^\mu X = (\partial_\mu X)( \partial^\mu X). $$ Total derivatives are usually written clearly as $$ \partial_\mu(\ldots). $$


2

The bounds of the integral have no dependence on any of the variables, and hence we may move the differential operator into the integrand, $$\frac{\delta}{\delta \eta (z)} \int \mathrm{d}^4 y \, S_F (z-y) \eta(y) = \int \mathrm{d}^4 y \, S_F (z-y) \delta^{(4)}(z-y)$$ Evaluating the integral using the standard delta distribution identity, we obtain your ...


0

The question is somewhat ill-defined. This answer is based on what I gather the OP is asking. No. Generically, the covariant derivative of a tensor is zero only if the entire tensor is zero. Note how the covariant derivative is defined (taking example of a 1-form) $$ (\nabla_\mu A)_\nu = \partial_\mu A_\nu - \Gamma^\lambda_{\mu\nu} A_\lambda $$ Now, ...


7

The result is sometimes called Flanders' lemma. The remarkable point is that it does not need that $f$ is analytic, but just that it is $C^\infty$. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge. It works in any open star-shaped neighborhood of points in $\mathbb R^n$. A set $A\subset ...


0

I'll share a one dimensional example and see if this helps to shed any light on the matter. Suppose $f(x)$ is an analytic function (i.e. the taylor series of $f$ converges to $f$). Then we can write, $$ f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2 + \cdots $$ $$ = f(a) + (x-a) \left[ f'(a) + f''(a)(x-a)/2 + \cdots \right] $$ $$ = f(a) + (x-a) H(x) $$ ...


1

Let us star with the fundamental theorem of calculus for $n=1$: $F(x) - F(a) = \int_{a}^{x} F'(s)ds$ Now use the substitution $s=t(x-a)+a$. This is because $s$ rescales the interval $[a,x]$ to $[0,1]$. Then $ds = dt(x-a)$, then we get: $F(x) - F(a) = (x-a)\int_{0}^{1} F'(t(x-a)+a)dt$, and this proofs $n =1$. Now define a vector $y^{\mu} = t(x^{\mu} ...


2

It is the Taylor expansion around a point $a$ of a multivariable function where $H_{\mu}$ is similar to the Hessian matrix but for order 1. You can find it in Vector Calculus of Marsden, Tomba, or just google it. This representation is written with the Einstein summation



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