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1

As Draksis said, the condition is that the integral over any volume has to be zero. If you want a formal proof, here you go: Let's call $f(\mathbf{x}) = \nabla \cdot \mathbf{B}$, and assume it is continuous. Suppose there's some $\mathbf{x}_0 \in \mathbb{R}^3$ with $f(\mathbf{x}_0) \neq 0$, and let's say that $f(\mathbf{x}_0) > 0$ (the proof for $f < ...


1

Intuitively, if the volume integral of a function is 0 over any arbitrary volume, the function itself must be 0 at all points in space. More concretely, consider a function for which $\int_V \, f \, \mathrm{d}x = 0$ for any volume $V$. Then, $\int_{V+dV} \, f \, \mathrm{d}x = 0$ for any infinitesimal addition to V. $$\int_{V+dV} \, f \, \mathrm{d}x - ...


0

An isometry of ${\cal{M}}$ is a transformation $\phi:{\cal{M}}\rightarrow{\cal{M}}$ such that the following condition hold: \begin{equation} g_{p}(v,v)=g_{\phi(p)}(\phi_{*}v,\phi_{*}v) \end{equation} where $v\in T_{p}{\cal{M}}$ and $\phi_{*}v\in T_{\phi(p)}{\cal{M}}$ and $\phi_{*}$ is the pushforward. Which is basically the statement that the the metric is ...


1

When you write $\omega(X,Y)$ this gives the impression that $\omega$ is a 2-form! It is $d\omega$ that is a 2-form, if $\omega$ is a 1-form. The action of the 1-form $\omega =\omega_\mu dx^\mu$ on the vector field $X = X^\mu \partial_\mu$ is simply $$\omega(X) = \omega_\mu X^\mu.$$ (Since you use index notation, you probably know about relativity and this ...


1

You're on the right track. Hints: Recall that under a diffeomorphism $f$, the tensor transformation law tells us that the metric transforms as $g\to g_f$ where \begin{align} (g_f)_{\mu\nu}(f(x)) = g_{\alpha\beta}(x)\partial_\mu (f^{-1})^\alpha(f(x))\partial_\nu (f^{-1})^\beta(f(x)) \end{align} which, sending $x\to f^{-1}(x)$ can be re-written as ...


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Write $$g^\mathfrak{ab} = x^\mathfrak{a}_\mu x^\mathfrak{b}_\nu g^{\mu\nu}.$$ Consider everything as a function of ${x'} = x + \xi$ and Taylor expand to first order.



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