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The existance of Killing fields even just for a small region is a special property of the metric. For general metrics you cannot expect to find Killing vectors. Notice that the Killing equation which should be written using covariant derivatives as $X_{\mu;\nu} + X_{\nu;\mu} = 0$ is 10 independent partial differential equations for only 4 field components so ...


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In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation $r \to \lambda r$, the geometry is only preserved by $R_S \to \lambda R_S$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation. I went to some effort (as an exercise for myself) ...


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You basically have the right idea: the existence of a covariantly constant vector field is a big restriction on the metric. You already discovered that $A^a$ has constant norm. The next thing we find is that $A^a$ is a Killing vector, because obviously $$\nabla_{(a}A_{b)}=0.$$ Furthermore, $A^a$ is geodesic, $A^a\nabla_a A^b=0$, and hypersurface ...


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Doesn't the position function r(t) imply that a particle is traveling through the curve from A to B, separate from the influence of the force field since that is how parameterization is defined? Furthermore, isn't the force field that acts on such a particle that is moving on it's fixed path or curve C merely just changing the energy and speed at which ...


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Given $\bar{x}^i(x^j)$ transformation law of coordinates. The vector components $X^i$ and the covector $X_i$ are defined to transform like: $$ \bar X^i = \frac{\partial\bar x_i}{\partial x_j} X^j,\quad\quad\quad \bar X_i = \frac{\partial x_i}{\partial\bar x_j} X_j $$ Both are simultaneously defined independent of the metric. And their definitions is ...


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This proof from Griffiths book introduction to electrodynamics Consider the vector function $$\vec{a}=\frac{1}{r^2}\hat{r}$$ At every location $\vec{a}$ is directed radially outward ; if ever there was a function that ought to have a large positive divergence, this is it. and yet, when you actually calculate the divergence, you will get ...


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Indeed the answer is not zero but $-4\pi\delta(r)$ (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at $r=0$ we must be careful. At any other points is easy to calculate it. It is ...



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