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2

In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric $g$ in the coordinates $x_1, x_2, \ldots$, the algorithm produces a list \begin{align} \Lambda & = \Lambda(x_i) \\ \Psi_k & = ...


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Start with the Minkowski metric $ds^2=-dx^2+dy^2$ and suppose $x=x(r,t),y=y(r,t)$ then you get $$ ds^2=-\left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial t}dt \right)^2+\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial t}dt\right)^2=-r^2dt^2+dr^2.$$ Now equating the coefficients at $dr^2, dt^2,drdt$ you get ...


1

If you were to Wick rotate $t \rightarrow i \theta$, the metric would be $ds^2 = dr^2 + r^2 d\theta^2$, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by $x=r\cos\theta$, $y=r\sin\theta$. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines ...


0

Let's be clear on the terminology here, because taken literally, your question is a bit like asking whether angular momentum is green or non-green. Polar coordinates exist in space. Curves of constant $r$ are circular and curves of constant $\theta$ are radial, in the usual way. Taking their tangent vectors and normalizing them, we get ...


0

Having pondered this for a while, I think the simplest answer must be that the formula $$\boldsymbol{a} = (\ddot{r} - r \dot{\theta}^2) \textbf{e}_r + (r \ddot{\theta} + 2\dot{r}\dot{\theta})\textbf{e}_\theta$$ can only be produced when we differentiate with respect to the inertial observer. It is only due to our differentiation of the terms in the inertial ...


1

The uncertainty principle tells us that when we make a measurement of $x$ and a measurement of $p_x$ (i.e. the $x$ component of the momentum) then no matter how ideal our measurement the inequality: $$ \Delta x\Delta p_x \ge \frac{\hbar}{2} $$ applies. The equation contains $x$ because we have arranged our axes to make the measurement in the $x$ direction. ...


3

For a freely falling observer the local geometry of spacetime is always flat i.e. described by the Minkowski metric. So the freely falling observer can never observe themself to fall through an event horizon, because that contradicts the requirement that spacetime be locally flat. In fact the freely falling observer will observe an apparent horizon that ...


1

Comments to the question (v3): Given a manifold $M$, if a smooth vector field $X\in \Gamma(TM)$ does not vanish in a point $p\in M$, then one may choose a local coordinate neighborhood $U\subseteq M$ of $p$, with local coordinates $(x^1, \ldots, x^n)$, so that $X=\frac{\partial}{\partial x^1}$. This procedure is sometimes called stratification or ...


4

@Phoenix87 is spot on, but I'll elaborate a bit. Definition 1 A spacetime $(M,g)$ is stationary if there exists a timelike Killing field $K$, i.e. a vector field $K$ such that $\langle K,K\rangle<0$ and $\mathcal{L}_Kg=0$. We shall show that Definition 1 implies the existence of local coordinates for which $g_{\mu\nu}$ is independent of time. ...


3

The rough idea: take the local flow of the vector field and use it to get a new "time" coordinate. In general this will work locally, so you have to patch your manifold with small enough open subsets where you can then define the new set of coordinates where now the Killing vector field corresponds to $\partial_t$.



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