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6

Comments to the post (v2): Note that $\sqrt{|g|}$ transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of $\sqrt{|g|}$ does not necessarily coincide with the partial derivative of $\sqrt{|g|}$. Here is a heuristic explanation using local coordinates. The Levi-Civita connection is ...


6

Parametrize the particle's worldline w.r.t. $t$: $$~x^\mu (t) = (t,1)$$ Its four-velocity is: $$u^\mu =\frac{d x^\mu}{d \tau}$$ To evaluate this we use the fact that: $$d \tau^2 = coshx~dt^2-dx^2$$ Also use $x=1$ and $dx=0$: $$d \tau^2 = cosh(1) dt^2$$ Therefore: $$u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = ...


6

Your intuition that the Einstein equations are equations for the metric tensor, not for the manifold is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your ...


4

$$ \partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right)= \phi_{t}^{*}\left(\it\unicode{xA3}_X\left(\rho_{t}\mathrm{d}\omega\right)\right)+\phi_{t}^{*}\left(\left(\partial_{t}\rho_{t}\right)\mathrm{d}\omega\right)\tag 1$$ This is quite easy, the fundamental reason for the result above is that $\quad\quad$ the function ...


3

Take a future-directed timelike curve $\gamma= \gamma(\tau)$, $\tau$ being the proper time along $\gamma$ in the spacetime $M$. Assume that $p = \gamma(0)$ is the initial point of $\gamma$. Fermi coordinates adapted to $\gamma$ are constructed this way. Consider an orthonormal basis of $T_pM$ with $e_0$ parallel to $\dot{\gamma}$. Transport the basis ...


3

Diffeomorphism Invariance Let $M$ be a smooth manifold. Let $\phi: M \to M$ be a diffeomorphism. A simple property of the Einstein equations is $$ g \in \otimes^2 TM \text{ is solution to vacuum Einstein equation} \implies \text{ so is } \phi^*g $$ To see that this is true, simply pull back both sides of the Einstein equation by $\phi$, and use the ...


2

The solution that you wrote in your last (not numbered) equation is not a basis of a Hilbert space of sections because the phase factor: $(-1)^n e^{2i\pi A}$ depends on $n$. The phase factor should not depend on $n$. Please see your (correct) equation (2) defining the boundary conditions, in which the phase factor does not depend on $n$. Thus there is no ...


2

You don't. Two given spacetimes can have their metrics written in the same way but may have different coordinate ranges. A simple example is just a spacetime with spatial coordinate identified , such as the cylinder spacetime : $$ds^2 = -dt^2 + d\theta$$ Identical to Minkowski space, which is its universal cover. Of course, two things to watch out for : ...


2

A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point $x$ and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing ...


2

I) Assuming that the variational problem for the action $S=\int \! d^nx~{\cal L}$ is well-posed (with appropriate boundary conditions), the field-theoretic Euler-Lagrange (EL) equations read in general $$\tag{1} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial ...


2

Here's a heuristic calculation: Let $\{E_i\}$ be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then $\mu$ is the canonical volume form $\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ iff $\mu(E_1,\dotsc, E_n)=1$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X ...


1

The square root of the determinant of the metric can be understood as a particular function of the components of the metric $g_{ab}$ $$\sqrt{-g} =f(g_{ab})$$ By the chain rule we of course have $$\nabla_a \sqrt{-g} = \nabla_a f(g_{bc}) = \frac{df}{d g_{bc}} \nabla_a g_{bc}$$ But we know that $\nabla_a g_{bc}=0$ so that of course $\nabla_a \sqrt{-g} =0$. This ...


1

Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...



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