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0

Note that $\int^{\infty}_{-\infty}f(t)dt=\int^{\infty}_{-\infty}f(-t)dt$ and also $\int^{\infty}_{-\infty}\frac{df(t)}{dt}dt=\int^{\infty}_{-\infty}\frac{df(-t)}{-dt}dt$


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You can't deduce this, because all metrics with different $\mu$ are invariant under Lorentz transformations. The best you can do is to choose the units in which you measure the invariant interval such that $\mu$ is equal to whatever you like it to be.


2

We know that the Levi-Civita connection satisfies $\nabla_a g_{bc} = 0$ and the product rule. The definition of the inverse metric $g^{ab}$ is $g^{ab}g_{bc} = \delta^a_c$. Therefore, we have: $$\begin{align} 0 &= \nabla_a \delta^b_c \\ &= \nabla_a (g^{bd}g_{dc}) \\ &= (\nabla_a g^{bd}) g_{dc} + g^{bd} \nabla_a g_{dc} \\ &= (\nabla_a g^{bd}) ...


2

The metric being a rank $(0,2)$ tensor transforms under general coordinate transformations $x^\mu \to x'^\mu(x)$ as $$ g'_{\mu\nu} (x') = \frac{ \partial x^\rho}{ \partial x'^\mu } \frac{ \partial x^\sigma }{ \partial x'^\nu } g_{\rho\sigma} (x) $$ Now set $x'^\mu (x) = x^\mu + \alpha k^\mu(x)$ in the above expression and take a limit of small $\alpha$. ...


3

The convention we pick here will interact with the convention we have for matrix multiplication in the following way: If we have matrices $A$ and $B$ and we use the usual convention that the matrix multiplication $AB$ multiplies the rows of $A$ with the columns of $B$ then we have either $$(AB)_{ij}=\sum_kA_{ik}B_{kj}\tag{#}$$ or ...


0

This is happening because your field depends on $t$, $\Psi = \Psi (x,t)$. Therefore, when you perform the Wick rotation $t = i\tau$, you also Wick rotate to your field, and obtain an action for $\Psi(x,i\tau)$. In the second case, you obtain an action for $\Psi(x,-i\tau)$. Those are not the same $\Psi$'s.


5

I believe it can be useful to define the following concepts (I won't be very formal here for pedagogical reasons): Any event can be described through four real numbers, which we take to be: the moment in time it happens, and the position in space where it takes place. We call this four numbers the coordinates of the event. We collect these numbers in a ...


4

You're right. $\eta_{\mu\nu}\rightarrow \eta_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\,\mu^{'}}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\,\nu^{'}}$ just says that the metric transforms as a tensor as you would expect from its indices; there's nothing special about that. Being invariant means that when you make the transformation you get back the same matrix: ...


0

In a gravity theory in spacetime, the metric has signature $− + +· · ·+$. That's a convention. Other conventions are that it has signature $+ - - -$. Concretely this means that the metric tensor $g_{μν}$ may be diagonalized The signature doesn't tell you that it is diagonalizable. The fact that $g_{\mu\nu}=g_{\nu\mu}$ tells you that it is ...


2

Let's go step by step as it seems you're missing some fundamentals. We know from (linear) algebra, that a symmetric bilinear form can be transformed to a diagonal matrix with elements $e$ on the main diagonal $e\in \{0,1,-1\}$. The tripel counting the amount of times each number appears is called signature. If you didn't know that, check this. Now, a ...


0

Thanks to Pythagoras, we know Euclidean space satisfies $\text{d}s_E^2=\text{d}x^2+\text{d}y^2=\text{d}z^2=\text{d}r^2+r^2\text{d}\theta^2+r^2\sin^2\theta\text{d}\phi^2$, where the last result is obtained by the chain rule. Clearly, any manifold with local coordinates satisfies a result of the form ...


2

The lapse function is not defined by the metric alone, but instead depends on both the metric $g_{ab}$ and its slicing into timelike hypersurfaces. One way to "slice" a spacetime $\mathcal{M}$ into timelike hypersurfaces is to define a timelike coordinate $f$, which is just a function $f: \mathcal{M} \to \mathbb{R}$ such that $\nabla_a f$ is a timelike ...


1

The metric measures lengths in various directions, and also angles between various directions. For example if $\vec{e}_{(1)}$ is the basis vector in the $x^1$-direction, it will have length given by $$ \lVert \vec{e}_{(1)} \rVert^2 = g(\vec{e}_{(1)}, \vec{e}_{(1)}) = g_{11}. $$ If we also have the basis vector $\vec{e}_{(2)}$ in the $x^2$-direction, then the ...


1

The metric is an important concept in general relativity. In GR, vectors correspond to weighted directions in spacetime (by "weighted", I mean any scalar multiple of a vector corresponds to the same direction, but weighted differently). The metric tensor can then tell us about the angle between two directions or the magnitude of a given vector, which gives ...


2

As has been pointed out in the comments, it's not entirely clear how you intend to specify a metric without the use of some set of coordinates. That said, a couple of common GR texts have non-standard approaches to the Schwarzchild metric that you might find interesting. Misner, Thorne, and Wheeler's Gravitation has a fairly detailed sidebar (Box 23.3, ...


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Space-time is not simply a manifold, but a differentiable and pseudo-Riemannian manifold. The topology of the manifold is already fixed by the (topological) manifold structure (that is: local homeomorphism to $\mathbb R^n$ without assuming any inner product). In this sense sufficiently small open sets from the manifold carry the topology of a subset of the ...


4

It is simply false, at least written as it stands. The point is that the relation between the topology and the metric is more complicated than in the Riemannian case, where the geodesical balls form a basis of the topology$^1$. As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a ...



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