# Tag Info

## New answers tagged metric-tensor

1

No, instead of the metric, the Vierbein enters, pulling back the gamma matrix, defined in the usual way in the tangent space, to the spacetime manifold. $$\bar \psi \, \gamma^\mu D_\mu \psi = \bar \psi \, \gamma^\alpha {e^\mu}_\alpha D_\mu \psi$$ This can be considered as the insertion of "half a metric", if one wishes. The Vierbein captures the ...

1

Yes. The short answer is you have one action you extremize to get Einstein's Field Equation $G_{\alpha\beta}=kT_{\alpha\beta}.$ Which you can think of as equations of motion for the gravitational metric $g_{\alpha\beta}.$ (They determine the second derivatives of the metric in terms of the matter fields and metric and the first derivatives of the metric.) ...

2

The easiest trick is to write (say, in three dimensions. For higher dimensions, add indices to the Levi-Civita symbol, and factors of g): $$g = \frac{1}{3!}\epsilon^{abc}\epsilon^{xyz}g_{ax}g_{by}g_{cz}$$ Then, the variation is easy. I'll leave it as an excersise to work out the variation, and how to translate the result into factors of $g_{ab}$ and $g$

2

Use the identity that if $M$ is invertible and $\delta M$ is "small" compared to $M$, then we have $$\det (M + \delta M) = \det(M) \det( 1 + M^{-1} \delta M) \approx \det(M) \left[ 1 + \text{tr} (M^{-1} \delta M) \right].$$ In the case of the metric, this implies that $$-\det(g + \delta g) \approx -\det(g) \left[ 1 + g^{ab} \delta g_{ab} \right]$$ and ...

1

Marek didn't really make a mistake, but you did. But Marek might have been unclear by jumping over some steps. So first let's clarify. The metric gives the differential squared interval $ds^2=-dt^2+dx^2$ from which you can get the differential proper time $d\tau=\sqrt{dt^2-dx^2}.$ So for the blue straight line you have $\tau$=$\int ... 0 Yes and no. Firstly projection is an operation that takes a vector and gives another vector that lives in a subspace. Projecting doesn't give a length (though you can talk about the length of s projection). And sometimes people talk about different kinds of projections but the first (and only) projection many people learn is the orthogonal projection, so ... 0 The Minkowski metric tensor$\mathbf \eta$takes two four-vectors as arguments and produces a real number, the inner product of the two vectors: $$\mathbf \eta(\vec u, \vec v) = \vec u \cdot \vec v = \langle\tilde u, \vec v\rangle$$ where the one-form$\tilde u$is given by $$\tilde u = \mathbf \eta (\vec u, )$$ Geometrically, this is pictured as ... 3 While I can't speak for the author, I find it very likely that this is a typo. The author probably meant the Schrödinger equation, not Klein-Gordon. 0 Like in most physically complicated cases, one would normally solve the dynamic system numerically. In the case of general relativity, the "3+1 decomposition" method or equivalent is used. There are some open source tools for aiding in numerical research. 1 First, Landau and Lifshitz stated that$ds$and$ds'$approach zero simultaneously, so that there is some hidden variable$x$such that, $$\lim_{x\to 0} ds(x) =0$$ and $$\lim_{x\to 0} ds'(x) =0,$$ assuming and$ds$and$ds'$are continuous functions of$x$. Next, the two are infinitesimals of the same ... 0 The metric tensor$g_{\mu\nu}\$ in 4 dimensions has 10 independent components. Each of these has the potential of being a dynamical degree of freedom, also know as a "mode." There are many different ways of parameterizing these 10 components. In general, of the 10 components, you can always pick one of them to correspond to the overall local scale of the ...

1

It is not the definition of an event horizon, and in fact you can choose coordinates that are regular near the event horizon. A common reason for coordinates that are irregular at the horizon is if the coordinate is primarily made to represent time far away. In that case, a timelike curve has a negative interval in your convention, so you can have time ...

1

It is an extremely complicated problem. Imposing restrictions on the stress energy tensor will give you different possible spacetimes (which spacetimes is also quite a complex problem), but if those restriction apply in all cases is quite hard to prove. The stress energy tensor still has to obey the field equations of the various matter fields that exist, ...

Top 50 recent answers are included