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53

To really understand this you should study the differential geometry of geodesics in curved spacetimes. I'll try to provide a simplified explanation. Even objects "at rest" (in a given reference frame) are actually moving through spacetime, because spacetime is not just space, but also time: apple is "getting older" - moving through time. The "velocity" ...

19

When the apple was detatched from the branch of the tree, it was stationary, so it did not have to follow any geodesic curve. Even when at rest in space, the apple still advances in space-time. Here is a visualization of the falling apple in distorted space-time: http://www.youtube.com/watch?v=DdC0QN6f3G4

11

No, general relativity is based on something called "intrinsic curvature", which is related to how much parallel lines deviate towards or away from each other. It doesn't require embedding space-time in a higher dimensional structure to work. You're thinking of something called "extrinsic curvature". In fact, many examples of extrinsic curvature - including ...

7

Two general methods come to mind: Prove that the Riemann tensor takes the form of equation 3.191, i.e. $$R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc})$$ If you are handed a metric, this should in principle be a straightforward calculation. If the metric is actually maximally symmetric, the calculation of the Riemann tensor usually turns out to ...

5

There is a conserved quantity for geodesics which come from the fact that the metric $g_{ab}$ is (trivially) a Killing tensor, i.e. $$\nabla_{(c}g_{ab)} = 0.$$ Any tensor $\xi_{ab}$ that satisfies $\nabla_{(c}\xi_{ab)}=0$ gives rise to the conserved quantity $\epsilon = u^a u^b\xi_{ab}$, which is preserved along geodesics for which $u^a$ is the tangent ...

5

Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$. I got the (possibly faulty) impression from reading the first lines in OP's question formulation (v18) that OP is asking: Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form ...

5

As to the first paragraph, gravity shows up as geodesic deviation; initially parallel geodesics do not remain parallel. Since, for a freely falling particle, the proper acceleration (the reading of an accelerometer attached to the particle) is zero, it is correct to say that a particle whose worldline is a geodesic has no proper acceleration. But it is not ...

4

For $p=1$, CTC's do not exist in Minkowski spacetime. In other $1+3$ spacetimes, in principle they are admitted in the absence of further requirements (like globally hyperbolicity) on the causal structure of the spacetime. They must be present if the spacetime is compact, for instance. For $p\geq 2$, the answer is obviously YES. Consider a manifold $M$ with ...

4

Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct. Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ ...

4

As a general rule, compactifaction on a Calabi-Yau $n$-fold results in a preservation of $2^{(1-n)}$ parts of the original supersymmetry. If you start with $\mathcal{N}=4$ and compactify on a 2-fold, you preserve $2^{1-2}=1/2$ of the original supersymmetry, i.e. $\mathcal{N}=2$. One now has to realize what a Calabi-Yau $n$-fold is: it is a Kähler manifold ...

3

Not everything needs to follow geodesic Spacetime curvature available to it. With external force, you can prevent a particle from following Spacetime curvature. Only "freely" falling particles follow Spacetime curvature available to them. So, when you see a stationary object not following Spacetime curvature, it's because an external force is preventing it ...

3

Nope, spacetime curvature says nothing about the dimensionality. Your intuition here is probably wrong because human imagination needs 'some dimension to bend into' in order for something to be curved (i.e. an embedding in a higher-dimensional space). This is just our lack of imagination showing, though.

3

I) Nick's answer already correctly explains that the partial derivative $\partial_{\lambda}T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$ of a $(p,q)$ tensor is in general not a tensor, in the sense that it does not transform covariantly under coordinate transformations. II) In a coordinate-independent formulation, a $(p,q)$ tensor $$\tag{1} T~\in~\Gamma ... 3 I) The vielbein e^a{}_{\mu} in the Cartan formalism is an intertwiner$$\tag{1} g_{\mu\nu}~=~e^a{}_{\mu} ~\eta_{ab} ~e^b{}_{\nu} $$between the curved (pseudo) Riemannian metric g_{\mu\nu} and the corresponding flat metric \eta_{ab}. Here \mu,\nu,\lambda, \ldots, are so-called curved indices, while a,b,c, \ldots, are so-called flat indices. ... 2 There is a really nice derivation of this identity using differential forms, and it completely avoids all the messiness of the Christoffel symbols. The nice thing about differential forms is that the exterior derivative can be computed using any derivative operator, so it allows us to compare the expressions we get using the covariant derivative to the ... 2 This is based on the observation that, given some vector V^\mu,$$\nabla_\mu V^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}V^\mu)$$We can show explicitly that this is true:$$\nabla_\mu V^\mu=\partial_\mu V^\mu +\Gamma^\mu_{\mu\lambda}V^\lambda$$Let's examine the last term:$$\Gamma^\mu_{\mu\lambda}=\frac{1}{2}g^{\mu\rho}(\partial_\mu ...

2

Given a pseudo-Riemannian manifold $(M,g)$, the Laplace-Beltrami operator acts on scalar functions. The formula for the Laplace-Beltrami operator follows from the formula $$\Gamma^{\nu}_{\mu\nu}=\partial_{\mu}\ln\sqrt{|g|}$$ for the Levi-Civita connection.

2

This is true in general, and there is a very nice geometrical reason why. First use that the Lie derivative satisfies the Leibniz rule, $$£_N(q_{ab} p^{ab})=(£_Nq_{ab})p^{ab}+q_{ab}£_Np^{ab}$$ to rewrite the integral as $$\int d^3x (£_N q_{ab})p^{ab}= \int d^3x\,£_N(q_{ab}p^{ab}) - \int d^3x\,(£_N p^{ab})q_{ab}$$ Now note that the first integrand on ...

2

I've always liked the interpretation you get from the Raychaudhuri equation. It shows you that the Ricci tensor tends to cause geodesics to focus together. If you begin with a family of geodesics with tangent vector $u^a$, you can define the expansion $\theta\equiv \nabla_a u^a$ which measures the rate at which geodesics are spreading out or converging ...

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