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Take a trace of Einstein equations (trace of $g_{\mu \nu}$ is $D$), you obtain $$R - \frac{D}{2} R + D \Lambda = 0$$ Or $$R=\frac{D \Lambda}{D/2-1}$$ Then substitute this expression for $R$ into full Einstein equations and you obtain trivially $$R_{\mu \nu } = \frac{\Lambda}{D/2 - 1} g_{\mu \nu}$$

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Yes. The tear is initiated at stress concentrations around the holes, where stress is highest. After initiation, the tear continues to propagate along the line of highest stress. Stress is a function of force and geometry ($\sigma_{n} = \frac {F}{A_{n}}$). In a piece of paper without holes, the stress is uniform, and the paper tears when stress exceeds ...

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In tensor notation, $\Lambda^\mu{}_\nu$ is the matrix that performs the Lorentz transformations. Now, since $p^\mu$ is a tensor, under an arbitrary Lorentz transformation, it transforms as $$p^\mu \to p'^\mu = \Lambda^\mu{}_\nu p^\nu$$ Consider now a particle moving entirely in the $x$ direction, with velocity $v_1$. Its 4-momentum is $$p^\mu = \gamma_1( ... 1 In the following, (\mathcal{M},g,\epsilon) is an oriented Lorentz manifold \mathcal{M} with metric tensor g_{ab} and volume form \epsilon=\sqrt{-\det g}\,\mathrm{d}x^0\wedge\dots\wedge\mathrm{d}x^3. Theorem. Given an antisymmetric tensor A_{ab} and a unit timelike vector u^a there exist unique vectors q^a,b^a such that ... 1 I think your confusion is arising from the fact that you are imagining operators as matrices. This is mostly fine, but in this case, the operator itself being a vector is what is causing the confusion - so let me elaborate. {\bf A} is a vector of operators. For example$$ {\bf A} = \pmatrix{ A_1 \\ A_2 \\ A_3} $$We can denote this collectively as A_i. ... 3 Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric:$$T^a T_a = \sum_{a,b} g_{ab} T^a T^b$$You have terms like ... 0 Since you are talking about layers of liquid, i am going to assume you are talking about incompressible, laminar pipe flow (i.e. \mathrm{Re}\ll2000). If it is, can anybody describe it somehow that matches my physical feeling? Shear stress is simply friction between layers of fluid. Imagine rubbing your flat hand across each other, it should be ... 0 It should help to go back to a definition here. Consider the diagram below: A material is enclosed between two parallel plates with surface area A and distance h apart from each other. Assume that a force F is applied to the top plate which now starts moving at constant speed v (the bottom plate is kept stationary). The equation of motion for ... 1 Given any \mathbb{R}-vector space V and its dual space V^\ast, a tensor of rank (k,l) is a map$$ T : V\times\dots V\times V^\ast\times\dots\times V^\ast \to \mathbb{R}\tag{1} where $V$ occurs $k$ times and $V^\ast$ occurs $l$ times, that is linear in each argument. Equivalently, it is an element of the tensor product $V^\ast\otimes\dots\otimes ... 2 This isn't really a physics question, but while it stays up I can give a quick response: Your are correct that a tensor is often defined in terms of its coordinate transformations, but this is not the only or most modern definition. Here is a coordinate-free definition: A tensor of type (r,s) on a vector space$V$and its dual$V^*$is a complex-valued ... 0 From a formal point of view, the reason for the tensor product rule is as follows: For any quantum system the Hilbert space of states$\mathcal{H}$is defined in terms of a complete set of degrees of freedom$S_1, S_2,…, S_n$that can be measured simultaneously. The corresponding quantum states, characterized by sets of values$\{s_1,s_2,…,s_n\}\$ and ...

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