# Tag Info

There's a trick which is often used in computergraphics to account for rotations + translations in one single matrix multiplication. If $R$ is you rotation matrix and $\vec{t}$ is your translation vector you construct the following rotation-translation-matrix: $$M = \left( \matrix{. . . \ \ |\ \ . \\ . R . \ \ |\ \ \vec{t} \\ ... 0 The Ricci tensor obviously a tensor that accepts two vectors and outputs a number. This number represents in some sense the "average" sectional curvature at a given location on the manifold M. In GR, we usually use spacetime manifolds. In his book Riemannian Geometry, Manfredo Do Carmo states the following on page 97: Let x = z_n be a unit vector ... 1 Some tensors have more intuitive pictures than others. For instance, simple antisymmetric tensors tensors can actually represent oriented subspaces with a magnitude. So that's the clear inheritor of the oriented-1-dimensional-subspace with magnitude. And then you can imagine little oriented planes, little oriented 3-volumes in 4d (for relativistic ... 1 I'll just throw out two simple examples of tensors that you already know: 1) the dot product, itself, is a tensor. It takes two vectors as input, and spits out a number 2) Now, consider a general fluid. It might have viscosity, and shear and everything. It's completely general. Now, consider yourself to be something living in this fluid. You move a ... 2 There is another definition. Let's say v and w are part of the vector spaces V and W. Now, let's consider bilinear maps f from V and W to vector spaces such as Z. Bilinear means that  f(v_1+v_2, w) = f(v_1, w) + f(v_2, w)  f(v, w_1 + w_2) = f(v, w_1) + f(v, w_2) f(cv, w) = cf(v, w) = f(v, cw) This maps are important for some reason. ... 1 That is only one definition my friend, but there is another. But to understand v \otimes w, you must first understand V \otimes W (where V and W are the vector spaces for v and w respectively.) First, consider, for all the vectors v and w, the abstract symbols v \otimes w, and use this as a basis for a free vector space. Now, using ... 0 Once ^* is the Hodge star operator, its definition in non-flat spacetime (\star \eta)_{i_1,i_2,\ldots,i_{n-k}} = \frac{1}{(k)!} \eta^{j_1,\ldots,j_k}\,\sqrt {|\det g|} \,\epsilon_{j_1,\ldots,j_k,i_1,\ldots,i_{n-k}} say that we only need the determinant of the metric tensor \bf{g} to change from \bf{F} to \bf{^*F}. Thanks for @ACuriousMind and ... 2 The definition suggested by joshphysics and clarified by Qmechanic already exists in the literature under then name of representation operator. This is discussed in, e.g., Sternberg's Group Theory and Physics, as well as the somewhat more elementary text An Introduction to Tensors and Group Theory for Physicists by Jeevanjee. 4 Because the equations for linear and angular motion are very symmetrical In Newtonian mechanics, linear momentum is a vector while angular momentum is pseudo-vector which hints at its true nature as a higher rank tensor object. In relativistic mechanics, four-momentum is a four-vector while angular momentum is a (rank 2) four-tensor. So, the ... 0$$ \frac{\eta^{\mu\nu}}{\Box}