# Tag Info

10

It is essentially impossible to answer the general question of "how does multilinearity come up naturally in physics?" because of the myriad of possible examples that make up the total answer. Instead, let me describe a situation that very loudly cries out for the use of tensor products of two vectors. Consider the problem of conservation of momentum for a ...

9

In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources. But this is rather illusory. In relativity, the equations look quite different: \begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= ... 8 The question seems to conflate many different things: the invariance of a mathematical quantity (usually a scalar such as ds^2 for the separation of two events in special relativity) covariance of tensors (the values of components of tensors may be calculated from those in another frame but they're not the same thing) universality of equations in ... 7 Are those square brackets standard notation in Physics? Yes. See, for example Sean Carroll notes. At least I can tell you from two other classic references using that notation, "General Relativity" by Wald (1984) and "A First Introducion to General Relativity" by Schutz (2009 for the most recent edition)   If I am in a non-curved \mathbb M ... 7 The tensor equations you mention are not invariant, they are covariant. Big difference. Both are differential equations, which transform linearly under nonlinear transformations from one manifold to another because they are differential equations at a point. The nonlinear transformation from one manifold to another induces a linear transformation of the ... 7 This is true - in fact you could define \nabla^\sigma = g^{\sigma\rho} \nabla_\rho. I assume this meant to say g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch g^{\sigma\rho} with \nabla_\nu, which you ... 7 The moment-of-inertia (MOI) tensor is real (no imaginary terms), symmetric, and positive-definite. Linear algebra tells us that for any (3x3) matrix that has those three properties, there's always a set of three perpendicular axes such that the MOI tensor can be expressed as a diagonal tensor in the basis of those axes. These are called the principal axes ... 7 The antisymmetric part is defined as$$ A_{[a_1 \cdots a_n]} = \frac{1}{n!} \sum\limits_{\sigma \in P(n)} \text{sgn}(\sigma)A_{a_{\sigma(1)} \cdots a_{\sigma(n)}} $$where P(n) is the set of all permutations of the set \{1,\cdots,n\}. \text{sgn}(\sigma) is called the sign of the permutation and is positive of \sigma is obtained from the identity ... 6 The answer to your question is affirmative in the following sense: In the Riemann normal coordinates at p the coefficients of the Taylor expansion of the metric g_{ij}(x) are polynomials in the Riemann tensor at p and its covariant derivatives at p. [Assuming the proof in this random thing I googled[a] is correct, starting at (5.1)]. I think this ... 5 The composition law for quantum systems is always a tensor product. Your problem arises from a confusion over what the tensor product is applied to: you are trying to tensor product the spatial coordinates together, when it is in fact the basis vectors of the Hilbert space you should be tensoring together. More formally, take two quantum systems A and B, ... 5 This depends on what you mean by "pass from one manifold to another". In General Relativity one generally considers a single manifold \mathcal{M} and diffeomorphisms \phi: \mathcal{M} \rightarrow \mathcal{M}. I think the idea you are trying to get at is that if you consider a geometry on \mathcal{M}, that is a pair (\mathcal{M} , g) where g is a ... 5 Yes, it's just the second derivative of some function, it doesn't matter that this function is organized as a component of a tensor, h_{\mu\nu}. The identity above – assuming the function is differentiable and smooth etc. (add some "niceness" conditions on the function) – follows from the rules of calculus and is formally proven by the ... 5 Symmetry of the canonical energy-momentum tensor can be related to the spin of the object(s) that contribute to it (in other words, the representation of the Lorentz group under the fields transform). Note that the canonical EM tensor is obtained by using the Noether's procedure for translational symmetry$$ T_{\mu\nu} = \sum\limits_r \frac{\delta {\cal ...

4

We start with the definition $$\tag{1} S^{\alpha \beta}~:=~u^\alpha v^\beta-u^\beta v^\alpha.$$ Indices are raised and lowered with the metric. Up to an overall factor, one has $$\tag{2} \bar{S}_{\alpha \beta}~\propto~ \epsilon_{\alpha \beta \gamma \delta} S^{\gamma \delta},$$ so that the matrix trace $$\mathrm {Tr} (\mathbf{\bar{S}\cdot S }) ~=~ ... 4 If p is a point of the manifold then F at p is equal to ϕ∗F at ϕ(p), since they are related by the tensor transformation law, and tensors are independent of coordinate choice. This is roughly true. Initially, there is no meaning when one says that tensors at different tangent spaces are equal. However, the diffeomorphism induces an isomorphism between ... 4 Every term contains one \lambda in the superscript and one in the subscript, so you sum over those. The only indices which don't appear in both superscript and subscript in the same term are \mu and \nu. Example:$$\Gamma_{\lambda\sigma}^\lambda\Gamma_{\mu\nu}^\sigma = \Gamma_{00}^0\Gamma_{\mu\nu}^0 + \Gamma_{01}^0\Gamma_{\mu\nu}^1 + \cdots + ...

4

$\partial_t\equiv\frac\partial{\partial t}$ and $\partial^\mu\equiv g^{\mu\nu}\frac\partial{\partial x^\nu}=\left(\sum_{\nu=0}^3g^{\mu\nu}\frac\partial{\partial x^\nu}\right)_{\mu=0}^3$ are differential operators. $\partial^\mu$ is formally contravariant (upper index) and obeys the corresponding transformation laws. $\partial_t$ has a lower index and is (up ...

4

The antisymmetric second-rank tensor being referenced is the electromagnetic field tensor. It is defined as follows. Let $\varphi$ be the electrostatic potential (a scalar field), and let $\underline{A}$ be the magnetic potential (a 3-vector) from classical E&M. Concatenate them into a 4-vector $\vec{A}$. Now define the tensor of interest as the exterior ...

4

Note; I'll use the summation convention throughout here. In the context of differential geometry, the indices on tensorial objects are raised and lowered with the metric on the space (manifold) being studied. So for example $$T^i_{\phantom i j} = g^{ik}T_{kj}$$ and $$T^{ij} = g^{ik}g^{jl}T_{kl}$$ Notice that if the metric is simply that of ...

4

First off, please don't use units with $c\ne 1$ in GR. It makes everything horribly messy. What we normally think of as a ruler or clock measurement is represented in GR by an upper index quantity like $\Delta x^\mu$. Therefore in a Cartesian coordinate system in the fluid's rest frame, we are guaranteed that $u^\mu=(1,0,0,0)$, not $(-1,0,0,0)$. This is ...

4

A variation of a tensor is always a tensor and the formula for the value above doesn't show otherwise. What you probably find surprising is that $\delta g_{\mu\nu}$ and $\delta g^{\rho\sigma}$ are not related to each other by simply raising the indices $\mu,\nu$ or lowering the indices $\rho,\sigma$. Indeed, they're not related in this way. In this case, ...

4

Just use the Jacobian of the coordinate system transformation. If your Cartesian coordinates are $\mu$ and $\nu$ and your cylindrical coordinates are $\mu', \nu'$, then there is a Jacobian ${f_\mu}^{\mu'}$ that allows you to write $$F^{\mu' \nu'} = F^{\mu \nu} {f_\mu}^{\mu'} {f_\nu}^{\nu'}$$ where the Jacobian is given by $${f_\mu}^{\mu'} = ... 4 1. Let p\in M be a point in the manifold M. A tensor of type (r,s) at p is an element of the tensor product between r copies of the tangent space at p and s copies of the cotangent space at p. To evaluate the tensor, you plug in the vectors from the frame and the coframe. For instance, if the frame is (e_i), and (e^i) is its dual, the ... 4 This is paraphrasing Wald - General Relativity, section 2.4. Antisymmetrizing n indices means summing over all permutations of the indices, times the sign of each permutation. Since there are n! permutations, it's a sane convention to divide by n! (not all authors do this). For your example, there are 3! = 6 permutations of (abc). The even ones ... 4 The author is most likely considering the following scenario: The tensor components T_{ij} are coordinates in a vector space V consisting of vectors$$T~=~\sum_{ij}T_{ij}~ e^i \otimes e^j$$with basis elements e^i \otimes e^j\in V. [Here the word vector is used in the sense of linear algebra (as opposed to e.g. the sense of a (1,0) contravariant ... 4 Start with the following form of the Bianchi Identities$$ \nabla^\mu R_{\mu\nu} = \frac{1}{2} \nabla_\nu R $$Contract both sides with K^\nu. We find$$ \frac{1}{2} K^\nu \nabla_\nu R = K^\nu \nabla^\mu R_{\mu\nu} = \nabla^\mu \left( K^\nu R_{\mu\nu} \right) - R_{\mu\nu} \nabla^\mu K^\nu $$The second term vanishes due to symmetry of R_{\mu\nu}. Now, ... 4 Self containing the discussion in the realm of quantum mechanics, a nice explaination could be the following: The state of a quantum mechanical system is described by a ray on a complex (separable) hilbert space \mathcal{H}. With this I mean that you fix a Hilbert space \mathcal{H} and one state of your system corresponds to a point in ... 3 The doublets – I assume that you mean 2-dimensional representations of SU(2)\equiv Spin(3) – are spin-1/2 representation. Tensor products mean the addition of the angular momentum. The tensor product is 4-dimensional and under SU(2), it decomposes to a j=0 multiplet, a scalar or singlet, and a j=1 multiplet, a vector:$${\mathbf 2}\otimes {\mathbf 2} ...

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