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## Hot answers tagged tensors

12

I can easily construct an example of smooth tensor field over a manifold whose "rank" changes depending on the point. My idea relies upon the following elementary proposition. I stress that the notion of "rank" used here is that introduced within the original question and not the standard one. Proposition. Consider a $n$-dimensional real vector space $V$ ...

11

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 ...

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Applying a force in the $x$-direction might change the shape of the material in the $y$-direction. The only way to capture such an effect is through a tensor. If you have a general force acting on your body $$\vec F = (F_x, F_y, F_z)^T$$ and you are interested in the reaction of the body by looking at its deformation $$\vec \epsilon = (\epsilon_x, ... 10 In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources. But this is rather illusory. In relativity, the equations look quite different:$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= ...

10

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 ...

10

Yes. The indices on gamma matrices can be treated like four-vector indices. In particular, indices on gamma matrices are commonly raised and lowered with the Minkowski metric $\eta_{\mu\nu}$ as you indicate; \begin{align} \gamma_\mu = \eta_{\mu\nu}\gamma^\nu. \end{align} Now, as user26143 writes in his comment above, the gamma matrices have the ...

9

In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ...

9

I) Let us for simplicity discuss tensors in the context of (finite-dimensional) vector spaces and multilinear algebra. [There is a straightforward generalization to manifolds and differential geometry.] II) Abstractly in coordinate-free notation, the Kronecker delta tensor, or tensor contraction, is the natural pairing $$\tag{1} V \otimes ... 8 The question seems to conflate many different things: the invariance of a mathematical quantity (usually a scalar such as ds^2 for the separation of two events in special relativity) covariance of tensors (the values of components of tensors may be calculated from those in another frame but they're not the same thing) universality of equations in ... 8 Are those square brackets standard notation in Physics? Yes. See, for example Sean Carroll notes. At least I can tell you from two other classic references using that notation, "General Relativity" by Wald (1984) and "A First Introducion to General Relativity" by Schutz (2009 for the most recent edition)   If I am in a non-curved \mathbb M ... 8 The antisymmetric part is defined as$$ 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)}} $$where P(n) is the set of all permutations of the set \{1,\cdots,n\}. \text{sgn}(\sigma) is called the sign of the permutation and is positive of \sigma is obtained from the identity ... 8 On any manifold we can define the differential df of a scalar f. The differential is a 1-form: something that eats vectors and spits out scalars, or even less formally, something with one down index. We have the following formula for the differential,$$df = \frac{\partial f}{\partial x^i} dx^i$$(sum over i implied). You can write it in index notation ... 7 The tensor equations you mention are not invariant, they are covariant. Big difference. Both are differential equations, which transform linearly under nonlinear transformations from one manifold to another because they are differential equations at a point. The nonlinear transformation from one manifold to another induces a linear transformation of the ... 7 This is true - in fact you could define \nabla^\sigma = g^{\sigma\rho} \nabla_\rho. I assume this meant to say$$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch g^{\sigma\rho} with \nabla_\nu, which you ... 7 Note that:$$ h^{\mu \nu} = \eta^{\mu \rho}\eta^{\nu \lambda} h_{\rho \lambda} Therefore, up to first order, we have: \begin{aligned} g^{\mu \nu}g_{\nu \sigma} & = (\eta^{\mu \nu} - h^{\mu \nu})(\eta_{\nu \sigma} + h_{\nu \sigma}) \\& =\eta^{\mu \nu}\eta_{\nu \sigma} + \eta^{\mu \nu}h_{\nu \sigma} - \eta_{\nu \sigma} h^{\mu \nu} + ... 7 I don't think the author should use the tensor product \otimes in\vec{S}^{(A)} \otimes \vec{S}^{(B)} = \frac{\hbar^2}{4}(\sigma_x \otimes \sigma_x + \sigma_y \otimes\sigma_y + \sigma_z \otimes \sigma_z)$$because he really doesn't mean tensor product. Rather, \vec S is a vector operator, that is, its components transform like the components of a ... 6 The composition law for quantum systems is always a tensor product. Your problem arises from a confusion over what the tensor product is applied to: you are trying to tensor product the spatial coordinates together, when it is in fact the basis vectors of the Hilbert space you should be tensoring together. More formally, take two quantum systems A and B, ... 6 The answer to your question is affirmative in the following sense: In the Riemann normal coordinates at p the coefficients of the Taylor expansion of the metric g_{ij}(x) are polynomials in the Riemann tensor at p and its covariant derivatives at p. [Assuming the proof in this random thing I googled[a] is correct, starting at (5.1)]. I think this ... 6 It is a quite famous theorem due to Cauchy. Consider an internal portion S of a continuous body C. There are two kinds of forces acting on it: Forces proportional to the mass, of the form$$\int_V \mu(x) \vec{f}(x) d^3x\tag{0}$$where \vec{f}(x) is the density of force acting on x \in V. And forces acting through the surface \partial V, the ... 6 See Edit below, the original answer is not completely correct. There is no gauge freedom in F. F is gauge invariant. In fact, F is completely measurable. It's components are the Electric and Magnetic fields, so you just go out with a set of test charges and measure E and B and you've got F. One hint that T and F do not contain the same ... 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 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 ... 5 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, ... 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 ...

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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 ...

5

What you said is only true if the hypersurface is space-like or time-like. If a non-null hypersurface is defined by $f(x) =$ constant, then the normal to the hypersurface is given by $$n_\alpha \propto \partial_\alpha f$$ The fact that the hypersurface is non-null implies $$g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0$$ ...

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There are both physical and formal reasons to introduce the spin connection. Physically, we know that there are spin 1/2 particles. A spin 1/2 field cannot be described by anything built from 4-vector fields. You can realize this for example by that 4-vector fields (and so anything built from them) returns to their original value after a $2\pi$ rotation ...

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