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## Hot answers tagged tensors

26

A (rank 2 contravariant) tensor is a vector of vectors. If you have a vector, it's 3 numbers which point in a certain direction. What that means is that they rotate into each other when you do a rotation of coordinates. So that the 3 vector components $V^i$ transform into $$V'^i = A^i_j V^j$$ under a linear transformation of coordinates. A tensor is a ...

21

Color charge is the representation of the SU(3) gauge group. The representation theory of SU(3) is described below: The basic representation is called the "3" or the fundamental, or defining, representation. It is a triplet of complex numbers $V^i$, which transform under a 3 by 3 SU(3) matrix by getting multiplied by the matrix. The value of "i" is ...

14

Strictly speaking, a matrix is not a tensor, it is a representation of a tensor in a particular basis. You can't tell whether a given matrix is a tensor using only its components. You would have to know how it transforms to different reference frames. For the electromagnetic field tensor, for example, you could write the equations for some physical ...

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

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 ...

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 ...

11

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 &= ... 11 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

The answer to your question depends quite a bit on what you consider to be fundamental and what you consider to be derived. A modern, manifestly relativistic treatment of E&M would define the electromagnetic field tensor as $$F^a = q\mathcal{F}^a_bv^b,$$ where $\mathcal{F}$ is the field tensor and $F$ is the four-force acting on a particle. By this ...

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 $$\tag{1} V \otimes ... 9 In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ... 8 On any manifold we can define the differential df of a scalar f. The differential is a 1-form: something that eats vectors and spits out scalars, or even less formally, something with one down index. We have the following formula for the differential,$$df = \frac{\partial f}{\partial x^i} dx^i$$(sum over i implied). You can write it in index notation ... 8 Would we have to prove that the electromagnetic field tensor transforms as a tensor by brute calculation? Not if you accept that the derivative of a tensor is a tensor with one higher covariant rank. In SR, the scalar electric and vector magnetic potential are components of a four-vector (a rank 1 tensor), the electromagnetic four-potential ... 8 The question seems to conflate many different things: the invariance of a mathematical quantity (usually a scalar such as ds^2 for the separation of two events in special relativity) covariance of tensors (the values of components of tensors may be calculated from those in another frame but they're not the same thing) universality of equations in ... 8 Are those square brackets standard notation in Physics? Yes. See, for example Sean Carroll notes. At least I can tell you from two other classic references using that notation, "General Relativity" by Wald (1984) and "A First Introducion to General Relativity" by Schutz (2009 for the most recent edition)   If I am in a non-curved \mathbb M ... 8 The antisymmetric part is defined as$$ 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)}} $$where P(n) is the set of all permutations of the set \{1,\cdots,n\}. \text{sgn}(\sigma) is called the sign of the permutation and is positive of \sigma is obtained from the identity ... 7 This is true - in fact you could define \nabla^\sigma = g^{\sigma\rho} \nabla_\rho. I assume this meant to say$$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch g^{\sigma\rho} with \nabla_\nu, which you ... 7 The tensor equations you mention are not invariant, they are covariant. Big difference. Both are differential equations, which transform linearly under nonlinear transformations from one manifold to another because they are differential equations at a point. The nonlinear transformation from one manifold to another induces a linear transformation of the ... 7 Yes, your confusion is wholly caused by you thinking classically ;) In a hand-wavy way, particles are certain localized excitations of the quantized fields. The QFT picture contains the particle picture in the perturbative approach known as Feynman diagrams (and, relatedly, the LSZ formalism). There, we are given the action of our theory dependent on some ... 7 Since torque is defined as the rate of change of angular momentum, the more fundamental question would be whether angular momentum is a vector in SR. The answer is no, because there is no vector cross product in four dimensions. Angular momentum is a rank-2 tensor. 7 Note that:$$ 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)$$because he really doesn't mean tensor product. Rather, \vec S is a vector operator, that is, its components transform like the components of a ... 6 At the risk of telling you how to "suck eggs" (your level in these things is not altogether clear), here goes. Ingredients: The essential ingredients to this explanation are: A physical "system" which evolves in and whose "events" happen in some space \mathcal{U} (ordinary Euclidean 3-space or Minkowsky spacetime, for example); in physics this space is ... 6 It is a quite famous theorem due to Cauchy. Consider an internal portion S of a continuous body C. There are two kinds of forces acting on it: Forces proportional to the mass, of the form$$\int_V \mu(x) \vec{f}(x) d^3x\tag{0}$$where \vec{f}(x) is the density of force acting on x \in V. And forces acting through the surface \partial V, the ... 6 I think that it is only necessary to use the cyclic identity. Contracting both sides with the Levi-Civita, we should have$$0 = (R_{abcd} + R_{adbc} + R_{acdb}) \varepsilon^{abcd} \tag{1}. Let $S = R_{abcd}\varepsilon^{abcd}$. Then $R_{adbc}\varepsilon^{abcd} = -R_{adbc}\varepsilon^{acbd} = R_{adbc}\varepsilon^{adbc} = S$ where the last step is renaming ...

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 ...

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

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

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 ...

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