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27

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 ...


20

Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor. Pressure is part of the stress tensor. The diagonal elements ...


15

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 ...


13

For each surface on a unit cube (see below), the stress on that surface can point in each of the three directions. (source) Since it is not necessarily the case that $\sigma_{11}=\sigma_{31}=\sigma_{21}$ (all pointing the in the same $\mathbf{e}_1$ direction)--or any of the other $\sigma_{ij}$ combinations, we need to have 9 components describing it, ...


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

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, ...


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

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

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 ...


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

Below follows a handful of excerpts from the book Introduction to the Classical Theory of Particles and Fields (2007) by B. Kosyakov. Controversial/misleading/wrong statements are marked in $\color{Red}{\rm red}$. We agree with OP that the statements marked in $\color{Red}{\rm red}$ are opposite standard terminology/conventions. Some (not all) correct ...


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 ...


9

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 ...


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 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

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 ...


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

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

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 ...


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{equation} \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 ...


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 ...


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

Let $X$ be a vector field on a semi-Riemannian manifold $(M,g)$, then in local coordinates, the vector fields $\partial_\mu = \partial/\partial x^\mu$ yield a basis for the tangent space of at every point on the manifold. This means that if $X\in TM$ is a vector field, then we can write it as a linear combination of the coordinate basis vector fields ...


6

OP is formally correct that $$\frac{\partial}{\partial x^{\mu}}~\in~ \Gamma(TM|_{U}) $$ is a vector field (defined in a local coordinate neighborhood $U$), and not a one-form. What Weinberg simply means by casually saying that The partial derivative operator $\partial/\partial x^\mu$ is a covariant vector, or in other words a 1-form,[...] is just ...



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