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2

Short and a little inaccurate answer: vector is one-dimensional tensor, matrix is a two-dimensional tensor. More details now: Tensors are multidimensional arrays which have certain properties. Not every multidimensional array is a tensor, check this discussion for more details. There are two types of one-dimensional tensors: vectors and co-vectors. Both ...


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OK, I got it. Menzel didn't spell it all the way out, but provides sufficient motivation to conclude that the dyadic dot-product is associative. Using the following, and assuming Einstein summation convention. $\mathfrak{A}=a^{ij}\hat{\mathfrak{e}}_{i}\hat{\mathfrak{e}}_{i}$ $\mathfrak{B}=b^{ij}\hat{\mathfrak{e}}_{i}\hat{\mathfrak{e}}_{j}$ ...


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A few hints and comments that may or may not amount to an answer: $b^1, b^2, b^3$ are not defined a priori. You are rather choosing them such that $\phi^{-1}$ is the inverse of $\phi$. So you are not assuming mutual orthogonality of the $b$s with the $B$, you are imposing it and showing that this is indeed necessary to obtain the inverse. At this point it ...


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Easy way Let me first state the straight-forward way to do this computation. $$ \langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d $$ First equality follows from compatibility, second equality uses ...


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From what I can see in your photo, it looks like you want to do this $$ h_{\nu}^{\alpha},_{\mu \alpha} $$ which is h_{\nu}^{\alpha},_{\mu \alpha} EDIT: I think @NowIGetToLearnWhatAHeadIs is correct, the comma is on the subscript level (with $\mu{}\alpha{}$). Additionally, separating the $\nu$ and the $\alpha$ is possible by using the $\{\}$ characters. ...


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That cannot be true, because $$ \left(\frac{1-\gamma_5}2\right)\left(\frac{1+\gamma_5}2\right)=0 $$ EDIT: The youtube video from the comments is actually wrong. The correct expression is $$ \bar\psi_L\gamma^\mu\psi_{R,\mu}=\bar\psi P_R\gamma^\mu P_R\psi_{,\mu} $$ From this, you should be able to show that this term vanishes.


4

The Einstein field equations $$ R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu} $$ for zero stress energy means that the Ricci Curvature $R_{\mu\nu}$ is proportional to the metric with $R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar $R~=~R_{\mu\nu}g^{\mu\nu}$ this is a spacetime of constant ...


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You can easily see this isn't the case by considering the special case of the stress-energy tensor equal to zero i.e. the vacuum solutions. These include the Minkowski metric, which is flat, but also the Schwarzschild and Kerr metrics and of course gravitational waves.


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As is written here the two remaining equations follow from the Bianchi identity which says that the anti-symmetrized derivative is zero, ie. $$ \partial_{[a} F_{bc]} = \partial_{a} F_{bc}+\partial_{b} F_{ca}+\partial_{c} F_{ab} = 0 $$ (remember the $F_{\mu\nu}$ is antisymmetric itself!)


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I believe $F^{\mu\nu}$ is better known as the Faraday tensor; Maxwell tensor is only the spatial part of it, 3x3 matrix. The argument goes like this: $F$ is a four-tensor, so if any quantity is constructed from it by multiplication of its components, this new quantity also has to transform as a tensor. The construct $$ ...


4

How does a scalar quantity transforms under a Lorentz transformation? I will show you here how this works for a vector (contravariant) and a covector. The transformation rule for these objects are $$u^a = { \partial x^a \over \partial x' ^b} u' ^b $$ $$u_a = { \partial x'^b \over \partial x^a}u' _b $$ Multiplying the two: $$ u^{a} u_{a} = { \partial ...


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Yes, it is not possible to construct a totally antisymmetric spin state with more than two electrons. This is just a statement of Pauli's exclusion principle.


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I think @Joce has the correct answer in his comment, but I'm going to expand their answer a little bit to provide some more detail. The stress power is specified per unit volume, but you have to specify which unit of volume. The stress power integrated over an entire body in the current configuration $\Omega_t$ is given by: \begin{align} \int_{\Omega_t} ...


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Write the right part of your last equation in vector form, the $T$ stands for transposition: $$\mathbf{y}= (\mathbf{{T^{-1}}})^T \cdot\mathbf{x}.$$ Take the transpose of this equation, ${(A\cdot B)}^T=B^T \cdot A^T$: $$\mathbf{y}^T= {{((\mathbf{{T^{-1}}})^T \cdot\mathbf{x})}}^T={\mathbf{x}^T}\cdot\mathbf{{\mathbf{T^{-1}}} }.$$ While the result is one ...


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I'll use a notation where I write vectors as $\vec{v}$ (because that's how \vec renders and I have no idea how to turn it into boldface in MathJax) and use primes to indicate the different basis: $v'^i$ is the components of $\vec{v}$ in the primed basis (some people use primes on the indices, which I find confusing). So given some basis ...


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You use the metric of your coordinate system; if that's Cartesian, then it's a Kronecker delta. Since permitivity is measured in Farads per meter- emphasis "per meter"-- it should transform accordingly.


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By convention, vectors are written as column vectors, whereas dual vectors are written as row vectors. This means that in principle, upper indices should index columns and lower indices should index rows. However, in practice, we normally translate rank-2 tensors to matrices by order of the indices, the first one indexing rows, the second one indexing ...


3

$$\Gamma^\mu{}_{\alpha\beta} y^\alpha y^\beta$$ is defined to mean $$\sum_{\alpha = 0}^3 \sum_{\beta = 0}^3 \Gamma^\mu{}_{\alpha\beta} y^\alpha y^\beta $$ that is, each repeated index is summed over independently.


1

$$ \nabla_\mu A^\nu = g^{\nu\alpha} \nabla_\mu A_\alpha = g^{\nu\alpha} ( \partial_\mu A_\alpha - \Gamma^\lambda_{\mu\alpha} A_\lambda ) = g^{\nu\alpha} \partial_\mu ( g_{\alpha\beta} A^\beta ) - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} A^\beta $$ This simplifies to $$ \nabla_\mu A^\nu = \partial_\mu A^\nu + ( g^{\nu\alpha} \partial_\mu ...


1

Explicitly working with the components of $g$ and $\Gamma$ can be messy. Here is a nicer way to derive the coordinate expression for the covariant derivative of a vector (or any tensor) without raising or lowering indices. Consider vector fields $A^\mu, B_\mu$. The covariant derivative of a scalar is the same as the coordinate derivative (in any coordinate ...


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I am reading Schutz on the subject of vectors, and I believe that he says that; given a manifold with two sets of coordinate systems that are related through a linear transformation; a pair of sets of numbers (with the same dimension as the manifold) is a vector if one set of numbers is related to the other set of numbers through the linear transformation.


2

Apologies for evincing magisterial cluelessness about what your diagrams represent and what you'd want to achieve, but I'd array the standard facts on tetraquarks avoiding Young diagrams, although they are self evident in the Dynkin labelling, which I also give, next to the tensor labelling. They may be useful to what you appear to be after--but I can't ...


4

Your problem is that the thing you want to express is only true in the particular basis you're working in. But equations in the summation convention are true in every basis. Your expression involving an explicit summation is fine and is probably what any reasonable mathematician would do. But if you want to show unreasonable devotion to the summation ...


2

The notation $(\vec S_1\otimes 1) \cdot (1\otimes \vec S_2)$ is interpreted as a contraction over spatial indices. Hence, $\vec S_1\cdot \vec S_2=\sum_{ij} \delta_{ij}S_{1i} \otimes S_{2j}$. By contrast, $(S_{1x}+S_{1y}+S_{1z})\otimes(S_{2x}+S_{2y}+S_{2z})=(\vec S_1\cdot [1,1,1])\otimes (\vec S_2\cdot [1,1,1])$.


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How do we formally define vectors in physics? An excerpt from chapter one, page 12 of "Mathematics of Classical and Quantum Physics" Originally, we introduced a vector as an ordered triple of numbers. The rule for expressing the components of a vector in one coordinate system in terms of its components in another system tells us that if we ...


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A physical quantity is a vector if it transforms in the same way as a position vector when the coordinate system undergoes a transformation.


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Well, you might have spared yourself confusion and grief by checking your peculiar language in SO(3), which any undergraduate is familiar with. Let me illustrate this for SO(3), before moving on to the much messier SU(3). For SO(3), Kronecker-composing two vectors (spin 1, so 3 s) yields a spin 2 quintet (call it φ, so 5), a triplet (π) and a singlet (s), ...


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Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...


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There are no easy to verify criteria which will work in general. One way to see it is by noting that for every classical StatMech model, we can define a PEPS with the same correlation functions (for which the tensors can be easily constructed from the StatMech model), see http://arxiv.org/abs/quant-ph/0601075. On the other hand, for StatMech models it is ...


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As a mathematician, I would like to point out the following technicality: What we (can) represent as a matrix are tensors with one index up and one index down, that is elements of $V\otimes V^*$ (where $V$ is a vector space, or a vector bundle over a manifold). A tensor with two indexes up is a bilinear form on $V$, and one with two indices down a bilinear ...


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Slogan: Matrices are a tool to compute sums; tensors tell you which sums make sense. When you convert between rank-2 tensors and matrices, the decision as to which index of the tensor labels the rows and which one labels the columns is purely conventional. Matrix multiplication is no more than a convenient way to write products of the form $$K(i,k) = ...


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As @AccidentalFourierTransform said, it's a mistake to try and think of tensors as matrices, as you rapidly end up going mad when you need to compute $A^{\alpha\beta\gamma\delta}B_{\beta\delta}$ or something. That said $A^{\alpha\beta}B_{\alpha\gamma}$ 'looks like' $A^TB$. The right way to think of tensors is as multilinear functions, because that is what ...


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Yes, their outer product is defined as you said. Further, the product of operators is given by $$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$ Therefore, $$ [A \otimes B, F \otimes 1_B] = [A,F] \otimes [B,I_B] = [A,F]\otimes {\bf 0} = 0 $$


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The coordinate invariant volume element/measure on a manifold with metric $g$ is $$ d^d x \sqrt{|g|} $$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$ d^d x' ...


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Take for example $q$ a vector field: $$ q^a=A^\mu $$ where $a=\mu$ is a vector index. The conjugate momentum is $$ \frac{\partial\mathcal L}{\partial A^\mu} $$ and, as it is an upper index in the denominator, it makes sense to write it as $\pi_\mu$. Also, you can use the definition of vectors and covectors to prove that $\pi_\mu$ transforms as a covector, ...



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