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## Hot answers tagged tensor-calculus

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The $4$-gradient is a $4$- vector. Formally, when $x^\mu\to x'^\mu=\Lambda^\mu{}_\nu x^\nu$ \begin{align*} \partial'_\mu &=\frac{\partial}{\partial x'^\mu}\\ &=\frac{\partial}{\partial (\Lambda^\mu {}_\nu x^\nu)}\\ \end{align*} $\therefore$ \begin{align*} \Lambda^\mu {}_\nu\partial'_\mu&=\partial_\nu\\ \end{align*} which makes $\... 4 In arbitrary spacetime dimensions, the Einstein tensor is $$G_{\mu\nu}~:=~R_{\mu\nu}(\Gamma)-\frac{1}{2}g_{\mu\nu}g^{\rho\sigma}R_{\rho\sigma}(\Gamma).\tag{1}$$ Let us also define the Einstein tensor-density $${\cal G}_{\mu\nu}~:=~\sqrt{|g|}G_{\mu\nu}. \tag{2}$$ It satisfies a functional Maxwell relation $$\frac{\delta {\cal G}_{\mu\nu}(x)}{\delta g^{\rho\... 3 The idea of generalizing laws to curved space-time is to notice that we actually live in a curved space-time ourselves. What we know as "flat space-time equations" are, in fact, equations in curved space-time derived/discovered in our local (almost-)inertial frame. We can then derive their curvilinear form by simply transforming to a general frame. This is ... 3 Consider the following expression :$$\epsilon_{i_1i_2...i_n}A^{i_1}_{1}A^{i_2}_{2}...A^{i_n}_{n}$$Now THIS is exactly the definition of the determinant of A. Why ? Well the determinant is defined as$$detA=\sum_{\sigma\in S_n} \epsilon(\sigma)A^{\sigma(1)}_{1}A^{\sigma(2)}_{2}...A^{\sigma(n)}_{n}$$Which, if you look hard enough, is exactly the same as ... 3 Vary$$S=\int \sqrt{\det g} g^{\alpha\beta} X_{\alpha\beta}$$with respect to g^{\alpha\beta} to get the index-down stress-energy tensor (assuming X does not depend on g)$$ \frac{\delta S}{\delta g^{\alpha\beta}}= \sqrt{\det g} \left(X_{\alpha\beta}-\frac{1}{2} g_{\alpha\beta} X_{\gamma\delta}g^{\gamma\delta}\right)$$so stress energy tensors for non-... 3 It's a standard result for bilinear maps E that$$E(u,u)=0\;\forall u\iff E(u,v) = -E(v,u)\;\forall u,v$$The part that's slightly less obvious is going from left to right, though it's still just a one-liner:$$ 0 = E(u+v,u+v) = E(u,u) + E(u,v) + E(v,u) + E(v,v) = E(u,v) + E(v,u) $$2 Why not simply write out how a vector transforms, how tensor transforms and do it by linear algebra. Let S:\{x^\alpha\}^{\alpha=1\dots N} be one set of coordinates and \tilde{S}:\{x^\alpha\}^{\alpha=1\dots N} be another set. If V_\alpha is a co-vector in S whilst \tilde{V}_\alpha is a co-vector in \tilde{S}, then, by definition: V_\alpha = \... 2 The "gradient" you wrote is not a four vector (and that's not what should be called a gradient). What you have written is a four co-vector (i.e. a dual vector, which is a linear functional over the space of four-vectors). What you have written is technically the differential of the field \phi which is an exact on-form:$$d\phi = \frac{\partial\phi}{\... 2 Per se, this doesn't have anything to do with relativity or curvature: The factor of$\sqrt{|g|}$comes in for a similar reason that has the determinant of the Jacobian pops up in the substitution formula for integration of multiple variables: When integrating, you need to account for the volume of the unit cell spanned by your coordinate frame. So if you're ... 2 Let's consider a simple explicit example of how contracting two upper Lorentz indices or two lower indices doesn't produce a Lorentz invariant. A particle has energy and momentum. The particle can be observed in two different inertial frames which might be moving relative to one another. In one frame the energy is$E$and the momentum is$\mathbf{p}=(p^x,p^...

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Why is this operation invalid? Or is it not invalid, it just doesn't return a meaningful object? The point is that a scalar isn't just any real number, it's a real number which transforms between frames appropriately. (Specifically, the correct transformation is that it shouldn't transform at all.) For example, "the energy of a ball" isn't a scalar, even ...

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Under coordinate transformations, vectors transform with the Jacobi matrix, whereas covectors transform with its inverse. If you contract an upper and a lower index, these operations cancel, and the result will be invariant under coordinate transformations. In contrast, summing over two indices in the same position generally won't give you an invariant. ...

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In tensor notation: $$v_{(ijk)}w_l = x_{(ijkl)} + \varepsilon_{ml(k} y_{ij)}^m + \underbrace{\varepsilon_{l(jk}z_{i)}}_{0}$$ $$y^i_{(ij)} = 0$$ Or: $$\mathbf{10 \otimes 3 = 15 \ \oplus \ 15^{\prime}}$$ Or by using Young tables:

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I don't know how familiar you are with differential geometry, so I'll try to give a self contained answer. First, as you might know, the metric tensor $g$ is a bilinear map taking two (tangent) vectors, e.g. velocity vectors. This object can be defined without choosing coordinates of our space(time). Now take coordinate functions $x^0,...,x^d$, then we can ...

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If one sacrifices mathematical rigour, then one can get away with not using the variational bicomplex or jet bundles. Assuming functional derivatives are defined with respect to variations with compact support, thus one can throw away all boundary terms, we have $$\frac{\delta}{\delta \phi^a(x)}\frac{\delta}{\delta \phi^b(y)}=\frac{\delta}{\delta \phi^b(y)}... 1 The field-theoretic version of (closed & exact) differential forms uses the variational bicomplex and jet bundles, see e.g. Refs. 1-3. References: P.J. Olver, Applications of Lie Groups to Differential Equations, 1993. I. Anderson, Introduction to variational bicomplex, Contemp. Math. 132 (1992) 51. G. Barnich, F. Brandt & M. Henneaux, Local BRST ... 1 The four gradient is a four vector but it transforms covariantly, rather than contravariantly. This makes it a "covector". It also has a contravariant form, obtained by multiplying it by the metric, which transforms like 4 four position or four momentum. 1 The buoyancy from the Archimedes principle results from the gradient of pressure on the liquid. Maybe you are thinking about effects resulting from gradient of scalar fields. Another example is heat flow, that results from the gradient of temperature in a wall. Or classical gravity acceleration, that comes from the gradient of gravity potential. In order ... 1 Suppose that to start with, the generators of the Lorentz group are defined in terms of raised indices and denoted by J^{\mu\nu}. Next, it is possible to define J_{\mu\nu} i.e. generators with lowered indices as$$J_{\mu\nu}=\eta_{\mu\sigma}\eta_{\nu\rho}J^{\sigma\rho}.$$As a passing remark, we note that since J^{\mu\nu} is antisymmetric, J_{\mu\nu} ... 1 Let$$V=V_{\mu} dx^{\mu}$$be the co-vector, and let$$T=T^{\mu\nu} \frac{\partial}{\partial x^{\mu}} \otimes\frac{\partial}{\partial x^{\nu}}$$be the tensor. And let the tensor product of T and V be$$T^{'}=T^{\mu\nu}V_{\nu} \frac{\partial}{\partial x^{\mu}} \otimes\frac{\partial}{\partial x^{\nu}}\otimes dx^{\nu}$$. Contraction (see "https:/... 1 I'm not sure I fully understand your question, but you're right that the subspace of transversely polarized states (aka the physical subspace) is the part of the bigger space which is annihilated by a_0 and a_3. Why a_3? It is just a convention to have the third polarization vector pointing in the same direction as the 3-momentum \vec{p} of the ... 1 Here is a more intuitive way to look at the electromagnetic field tensor: The electromagnetic field tensor is correlated to the Lorentz force law$$\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = e (\mathbf{E} + \mathbf{v} \times \mathbf{B}). To problem with this equation is that it is not a "geometric equation", i.e. neither the righthand nor the lefthand side ...

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Let $F$ and $G$ be a finite sets like for instance $F=[m]:=\{1,2,\ldots,m\}$ and $G=[n]$. Let $V$ be the vector space of all functions $f:F\rightarrow \mathbb{R}, x\mapsto f(x)$. Yes I used "$x$" for an element of $\{1,2,\ldots,m\}$, instead of $i,j,\ldots$ So what. Likewise, let $W$ be the vector space $g:G\rightarrow \mathbb{R}, y\mapsto g(y)$. If all we ...

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