# Tag Info

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For question iii) you can see my question that should provide the answer: Coordinate-free proof of the constancy of the scalar product of a Killing vector with the geodesic tangent vector along the geodesic For iv) the essential ingredients are of course the definition equation of the killing vectors, the definition of the curvature tensor and the first ...

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You've intersected with the standard approach with which engineering students learn about stress and strain in college. The framework is covered in Beer & Johnston's Mechanics of Materials, for one, among several other standard textbooks on the so-called mechanics of deformable materials. Since you're well accustomed to self-education, I'll just provide ...

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I think you know you have a hard job ahead of you. Wishing you success. In order to know the things that physicists need in this area, I would recommend you first see these two packages: xAct (in Wolfram Mathematica) and GRtensor (in Maple). These are really great. Here I summarize some of the most important things (with appropriate links) that come to my ...

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Metric $ds^2$ in Cartesian/Spherical/... coordinates Inverse of the metric Angle between $d^{(1)}x^{\alpha}$ and $d^{(2)}x^{\alpha}$ Christoffel symbols Geodesic equations Geodesic equations in Newtonian limit Components of generalized momentum Riemann tensor Ricci tensor Traceless Ricci tensor Ricci scalar Einstein tensor Weyl tensor Some of the identities (...

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This answer has been directly copied from my answer to Finding the irreducible components of a rank 3 tensor on the Mathematics Stack Exchange site. Using the bold-number notation to denote the representation of dimension $2\ell+1$ (with total angular momentum number $\ell$), the decomposition you give of rank-2 tensors is equivalent to the decomposition $$... 0 Let's start with this expression, -\ast \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m} dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m}. Now, $$\ast dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m}=\sqrt{|g|}\epsilon^{\nu\nu_2\cdots\nu_m}.$$ ... 1 If you have the values a list in Python, then you can simply use Total_list = [38.67, 38.72, ...] Total_var = np.var(Total_list) which is equal to Total_var = np.dot(Total_list,Total_list)/len(Total_list) - (np.sum(Total_list)/len(Total_list))**2 The first term is$$ \langle A^2\rangle = \frac{1}{N}\sum_{i=1}^N A^2_i $$and the negative term is$$ \langle ...

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It might be simple to get the minus sign difference between first index raised and second index raised curvature tensor from components definition. For first index raised curvature, $R^{\kappa}{}_{\lambda\mu\nu}$ is defined as $$\langle dx^\kappa, R(e_\mu,e_\nu)e_\lambda \rangle = \langle dx^\kappa, \nabla_\mu\nabla_\nu e_\lambda-\nabla_\nu\nabla_\mu e_\... 0 The strain tensor does not have to be positive definite, as mentioned before. On the other hand, you are asking what type of strain tensor are positive definite. A positive tensor would have positive principal strains. That implies a tensor where you have elongation in 3 different directions. 2 In differential geometry, checking that (1) does not depend on the coordinate chart chosen is enough to show that it defines a tensor \rm{d}X which is independent of the chart used. This is the "right" definition for the exterior derivative. However, it will not keep the same form in a frame which is not a coordinate basis (which is to be ... 1 As you said, the contraction is a scalar so that$$ \begin{align} \nabla_{\gamma} (g^{\mu \nu} A_{\mu \nu}) &= \partial_{\gamma}(g^{\mu \nu}A_{\mu \nu}) \\ &= A_{\mu \nu} \, \partial_{\gamma}g^{\mu \nu} + g^{\mu \nu} \partial_{\gamma} A_{\mu \nu} \ \end{align} $$and the same thing for \nabla_{\gamma} (g_{\mu \nu} A^{\mu \nu}). Your equations are ... 3 What is true is that$$\oint P \, dV = - \oint V \, dP$$because of the area argument you made, noting that the signs are opposite. From this, you can't remove the integral sign to conclude that P \, dV = -V \, dP, which is certainly not true. The integral result is easy to show in general, as$$\oint P \, dV + \oint V \, dP = \oint d (PV)$$and the net ... 1 I don't actually understand what it means to "draw a geometrical picture of the four-dimensional curved space-time to write an appropriate Christoffel symbol formula." The answer to the title question, Is it incorrect to assume the Christoffel symbol symmetry (with respect to the lower indices) for a curved space-time? is no. The Christoffel ... 0 Maybe the best way is to think about \vec{B} in terms of the Biot-Savart law. Imagine a loop carrying a current I in a plane that is perpendicular to a mirror. The Biot-Savart law says that the B-field at position \vec{r} is given by$$\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\, \oint \frac{I\, d\vec{l} \times \vec{r'}}{|\vec{r'}|^2}\ dl, $$where \vec{r'}... 0 Bundle the indices into pairs$$ 11\to 1\\ 12\to 2\\ 13\to 3\\ 21\to 4\\ 22\to 5\\ \vdots $$and then your rank four object becomes a rank-two matrix on a larger space. This will be 9 dimensional in the case i or j= 1,2,3 I was writing above. Such packaging is what one does for linear operators on tensor products of representations when one is ... 0 I just heard from the author of the textbook who confirms that it is indeed a typo. So equations (7) and (8) in my question are the correct ones. 3 Hint:$$ \epsilon^{\rm symbol}_{a_1\ldots a_d} e^{a_1}_{\mu_1}\ldots e^{a_d}_{\mu_d} ~=~ \det(e^a_{\mu})\epsilon^{\rm symbol}_{\mu_1\ldots \mu_d} ~=~ \sqrt{|\det(g_{\mu\nu})|}\epsilon^{\rm symbol}_{\mu_1\ldots \mu_d} ~=~ \epsilon^{\rm tensor}_{\mu_1\ldots \mu_d}. $$2 Observe that the definition of the determinant and its change of sign under interchanging rows or columns tells us that$$ \epsilon^{a_1\ldots a_d} e^{\mu_1}_{a_1}\ldots e^{\mu_d}_{a_d}= \epsilon^{\mu_1\ldots\mu_d} {\rm det}[e^\mu_a], $$and remember that there is a connection between the e^\mu_a and the metric. 2 The situation that I am familiar with is when you combine the energy momentum tensor "covariant conservation"$$ 0=\nabla_\mu T^{\mu\nu} $$equation with a Killing vector \xi obeying \nabla_\mu\xi_\nu+\nabla_\nu \xi_\mu=0 to get$$ 0=\nabla_\mu (T^{\mu\nu}\xi_\nu)=0. $$Then T^{\mu\nu}\xi_\nu is a vector, so$$ 0=\nabla_\mu (T^{\mu\nu}\...

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However I am quite confused by this because all the examples of tensor products I have seen involve matrices and not integers. My brain is also interpreting that as 0*1=1, which is obviously not the case. You are misreading your Kronecker products. A Kronecker product of two operators (here, isospin matrices) is a composition of two matrices of different ...

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I think its more common to use the dimension of the representation ($2I+1$), rather then the quantum number, $I$. With that, the isoscalar is one-dimensional (trivial representation): $${\bf 1}$$ and the isovector is 3 dimensional (adjoint representation): $${\bf 3}$$ so that: $${\bf 1}\otimes{\bf 3}={\bf 3}$$ A more active example is 2D spinor-rep (up ...

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In the reference that you provided, you have, $$\frac{i}{2}\omega_{\mu\nu}\xi_{\alpha}\left[J^{\mu\nu},P^{\alpha}\right]=\omega_{\mu\nu}\xi_{\alpha}\eta^{\nu\alpha}P^{\mu}$$ where $\xi_{\alpha}$ are arbitrary numbers, $\omega_{\mu\nu}$ are antisymmetric, but otherwise arbitrary and numeric, and $J^{\mu\nu}$ are antisymmetric in $\mu$ and $\nu$, never mind ...

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I think the material you found is just wrong. A trivial counter example is zero displacement = zero strain, which is not positive definite. Nor do I see where such a property would be particularly useful when calculating anything. It's a different story when talking about the other tensors we deal with (such as the elasticity tensor), where it definitely is ...

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