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4

Potential energy has absolutely nothing to do with stress-energy or pressure. The following reference is a good source about the origin of the pressure term in the stress-energy tensor: "Momentum due to pressure: A simple model" by Kannan Jagannathan in American Journal of Physics 77, 432 (2009);  http://dx.doi.org/10.1119/1.3081105 Potential energy ...


1

I can understand your confusion. Let me start by saying that this is the "correct" way of writing down what you want to have in physics notation: $$ \langle e_i|\otimes \mathbf{1} \sum_j |e_j\rangle\otimes |w_j\rangle = \sum_j \delta_{ij} |w_j\rangle $$ Note the absence of the tensor product sign. Otherwise, you will never have this reduction of dimension, ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\o}{\mathbf 1}$ Physicists are lazy people we all are! When you see something like $S_{1z}+S_{2z}$ you should really think of the following: $$S_{1z}+S_{2z} \equiv S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z}$$ Since you get tired of writing it over and over you just shorten it by an addition ...


1

Assuming that the expressions given for a 2D electron gas here are valid, there really is no contradiction. Remember that there is an implicit sum in a repeated index (Einstein summation convention). Equivalently, an inverse matrix need not be the component-by-component inverse of the original matrix. For example, we get: $$[\sigma\cdot\rho]_{xx} = ...


2

how on earth.....? A possible way to look at it goes like this, let's consider a litle cube then the stress force in the ith direction acting on jth surface element is $-\sigma_{ji}(x_i,x_j,x_k)*dA$ where $dA=l^2$, the force in the ith derection acting on the other surface element paralell to the first is $\sigma_{ji}(x_i+dx_i,x_j,x_k)*dA$, so the total ...


0

In modern mathematical terminology, a functor is called covariant when it preserves the direction of the morphisms, contravariant if it reverses it. For a given differentiable map between manifolds (of which a special case would be open sets within the same manifold), the derivative is a map between the associated tangent bundles. This defines a covariant ...


2

A full introduction to the DMRG algorithm definitely does not fit here, and you can find many well-written introductory materials online. DMRG has been applied to simulate perturbed toric code model, which is the simplest example of string-net model, see http://arxiv.org/pdf/1205.4289.pdf. Generally speaking, DMRG for 2D spin models is indeed much more ...


0

Yes, for sure. Indeed for 1 rank tensor we have that $A^{i} = \frac{\partial x^i}{\partial x^{\prime j} } A^{\prime j} $. Even looking at this formula we can say that there is no such constrains on coordinates system, as you assumed. Problem occur when you have non-differentiable functions (derivative does not exist ...)


0

... Another way to look at it is to use geometry which includes the vertical time axis, the horizontal space axis, and the stacking of both motion vectors( c, v ) and length scalars. Rotation determines the direction of travel in space-time, thus determines the velocity across space and the rate of the ticking of time. Here, twin spaceship A is a rest in ...


2

You probably already understand that merely rotating a coordinate system does not change the physical system you're modeling. Given one set of axes, any rotation of those axes just represents some freedom of choice to do math how you see fit. So with that in mind, if you were pointing northeast (and chose a pair of coordinate axes to point northeast and ...


2

Bernhard Schutz discusses this reasonably well in his book A First Course in General Relativity. Consider sending a light beam horizontally along an $x$ axis and then receiving it back again. A space time plot of this would look like Here's an example of the rotation you are describing And this is how everything becomes distorted when you create such ...


2

Adding to Lubos' answer, let me address this part more specificially: I am very confused about this - actually - I'm guessing that transformations that can be written $A⊗B$ are a subset of all possible transformations with all 16 coefficients free. This is correct. First some notation: Let $H_1$ and $H_2$ be two (finite-dimensional) Hilbert spaces for ...


6

First of all, the equation $$ \begin{equation}A\otimes B=A\otimes \mathbb{1}+\mathbb{1}\otimes B,\end{equation} $$ is a claim about an identity, and this claim is incorrect. Note that for $1\times 1$ matrices, the matrices are numbers and the equation above reduces to $$ a\cdot b = a\cdot 1 + 1 \cdot b$$ which is clearly wrong because the addition (the right ...


1

Yes. $$ f(A,B) \pm (A\leftrightarrow B) ~:=~ f(A,B)\pm f(B,A). $$ The notation is useful as a shorthand, or to convey a symmetry/antisymmetry that may often otherwise be less apparent. See e.g. the last equation in my Phys.SE answer here for a nested example.


0

I've never seen it either. It looks like it means "a term just like the one I just wrote, but with the indicies $\sigma$ and $\rho$ exchanged."


0

To add to what cesaruliana wrote, I think the confusion lies in the fact that apart from the gravitational NP scalars $\Psi_0$ through $\Psi_4$, the NP formalism also defines analogous electromagnetic scalars, $\Phi_0$ through $\Phi_2$. These are contractions of the tetrad vectors with the electromagnetic field tensor $F_{\mu \nu}$. In this case, with the ...


2

Basically, vectors are called contravariant because their components transform oppositely to the basis vectors: if our change of coordinates is such that $$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}$$ then if we have a vector $\mathbf{V}$, its components $V^i_x$ in the $x$ coordinates are related to its ...


1

Are you curious about the case where the $a_j, b_k$ are matrices or operators? As in gautam1168's answer. $$(\mathbf{a}\times \mathbf{b})_i = \epsilon_{ijk} a_j b_k$$ So taking the dagger: $$(\mathbf{a}\times \mathbf{b})_i^{\dagger} = (\epsilon_{ijk})^\dagger b_k^\dagger a_j^\dagger,$$ because $(\epsilon_{ijk})^\dagger$ is just a number which commutes with ...


1

I want to add, that the situation changes if the elements of these vectors are operators. Since this is what your notation might imply. Taking the definition of what a vector product for these quantities should mean, i.e. $$(\mathbf{a}\times\mathbf{b})_k = \epsilon_{ijk} a_i b_j \hspace{1cm} (1)$$ then gives by definition (note that $(AB)^\dagger = ...


1

As ACuriousMind said your expression is not correct: $$(\mathbf{a}\times \mathbf{b})_i = \epsilon_{ijk} a_j b_k$$ So taking the dagger: $$(\mathbf{a}\times \mathbf{b})_i^{\dagger} = (\epsilon_{ijk})^\dagger a_j^\dagger b_k^\dagger\\ =\epsilon_{ijk} a_j^* b_k^*$$ because $\epsilon_{ijk}$ is real scalar, and $a_j$ and $b_k$ are complex scalars. (for scalars ...



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