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6

It depends on how the quantity in question transforms. Almost always, densities in the form of "stuff per unit volume" and generally the "stuff" (like a charge) is a scalar (a number of things - number of elementary charges), but the volume it is contained in is observer dependent, owing to the Lorentz contraction. Therefore the density is ...

1

The answer: $$2\partial^\sigma A^\rho.$$ It's not the chain rule but the product rule you use. It is as if the tensor $$\partial_\mu A_\nu$$ were the variable you are differentiating with respect to. To understand the indices, consider a simpler example: $$\frac{\partial}{\partial x^i}(x^j x_j) = \frac{\partial}{\partial x^i}(x^j x^k g_{jk})= \delta^j_i x_j ... 3 http://arxiv.org/abs/quant-ph/9607007 discusses necessary conditions on T (more precisely, on its singular values) for \rho to be positive. They don't seem to derive sufficient conditions, however. The basic idea is that one can perform a rotation U_A and U_B on the two qubits, respectively, which correspondingly transforms r\mapsto O_Ar, ... 1 Mathematicians don't talk about rotations or isometries or whatnot because they already know if they're talking about a scalar or something else; a physicist has to determine whether a physical quantity has the properties of a scalar or a vector or something else. The easiest way to do that is to look at transformation laws--to devise some experiment (real ... 1 When talking about scalars, mathematicians usually use your definition, that is, something which doesn't vary with coordinate changes. (Basically, that there's some mapping to the actual points in space, in which the scalar is well defined) When physicists talk about scalars, we usually refer to Lorentz scalars, which requires two things: Invariance under ... 1 Let u^{\mu} be the 4-velocity of some observer. Then F_{\alpha\beta} = 2E_{[\alpha}u_{\beta]} + \epsilon_{\alpha\beta\gamma\delta}u^{\gamma}B^{\delta} where E^{\alpha}, B^{\alpha} are the electric and magnetic fields relative to this observer and brackets denote antisymmetrization. By definition, u^{\alpha}E_{\alpha} = u^{\alpha}B_{\alpha} = 0. ... 3 Don't use indices. Use a sophisticated and efficient notation for tensor manipulations: through clifford algebra. Let \gamma^0, \gamma^1, \gamma^2, \gamma^3 be basis vectors. Under the clifford product operation, they obey the following:$$\gamma^\mu \gamma^\nu = \begin{cases}\eta^{\mu \nu} & \mu = \nu \\ -\gamma^\mu \gamma^\nu & \mu \neq ...

0

Here's a systematic way to do this: first recall that the electromagnetic tensor is given by $$F^{\alpha\beta}=\partial^\alpha A^\beta-\partial^\beta A^\alpha$$ where $$F_{\alpha \beta}=\eta^{\mu \nu} F^{\alpha \beta}\eta^{\eta \nu}$$ and $$\partial^\mu=-\nabla,\frac{1}{c}\frac{\partial}{\partial t}$$ $A$ is the magnetic vector potential and $A^0$ is ...

2

In one dimension, MERA naturally capture critical systems (i.e., systems with power-law decaying correlations and a log-divergence in the entanglement entropy). MPS (i.e., one-dimensional PEPS), one the other hand, have exponentially decaying correlations and a constant entanglement entropy. (Note: This is for a constant bond dimension and does not preclude ...

9

Section A : The connection of the transformations of complex $\:3\times 3\:$ antisymmetric tensors and their representative complex $\:3$-vectors. Let $\:U\:$ be a special unitary transformation in $\:SU(3)\:$ represented by the $\:3\times 3\:$ complex matrix U= \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ u_{21} & u_{22} ...

3

If $$\mathcal{L} = \boldsymbol{a}\dot{\boldsymbol{q}} + \tfrac{1}{2}\dot{\boldsymbol{q}}^t\mathsf{\boldsymbol{T}}\dot{\boldsymbol{q}} - U(\boldsymbol{q})$$ with some constant vector $\boldsymbol{a}$ and constant symmetric tensor $\mathsf{\boldsymbol{T}}$, then ...

1

Thanks to MBN for the comment - which is also pretty much the answer. Which is basically that I can not make the two equations give the same result, because the $S$ which I was using is invalid. I was thinking of $S$ in terms of a matrix of coordinate changes in a vector tangent space. The transformation I was trying to achieve was: $$x' = x, y' = xy$$ ...

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