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13h
answered A question about the Fermi acceleration
16h
comment Dipole moment of a single point charge
If the moments up to order $n-1$ vanish, the moment of order $n$ is independent of coordinate system. Perhaps you should add "in general" to the first sentence?
1d
comment Converting between matrix multiplication and tensor contraction
@tfb Some months ago I came across this, which reminds me of your story.
1d
comment Unitary operators evolving the set of Pauli matrices
The paper mixes two conventions and probably defines $\vec{\sigma}(t) = \sigma_1(t) \vec{e}_1 + \sigma_2(t) \vec{e}_2 + \sigma_3(t) \vec{e}_3$, where the vector space is defined by the Pauli matrices, i.e. $\vec{e}_1 = \sigma_x$, etc.
1d
comment Unitary operators evolving the set of Pauli matrices
Check Frobenius's answer below, it's more explicit. Does the paper define $\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$?
1d
comment Unitary operators evolving the set of Pauli matrices
For example, $\vec{\sigma}_{11}$ is $\vec{e}_z$. Probably, the notation confuses you because you interpret $\vec{e}_i$ in terms of coordinates. Vectors are elements of a vector space and it is an extra step to assign coordinates to them. The vector spaces where $\vec{e}_i$ and $\sigma_{i}$ belong are not the same in general!
1d
comment Unitary operators evolving the set of Pauli matrices
$\sigma_{ij}$ is the $i,j$-th element of the matrix $\vec{\sigma}$. A common convention is to interpret $i$ as the row and $j$ as the column.
1d
comment Unitary operators evolving the set of Pauli matrices
This notation can be used even if the coefficient matrices are linearly dependent!
1d
comment Unitary operators evolving the set of Pauli matrices
$\vec{\sigma}$ is a $2\times 2$ matrix, $\sigma_x \vec{e}_x + \sigma_y \vec{e}_y + \sigma_z \vec{e}_z$.
2d
answered Is the electric field of a volume charge distribution well defined?
Apr
10
awarded  Student
Mar
22
awarded  Enlightened
Mar
22
awarded  Nice Answer
Dec
14
answered Ernst potential from Kaluza-Klein reduction of axisymmetric space-time
Nov
23
awarded  Yearling
Nov
15
comment Why do we obtain classical physics by taking the limit of Planck's constant to zero?
Well, one obvious reason is that $h$ does not appear in the equations of classical physics. If $[x,p]=0$, every state is defined simultaneously by position and momentum, which differentiates between the classical and the quantum theory. This answer does not explain certain details, which appear in chapter VI, §1 of Quantum mechanics by Messiah.
Nov
13
comment Conservation of energy and Killing-field
Also, it is not necessary that the Killing vector is timelike, it is necessary that it is asymptotically timelike, and this is for the reason described in my answer below, namely that the scalar defined using it has the appropriate asymptotic behaviour. In Kerr space-time, $K^a$ may be space-like!
Nov
13
comment Conservation of energy and Killing-field
"Physically, asymptotically flat space-times represent isolated systems", cf. Robert Geroch and Jeffrey Winicour. Linkages in general relativity. Journal of Mathematical Physics, 22(4):803-812, 1981.
Nov
13
revised Conservation of energy and Killing-field
Probably used the wrong word "for" instead of "in". The former might confuse someone in that the answer refers to the energy of the spacetime.
Nov
12
answered Curvature of Conical spacetime