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-1

Comments to the question (v1): As usual, be prepared that different authors use different conventions and notations. E.g. what some authors call a vielbein might be what other authors call a transposed vielbein. A curved index (aka. as coordinate index) is raised and lowered vertically with the curved metric tensor, while a flat index (aka. as tetrad ...


1

On a two-index tensor, swapping the two indices is equivalent to transposing a matrix. You may not see many authors spending a lot of effort on this issue simply because an awful lot of the tensors we deal with are symmetric. This includes the metric, Ricci tensor, Einstein tensor, and stress-energy tensor. Therefore there is no special interest in ...


1

Comments to the question (v1): Indices are raised and lowered vertically by the pertinent metric tensor of the theory. The horizontal position of indices is important for a tensor that is not totally symmetric, e.g., the EM field strength $F_{\mu\nu}$ or the Riemann curvature tensor $R_{\mu\nu\lambda\kappa}$, etc, in order to properly identify which ...


0

The Navier-Stokes equation includes a term of the form \begin{equation} (\vec{u}\cdot\nabla)\vec{u}\end{equation} which in index notation is written as \begin{equation} u_i \partial_i u_j \end{equation} Now consider the quantity $\nabla \cdot (\vec{u}\vec{u})$. If we express this (rather less confusingly) in index notation and expand by the product rule we ...


2

The divergence is a vector operator. This simply means that it is a differential operator that acts only on vectors. In this particular case, $$ {\rm div}\equiv\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z} $$ Which, assuming an implicit summation, $$ {\rm div}\equiv\frac{\partial}{\partial ...


0

i think this reference about index notation (and vectors) should help (another one) BTW $$\frac{\partial u_i}{\partial x_j}u_i$$ does not mean $${\rm div} uu$$ (as one can see in the reference) unless sth is a typo somewhere. UPDATE: Tha vector form of Navier-Stokes equations (general) is: The term: $$v \cdot \nabla v$$ in index notation is ...


0

The $\Delta$ is a quartet of particles with isospin 3/2: $$ \Delta^-, \Delta^0, \Delta^+, \Delta^{++} $$ I would expect the anti-$\Delta$ to be written $\bar\Delta$, with the four isospin projections $$ \bar\Delta^{--}, \bar\Delta^-, \bar\Delta^0, \bar\Delta^+ $$ In this case the antiparticle of the $\Delta^+$ would be the $\bar\Delta^-$. If you'd like a ...


4

It does matter, and the product comes first. In general any sort of multiplication is understood to have higher precedence than any sort of addition. Thus $$ \vec{E} + \vec{v} \times \vec{B} \equiv \vec{E} + (\vec{v} \times \vec{B}). $$


2

Excluding F, G, M, and m (you've already used those names in this expression), you could label that distance any letter from a to z or from A to Z or from $\alpha$ to $\omega$. Or whatever. It doesn't matter. It's a variable. That said, there are conventions. It's best not to call that distance v, for example. The symbol v usually means a velocity or speed, ...


2

For physicists it can be very annoying that our historically evolved units of measurement cause the speed of light $c$ to differ from unity. So physicists often apply a trick to avoid distracting conversion factors corresponding to the numerical value of (powers of) the speed of light popping up in their equations. That trick is simply to define your own ...


1

A sample unit conversion for the second half of your question: \begin{alignat}{2} 0.511\,\mathrm{MeV}/c^2 &= 0.511\,\mathrm{MeV}/c^2 \times \frac{10^6\,\mathrm{eV}}{1\,\mathrm{MeV}} \times \frac{1.60\times10^{-19}\,\mathrm{joule}}{1\,\mathrm{eV}} \\ &\quad\qquad \times \frac{1\,\mathrm{kg\cdot m^2/s^2}}{1\,\mathrm{joule}} \times \left( ...


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Capital $\mathrm{C}$, in upright font, is the symbol for the coulomb. Lowercase $c$, italicized, is the speed of light in vacuum. Thanks to Einstein's equation, we can switch between mass and energy ($\mathrm{MeV}$ is a unit of energy) by using factors of $c^2$, and sometimes it's more convenient to know the energy equivalent of a particle's mass rather than ...


0

It's unclear precisely which notation you're asking about, but I'm going to guess it's about the bra-ket notation. The things next to the bra (which is $\langle \text{this}|\ $) and the ket (which is this $|\text{this}\rangle\ $) are typically either complex numbers or quantum mechanical operators. The bras and kets themselves represent quantum mechanical ...


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It is just a matter of definition (dont be led too astray by this). Well a complex conjugate form acts on the dual space of the space where the normal (non-conjugate) form acts (in your example the eigenbra space). Of course for Hilbert spaces, which are usually self-dual, the difference is almost none. The rest is just a matter of notational definiton.


-2

$\bar x$ is the sum of the $x$ divided by $n$, the sample size. Similarly, $\bar y$ is the sum of the $y$ divided by $n$, the sample size. i.e. \begin{align} x_1=& 5,\, x_2=3,\, x_3= 1 \\ \bar x &=\frac{ 5 + 3 + 1}3 \\ \bar x &= 3 \end{align}



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