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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 ...


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

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 ...


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 ...


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, ...


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( ...


1

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.



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