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6

Force is indeed a vector. Technically you should write $|\overrightarrow{F}| = 30N$, however there is usually context given that let you omit this. If you are working in one dimension, then the vector-like direction is all encapsulated in the sign once you've defined your coordinate system (e.g. -30N is 30N downwards.) Beyond that, it is typically just a ...


5

Either. It's context dependent. Chemists generally mean the whole atoms, nuclear physicists usually mean the nucleus, and people not in those categories could mean either. And there are exception to all those rules or thumb. And the distinctions is important when people start throwing masses around because the mass of an electron is almost on the same ...


5

Use: $$ u_{\alpha}= g_{\alpha\beta}u^{\beta} $$ where $g_{\alpha\beta}$ is the metric tensor.


5

Two conventions. First - use a capital M - make sure you make it big and pointy, so it cannot be confused with lower case: When it is right next to the lower case 'm', the difference should stand out clearly. Second - some people use the "computer short hand" E6: 1.7E6 m This is generally understood to mean (but quicker to write than) $1.7\cdot ...


3

Before going further, I would suggest you to read Chapter 13 ("Spinors") of R.Wald's book "General Relativity". In that chapter, you will see that 2-spinors are simply vectors living in a two-dimensional complex vector space. The capital letters in the indices are simply the abstract index notation for these vectors (see Section 2.4 in Chapter 2 of the same ...


3

Some people use $\mathbf{F}$ instead of $\vec{F}$ or even $\overrightarrow{F}$. I agree that often $F=\| \vec{F} \|$ is a convenient shortcut. So for example A force $\mathbf{F}=(10 \mbox{ N},0,0)$ has magnitude $\|\mathbf{F}\|=10 \mbox{ N}$. The components of $\mathbf{F}$ are $F_x = 10\mbox{ N}$, $F_y=0$ and $F_z=0$ So the subscript is used to ...


2

No, you should not write "$\left|\vec{F}\right| = 30\textrm{N}$", because it's no better than "$F=30\textrm{N}$" Since force is a vector, you could write out the list of components, either as a parenthetical list or a column vector: $$\vec{F} = \left(30\textrm{N}\right) = \left[ 30\textrm{N}\right]$$ You could also write the one component as a scalar: ...


2

The notation is that of one specific isotope (isotopes are nuclides with the same number of protons) of the chemical element Pu. 94 is the number of its protons, which is also the total charge, 240 is the total number of nucleons (protons and neutrons). In a neutral Pu atom there will always be 94 electrons to offset the charge of the protons in the nucleus. ...


2

You're probably used to the convention where a hat is used to denote that something is an operator. But that convention is not universal. In many cases, when it's clear from the context whether something is an operator or not, we just write it without a hat either way. For this case in particular, $\boldsymbol{J}$ is defined to be an operator. The fact that ...


2

$\newcommand{\ket}[1]{\lvert #1 \rangle} \newcommand{\bra}[1]{\langle #1 \rvert}$The states of a quantum system are nothing else than the abstract vectors in the Hilbert space of states $\mathcal{H}$. For one particle, given a basis of position eigenkets $\ket{x}$ with $\hat{x}\ket{x_0} = x_0\ket{x_0}$ and a state $\ket{i}\in\mathcal{H}$, the wavefunction is ...


2

First I want to point out that most of these questions do not bring up issues specific to Bohmian mechanics. That's not a criticism, I'm just pointing out that these notations and concepts are already employed in standard quantum mechanics, or even in classical mechanics. I am going to answer this a little casually, and then make my answer "community ...


2

People who work with neutrons frequently find themselves discussing mega-electronvolts (MeV, typical nuclear energy) and milli-electronvolts (meV, typical room-temperature thermal energy) in the same sentence. It is mostly not a problem to use MeV and meV when writing. When speaking, some people will say "big em ee vee" or "little em ee vee", or pronounce ...


1

It seems to me that you are making some confusion. The problem with the passage [1] (and [2]) that you outline is that you are not allowed to do that (on a rigorous level) if the operator has continuous spectrum, for there are no corresponding eigenvectors on the Hilbert space (and it is wrong also on a non-rigorous level as pointed out by others). Anyways, ...


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.


1

The notation whether it be d or delta doesn't matter as long as it describes an element (a minute amout) of the quantity. Please keep in mind that this is NOT a ratio. So you can't write dq = I. dt This is mathematically wrong. As differentiation is an operation and not a mere ratio. It is like a machine and you can't separate it's parts or the machine ...


1

If we know there is a particle(s) in state $i_1$ why do we need $r_1$? Does the state $i_1$ not specify position? The fact that the particle is in a state $|\psi\rangle$ does not specify the particle's position - it specifies the wave-function $$ \langle x | \psi \rangle = \psi(x) $$ from which one can find $$|\psi(x)|^2$$ This is the probability ...


1

You can always transform to the coordinates of the traveler which experiences no motion in space but only in time (with proper time!). That way $ d\tau=dt $ and $ dx=dy=dz=0 $ . $ |u|^2 = \eta_{\alpha \beta} u^{\alpha} u^{\beta}=-1 $ - which is a scalar. Scalars are invariant under any coordinate transformation, so even after you return to your original ...



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