# Tag Info

2

The triple dots denote a tensor-contraction over three indices. This is a generalization of the notation for the scalar product (which is contraction over one index). The three adjacent terms $E$ are implied to form a tensor product. $\chi^{(3)}$ is third order term of the perturbation expansion of the full (non-linear) susceptibility (more specifically ...

1

With the help from the comments this now makes sense. $$\int \delta(E^2-p^2-m^2)dE$$ With $$E_p^2-p^2-m^2=0$$ Use substitutions $$f(E)=E^2-p^2-m^2\quad df=2EdE$$ $$\int \delta(f)\frac{df}{2E(f)}=\frac{1}{2E_p}$$ $E(f)$ is easily found by inverting $f$ Thanks!

2

The meaning of $\sigma_{ij}$ is force in direction $j$ applied in a surface whose normal is in the direction $i$. Therefore $\sigma_{xx}$ is an x-directed force applied in a surface whose normal is in the x direction, which we interpret as pressure. When $i\neq j$ we call it shear, but the idea is the same. This drawing from Wikipedia might be helpful:

3

We do not "swallow" the index. You must distignuish between the geometrical object and its components, and that is not unique to forms, but occurs for all vectors: If you have a vector $v\in V$, where $V$ is some vector space, it has no indices. It's just an element. Now, if you choose a basis $e_1,\dots,e_n$ of the space, you write $$v = v^\mu e_\mu$$ and ...

0

As Danu said, indices are not "natural" part of tensor fields, they are just a pretty outdated formalism of dealing with them. Unfortunately or fortunately, it is also a pretty well-working and efficient formalism, at least in general relativity. Electrodynamics and classical mechanics would be better off using differential form notation imo... Anyways, the ...

2

In the context of the mathematics of differential geometry, the concept of differential forms is made more precise. It is an invariant object, and does not transform under coordinate changes. In particular, they are not objects that 'inherently' come with indices, so a 2-form like the electromagnetic field tensor would be expressed by mathematicians as ...

0

Voltage and volume are not SI-units (Volt and $m^3$ are the SI units) Both can be abbreviated as V but this in not obligatory: one can choose any letter as long as it is clear what the meaning is. By the way,the letter "U" is quite frequently used for voltage instead of V, so you can avoid the problem.

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 ...

5

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

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: ...

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 ...

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 ...

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. ...

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 ...

1

The entire atom is referred to by Pu.

0

While $[2]$ is "legal", $[1]$ definitely isn't. We are able to write: $$\int dx\ \psi^\ast H\psi = \int dx\ \psi^\ast E\psi = E$$ only if $H\psi = E\psi$ i.e. when $\psi$ is an eigenfunction of $H$. In the general case, it is not, and we generally have $\psi = \sum_n c_n \psi_n$, where $\psi_n$ are the eigenstates of $H$, with $H\psi_n = E_n\psi_n$. The ...

0

Your step $[1]$ is not OK. Just because $\hat H$ has eigenvectors whose eigenvalue is $E$ does not mean that any vector will be an eigenvector of $\hat H$ with eigenvalue $E$. In particular, the position eigenstates are emphatically not eigenstates of the hamiltonian. To go beyond $$⟨\hat{H}⟩ = \int dx'\int dx⟨\psi|x'⟩⟨x'|\hat{H}|x⟩⟨x|\psi⟩$$ you need to ...

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, ...

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 ...

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) $1.7\cdot ... 2 You're probably used to the convention where a hat is used to denote that something is an operator. But that convention is not universal. In many cases, when it's clear from the context whether something is an operator or not, we just write it without a hat either way. For this case in particular,$\boldsymbol{J}\$ is defined to be an operator. The fact that ...

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 ...

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