# Tag Info

27

The thing is that $\mathrm{dm}$ is a single symbol, not a combination of two symbols. Yes, it can be understood in terms of a prefix and a base indicator, but it is still a single symbol. An analogy to the concatenation of variable is inappropriate. Reference to an authoritative statement: The grouping formed by a prefix symbol attached to a unit ...

26

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

18

The statement "$A$ happens given that $B$" is equivalent to "If $B$, then $A$", which is symbolically represented as an implication $$B \Rightarrow A$$ or, if you want to preserve the order of $A$ and $B$ in the original statement $$A \Leftarrow B$$

15

Surely there is a story behind the print on Feynman's tee, these are the CM-1/CM-2 T-shirts. Quoting Tamiko Thiel: The geometric boxes and their 'hard' connections represent the 12-dimensional 'cube of cubes' that forms the internal hardware network connecting all processor chips with each other in a maximum of 12 steps. Feynman is the one who suggested ...

13

Bra ket notation has nothing to do with integrals. If I have two vectors in 3D space, $\vec{v}$ and $\vec{w}$ , I can write them as $|v\rangle$ and $|w\rangle$ if I want. In that case, I would write their dot product (a.k.a. inner product) as $\langle v | w \rangle$. This dot product has nothing to do with integration. When your vectors are functions then ...

13

If I saw the word "amp" written as such in a paper in my field (astrophysics) it would strike me as a bit informal. I would expect to see the full "ampere" written. That said, it is rare to actually write out the full name of a unit; usually it follows a number and is given its standard abbreviation. When abbreviated to e.g. "$5\ \mathrm{A}$", I would ...

11

Technically, apparently, your teacher is correct. BIPM and NIST In the official brochure from the Bureau international des poids et mesures (BIPM, the keepers of SI units) in §5.1 Unit symbols we find: It is not permissible to use abbreviations for unit symbols or unit names, such as sec (for either s or second), sq. mm (for either mm2 or ...

10

The wedge product has its roots in exterior algebra. Exterior algebra lets you talk about objects like planes or volumes as algebraic elements of their own, separate from ordinary vectors, but still obeying the same notions of being "vectors" in their own vector spaces. The wedge product of two vectors is a bivector, and many concepts you may have been ...

10

Quite often everything inside bra or ket is just a label. In this particular case the meaning of $|λ,m_l⟩$ is "a state with the square of the angular momentum being equal to $λ$ (in atomic units, where $\hbar=1$) and with the projection of the angular momentum in some direction ($z$-axis conventionally) being equal to $m_l$". That is, $|λ,m_l⟩$ state is ...

10

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 ...

10

The notation $\lvert \rangle$ is meant to imply that $\lvert \text{anything here you want to put here} \rangle$ is a vector in a Hilbert space. If you have got some wavefunction $\psi(x)$, then you often denote the abstract vector (instead of the concrete realisation in a basis like $\psi(x)$) it represents by $\lvert \psi \rangle$. If you have got only a ...

9

In physicist jargon, we talk about group representations of $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$ by denoting an irreducible representation whose representation vector space has dimension $N$ by $\mathbf{N}$. Hence, the statement $\mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{1} \oplus \mathbf{8}$ is the statement that the tensor product of the ...

9

If you want to make statements over (discrete) time, linear temporal logic may be worth looking at. For instance, $\qquad\displaystyle \Box (A \implies B)$ means that whenever $A$ holds, $B$ has to hold at the same time. $\qquad\displaystyle \Box (A \implies \Diamond B)$ means that whenever $A$ holds, $B$ will hold at some point in the future. Or yet ...

9

They're not used because it's ugly to read such texts with parentheses and it's time-consuming to write it down. A decimeter is indeed a "product" of "deci" and a meter, so the origin is analogous to the product of two real numbers $ab$. But once we define the new derived unit ${\rm dm}$, we treat it as a single object, so it really means what you would ...

8

Canonically, the wedge product is distinct from the cross product and should not be confused, however in three dimensions they are inextricably linked. The wedge product (or outer product) comes from exterior algebra, first due to Grassmann who generalized vector products to arbitrary dimensions. This would later be extended by Clifford into the Clifford ...

8

A physicist would write your first equation $x^a = x^\mu e_\mu^a$. The notation $x^a$ is invariant in your terminology. The $a$ is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that $x$ is a vector (i.e., rank 1,0 tensor.) Similarly for each $\mu$, $e^a_\mu$ is ...

7

Those Greek letters are indices indexing the components of $g$. Generally if one expresses a rank-2 tensor like $g$ as a matrix, the first index indexes the rows, the second the columns. In your example, we have $g_{rr} \equiv g_{11} = 1$, $g_{\theta\theta} \equiv g_{22} = r^2$, $g_{r\theta} \equiv g_{12} = 0$, etc. As you can see, we sometimes use numbers ...

7

$(Q\cdot Q)_{ij}=Q_{im}Q_{mj}$ $(Q^T\cdot Q)_{ij}=(Q^T)_{im}Q_{mj}=Q_{mi}Q_{mj}$ where we use that $(Q^T)_{im}=Q_{mi}$

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 ...

6

Given the way that you've presented your table, I would personally put a "-" rather than a 1 in the units column. This to me would signify that units such as "g, km, s, A" etc. do not apply here. In terms of your symbols, in many branches of physics it is common to use a "hat", "tilde" or "star" notation above a symbol to indicate that it is a ...

6

It is a mnemonic notation that indicates that $\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ is the object whose square root is to be used as the infinitesimal line element, traditonally denoted $\mathrm{d}s$, when determining the lengths of worldlines $x : [a,b] \to \mathcal{M}$ by integrating the line element along them as \begin{align*} ... 6 What Goldstein means by \nabla_iV_i is \nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i $$which is indeed a vector. Here, \mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i}) is the position of the ith particle (with respect to the origin), ... 6 The Lane-Emden is really a non-dimensional form of the Poisson's equation with spherical symmetry:$$ \nabla^2f(r)\equiv\frac1{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right) $$This should be clear to all that, since we have two factors of d/dr on the right hand side of the above, it must be a 2nd order differential equation. This 2nd-order derivative ... 5 I will try to answer this in the more general case where the configuration space is \mathbb R^n. In this case the Hilbert space of the quantum theory is L^2(\mathbb R^n) with Lebesgue measure, and the inner product has the representation$$(f,g) = \int_{\mathbb R^n}\overline{f(x)}g(x)\ \text d\lambda(x)$$where \lambda is the Lebesgue measure. Let U ... 5 The equation is that of the relativistic energy of a 'particle' with a nonzero mass.$$ E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} $$The symbol itself I've never seen before. However, I think it's important to mention that 1/\sqrt{1-\frac{v^2}{c^2}} is represented by \gamma (\text{gamma}). The factor \gamma is often used in special relativity in ... 5 Here is a link to the Einstein archives online. You will see the same cursive letter E used in his signature. 5 The way I imagine it is that the left side of the direct product is exclusively reserved for hilbert space 1 and the right side is for Hilbert space 2. So that the total hilbert space you are working in is written as:$$ H=H_1⊗H_2 $$And so when you have a wavefunction in H you write:$$|\psi\rangle = |r_1\rangle \otimes |r_2 \rangle $$And then: ... 5 Each set of coordinates comes with its own set of coordinate differentials and its own coordinate expression for the metric tensor: Given coordinates (t,x,y,z), the coordinate differentials are dt,dx,dy,dz and$$ (\eta_{\mu\nu})=\begin{pmatrix} c^2&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix} $$... 5 This is the same notation that you'll find in Weinberg's books.$$(\psi, \chi)$$is the inner product of the two states \psi and \chi, and corresponds to$$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to$$ \frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$This new object is just the normalized ... 5 The square brackets mean antisymmetrization. That is:$$ X_{[a_1a_2\dots a_n]} = \frac{1}{n!}\sum_{P\in S(n)} \text{Sign}(P) X_{a_{P(1)}a_{P(2)}\dots a_{P(n}}  where $S(n)$ is the set of permutations of $n$ elements, and $\text{Sign}(P)$ is the sign of the permutation $P$, that is, $\text{Sign}(P)=-1$ if you need an odd number of element exchanges, and ...

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