# Tag Info

25

Straight from the horse's mouth: Source: Bureau International des Poids et Mesures (Search for "dimensionless" for all guidelines.) The International Bureau of Weights and Measures (French: Bureau international des poids et mesures), is an international standards organisation, one of three such organisations established to maintain the ...

17

The symbol $\Delta$ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small. The symbols $d,\delta$ refer to infinitesimal variations or numerators and denominators of derivatives. The difference between $d$ and $\delta$ is that $dX$ is only used if $X$ without the $d$ is an actual quantity that may be ...

11

It's c for constant or celeritas, which means speed in Latin. Everyone uses it because it's convention. You could use $\xi$ or $\zeta$ or $\gamma$ or any other symbol you wanted, but then you'd have to explain what it meant, and people would have to go through the trouble to remember this every time they read your papers. Better to go with convention and ...

8

Each of the indices in a tensor have a particular left-right ordering. This ordering cannot be changed unless the tensor has some particular symmetry that permits it (or rather, that equates different components on interchange). The up-down positions of indices tells us about whether the index is associated with using a basis vector (up) or a basis ...

7

The antisymmetric part is defined as $$A_{[a_1 \cdots a_n]} = \frac{1}{n!} \sum\limits_{\sigma \in P(n)} \text{sgn}(\sigma)A_{a_{\sigma(1)} \cdots a_{\sigma(n)}}$$ where $P(n)$ is the set of all permutations of the set $\{1,\cdots,n\}$. $\text{sgn}(\sigma)$ is called the sign of the permutation and is positive of $\sigma$ is obtained from the identity ...

7

It is just a matter of notations. For some reasons, physicists tends to note position using a function notation $\hat \psi(x)$, and momentum with a subscript $\hat a_k$. It is just a matter of taste $\hat \psi(x)=\hat \psi_x$. You seem to be confused by the use of a continuous value for the "index" $x$. If you prefer (and I think that is what mathematicians ...

6

The convention I have seen in journal articles, and that I prefer, is to simply omit any mention of units for dimensionless quantities. EDIT: I also see the style Emilio Pisanty recommends, particularly in tables and graphs. For a graph, the idea is that the datapoints you are plotting are actually numbers, so you want to divide them by the relevant base ...

6

Refer to the nice complement on coherent states in the book by Cohen-Tannoudji, Diu and Laloë, volume 1. It starts off defining coherent states as neither of the ones you mention, and then derives all properties. To answer the question, if you start with definition 2, you can easily show 1, and then from 2, 3. First expand the exponential using ...

5

It is the second, $\psi(x) = \langle x|\psi\rangle$ which is correct. The first, if $x$ is the position operator, is just the position operator acting on the state $|\psi\rangle$. The abstract state $|\psi\rangle$ can be expanded in any basis, using a completion relation: $$|\psi\rangle = \underbrace{\sum_i |i\rangle \langle i|}_{1~=~identity}\psi\rangle$$ ...

5

An easy way to see that they are distinct is to consider what happens upon raising (or lowering) all indices. For example, upon lowering, $$T_{ab}{}^{cde}$$ becomes $T_{abcde}$, whereas $$T_{a}{}^{cd}{}_{b}{}^{e}$$ becomes $T_{acdbe}$, and similarly $$T_{a}{}^{cde}{}_{b}$$ becomes $$T_{acdeb}.$$ You need to "slant" the indices so as to keep track ...

5

There is no significance in the choice between upper- and lower-case $\psi$ (or $\Psi$) to denote a system's wavefunction. The two are used interchangeably and it is the author's discretion to use either symbol. (On the other hand, of course, one shouldn't use the two symbols interchangeably within the same text; if both are used they would refer to ...

5

It's purely notation. Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) ...

5

Is this based off some new, obscure research in string theory? No. This is fringe science. There is no mathematical connection to string theory. It is remarkably bad behavior for this man to set up himself up as a public intellectual -- giving TED talks, making a website juxtaposing himself with Einstein, writing a popular science book -- when he has no ...

5

The (anti)symmetrization simply acts on all the enclosed indices (at the same "height" which are really enclosed between the brackets), regardless of their belonging to the same tensor or different tensors. For example, $$\delta^{[\alpha}{}_{[\gamma} R^{\beta]}{}_{\delta]} = \frac 12 \left(\delta^{[\alpha}{}_{\gamma} R^{\beta]}{}_{\delta} - ... 4 There are two aspects. One is sort of trivial and comprehensible; the other is a bit technical. The trivial reason is that \tilde t \bar{\tilde t} has two "accents" on top of each other and the symbol therefore occupies too much vertical space which is undesirable because we may get overlapping characters and/or non-uniform spacing between lines. The ... 4 Written explicitly, (assuming summation over indices from 0 to 3)$$a^{ij}b_{ij} = \sum_{i=0}^3 \sum_{j=0}^3 a^{ij}b_{ij}$$You can expand this to$$a^{ij}b_{ij} = \sum_{i=0}^3 \left( a^{i0}b_{i0} + a^{i1}b_{i1} + a^{i2}b_{i2} + a^{i3}b_{i3} \right) \implies a^{ij}b_{ij} = a^{00}b_{00} + a^{01}b_{01} + a^{02}b_{02} + a^{03}b_{03} + a^{10}b_{10} + ...

4

This is paraphrasing Wald - General Relativity, section 2.4. Antisymmetrizing $n$ indices means summing over all permutations of the indices, times the sign of each permutation. Since there are $n!$ permutations, it's a sane convention to divide by $n!$ (not all authors do this). For your example, there are $3! = 6$ permutations of $(abc)$. The even ones ...

4

The books are correct. The statement is a definite relation that is being imposed between the 'old' metric structure and the transformed one, for the transformation to be conformal. The equation you're unhappy about, $$g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x)$$ states that the transformed metric $g_{\mu \nu}'$ at the transformed point $x'$ can be ...

4

As Kyle says, $\nu$ is just a (free) index. You can use any letter. More precisely, $x^\mu$ is the $\mu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. And $x^\nu$ is the $\nu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. So you can see that the vector is the same, $\mathbf{x}$, you just name the components with a different ...

4

There is a unitary operator, called the spatial translation operator, that implements translations in precisely the way you want. In fact, for any $a$, it is defined as \begin{align} T_a = e^{-iaP/\hbar} \end{align} where $P$ is the momentum operator, and we are here using the operator exponential. This operator translates position basis elements: ...

4

Well, you can of course simply define a new symbol and use that notation, but no one does that because raising and lowering indices is an operation that has a well-defined, coordinate-free meaning on tensors (it has to do with something called the tangent-cotangent isomorphism), but the connection coefficients are not the components of a tensor. Addendum. ...

3

The 'five' in $\gamma_5$ is not a Lorentz index, so it doesn't make sense to lower or raise it. It can be defined in different ways, one convention is: $$\gamma_5 = \frac{i}{24}\epsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} = \frac{i}{24}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma}$$, where ...

3

This is a standard notation for the (reducible) representation the field transforms under. Usually, the first number is the dimension of the representation for $SU(3)_c$. So a $\mathbf 1$ represents a Lepton (the one-dimensional representation is the trivial representation), a $\mathbf 3$ is a quark. In GUT physics one needs the right-handed fields to ...

3

One of the best answers to your question is due to the painter René Magritte : http://www.wikipaintings.org/en/rene-magritte/the-treachery-of-images-this-is-not-a-pipe-1948 . It says: "This is not a pipe." There are several ways for interpreting that statement, questionning the language, the image, or the role of representations. Another answer is given ...

3

Of course, in principle it makes no difference, however, I think there is an important point to be made: Order the units for maximum physical sense. Take a simple example, of 'speed'. The units would normally be expressed (in SI) as $\text{ms}^{-1}$, not $\text{s}^{-1}\text{m}$. This is because we normally think of speed as "how far something goes per ...

3

It makes no difference as long as you are clear what the units are. The standard for many people is kg m, but you may see in a lot of places m kg. In general, people usually write it thusly in SI units: [charge][mass][length][time][temperature] Unusual units generally go toward the beginning such as this: ...

3

So to address your 2nd question regarding $F = ma$ and its simple transformation through subtraction $F - ma = 0$. In this case $\vec F$ and $\vec a$ are vectors, $m$ is simply a scalar multiplier on $a$ that is along for the ride. When you subtract one vector from another this requires that you flip the direction of the vector in order to do the ...

3

The problem arises when one naively takes the limit of an expression as a constant, such as $c$ or $\hbar$, goes to a value (or infinity). What these limits physically mean is that a dimensionless ratio between a characteristic magnitude and that constant goes to certain value (or infinity). Special relativity The so-called non-relativistic limit (the ...

3

To formalize the comments as an answer: The difference between requiring $$(\alpha u,v)=\alpha(u,v)\quad\text{ (mathematician's definition)}$$ and $$\langle u, \alpha v\rangle=\alpha\langle u,v\rangle\qquad\quad\,\,\text{ (physicist's definition)}$$ is purely one of convention, and the two definitions are equivalent as $(u,v)=\langle v,u\rangle$. There's no ...

3

Repeated indices will be summed throughout. Recall that given any two matrices $A = (A_{ij})$ and $B = (B_{ij})$, the matrix product is a new matrix $AB = (C_{ij})$ defined as follows: \begin{align} C_{ij} = A_{ik}B_{kj} \end{align} In particular, if we think of the first index as the row index and the second index as the column index, then we see that in ...

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