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22

This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...


18

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 ...


13

It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ...


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

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

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) ...


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

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

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} - ...


5

There are two more points that can be made here. Sorry if I repeat someone. In a way you are right that if you have a vector space and its dual there is no intrinsic way to say which space is the original and which is the dual. This is because there is a canonical isomorphism between a vector space and the dual of its dual. In other words if $V$ is a vector ...


5

For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over $\mathbb{H}$) convention really. Don't forget that this is shorthand Einstein convention. When you write $$ \mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk} $$ you are writing a shorthand version in ...


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

This is what physicists call abuse of notation. In mathematics, as you say, one function symbol, such as $f$, denotes one particular function $f: A \rightarrow B$ and the letter we use to denote its argument is irrelevant. In physics, one could say that the letter used to denote a function's argument is also part of that function's name. This is especially ...


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 ...


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

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

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

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 ...



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