# Tag Info

1

Very often in a wave propagation analysis, superscript '+' indicates the fields associated with the wave propagating in the positive direction and superscript '-' indicates the fields associated with the wave propagating in the negative direction. When calculating the impedance, you would take the ratio of the E and H fields for the wave travelling in one or ...

1

which of the following is not a correct representation for prefixes: (A) 1mm, (B) 10km, (C) 1000 micrometer, (D) Both A and B. The question is about prefixes and has nothing to do with the numbers. The three prefixes here are “m” for milli, “k” for kilo, and “micro” for micro. The proper representation of micro is “$\mu$”, not “micro”.

0

Writing $1000\:\mu \rm m$ is not wrong as such, in the sense that there are plenty of contexts in which it can be used, or might even be the best representation for that length. (As a rule, the best notation is the one that adds the most clarity to the text.) Presumably, your exam has singled out that notation because the length can also be expressed as $\rm ... 0 It is more a matter of convention. For example, it is quite conventional to speak about pressures of 1000 hPa. There are historical reasons for that. -1 1000 micrometers is one millimeter 0 I suppose that you consider the Lieb-Liniger model which describes a one-dimensional gas of bosons with a contact interaction, i.e. particles feel each other only when they are at the same point. In quantum mechanics, we are interested in the stationary states which are solutions of the Schrödinger equation. In the absence of any interaction between ... 3 It doesn't matter, because$\langle j|V\rangle$is just a number. It's written that way to reflect the fact that$\sum_i |i\rangle\langle i|$is the identity operator. Acting with it on$|V\rangle$from the left then gives an expansion of$|V\rangle$in the basis$|i\rangle$. 5 So you aren't doing the same thing as your teacher here. Notice how in what your teacher did they have$\Vert d \mathbf r \Vert =ds$, so they did not say that $$\frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {dr}{dt}\right)^2$$ this would not be true in say circular motion, where the position vector is changing but its magnitude is not. However, ... 0 I think that in the first case, a step is missing: $$\frac{d\mathbf r}{dt}.\frac{d\mathbf r}{dt} = \frac{d\mathbf r}{ds}\frac{ds}{dt}.\frac{d\mathbf r}{ds}\frac{ds}{dt}$$$\frac{d\mathbf r}{ds}$is an unit vector in the direction of$\frac{d\mathbf r}{dt}$. So the dot product:$\frac{d\mathbf r}{ds}.\frac{d\mathbf r}{ds} = 1$We can't do the same for$\frac{...

3

Yes, $\vec v=\frac{ds}{dt}\hat{t}$ So, $$d\vec{v}\cdot d\vec {v}=(d^2s\hat t +ds\,d\hat t)\cdot(d^2s\hat t+ds\,d\hat t)=(d^2s)^2+(ds)^2$$ But ${dv}^2=(d^2s)^2$

1

This notation is known as the Feynman or Dirac slash notation. The symbol under the slash must be a Lorentz four-vector, and the slash implies that this four-vector should be contracted with the four-vector of Dirac gamma matrices: $${A\!\!\!/} = \gamma ^{\mu }A_{\mu }.$$ (This identity uses the Einstein summation convention, so repeated indices are ...

3

Are you talking about things like $\frac{d}{dt}$? $\frac{dx}{dt}$ is meant to be interpreted as $\frac{d}{dt}$ of $x$. $x$ is a function on which, when $\frac{d}{dt}$ is applied, it gives another function. The two functions are related in the sense that the function $\frac{dx}{dt}$ represents the rate at which the function $x$ changes as time passes. If $x$ ...

2

"d" stands for derivative. It's close to what you refer to as "delta" above ie., $\Delta$ as in $\frac{\Delta x}{\Delta t}$, but in the limit as these quantities become infinitesimally small. If you drew a graph of $x$ versus $t$, then the quantity $\frac{\Delta x}{\Delta t}$ will be the gradient. Now think about wanting to know what the ...

0

In many discussions of Special Relativity, there are three velocities: the velocity of an object in some reference frame, the velocity of the same object in some other reference frame (one that is moving relative to the first frame), and the relative velocity between the two frames. These are often denoted as $\mathbf u$, $\mathbf u’$, and $\mathbf v$, or $\... 1 Here is another proof. The covariant derivatives satisfy the Jacobi identity $$[\nabla_\mu,[\nabla_\nu,\nabla_\kappa]]+ [\nabla_\nu,[\nabla_\kappa,\nabla_\mu]]+[\nabla_\kappa,[\nabla_\mu,\nabla_\nu]]=0.$$ This can be verified directly, but it is also known that pretty much any associative algebra will satisfy the Jacobi identity, and the elements$\nabla_1,...

1

You can show that in two steps. Show that if a Lorentz tensor vanishes in one Lorentz frame, it vanishes in all Lorentz frames. This is quite simple so I won't do it. This means that you switch to any frame and calculate this identity since what you want to get is zero, which, if true, is valid in all reference frames. Thus, switch to the normal ...

2

Starting from your last equation, it should be clear that $$g_{\mu\nu} v^\mu f^\nu = q \delta^\alpha_\mu F_{\alpha \beta} v^\beta v^\mu = q F_{\alpha \beta} v^\beta v^\alpha,$$ since only the terms where $\alpha = \mu$ will be non-zero because of the Kronecker delta. From here it's quite trivial, since $F_{\alpha \beta}$ is an antisymmetric tensor, and $v^\... 2 @Qmechanic 's impeccable answer above may be illustrated by the basic rule of normal ordering, which is not a functor. That is, normal orderings of commutators vanish; the objects inside : : are commutative entities, "(Weyl) symbols", and not operators, so are reprieved through commutative rules, as Sebastiano Peotta's answer reviews. So linearity ... 0 You are in two dimensions, so the free Schroedinger equation has$E=k^2$solutions $$\psi(r,\theta) =e^{il\theta}J_l(kr),\quad l\in {\mathbb Z}$$ The radial equation is Bessel's equation which has$l^2/r^2$rather than$l(l+1)/r^2$. 3 The main point is that the normal ordering procedure$:~:$does not take operators to operators, but symbols/functions to operators, cf. e.g. this & this Phys.SE posts. With this understanding, the normal ordering procedure$:~:$becomes a linear map, cf. OP's first bullet point. A commutator of symbols/functions is manifestly zero, which resolves OP's ... 2 Apply the Killing equation$\xi_{\alpha;\,\beta}=-\xi_{\beta;\,\alpha}to the third, fifth and sixth terms in (2):\begin{align}0&=\xi_{\sigma;\rho\mu}-\xi_{\sigma;\mu\rho}+\xi_{\mu;\sigma\rho}-\xi_{\mu;\rho\sigma}+\xi_{\rho;\mu\sigma}-\xi_{\rho;\sigma\mu}\\&=\xi_{\sigma;\rho\mu}-\xi_{\sigma;\mu\rho}-\xi_{\sigma;\mu\rho}-\xi_{\mu;\rho\sigma}-\xi_{\... 5 It's just a notation for vectors, like \vec{F} or \mathbf{F}. It's not that common, but I had at least one instructor use it for a class when I was in undergrad. If you look again carefully, you'll notice the work W is not underscored, because it's a scalar. 4 You should use the definition of the curvature tensor[\nabla_\mu, \nabla_\nu]\xi_\kappa = R_{\mu\nu\kappa}^{\ \ \ \ \ \ \lambda}\xi_\lambda.$$Then by differentiating the Killing condition we obtain$$\nabla_\kappa\nabla_\mu\xi_\nu + \nabla_\kappa\nabla_\nu\xi_\mu = 0.$$One can then add a term to both sides to get a commutator and insert the above ... 1 If \xi^a is a Killing vector then Killing's equation says that \xi_{a;b} is antisymmetric. It follows that, when completely contracted with a symmetric tensor, it will give zero. But \xi^a \xi^b is a symmetric tensor. Thus I do both steps in one go. But if you want to separate the two steps, then the argument is as follows. Step 1: any second rank ... 3 Dirac is a brilliant writer and this is a nice paper. But, in modern physics (at least in my experience), it is not particularly common to use |\rangle or |\rangle_\Psi to refer to a state. Looking through the paper, I think that in the language of the time (pre-Dirac notation), one would use \Psi or \psi as special symbols referring to the state. So ... 2 It is generally very unusual for uppercase Greek letters to be typeset in italics, though this depends on where you are. Quoting from the Wikipedia page Typographical conventions in mathematical formulae, as far as 'international' recommendations (specifically IUPAC, NIST and ISO) go, anything that represents a variable (for example, h for a patient's ... 0 You can verify these formulae satisfy \Lambda_\mu^{\:\nu}=\eta_{\mu\rho}\eta^{\nu\sigma}\Lambda_{\:\sigma}^\rho, so the equation you ask about can be rewritten as \delta_\alpha^\beta=\Lambda^\mu_{\:\alpha}\eta_{\mu\rho}\eta^{\beta\sigma}\Lambda^{\:\rho}_\sigma, which uses only the first version of the Lorentz transformation. You could similarly switch to ... 0 In general, I find \nabla\vec u to be slightly ambiguous notation, which should be either avoided or explained when it is first introduced (which can be done). Usually, however, it refers to a matrix whose entries are$$ (\nabla\vec u)_{ij} = \frac{\partial u_j}{\partial x_i}, and the notation \vec u \cdot \nabla \vec u is usually the left-hand vector-... 1 There is no physics behind this nomenclature it's purely incidental. Although the user[[Pieter]] mentioned the intensity related stuff although there is no such evident fact but still you can use it for your better understanding. 1 It is based on intensity: K_\alpha is the strongest line, K_\beta the second strongest, K_\gamma the third strongest. Yes, in this case it is from the next higher shell (from 2p, 3p, 4p etcetera). For L_\alpha and L_\beta, again L_\alpha is the stronger one, but the intensities are more similar than between the two strongest K lines. And ... 3 Here \vec{u} is a vector field \vec{u} = (u_{x},u_{y}, u_{z})^T where u_{x}, u_{y} and u_{z} are scalar fields. So writing things out for clarity we have: \begin{align} (\vec{u}\cdot\nabla)\vec{u} &= (u_x\frac{\partial}{\partial x} +u_y\frac{\partial}{\partial y}+u_z\frac{\partial}{\partial z})\vec{u} \\ &= (\vec{u}\cdot\nabla u_{x},\vec{u}\... 1 The two expressions are equivalent. For the second form, you left-multiply by the vector at the end to return to a vector result. The first form is simpler, because it avoids the confusion you highlighted, and makes the physical meaning of the term (looking for the change along the direction of u, which is what u\cdot\nabla projects out). 1 If you want to expand the Born-Infeld action^1 using\det(\mathbb{1}+M)~=~\exp({\rm tr}\ln(\mathbb{1}+M)),\tag{A} $$then you need the logarithm of a whole matrix$$L~:=~\ln(\mathbb{1}+M)~=~-\sum_{n=1}^{\infty}\frac{1}{n} (-M)^n,\tag{B} $$as opposed to the logarithm of a single matrix element (2). Then$$ L^a{}_b~=~M^a{}_b -\frac{1}{2}M^a{}_cM^c{}_b +{\... 1\renewcommand{\lag}{\mathcal{L}}\renewcommand{\pd}{\partial}\renewcommand{\d}{\mathrm{d}}$$\pd^\mu is defined as \pd^\mu := g^{\mu\nu}\pd_{\nu}, where I use the convention that all repeated indices are summed and g^{\mu\nu} are the components of the inverse metric tensor. Thus your Lagrangian can be rewritten as$$\lag=\tfrac12g^{\mu\nu}(\pd_\mu\phi)(\...

0

Your definition of the Taylor series in one variable to first-order derivatives is $$f(x) = f(a) + \frac{\partial f}{\partial x}(a)(x-a)$$ Notice we neglect higher-order terms. In two variables this would look like $$f(x_1,x_2) = f(a_1,a_2) + \frac{\partial f}{\partial x_1}(a_1,a_2)(x_1-a_1)+ \frac{\partial f}{\partial x_2}(a_1,a_2)(x_2-a_2)$$ Let us ...

4

There is not one single special potential function, rather the opposite. The potential function is a placeholder that takes a different functional form depending on what kind of physical situation you want to model. The physics and the system that we want to describe goes into the Schrödinger equation via this potential function. The only information that ...

0

The potential energy in Schrödinger equation is the electrostatic one. Here are a few points to note: Since the Schrödinger equation is used on a microscale, we exclude the "microscopic" kinds of potential energy familiar from Newtonian mechanics, such as, e.g., the "elastic potential energy", which are really result of the electrostatic ...

5

It's just a normal potential energy function. It could be gravitational potential energy, electric potential energy, or any other kind of potential energy from classical mechanics. The way that the particle reacts to the potential energy will be different, but the form of $V(x)$ is exactly the same. For the infinite square well, we are not really concerned ...

3

From a "physical" point of view, this means that we are only interested in the variation of $S$ as $E$ changes, even though probably changing $E$ might also (if $N$ depends on $E$ in some way) somewhat affect $N$ and therefore change $S$ not directly by the change of $E$ but rather through $N$. The partial derivative you wrote ignores this "...

7

Note that you have too many $\mu$'s floating around in your expression. The correct way to differentiate is to note that by definition, $\partial^\mu \phi = \eta^{\mu\nu} \partial_\nu \phi$. Therefore, \frac{\partial }{\partial (\partial_\sigma\phi)}\left[ \eta^{\mu\nu}(\partial_\mu \phi)(\partial_\nu\phi)\right] = \eta^{\mu\nu} \delta^\sigma_\mu (\...

Top 50 recent answers are included