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51

It isn't a problem because two of the eight equations are constraints and they're not quite independent from the remaining six. The constraint equations are the scalar ones, $${\rm div}\,\,\vec D = \rho, \qquad {\rm div}\,\,\vec B = 0$$ Imagine $\vec D=\epsilon_0\vec E$ and $\vec B=\mu_0\vec H$ everywhere for the sake of simplicity. If these equations are ...

25

The energy levels of a diatomic molecule are $E = 2B, 6B, 12B$ and so on, where $B$ is: $$B = \frac{\hbar^2}{2I}$$ Most of the mass of the molecule is in the nuclei, so when calculating the moment of inertia $I$ we can ignore the electrons and just use the nuclei. But the size of the nuclei is around $10^{-5}$ times smaller than the bond length. This ...

24

I) Let us just for fun generalize OP's question to $n$ spacetime dimensions, and check how the counting of eqs. and degrees of freedom (d.o.f.) work out in this general setting. We shall use Lubos Motl's answer as a template for this part. Also we shall use a special relativistic $(-,+,\ldots,+)$ notation with $c=1$, where $\mu,\nu\in\{0,\ldots,n-1\}$ denote ...

18

This follows from the equipartition theorem. The equipartition theorem states that in thermal equilibrium, the average energy of each degree of freedom (each independent way the system can move) is $k_B T/2$, where $T$ is the temperature and $k_B$ (or just $k$) is called the Boltzmann constant. There are three independent directions in which a gas particle ...

16

First of all, Sean Carroll is a relativist so his treatment of the diffeomorphism symmetry as a gauge symmetry should be applauded because it's the standard modern view preferred by particle physicists – its origin is linked to names such as Steven Weinberg, it is promoted by physicists like Nima Arkani-Hamed, and naturally incorporated in string theory so ...

14

We do not treat $\dot q$ as an independent variable in the derivation of the Euler-Lagrange equations. The rough answer is that $q$ and $\dot q$ are independent as inputs to the Lagrangian, but become linked once we specify a path through configuration space - I expand on this in points 5 and 6. I'll be quite formal in what follows, but perhaps the ...

12

You've duplicated constraints because if any one particle is constrainined in all three dimensions with all the other particles this constrains all the particles. The number of constraints is 3(N - 1). To give an example, take three particles a, b and c. If a is fixed relative to b and is also fixed relative to c, then b and c are fixed relative to each ...

12

Bare with me, I don't remember every little step, but I hope this derivation helps you. First remember how a wave travels through a waveguide (dielectric). $$E(x,y,z) = E^{0}(x,y)e^{-\gamma z}$$ $$H(x,y,z) = H^{0}(x,y)e^{-\gamma z}$$ Then consider Ampere's and Faraday's Laws for a source-free region. $$\triangledown \times H = j\omega\epsilon E$$ $$\... 11 The above equation solves for the average kinetic energy of a gaseous particle at a given temperature. k is known as Boltzmann's constant, k_B = 1.3806503 \times 10^{-23}~\mathrm{\frac{m^2kg}{s^2K}}  and is equal to the ideal gas constant divided by Avagadro's number, \frac{R}{N_A}. So where does the equation come from? The short answer: The equation ... 11 Just an addition to John Rennie's answer. The equipartition theorem can only be derived in classical statistical physics. In quantum statistics it is not correct. For each degree of freedom there is a characteristic temperature below which the quantum effects are significant. This temperature is very high for rotation around the axis of the molecule; I guess ... 10 If \phi is non-zero, fixing the phase of \phi is a perfectly valid gauge condition. It's used frequently in Standard Model calcuations involving the Higgs field, where it goes by the name unitarity gauge. This is a nice gauge in some ways, because it makes manifest the fact that there's a massive vector field in the system. Edit: Some caution is ... 10 The potential energy for a diatomic molecule is not$$ U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} |\vec{q}_1 - \vec{q}_2|^2 $$but is instead$$ U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} (|\vec{q}_1 - \vec{q}_2| - r_0)^2, $$where r_0 is the equilibrium bond distance. The important difference here is that in your version, any displacement of the vector \... 9 Each particle that makes up a mechanical system, can be located by three independent variables labelling a point in space. You can choose any particle in the rigid body to start with and move it any where you want, giving three independent variables needed to specify its location. Choosing a second particle, you choose another set of three independent ... 9 1L) The (generalized) position q and (generalized) velocity v are independent variables of the Lagrangian L(q,v,t). 1H) The position q and momentum p are independent variables of the Hamiltonian H(q,p,t). 2L) The position path q:[t_i,t_f] \to \mathbb{R} and velocity path \dot{q}:[t_i,t_f] \to \mathbb{R} are not independent in the Lagrangian ... 9 Since the metric g_{\mu\nu}=g_{\nu\mu} is symmetric, we must demand that$$\tag{1} \delta g_{\mu\nu}~=~\delta g_{\nu\mu}~=~\frac{1}{2}\left(\delta g_{\mu\nu}+\delta g_{\nu\mu}\right)~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta g_{\alpha\beta},$$and therefore$$\tag{2} \frac{\...

9

First, note that the equation you use is only valid when all relativistic particles are in thermal equilibrium. The more general equation, which allows for particles with different temperatures, is $$g(T) = \sum_B g_B\left(\frac{T_B}{T}\right)^4 + \frac{7}{8}\sum_F g_F\left(\frac{T_F}{T}\right)^4$$ where $T$ is the photon temperature and $T_B$, $T_F$ are ...

9

OK, as per your request…. My sense is you want to learn everything about integrability from here, and combine issues which confuses them, instead of separating them…. How about you supplement L&L with Arnold’s book? The seven additive integrals of L&L are the additive conservation laws of the isolated center of mass system, and standard center of ...

9

$\det {e^A} = e^{tr A}$, thus your matrix $H$ must have vanishing trace: one of the $n$ real elements of its diagonal is fixed by the remaining elements. You therefore have $n^2-1$ real degrees of freedom.

8

Hints: We are evidently only supposed to solve for $z$-dependence (as opposed to $x$- and $y$-dependence). Note that the two variables $E_z$ and $H_z$ can be eliminated. In the reduced coupled ODE system of four first-order ODEs and four variables $(E_x,E_y,H_x,H_y)$, note that the variables couple two and two together. Which pairs? Within one such pair, it ...

8

What you have written even without the time derivative is wrong; remember the divergence is, $$\nabla \cdot \vec F = \partial_x F_x + \partial_y F_y + \partial_z F_z$$ and not the derivatives acting on all of $\vec F$ as you wrote. Now, if we want to include a time derivative in the divergence, it only makes sense if $\vec F$ has a time component. This ...

8

This is a nice question, glossed over in many textbooks! Let's start with the electromagnetic field Hamiltonian, $$H \sim \frac12 \left(E^2 + B^2\right).$$ Naively one would say that $\mathbf{E}$ and $\mathbf{B}$ are the two degrees of freedom, giving the factor of two. But as you noted, this isn't correct. The fields $\mathbf{E}$ and $\mathbf{B}$ are not ...

7

The central charge counts the number of degrees of freedom only for matter fields living on a flat manifold (or supermanifold in the case of superstrings). An example where this counting argument fails for matter fields is the case of strings moving on a group manifold $G$ whose central charge is given by the Gepner-Witten formula: $c = \frac{k\mathrm{dim}(... 7 Assuming that the$N$argon atoms are being treated as a classical system, then there are 3 degrees of freedom per atom. That assumes that we are neglecting electronic degrees of freedom, which is OK since one needs a fairly high temperature to thermally excite the electrons in argon. Now if there are$N$atoms, then there are$3N$degrees of freedom. No ... 7 For a degree of freedom whose energy is quadratic in just momentum (but flat in position, or flat with hard walls), the average energy classically is$kT/2$. That is the basic equipartition theorem for an ideal gas. However a lesser known result is that a classical degree of freedom with energy quadratic in both momentum and position has an average energy of ... 7 Degrees of freedom can be defined as the number of independent ways in which the space configuration of a mechanical system may change. Suppose I place an ant on a table with the restriction that the ant can move only through a tube on a line along x-axis. Then the ant will have only one degree of freedom in three dimensional space. However if I allow ... 7 It is natural to generalize to an Abelian$p$-form gauge field $$A~=~\frac{1}{p!} A_{\mu_1\mu_2\ldots\mu_p} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_p}\tag{1}$$ with$\begin{pmatrix} D \cr p \end{pmatrix}$real component fields$A_{\mu_1\mu_2\ldots\mu_p}$in a$D\$-dimensional spacetime. I) Massless case: There is a gauge symmetry $$\... 6 Every rigid body has 3 translational dof. In addition, there are 0, 2, or 3 rotational dof, depending on the geometry, giving a total of 3, 5, or 6 dof. A spherically symmetric rigid body has no rotational dof. A rigid body with rotational symmetry around an axis has 2 rotational dof, namely two angles for orienting the symmetry axis along a direction. ... 6 Your N^2-N^2 calculation is naive, well, it is incorrect because not all N^2 generators of U(N) are changing the Hermitian matrix M. If a generic Hermitian matrix M is transformed to$$ M\to U M U^{-1} ,\quad U U^\dagger = 1,$$then N directions in U i.e. in U(N) don't change M at all. This is easily seen in the basis in which M is ... 6 So, let's take your formula, and set n=3. This gives you N = \frac{9\cdot 8}{12} = 6. Well, how many independent components of the Ricci tensor do you have? Well, since it's a 3x3 symmetric tensor, you've got six independent components. Therefore, there is no room in the Riemann tensor to have additional components. Since, for n > 3, you will ... 6 I would like to know how exactly the equations of motion in the Lorenz gauge removes the second degree of freedom. In the Lorenz 'gauge', we have$$\Box A^{\mu} = \mu_0j^{\mu}$$If A^{\mu} is a solution, then so is A^{\mu} + N\epsilon^{\mu}e^{-ik\cdot x} if$$\Box (N\epsilon^{\mu}e^{-ik\cdot x}) = 0$$Consistency with the Lorenz condition$$\...

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