## 物理代写|电动力学代写electromagnetism代考|PHYSICS2534

2023年3月29日

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## 物理代写|电动力学代写electromagnetism代考|Electromagnetic Radiation in Thermal Equilibrium

The quantum mechanical picture of blackbody radiation in a cavity of volume $\Omega$ is that it consists of a ‘gas’ of photons in thermal equilibrium with their surroundings. The photon has no electric charge, and at non-relativistic thermal energies there can be no photon-photon interactions; consequently, the photons cannot equilibrate among themselves, and thermal equilibrium is achieved through continuous absorption and emission of photons by matter in the cavity. The distinguishing feature of this description compared to the familiar thermodynamics of the ‘Ideal Gas’ (a non-interacting atomic gas) is that the particle number is not a conserved quantity; hence we must take the chemical potential $\mu$ to be zero, and the Gibbs free energy of the photon gas is zero. The (Helmholtz) free energy at constant volume is, however, non-trivial and can be used in the usual way to calculate all the customary thermodynamic parameters.
In thermal equilibrium at temperature $T$, the probability that the mode with wave vector $\mathbf{k}$ is occupied with $n_{\mathbf{k}}$ photons is given by Boltzmann’s formula,
$$p\left(n_{\mathbf{k}}\right)=\frac{e^{-E_{n_{\mathbf{k}}} / k_B T}}{Z} .$$
where $E_{n_{\mathrm{k}}}$ is the quantised mode energy, (7.5), and $Z$ is the partition function for the gas,
$$Z=\sum_{n_{\mathrm{k}}} e^{-E_{n_{\mathrm{k}}} / k_B T}$$
Writing
$$x=\exp \left(-\hbar \omega / k_B T\right)$$
we obtain
$$p\left(n_{\mathbf{k}}\right)=(1-x) x^{n_{\mathbf{k}}}$$
since $Z$ is a simple geometric series in $x$ with sum $(1-x)^{-1}$.5
The average energy of the mode at temperature $T$ calculated with this probability can then be put in the form [6]
$$\bar{E}{\mathbf{k}}=\bar{n}{\mathbf{k}} \hbar \omega$$
where the mean photon number is
$$\bar{n}_{\mathbf{k}}=\frac{1}{\exp \left(\hbar \omega / k_B T\right)-1}$$

## 物理代写|电动力学代写electromagnetism代考|Angular Momentum and Polarisation

In Chapter 4 we derived expansions of the vector potential $\mathbf{A}(\mathbf{x})$ in terms of a complete orthogonal set of transverse vector functions with definite angular momentum $(J M)$ and definite parity, (4.126), (4.127). These functions are composed of vector spherical harmonics and appropriately chosen radial functions; such expansions can be quantised in exactly the same way as the expansion of the field in terms of plane waves, $\S 7.3$. Instead of a box of volume $V$ we quantise the field in a large sphere of radius $R$. The coefficients of the orthogonal vector functions can be interpreted as annihilation and creation operators for photons with definite angular momentum and parity. For each frequency $\omega$ and fixed parity we have a commutation relation that replaces (7.119),
$$\left[\mathrm{c}{\omega / M}^{+}, \mathrm{c}{\omega J^{\prime} M^{\prime}}\right]=\delta_{J J^{\prime}} \delta_{M M^{\prime}}$$
with all other commutators vanishing. For a definite value of $J$ the energy $E_J$ of the field is determined by an integral over the radial function, (4.135), and this can be used to fix the normalisation of the expansion functions by setting $E_J$ equal to the energy $\hbar \omega$ of a photon.

We thus arrive naturally at the idea that the photon can carry a certain angular momentum. There are, however, some special features in such a description which arise because the photon is a particle with zero mass. Such a particle must be considered to travel at the speed of light in all reference frames and hence has no rest frame. The conventional identification of the quantum mechanical spin of a particle as its angular momentum in its rest frame is therefore inapplicable. Furthermore, for such a relativistic particle there is no consistent definition of a (local) position operator, and so the customary description of orbital angular momentum based on the classical formula $\mathbf{L}=\mathbf{r} \wedge \mathbf{p}$ is not available. The momentum of a photon is, however, a welldefined quantity and can serve equally well for the formulation of an account of photon angular momentum. The orbital angular momentum operator in this ‘momentum’ representation can be taken to be
$$\mathbf{L}=-i \hbar \mathbf{k} \wedge \boldsymbol{\nabla}_{\mathbf{k}}$$
and therefore differs from the orbital angular momentum operator in the position representation only in that $\mathbf{x}$ is replaced by $\mathbf{k}$ [19].

# 电动力学代考

## 物理代写|电动力学代写electromagnetism代考|Electromagnetic Radiation in Thermal Equilibrium

$$p\left(n_{\mathbf{k}}\right)=\frac{e^{-E_{n_{\mathbf{k}}} / k_B T}}{Z}$$

$$Z=\sum_{n_{\mathrm{k}}} e^{-E_{n_{\mathrm{k}}} / k_B T}$$

$$Z=\sum_{n_{\mathrm{k}}} e^{-E_{n_{\mathrm{k}}} / k_B T}$$

$$x=\exp \left(-\hbar \omega / k_B T\right)$$

$$p\left(n_{\mathbf{k}}\right)=(1-x) x^{n_{\mathbf{k}}}$$

$$\bar{E} \mathbf{k}=\bar{n} \mathbf{k} \hbar \omega$$

$$\bar{n}_{\mathbf{k}}=\frac{1}{\exp \left(\hbar \omega / k_B T\right)-1}$$

## 物理代写|电动力学代写electromagnetism代考|Angular Momentum and Polarisation

$$\left[\mathrm{c} \omega / M^{+}, \mathrm{c} \omega J^{\prime} M^{\prime}\right]=\delta_{J J^{\prime}} \delta_{M M^{\prime}}$$

$$\mathbf{L}=-i \hbar \mathbf{k} \wedge \boldsymbol{\nabla}_{\mathbf{k}}$$因此，与位置表示中的轨道角动量算子的不同之处仅在 于 $\mathbf{x}$ 被替换为 $\mathbf{k}$ [19].

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## MATLAB代写

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