### 物理代写|电动力学代写electromagnetism代考|PHYS3040

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|电动力学代写electromagnetism代考|Mass Renormalisation

The relationship between the mechanical mass $m$ and the observed mass $m^{\text {obs }}$ is the basis for mass renormalisation. We take account explicitly of the contribution to the mass of the charged particle due to the electromagnetic self interaction, so that the ‘structure’ parameter does not appear in the equations of motion. For the theory based on an extended charge distribution, this is achieved by extracting from the equation of motion a term simply proportional to $\dot{\mathbf{q}}(t)$ and identifying its coefficient as the mass correction due to the self-interaction. Clearly this is not possible if the point charge limit is taken first; historically, mass renormalisation was devised within the point charge model and had to proceed by quite different means [44].

We use integration by parts on the $t^{\prime}$ integration in (3.317), choosing the ‘ $\mathrm{d} v$ ‘ factor as $\sin \left[k c\left(t-t^{\prime}\right)\right]$. The boundary term is easily evaluated since it vanishes in the far past and the exponential and cosine terms simply give 1 at $t^{\prime}=t$. Hence, after the remaining elementary integration over $\mathbf{k}$, this contribution to the vector potential reduces to
$$u v \mid=\left(\frac{\Delta m}{e}\right) \dot{\mathbf{q}}(t) \text {. }$$
The integrated part does not simplify and can probably only be usefully evaluated in some approximation. The renormalised equation of motion for the coordinate $\mathbf{q}$ is therefore
\begin{aligned} & m^{\mathrm{obs}} \dot{\mathbf{q}}=\mathbf{p} \ & -\left(\frac{e C}{c}\right) \iint_{-\infty}^t\left(\frac{\chi_a^2(k)}{k^2}\right) \cos \left[k c\left(t-t^{\prime}\right)\right] \frac{\mathrm{d} \boldsymbol{\varepsilon}\left(\mathbf{k}, t^{\prime}\right)}{\mathrm{d} t^{\prime}} \mathrm{d} t^{\prime} \mathrm{d}^3 \mathbf{k} \end{aligned}
where we have put
$$m^{\mathrm{obs}}=m+\Delta m,$$
and
$$\boldsymbol{\varepsilon}\left(\mathbf{k}, t^{\prime}\right)=\left(\left(1+K_{\mathbf{q}}\left(\mathbf{k}, t^{\prime}\right)\right)(\mathbf{1}-\hat{\mathbf{k}} \hat{\mathbf{k}}) \cdot \dot{\mathbf{q}}\left(t^{\prime}\right)\right)$$

## 物理代写|电动力学代写electromagnetism代考|The Point Charge Model

An important limiting case of the calculation just described is the point charge limit with $\chi_0(k)=1$. In this limit we have $\mathbf{Q}_{t^{\prime} t}=0$, and the coefficient of $\Delta m$ is simply proportional to $\ddot{\mathbf{q}}$ [42]. Strictly speaking, we can no longer take the particle momentum to be constant in time, since the homogeneous field $\mathbf{A}(\mathbf{q}, t)_h$ contributes $^{17}$ also to (3.315), so we write the equation of motion for a point charge as
$$\ddot{\mathbf{q}}(t)-\omega_0 \dot{\mathbf{q}}(t)=-\left(\frac{\omega_0}{m}\right) \mathbf{p}(t)+\left(\frac{e}{m}\right) \mathbf{A}(\mathbf{q}, t)_h,$$
where
$$\omega_0=\left(\frac{c}{2 a}\right)\left(\frac{m}{\Delta m}\right)$$

Let
$$\dot{\mathbf{q}}(t)=\mathbf{z}(t),$$
so that
$$\mathbf{q}(t)=\int^t \mathbf{z}\left(t^{\prime}\right) \mathrm{d} t^{\prime}+\mathbf{q}0$$ where $\mathbf{q}_0$ is an integration constant. The solution for the velocity is $$\mathbf{z}(t)=e^{\omega_0 t}\left[e^{-\omega_0 t_0} \mathbf{z}\left(t_0\right)+\int{t_0}^t e^{-v \omega_0}\left(\left(\frac{\omega_0}{m}\right) \mathbf{p}(v)+\left(\frac{e}{m}\right) \mathbf{A}(\mathbf{q}, v)_h\right) \mathrm{d} v\right],$$
which in general shows runaway behaviour, $\mathbf{z}(+\infty)=\infty$; the omission of the free-field vector potential does not alter this conclusion. Since $\omega_0$ contains $e^{-2}$, the coordinate has an essential singularity at $e=0$, so this is a non-perturbative solution.

The situation can be ‘saved’ if we allow the specification of a particular value for the velocity $\mathbf{z}$ at the instant $t_0$ as an extra initial condition. This is contrary to the spirit of Hamilton’s equations which are a pair of coupled first-order differential equations to be solved with initial data $\mathbf{q}\left(t_0\right), \mathbf{p}\left(t_0\right)$. We chose $\mathbf{z}\left(t_0\right)$ so that ${ }^{18}$
$$e^{-\omega_0 t_0} \mathbf{z}\left(t_0\right)=\int_{t_0}^{\infty} e^{-\omega_0 v}\left(\frac{\omega_0}{m} \mathbf{p}(v)-\left(\frac{e}{m}\right) \mathbf{A}(\mathbf{q}, v)_h\right) \mathrm{d} v .$$
Substitution of this choice for $\mathbf{z}\left(t_0\right)$ in (3.352) yields the velocity as
\begin{aligned} \mathbf{z}(t) & =\int_t^{\infty} e^{\omega_0(t-v)}\left(\frac{\omega_0}{m} \mathbf{p}(v)-\left(\frac{e}{m}\right) \mathbf{A}(\mathbf{q}, v)_h\right) \mathrm{d} v \ & =\int_0^{\infty} e^{-\omega_0 \tau}\left(\frac{\omega_0}{m} \mathbf{p}(\tau+t)-\left(\frac{e}{m}\right) \mathbf{A}(\mathbf{q}, \tau+t)_h\right) \mathrm{d} \tau . \end{aligned}

# 电动力学代考

## 物理代写|电动力学代写electromagnetism代考|Mass Renormalisation

$$u v \mid=\left(\frac{\Delta m}{e}\right) \dot{\mathbf{q}}(t)$$

$$m^{\mathrm{obs}} \dot{\mathbf{q}}=\mathbf{p} \quad-\left(\frac{e C}{c}\right) \iint_{-\infty}^t\left(\frac{\chi_a^2(k)}{k^2}\right) \cos \left[k c\left(t-t^{\prime}\right)\right] \frac{\mathrm{d} \varepsilon\left(\mathbf{k}, t^{\prime}\right)}{\mathrm{d} t^{\prime}} \mathrm{d} t^{\prime} \mathrm{d}^3 \mathbf{k}$$

$$m^{\text {obs }}=m+\Delta m$$

$$\varepsilon\left(\mathbf{k}, t^{\prime}\right)=\left(\left(1+K_{\mathbf{q}}\left(\mathbf{k}, t^{\prime}\right)\right)(\mathbf{1}-\hat{\mathbf{k}} \hat{\mathbf{k}}) \cdot \dot{\mathbf{q}}\left(t^{\prime}\right)\right)$$

## 物理代写|电动力学代写electromagnetism代考|The Point Charge Model

$$\mathbf{z}(t)=\int_t^{\infty} e^{\omega_0(t-v)}\left(\frac{\omega_0}{m} \mathbf{p}(v)-\left(\frac{e}{m}\right) \mathbf{A}(\mathbf{q}, v)_h\right) \mathrm{d} v \quad=\int_0^{\infty} e^{-\omega_0 \tau}\left(\frac{\omega_0}{m} \mathbf{p}(\tau+t)-\left(\frac{e}{m}\right)\right.$$

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