### 物理代写|电动力学代写electromagnetism代考|ELEC3104

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

In the linear approximation the momentum is a constant of the motion, $\mathbf{p}_0$, so that we need only consider the equation of motion for the coordinate. We make the following substitutions in the linear part of the vector potential, $\mathbf{A}^L$
$$x=\left(\frac{2 a}{\pi}\right) k, \quad y=\left(\frac{\pi c}{2 a}\right)\left(t-t^{\prime}\right) .$$
18 This idea is due to Dirac in a slightly different context [45].

The equation of motion derived from (3.317) in this approximation with $\Delta m$ as before is then
$$m \dot{\mathbf{q}}(t)=\mathbf{p}_0-\Delta m \int_0^{\infty} \int_0^{\infty} x \chi_a^2\left(\frac{\pi x}{2 a}\right) \sin (x y) \dot{\mathbf{q}}\left(t-\frac{2 a y}{\pi c}\right) \mathrm{d} y \mathrm{~d} x$$
which is a linear integro-differential equation with a delay. We define a linear operator $\mathrm{L}$ by the relation
$$\mathrm{L}(\phi(t))=m \phi(t)+\Delta m I_a(\phi(t))$$
where
$$I_a(\phi(t))=\int_0^{\infty} \int_0^{\infty} x \chi_a^2\left(\frac{\pi x}{2 a}\right) \sin (x y) \phi\left(t-\frac{2 a y}{\pi c}\right) \mathrm{d} y \mathrm{~d} x$$
so that (3.356) is concisely expressed as
$$\mathrm{L}(\dot{\mathbf{q}}(t))=\mathbf{p}_0$$
If we can solve this equation, the orbit of the particle will again be (3.351).
The linear equation (3.356) can be solved by the method of characteristic functions [43]. The characteristic equation of $L$ is found directly by studying its action on the exponential function $e^{s t}$, where $s$ is a parameter that will determine the solutions, if any exist; in general $s$ will be a complex number. Consider then
$$\mathrm{L}\left(e^{s t}\right)=m e^{s t}+\Delta m I_a\left(e^{s t}\right)$$
The $y$ integration is elementary and there results
$$\mathrm{L}\left(e^{s t}\right)=e^{s t}\left[m+\Delta m \int_0^{\infty} \frac{x^2 \chi_a^2\left(\frac{\pi x}{2 a}\right)}{x^2+\left(\frac{2 a s}{\pi c}\right)^2} \mathrm{~d} x\right]$$

## 物理代写|电动力学代写electromagnetism代考|Classical Hamiltonian Electrodynamics Revisited

A fundamental result in the Hamiltonian formulation of mechanics is that the time evolution of the system can be regarded as the unfolding of a sequence of infinitesimal canonical transformations for which the Hamiltonian itself is the generator. Recall that if $G$ is the generator of such a transformation, the change in any dynamical variable $\Omega$, a function of the canonical variables, is given by the P.B. relation
$$\delta \Omega={\Omega, G}$$

If we choose $H \mathrm{~d} t$ as the generator, we get the following relations for the result of transformation of the basic phase space variables $\left(q(t)n, p(t)_n\right)$, \begin{aligned} & Q(t)_n=q(t)_n+\frac{\partial H}{\partial p_n} \mathrm{~d} t=q(t)_n+\dot{q}_n \mathrm{~d} t=q(t+\mathrm{d} t)_n \ & P(t)_n=p(t)_n-\frac{\partial H}{\partial q_n} \mathrm{~d} t=p(t)_n+\dot{p}_n \mathrm{~d} t=p(t+\mathrm{d} t)_n \end{aligned} corresponding to the ‘passive’ interpretation (LHS) in terms of transformation to new variables, and an ‘active’ interpretation (RHS) in terms of the time evolution of $q(t)_n, p(t)_n$. A transformation from old $\left(q_n, p_n\right)$ to new $\left(Q_n, P_n\right)$ variables is canonical if the P.B. relations are preserved by the transformation, \begin{aligned} & \left{q_n, q_m\right}=\left{p_n, p_m\right}=0, \quad\left{q_n, p_m\right}=\delta{n m} \ & \quad \rightarrow\left{Q_n, Q_m\right}=\left{P_n, P_m\right}=0, \quad\left{Q_n, P_m\right}=\delta_{n m} \end{aligned}
Now it is easily seen that if we take the Hamiltonian for a charge interacting with its own electromagnetic field, the above relations are not satisfied. The velocity in (3.316) is the gauge-invariant quantity defined by (3.244) which by (3.259) has components which no longer have vanishing P.B.s with each other. $\mathbf{q}$ also occurs in the infinitesimally time-translated field variables, and so the field and particle variables will have some non-zero P.B.s, contrary to the original assumptions. There is therefore a fundamental problem with the conventional classical Hamiltonian formulation which amounts to an incomplete specification of the set of dynamical variables; in other words, we need to identify additional variables such that we can make independent variations in the action integral. In the point particle limit the vector potential for the interacting system is proportional to the particle acceleration [42]. If such a Hamiltonian is to be derived from a Lagrangian, it too must involve the particle acceleration.

# 电动力学代考

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

$$x=\left(\frac{2 a}{\pi}\right) k, \quad y=\left(\frac{\pi c}{2 a}\right)\left(t-t^{\prime}\right)$$
18 这个想法是由于狄拉克在稍微不同的背景下提出的 [45]。

$$m \dot{\mathbf{q}}(t)=\mathbf{p}_0-\Delta m \int_0^{\infty} \int_0^{\infty} x \chi_a^2\left(\frac{\pi x}{2 a}\right) \sin (x y) \dot{\mathbf{q}}\left(t-\frac{2 a y}{\pi c}\right) \mathrm{d} y \mathrm{~d} x$$

$$\mathrm{L}(\phi(t))=m \phi(t)+\Delta m I_a(\phi(t))$$

$$I_a(\phi(t))=\int_0^{\infty} \int_0^{\infty} x \chi_a^2\left(\frac{\pi x}{2 a}\right) \sin (x y) \phi\left(t-\frac{2 a y}{\pi c}\right) \mathrm{d} y \mathrm{~d} x$$

$$\mathrm{L}(\dot{\mathbf{q}}(t))=\mathbf{p}_0$$

$$\mathrm{L}\left(e^{s t}\right)=m e^{s t}+\Delta m I_a\left(e^{s t}\right)$$

$$\mathrm{L}\left(e^{s t}\right)=e^{s t}\left[m+\Delta m \int_0^{\infty} \frac{x^2 \chi_a^2\left(\frac{\pi x}{2 a}\right)}{x^2+\left(\frac{2 a s}{\pi c}\right)^2} \mathrm{~d} x\right]$$

## 物理代写|电动力学代写electromagnetism代考|Classical Hamiltonian Electrodynamics Revisited

$$\delta \Omega=\Omega, G$$

$$Q(t)_n=q(t)_n+\frac{\partial H}{\partial p_n} \mathrm{~d} t=q(t)_n+\dot{q}_n \mathrm{~d} t=q(t+\mathrm{d} t)_n \quad P(t)_n=p(t)_n-\frac{\partial H}{\partial q_n} \mathrm{~d} t=p(t)_n$$

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## MATLAB代写

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