2018级电动力学期末试题 | 中山大学物理学院期末试卷分享&讨论

[physchtest]

https://physchtest.github.io/2021/07/24/2018%E7%BA%A7%E7%94%B5%E5%8A%A8%E5%8A%9B%E5%AD%A6%E6%9C%9F%E6%9C%AB%E8%AF%95%E9%A2%98/

[Gertese]

等我做完了一定传上来

[Gertese]

(1)
$$\varphi = const \rightarrow \frac{\partial \varphi}{\partial t} = 0$$

洛伦兹规范为：
$$\nabla A + \frac{1}{c^2}\frac{\partial \varphi}{\partial t} = \frac{\partial \sum [a_k(t)e^{ikz}+a_k^*(t)e^{-ikz}]}{\partial x} = 0$$

(2) 此时电矢势的波动方程可以写为：
$$- \sum k^2[a_k(t)e^{ikz}+a_k^(t)e^{-ikz}] \hat{x} - \frac{1}{c^2}\sum[a_k^{''}(t)e^{ikz}+a_k^{''}(t)e^{-ikz}] \hat{x} = 0$$

可以继续写成$$- \sum k^2a_k(t)e^{ikz} - \frac{1}{c^2}\sum a_k^{''}(t)e^{ikz} = 0 \text{ and } - \sum k^2a_k^(t)e^{-ikz} - \frac{1}{c^2}\sum a_k^{''}(t)e^{-ikz} = 0$$

由于初始条件的存在，可以写成$$- k_0^2a_{k_0}(t)e^{ik_0z} - \frac{1}{c^2} a_{k_0}^{''}(t)e^{ik_0z} = 0$$

可以解出$$a(t) = a_0 e^{ick_0}$$

(3)矢势为$$A = a_0 (e^{ick_0}e^{ik_0z} + e^{-ick_0}e^{-ik_0z}) \hat{x}$$

磁场为：$$B = \nabla \times A = a_0 (ik_0e^{ick_0}e^{ik_0z} - ik_0 e^{-ick_0}e^{-ik_0z}) \hat{y}$$

电场为$$E = \frac{ic}{k} \nabla \times B = cB \times \hat{z} = -ca_0 (ik_0e^{ick_0}e^{ik_0z} - ik_0 e^{-ick_0}e^{-ik_0z}) \hat{x}$$

(1)
在惯性系$\sum$中沿$\hat{y}$传播的平面波$$k_\mu = (0,k_y,0,i\frac{\omega}{c})$$

有关系$$k_1^2+k_2^2+k_3^2 = -k_4^2$$

(2) $k_\nu = (-\gamma v \omega ,k_y,0,i\gamma \frac{\omega}{c})$

(3) $xy$平面第二象限

(4) $sin \theta = \frac{k_y}{\sqrt{k_y^2+v^2\gamma^2\omega^2}}$

(1)
$E = \frac{m_0c^2}{\sqrt{1 - \frac{v^2}{c^2}}} \rightarrow v =c\sqrt{1-\frac{m_0^2c^4}{E^2}}$

(2) $p = \gamma m_0 v$

(3)
$qvB = \frac{\gamma m_0 v^2}{R} \rightarrow R = \frac{\gamma m_0 v^2}{qvB}$，$\omega = vR$

(4) $\int_{0}^{\frac{2\pi R}{v}} P(t)dt$