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如何用四维矢量来解决狭义相对论问题（电磁学与动力学）

#Physics#STR11
AI-generated summary
This article builds on the previous discussion of four-dimensional vectors and spacetime transformations, focusing on the four-dimensional representation of electromagnetic laws. Key points include: 1. **Four-Dimensional Charge**: Electric charge is identified as a four-dimensional scalar, with the four-current density vector expressed as \( j_p = (j_x, j_y, j_z, ic\rho) \). 2. **Lorenz Condition**: The Lorenz condition is formulated as \( \nabla \cdot A + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0 \), leading to the expression \( \partial_\mu A_\mu = 0 \). 3. **D'Alembert Equation**: The D'Alembert equation relates the electromagnetic potentials and currents, resulting in the four-dimensional wave operator \( \square A_\mu = -\mu_0 J_\mu \). 4. **Electromagnetic Field Tensor**: The electromagnetic field tensor is constructed from the potentials, allowing for transformations under Lorentz transformations, yielding modified electric and magnetic fields. 5. **Maxwell's Equations**: These equations describe the behavior of electric and magnetic fields, with the relationship \( \partial_\nu F_{\mu\nu} = \mu_0 J_\mu \) leading to the conservation law \( \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \). 6. **Electromagnetic Force Density**: The article discusses the need for a four-dimensional vector for electromagnetic force to ensure covariance under Lorentz transformations, leading to the expression for four-dimensional momentum and force. In conclusion, the article emphasizes the importance of four-dimensional representations in understanding electromagnetic phenomena in the context of special relativity.

在上一篇文章中，我们已经介绍了基础的四维矢量与时空变换，这篇文章将在上一篇的基础上进行叙述。

电磁规律的四维矢量表示#

电量是四维标量#

在我们之前的介绍中我们已经得到了四维电流密度矢量
jp=(jx , jy , jz , icρ)j_p=(j_x \ ,\ j_y \ , \ j_z \ , \ ic\rho)
又因为
ρ=γρ′ , dV=dV′γ\rho = \gamma \rho' \ , \ dV = \frac{dV'}{\gamma}
可以得到
ρdV=ρ′dV′\rho dV = \rho' dV'
从中可以看出电量是四维标量

洛伦兹条件#

由洛伦兹条件
∇⋅A+1c2∂ϕ∂t=0\nabla \cdot A + \frac{1}{c^2} \frac{\partial \phi}{\partial t} = 0
即

达朗贝尔方程#

由达朗贝尔方程
∇2A−1c2∂2A∂t2=−μ0j , ∇2φ−1c2∂2φ∂t2=−ρε0\nabla^2 \mathbf{A} - \frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} = - \mu_0 \mathbf{j} \ , \ \nabla^2 \varphi - \frac{1}{c^2} \frac{\partial^2 \varphi}{\partial t^2} = - \frac{\rho}{\varepsilon_0}
且四维标量算符
□=∂μ∂μ=∇2−1c2∂2∂t2\square = \partial_\mu \partial_\mu = \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}
可以得到
□Aμ=−μ0Jμ\square A_\mu = - \mu_0 J_\mu

电磁场张量#

由电磁场与电磁势的微分关系
B=∇×A , E=−∇φ−∂A∂t\mathbf{B} = \nabla \times \mathbf{A} \ , \ \mathbf{E} = - \nabla \varphi - \frac{\partial \mathbf{A}}{\partial t}
可以构造四维电磁场张量
Fμν=∂μAν−∂νAμF_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu
即有

0 & B_3 & -B_2 & -iE_1/c \\ -B_3 & 0 & B_1 & -iE_2/c \\ B_2 & -B_1 & 0 & -iE_3/c \\ iE_1/c & iE_2/c & iE_3/c & 0 \end{bmatrix}$$ 我们会有变换 $$\mathbf{F}' = \mathbf{LFL}^\mathrm{T}$$ 容易得到

\begin{gathered}
\mathbf{E}' = \gamma(\mathbf{E} + \mathbf{v} \times \mathbf{B}) - (\gamma - 1)\mathbf{E}{\parallel} \
\mathbf{B}' = \gamma\left(\mathbf{B} - \frac{1}{c^2}\mathbf{v} \times \mathbf{E}\right) - (\gamma - 1)\mathbf{B}{\parallel}
\end{gathered}

### 麦克斯韦方程 由麦克斯韦方程

\begin{gathered}
\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} \ , \ \nabla \times \mathbf{B} - \varepsilon_0\mu_0 \frac{\partial \mathbf{E}}{\partial t} = \mu_0 \mathbf{j} \
\nabla \cdot \mathbf{B} = 0 \ , \ \nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0
\end{gathered}

且 $$\partial_\nu F_{\mu\nu} = \mu_0 J_\mu$$ 可以得到 $$\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0$$ ### 电磁力密度和电磁场动量与能量 狭义相对论的核心原则是物理定律在所有惯性系中形式相同。经典的洛伦兹力公式 $$F=q(E+v×B)\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})F=q(E+v×B)

中的
E、B和v\mathbf{E}、\mathbf{B} 和 \mathbf{v}
在洛伦兹变换下并不是简单的矢量变换。为了确保力的定律也是协变的，我们需要引入电磁力的四维矢量。

对于四维动量，由动量定理可以得到\begin{gathered} \text{对于四维动量，由动量定理可以得到} \end{gathered}
K=ddτ(P)=ddt(P)⋅dtdτ=γ⋅ddt(P)=γ⋅(dp⃗dt , 1c⋅dEtotdt)=γ⋅(f⃗ , 1c⋅f⃗⋅u⃗)=γ⋅q⋅(E⃗+u⃗×B⃗ , 1c⋅E⃗⋅u⃗)=qc⋅F⋅U\begin{gathered} K = \frac{d}{d\tau}(P) = \frac{d}{dt}(P) \cdot \frac{dt}{d\tau} = \gamma \cdot \frac{d}{dt}(P) = \gamma \cdot (\frac{d\vec{p}}{dt} \ , \ \frac{1}{c} \cdot \frac{dE_{tot}}{dt}) \\ = \gamma \cdot (\vec{f} \ , \ \frac{1}{c} \cdot \vec{f} \cdot \vec{u}) = \gamma \cdot q \cdot (\vec{E} + \vec{u} \times \vec{B} \ , \ \frac{1}{c} \cdot \vec{E} \cdot \vec{u}) = \frac{q}{c} \cdot F \cdot U \end{gathered}
所以可以得到\begin{gathered} 所以可以得到 \end{gathered}
K′=L⋅K=qc(L⋅F⋅L−1)⋅(L⋅U)=qcF′⋅U′\begin{gathered} K' = L \cdot K = \frac{q}{c} (L \cdot F \cdot L^{-1}) \cdot (L \cdot U) = \frac{q}{c} F' \cdot U' \end{gathered}
我们发现K确实是满足洛伦兹协变性的四维矢量\begin{gathered} 我们发现K确实是满足洛伦兹协变性的四维矢量 \end{gathered}
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