note-math

action-point-particle-relativity_(tag) Action

\int 𝑑 𝑙 (𝑚 𝑐) = \int 𝑑 𝜏 (𝑚 𝑐 | \dot{𝑥} |)

The result is a geodesic

Using the spacetime $ℝ^{1, 3}$ 's metric volume form $𝑑 Vol = | det 𝑔 |^{\frac{1}{2}} 𝑑 𝑥$ restricted to a one-dimensional path, we obtain the length $𝑑 𝑙$ , which uses the square root of the quadratic form, rather than the quadratic form alone

For a path, in the "time coordinate" $𝑥^{0} (𝑡) = 𝑐 𝑡$ , let $𝑣 = \frac{𝑑}{𝑑 𝑡} 𝑥 . space$ . Action

𝑑 𝑙 = 𝑚 𝑐^{2} {(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}} 𝑑 𝑡

equation-point-particle-relativity_(tag) let $𝐿 (𝑥, 𝑣) = 𝑚 𝑐^{2} {(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}$ . Similar to #link(<point-particle-Lagrange-equation>)[the non-relativistic case], the equation of action

\frac{\partial 𝐿}{\partial 𝑥} - \frac{𝑑}{𝑑 𝑡} \frac{\partial 𝐿}{\partial 𝑣} ⟺ \frac{𝑑}{𝑑 𝑡} \frac{𝑚 𝑐^{2} 𝑣}{{(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}} = 0

point-particle-relativity-approximate-to-non-relativity_(tag) The relativistic action "approximates" to the non-relativistic action

𝑚 𝑐^{2} {(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}} = 𝑚 𝑐^{2} - \frac{1}{2} 𝑚 𝑣^{2} + {𝑂 (\frac{𝑣}{𝑐})}^{2}

Then the constant value $𝑚 𝑐^{2}$ will vary to zero $0$

这种非相对论近似极限的方式是坐标依赖的. 在弯曲流形上, 由于可能需要多个坐标覆盖整个流形, 非相对论近似极限的的定义问题会更困难

Symmetry and conserved quantities

The symmetry group of $ℝ^{1, 3}$ spacetime is the isometry $SO (1, 3) ⋊ ℝ^{1, 3}$ alias Poincare group

Translation

Using time coordinates. Similar to the non-relativistic case, the relativistic versions of #link(<energy-point-particle-non-relativity>)[energy] and #link(<momentum-point-particle-non-relativity>)[momentum] are energy-momentum-point-particle-relativity_(tag)

\begin{matrix} 𝐸 & = & \frac{\partial 𝑓}{\partial 𝑣} \cdot 𝑣 - 𝑓 & 𝑝 & = & \frac{\partial 𝑓}{\partial 𝑣} \\ 𝐸 & = & \frac{𝑚 𝑐^{2}}{{(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}} & 𝑝 & = & \frac{𝑚 𝑣}{{(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}} \end{matrix}

Denoted as 4-momentum

𝑚 𝑐 \dot{𝑥} = \frac{𝑚 (𝑐, 𝑣)}{{(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}} = (\begin{matrix} \frac{𝐸}{𝑐} \\ 𝑝 \end{matrix})

The relativistic Lagrangian $| \dot{𝑥} |$ is invariant under $SO (1, 3)$ , but the boost still changes the time and space endpoints of the path i.e. changes the action $\int 𝑑 𝜏 (𝑚 𝑐 | \dot{𝑥} |)$

Rotation

Similar to the non-relativistic case, the relativistic version of #link(<rotation-momentum-point-particle-non-relativity>)[momentum-point-particle-non-relativity] is rotation-momentum-point-particle-relativity_(tag)

𝑥 \times 𝑝 = \frac{𝑥 \times 𝑚 𝑣}{{(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}}

boost

boost by #link(<hyperbolic-angle>)[hyperbolic angle]

exph 𝜃 i_{split} = (\begin{matrix} cosh 𝜃 & sinh 𝜃 \\ sinh 𝜃 & cosh 𝜃 \end{matrix})

So δ boost by hyperbolic angle, is

𝜃 i_{split} = (\begin{matrix} 𝜃 \\ 𝜃 \end{matrix})

In a coordinate of $ℝ^{1, 3}$ , let the spatial vector $𝑛 \in ℝ^{3}$ , $| 𝑛 | = 𝜃$ , corresponding to δ boost, define the hyperbolic cross product $𝑛 \times (\begin{matrix} 𝑐 𝑡 \\ 𝑥 \end{matrix}) = (\begin{matrix} 𝑐 𝑡 + 𝑛 \cdot 𝑥 \\ 𝑥 + 𝑐 𝑡 𝑛 \end{matrix})$

Similar to the case of energy, boost also changes the action

The calculation result of boost momentum will have 4-momentum, thus energy $𝐸$ will appear

boost-momentum-point-particle-relativity_(tag)

\begin{matrix} (\begin{matrix} \frac{𝐸}{𝑐} \\ 𝑝 \end{matrix}) \cdot (𝑛 \times (\begin{matrix} 𝑐 𝑡 \\ 𝑥 \end{matrix})) & = & 𝑛 \cdot (\begin{matrix} 𝐸 𝑡 - 𝑝 \cdot 𝑥 \\ 𝑐 𝑡 𝑝 - \frac{1}{𝑐} 𝐸 𝑥 \end{matrix}) \\ spatial-part & = & 𝑛 \cdot (\begin{matrix} 𝑐 𝑡 𝑝 - \frac{1}{𝑐} 𝐸 𝑥 \end{matrix}) \end{matrix}

Note that the $(1, 3)$ spacetime metric has a negative definite spatial metric

Spatial $ℝ^{3}$ vector

𝑐 𝑡 𝑝 - \frac{1}{𝑐} 𝐸 𝑥 = \frac{𝑚 𝑐 (𝑡 𝑣 - 𝑥)}{{(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}}}

Also called boost momentum

Because $ℝ^{1, 3}$ coordinates are used to separate time and space, although rotational momentum and boost momentum are $SO (1, 3)$ invariant, the representations $𝑥 \times 𝑝$ and boost momentum $𝐸 𝑥 - 𝑡 𝑝$ are not invariant

Combined, it can be written as angular momentum $𝐿_{𝜇 𝜈} = [𝑥_{𝜇}, 𝑝_{𝜈}]$

Particle system

potential $𝑈 (| 𝑥 |), \sum_{𝑖 < 𝑖^{'}} 𝑈 (𝑥_{𝑖} - 𝑥_{𝑖^{'}}), \sum_{𝑖 < 𝑖^{'}} 𝑈 (| 𝑥_{𝑖} - 𝑥_{𝑖^{'}} |)$

potential $𝐴 (𝑥) (\dot{𝑥}) = 𝐴_{0} (𝑥) {\dot{𝑥}}^{0} + \dots + 𝐴_{3} (𝑥) {\dot{𝑥}}^{3}$

point particle in Lorentz-manifold

For the action

\int 𝑑 𝑙

and conserved quantities, metric-connection and δ-isometry are needed

Example

$ℝ^{1, 3}$ 相对论点粒子和规范场的耦合. 作用量

\begin{matrix} \int 𝑑 𝜏 (𝑚 𝑐 | \dot{𝑥} | + \frac{𝑒}{𝑐} \cdot 𝐴 (𝑥) \cdot \dot{𝑥}) \\ = & \int 𝑑 𝑡 (𝑚 𝑐^{2} {(1 - {(\frac{𝑣}{𝑐})}^{2})}^{\frac{1}{2}} + 𝑒 (ϕ - A \cdot 𝑣)) \end{matrix}

Question

隐藏的 $U (1)$ 规范对称性

场相互作用中使用的规范变换 $𝑒^{𝜃}$ 会导致联络的变换 $𝐴 = 𝐴^{'} + 𝑑 𝜃$ . 对于点粒子和电磁场的作用量, $𝑑 𝜃$ 是散度量 $𝑑 𝜃 (𝑥) \cdot \dot{𝑥} = \frac{𝑑}{𝑑 𝜏} 𝜃 (𝑥 (𝜏))$ , 用边界是零, 得到变分是零

尽管 invariant 的是方程而不是作用量

这不同于例如纯量场的情况是, 作用量也 invariant, 而方程的 invariant 通过协变导数的定义

current-gauge-particle_(tag) 这种隐藏的 $U (1)$ 规范对称性是否能给出点粒子的守恒 4-电流? $(𝜌, j) = 𝑗 = 𝜌 (1, v) = 𝜌 𝑣$