note-math

cf. #link(<Klein--Gordon-Lagrangian>)[Action of scalar field]

Symmetry and conserved currents

Spacetime translation

Spacetime translation does not fix the $ℝ^{1, 3}$ "boundary". The variation is non-zero. Similar to the case of time translation for a point particle

\frac{𝑑}{𝑑 𝑠} \int_{ℝ^{1, 3} + 𝑠 𝑎} 𝑑 𝑥 (𝑓 (𝑥)) = \int_{ℝ^{1, 3}} 𝑑 𝑥 (\partial_{𝑥} 𝑓 (𝑥) \cdot 𝑎)

Generally, region change is given by $exp (𝑎 (𝑥))$ #link(<vector-field-as-δ-diffeomorphism>)[by δ diffeomorphism] $𝑎 (𝑥)$

\frac{𝑑}{𝑑 𝑠} \int_{exp (𝑠 𝑎 (𝑥)) 𝑈} 𝑑 𝑥 (𝑓 (𝑥)) = \int_{𝑈} 𝑑 𝑥 (\partial_{𝑥} 𝑓 (𝑥) \cdot 𝑎 (𝑥))

Using the change of variable formula

\int_{ℝ^{1, 3} + 𝑠 𝑎} 𝑑 𝑥 (𝑓 (𝑥)) = \int_{ℝ^{1, 3}} 𝑑 𝑥 (𝑓 (𝑥 + 𝑠 𝑎))

Swap differentiation and integration

\frac{𝑑}{𝑑 𝑠} \int_{ℝ^{1, 3}} 𝑑 𝑥 (𝑓 (𝑥 + 𝑠 𝑎)) = \int_{ℝ^{1, 3}} 𝑑 𝑥 \frac{𝑑}{𝑑 𝑠} 𝑓 (𝑥 + 𝑠 𝑎)

Apply it to

\begin{matrix} 𝑓 (𝑥) & = & 𝐿 (ϕ (𝑥), \partial ϕ (𝑥)) \end{matrix}

Consider the derivative of the action for a variation that is a translation in the $𝜈 \in ℝ^{1, 3}$ direction. let $Δ ϕ = \partial_{𝜈} ϕ$ , first order derivative $Δ 𝑆 =$

\begin{matrix} \int_{ℝ^{1, 3}} 𝑑 𝑥 (\partial_{𝜈} 𝐿) & = & \int_{ℝ^{1, 3}} 𝑑 𝑥 (\frac{\partial 𝐿}{\partial ϕ} \cdot Δ ϕ + \frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} \cdot \partial_{𝑥} Δ ϕ) \end{matrix}

use product rule

\partial_{𝑥}^{†} (\frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} \cdot Δ ϕ) = (\partial_{𝑥}^{†} \frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)}) \cdot Δ ϕ + \frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} \cdot \partial_{𝑥} Δ ϕ

use Lagrange-equation

\frac{\partial 𝐿}{\partial ϕ} - \partial_{𝑥}^{†} \frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} = 0

use divergence + zero boundary, to get

\begin{matrix} \int_{ℝ^{1, 3}} 𝑑 𝑥 (\partial_{𝜈} 𝐿) & = & \int_{ℝ^{1, 3}} 𝑑 𝑥 (\partial_{𝑥}^{†} (\frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} \cdot Δ ϕ)) \end{matrix}

energy-momentum-tensor-KG_(tag)

𝑇 = 𝑇_{𝜈}^{𝜇} = {\begin{matrix} \frac{\partial 𝐿}{\partial (\partial_{𝜇} ϕ)} \cdot \partial_{𝜈} ϕ & if 𝜇 \neq 𝜈 \\ \frac{\partial 𝐿}{\partial (\partial_{𝜈} ϕ)} \cdot \partial_{𝜈} ϕ - 𝐿 & if 𝜇 = 𝜈 \end{matrix}

get

0 = \int_{ℝ^{1, 3}} 𝑑 𝑥 (\partial^{†} 𝑇 \cdot \partial_{𝜈} ϕ)

for all $\partial_{𝜈} ϕ$ as coordinate-frame, so $\partial^{†} 𝑇 = \partial_{𝜇} 𝑇_{𝜈}^{𝜇} = 0$

for $𝐿 (ϕ, \partial_{𝑥} ϕ) = \frac{1}{2} (\partial ϕ^{*} \cdot \partial ϕ - 𝑚^{2} ϕ^{*} ϕ)$ calculate

\frac{\partial 𝐿}{\partial (\partial_{𝜇} ϕ)} \cdot \partial_{𝜈} ϕ = Re (\partial^{𝜇} ϕ^{*} \partial_{𝜈} ϕ)

or $\frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} = Re (\partial_{𝑥} ϕ^{*} \cdot)$

It can be seen that this EM tensor, after the index is lowered, $𝑇_{𝜇 𝜈} = Re (\partial_{𝜇} ϕ^{*} \partial_{𝜈} ϕ)$ is symmetric, so $𝑇_{𝜇 𝜈} = 𝑇_{𝜈 𝜇}$

In addition, the KG action $𝐿$ is a real value, so its EM tensor is a real value

Assume that the EM tensor of $ℝ^{1, 3}$ can be integrated over $ℝ^{3}$ . Use the notation $(𝑥_{0}, 𝑥) \in ℝ^{1, 3}$ . energy

𝐸 = \int_{ℝ^{3}} 𝑑 𝑥 (𝑇_{0}^{0}) = \int_{ℝ^{3}} 𝑑 𝑥 \frac{1}{2} (| \partial_{0} ϕ |^{2} + | \partial_{𝑥} ϕ |^{2} + 𝑚^{2} | ϕ |^{2})

General potential $𝑚^{2} | ϕ |^{2}$ => $𝑉 (| ϕ |^{2})$

The energy of a relativistic scalar field is real and positive

conserved-spatial-integral-energy-KG_(tag)

Fixing the $ℝ^{1, 3}$ coordinates, assume $𝑇_{𝜇}^{0}$ is a quantity integrable over $ℝ^{3}$

\begin{matrix} \partial_{0} \int_{ℝ^{3}} 𝑑 𝑥 (𝑇_{𝜇}^{0}) & = & \int_{ℝ^{3}} 𝑑 𝑥 (\partial_{0} 𝑇_{𝜇}^{0}) \\ = & - \int_{ℝ^{3}} 𝑑 𝑥 (\partial_{1} 𝑇_{𝜇}^{1} + \partial_{2} 𝑇_{𝜇}^{2} + \partial_{3} 𝑇_{𝜇}^{3}) by \partial^{†} 𝑇_{𝜇} = 0 \\ = & - lim_{𝑟 \to \infty} \int_{𝔹^{3} (𝑟)} 𝑑 𝑥 (div 𝑇_{𝜇}^{𝑥}) \\ = & - lim_{𝑟 \to \infty} \int_{𝕊^{2} (𝑟)} 𝑑 Vol (𝑇_{𝜇}^{𝑥} \cdot \frac{𝑥}{| 𝑥 |}) \end{matrix}

As long as we assume that the flux density $𝑟 \to \infty ⟹ 𝑇_{𝜇}^{𝑥} \cdot \frac{𝑥}{| 𝑥 |} \to 0$ , then we have time invariance of $\int_{ℝ^{3}} 𝑑 𝑥 (𝑇_{𝜇}^{0})$

$𝑇_{0}^{0} = \frac{1}{2} (| \partial_{0} ϕ |^{2} + | \partial_{𝑥} ϕ |^{2} + 𝑚^{2} | ϕ |^{2})$ . Field energy conservation $\partial_{0} 𝐸 = 0$
$𝑇_{𝑥}^{0} = Re (\partial_{0} ϕ^{*} \partial_{𝑥} ϕ)$ . Field momentum (?) conservation

Other $𝑇$ components, e.g. $\int_{ℝ^{1, 3}} 𝑑 𝑥 (𝑇_{𝜇}^{1})$ , are invariant along the $𝑥_{1}$ direction. Use $\partial_{1}$ , the integral of $ℝ^{1, 2}$ and its $div$ . The limit of region approximation $lim_{𝑟 \to \infty}$ is for hyperbolic geodesic spheres (multi-radius)

Example For plane wave expansion

ϕ (𝑥) = \int_{\begin{matrix} {ℍ 𝕪}^{3} \\ 𝕊 (Im ℂ) \end{matrix}} 𝑑 𝑝 𝑑 i (𝑎 (𝑝, i) 𝑒^{𝑝 𝑥 i})

Energy is (Question)

𝐸 = \int_{\begin{matrix} {ℍ 𝕪}^{3} \\ 𝕊 (Im ℂ) \end{matrix}} 𝑑 𝑝 𝑑 i (𝑝_{0}^{2} {| 𝑎 (𝑝, i) |}^{2})

Rotation and boost

For fields, whether spatial rotation or boost, even if the Lagrangian is invariant, the action still changes

Now use the notation $(𝑥_{0}, 𝑥), (\partial_{0}, \partial_{𝑥}) \in ℝ^{1, 3}$

Spatial rotation of $𝑖, 𝑗$ . If $(𝑥_{𝑖}, 𝑥_{𝑗}) = 𝑟 (cos 𝑡, sin 𝑡)$ , then $\frac{𝑑}{𝑑 𝑡} = 𝑟 (- sin 𝑡, cos 𝑡) = (- 𝑥_{𝑗}, 𝑥_{𝑖})$ , so the tangent vector is $- 𝑥_{𝑗} \partial_{𝑖} + 𝑥_{𝑖} \partial_{𝑗}$

Let $𝑛 \in so (3) ≃ ℝ^{3}$ be the spatial rotation axis, then the tangent vector will be $𝑛 \cdot (𝑥 \times \partial)$
Boost of $0, 𝑖$ . If $(𝑥_{0}, 𝑥_{𝑖}) = 𝑟 (cosh 𝑡, sinh 𝑡)$ , then $\frac{𝑑}{𝑑 𝑡} = 𝑟 (sinh 𝑡, cosh 𝑡) = (𝑥_{𝑖}, 𝑥_{0})$ , so the tangent vector is $𝑥_{𝑖} \partial_{0} + 𝑥_{0} \partial_{𝑖}$

Let $𝑛 \in ℝ^{3}$ be the spatial boost axis, then the tangent vector will be $𝑛 \cdot (𝑥_{0} \partial - 𝑥 \partial_{0})$ (The $ℝ^{1, 3}$ spacetime metric has negative definite space)

Now use the notation $𝑥, \partial_{𝑥} \in ℝ^{1, 3}$ . The tangent vectors for rotation and boost are collectively denoted as $[𝑥, \partial_{𝑥}]$ , acting as δ spacetime rotation $so (1, 3)$ on the field $ϕ$ , as a δ diffeomorphism

\int_{ℝ^{1, 3}} 𝑑 𝑥 ([𝑥, \partial] 𝐿) = \int_{ℝ^{1, 3}} 𝑑 𝑥 (\frac{\partial 𝐿}{\partial ϕ} \cdot [𝑥, \partial_{𝑥}] ϕ + \frac{\partial 𝐿}{\partial (\partial_{𝑥} ϕ)} \cdot \partial_{𝑥} [𝑥, \partial_{𝑥}] ϕ)

Using the KG equation, rearranging terms, we get the zero-divergence conserved current

angular-momentum-KG_(tag) let $𝑇$ be the energy-momentum tensor of the KG field. The angular momentum of the field

𝑀 = [𝑥, 𝑇]

𝑀_{𝜇 𝜈}^{𝜆} = [𝑥_{𝜇}, 𝑇_{𝜈}^{𝜆}]

Current of KG field under gauge field

let $ϕ (𝑥)$ be the solution of KG eq. The phase change $𝑒^{𝜃 (𝑥)} ϕ (𝑥)$ and its δ change $𝜃 ϕ$ belong to variations near the solution with fixed boundaries, so

\begin{matrix} 0 & = & \int_{ℝ^{1, 3}} 𝑑 𝑥 \frac{1}{2} (\partial (- 𝜃 ϕ^{*}) \cdot \partial ϕ + \partial ϕ^{*} \cdot \partial (𝜃 ϕ)) \\ = & \int_{ℝ^{1, 3}} 𝑑 𝑥 \frac{1}{2} (- ϕ^{*} \partial ϕ + ϕ \partial ϕ^{*}) \cdot \partial 𝜃 \end{matrix}

Using product rule + divergence + Stokes' theorem + zero boundary

0 = \int_{ℝ^{1, 3}} 𝑑 𝑥 (\frac{1}{2} \partial^{†} (- ϕ^{*} \partial ϕ + ϕ \partial ϕ^{*}) 𝜃)

for all $Im (ℂ)$ valued function $𝜃 (𝑥)$ , therefore

\forall 𝑥 \in ℝ^{1, 3}, \partial^{†} (- ϕ^{*} \partial ϕ + ϕ \partial ϕ^{*}) = 0

$- ϕ^{*} \partial ϕ + ϕ \partial ϕ^{*}$ is called the 4-current of the KG field current-gauge-KG_(tag)

$- ϕ^{*} \partial ϕ + ϕ \partial ϕ^{*} = - Im (ϕ^{*} \partial ϕ)$

Fixing the $ℝ^{1, 3}$ coordinates, and considering the 4-current components as quantities integrable over $ℝ^{3}$ , then the zero component, i.e., charge conservation, holds conserved-spatial-integral-charge-KG_(tag)

0 = \partial_{𝑡} \int_{ℝ^{3}} 𝑑 𝑥 (- ϕ^{*} \partial_{𝑡} ϕ + ϕ \partial_{𝑡} ϕ^{*}) = \partial_{𝑡} 𝑄

note that it's non positive defnite, $(ϕ, 𝜓) ⇝ - 𝜓^{*} \partial_{𝑡} ϕ + ϕ \partial_{𝑡} 𝜓^{*}$ is anti-Hermitian