エネルギー運動量テンソル

場の変分は$\delta\phi_{r}(x) \equiv \phi^{\prime}_{r}(x) - \phi_{r}(x)$と定められる。 加えて，以下の座標変換を考える。 $$x^{\prime}_{\mu} = x_{\mu} + \varepsilon_{\mu\nu}x^{\nu} + a_{\mu}$$ 次に，座標変換後の場ともともとの場の変分を$\delta_{T}\phi_{r}(x)$とすると， $$\begin{align*} \delta_{T}\phi_{r}(x) &= \phi^{\prime}_{r}(x^{\prime}) - \phi_{r}(x) \\ &= \phi^{\prime}_{r}(x^{\prime}) - \phi_{r}(x^{\prime}) + \phi_{r}(x^{\prime}) - \phi_{r}(x) \\ &= \delta \phi_{r}(x^{\prime}) + \partial_{\mu} \phi_{r}(x) \delta x^{\mu} \end{align*}$$ となる。ただし，無限小変換を考えている限り，$\delta\phi_{r}(x^{\prime})=\delta\phi_{r}(x)$ となるので， $$\delta_{T}\phi_{r}(x) = \delta \phi_{r}(x) + \partial_{\mu} \phi_{r}(x) \delta x^{\mu}$$ が得られる。ここで，座標変換まで含んだ作用変分 $$\delta_{T}S = \int d^{4}x^{\prime} \mathcal{L}(\phi_{r}^{\prime}(x^{\prime}),\partial_{\mu}^{\prime}\phi_{r}^{\prime}(x^{\prime})) - \int d^{4}x \mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x))$$ を考える。積分変数を揃えるために第1項目を書き換える。 $$d^{4}x^{\prime} = \dfrac{\partial(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})} {\partial(x^{0},x^{1},x^{2},x^{3})}d^{4}x = \begin{vmatrix} \partial x^{\prime 0}/\partial x^{0} & \partial x^{\prime 0}/\partial x^{1} & \partial x^{\prime 0}/\partial x^{2} & \partial x^{\prime 0}/\partial x^{3} \\ \partial x^{\prime 1}/\partial x^{0} & \partial x^{\prime 1}/\partial x^{1} & \partial x^{\prime 1}/\partial x^{2} & \partial x^{\prime 1}/\partial x^{3} \\ \partial x^{\prime 2}/\partial x^{0} & \partial x^{\prime 2}/\partial x^{1} & \partial x^{\prime 2}/\partial x^{2} & \partial x^{\prime 2}/\partial x^{3} \\ \partial x^{\prime 3}/\partial x^{0} & \partial x^{\prime 3}/\partial x^{1} & \partial x^{\prime 3}/\partial x^{2} & \partial x^{\prime 3}/\partial x^{3} \end{vmatrix} d^{4}x$$ 無限小変換の場合 $$\dfrac{\partial(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})} {\partial(x^{0},x^{1},x^{2},x^{3})} = 1 + \partial_{\nu}\delta x^{\nu}$$ となるから， $$\begin{align*} \delta_{T} S&=\int d^{4}x\bigl[ \mathcal{L}(\phi^{\prime}(x), \partial_{\mu}^{\prime}\phi_{r}^{\prime}(x^{\prime})) (1 + \partial_{\nu}\delta x^{\nu}) - \mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x))\bigr] \\ &= \int d^{4}x \biggl[\biggl\{ \mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x)) + \biggl(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\biggr)\delta_{\rm T}\phi_{r} + \biggl(\dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{\rm T}\partial_{\mu}\phi_{r}\biggr\} (1 + \partial_{\nu}\delta x^{\nu}) - \mathcal{L}(\phi{r}(x),\partial_{\mu}\phi_{r}(x)) \biggr] \\ &= \int d^{4}x \biggl[\biggl\{\mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x))\partial_{\nu}\delta x^{\nu} + \biggl\{\partial_{\mu}\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{\rm T}\phi_{r} + \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{\rm T}\partial_{\mu}\phi_{r} \biggr\} \biggr] \end{align*}$$ となる。ここで，作用積分$S$が不変であれば $$\begin{align} \biggl(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\biggr)\delta_{T}\phi_{r}(x) + \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{T}\partial_{\mu}\phi_{r}(x) + \mathcal{L}\partial_{\nu}\delta x^{\nu}=0 \end{align}$$ となる。また， $$\begin{align} \delta\phi_{r}(x) = \delta_{\rm T}\phi_{r}(x) - \partial_{\mu}\phi_{r}(x)\delta x^{\mu} \end{align}$$ であり，変分$\delta$は微分と交換可能なことから $$\begin{align*} \delta_{T}\partial_{\mu}\phi_{r}(x) &= \delta\partial_{\mu}\phi_{r}(x) + \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu}\\ &= \partial_{\mu} \delta\phi_{r}(x) + \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu} \\ &= \partial_{\mu}[\delta_{T}\phi_{r}(x) - \partial_{\nu}\phi_{r}(x)\delta x^{\nu}] + \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu} \\ &= \partial_{\mu}\delta_{T}\phi_{r}(x) - \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu} - \partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu} + \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu}\\ &= \partial_{\mu}\delta_{\rm T}\phi_{r}(x) - \partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu} \end{align*}$$ となることを用いると，作用変分は $$\begin{align*} \delta_{\rm T}S &= \int d^{4}x \biggl[ \mathcal{L}\partial_{\nu}\delta x^{\nu} + \biggl(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\biggr) \delta_{T} \phi_{r}(x) +\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr) \bigl\{\partial_{\mu}\delta_{\rm T}\phi_{r}(x) - \partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu}\}\biggr] \\ &= \int d^{4}x \biggl[ \mathcal{L}\partial_{\nu}\delta x^{\nu} + \partial_{\mu}\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr) \delta_{T} \phi_{r}(x) +\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\mu}\delta_{\rm T}\phi_{r}(x) - \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu}\biggr] \\ &= \int d^{4}x \biggl[ \partial_{\mu}\{\mathcal{L}\delta x^{\mu}\} - \partial_{\mu}\mathcal{L}\delta x^{\mu} + \partial_{\mu}\biggl\{\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{T}\phi_{r}(x)\biggr\} - \partial_{\mu}\biggl\{\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\nu}\phi_{r}(x)\delta x^{\nu}\biggr\} + \partial_{\mu}\biggl\{\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\nu}\phi_{r}(x)\biggr\}\delta x^{\nu} \biggr]\\ &= \int d^{4}x \biggl[ \partial_{\mu}\biggl\{\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} \delta_{T}\phi_{r}(x) - \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial_{\nu}\phi_{r}(x)\delta x^{\nu} + \mathcal{L}(\phi_{r},\partial_{\nu}\phi_{r})\delta x^{\mu}\biggr\} \\ &\qquad\qquad + \biggl\{\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\partial_{\nu}\phi_{r}(x) + \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial_{\mu}\partial_{\nu}\phi_{r}(x) - \partial_{\mu}\mathcal{L} \biggr\}\delta x^{\mu} \biggr] \\ &= \int d^{4}x \biggl[ \partial_{\mu}\biggl\{ \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi{r})}\delta_{T}\phi_{r}(x) - \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial_{\nu}\phi_{r}(x) - g^{\mu\nu}\mathcal{L}\biggr)\delta x^{\nu} \biggr\} \biggr] \end{align*}$$ が得られる。これにより，最終的に $$\begin{align} \delta_{T}S &= \int d^{4}x \partial_{\mu}\biggl[ \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\delta_{T}\phi_{r}(x) - \mathcal{T}^{\mu\nu}\delta x_{\nu}\biggr] \end{align}$$ を得る。ここで， $$\begin{align} \mathcal{T}^{\mu\nu} &= \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial^{\nu}\phi_{r} - g^{\mu\nu}\mathcal{L} \end{align}$$ はエネルギー運動量テンソルであり，保存則の式として $$\begin{align} \partial_{\mu}\biggl[ \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\delta_{T}\phi_{r}(x) - \mathcal{T}^{\mu\nu}\delta x_{\nu}\biggr] = 0 \end{align}$$ が得られ，場の理論におけるNoether カレントの保存則を表す。 \subsection{時間と空間の平行移動に対する保存量} 時間と空間の並進について考えると，$x^{\prime\mu} = x^{\mu} + \varepsilon^{\mu}$ であり，$\delta_{T}\phi_{r} = 0$となっているので，保存カレントは$\mathcal{T}^{\mu\nu}$ となり，$\partial_{\mu}\mathcal{T}^{\mu\nu} = 0$が成り立つ。これより，保存量は $$\begin{align} P^{\mu} \equiv \int d^{3}\bm{x} \mathcal{T}^{\mu0} = \it d^{3}\bm{x} \bigl\{ \pi_{r}(x)\partial^{\nu}\phi_{r}(x) - \mathcal{L}g^{\mu0}\bigr\} \end{align}$$ となる。 時間成分と空間成分をそれぞれ書くと， $$\begin{align} P^{0} &= \int d^{3}\bm{x}\{ \pi_{r}(x)\dot{\phi}_{r}(x) - \mathcal{L}(\phi_{r},\partial_{\mu}\phi)\} = \int d^{3}\bm{x} \mathcal{H} = H \\ P^{j} &= \int d^{3}\bm{x} \{\pi_{r}(x) \partial^{j}\phi_{r}(x)\} \end{align}$$ となり，ハミルトニアンと運動量であることがわかる。 \subsection{ローレンツ変換に対する保存量} 無限小ローレンツ変換は$\varepsilon_{\mu\nu} = -\varepsilon_{\nu\mu}$を用いて $$\begin{align} \delta x_{\mu} = \varepsilon_{\mu\nu} x^{\nu},\quad \delta_{T}\phi_{r}(x) = \dfrac{1}{2}\varepsilon_{\mu\nu}(S^{\mu\nu})_{rs} \phi_{s}(x) \end{align}$$ $S^{\mu\nu}$はローレンツ変換の生成子の一つの表現で $$\begin{align} \begin{cases} \text{スピン0(スカラー場)なら} S^{\mu\nu}= 0\\ \text{スピン$\dfrac{1}{2}$(フェルミオン場)なら} S^{\mu\nu}= \dfrac{1}{4}[\gamma^{\mu},\gamma^{\nu}] \text{スピン1(ベクトル場)なら} S^{\mu\nu}= S^{\mu\nu} = L^{\mu\nu} \end{cases} \end{align}$$ ベクトル場の場合， $$\begin{align*} \delta x_{\mu} = \varepsilon_{\mu\nu}x^{\nu} = \dfrac{1}{2}\varepsilon_{\alpha\beta} (S^{\alpha\beta})_{\mu\nu} x^{\nu} \end{align*}$$ より， $$\begin{align*} \varepsilon_{\mu\nu} x^{\nu} = \varepsilon_{\alpha\beta}\delta^{\alpha}_{\;\mu} \delta^{\beta}_{\;\nu} x^{\nu} = \dfrac{1}{2}\varepsilon_{\alpha\beta}(S^{\alpha\beta})_{\mu\nu}x^{\nu} \end{align*}$$ また， $$\begin{align*} \varepsilon_{\mu\nu}x^{\nu} = \dfrac{1}{2}\varepsilon_{\beta\alpha}(-S^{\alpha\beta})_{\mu\nu} x^{\nu} \end{align*}$$ より， $$\begin{align*} \varepsilon_{\mu\nu}x^{\nu} = -\varepsilon_{\beta\alpha} \delta^{\beta}_{\;\alpha} \delta^{\alpha}_{\;\nu} x^{\nu} = \dfrac{1}{2} \varepsilon_{\beta\alpha} (-S^{\alpha\beta})_{\mu\nu} x^{\nu} \end{align*}$$ よって， $$\begin{align*} \dfrac{1}{2}(S^{\alpha\beta})_{\mu\nu} &= \delta^{\alpha}_{\;\mu} \delta^{\beta}_{\;\nu} \\ -\dfrac{1}{2}(S^{\alpha\beta})_{\mu\nu} &= \delta^{\beta}_{\;\mu} \delta^{\alpha}_{\;\nu} \end{align*}$$ 以上より， $$\begin{align} (S^{\alpha\beta})_{\mu\nu} = \delta^{\alpha}_{\;\mu} \delta^{\beta}_{\;\nu} - \delta^{\beta}_{\;\mu} \delta^{\alpha}_{\;\nu} \end{align}$$ 保存則の式に代入すると， $$\begin{align*} &\dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} \delta_{T}\phi_{r}(x) - \mathcal{T}^{\mu\nu}\delta x_{\nu} \\ &= \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} \dfrac{1}{2}\varepsilon_{\nu\lambda} (S^{\nu\lambda})_{rs} \phi_{s} - \mathcal{T}^{\mu\nu} \varepsilon_{\nu\lambda} x^{\lambda} \\ &= \dfrac{1}{2}\varepsilon_{\nu\lambda} \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} (S^{\nu\lambda})_{rs}\phi_{s} - \dfrac{1}{2}\varepsilon_{\nu\lambda} \mathcal{T}^{\mu\nu} x^{\lambda} + \dfrac{1}{2}\varepsilon_{\lambda\nu} \mathcal{T}^{\mu\lambda} x^{\nu} \\ &= \dfrac{1}{2} \varepsilon_{\nu\lambda} \left[ \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} (S^{\nu\lambda})_{rs} \phi_{s} - \mathcal{T}^{\mu\nu} x^{\lambda} + \mathcal{T}^{\mu\lambda} x^{\nu} \right] \end{align*}$$ これより，Noetherカレントは $$\begin{align} \mathcal{M}^{\mu\nu\lambda} = \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} (S^{\nu\lambda})_{rs} \phi_{s} - \mathcal{T}^{\mu\nu} x^{\lambda} + \mathcal{T}^{\mu\lambda} x^{\nu} \end{align}$$ となる。この保存則は$\partial_{\mu}\mathcal{M}^{\mu\nu\lambda}$と書かれるので， 保存量は $$\begin{align*} M^{\mu\nu} &= \int d^{3}\bm{x} \mathcal{M}^{0\mu\nu} \\ &= \int d^{3}\bm{x} \left[ \dfrac{\partial \mathcal{L}}{\partial(\partial_{0}\phi_{r})}(S^{\mu\nu})_{rs}\phi_{s}(x) - \mathcal{T}^{0\mu}x^{\nu} + \mathcal{T}^{0\nu} x^{\mu} \right] \\ &= \int d^{3}\bm{x}\left[\pi_{r}(x)(S^{\mu\nu})_{rs}\phi_{s}(x) - \mathcal{T}^{0\mu}x^{\nu} + \mathcal{T}^{0\nu} x^{\mu} \right] \end{align*}$$ 特に，空間成分については $$\begin{align*} M^{ij} &= \int d^{3}\bm{x} \left[ \pi_{r}(x) (S^{ij})_{rs} \phi_{s}(x) - p^{i}x^{j} + p^{j}x^{i}\right] \end{align*}$$