- #1

ergospherical

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Let ##g_{\mu \nu}(x)## be a time-independent metric. A photon following a curve ##\Gamma## has action\begin{align*}

I[x,e]= \dfrac{1}{2} \int_{\Gamma} e^{-1}(\lambda) g_{\mu \nu}(x)\dot{x}^{\mu} \dot{x}^{\nu} d\lambda

\end{align*}with ##e(\lambda)## an independent function of ##\lambda## (an einbein). The canonical momentum is ##\partial L / \partial \dot{x}^{\mu} = e^{-1}(\lambda) \dot{x}_{\mu}## which yields a conserved energy ##-E \equiv -e^{-1}(\lambda) \dot{t}## since the Lagrangian does not depend on time. The Hamiltonian is\begin{align*}

H = \dfrac{1}{2} e^{-1}(\lambda) g_{\mu \nu}(x) \dot{x}^{\mu} \dot{x}^{\nu} = \dfrac{1}{2}e(\lambda)\left( - E^2 + \mathbf{p}^2 \right)

\end{align*}From this follows the mass-shell equation ##E^2 = \mathbf{p}^2##. Why is this? The einbein field ##\dfrac{1}{2}e(\lambda)## is supposedly acting as a Lagrange multiplier but my variational calculus is rusty. (The next step will be to Legendre transform the cyclic variables only to determine the Routhian).

I[x,e]= \dfrac{1}{2} \int_{\Gamma} e^{-1}(\lambda) g_{\mu \nu}(x)\dot{x}^{\mu} \dot{x}^{\nu} d\lambda

\end{align*}with ##e(\lambda)## an independent function of ##\lambda## (an einbein). The canonical momentum is ##\partial L / \partial \dot{x}^{\mu} = e^{-1}(\lambda) \dot{x}_{\mu}## which yields a conserved energy ##-E \equiv -e^{-1}(\lambda) \dot{t}## since the Lagrangian does not depend on time. The Hamiltonian is\begin{align*}

H = \dfrac{1}{2} e^{-1}(\lambda) g_{\mu \nu}(x) \dot{x}^{\mu} \dot{x}^{\nu} = \dfrac{1}{2}e(\lambda)\left( - E^2 + \mathbf{p}^2 \right)

\end{align*}From this follows the mass-shell equation ##E^2 = \mathbf{p}^2##. Why is this? The einbein field ##\dfrac{1}{2}e(\lambda)## is supposedly acting as a Lagrange multiplier but my variational calculus is rusty. (The next step will be to Legendre transform the cyclic variables only to determine the Routhian).

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