Heidi said:
Hi Pf
i am accustomed with the action $$mc \int ds $$ for relativistic particle.
i found
a paper by Andrew Wipf with another lagrangian. please look at the beginning of chapter seven (7.1) Is it possible to deduce from it the shape of its trajectory? I have a lot of question about this chapter...
Let (\mathcal{M}^{1} , e) be a 1-dimensional “space-time” with coordinate \tau and metric d\sigma^{2} = e(\tau) d\tau d\tau , with e(\tau) \equiv e_{\tau \tau}(\tau) being the component of the (obviously symmetric) metric tensor. Under a general coordinate transformation (or, which is the same thing, diffeomorphism or gauge transformation) on \mathcal{M}^{1}, \tau \to \tau^{\prime} = \tau^{\prime}(\tau), the (component) of the metric tensor transforms as e^{\prime} (\tau^{\prime}) = \left( \frac{d\tau}{d \tau^{\prime}}\right)^{2} \ e(\tau) . This means that d\tau \sqrt{e(\tau)} is a diffeomorphism-invariant measure on \mathcal{M}^{1} (prove it). Next, we think of \left(\mathcal{M}^{1}, e(\tau)\right) as the world-line x^{\mu}(\tau) of particle in the 4-dimentional (pseudo Riemannian) space-time (\mathcal{M}^{(1,3)}, g_{\mu\nu}(x)). In other words, we define e(\tau) to be a metric on the world-line x^{\mu}(\tau), or (equivalently) define x^{\mu}(\tau) to be 4 scalar fields on the 1-dimensional “space-time” \left(\mathcal{M}^{1}, e(\tau)\right).
We now try to formulate “general relativity”- type theory on the world line, i.e., we try to construct a diffeomorphism-invariant action-integral on \mathcal{M}^{1} using the metric e(\tau) and the scalar fields x^{\mu}(\tau). From the invariant measure on \mathcal{M}^{1}, we form the following cosmological-constant-type (gauge invariant) action S[e] = - \frac{m^{2}}{2} \int d \tau \ \sqrt{e(\tau)} , where m is a constant of mass dimension. For the scalar fields x^{\mu}(\tau), we seek a scalar (Lagrangian) that plays the part of the Ricci scalar in General Relativity, i.e., we would like to find an action of the form S[x] = \int d\tau \sqrt{e(\tau)} R(\mathcal{M}^{1}), where R(\mathcal{M}^{1}) is some scalar depending on the fields e(\tau) and x^{\mu}(\tau). Let us define the function \mathcal{L} (x) = \frac{1}{2} g_{\mu\nu}(x) \frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}, and examine its behaviour under the diffeomorphism \tau \to \bar{\tau} = \bar{\tau}(\tau) (infinitesimally, this is written as \bar{\tau} = \tau + \epsilon (\tau), but we don’t need this in here). Since \bar{x}^{\mu}(\bar{\tau}) = x^{\mu}(\tau), we find (I invite you to do the simple algebra)\mathcal{L}(\bar{x}) = \mathcal{L}(x) \left(\frac{d\tau}{d\bar{\tau}}\right)^{2} . From this, we can identify the Ricci scalar on \mathcal{M}^{1} by R(\mathcal{M}^{1}) \equiv e^{-1} (\tau) \mathcal{L}(x) = \bar{e}^{-1}(\bar{\tau}) \mathcal{L}(\bar{x}) . Thus, we have the following “Einstein-Hilbert” action on the world-line S[x] = \int d\tau \sqrt{e} \ R(\mathcal{M}^{1}) = \frac{1}{2} \int d\tau \sqrt{e(\tau)} e^{-1}(\tau) g_{\mu\nu}(x)\dot{x}^{\mu}(\tau)\dot{x}^{\nu}(\tau) . Therefore, for the total action, we have S[e,x] = \frac{1}{2} \int d\tau \left( \frac{1}{\sqrt{e}}g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu} - m^{2} \sqrt{e}\right) . This action is suitable for massive and massless particles. Let us find the equations of motion for the fields (Please, fill in the algebraic details!): For e(\tau) \frac{\delta S}{\delta e} = 0 \ \Rightarrow \ e^{-1}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} + m^{2} = 0 . This is a constraint equation. For m \neq 0, we can write e = - \frac{g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}{m^{2}} . Substituting this back in the total action, we find the familiar action for massive particle S_{m}[x] = -m \int d\tau \ \sqrt{g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}} = -m \int \ ds.
Exercise(1): Calculate the conjugate momentum p_{\mu} = \frac{\partial L}{\partial \dot{x}^{\mu}} = \frac{1}{\sqrt{e}} g_{\mu\nu}(x)\dot{x}^{\nu}, and show that the constraint equation is nothing but the dispersion relation g_{\mu\nu}p^{\mu}p^{\nu} + m^{2} = 0. Choose \tau to be the proper time and show that p^{\mu} = m\dot{x}^{\mu}.
Exercise(2): Show that \frac{\delta}{\delta x^{\mu}}S[e,x] = 0 \ \Rightarrow \ \sqrt{e} \frac{d}{d\tau} \left( \frac{1}{\sqrt{e}}\dot{x}^{\mu}\right) + \Gamma^{\mu}_{\rho \sigma}(x) \dot{x}^{\rho}\dot{x}^{\sigma} = 0. This can be written as \ddot{x}^{\mu} + \Gamma^{\mu}_{\rho\sigma} \dot{x}^{\rho}\dot{x}^{\sigma} - \frac{1}{2} \frac{\dot e}{e} \dot{x}^{\mu} = 0. What is the geometric meaning of the factor \frac{1}{2}e^{-1}\dot{e} in the last term of the above equation of motion? Why is it always possible to choose \tau such that e(\tau) = 1? Of course, in this case you will have p_{\mu} = g_{\mu\nu}\dot{x}^{\nu}, p_{\mu}\dot{x}^{\mu} + m^{2} = 0 and \ddot{x}^{\mu} + \Gamma^{\mu}_{\rho \sigma}\dot{x}^{\rho}\dot{x}^{\sigma} = 0.
Exercise(3): Investigate the case when m = 0.
