# Applying Noether's Theorem for a Relativistic Point Particle

Let us consider a relativistic point particle action $$$$S= -mc\int \sqrt{-\frac{dz^\mu}{d\sigma}\frac{dz_\mu}{d\sigma}} d\sigma$$$$ for some arbitrary curve parameter $$\sigma$$ and we use the mostly plus metric. My aim is to find the total energy through Noether's theorem by considering $$\delta z^\mu$$ translations restricted to the symmetry variation : $$\delta z^0 = c\epsilon~,\delta z^i=0$$. The calculation consists of three parts : 1) Find the total boundary term in the variation of the action, and 2) Find how the Lagrangian transforms under the symmetry variation. 3) Construct the Noether charge. So we have :

1. Total variation : $$$$\delta S = -mc \int d\sigma \frac{d}{d\sigma} \left[\frac{z^\mu}{\sqrt{-\dot{z}_\mu\dot{z}^\mu}}\right] \delta z_\mu$$$$ and this brings the boundary term $$$$Q= -\epsilon \frac{mc^3}{\sqrt{-\dot{z}^\mu\dot{z}_\mu}}=-\epsilon \frac{mc^2}{\sqrt{1-\frac{\dot{z}_i^2}{c^2}}}=-\epsilon( \gamma mc^2)$$$$

2. Lagrangian variation : Lagrangian is $$L=-mc\sqrt{-\dot{z}^\mu\dot{z}_\mu}$$ which brings $$$$\delta L = \frac{\partial L}{\partial \dot{z}^\mu}\delta{\dot{z}^\mu}= mc \frac{\dot{z}^\mu \ddot{z}_\mu}{\sqrt{-\dot{z}^\mu\dot{z}_\mu}}\epsilon = \epsilon \frac{d}{d\sigma}L$$$$ where we used $$\delta \dot{z}^\mu = \epsilon\ddot{z}^\mu$$ and hence the boundary term $$$$K= \epsilon L$$$$

3. Noether charge : $$$$E= K-Q = -mc^2\sqrt{1-\dot{z}_i^2/c^2} + mc^2\frac{1}{\sqrt{1-\dot{z}_i^2/c^2}} = \frac{m\dot{z}_i^2}{\sqrt{1-\dot{z}_i^2/c^2}} = \gamma m \dot{z}_i^2$$$$

Physically, I would expect that the Noether should be equal to total relativistic energy $$$$E=\gamma mc^2$$$$ but not $$\gamma m \dot{z}_i^2$$ which seems as a non-sense quantity to me. So, what's going wrong in the above calculation ?

1. The 4-position $$x^{\mu}$$ is a cyclic variable for the Lagrangian for massive relativistic point particle is$$^1$$ \begin{align} L_0~=~&-m_0c\sqrt{-\dot{x}^2}, \cr \dot{x}^2~:=&~\dot{x}^{\mu}\eta_{\mu\nu}\dot{x}^{\nu}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}.\end{align}\tag{1}
2. Noether's theorem then yields that the corresponding Noether charge is the 4-momentum $$p_{\mu}~=~\frac{\partial L}{\partial \dot{x}^{\mu}}~=~\frac{m_0c\dot{x}_{\mu}}{\sqrt{-\dot{x}^2}}.\tag{2}$$
3. In the static gauge $$\lambda=t=\frac{x^0}{c}$$, eq. (2) is the standard expression for the 4-momentum. In particular the 0-component (times $$c$$) $$p^0c~=~m_0\gamma c^2\tag{3}$$ is the total energy, cf. OP's question.
$$^1$$ We use the sign convention $$(-,+,+,+)$$ for the Minkowski metric $$\eta_{\mu\nu}$$.