1
$\begingroup$

Let us consider a relativistic point particle action $$ \begin{equation} S= -mc\int \sqrt{-\frac{dz^\mu}{d\sigma}\frac{dz_\mu}{d\sigma}} d\sigma \end{equation} $$ 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 : $$ \begin{equation} \delta S = -mc \int d\sigma \frac{d}{d\sigma} \left[\frac{z^\mu}{\sqrt{-\dot{z}_\mu\dot{z}^\mu}}\right] \delta z_\mu \end{equation} $$ and this brings the boundary term $$\begin{equation} Q= -\epsilon \frac{mc^3}{\sqrt{-\dot{z}^\mu\dot{z}_\mu}}=-\epsilon \frac{mc^2}{\sqrt{1-\frac{\dot{z}_i^2}{c^2}}}=-\epsilon( \gamma mc^2) \end{equation} $$

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

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

Physically, I would expect that the Noether should be equal to total relativistic energy $$ \begin{equation} E=\gamma mc^2 \end{equation} $$ 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 ?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

For what it's worth, here is a straightforward derivation:

  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 $\eta_{\mu\nu}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.