2
$\begingroup$

I'm studying the ninth section of The Classical Theory of Fields by Landau & Lifshitz, where they introduce four-momentum through the principle of least action. I can understand the derivation until the point in which they say (I'll use Greek indices instead of the Latin ones in the book) \begin{equation*} \delta S=-mc\eta_{\mu\nu}u^\nu\delta x^\mu. \tag{9.11} \end{equation*} Now, here's my doubt: I know that \begin{equation*} \delta S=\frac{\partial S}{\partial x^\mu}\delta x^\mu, \end{equation*} from which we obtain that \begin{equation*} \frac{\partial S}{\partial x^\mu}=-mc u_\mu, \end{equation*} but now L&L say that \begin{equation*} p_\mu~=~\color{red}{-}\frac{\partial S}{\partial x^\mu} \tag{9.12} \end{equation*} is the four-momentum. In their Mechanics book, however, canonical momentum is derived from the action as \begin{equation*} p_i~=~\color{red}{+}\frac{\partial S}{\partial q^i}, \tag{43.3} \end{equation*} so where does that minus sign come from? I thought that the canonical coordinates $q^i$ corresponded to the contravariant spacetime $x^\mu$ coordinates, i.e. $(ct,q^1,q^2,q^3)\leftrightarrow(x^0,x^1,x^2,x^3)$. This would mean that \begin{equation*} \frac{\partial S}{\partial t}=c\frac{\partial S}{\partial x^0} \quad\text{and}\quad \frac{\partial S}{\partial q^i}=\frac{\partial S}{\partial x^i} \quad (i=1,2,3) \end{equation*} but this means that my results have a wrong sign. So I ask you, where is my error?

$\endgroup$
8
$\begingroup$

TL;DR: The minus sign comes from the Minkowski sign conventions.

  1. Ref. 1 uses only the Minkowski signature convention $(+, -, -, -)$, but we shall show both conventions for reference/clarity. Let us also put $c=1$ for simplicity. Ref. 1 defines the metric tensor $$ g_{\mu\nu}~=~{\rm diag} (\mp 1, \pm 1, \pm 1, \pm 1) ,\tag{6.5} $$ the $4$-velocity $$ u^{\mu}~:=~\frac{dx^{\mu}}{d\tau}~=~\gamma\frac{dx^{\mu}}{dt}, \qquad \frac{dx^{\mu}}{dt}~=~(1,{\bf v}), \tag{7.1/2} $$ $$ \frac{d\tau}{dt}~=~\frac{1}{\gamma} ~=~\sqrt{ \mp g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt} } ~=~\sqrt{1-v^2}, \tag{7.1b} $$ the off-shell action functional $$ S[x]~=~\int_{t_i}^{t_f} \! dt~ L ~=~-m_0\int_{\lambda_i}^{\lambda_f} \! d\lambda~\sqrt{ \mp g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda} } ~=~-m_0 \Delta \tau, \qquad \Delta \tau~:=~\tau_f-\tau_i,\tag{8.1} $$ and the Lagrangian $$ L~=~-\frac{m_0}{\gamma}. \tag{8.2} $$ We should point out that the overall normalization of the Lagrangian (8.2) is not arbitrary, but follows from the need to reproduce the correct non-relativistic formula $$L~=~\frac{1}{2}m_0v^2 -(\text{rest energy})+ {\cal O}(v^4)\qquad\text{for}\qquad v~:=~|{\bf v}|~\ll~ 1. $$

  2. Ref. 1 concludes that the Dirichlet on-shell action function $S(x_f,x_i)$ satisfies $$ \delta S ~\stackrel{(8.1)}{=}~ \pm m_0\int_{\lambda_i}^{\lambda_f} \! d\lambda~\frac{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{d\delta x^{\nu}}{d\lambda}}{\sqrt{ \mp g_{\rho\sigma}\frac{dx^{\rho}}{d\lambda}\frac{dx^{\sigma}}{d\lambda} }} ~=~\pm m_0\int_{t_i}^{t_f} \! dt~u_{\mu}\frac{d\delta x^{\mu}}{dt} $$ $$ ~\stackrel{\text{int. by parts}}{=}~\pm m_0 \left[ u_{\mu} ~ \delta x^{\mu}\right]_{t=t_i}^{t=t_f} ~\mp m_0\int_{t_i}^{t_f} \! dt~\underbrace{\frac{du_{\mu}}{dt}}_{\text{EOM}}~ \delta x^{\mu}\tag{9.10}$$ $$~\stackrel{\text{EOM}}{\approx}~\pm m_0 \left( u_{\mu}^f ~ \delta x^{\mu}_f - u_{\mu}^i ~\delta x^{\mu}_i\right), \qquad u_{\mu}~:=~g_{\mu\nu} u^{\nu}, \tag{9.11} $$ cf. e.g. my Phys.SE answers here & here. [Here the $\approx$ symbol means equality modulo EOM. The words on-shell and off-shell refer to whether EOM are satisfied or not.]

  3. Up until now there has been no room for different conventions. At this point Ref. 1 chooses the contravariant $4$-momentum to be $$ (E,{\bf p}) ~=~p^{\mu}~=~m_0 u^{\mu}, \tag{9.13/14}$$ which means that the covariant $4$-momentum then reads $$ (\mp E,\pm {\bf p}) ~=~p_{\mu}~=~m_0 u_{\mu}. $$

  4. For eqs. (9.11) & (9.13/14) to both hold, we then have to define $$ {\bf p}~:=~ \frac{\partial L}{\partial {\bf v}},\tag{9.1} $$ $$ p^f_{\mu}~:=~\color{red}{\pm} \frac{\partial S}{\partial x^{\mu}_f}, \tag{9.12}$$ $$ p^i_{\mu}~:=~\mp \frac{\partial S}{\partial x^{\mu}_i}. $$

References:

  1. L.D. Landau & E.M. Lifshitz, Vol.2, The Classical Theory of Fields, $\S$9.
$\endgroup$
  • $\begingroup$ Oh, maybe I understood my blunder: if $p_\mu=-\frac{\partial S}{\partial x^\mu}$, then $p^i=\frac{\partial S}{\partial x^i}$ for $i=1,2,3$ and $p^0=-\frac{\partial S}{\partial x^0}$ and everything checks out as I wanted. Is this right? $\endgroup$ – yellowquark Aug 8 '17 at 22:04
  • $\begingroup$ Yes, you can say it like that. $\endgroup$ – Qmechanic Aug 8 '17 at 22:10
  • $\begingroup$ I'm sorry, but I have another doubt: why can L&L choose what the contravariant components of the four-momentum look like? Maybe we could choose the sign of the spatial components so that for $v/c\ll 1$ they reduce to non-relativistic momentum. But what about the time component? $\endgroup$ – yellowquark Aug 9 '17 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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