# On Landau-Lifshitz's derivation of four-momentum

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?

## 1 Answer

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.
• 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? – yellowquark Aug 8 '17 at 22:04
• Yes, you can say it like that. – Qmechanic Aug 8 '17 at 22:10
• 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? – yellowquark Aug 9 '17 at 17:18