Let's take metric tensor $η_{μν}=(+,-,-,-)$ and Lagrangian $L=-m\sqrt{\dot{x}_{\mu }\dot{x}^{\mu }}$ .By the definition of momentum

$$p_{\alpha }=\frac{\partial L}{\partial \dot{x}^{\alpha }}=-\frac{m\dot{x}% _{\alpha }}{\sqrt{\dot{x}_{\mu }\dot{x}^{\mu }}}$$ or $p^{\alpha}=-\frac{m\dot{x}^{\alpha }}{\sqrt{\dot{x}_{\mu }\dot{x}^{\mu }}}$ then we fix $x^{0}=t$ and finaly we get

$$p^{0}=E=-\frac{m}{\sqrt{1-v^{2}}}$$

$$\vec{p}=-\frac{m\vec{v}}{\sqrt{1-v^{2}}}$$

So there is extra minus sign in the definition energy-momentum. Should we define the momentum with minus sign?

$$p_{\alpha }=-\frac{\partial L}{\partial \dot{x}^{\alpha }}$$

• I choose $c=1$ . Nov 28 '17 at 18:47
• Dot is derivative by $t$ . I fix gauge $x^{0}=t$ Nov 28 '17 at 18:59
• Why the Lagrangian needs a minus sign? Nov 28 '17 at 19:08
• @Mauricio the minus sign gives the correct non relativistic expression.
– CAF
Nov 28 '17 at 19:26

Yes, OP is right. In the Minkowski sign convention $(\color{red}{\mp}, \color{red}{\pm}, \color{red}{\pm}, \color{red}{\pm})$, the momentum 4-covector is defined as
$$p_{\mu}~:=~ \color{red}{\pm} \frac{\partial L}{\partial \dot{x}^{\mu}},$$