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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?

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

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

$$ p_{\mu}~:=~ \color{red}{\pm} \frac{\partial L}{\partial \dot{x}^{\mu}}, $$

respectively. See also this related Phys.SE post.

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