Deriving the $0$-component of 4-momentum using the relativistic Lagrangian

Question

My question arises from Susskinds book on Special Relativity and Classical Field Theory. (page 102 equation 3.29 to 3.30 and page 105 equation 3.34 to 3.36.)

The relativistic Lagrangian for a free particle is given by the following equation. $$ L = -mc^2\sqrt{1-\frac{v^2}{c^2}} = \frac{-mc^2}{\dot{X}^0},\tag{1} $$ where dot means differentiation with respect to the proper time. The $i^{th}$ component of momentum is given by ($i=1, 2, 3$), $$ P_{i} = \frac{\partial L}{\partial \dot{X^{i}}}.\tag{2} $$

This definition works perfectly fine for the 3 spatial components of relativistic momentum and gives $$P_{i} = m\dot{X^{i}}.\tag{3}$$

However, for the time component of 4-momentum, Susskind uses the relativistic Hamiltonian to derive $$P_{0} = m\dot{X^{0}}.\tag{4}$$

I am aware that the time component of 4-momentum corresponds to the energy, but I would like to know why we can't use the Lagrangian definition: $$P_{0} = \frac{\partial L}{\partial \dot{X^{0}}}\tag{5}$$ here.

I am new to this subject and would be really grateful for any help or insights.

Answer 1

That's a good question.

1. Note first of all that it is inconsistent to use proper time $\tau$ as the world-line (WL) parameter $\lambda$ for the principle of stationary action (PSA). The point is that the WL parameter $\lambda$ is never varied in the PSA, but the action $S$ happens to be proportional to $\tau$, which we are trying to maximize. In particular, the rightmost expression $-m_0c^2\left(\frac{dx^{0}}{d\tau}\right)^{-1}$ in OP's eq. (1) cannot be used as an off-shell formula for the Lagrangian $L$, although correct in value. The same issue is discussed in my Phys.SE answers here & here using slightly different words.

2. In Ref. 1 the WL parameter $\lambda=t\equiv \frac{x^0}{c}$ is instead the laboratory time, i.e. it uses the static gauge where $\dot{x}^0=c$. (In this answer dot means differentiation wrt. $\lambda$.) Conceptually this is the easiest route. However, this destroys manifest (but not actual) Lorentz covariance, so the derivative $\frac{\partial L}{\partial \dot{x}^0}$ does not make sense. Ref. 1 therefore obtains the 0-component $p_0$ in a roundabout manner, which is equivalent to my Phys.SE answer here.

3. Finally, let us return to OP's question: Yes, there exists a manifest Lorentz covariant formulation where $p_0=\frac{\partial L}{\partial \dot{x}^0}$, but it involves gauge symmetry and constraints, and is conceptually more challenging, cf. e.g. my Phys.SE answers here & here.

References:

1. L. Susskind & A. Friedman, Special Relativity and Classical Field Theory: The Theoretical Minimum, 2017; p. 102-106.

Answer 2

$X_i$ is being differentiated with respect to the proper time $X_0$ only. So if you consider the derivative of $X_0$ with respect to $X_0$, that's one, and hence $d(\dot{X_0})$ is identically zero! However, if you wish to use the Lagrangian only to calculate energy, you can appeal to Noether's theorem and calculate the Noether's charge corresponding to time translations. I hope this helps.