2

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.

PNS
  • 2,156
a_point_particle
  • 95

2 Answers2

6

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.
Qmechanic
  • 201,751
1

$X_i$ 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.

PNS
  • 2,156
Ayush Raj
  • 469