I was reading about the formulation of mechanics in special relativity and found that the action for a massive free point-particle as $$ S = -mc\int_a^b ds $$ So, I did a few observations, ie. $$ S = -mc^2 \int_a^b dt $$ and being $$u_{\mu}u^{\mu} = c^2$$ I wrote the action as $$ S = -\int_a^b m u_{\mu}u^{\mu} dt $$ which in my opinion resembles more to the classical kinetic energy.

Now, I never saw something like this in a book or text on internet. Everyone seems to work with the classical speed and the old good Euler-Lagrange equations with time as parameter.

So my question is: it's possible to derive the correct equations of relativistic dynamics from this action?

Attempt of solution

The action is $$ S = \int_a^b L(x,v,t) dt $$ and it should be Lorentz invariant (the action or the lagrangian?). Now, changing the parametrization of the path with $d\tau$ instead of $dt$ shouldn't change the integral (with the condition that everything changes accordingly). So I rewrite the action as $$ S = \int_{\alpha}^{\beta} L(x_{\mu}, \frac{dx_{\mu}}{d\tau}, \tau)d\tau $$ and do the variation of this and minimize it: $$ \delta S = \int_{\alpha}^{\beta} \left[ \frac{\partial L}{\partial x_{\mu}}\delta x_{\mu} + \frac{\partial L}{\partial u_{\mu}}\delta u_{\mu}\right] d\tau = 0 $$ Then, using the old manipulations it yields $$ \frac{d}{d\tau}\left( \frac{\partial L}{\partial u_{\mu}} \right) - \frac{\partial L}{\partial x_{\mu}} = 0 $$

This applied to the lagrangian I wrote $$ L = m \eta_{\mu\nu}u^{\mu} u^{\nu} $$ seems to yield the correct 4-impulse $$ p_{\nu} = \frac{\partial L}{\partial u_{\mu}} = m \eta_{\mu\nu}u^{\nu} $$

but the hamiltonian (total energy in the text I'm reading) is zero:

$$ H = p_{\mu}u_{\mu} - L = mu_{\mu}u^{\mu} - mu_{\mu}u^{\mu} = 0 $$

I think I did wrong the derivation of the Euler-Lagrange equations, but not sure.

  • 1
    $\begingroup$ That the Hamiltonian is zero is completely correct - the action is time-reparametrization invariant, and time-reparametrization invariant actions generically yield zero Hamiltonians (which then do not correspond to the energy). Is there a question here besides "Is this correct?", and if yes, can you make it clearer? $\endgroup$
    – ACuriousMind
    Jun 4, 2016 at 9:57
  • 3
    $\begingroup$ However, note that you did something rather questionable: You used $u^2 = 1$, which is a property of the solutions to the equation of motion, but not of a generic four-vector $u$ on the level of the action where the equations of motion are usually not assumed to hold. $\endgroup$
    – ACuriousMind
    Jun 4, 2016 at 10:00
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    $\begingroup$ Comments to the post (v2): 1. OP's non-square root Lagrangian is discussed e.g. in Goldstein, Classical Mechanics, 2nd edition, Sections 7.9 & 8.4. 2. Note that there should be a half in the Lagrangian $L = \frac m 2\eta_{\mu\nu}u^\mu u^\nu$. Else the canonical Lagrangian momentum $p_\mu = \frac{\partial L}{\partial u^\mu}$ becomes twice $m\eta_{\mu\nu}u^\nu$. 3. This implies that the Hamiltonian is not zero. In fact, it is equal to the Lagrangian, in value. 4. In contrast, the Hamiltonian corresponding to the square root Lagrangian vanishes due to world-line reparametrization invariance. $\endgroup$
    – Qmechanic
    Jun 4, 2016 at 10:10
  • $\begingroup$ @ACuriousMind yes you are right. I forgot to say that $u$ is the 4-velocity of the particle. My question in reality is "Given this lagrangian, how are the Euler-Lagrange equations and if it's possible to obtain the usual dynamical parameters as momentum, energy, etc. using only vectors in Minkowski space and no vectors in old euclidean space". $\endgroup$
    – user71714
    Jun 4, 2016 at 10:39
  • $\begingroup$ @Qmechanic I went to rent the book and understood a lot. I was more or less in the correct way, but now I don't understand why the 1/2 factor. I mean, it works and all, but I derived the lagrangian with the quadratic form on the 4-velocities directly from the lagrangian in the book (the first eq I wrote). $\endgroup$
    – user71714
    Jun 4, 2016 at 12:39

1 Answer 1


Comments to the question (v2):

  1. The Minkowski spacetime can be generalizes to a Lorentzian manifold $(M,g)$. We choose the Minkowski signature $(-,+,+,+)$.

  2. OP evidently knows that the action $$S = -E_0~ \Delta \tau\tag{1}$$ for a massive point particle is minus the rest energy $E_0=m_0c^2$ times the change $\Delta \tau$ in proper time, cf. e.g. my Phys.SE answer here. In more details, eq. (1) is the square root action $$S ~=~ \int_{\lambda_i}^{\lambda_f} \!d\lambda~ L, \qquad L~:=~- m_0c\sqrt{- \dot{x}^2}, $$ $$ \dot{x}^2~:=~g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda}, \tag{2} $$ where $\lambda$ is a world-line parameter.

  3. The canonical Lagrangian $4$-momentum is precisely the mechanical $4$-momentum $$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}} ~=~\frac{m_0c\dot{x}_{\mu}}{\sqrt{-\dot{x}^2}}, \qquad \dot{x}_{\mu}~:=~g_{\mu\nu}~\dot{x}^{\nu}. \tag{3}$$

  4. The Lagrangian energy function $$h~:=~p_{\mu}\dot{x}^{\mu}-L~=~0 \tag{4}$$ vanish identically. This is related to the fact that the square root action (2) has world-line reparametrization invariance, which is a gauge symmetry. Note that the $4$-momentum (3) is reparametrization invariant. One often uses the static gauge $$x^0~=~c\lambda.\tag{5}$$

  5. OP is essentially pondering if one instead of the square root action (2) could use the non-square root action $$\tilde{S} ~=~ \int_{\lambda_i}^{\lambda_f} \!d\lambda~ \tilde{L}, \qquad \tilde{L}~:=~ \frac{m_0}{2}\dot{x}^2~~? \tag{6} $$ The answer is Yes. The corresponding Euler-Lagrange (EL) equations are in both cases the geodesic equation, cf. e.g. this Phys.SE post.

  6. Note that the non-square action (6) does not have world-line reparametrization invariance. Moreover, for a solution to the EL equations, the world-line parameter $\lambda$ and the proper time $\tau$ are always affinely related, cf. my Phys.SE answer here.

  7. The canonical Lagrangian $4$-momentum $$ \tilde{p}_{\mu}~:=~\frac{\partial \tilde{L}}{\partial \dot{x}^{\mu}} ~=~m_0\dot{x}_{\mu}\tag{7}$$ is the mechanical $4$-momentum if we identify the world-line parameter $\lambda$ and the proper time $\tau$. (We stress that it is not possible to make this identification $\lambda=\tau$ before varying the action. The identification $\lambda=\tau$ is only possible on-shell.)

  8. The Lagrangian energy function $$\tilde{h}~:=~\tilde{p}_{\mu}\dot{x}^{\mu}-\tilde{L}~=~\tilde{L} \tag{8}$$ is just the Lagrangian itself.

  9. The non-square root Lagrangian (6) and its corresponding Hamiltonian is discussed in Refs. 1 and 2.


  1. H. Goldstein, Classical Mechanics, 2nd edition, Sections 7.9 & 8.4.

  2. H. Goldstein, Classical Mechanics, 3rd edition, Sections 7.10 & 8.4.


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