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What is the relativistic action of a massive particle?

one answer suggests that the action would be (negative of) the rest energy times the change in proper time.

$$I = -m_0c^2d\tau = -E_0d\tau$$

Why is this so?

It makes sense dimensionally, but classically we think of action as the Lagrangian integrated over the time parameter. Are we just assuming $V = 0$ in the Lagrangian since the potential is realised in the curvature of spacetime?

As in, where does the $m_0c^2$ come from?

  • 1
    $\begingroup$ You are missing the integration symbol. The action is $I = - m c^2 \int d\tau = - m c^2 \Delta \tau$. $\endgroup$
    – Prahar
    Feb 25, 2021 at 11:34

2 Answers 2


Let's start with the free particle. For a free particle, it's natural to assume that the action is proportional to its proper time, $$ Ldt=\alpha\left(m,\boldsymbol{x},\dot{\boldsymbol{x}}\right)d\tau $$ where $\alpha$ is some function that should take care of how the Lagrangian of the free particle depends on position and velocity. But a little reflection shows you that $\alpha$ (a universal function) should not depend either on position (otherwise the particle wouldn't be free) or on its frame-dependent velocity (otherwise the principle of relativity wouldn't be satisfied). So you're left with, $$ Ldt=\alpha\left(m\right)d\tau $$ Why $\tau$, proper time? Because that's essentially the only relativistic invariant that's proportional to coordinate time. It's the version of time elapsed that all inertial observers agree upon. You can now do a Taylor series expansion in powers of $v/c$, while keeping $\alpha$ there, and check that it must be $-mc^2$ (with the sign included) if you want the action to fit $L=\text{K.E.}$ (non-relativistic kinetic energy) for the free particle. Instead, I'm just going to plug in that value: $$ L=-mc^{2}\sqrt{1-v^{2}/c^{2}} $$ $$ \left\Vert \boldsymbol{v}\right\Vert \ll c\Rightarrow-mc^{2}\sqrt{1-v^{2}/c^{2}}\approx-mc^{2}\left(1-\frac{v^{2}}{2c^{2}}+\cdots\right)= $$ $$ =-mc^{2}+\frac{1}{2}mv^{2}+\cdots= $$ The only difference with the classical non-relativistic action is this rest-mass term, but that doesn't affect the equations of motion. The other part of the Lagrangian (the interaction term) is just conjectured to be the way it's usually assumed, because the natural extrapolation to interacting particles is, $$ -m_{1}c^{2}d\tau_{1}-m_{2}c^{2}d\tau_{2}+L_{\textrm{int}}dt $$ The interaction term you write in terms of coordinate time because it's the only one that both particles "share". But it's not a relativistic invariant. This is the most delicate point of the whole story. So you would have to check that $L_{\textrm{int}}dt$ is overall an invariant. The whole discussion gets complicated at this point, so let's assume that subsystem 2 acts as a field source for subsystem 1, so that it remains unaffected by it. Then you can ignore the motion of 2, assume that the interaction term goes like, $$ L_{\textrm{int}}=f\left(\boldsymbol{x}_{1}-\boldsymbol{x}_{2},\boldsymbol{v}_{1}-\boldsymbol{v}_{2}\right) $$ And this is what we call (minus) the potential energy. The fact that it must depend only on differences of coordinates (and their derivatives, but in a very restricted way) is down to the fact that space is symmetric and it should be impossible to detect where I put the whole thing. I hope that was clear, and you're welcome to ask for further clarification.

  • $\begingroup$ Right now I'm busy, but I left something out that connects with @Andrew Steane's use of the word "simplicity" and with OP's concern with GR. I don't want to give the impression that the above is the only way to define an action principle for relativistic particles. Of course, it's not. It's just a no-nonsense, mainstream set of assumptions that lead to a relativistic generalisation of the AP for non-relativistic particles. But I deliberately left out GR and diffeomorphism invariance, as well as other possible action-principle formulations (I'm thinking the Jacobi-Maupertuis AP). Maybe later. $\endgroup$
    – joigus
    Feb 25, 2021 at 14:36

To ask about an action is to ask about a Lagrangian. And to ask about a Lagrangian, at least in particle physics, is to reach down to the basic premises of the subject. There is not anywhere deeper to go to explain why this or that Lagrangian is the right one, except by appeal to issues of simplicity and symmetry. In the present example one wants a Lagrangian which is as simple as possible while still leading to some sort of interesting behaviour, and which also is invariant with respect to translation in space and time (if one is considering an isolated particle) and, if one is adopting proper time in the action integral, then one wants a Lorentz scalar. So one considers the amazingly simple ${\cal L} = m c^2$. One discovers that to get the right momentum one needs ${\cal L} = -m c^2$. And hey-presto! there it is: conjured out of nothing but simplicity, symmetry, and covariance. The "proof" that it is right is that it leads to dynamics that are consistent with experiment. To discuss dynamics more fully one needs other terms in the Lagrangian, such as interaction terms, but even with this simple Lagrangian one can treat energy and momentum conservation and thus get insight into particle collisions.

Added remark

After a helpful comment exchange with my2cts, I realised the above is perhaps a little too brief to be really helpful. The more full statement of the Lagrangian in a manifestly covariant approach is $$ {\cal L} = - mc(-u^\mu u_\mu)^{1/2} $$ which previously I abbreviated to $-mc^2$ because that is indeed its value along the worldline which the particle actually follows. However, when using this in the Euler-Lagrange method you need to know its dependence on the 4-velocity for other paths, and this is why the full statement (just given) is needed. The action is then $S = \int {\cal L} d\tau$ and the Euler-Lagrange equation is $$ \frac{d}{d\tau} \left( \frac{\partial {\cal L}}{\partial u^a} \right) = \frac{\partial {\cal L}}{\partial x^a}, $$ where $u^a = dx^a/d\tau$.

However, for anyone learning the subject for the first time I think there are good arguments to introduce the treatment in terms of coordinate time in the first instance. Such a treatment adopts $$ \tilde{\cal L} = - m c^2 / \gamma $$ and $S = \int \tilde{\cal L} dt$, leading to Euler-Lagrange equation $$ \frac{d}{d t} \left( \frac{\partial \tilde{\cal L}}{\partial \dot{x}^a} \right) = \frac{\partial \tilde{\cal L}}{\partial x^a}, $$ where $t$ is coordinate time in some given inertial frame, and the dot denotes $d/dt$.

  • $\begingroup$ This answer is not clear to me. Is m the relativistic mass? This is an obsolete concept and if indeed the case then this is not the correct Lagrangian. $\endgroup$
    – my2cts
    Feb 25, 2021 at 16:43
  • $\begingroup$ @my2cts aargh NEVER use $m$ for anything other than rest mass. That is what it is in any modern treatment of relativity or particle physics. That is what it is. That is what it evermore shall be. So you never need the subscript zero. $\endgroup$ Feb 25, 2021 at 17:17
  • $\begingroup$ I agree , but then a factor $\sqrt{1-v^2}$ is missing. $\endgroup$
    – my2cts
    Feb 25, 2021 at 17:54
  • $\begingroup$ @my2cts There are two ways to formulate Lagrangian method in relativity. If you take an action integral with respect to coordinate time then you need the $\gamma$ factor in the Lagrangian. This is a perfectly good method but I think the questioner wished to take an action integral with respect to proper time. If you take that approach then the Lagrangian has to be a Lorentz scalar, and it does not have a $\gamma$. The two approaches can be related by using $dt/d\tau = \gamma$. $\endgroup$ Feb 25, 2021 at 18:04
  • $\begingroup$ I'm lost. What is the Euler-Lagrange equation in terms of proper time? $\endgroup$
    – my2cts
    Feb 25, 2021 at 18:09

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