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The action of a relativistic free particle is

$$\mathcal{S}=\int^{t_{1}}_{t_{0}} L dt\tag{1},$$

for

$$L=-\frac{mc^{2}}{\gamma}.\tag{2}$$

I understand that a particle will follow the trajectory of stationary action. I've read that from this Lagrangian, one can deduce that the trajectory of a particle is one which minimises the Minkowski distance (i.e. a geodesic). I'm not entirely sure how we can deduce that here. Classically I can see this, since for a free particle, with velocity in the $x$-direction, we can say that $L=\frac{1}{2}m\dot{x}^{2}$. Therefore

$$\begin{split} \mathcal{S} &=\int^{t_{1}}_{t_{0}}\frac{1}{2}m\dot{x}^{2}dt \\ &=\int^{x_{1}}_{x_{0}}\frac{1}{2}m(\frac{dx}{dt})^{2}\frac{dt}{dx}dx \\&=\int^{x_{1}}_{x_{0}}\frac{1}{2}mvdx \\&=\frac{1}{2}mv(x_{1}-x_{0}) \end{split} \tag{3}$$

Hence, classically, minimising the distance travelled, $x_{1}-x_{0}$, will minimise the action. Unfortunately, it's not clear to me why relativistically a particle will travel the minimum minkowski distance. Also, it's not clear to me whether or not the Minkowski distance is minimised for the trajectory of a general particle, or just that of a free particle. Any help is really appreciated :)

Shiki Ryougi
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Adrien Amour
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2 Answers2

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  1. It is straightforward to derive that OP's action (1) is $$S~=~ - m_0c ~ \Delta s, $$ where $$\Delta s~=~c\Delta\tau $$ is the spacetime distance $$\Delta s ~=~\int \!ds$$ in the $(+,-,-,-)$ sign convention, $$(ds)^2~=~g_{\mu\nu}dx^{\mu} dx^{\nu},$$ cf. e.g. this Phys.SE post.

  2. Therefore a minimum for the action $S$ is the same as a maximum for the spacetime distance $\Delta s$.

  3. The fact that the straight line of a free particle maximizes (rather than minimizes) the spacetime distance $\Delta s$ follows from the reversal of the triangle inequality in Minkowski spacetime.

Qmechanic
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Let's just justify this physically without making an explicit appeal to the Lagrangian math.

Assume the Minkowski metric and $c=1$ units. We already know, if the theory is going to make any sense at all, that the "stationary" path of a particle not moving through space, and going between the points $(t_1, 0, 0,0)$ and $(t_2, 0,0,0)$ has to be a path, and we also know, being a physical path that can be realized in the absence of any external forces, that it has to be an extremum of the relevant action, with length $s_{0}^{2} = -(t_{2} - t_{1})^{2}$. Is it a minimum or a maximum?

Well, now, choose another path, where the particle moves out at some speed $v < 1$ (remember, we've chosen $c=1$ units) and turns around at time $\frac{1}{2} (t_1 + t_2)$, to end up back at the origin at time $t_2$. then, our path is two segments:

$$(t_1 ,0,0,0) \rightarrow (\frac{1}{2}(t_1 + t_2), \frac{1}{2}v(t_2 -t_{1}),0,0)\;\;\; {\rm and}\;\;\; (\frac{1}{2}(t_1 + t_2), \frac{1}{2}v(t_2 -t_{1}),0,0) \rightarrow (t_{2}, 0,0,0)$$

so, we get the total Lorentz distance is (the factor of two on the LHS comes from the fact that the two segments have the same s, and we can just halve $s$ and have only one segment on the RHS:

$$\begin{align} (\frac{1}{2}s)^{2} &= -(\frac{1}{2}(t_{2} - t_{1})^{2} + (\frac{1}{2}v(t_{2}-t_{1})^{2})\\ s^{2}&= - (t_2 - t_{1})^{2}(1 - v^{2}) \\ &= s_{0}^{2}(1 - v^{2}) \end{align}$$

since $v$ must always be less than 1, the right hand side always indicates a timelike direction, and also, since the factor $(1 - v^{2})$ is always between 0 and 1, $s$ must always be less than or equal to $s_{0}$ for any path of this form. Therefore, inertial particle paths must maximize the Minkowski distance between two spacetime points, not minimize it. Note that the result just follows from the physics telling us that the stationary path is an extremum, and deriving one type of perturbation from that path that clearly has a shorter spacetime distance.

Zo the Relativist
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