The action of a relativistic free particle is

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



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 :)

  • $\begingroup$ It is the same thing actually, but instead of working with vectors, you have to work with four-vectors and the derivative will be with respect to the proper time. I.e. substitute x with $x^{\mu}$ and $d/dt$ with $d/d\tau$ $\endgroup$
    – schris38
    May 14 at 16:30
  • $\begingroup$ Also, I think what is meant by the term "Minkowski distance" is actually a spacetime distance instead of just spatial distance $\endgroup$
    – schris38
    May 14 at 16:32
  • $\begingroup$ The only place that $x^{\mu}$ will appear in the lagrangian is in the $\gamma$ factor. Is this what you mean? $\endgroup$ May 14 at 16:33
  • 1
    $\begingroup$ Yeh that's what i meant by Minkowski distance, sorry. $\endgroup$ May 14 at 16:33
  • $\begingroup$ Emmm I think no. The action, as you correctly point out is given by $S=-m\int ds$, where $ds$ is some infinitesimal line element in that spacetime (i.e. Minkowski in your case). The latter line element is can be given in terms of the metric, $ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$. So there you have your four vectors. It is usually re-written in the following form as well: $ds=\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}} d\tau$, where $\tau$ is the proper time (time experienced by the particle, whose action is the one we study. $\endgroup$
    – schris38
    May 14 at 16:37

2 Answers 2

  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.

  • $\begingroup$ This almost makes sense, i just want to clarify something. When you obtained $\Delta \tau$ by simply integrating over $\int d\tau$. The reason we can do this is because $\tau$ is a Lorentz scalar that neither $\gamma$ nor $m$ depend on. I think i was getting confused because $\gamma$ depends on speed terms $\frac{dx}{dt}$. I thought that we would have to integrate $\gamma$ because $t$ isn'y independent of s $\endgroup$ May 14 at 20:41
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    May 15 at 5:37
  • $\begingroup$ Makes more sense now. So in the (+,-,-,-) convention, we are maximising the spacetime distance, which minimises the spatial distance in a given frame (as you would expect classically. The same is true in the (-,+,+,+) convention except this corresponds to minimising the spacetime distance. They both have the same magnitude but opposite sign, so essentially the trajectory of a free particle will minimise the magnitude of the spacetime distance |$ds$|. $\endgroup$ May 15 at 10:13
  • $\begingroup$ In the $(-,+,+,+)$ convention spacetime distance (of a point particle) becomes imaginary. $\endgroup$
    – Qmechanic
    May 15 at 10:18

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.

  • $\begingroup$ Thanks, this gave me a different perspective. I suppose if you make the assumption that the spacetime distance propagated in the restframe ($t_{2}$-$t_{1}$^{2}) [(+,-,-,-) convention] is a an extremal point of the action then since spacetime distance is a Lorentz scalar we can say it's minimised in all frames? $\endgroup$ May 15 at 10:32
  • $\begingroup$ @AdrienAmour the spacetime distance for any path is the same in all frames. The above is the computation for two different paths between $t_{1}$ and $t_{2}$ $\endgroup$ May 16 at 13:49

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