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It is usual to write the "kinetic" part of the SR action as the Minkowski space-time interval, here $(-,+,+,+)$, times $mc$ $$ S_{kin} = -\int_{\tau_1}^{\tau_2}mc\sqrt{-\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}d\tau $$

yielding the eom $$ \dfrac{d}{d\tau}\left(mc\dfrac{\eta_{\alpha\beta}\dot{x}^{\beta}}{\sqrt{-\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}}\right)=0 $$

which brings forth the constancy of the velocity (principle of inertia).

But actually, any (scalar) function of $\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$ will prove valid to satisfy the principle of inertia. For instance $$ S_{kin} = -\int_{\tau_1}^{\tau_2}\dfrac{m}{2}f(-\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu})d\tau $$

results in the eom $$ \dfrac{d}{d\tau}\left(mf'(-\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}=c^2)\eta_{\alpha\beta}\dot{x}^{\beta}\right)=0 $$

which again agrees with the principle of inertia.

We can take this even further, and use any old constant (covariant) matrix $a_{\mu\nu}$ $$ S_{kin} = \int_{\tau_1}^{\tau_2}\dfrac{m}{2}f(a_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu})d\tau $$

to arrive at $$ \dfrac{d}{d\tau}\left(mf'(a_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu})a_{\alpha\beta}\dot{x}^{\beta}\right)=0, $$

and doesn't this give way to the principle of inertia as well? So why does the Minkowski matrix enter the action? The Lorentz invariance property that $\Lambda^{\mu}_{\alpha}\eta_{\mu\nu}\Lambda^{\nu}_{\beta} = \eta_{\alpha\beta}$ is not used anywhere.

Is there any physical argument why this matrix should be the one inside the action? Perhaps only noticeably in the presence of interactions?

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2 Answers 2

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In the context of relativity, the actions must be Lorentz invariant. Let's set $c=1$. Then it is given by the following formula $$S=-m\int ds$$ where $s$ is the propr time. In a fixed frame: $$S=-m\int dt\sqrt{-\eta_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}}=- m\int dt\sqrt{1- \dot{\vec{x}}\cdot \dot{\vec{x}}}$$ Obviously, in $d=3+1$, there are 3 degrees of freedom: $x_i(t): i=1,2,3$

However, we can rewrite the above actions in a general frame by introducing the world line parameter $\alpha$ $$S=-m\int d\alpha\sqrt{-\eta_{\mu\nu}\frac{dx^{\mu}}{d\alpha}\frac{dx^{\nu}}{d \alpha}}$$ It looks like there are 4 degrees of freedom: $x^{0}(\alpha)$ and $x^{i}(\alpha): i=1,2,3,$. This is not true! Because the above action has one more property (compared to the previous one written in a fixed frame): $$\textbf{reparametrization invariance}$$ which is the statement that the action is invariant under $\alpha\rightarrow \tilde{\alpha}=\tilde{\alpha}(\alpha)$ (easy to check). You can then show that fixing this redundancy is equivalent to reducing the degrees of freedom from 4 to 3. In this way the two actions I wrote are equivalent.

However, in your case, although the actions you propose are Lorentz invariant, they are not reparameterization invariant. Therefore, your action describes the dynamics of 4 degrees of freedom, which is not equivalent to the problem you started with.

Hope this helps.

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    $\begingroup$ See also David Tong Lectures on String Theory: damtp.cam.ac.uk/user/tong/string/string.pdf $\endgroup$
    – Navid
    Nov 1, 2023 at 3:56
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    $\begingroup$ Very interesting! Is there any physical reason why the action should be reparametrization invariant? $\endgroup$
    – K. Pull
    Nov 1, 2023 at 7:58
  • $\begingroup$ Also why use the $\eta_{\mu\nu}$? Wouldn't another tensor $a_{\mu\nu}$ be valid to make the action reparametrization invariant plus Lorentz invariant? $\endgroup$
    – K. Pull
    Nov 1, 2023 at 8:09
  • $\begingroup$ Rep. invariance is not necessary! The action in the fixed frame also works well. However, you don't see all the symmetries obviously. In fact, rep. invariance in our case is similar to gauge symmetry in EM. Then writing action in a fixed frame, is similar to the electromagnetic action in a special gauge. This works, but you can't change the gauge, so we prefer to keep it gauge invariuant (here, rep. invariant); then it has a nice advantage: we can clearly see Lorentz symmetries (and even Poincaré symmetries) of the particle. $\endgroup$
    – Navid
    Nov 1, 2023 at 8:22
  • $\begingroup$ What is a fixed frame? $\endgroup$
    – basics
    Nov 1, 2023 at 9:00
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  1. OP seems to suggest to include another metric tensor $a$. Although bi-metric theories are studied in the literature, there has so far been no physical/experimental evidence of this.

    Here we shall only considering the standard Minkowski spacetime $(\mathbb{R}^{3,1},\eta)$ endowed with the standard Minkowski metric tensor $\eta$.

  2. The Minkowski metric tensor $\eta$ is used to construct Lorentz invariants. Lorentz covariance of the EOM suggests that the Lagrangian one-form should be Lorentz invariant, i.e of the form $$\mathbb{L}~=~ f(-\dot{x}^2)\mathrm{d}\lambda, \qquad \dot{x}^2~:=~\eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}, \qquad x^0~\equiv~ct. \tag{1}$$ Here $\lambda$ denotes a world-line (WL) parameter, and a dot denotes differentiation wrt. $\lambda$. This answers OP's title question.

  3. If we furthermore impose that $\mathbb{L}$ should be WL reparametrization invariant then the function $$f~\propto ~\sqrt{\cdot}\tag{2}$$ is proportional to a square root. See also e.g. my related Phys.SE answer here.

  4. Concerning the interesting fact that the stationary path does not depend on the function $f$, see also e.g. my related Math.SE answer here.

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  • $\begingroup$ Thank you for your answer. 1. Is there any physical principle why the action should be reparametrization invariant? 2. And wouldn't any other non-singular matrix $a_{\mu\nu}$ serve the same purpose as the Minkowski metric? $\endgroup$
    – K. Pull
    Nov 1, 2023 at 8:02
  • $\begingroup$ 1. Well, reparametrization invariance can be viewed as a physical principle. 2. No, not if the matrix $a_{\mu\nu}$ is supposed to be constant in all inertial frames. However, the SR construction with the Minkowski metric $\eta_{\mu\nu}$ can be generalized to a curved spacetime metric $g_{\mu\nu}(x)$ in GR. $\endgroup$
    – Qmechanic
    Nov 1, 2023 at 9:04
  • $\begingroup$ 2. But why not? $a_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$ is a scalar just like $\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}$, as long as the $a_{\mu\nu}$ transform as a covariant tensor $\endgroup$
    – K. Pull
    Nov 1, 2023 at 9:11
  • $\begingroup$ 2. A non-degenerate symmetric tensor $a_{\mu\nu}$ would be a second metric tensor. Mathematically, one may consider this, but physically it is not part of standard SR. It would be extra geometric data. However, bi-metric theories are studied in the literature. $\endgroup$
    – Qmechanic
    Nov 1, 2023 at 9:52

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