Why does the Minkowski matrix appear in the free particle action?

Question

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?

Answer 1

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.

Answer 2

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.