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There are a few questions about a Lagrangian for massless relativistic particles, notably here, here and here, regarding free particles in particular.

In the case of the square-root Lagrangian for massive particles, adding a potential term (such as a Coulomb potential energy) was no different to the non-relativistic case.

Is the same still true in the case of the Lagrangian without the square-root? My instinct is no, due to the difference in parameterisation and presence of the einbein. If not, is there a way to determine how the potential should be expressed?

Qmechanic
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tomdodd4598
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1 Answers1

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  1. Let us parametrize the point particle by an arbitrary world-line (WL) parameter $\tau$ (which does not have to be the proper time).

  2. Let us now address OP's question: Yes, even for the non-square-root Lagrangian $$\begin{align}L~=~&\frac{\dot{x}^2}{2e}-\frac{e (mc)^2}{2} -V, \cr \dot{x}^2~:=~&g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0,\cr \dot{x}~:=~&\frac{dx^{\mu}}{d\tau}, \end{align}\tag{A}$$ with the einbein field $e>0$ one just subtracts the potential $V$ as usual.

  3. As a consistency check (in the massive case $m>0$ and assuming that $V$ does not depend on $e$), we can integrate out the einbein field $e$ to obtain the usual square-root Lagrangian $$ L_0~=~ -mc\sqrt{-\dot{x}^2}-V,\tag{B}$$ cf. e.g. this Phys.SE post.

  4. Geometrically, we should point out that it is implicitly assumed that the 1-forms $$ e\mathrm{d}\tau, \qquad V \mathrm{d}\tau, \qquad L_0\mathrm{d}\tau, \qquad L\mathrm{d}\tau,\tag{C}$$ (and the corresponding actions) are invariant under WL reparametrizations
    $$ \tau\longrightarrow \tau^{\prime}=f(\tau). \tag{D}$$ In other words, the WL reparametrization (D) is a gauge symmetry. This means that one can choose a gauge-fixing condition, cf. e.g. this, this, this & this related Phys.SE posts.

  5. Concerning coexistence of square-root & non-square-root Lagrangians, see also e.g. this Phys.SE post.

Qmechanic
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