3
$\begingroup$

Landau and Lifshitz (Mechanics, Vol. 1) derive the form of the Lagrangian for a free particle by requiring invariance of action under Galilean transformation and assuming homogeneity and isotropy of the space-time. The same can be done for a relativistic free particle (see, here). Now, assuming the form of the Lagrangian to be $L(\mathbf{v}^2)$ is motivated from homogeneity and isotropy of spacetime. For example, $\mathbf{v}^T\boldsymbol{\cdot}\mathbf{v}$ is invarint under rotation (isotropy) and $\mathbf{v}$ itself is invariant under translation $\mathbf{x}\rightarrow\mathbf{x^\prime}=\mathbf{x}+\mathbf{a}$ (homogeneity). In four dimensions one must demand that the action is invariant under 3 boosts, 3 rotations and 4 translations (Poincare transformations) to derive the form of the Lagrangian. Having done that, I want to motivate the Lagrangian for a particle moving in an underlying potential $V(\mathbf{x})$. However, when a potential function is introduced $L(\mathbf{v}^2,\mathbf{x})$, the symmetry under translation (and the rotation as well) cannot be guaranteed. Then, does the introduction of the potential function breaks the Poincare invariance down to a invariance under Lorentz boosts?

Can even the invariance under boosts be guaranteed? If not, then why do we again use the free particle Lagrangian? Basically what I am trying to get at is that how can we argue that the position dependence should come as an addition to the free Lagrangian, which satisfies a symmetry under a bigger group? What justifies the separability: $L(\mathbf{v}^2,\mathbf{x})=L_{\text{free}}(\mathbf{v}^2)-V(\mathbf{x})$? For example, why not use exponential $L(\mathbf{v}^2,\mathbf{x})=L_{\text{free}}(\mathbf{v}^2)e^{-V(\mathbf{x})}$? For $V(\mathbf{x})\rightarrow 0$, this would reduce to $L_{\text{free}}(\mathbf{v}^2)$. (Occum's razor?)

If the case for relativistic particles is described in some source in details in a similar manner as L&L does for non-relativistic case, let me know that as well.

Edit: I realize that there can be potential functions of the form $V(|\mathbf{x}-\mathbf{y}|)$ and this type of functions would respect both homogeneity and isotropy conditions. However, the there could be plenty other potentials as well which are not of this form and may break both homogeneity and isoropy.

$\endgroup$
1
  • $\begingroup$ I mean, my gut instinct would be that just like not all potentials are rotationally symmetric, not all of them are going to be boost symmetric. You imagine, for example, a constant $1/r$ potential centered at the origin: after a boost this is going to look like a potential moving across space, so that doesn't seem like a symmetry. But a $1/\sqrt{r^2-c^2 t^2}$ looks a little more likely. As for why we add Lagrangians, I am probably not the right person to ask. It does not seem to be absolutely necessary, I think for example of electromagnetism where we modify the kinetic energy part. $\endgroup$ – CR Drost Nov 4 '20 at 5:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.