-3

enter image description here

I can provide the resource for where this is from. Can somebody explain how to get this expression?

Qmechanic
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jerrymor
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    "I can provide the resource for where this is from" Please do. It's of no use to state that you can provide the source and then not provide it. – ACuriousMind Sep 01 '18 at 17:31

2 Answers2

1

it is not "derived" from anywhere. It is coooked up so that the resulting equations of motion are the ones that are desired. The action is not unique.

mike stone
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1

  1. OP's action (1.2) can be viewed as the "point-particle Polyakov action" in the $e=1/m$ gauge, cf. this Phys.SE. (Here $g^{\text{OP}}_{\mu\nu}=m^2g_{\mu\nu}$.)

  2. Alternatively, one can argue that the Lagrangian for a massive point particle/geodesic can be defined with or without a square root, cf. this & this Phys.SE posts.

Qmechanic
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