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In classical physics, the action $S$ is defined as the time integral of the Lagrangian $L$, i.e.

$$ S = \int_{t_0}^{t_1} L\, dt. $$

What is the relation between the action, for a point particle, and the Lagrangian in general relativity? If we integrate $L$, do we integrate it over the time $t$ as measured by some observer, the proper time $\tau$, or can we even use any parameter $\lambda$ that is used to parameterize the worldline(s) of the object(s) in the system?

Qmechanic
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HelloGoodbye
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  • @G.Smith The action for a point particle is what I need to know now. But I wouldn't mind learning how the action for a field is defined as well. I know how it is defined in QFT for flat spacetime (it's just the integral of the Lagrangian density over d^4x), but I don't know whether it differs if your space-time is curved. – HelloGoodbye Apr 11 '21 at 19:17
  • Possible duplicates: https://physics.stackexchange.com/q/618080/2451 and links therein. – Qmechanic Apr 11 '21 at 19:26
  • The important thing is that the integrand be a scalar. If you integrate over the proper time, then $L$ will also be a scalar. If you integrate over $t$, it won’t be. – G. Smith Apr 11 '21 at 19:29
  • @G.Smith Why wouldn't $L$ be a scalar if you integrate over $t$? Since no (time-like) particle can travel backwards in time, there is a one-to-one correspondence between $t$ and $\tau$, so it seems to me like you would just be able to change integration variable from $\tau$ to $t$ by multiplying the integrand by $\displaystyle\frac{d\tau}{dt}$, which doesn't make the integrand non-scalar? – HelloGoodbye Apr 11 '21 at 19:52
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    $L,dt$ must be a scalar, but $dt$ is not a scalar. If $L$ were a scalar, $L,dt$ would not be. – G. Smith Apr 11 '21 at 20:03
  • @G.Smith If $dt$ is not a scalar, then what is it? I thought it was an infinitesimal increase of the time component of the four-vector of the particle's spacetime position, $x^0=t$. Isn't this a scalar, or are you talking about some other $t$? – HelloGoodbye Apr 11 '21 at 21:15
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    In relativity, as in other areas of physics, “scalar” doesn’t mean “one number”; it means “invariant under coordinate transformations”. Components of four-vectors are not invariant; the four-vector has a transformation rule. – G. Smith Apr 11 '21 at 22:19
  • @G.Smith Ah, I get it. So why does $L,dt$ have to be invariant under coordinate transformations (I take that to mean Lorentz invariant)? – HelloGoodbye Apr 12 '21 at 01:58
  • If the action is invariant, the equations of motion will have the same form in all reference frames, which is what we want for fundamental laws. See https://physics.stackexchange.com/q/144389/ – G. Smith Apr 12 '21 at 03:02

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