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Let

$$ S= m \int_a^b \dot{x}dt $$

Using the relation $L\to L^2/2$, (see Geodesic Equation from variation: Is the squared lagrangian equivalent?)

I obtain

$$ S=m\int_a^b\frac{1}{2}(\dot{x})^2dt $$

Adding a potential,

$$ S=m\int_a^b\left( \frac{1}{2}(\dot{x})^2-V \right)dt $$

which is the well-known formulation $L=T-V$

So the Lagrangian of a classical non-relativistic free particle is just $\dot{x}$?


In multiple dimensions

$$ S=m\int_a^b \sqrt{\dot{x}^2+\dot{y}^2}dt $$

with $L \to L^2/2$

$$ S=m\int_a^b \left( \frac{1}{2}\dot{x}^2+ \frac{1}{2}\dot{y}^2 - V \right) dt $$


As a special case of general coordinates:

$$ S=m\int_a^b \sqrt{ g_{\mu\nu}\partial X^\mu \partial X^\nu }dt $$

in 0+1D, I get

$$ g=\pmatrix{1} \implies S=m\int_a^b \sqrt{ (\partial X^0)^2 }dt $$

which is the equation I started with.

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    $\begingroup$ Also, all SE posts are version controlled, so please do not make your post look like a revision table, instead just seamlessly integrate the new material into the post. There is an edit history button at the bottom of the post for those interested in seeing what changed. $\endgroup$ – Kyle Kanos Aug 23 '19 at 11:57
  • $\begingroup$ Did you try to derive the equation of motion for your suggested Lagrange function $L\sim \dot x$? $\endgroup$ – Toffomat Aug 23 '19 at 12:09
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    $\begingroup$ @Toffomat I just did and yes it fails - thanks for the suggestion. Can you look at my last edit and explain why it fails when $g=[1]$ (a $1\times 1$ matrix). I would have thought $L=\int_a^b \sqrt{g_{\mu\nu} \partial X^\mu \partial X^\nu} dt$ worked with any metric. But clearly, it fails when $g=[1]$. $\endgroup$ – Anon21 Aug 23 '19 at 12:22
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    $\begingroup$ Just a remark on terminology: the Lagrangian is the integrand of the expression you write, whereas the integral you denote by $L$ is the action. You actually (correctly) apply $L\mapsto L^2/2$ to the integrand. $\endgroup$ – doetoe Aug 23 '19 at 16:07
  • $\begingroup$ You aren't allowed to just square a Lagrangian. $\endgroup$ – knzhou Aug 24 '19 at 17:33
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  1. No, a Lagrangian term $$L~\sim~ \dot{x}\tag{1} $$ is a total time derivative term, and hence doesn't contribute to the Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post. Moreover, such term, as written, only makes sense in 1D.

  2. If we instead consider speed $$L~\sim~ |\dot{\bf x}|~\equiv~\sqrt{\dot{\bf x}^2},\tag{2}$$ i.e. the length of the velocity vector, the non-relativistic Lagrangian (2) wouldn't be differentiable at zero velocity, cf. e.g. this Phys.SE post. (The square root is not an issue for differentiability of the corresponding relativistic Lagrangian, since a massive point particle has a timelike 4-velocity.)

  3. Note that generically, one is not allowed to replace a Lagrangian with its square (or square root), cf. e.g. this Math.SE post.

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    $\begingroup$ I would have thought $L=\int_a^b \sqrt{g_{\mu\nu} \partial X^\mu \partial X^\nu} dt$ was valid for any metric. But, it fails when $g=[1]$ (a $1 \times 1$ matrix)? This is surprising to me. $\endgroup$ – Anon21 Aug 23 '19 at 12:28
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    $\begingroup$ @Alexandre that metric is the spacetime metric, so you need at least two dimensions. $\endgroup$ – Javier Aug 23 '19 at 12:52
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The first integral you write down is proportional to the Euclidean path length in 1D (for monotonous paths), or in general if you would replace $\dot x$ by $\|\dot x\|$. The integration parameter is just that, a parameter for the path, and it has no relation with time, so this Lagrangian doesn't encode any dynamics. Any reparameterization will preserve the path length. In higher dimensions it will still give you the correct path, but there is no relation with time.

For a Lagrangian $|\dot x|$ in 1D, one might want to say that the fact that the Euler-Lagrange equation is not well-defined in 0, and vacuous outside of 0, reflects that every path gives you a minimal path as long as it doesn't change direction, i.e. $\dot x$ doesn't change sign.

The second expression integrates the kinetic energy, and indeed, a path that minimizes the kinetic energy minimizes the length, but the other way around doesn't have to hold, and in fact, it depends on the parameterization. The latter does encode dynamics.

When we are in general relativity, things are a bit different. Now time is one of the coordinates, and not the parameter of the path (one could always reparameterize of course to give the parameter meaning). In this case, any extremal of the path length will give you the correct solution, but now the time dependence is in the coordinates, not in the parameter. If in this situation we replace the path-length functional by the energy functional, we find true geodesics. The curves are the same, but now the parameter does have a meaning: for timelike paths it is the proper time.

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