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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.

Qmechanic
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Anon21
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2 Answers2

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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.

Qmechanic
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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.

doetoe
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