## Why the Lagrangian is kinetic minus potential energy (at least for gravity)

Warning: This is a higher level post, assuming some familiarity with Lagrangian mechanics and general relativity.

When the aspiring physics student is first exposed to Lagrangian mechanics, which differs from Newtonian mechanics in that it focuses on scalars and energies rather than vectors and forces, the question always and everywhere arises: “But why is it kinetic minus potential energy?”

The answers usually provided are deeply unsatisfactory. Things like “Well, we’ll soon show that it’s equivalent to Newton’s Laws, so if you accept those than that should be enough.” The initial construction[1] just seemed so deep and careful, where we posit that space is homogeneous (to remove absolute position from the Lagrangian) and isotropic (so only the magnitude of velocity matters, not it’s direction) and that time is homogeneous (to remove time from the Lagrangian) and the principle of relativity which requires that the Lagrangian vary only linearly with the quantity $\displaystyle v^2$. But when the time comes to include other particles, we get a big, dirty, mysterious minus sign.

When we move up to a higher plane though, things start to become a lot cleaner. The action for a relativistic particle is
$\displaystyle S = -mc^2 \int d\tau$
which is a lot more aesthetically pleasing. Something needs to be extremized, an the proper time is a natural fit. When we move to 4-vectors we find that $\displaystyle c^2$ is a naturally invariant representation carrying the right units we require.

The Schwarzschild metric for a spherically symmetric mass like the Earth is
$\displaystyle d\tau^2 = \left( 1 - \frac{2GM}{rc^2}\right) dt^2 - \frac{1}{c^2}\left( 1 - \frac{2GM}{rc^2}\right)^{-1} dr^2 -\frac{r}{c^2} \, d\Omega^2.$
For an object near Earth’s surface, we can simplify this expression by neglecting $\displaystyle d\Omega^2$ since anything we happen to drop will move radially. Moreover, the quantity $\displaystyle 2GM/rc^2$ is very small so it can be simplified further to (and don’t let this bother you, since classical mechanics is itself a simplification)
$\displaystyle d\tau^2 = \left( 1 - \frac{2GM}{rc^2}\right) dt^2 - \frac{1}{c^2}dr^2.$
Why did we simplify just for the space part and not the time part? The space part had the additional $\displaystyle 1/c^2$ making it very small.

So let’s plug this expression into our relativistic action. We get
$\displaystyle S = -mc^2 \int \sqrt{\left( 1 - \frac{2GM}{rc^2}\right) dt^2 - \frac{1}{c^2}dr^2} = -mc^2 \int \sqrt{\left( 1 - \frac{2GM}{rc^2}\right) - \frac{v^2}{c^2}}dt.$
We can use the binomial approximation to simplify this to
$\displaystyle S \approx -mc^2 \int 1 - \left(\frac{GM}{rc^2} + \frac{v^2}{2c^2}\right)dt$
or multiplying the prefactor in
$\displaystyle S \approx \int -mc^2 + \left(\frac{GMm}{r} + \frac{1}{2}mv^2\right)dt.$
But constant terms in the Lagrangian don’t matter, so we can toss out the $\displaystyle -mc^2$ and further note that the gravitational potential energy is
$\displaystyle U = -\frac{GMm}{r}$
so we’ve recovered the classical
$\displaystyle S = \int \left(T - U\right)dt.$
We can now trace back where the mysterious minus sign came in. The various flips came from the difference in sign between the space and time components of the metric, the square root approximation, and the fact that gravity has an attractive potential.

Hopefully, the classical Lagrangian is now at least one step less mysterious.

References:
[1] See for example Landau-Lifshitz