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The Lagrangian of a system is the difference between its kinetic energy $T$ and potential energy $V$, and is relativistically invariant:

$L = T - V$

The Hamiltonian of the same system is the sum of the kinetic and potential energy, but is not relativistically invariant:

$H = T + V$

In special relativity, setting $c=1$ and $q=\sqrt{x^2+y^2+z^2}$ allows the Minkowski hyperbolic distance (interval) between two points in spacetime to be expressed as the relativistically invariant expression:

$s^2 = q^2 - t^2$

While seldom used, the Euclidean distance between two points in spacetime can be defined for classical situations, but of course is not relativistically invariant:

$s_e = q^2 + t^2$

For the last couple of weeks I've been mulling off and one whether these two pairings might be directly related to each other.

That is, is it possible that the relativistic Lagrangian "belongs" with the SR-compliant hyperbolic interpretation of spacetime, while the non-relativistic Hamiltonian "belongs" with less common and non-compliant Euclidean interpretation of spacetime?

Opinions, anyone?

Terry Bollinger
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1 Answers1

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You are confusing Lagrangian with Lagrangian density. The latter is often abbreviated as Lagrangian as well, but actually means a different thing.

The former is in fact not Lorentz invariant. The simplest case: a relativistic free particle has Lagrangian $L=-mc^2 \sqrt{1-\frac{v^2}{c^2}}$ (which, you might notice, is not $T-V$ either). It clearly depends on $v$, which is not invariant.

Lagrangian density, on the other hand, are Lorentz invariant.

Siyuan Ren
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