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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?

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A Lagrangian isn't automatically Lorentz invariant, you have to construct it to be so. So I'd lean toward the answer being "no," though for an actual answer I will defer to someone who can say so more definitively. –  David Z Jun 7 '12 at 1:16
    
Thanks! I'll try to rephrase the question later when I get the time. I'd intended to focus more on whether there exist functions that can be derived in parallel from the hyperbolic and Euclidean views, and if some of those are well-known. –  Terry Bollinger Jun 7 '12 at 15:03
    
Related: physics.stackexchange.com/q/50075/2451 –  Qmechanic May 26 '13 at 17:49
    
Qmechanic, thanks, that is a good reference. –  Terry Bollinger May 26 '13 at 17:52
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1 Answer 1

up vote 1 down vote accepted

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.

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Useful, thanks! I'll try to rephrase after reading up. Any comments on hyperbolic/Euclidean aspect of the question? –  Terry Bollinger Jun 7 '12 at 14:59
    
Karsus Ren, thanks, I was throwing that one together quickly and your reference (and also @DavidZaslavsky's comment) are helpful. I'm accepting your answer rather than trying to rephrase, and will post a new question if there's something there for me to ask. –  Terry Bollinger Jun 8 '12 at 0:52
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