Special Relativity Experts?

Question

Explain why the relativistic Lagrangian is expressed as L = T' - U, where

T' = - m0 c² √(1-(v/c)²)

instead of L = T - U, where T is the relativisitc kinetic energy,

T = (m0 c²) / √(1-(v/c)²) - m0 c²

In particular, what's the "analytic continuation" from Newtonian to Relativistic in this case?

Yeah, it's the "why not?" that I like to know about. I'm looking for any answers talking about the "why" and "why not".

Answer 1

UPDATED:
Basically, because when you use your guess for the Lagrangian and plug it into the Euler-Lagrange equation, it doesn't give the right answer for the equations of motion, which we know by "relativizing" Newton's equations.

Why not? I don't know if there's an answer to that question.

Update_1:

Jackson's "Classical Electrodynamics" gives an elementary approach to the relativistic Lagrangian: He requires the action integral to be an invariant, therefore
Integral (L) dt must be invariant, therefore
Integral (L*gamma) dτ must be invariant, therefore
L*gamma must be invariant

where gamma = 1/sqrt(1 - (v/c)^2)

So therefore:
L must be proportional to 1/gamma = sqrt(1 - (v/c)^2)
which gives you everything except the rest-mass factor; but that's a constant anyway.

If you demand that the classical method of deriving the Lorentz force works ( by substitution of (p - eA/c) for p in the Lagrangian), you end up with the U term for the case of electrodynamics at least.

If you don't find this argument compelling, keep in mind that physics is not mathematics: It is not a completely deductive structure. Relativity really IS a different theory than Newtonian mechanics. So if it shows up in the Lagrangian as a difference in functional dependence, that's just the way the cookie crumbles, sometimes.

It's ironic, because in particle physics everybody works with the Lagrangian, it's sacred in transitioning from ordinary QM to quantum field theory, etc. But they start from the relativistic Lagrangian. Getting to the relativistic Lagrangian doesn't seem so self-evident.

Good luck and have fun looking for a further explanation.

Answer 2

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