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As argued in Jackson p. 580, the quantity [tex] \gamma L [/tex] is invariant. So imagine a free particle. In the particle's frame, the particle can be treated non-relativistically since its v << c (it's zero). But non-relativistically we define the [tex] L = T - V [/tex]. In the particle's frame, this is zero. So invariance would say L would be zero in all frames.

I think the way out of this is to argue that we can only define a Lagrangian up to a total time derivative of some function. In this case the function is [tex] -t mc^2 [/tex]. Consequently, in the particle's frame [tex] L = T - V - mc^2 [/tex]. Doing this we get the the standard relativistic Lagrangian for a free particle.

My Questions:

Why is the added function mc^2? This argument seems rather ad hoc. Also, in introductory physics courses why don't we just define the Lagrangian as [tex] L = T - V - mc^2 [/tex]. Thanks.

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# Relativistic Lagrangian of a Free Particle

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