Why Must the Term dL/d(v^2) v.e Be Linear in v to Be a Total Time Derivative?

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SUMMARY

The discussion centers on the necessity for the term dL/d(v^2) v.e to be linear in v to qualify as a total time derivative in the context of Lagrangian mechanics, as outlined in section 4 of Landau and Lifgarbagez. The linearity ensures that the equations of motion remain invariant under the transformation. The participants emphasize that the term's dependence on velocities is crucial, as it directly influences the derivation of kinetic energy without altering the dynamics of the system.

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rc75
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In section 4 of Landau and Lifgarbagez they derive the expression for the kinetic energy by expanding the Lagrangian around v+e. The resulting expression has a term which must be a total time derivative so that the equations of motion are unaffected. The text claims that the term dL/d(v^2) v.e must be linear in v to be a total time derivative, but I don't understand why this is.
 
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I just read that section.
I think it would have helped if they stated that the 2nd term is a total time derivative 'of a function of coordinates and time' ...
df(x,t) / dt = df/dx * dx/dt + df/dt (partial d's now)
Since f does not depend on the velocities, df/dx and df/dt don't, and the overall dependence of df/dt on v=dx/dt is linear.
 

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