How Does Homogeneity of Space and Time Affect Lagrangian Mechanics?

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SUMMARY

The discussion centers on the implications of the homogeneity of space and time in Lagrangian mechanics. It establishes that the Lagrangian, denoted as L, must not explicitly depend on the radius vector r or time t, leading to the conclusion that L is a function of velocity v only. The participant clarifies that while the general definition of the Lagrangian is L = ∫ L(ṡ, q, t) and includes velocity, in specific cases such as a free particle, L can be expressed as L(v²), indicating a dependence solely on the square of the speed. This distinction highlights that L(v²) is a specific case rather than a contradiction to the general form.

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Andrea Vironda
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Hi,
i know that The homogeneity of space and time implies that the Lagrangian cannot contain
explicitly either the radius vector r of the particle or the time t, i.e. L must be a function of v only

but the lagrangian definition is ##L=\int L(\dot q,q,t)##, so velocity appears in the definition and it's in contrast with ##L=L(v^2)##
why?
 
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The definition is general, and not just for homogeneous space and time. For example, in a uniform gravitational field, the Lagrangian of a particle does depend on position, through the potential energy (mgy). The Lagrangian of a free particle does not depend on position. It also does not depend on direction (isotropy) and depends only on the square of the speed.
 
Why should ##L(q,\dot{q},t)## and ##L(v^2)## be in contrast at all? The latter sais, that ##L## in that specific case depends on ##v=\dot{q}## only via ##v^2## and that the dependence on ##q## and ##t## drops, so ##L(v^2)## is a specific restriction of the most general case ##L(q,\dot{q},t)##, but not in contrast with it.
 
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