Time dependence of kinetic energy in Lagrangian formulation

  • #1
Ahmed1029
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Could kinetic energy possibly depend explicitly on time in the lagrangian for some arbitrary set of generalized coordinates?
 

Answers and Replies

  • #2
vanhees71
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Yes, if the inertial Cartesian coordinates as functions of the generalized coordinates depend explicitly on time (describing the motion in a non-inertial frame of reference) you get from
$$\vec{x}=\vec{x}(q^k,t), \quad k \in \{1,\ldots,f \}$$
the time derivative (Einstein summation convention applies)
$$\dot{\vec{x}}=\dot{q}^k \partial_k \vec{x} + \partial_t \vec{x}$$
and thus
$$T=\frac{m}{2} \dot{\vec{x}}^2 = \frac{m}{2} \left (\dot{q}^k \partial_k \vec{x} + \partial_t \vec{x} \right)^2,$$
which is in general explicitly time dependent.
 

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