# Time dependence of kinetic energy in Lagrangian formulation

• I
• Ahmed1029
In summary, if the inertial Cartesian coordinates as functions of the generalized coordinates depend explicitly on time, the kinetic energy in the Lagrangian will also be explicitly time dependent. This is because the time derivative of the position vector includes a term involving the partial derivative with respect to time.

#### Ahmed1029

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

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.