- #1
DuckAmuck
- 236
- 40
If a Lagrangian has explicit time dependence due to the potential changing, or thrust being applied to the object in question, how does calculus of variations handle this?
It's easy to get the Lagrange equations from:
δL = ∂L/∂x δx + ∂L/∂ẋ δẋ
What is not clear is how this works when t is an explicit variable in L
δL = ∂L/∂x δx + ∂L/∂ẋ δẋ + ∂L/∂t δt
How does this still result in:
∂L/∂x = d/dt ∂L/∂ẋ ?
It's easy to get the Lagrange equations from:
δL = ∂L/∂x δx + ∂L/∂ẋ δẋ
What is not clear is how this works when t is an explicit variable in L
δL = ∂L/∂x δx + ∂L/∂ẋ δẋ + ∂L/∂t δt
How does this still result in:
∂L/∂x = d/dt ∂L/∂ẋ ?