How does calculus of variations handle explicit time dependence in Lagrangian?

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DuckAmuck
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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/∂ δ

What is not clear is how this works when t is an explicit variable in L

δL = ∂L/∂x δx + ∂L/∂ δ + ∂L/∂t δt

How does this still result in:

∂L/∂x = d/dt ∂L/∂ẋ ?
 
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wrobel said:
if you mean the standard Least Action Principle then this principle holds for time dependent Lagrangians by the same way as for time independent Lagrangians

Can you show why? How is the term ∂L/∂t δt handled in Least Action?
 
Just calculate this: $$\frac{d}{d\epsilon}\Big|_{\epsilon=0}\int_{t_1}^{t_2}L(t,q(t)+\epsilon h(t),\dot q(t)+\epsilon \dot h(t))dt$$ for ##h(t)## such that ##h(t_i)=0,\quad i=1,2##
you will not get that term
 
That term comes from the chain rule of δL
I have seen the least action principle shown as 0 = δS = ∫δL dt, which I guess is misleading.
I have seen the form you have, and that makes more sense. You are explicitly minimizing with respect to epsilon.
 
By the way the Action functional is not compelled to attain a minimum for real motion, real motion is just a critical point of this functional . It can attain maximum for real motion
 
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DuckAmuck said:
That term comes from the chain rule of δL
I have seen the least action principle shown as 0 = δS = ∫δL dt, which I guess is misleading.
I have seen the form you have, and that makes more sense. You are explicitly minimizing with respect to epsilon.
By definition time is not varied in the usual Hamilton principle!