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

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The discussion focuses on how calculus of variations addresses explicit time dependence in Lagrangians. It clarifies that when time is an explicit variable in the Lagrangian, the term ∂L/∂t δt does not appear in the Least Action Principle because the endpoints of the interval are not varied. The derivation involves calculating the variation of the action integral without including that term, which arises from the chain rule of δL. The principle remains valid for time-dependent Lagrangians, as real motion is treated as a critical point of the action functional rather than requiring a minimum. Overall, the treatment of time in the Least Action Principle is nuanced and distinct from the variation of spatial coordinates.
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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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
 
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?
 
DuckAmuck said:
How is the term ∂L/∂t δt handled in Least Action?
the is no such a term in Least Action because you do not vary ends of the interval
 
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
 
Last edited:
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!
 

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