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

In summary, the Lagrangian with explicit time dependence may be handled in calculus of variations using the same principles as for time independent Lagrangians.
  • #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/∂ 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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  • #2
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
 
  • #3
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?
 
  • #4
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
 
  • #5
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
 
  • #6
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.
 
  • #7
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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  • #8
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!
 

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

1. How is explicit time dependence represented in the Lagrangian for calculus of variations?

Explicit time dependence in the Lagrangian is represented by including the time variable, t, as a parameter in the Lagrangian function. This allows for the variation of the system with respect to time.

2. Can explicit time dependence be included in all types of Lagrangian systems?

Yes, explicit time dependence can be included in all types of Lagrangian systems, including mechanical, electrical, and thermodynamic systems. It is a fundamental aspect of the calculus of variations method.

3. How does the inclusion of explicit time dependence affect the Euler-Lagrange equation?

Including explicit time dependence in the Lagrangian results in an additional term in the Euler-Lagrange equation, known as the "time derivative term". This term accounts for the variation of the system with respect to time.

4. What is the significance of explicit time dependence in the Lagrangian for physical systems?

Explicit time dependence in the Lagrangian allows for the consideration of time-varying systems, which are common in physical systems. It allows for a more accurate and comprehensive analysis of the system's behavior.

5. Are there any limitations to including explicit time dependence in the Lagrangian for calculus of variations?

One limitation is that the system must be autonomous, meaning that it does not explicitly depend on time in any other way besides through the Lagrangian. Additionally, the system must have a continuous and well-defined Lagrangian function.

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