• Support PF! Buy your school textbooks, materials and every day products Here!

Lagrangian -> Lagrangian'

  • Thread starter robb_
  • Start date
  • #1
340
0
Lagrangian --> Lagrangian'

It is said that the equations of motion remain unchaged if one switches from a known lagrangian to a new lagrangian plus a total time derivative of a function. IIRC, the function is a function of q and t only. I have shown this to be so mathematically, but am trying to understand what it implies conceptually. Thanks.
 

Answers and Replies

  • #2
150
0
A simple example would be to add a constant to the potential energy. Here, the value of the potential energy has some arbitrariness to it, but the equations of motion are unaffected. This makes sense because it's that change in potential that affects motion and not the assigned value. The property you mentioned generalizes this further.
 
  • #3
quasar987
Science Advisor
Homework Helper
Gold Member
4,773
8
This "gauge freedom" in the lagrangian will also reveal itself to be very key. It is at the heart of Noether's thm and canonical transformations in the hamiltonian formulation.
 
  • #4
340
0
Thanks. I can understand the example of the potential.
Hopefully Goldstein bears this out more in later chapters.
Any other thoughts?
 
  • #5
5
0
I guess the lesson is that the Lagrangian isn't a physical quantity. But the derivatives (both with time and space) of the Lagrangian are physical quantities.

If you're familiar with the action, you know that it's the integral of the Lagrangian over time. The path of minimum action is the physically realized path of a system with that Lagrangian. This path will be unchanged when you add a total time derivative to the Lagrangian. The reason is pretty clear. You're simply adding a constant to the action for every path, so it won't change which path minimizes the action--it will only change the value of the action over that path. But, again, the action isn't a physical quantity--but the path of least action is!
 
  • #6
340
0
So it sounds very much like a gauge invariance. Why can the function only depend on the coords and time? No velocity, etc. I am trying to imagine how that might change the minimum value of the action over a path.
thanks again
 

Related Threads on Lagrangian -> Lagrangian'

Replies
7
Views
7K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
519
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
712
  • Last Post
Replies
1
Views
672
Top