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Can The Lagrangian L=T-V Be Derived?

  1. Jun 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Thank you for answering my question about setting the Euler-Langrangian expression to zero separately for each coordinate (ehild ans.=yes). Now my question is: Can the Lagrangian be derived, or is it the expression, when inserted into the Euler-Lagrange equation(s), that gives the correct equation(s) of motion? In other words, you have to be someone like Lagrange to have the intuition from the beginning what to enter into the E-L equations for L?


    2. Relevant equations
    L=T-V


    3. The attempt at a solution
    I think there is no derivation, but I want to be sure. Thank you for reading my question. BTW, I read in the Wikipedia article on 'Lagrangian' that the generalized coordinates used in writing the expression L=T-V do not have to be orthogonal.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 1, 2013 #2
    To answer your question directly, no, the Lagrangian cannot be derived. It is defined as T - U. However, I think the better question is to ask why the quantity T - U should be of any interest at all. The Lagrangian turns out to be intersting simply because it works - I guess there is a sense in which Lagrange had to have some sense of intuition here.
     
  4. Jun 1, 2013 #3
    Thank you for answering my question about the Lagrangian L=T-V. Yes, there does not seem to be anything intuitive about it.
     
  5. Jun 2, 2013 #4

    dextercioby

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    Let's just say the the 'U' is the only sensitive point of L = T-U, because it makes sense iff the classical system has time-independent interactions. Then the observable total energy is conserved which makes us conclude that U is the potential energy.
     
  6. Jun 2, 2013 #5
    Thank you, dextercioby, but I did not understand your explanation. Am working at a very basic level here.
    Regards, Ted.
     
  7. Jun 2, 2013 #6
    I believe Dextercioby was pointing out that when our coordinates are "natural", that is, the relation between the generalized coordinates and underlying cartesian coordinates are independent of time, the observable total energy H = T + U. Additionally, if H is indep. of t, then it is conserved. Thus the U, in H = T + U, will correspond exactly to the potential energy in L = T - U. This is all derivable from Hamilton's equations. Classical Mechanics by Taylor, ch 13 explains this very well if you are curious.
     
  8. Jun 2, 2013 #7
    OK, tannerbk. Thank you for the reply.
     
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