Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Additivity of lagrangian and constraints on multiplication by arbitrary const

  1. Aug 17, 2010 #1

    I am using Landau's mechanics Vol I for classical mechanics. On page 4 he mentions for Lagrangian of a system composed of two systems A and B which are so far away so that their interactions can be neglected.

    then for the combined system we have L = LA + LB

    I'm trying to understand how this additivity implies only simultaneous multiplication of LA and LB by an arbitrary constant.

    I think to establish it we might have to consider the difference in Lagrangian when A & B are close by and when they are far away.

    Please guide. Any help will be appriciated.
  2. jcsd
  3. Aug 19, 2010 #2
    I think that's a typical example of a "Landau-type" reasoning: there is certainly a very profound idea behind every single thing he expresses, but he's certainly not good at all in exposing these ideas. In reading Landau's books, one spends more time trying to interpret what he means, than on actually understanding the physical concepts. That's why, in my opinion, one has to read Landau's books only after he is very familiar with the subject: you don't learn with it, but you certainly get a deeper point of view.

    In the specific case of the additivity of the lagrangians, I think he wants to say that exactly separate systems actually don't exist, so in principle you cannot define separate lagrangians. Only the total lagrangian exists, and you can multiply it by a constant. When you take the systems far away, you observe that the lagrangian tends to a sum of two lagrangians, and if you started with a total lagrangian multiplied by a constant, you end up with the two lagrangians multiplied by the same constant.
  4. Aug 28, 2010 #3
    Hey thanks. Does that mean it is meaningless to define the lagrangian of a system without refering to rest of the universe. We actually do it all the time by considering isolated systems and then deriving lagrangian by consideration of symmentries. In fact lagrangian of a two particle is written as the sum of two terms. One representing the lagrangian in case the particles are separated far enough. The other term representing the interaction of the two particles.
  5. Aug 28, 2010 #4
    Yes, and introducing an interaction generally breaks the symmetry. For example, invariance under translations causes each individual momentum to be conserved, but when you introduce a translation-invariant interaction only the total momentum is conserved, because the interaction causes the system to be invariant only under a global translation.
  6. Aug 28, 2010 #5
    Ok, so, if you read what is the point of Lagrangians, you would have understood that they generate the equations of motion for the system:

    \frac{d}{d t} \frac{\partial L}{\partial \dot{q}_{i}} - \frac{\partial L}{\partial q_{i}} = 0

    Now, two non-interacting system would mean that the equations of motion of system A do not contain any coordinates and speeds of system B and vice versa.

    This is only possible if the total Lagrangian of the system is:

    L(q, \dot{q}) = L_{A}(q_{A}, \dot{q}_{A}) + L_{B}(q_{B}, \dot{q}_{B})
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook