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In Landau Mechanics (third edition page 4), why does adding Lagrangians of two non interacting parts remove the indefiniteness of multiplying each Lagrangian by a different constant?

If both systems are completely non-interacting, I can perfectly create a new Lagrangian L = a*L1 + b*L2 and still maintain validity of the Euler-Lagrange equations and make the action integral perfectly stationary, with a and b different. Can't I?

Here's the original statement:

It is evident that the multiplication of the Lagrangian of a mechanical system by an arbitrary constant has no effect on the equations of motion. From this, it might seem, the following property of arbitrariness can be deduced: the Lagrangians of different isolated mechanical systems may be multiplied by different arbitrary constants. The additive property, however, removes this indefiniteness, since it admits only the simultaneous multiplication of the Lagrangians of all the systems by the same constant.

I kind of understand that, of course, after you create L = L1 + L2, xL = xL1 + xL2 but nothing stops me from multiplying each Lagrangian of each non interacting part by different constants a and b and then summing, like: xL = xaL1 + xbL2

So the statement seems without meaning for me. Could anyone clarify? Thanks!

One addition: I see that, if both systems are interacting externally, the constants must be equal (and the proportion of masses becomes relevant). My goal is exactly that: to understand, conceptually, how he derives the role of ration of masses later on in page 7. Going back to the text, what confused me is that the whole thing starts with the assumption that both system are closed so any external interaction is dismissed:

Let a mechanical system consist of two parts which, if closed (..), and the interaction between them may be neglected, the Lagrangian of the whole system tends to: lim L = La + Lb.

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More on the first pages of the textbook Mechanics by Landau and Lifshitz: physics.stackexchange.com/q/23098/2451 –  Qmechanic Oct 25 '12 at 12:02
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2 Answers

The implicit assumption in Landau and Lifschitz is that you can interact with the systems externally, to add or remove energy. The energy/momentum of the system is determined from the Lagrangian, and removing energy from one system, and adding the same energy to the other will reveal the proportionality constant between the two.

The reason this is left implicit is because the Lagrangian formulation already has an implicit energy law inside, since the Lagrangian is constructed from energy concepts (although technically speaking it is logically independent, with energy conservation only following for a time independent system). So when you have the sum Lagrangian, you know the ratio of kinetic and potential energies in the two systems, something which is physically relating the two coefficients in front to each other.

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Thanks Ron. I understand that, if both systems are interacting externally, the constants become important (and therefore the proportion of masses becomes relevant). My goal is exactly that: to understand, conceptually, how he derives the role of ration of masses later on in page 7. But going back to the start of the text, what confused me is that the whole thing starts with the assumption that both system are not only non interacting, they are also closed. Here it is: –  Gilberto Oct 25 '12 at 11:20
    
The quote from page 4 follows: Let a mechanical system consist of two parts which, if closed (..), and the interaction between them may be neglected, the Lagrangian of the whole system tends to: lim L = La + Lb. –  Gilberto Oct 25 '12 at 11:26
    
@Gilberto: As I explained, Landau/Lifschitz are making the assumption that energy additivity is physics. This is justified. –  Ron Maimon Oct 25 '12 at 12:54
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Since the Euler-Lagrange equations are homogeneous we can multiply the entire Lagrangian by a constant (or add a constant, incidentally) and not change the equations of motion. Not so when we attempt to multiply separate terms of the Lagrangian by different constants. Your equations of motion will not reduce to Newton's law, $\mathbf{F} = m \mathbf{a}$, among other problems (total energy will be wrong, angular momentum, etc.).

As a test case you can consider two particles in an external potential well (without mutual interactions). If you have your $a$ and $b$ above such that $a\ne b$, you'll deduce an equation of motion that's wrong.

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This is false--- you can do it without affecting the equation of motion. –  Ron Maimon Oct 25 '12 at 3:27
    
Yes, you're right. Thanks. –  MarkWayne Oct 28 '12 at 13:42
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