# Negligible terms and Lagrangians

Gold Member
Consider a theory with the Lagrangian $\mathcal L=\mathcal L_{free} + \mathcal L_{int}$. I think if we say $\mathcal L_{free} \gg \mathcal L_{int}$, this means that the equations of motion will be much near to the free equations. But I'm not sure that we can prove that if in an equation of motion, we can neglect interaction terms, then $\mathcal L_{free} \gg \mathcal L_{int}$! I know, It may not seem a clear question but my mind's not clear about it too so I'll appreciate any clarification(of the question and my mind) and any ideas.
Thanks

jedishrfu
Mentor
I think you'd have to consider length of time too, since over time the interactive effects will begin to appear.

No, that is not true. Consider some Lagrangian, then a sum of that Lagrangian with an arbitrarily large constant. Obviously, the equations of motion are the same.

Gold Member
I think you'd have to consider length of time too, since over time the interactive effects will begin to appear.

But sometimes we just dismiss some terms in the equation of motion because they're negligible compared to other terms. Your objection is applicable there too. But in those cases we're considering the equation in only a time interval which is shorter than the full time interval that the equation is applicable in. So here, we can take the action integral only in that subinterval where the approximation is a good one.

No, that is not true. Consider some Lagrangian, then a sum of that Lagrangian with an arbitrarily large constant. Obviously, the equations of motion are the same.
Well, I can take only terms which contain a dynamical variable!

Well, I can take only terms which contain a dynamical variable!

The instead of a constant term, consider a large but very slowly varying term.

Gold Member
The instead of a constant term, consider a large but very slowly varying term.
So you're saying that we may have a term in the Lagrangian that is negligible compared to other terms but is varying fast. Then it doesn't matter that its magnitude is small?

I would express that differently. It is not magnitude alone that matters.

Gold Member
I would express that differently. It is not magnitude alone that matters.
There are many famous approximations in physics. Of course some of them are in parts of physics that have a well established variational formulation. I think if I see one example of an approximation implemented in the variational language, I'll be OK. Do you know such an example?(I mean applying the approximation before getting the equations of motion, applying it to the Lagrangian!)

The classical example is given by small oscillations near an equilibrium at ##q = q_0##.

Say kinetic energy is ## {f(q) \over 2} \dot q^2 ## and potential energy is ## g(q) ##. Then kinetic energy is approximated with ## {f(q_0) \over 2} \dot q^2 ##, and potential energy with ## {g''(q_0) \over 2} q^2 ## (remember, ## g'(q_0) = 0 ## because ##q = q_0## is an equilibrium). So the approximated Lagrangian is ## {f(q_0) \over 2} \dot q^2 - {g''(q_0) \over 2} q^2 ##, whose Euler-Lagrange equation is ## f(q_0) \ddot q + g''(q_0) q = 0 ##, i. e., that of a harmonic oscillator.