# When is numerical value of Lagrangian evaluated on-shell a full differential?

I noticed recently that for many field equations, Lagrangian evaluated on-shell (i.e. using equations of motions) is a full derivative- a divergence or something, or in other words a boundary term. This is true for Weyl, Dirac, Klein-Gordon (without external potential), Maxwell, classical Schrodinger. I am wondering if these are special cases of some general pattern or principle.

• Ironically perhaps, for a non-relativistic point particle in a uniform gravitational field $L=\frac{m}{2}\dot{y}^2 -mgy$ (which is typically one of the first systems that one learns about), the action $S=\int_{t_i}^{t_f}\!dt~L \approx -\frac{1}{2}mg\int_{t_i}^{t_f}\!dt~y$ is not a boundary term on-shell. (Of course, this Lagrangian is not homogeneous.) – Qmechanic May 7 '16 at 13:46

Theorem: let $L$ be a homogeneous function of degree $k$; then the on-shell lagrangian is a total derivative.

Proof: according to the Euler's homogeneous function theorem, $$k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1}$$

On the other hand, because of the Euler-Lagrange equations, $$(1)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\stackrel{\mathrm{EL}}{=}q\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial \dot q}\right)+\dot q\frac{\partial L}{\partial \dot q} \tag{2}$$

Finally, integrating by parts, $$(2)=q\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial \dot q}\right)+\dot q\frac{\partial L}{\partial \dot q}\stackrel{\mathrm{IBP}}{=}-\dot q\frac{\partial L}{\partial \dot q}+\dot q\frac{\partial L}{\partial \dot q}+\text{total derivative}=\text{total derivative}\tag{3}$$ which is what we wanted to prove. The generalisation to field theory is straightforward.

As for an example of a non-homogeneous lagrangian, take $$L=\frac{1}{2}\dot q^2+\mathrm e^q\tag{4}$$ which, when evaluated on-shell, is $$L=\mathrm e^q\left(1-\frac{q}{2}\right)\tag{5}$$ i.e., this is not a total derivative. For a more realistic example, see QMechanics comment above ("a non-relativistic point particle in a uniform gravitational field").

• @Blazej I'm glad I could help :-) – AccidentalFourierTransform May 7 '16 at 11:16
• It is in interesting observation that this implies that for all theories with quadratic Lagrangians one can choose it such that Lagrangian vanishes on-shell. However, one might lose gauge invariance (example: electrodynamics) or reality of action (example: Shrodinger equation) in this process. – Blazej May 7 '16 at 11:27