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In this topic Why my 4-divergence term added to a Lagrangian modifies the equation of motion? I understood that if I have :

$$\mathcal{L}_1(\phi,\partial \phi)=\mathcal{L}_2(\phi,\partial \phi) + \partial_\mu V^\mu(\phi)$$

then a stationnary action of $S_1$ associated to $\mathcal{L}_1$ is equivalent to stationnary action of $S_2$ associated to $\mathcal{L}_2$.

In practice it means :

$$\frac{\partial \mathcal{L_1}}{\partial \phi}-\partial_\mu \frac{\partial \mathcal{L_1}}{\partial_\mu \phi}=0 \Leftrightarrow \frac{\partial \mathcal{L_2}}{\partial \phi}-\partial_\mu \frac{\partial \mathcal{L_2}}{\partial_\mu \phi}=0$$

Thus :

$$\frac{\partial \mathcal{L_1}}{\partial \phi}-\partial_\mu \frac{\partial \mathcal{L_1}}{\partial_\mu \phi}=0=\frac{\partial \mathcal{L_2}}{\partial \phi}-\partial_\mu \frac{\partial \mathcal{L_2}}{\partial_\mu \phi}=0 $$

My question is : How do that mean that we have the same equations of motion and not "extra information" from $\mathcal{L_2}$ ?

For example, if I note $f_1(\phi(x))=\frac{\partial \mathcal{L_1}}{\partial \phi}-\partial_\mu \frac{\partial \mathcal{L_1}}{\partial_\mu \phi}$ and $f_2(\phi(x))=\frac{\partial \mathcal{L_2}}{\partial \phi}-\partial_\mu \frac{\partial \mathcal{L_2}}{\partial_\mu \phi}$ (we imagine that these functions only depends on $\phi$ to simplify).

Could I have : $f_1(\phi(x))=3\phi(x)-1$ and $f_2(\phi(x))=\phi(x)$ ? Because if it is the case I would have : $\phi(x)=0=1/3$ which is an uncompatible equation.

So if it is the case I could have uncompatible equations with the two actions by having : $$\mathcal{L}_1(\phi,\partial \phi)=\mathcal{L}_2(\phi,\partial \phi) + \partial_\mu V^\mu(\phi)$$

But what does that physically mean ? Can I say that if I have a lagrangian $\mathcal{L}_1$ and if I can find $\mathcal{L}_2$ verifying this equation such as the equation of motion are uncompatible, then $\mathcal{L}_1$ is not a "physical" lagrangian ?

Other question : can I have two compatibles equation such as the lagrangian $\mathcal{L_2}$ gives us "more" information on the field than $\mathcal{L_1}$ alone ?

StarBucK
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1 Answers1

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One may prove that when the Euler-Lagrange (EL) operator $$ \frac{\partial}{\partial\phi^{\alpha}}-d_{\mu} \frac{\partial}{\partial\phi^{\alpha}_{,\mu}} +\ldots \tag{1}$$ acts on a total derivative/divergence term $d_{\mu}f^{\mu}$, the result vanishes identically. [The ellipsis $\ldots$ in eq. (1) denotes possible higher-derivative terms.] Therefore if two Lagrangian densities $${\cal L}_1- {\cal L}_2~=~d_{\mu}f^{\mu}\tag{2}$$ differ by a total divergence term, their EL equations are identical. See also this Phys.SE post.

Qmechanic
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