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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 ?

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    $\begingroup$ You've stated yourself that the euler lagrange equations (i.e., the equations of motion) are the same. And the euler lagrange equations are the equations of motion. So I do not understand your first question as you've asked it. The second part of the question I honestly do confess to not understand, but maybe someone else will :) $\endgroup$ – Sanya Apr 20 '17 at 13:32
  • $\begingroup$ What I mean is that if I have $f1'+f1=0$ and $f2'+f2=0$ : the equations have the same form (it is analog to my equations of motion on $\mathcal{L}_1$ and $\mathcal{L}_2$). But if $f_1$ and $f_2$ are different, I have some extra information that can lead to incompatible results. It is the exact same thing I wanted to explain with the equations of motion with lagrangians. I have the same form of Euler Lagrange equations with $L_1$ and $L_2$ but do I have some extra info from $L_2$ in practice. I hope it is more understandable now ? :o $\endgroup$ – StarBucK Apr 20 '17 at 13:41
  • $\begingroup$ If $f_1 ' + f_1 = 0$ and $f_2 ' + f_2 = 0$, this indeed can allow that $f_1$ and $f_2$ differ - but in a field theory, the only thing that has physical meaning are in the end the equations of motion, not the lagrangian itself (much like in electrodynamics, there is gauge invariance and thus, the vector potential itself has no physical meaning apart from its various derivatives) $\endgroup$ – Sanya Apr 20 '17 at 19:46
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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.

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  • $\begingroup$ Is the converse of your first sentence true? i.e., if the EL operator acting on $A$ is zero, then $A$ must be a total derivative? [or: perhaps I should ask this as a separate question?] $\endgroup$ – AccidentalFourierTransform Apr 20 '17 at 14:02
  • $\begingroup$ @AccidentalFourierTransform: The answer is Yes, modulo topological obstructions. See this Phys.SE post. $\endgroup$ – Qmechanic Apr 20 '17 at 14:06

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