Equivalent Klein-Gordon Lagrangians and equations of motion

  • #1
580
21

Main Question or Discussion Point

Suppose one starts with the standard Klein-Gordon (KG) Lagrangian for a free scalar field: $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}$$ Integrating by parts one can obtain an equivalent (i.e. gives the same equations of motion) Lagrangian $$\mathcal{L}=-\frac{1}{2}\phi\left(\Box +m^{2}\right)\phi$$ where ##\Box:=\partial_{\mu}\partial^{\mu}##.

My question is, what is the correct form for the Euler-Lagrange equations in this second case? Naively, I find that $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=-m^{2}\phi-\frac{1}{2}\Box\phi +0=-m^{2}\phi-\frac{1}{2}\Box\phi=0$$ which clearly isn't correct. Should the Euler-Lagrange equation be modified to $$\frac{\partial\mathcal{L}}{\partial\phi}-\Box\left(\frac{\partial\mathcal{L}}{\partial(\Box\phi)}\right)=0\;?$$ such that $$\frac{\partial\mathcal{L}}{\partial\phi}-\Box\left(\frac{\partial\mathcal{L}}{\partial(\Box\phi)}\right)=-m^{2}\phi-\frac{1}{2}\Box\phi
-\frac{1}{2}\Box\phi=-(\Box+m^{2})\phi =0$$
Apologies for a potentially silly question, but I'm a bit stuck on this issue.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,416
6,239
The EL equation you quote is only valid when the Lagrangian only contains first order derivatives. For the case when the Lagrangian contains higher order derivatives, you need to do additional partial integrations when deriving the corresponding equations of motion. I suggest going through the entire derivation of the EL equations in this case to see the difference.
 
  • #3
580
21
The EL equation you quote is only valid when the Lagrangian only contains first order derivatives. For the case when the Lagrangian contains higher order derivatives, you need to do additional partial integrations when deriving the corresponding equations of motion. I suggest going through the entire derivation of the EL equations in this case to see the difference.
Ah ok. So even if they are equivalent Lagrangians, their Euler-Lagrange equations will be different because the former contains up to first-order derivatives whereas the latter contains up to second-order derivatives.

I've had a go at deriving the Euler-Lagrange equations for a Lagrangian containing up to second-order derivatives and the form I get is: $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)+\partial_{\mu}\partial_{\nu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\right)=0$$ assuming that ##\delta\phi=0=\delta(\partial_{\mu}\phi)## on the boundary of the spacetime volume.
Would this be correct?
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,416
6,239
Ah ok. So even if they are equivalent Lagrangians, their Euler-Lagrange equations will be different because the former contains up to first-order derivatives whereas the latter contains up to second-order derivatives.
The resulting EL equations will be exactly the same if you include the additional terms that you need to include when there are second order derivatives in the Lagrangian. What terms in the EL equations that are non-zero will depend on which form of the Lagrangian you use - but the end result will be the same.

I've had a go at deriving the Euler-Lagrange equations for a Lagrangian containing up to second-order derivatives and the form I get is: $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)+\partial_{\mu}\partial_{\nu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\right)=0$$ assuming that ##\delta\phi=0=\delta(\partial_{\mu}\phi)## on the boundary of the spacetime volume.
Would this be correct?
Yes. Note that this will give you exactly the same equation of motion in the end (the KG equation).
 
  • #5
580
21
The resulting EL equations will be exactly the same if you include the additional terms that you need to include when there are second order derivatives in the Lagrangian. What terms in the EL equations that are non-zero will depend on which form of the Lagrangian you use - but the end result will be the same.



Yes. Note that this will give you exactly the same equation of motion in the end (the KG equation).
Ok cool, thanks.

How does one compute the second-order variation? Naively, if I expand the action around the stationary solution, ##\varphi## to second order, I have: $$S[\phi]=S[\varphi]+\int\,d^{4}x\frac{\delta S[\phi]}{\delta\phi(x)}\delta\phi(x)
+\int\,d^{4}xd^{4}y\frac{\delta^{2} S[\phi]}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)+\cdots$$ The first-order term is just ##\frac{\delta S[\phi]}{\delta\phi(x)}=\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\right)=(\Box +m^{2})\varphi=0##, however, what is the form of the second-order term? I assume that the second-order variation of the Lagrangian is $$\delta^{2}\mathcal{L}=\frac{\delta^{2}\mathcal{L}}{\delta\phi(x)\delta\phi(y)}\delta\phi(x)\delta\phi(y)+\frac{\delta^{2}\mathcal{L}}{\delta\phi(x)
\delta(\partial_{\mu}\phi(y))
}\delta\phi(x)\delta(\partial_{\mu}\phi(y))
+\frac{\delta^{2}\mathcal{L}}{
\delta(\partial_{\mu}\phi(x))
\delta(\partial_{\mu}\phi(y))
}
\delta(\partial_{\mu}\phi(x))
\delta(\partial_{\mu}\phi(y))$$ but I'm not sure.
 

Related Threads for: Equivalent Klein-Gordon Lagrangians and equations of motion

Replies
4
Views
2K
  • Last Post
Replies
1
Views
711
  • Last Post
Replies
3
Views
763
  • Last Post
Replies
14
Views
1K
Replies
6
Views
1K
Replies
7
Views
1K
Replies
6
Views
2K
Top