- #1
spaghetti3451
- 1,344
- 33
The Euler-Lagrange equation obtained from the action ##S=\int\ d^{4}x\ \mathcal{L}(\phi,\partial_{\mu}\phi)## is ##\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\big)=0##.
My goal is to generalise the Euler-Lagrange equation for the action ##S=\int\ d^{4}x\ \mathcal{L}(\phi,\partial_{\mu}\phi, \partial_{\mu}\partial_{\nu}\phi)##.
The variation of the Lagrangian density this time contains the extra term ##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta(\partial_{\mu}\partial_{\nu}\phi)##. Now,
##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta(\partial_{\mu}\partial_{\nu}\phi)##
##=\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\nu}(\delta\phi)\Big]-\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big]\partial_{\nu}(\delta\phi)##
##=\partial_{\mu}\Big[\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta\phi\big)-\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\big)\delta\phi\Big]-\Big[\partial_{\nu}\Big(\partial_{\mu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\big)\delta\phi\Big)-\partial_{\nu}\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big)\delta\phi\Big]##
Now, in the fourth term, ##\partial_{\nu}\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big)=0## for arbitrary ##\delta\phi## - this is the new term in the Euler-Lagrange equation.
The second and third terms must give zero, by Gauss' theorem and under the assumption that the field vanishes at spatial or temporal infinity.
What do I do about the first term? Should I have to introduce a constraint on the derivative ##\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta\phi\big)## - I don't want to, because it's not physical.
My goal is to generalise the Euler-Lagrange equation for the action ##S=\int\ d^{4}x\ \mathcal{L}(\phi,\partial_{\mu}\phi, \partial_{\mu}\partial_{\nu}\phi)##.
The variation of the Lagrangian density this time contains the extra term ##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta(\partial_{\mu}\partial_{\nu}\phi)##. Now,
##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta(\partial_{\mu}\partial_{\nu}\phi)##
##=\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\nu}(\delta\phi)\Big]-\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big]\partial_{\nu}(\delta\phi)##
##=\partial_{\mu}\Big[\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta\phi\big)-\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\big)\delta\phi\Big]-\Big[\partial_{\nu}\Big(\partial_{\mu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\big)\delta\phi\Big)-\partial_{\nu}\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big)\delta\phi\Big]##
Now, in the fourth term, ##\partial_{\nu}\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big)=0## for arbitrary ##\delta\phi## - this is the new term in the Euler-Lagrange equation.
The second and third terms must give zero, by Gauss' theorem and under the assumption that the field vanishes at spatial or temporal infinity.
What do I do about the first term? Should I have to introduce a constraint on the derivative ##\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta\phi\big)## - I don't want to, because it's not physical.