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I am varying a scalar field density with the term

${\cal L}~=~-\frac{1}{2}(\partial _\mu\phi)^2$

w.r.t the scalar field $\phi$.

First of all i want to know if its true that:

${\cal L} = -\frac{1}{2}(\partial_\mu \phi)(\partial_\mu \phi)$.

Secondly i also want to show that the variation of $\cal L$ w.r.t $\phi$ gives me the equation

$\delta \phi \nabla^2\phi = -\frac{1}{2}\delta(\partial_\mu\phi)^2$.

Do i have to use partial integration of the left hand side to show this?

Reference:

  1. https://doi.org/10.1103/PhysRevLett.126.011104, there the action is shown in eq. (1) and the result of the variation w.r.t $\phi$ is given in eq. (2).
Qmechanic
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Minoso
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1 Answers1

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The Lagrangian:

$${\cal L} = -\frac{1}{2}(\partial_\mu \phi)(\partial_\mu \phi)$$

is not a scalar, i assume that you want the kinetic term of a scalar field:

$${\cal L_{\text{kinetic}}} = -\frac{1}{2}(\partial^\mu \phi)(\partial_\mu \phi)$$

You have to vary with respect to $\phi$

$$ δ(g^{μν}\nabla_{μ}\phi\nabla_{ν}\phi) = g^{μν}δ(\nabla_{μ}\phi)\nabla_{ν}\phi + g^{μν}δ(\nabla_{ν}\phi)\nabla_{μ}\phi = 2\nabla_{μ}(δ\phi)\nabla^{μ}\phi$$

, then make use of Leibniz rule (integration by parts):

$$ \nabla_{μ}(δ\phi\nabla^{μ}\phi) = \nabla_{μ}(δ\phi)\nabla^{μ}\phi + δ\phi\nabla_{μ}\nabla^{μ}\phi $$

and you're there.

The action of the referenced paper also contains a potential for the scalar field coupled to the Gauss-Bonnet invariant. For that term you have to use the chain rule on $\delta(h(\phi) R^2_{GB})$

The expressions $(\partial \phi)^2$ and $(\partial^\mu \phi)(\partial_\mu \phi)$ are equivalent.

Noone
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