# 4-derivative kinetic term Lagrangian

## Homework Statement

Show that $$L=\phi\Box^2\phi$$ generates negative energy density.

## The Attempt at a Solution

The energy density is $$E=\frac{\partial L}{\partial \dot{\phi}}\dot{\phi}-L$$ Also the Lagrangian can be rewritten (using divergence theorem) as $$L=-\partial_\mu\phi\partial_\mu(\Box\phi)$$ So I would get $$E=-\partial_0\phi\partial_0(\Box\phi)+\partial_\mu\phi\partial_\mu(\Box\phi)$$ $$E=-\partial_x\phi\partial_x(\Box\phi)-\partial_y\phi\partial_y(\Box\phi)-\partial_z\phi\partial_z(\Box\phi)$$. Why is this negative necessarily? Thank you!

## Answers and Replies

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Orodruin
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Also the Lagrangian can be rewritten (using divergence theorem) as
L=−∂μϕ∂μ(□ϕ)​
No, it cannot. You need to remove the d’Alembertian from this expression.

No, it cannot. You need to remove the d’Alembertian from this expression.
I am not sure what you mean. I used $$\int(\phi\Box^2\phi)=\int\partial_\mu(\phi\partial_\mu \Box \phi)-\int(\partial_\mu\phi\partial_\mu \Box \phi)$$ What is wrong? What do you mean by getting rid of the d'Alembertian?

Orodruin
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I am not sure what you mean. I used $$\int(\phi\Box^2\phi)=\int\partial_\mu(\phi\partial_\mu \Box \phi)-\int(\partial_\mu\phi\partial_\mu \Box \phi)$$ What is wrong? What do you mean by getting rid of the d'Alembertian?
This is simply not true. What you are looking for is
$$\phi \Box \phi = \partial_\mu (\phi \partial^\mu \phi) - (\partial_\mu \phi)(\partial^\mu\phi)$$

Orodruin
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Ah, sorry. You have $\Box^2$, not $\Box$. In that case you cannot use the expression for the energy that you have given since your Lagrangian will contain derivatives of order higher than one.