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Let us consider Lagrangian

$$ \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2 $$

with $\phi$ being a scalar field, and Minkowski signature $(+,-,-,-)$. My question is concerning the calculation of the energy density, which is given by

$$ \mathcal{E} = \frac{\partial \mathcal{L}}{\partial(\partial_t \phi)} \partial_t\phi - \mathcal{L}. $$

How is the derivative of the time derivative applied on the 4-derivative of the Lagrangian?

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    $\begingroup$ The Lagrangian is a function of $\phi$ and its derivatives. The derivative in the energy density is the derivative of the Lagrangian with respect to time derivative of $\phi$. $\endgroup$ Jun 5 '20 at 18:46
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Remember that

$$\partial_\mu \phi \partial^\mu \phi = \dot \phi^2 - (\nabla \phi)^2$$ The Lagrangian density can therefore be written $$\mathcal L = \frac{1}{2}\dot \phi^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2$$ at which point taking the derivative with respect to $\dot \phi$ should be straightforward.

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