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I have a question regarding the red term below. This is the integration by parts during the derivation of the Euler-Lagrange equation for continuous systems. Why is this not the time derivative $d/dt$.... why do we use the partial? I have seen the equation expressed in both ways and wondered what the physicality's are.

\begin{equation} \int^{t_2}_{t_1}dt\frac{\partial \mathcal L}{\partial (\partial _t\phi)}\frac{\partial }{\partial t}\delta \phi =\frac{\partial \mathcal L}{\partial (\partial _t\phi)}\delta \phi \bigg|^{t_2}_{t_1}-\int^{t_2}_{t_1}\color{red}{\frac{\partial }{\partial t}}\bigg(\frac{\partial\mathcal L}{\partial (\partial _t\phi)}\bigg)\delta \phi dt \end{equation}

This particular Lagrange density is a function of $\mathcal L=\mathcal L(\phi,\dot \phi ,\partial _x\phi)$

Qmechanic
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AngusTheMan
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    It is to be interpreted as a total derivative here - it is not just a derivative with respect to explicit time-dependence. Some authors are sometimes a bit sloppy with the notation. The reason you can see this is that he uses integration by parts (which requires $\partial_t$ to be the total derivative). – Winther Sep 03 '14 at 21:40
  • Ah so really $\partial _t \phi$ is actually $d\phi dt$ the whole time? – AngusTheMan Sep 03 '14 at 21:48
  • Related: https://physics.stackexchange.com/q/321668/2451 and links therein. – Qmechanic Sep 03 '14 at 21:52
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    For $\delta\phi$, which is just a function of $t$ (and other coordinates, but not a function of fields), it does not matter: $\frac{\partial\delta\phi}{\partial t} = \frac{d\delta\phi}{dt}$. You will find this alot in physics text and papers where the true meaning of a notation must be intrepreted from the context it is given in (especially when it comes to $\partial$ vs $d$). – Winther Sep 03 '14 at 22:17

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