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## Main Question or Discussion Point

Hi guys, so this is a pretty generic question.

Starting off with the classical Lagrangian in a case where there is no interaction or explicit time dependence, the functional form is

[itex]L=L(x,\dot{x})=L(x,\partial_{t}x)[/itex].

Now when we look at the Lagrangian density in field theory, the functional dependence is

[itex]\mathcal{L}=\mathcal{L}(\varphi,\partial_{\mu}\varphi)[/itex].

And there's my question. Why does the Lagrangian density depend on [itex]\partial_{\mu}\varphi[/itex] and not only [itex]\partial_{t}\varphi[/itex]? i mean why the four-derivative?

I have a few ideas but I'm not sure if they are correct. I'm thinking along the lines that since [itex]\partial_{\mu} = \nabla + \partial_{t}[/itex], maybe the [itex]\partial_{\mu}[/itex] just appears to merge the two together and make it more compact as this one term includes possible interactions and also the field velocity?

Please help! thank you :)

Starting off with the classical Lagrangian in a case where there is no interaction or explicit time dependence, the functional form is

[itex]L=L(x,\dot{x})=L(x,\partial_{t}x)[/itex].

Now when we look at the Lagrangian density in field theory, the functional dependence is

[itex]\mathcal{L}=\mathcal{L}(\varphi,\partial_{\mu}\varphi)[/itex].

And there's my question. Why does the Lagrangian density depend on [itex]\partial_{\mu}\varphi[/itex] and not only [itex]\partial_{t}\varphi[/itex]? i mean why the four-derivative?

I have a few ideas but I'm not sure if they are correct. I'm thinking along the lines that since [itex]\partial_{\mu} = \nabla + \partial_{t}[/itex], maybe the [itex]\partial_{\mu}[/itex] just appears to merge the two together and make it more compact as this one term includes possible interactions and also the field velocity?

Please help! thank you :)