Why does field Lagrangian depend on four-derivative?

  • Thread starter Dixanadu
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  • #1
Dixanadu
254
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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 :)
 

Answers and Replies

  • #2
HomogenousCow
735
206
[itex]\partial_{\mu} = \nabla + \partial_{t}[/itex]

That's not true.
The field is assumed to be lorentz invariant, hence derivative terms must be built out of scalars involving the whole four-gradient. (e.g. box operator).
 
  • #3
Dixanadu
254
2
Okay thank you :) I do have a question though - why is it wrong? is the sign of the nabla meant to be negative or is it just completely wrong?
 
  • #4
HomogenousCow
735
206
Okay thank you :) I do have a question though - why is it wrong? is the sign of the nabla meant to be negative or is it just completely wrong?

Didn't see this sorry.
The partial derivative with the greek subscript just refers a derivative with respect to x,y,z or t.
 

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