- #1

Paulpaulpa

- 6

- 2

- Homework Statement
- Find the variation of a general Lagrangian density, independent of spacetime coordinates, under a translation in space time.

- Relevant Equations
- The space time translation 4-vector is ##e^{\mu}## and we have ##\frac{\partial \mathcal{L}}{\partial x}=0## with ##\mathcal{L}## the Lagrangian density

In Sydney Coleman Lectures on Quantum field Theory (p48), he finds : $$D\mathcal{L} = e^{\mu} \partial _{\mu} \mathcal{L}$$

My calulation, with ##\phi## my field and the variation of the field under space time tranlation ##D\phi = e^{\mu} \frac{\partial \phi}{\partial x^{\mu}}## :

$$D\mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}D\phi + \frac{\partial \mathcal{L}}{\partial \partial _{\nu} \phi} \partial _{\nu} D\phi$$

Then using the Euler Lagrange equations ##

\frac{\partial\mathcal{L}}{\partial\phi}=\partial_\mu

\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)##

I find $$D\mathcal{L} =

\partial_\mu

\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)

D\phi + \frac{\partial \mathcal{L}}{\partial \partial _{\nu} \phi} \partial _{\nu} D\phi$$

Then using the fact that ## \partial _{\mu} (ab) = \partial _{\mu} (a)b + a\partial _{\mu} (b)##

$$D\mathcal{L} = \partial _{\nu} (\frac{\partial \mathcal{L}}{\partial \partial _{\nu} \phi} D\phi)$$

I don't understand where I am wrong. Moreover how can ##\partial _{\mu} \mathcal{L}## be any different than zero since the lagrangian doesn't depend explicitely on spacetime ?

My calulation, with ##\phi## my field and the variation of the field under space time tranlation ##D\phi = e^{\mu} \frac{\partial \phi}{\partial x^{\mu}}## :

$$D\mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}D\phi + \frac{\partial \mathcal{L}}{\partial \partial _{\nu} \phi} \partial _{\nu} D\phi$$

Then using the Euler Lagrange equations ##

\frac{\partial\mathcal{L}}{\partial\phi}=\partial_\mu

\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)##

I find $$D\mathcal{L} =

\partial_\mu

\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)

D\phi + \frac{\partial \mathcal{L}}{\partial \partial _{\nu} \phi} \partial _{\nu} D\phi$$

Then using the fact that ## \partial _{\mu} (ab) = \partial _{\mu} (a)b + a\partial _{\mu} (b)##

$$D\mathcal{L} = \partial _{\nu} (\frac{\partial \mathcal{L}}{\partial \partial _{\nu} \phi} D\phi)$$

I don't understand where I am wrong. Moreover how can ##\partial _{\mu} \mathcal{L}## be any different than zero since the lagrangian doesn't depend explicitely on spacetime ?

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