Spacetime translations and general Lagrangian density for Field Theory

[Paulpaulpa]

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 ?

 

[LCSphysicist]

$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
As you pointed, first term is zero
$$\delta L = \frac{\partial L}{\partial (\partial_{\mu} \phi)} (\partial_{\mu} \delta \phi)$$
$$\delta \phi = e^{\mu} \frac{\partial \phi}{\partial x^{\mu}}$$

$$\delta L = \frac{\partial L}{\partial (\partial_{\mu} \phi)} \partial_{\mu} (e^{\mu}{\partial_{\mu} \phi}) = e^{\mu} \frac{\partial L}{\partial (\partial_{\mu} \phi)} \partial_{\mu}{\partial_{\mu} \phi} = e^{\mu} \partial_{\mu} L$$

I used chain rule.

 

[Paulpaulpa]

$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
As you pointed, first term is zero
$$\delta L = \frac{\partial L}{\partial (\partial_{\mu} \phi)} (\partial_{\mu} \delta \phi)$$
$$\delta \phi = e^{\mu} \frac{\partial \phi}{\partial x^{\mu}}$$

$$\delta L = \frac{\partial L}{\partial (\partial_{\mu} \phi)} \partial_{\mu} (e^{\mu}{\partial_{\mu} \phi}) = e^{\mu} \frac{\partial L}{\partial (\partial_{\mu} \phi)} \partial_{\mu}{\partial_{\mu} \phi} = e^{\mu} \partial_{\mu} L$$

I used chain rule.

Thanks for you answer. The first term is zero because of the equation of motion and the fact that
$\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) = 0$ since $\mathcal{L}$ do not depend on x. Is that right ? But why is $\partial_\mu \mathcal{L}$ different than zero ? $\mathcal{L}$ is supposed to be independant of x. Or should I interpret $\partial _{\mu} \mathcal{L}$ as a total derivative and not a partial one ?

 

[samalkhaiat]

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 ?

1) \partial_{\nu}\mathcal{L} = \frac{\partial \mathcal{L}}{\partial x^{\nu}}|_{\phi , \partial \phi} + \frac{\partial \mathcal{L}}{\partial \phi}\partial_{\nu}\phi + \frac{\partial \mathcal{L}}{\partial \partial_{\mu}\phi}\partial_{\nu}\partial_{\mu}\phi . The first term on the RHS vanishes when \mathcal{L} has no explicit dependence on x^{\mu}.

2) Don’t use the equation of motion when you want to find the change in the Lagrangian under a specific (given) transformation of the coordinates. Under infinitesimal translation, x \to x - e, the field changes according to \delta \phi = e^{\nu}\partial_{\nu}\phi . This induces the following (infinitesimal) change in the Lagrangian: \delta \mathcal{L} = e^{\nu} \left(\frac{\partial \mathcal{L}}{\partial \phi}\partial_{\nu}\phi + \frac{\partial \mathcal{L}}{\partial \partial_{\mu}\phi}\partial_{\nu}\partial_{\mu}\phi \right) . Since \mathcal{L} does not depend (explicitly) on x, the above is written as \delta \mathcal{L} = e^{\mu}\partial_{\mu}\mathcal{L} . \ \ \ \ \ \ (2) This is called “off-shell” change in the Lagrangian, meaning that Eq(2) has been obtained without using the equation of motion.

3) An arbitrary (infinitesimal) change in the field \delta \phi induces the following infinitesimal change in the Lagrangian: \delta \mathcal{L} = \left[ \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} \right) \right] \delta \phi + \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \delta \phi \right) . When you use the equation of motion, you get the so-called “on-shell” change in the Lagrangian: \delta \mathcal{L} = \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \delta \phi \right) . \ \ \ \ \ \ \ \ \ (3)

4) In this method, the (Noether) symmetry current is obtained by equating the off-shell variation Eq(2) with the on-shell variation Eq(3) with \delta \phi = e^{\nu}\partial_{\nu}\phi: \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi )} \partial_{\nu}\phi - \delta^{\mu}_{\nu} \mathcal{L} \right) = 0 .

 

[Paulpaulpa]

1) \partial_{\nu}\mathcal{L} = \frac{\partial \mathcal{L}}{\partial x^{\nu}}|_{\phi , \partial \phi} + \frac{\partial \mathcal{L}}{\partial \phi}\partial_{\nu}\phi + \frac{\partial \mathcal{L}}{\partial \partial_{\mu}\phi}\partial_{\nu}\partial_{\mu}\phi . The first term on the RHS vanishes when \mathcal{L} has no explicit dependence on x^{\mu}.

Thank you very much this is much more clear. I always noted total derivative $d \mathcal{L}$ so the $\partial _{\nu}\mathcal{L}$ confused me a lot. Your explanation of Noether's Current is more understandable than the one in my lecture.

 

[realanswers]

$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
As you pointed, first term is zero
$$\delta L = \frac{\partial L}{\partial (\partial_{\mu} \phi)} (\partial_{\mu} \delta \phi)$$
$$\delta \phi = e^{\mu} \frac{\partial \phi}{\partial x^{\mu}}$$

I used chain rule.

Where was the first term shown to be 0 / why would it be 0, this seems incorrect...?

I get your second reply specifying the off-shell condition but I do not see where you showed that $\frac{\partial L}{\partial \phi} \delta \phi= 0$