Spacetime translations and general Lagrangian density for Field Theory

  • #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 ?
 
Last edited:

Answers and Replies

  • #2
LCSphysicist
636
153
$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
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.
 
  • #3
Paulpaulpa
6
2
$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
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 ?
 
  • #4
samalkhaiat
Science Advisor
Insights Author
1,791
1,188
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) [tex]\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 .[/tex] The first term on the RHS vanishes when [itex]\mathcal{L}[/itex] has no explicit dependence on [itex]x^{\mu}[/itex].

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, [itex]x \to x - e[/itex], the field changes according to [tex]\delta \phi = e^{\nu}\partial_{\nu}\phi .[/tex] This induces the following (infinitesimal) change in the Lagrangian: [tex]\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) .[/tex] Since [itex]\mathcal{L}[/itex] does not depend (explicitly) on [itex]x[/itex], the above is written as [tex]\delta \mathcal{L} = e^{\mu}\partial_{\mu}\mathcal{L} . \ \ \ \ \ \ (2)[/tex] 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 [itex]\delta \phi[/itex] induces the following infinitesimal change in the Lagrangian: [tex]\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) .[/tex] When you use the equation of motion, you get the so-called “on-shell” change in the Lagrangian: [tex]\delta \mathcal{L} = \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \delta \phi \right) . \ \ \ \ \ \ \ \ \ (3)[/tex]

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 [itex]\delta \phi = e^{\nu}\partial_{\nu}\phi[/itex]: [tex]\partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi )} \partial_{\nu}\phi - \delta^{\mu}_{\nu} \mathcal{L} \right) = 0 .[/tex]
 
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  • #5
Paulpaulpa
6
2
1) [tex]\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 .[/tex] The first term on the RHS vanishes when [itex]\mathcal{L}[/itex] has no explicit dependence on [itex]x^{\mu}[/itex].
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.
 
Last edited:
  • #6
realanswers
13
0
$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
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##
 

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