Spacetime translations and general Lagrangian density for Field Theory

In summary, Sydney Coleman discusses the Euler-Lagrange equations and the variation of fields under space-time translation in his lectures on Quantum Field Theory. He finds that the change in the Lagrangian can be expressed as a total derivative, which is known as the off-shell change. This is different from the on-shell change, which is obtained when using the equation of motion. The Noether symmetry current is obtained by equating these two variations.
  • #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 ?
 
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  • #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.
 
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  • #3
LCSphysicist said:
$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
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
Paulpaulpa said:
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
samalkhaiat said:
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.
 
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  • #6
LCSphysicist said:
$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta (\partial_{\mu} \phi)$$
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##
 

Related to Spacetime translations and general Lagrangian density for Field Theory

1. What is spacetime translation in field theory?

Spacetime translation is a concept in field theory that refers to the shifting of coordinates in space and time. In other words, it is a transformation that moves the origin of a coordinate system to a new location in spacetime. This is a fundamental symmetry in physics, meaning that the laws of nature remain unchanged under such transformations.

2. How does spacetime translation affect the Lagrangian density in field theory?

Spacetime translation affects the Lagrangian density by introducing a term called the Noether current. This current is conserved, meaning that it remains constant over time, and is related to the symmetries of the system. The Noether current is used to derive the equations of motion for a field, making it an essential concept in field theory.

3. What is the general Lagrangian density in field theory?

The general Lagrangian density is a mathematical expression that describes the dynamics of a field in a given system. It is a function of the field variables and their derivatives, and it encapsulates all the information about the interactions and symmetries of the system. The equations of motion for the field can be derived from the Lagrangian density using the principle of least action.

4. What is the role of the Lagrangian density in field theory?

The Lagrangian density plays a crucial role in field theory as it provides a framework for understanding the dynamics of a system. It allows us to calculate the equations of motion for a field and predict its behavior under different conditions. Additionally, the Lagrangian density is used to define the energy and momentum of a field, making it a fundamental concept in theoretical physics.

5. How are spacetime translations and the Lagrangian density related in field theory?

Spacetime translations and the Lagrangian density are intimately connected in field theory. The Lagrangian density is invariant under spacetime translations, meaning that it remains unchanged when coordinates are shifted. This symmetry leads to the conservation of the Noether current, which is related to the Lagrangian density. Therefore, spacetime translations are a fundamental concept in understanding the dynamics of a field described by the Lagrangian density.

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