Field Theory: Prove transformations are a symmetry

AI Thread Summary
The discussion centers on proving that a specific transformation of the Lagrangian is a symmetry when the mass m is zero. The transformation involves shifting the fields φ and φ* by constants a and a*, respectively. Participants clarify that the invariance of the Lagrangian requires it to be expressible as a total derivative, which leads to the conclusion that a and a* must be constants independent of spacetime. This independence is crucial for simplifying the Lagrangian and confirming the symmetry, particularly in the absence of gravity. The conversation concludes with the acknowledgment that while the notation does not explicitly indicate the dependence of φ on spacetime, the nature of the transformation implies that a must be constant.
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Homework Statement
Consider the lagrangian L=\delta_\mu \phi \delta^\mu \phi^* - m^2 \phi \phi^*

Show that the transformation:
\phi \rightarrow \phi + a \,\,\,\,\,\,\,\,\,\, \phi^* \rightarrow \phi^* + a^*
is symmetry when m=0.

The attempt at a solution
Substituting the transformation into the lagrangian gives:
L=\delta_\mu ( \phi + a ) \delta^\mu ( \phi^* + a^* )- m^2 (\phi + a) ( \phi^* + a^*)
= (\delta_\mu \phi + \delta_\mu a) ( \delta^\mu \phi^* + \delta^\mu a^*) - m^2 ( \phi \phi^* + a^* \phi + a \phi^* + aa^*)

I know that this is a symmetry when the action is invariant, so the langrangian is invariant up to a total derivative. L \rightarrow L + \delta_\mu J^\mu

I can't see how I could rewrite the above expression so that L \rightarrow L + \delta_\mu J^\mu . Obviously the second term vanishes with m=0, but expanding the first gives:
L = \delta_\mu \phi \delta^\mu \phi^* + \delta_\mu a \delta^\mu a^* + \delta_\mu a \delta^\mu \phi^* + \delta_\mu \phi \delta^\mu a^*
meaning the total derivative would have to be
\delta_\mu J^\mu = \delta_\mu a \delta^\mu a^* + \delta_\mu a \delta^\mu \phi^* + \delta_\mu \phi \delta^\mu a^*
which is a form I can't rewrite it into (what would J^\mu be?)

The only solution I can see would be for a & a* to be independent of spacetime, so that \delta_\mu a = \delta^\mu a^* = 0 , but other than answering the question I don't know if there is any justification in saying that.

So, are a/a* independent of spacetime, or should I be able to rearrange the above to form a total derivative term? Or am I missing something else completely?

Thanks.
 
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gu1t4r5 said:
Homework Statement
Consider the lagrangian L=\delta_\mu \phi \delta^\mu \phi^* - m^2 \phi \phi^*

Show that the transformation:
\phi \rightarrow \phi + a \,\,\,\,\,\,\,\,\,\, \phi^* \rightarrow \phi^* + a^*
is symmetry when m=0.

The attempt at a solution
Substituting the transformation into the lagrangian gives:
L=\delta_\mu ( \phi + a ) \delta^\mu ( \phi^* + a^* )- m^2 (\phi + a) ( \phi^* + a^*)
= (\delta_\mu \phi + \delta_\mu a) ( \delta^\mu \phi^* + \delta^\mu a^*) - m^2 ( \phi \phi^* + a^* \phi + a \phi^* + aa^*)

I know that this is a symmetry when the action is invariant, so the langrangian is invariant up to a total derivative. L \rightarrow L + \delta_\mu J^\mu

I can't see how I could rewrite the above expression so that L \rightarrow L + \delta_\mu J^\mu . Obviously the second term vanishes with m=0, but expanding the first gives:
L = \delta_\mu \phi \delta^\mu \phi^* + \delta_\mu a \delta^\mu a^* + \delta_\mu a \delta^\mu \phi^* + \delta_\mu \phi \delta^\mu a^*
meaning the total derivative would have to be
\delta_\mu J^\mu = \delta_\mu a \delta^\mu a^* + \delta_\mu a \delta^\mu \phi^* + \delta_\mu \phi \delta^\mu a^*
which is a form I can't rewrite it into (what would J^\mu be?)

The only solution I can see would be for a & a* to be independent of spacetime, so that \delta_\mu a = \delta^\mu a^* = 0 , but other than answering the question I don't know if there is any justification in saying that.

So, are a/a* independent of spacetime, or should I be able to rearrange the above to form a total derivative term? Or am I missing something else completely?

Thanks.
Yes, "a" is a constant (independent of x). In the absence of gravity, that's all we can do. If we include gravity (by working withe formalism of GR and introducing the metric), then one may make space-time transformations local.
 
Great, thanks.
Is there a way to justify that a is a constant from the notation? (I assume not as there is no notation here to signify the dependence of \phi on X, but the question does not explicitly say a has no dependence)
 
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