Noether current

  • #1
LCSphysicist
636
153
Homework Statement:
I can't see why the expression gives by the author is right.
Relevant Equations:
.
1624046405408.png

I just don't understand what happened after (2.11). That' is, the second term is zero, so we have
$$\alpha \Delta L = \alpha \partial_{\mu} ( \frac{\partial L}{\partial (\partial_{\mu}\phi)} \Delta \phi )$$
So, second (2.10), isn't ##\Delta L = \alpha \partial_{\mu} J^{\mu}(x)##? So shouldn't the final equation reduce to this?:

##\alpha \alpha \partial_{\mu} J^{\mu}(x) = \alpha \partial_{\mu} ( \frac{\partial L}{\partial (\partial_{\mu}\phi)} \Delta \phi )##
##\partial_{\mu} (\frac{\partial L}{\partial (\partial_{\mu}\phi)} \Delta \phi - \alpha J^{\mu}(x) )##
 

Answers and Replies

  • #2
Gaussian97
Homework Helper
683
406
As defined in 2.11 ##\Delta \mathcal{L} = \partial_\mu \mathcal{J}^\mu##, notice that the ##\alpha## factor is taken into account separatelly.
 

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