Noether Current: Understanding 2.10 & 2.11

  • Thread starter Thread starter LCSphysicist
  • Start date Start date
  • Tags Tags
    Current Noether
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
LCSphysicist
Messages
644
Reaction score
163
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) )##
 
Physics news on Phys.org