Noether Current: Understanding 2.10 & 2.11

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The discussion centers on the confusion surrounding the transition from equation (2.11) to the interpretation of the Noether current. It highlights that the second term in the equation is zero, leading to the expression involving the derivative of the Lagrangian. The participants question whether the equation should simplify to a form that includes the Noether current, J^{\mu}(x). There is a focus on the separation of the alpha factor in the final equation, emphasizing its role in the derivation. The conversation aims to clarify the mathematical steps and implications of these equations in the context of Noether's theorem.
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I can't see why the expression gives by the author is right.
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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) )##
 
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As defined in 2.11 ##\Delta \mathcal{L} = \partial_\mu \mathcal{J}^\mu##, notice that the ##\alpha## factor is taken into account separatelly.
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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