Undergrad Why is ∂σ(δxσ) not equal to zero?

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The discussion centers on the derivation of Noether's theorem, specifically the equation J=1+∂σ(δxσ). The confusion arises from the assumption that ∂σ(δxσ) can be simplified to δ(∂σ(xσ)), leading to the conclusion that it equals zero. However, δxσ is a function of x, which means it cannot be treated as a constant during differentiation. Participants suggest consulting specific sections of relevant literature for clarification on the transformation properties involved in the theorem. Understanding the distinction between the variables and their roles in the symmetry transformation is crucial for resolving the confusion.
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I'm studying Noether theorem.In the derivation process,I saw a equation:J=1+∂σ(δxσ),where J is the Jacobian,the second σ is superscript...Since δ has the property to commute with differentiation,why is ∂σ(δxσ) not equal to δ(∂σ(xσ))=δ (1+1+1+1) =δ (4) =0?
 
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Could you please use LaTeX? I really can't read your formulae :-(. The derivation of Noether's theorem for fields, given in the stackexchange article is also in my above quoted FAQ article.

To get the determinant of a matrix ##\hat{A}=\hat{1} + \delta \hat{\omega}## to first order in ##\delta## just use the definition of the determinant using the Levi-Civita symbol (Einstein summation convention applies)
$$\mathrm{det} \hat{A} = \epsilon_{j_1 j_2 \cdots j_n} A_{1j_1} A_{2 j_2} \cdots A_{n j_n}.$$
It's also clear that all products occurring in this sum are of order ##\mathcal{O}(\delta^2)## or higher except the product of the diagonal elements, i.e., (summation convention doesn's apply in the next formula)
$$\mathrm{det} \hat{A} =\prod_{j} A_{jj} + \mathcal{O}(\delta^2) = 1 + \sum_{j} \delta \omega_{jj} + \mathcal{O}(\delta^2) = 1 + \mathrm{Tr} \delta \hat{\omega} + \mathcal{O}(\delta^2).$$
 
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vanhees71 said:
Could you please use LaTeX? I really can't read your formulae :-(. The derivation of Noether's theorem for fields, given in the stackexchange article is also in my above quoted FAQ article.

To get the determinant of a matrix ##\hat{A}=\hat{1} + \delta \hat{\omega}## to first order in ##\delta## just use the definition of the determinant using the Levi-Civita symbol (Einstein summation convention applies)
$$\mathrm{det} \hat{A} = \epsilon_{j_1 j_2 \cdots j_n} A_{1j_1} A_{2 j_2} \cdots A_{n j_n}.$$
It's also clear that all products occurring in this sum are of order ##\mathcal{O}(\delta^2)## or higher except the product of the diagonal elements, i.e., (summation convention doesn's apply in the next formula)
$$\mathrm{det} \hat{A} =\prod_{j} A_{jj} + \mathcal{O}(\delta^2) = 1 + \sum_{j} \delta \omega_{jj} + \mathcal{O}(\delta^2) = 1 + \mathrm{Tr} \delta \hat{\omega} + \mathcal{O}(\delta^2).$$
Thank you very much!I saw##J=1+\partial_\sigma \delta x^\sigma##in that thread(the first answer)But δ has the property to commute with differentiation,so I think it should continue to be equal to##1+\delta \partial_\sigma x^\sigma=1+\delta(1+1+1+1)=1+\delta(4)=1+0=1##...What did I do wrong?
 
No, ##\delta x^{\sigma}## is a given function of ##x##, defining the transformation of the space-time coordinates, which is part of a symmetry transformation (e.g., Poincare transformations) of a field theory. See Sect. 3.1 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
 
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vanhees71 said:
Could you please use LaTeX? I really can't read your formulae :-(. The derivation of Noether's theorem for fields, given in the stackexchange article is also in my above quoted FAQ article.

To get the determinant of a matrix ##\hat{A}=\hat{1} + \delta \hat{\omega}## to first order in ##\delta## just use the definition of the determinant using the Levi-Civita symbol (Einstein summation convention applies)
$$\mathrm{det} \hat{A} = \epsilon_{j_1 j_2 \cdots j_n} A_{1j_1} A_{2 j_2} \cdots A_{n j_n}.$$
It's also clear that all products occurring in this sum are of order ##\mathcal{O}(\delta^2)## or higher except the product of the diagonal elements, i.e., (summation convention doesn's apply in the next formula)
$$\mathrm{det} \hat{A} =\prod_{j} A_{jj} + \mathcal{O}(\delta^2) = 1 + \sum_{j} \delta \omega_{jj} + \mathcal{O}(\delta^2) = 1 + \mathrm{Tr} \delta \hat{\omega} + \mathcal{O}(\delta^2).$$
vanhees71 said:
No, ##\delta x^{\sigma}## is a given function of ##x##, defining the transformation of the space-time coordinates, which is part of a symmetry transformation (e.g., Poincare transformations) of a field theory. See Sect. 3.1 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
I get it!I appreciate your help very much
 
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