Why is ∂σ(δxσ) not equal to zero?

[GR191511]

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?

 

[vanhees71]

From which book are you studying from? I cannot read your equations? Is it about Noether for fields? Then maybe Sec. 3.1 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

is of some use.

 

[GR191511]

From which book are you studying from? I cannot read your equations? Is it about Noether for fields? Then maybe Sec. 3.1 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

is of some use.

https://physics.stackexchange.com/questions/534699/noethers-theorem-derivation-for-fields
the first answer said"the Jacobian J=1+∂σ(δxσ)"But I think：1+∂σ(δxσ)=1+δ(∂σxσ)=1+δ (1+1+1+1) =1+δ (4) =1+0=1 ...I don't know what I did wrong.Thank you.

 

[vanhees71]

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)
$$\mathrm{det} \hat{A} =\prod_{j} A_{jj} + \mathcal{O}(\delta^2) = 1 + \sum_{j} \delta \omega_{jj} + \mathcal{O}(\delta^2) = 1 + \mathrm{Tr} \delta \hat{\omega} + \mathcal{O}(\delta^2).$$

 

[GR191511]

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?

 

[vanhees71]

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

 

[GR191511]

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)
$$\mathrm{det} \hat{A} =\prod_{j} A_{jj} + \mathcal{O}(\delta^2) = 1 + \sum_{j} \delta \omega_{jj} + \mathcal{O}(\delta^2) = 1 + \mathrm{Tr} \delta \hat{\omega} + \mathcal{O}(\delta^2).$$

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