The discussion revolves around the mathematical expression ∂σ(δxσ) in the context of Noether's theorem, particularly regarding its derivation and implications in field theory. Participants explore the properties of differentiation and transformations related to symmetries in physics.
Participants do not reach a consensus on the treatment of the expression ∂σ(δxσ) and its implications. There are competing views on the interpretation of δ and its properties in this context.
Some participants highlight the need for clarity in mathematical notation and the assumptions underlying the transformations discussed. The discussion includes references to specific sections of literature that may contain relevant information.
https://physics.stackexchange.com/questions/534699/noethers-theorem-derivation-for-fieldsvanhees71 said: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.
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 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: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)
$$\mathrm{det} \hat{A} =\prod_{j} A_{jj} + \mathcal{O}(\delta^2) = 1 + \sum_{j} \delta \omega_{jj} + \mathcal{O}(\delta^2) = 1 + \mathrm{Tr} \delta \hat{\omega} + \mathcal{O}(\delta^2).$$
I get it!I appreciate your help very muchvanhees71 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