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
This is because the partial derivative (∂) measures the rate of change of a function with respect to one of its variables, while the variable δxσ represents an infinitesimal change in the function. Therefore, ∂σ(δxσ) represents the rate of change of an infinitesimal change, which is not necessarily equal to zero.
∂σ(δxσ) represents the rate of change of an infinitesimal change in a function. It is a way to measure the sensitivity of a function to small changes in one of its variables.
Yes, it is possible for ∂σ(δxσ) to be equal to zero. This occurs when the function is not sensitive to small changes in the variable δxσ, meaning that the function does not change significantly as δxσ changes.
∂σ(δxσ) can be thought of as the limit of the function as δxσ approaches zero. This is because as δxσ becomes smaller and smaller, the rate of change of the function with respect to δxσ becomes closer and closer to the limit of the function.
No, ∂σ(δxσ) is not the same as the derivative of the function. The derivative is a measure of the instantaneous rate of change of a function with respect to its variable, while ∂σ(δxσ) is a measure of the rate of change of an infinitesimal change in the function. However, in some cases, the derivative can be calculated using ∂σ(δxσ).