1
$\begingroup$

I am trying to understand the continuum version of Noethers theorem from this source (p 15- 17) however I am stuck on a couple of points. I will go through what I have so far and then ask my questions at the end.

A tensor field of rank $(p,q)$ is written $\phi ^{i_1\dots i_p}_{j_1\dots j_q}$ and it follows from the mathematics of tensors that under a coordinate transformation $x^i\mapsto x'^i(x^i)$ the tensor field will transform as below $\phi^{i_1\dots i_p}_{j_1\dots j_q}\mapsto \phi '^{k_1\dots k_p}_{l_1\dots l_q}(x')$ under a general coordinate transformation.

\begin{equation} \phi '^{k_1\dots k_p}_{l_1\dots l_q}(x')=\frac{\partial x'^{k_1}}{\partial x^{i_1}}\dots \frac{\partial x'^{k_p}}{\partial x^{i_p}}\frac{\partial ^{j_1}}{\partial x'^{l_1}}\dots \frac{\partial x^{j_q}}{\partial x'^{l_q}}\phi^{i_1\dots i_p}_{j_1\dots j_q }(x) \end{equation} In the infinitesimal case we can transform the coordinates $x\mapsto x+\delta x$ and Taylor expand $\phi'_I(x')=\phi '_I(x+\delta x)$. \begin{equation} \phi '_I(x+\delta x)=\phi '_I(x)+\frac{\partial }{\partial x^k}\phi '_I(x)\delta x^k +\dots \end{equation} We now define these two quantities because $\delta $ and $\partial /\partial x$ don't commute(?). \begin{equation} \delta x^i=\sum _{n=1}^{d}X^i_n\delta \omega _n,\ \ \ \ \ \ \ \ \ \delta \phi _I(x)=\sum _{n=1}^{d}\Phi _{I,n}\delta \omega _n \end{equation} Allowing us to write the variation of the form of a field (?) as below. \begin{equation} \bar \delta \phi _I(x)=\phi '_I(x)-\phi _I(x)=(\Phi _{I,n}-\partial _k \phi _I X^k_n)\delta \omega _n \end{equation}

So my questions are: 1) Why do we break these variations up into (what to me seems like) components and basis. 2) What is this variation of the form of a field? I don't understand its form either, I can't see how it follows from the above definitions (sorry if it's obvious). I am really struggling with this derivation so your help is really appreciated!

$\endgroup$
2
$\begingroup$

First of all $\delta$ and $\frac{\partial}{\partial x}$ don't commute because $$ \delta \left(\frac{\partial}{\partial x}\phi\right) = \delta\left( \frac{\partial}{\partial x}\right)\phi + \frac{\partial}{\partial x}\left(\delta\phi\right) $$ Second we divide $\delta\phi_I = \bar{\delta}\phi_I + X^k_n \delta \omega_n \partial_k \phi_I$ because the total variation of the field contains two contributions. One comes from how the components of the field transform because of it's tensor nature which, and the other coming from the variation of the $x^\mu$ argument. Notice in your first equation you had $\phi'_I(x') =S^J_I(x) \phi_J(x)$ (i changed notation for brevity), so that $S^J_I(x)$ is the transformation of the form of the field, (remember also that $S^I_J(x) = \delta^I_J + \delta\omega \ldots$) \begin{align*} S^J_I(x) \phi_J(x) &= \phi'_I(x') = \phi'_I (x + X\cdot\delta\omega)\\ \Rightarrow \phi'_I(x) &= S^I_J(x - X\cdot\delta\omega)\phi_J(x - X\cdot\delta\omega) \\ &= S^I_J(x)\phi_J(x - X\cdot\delta\omega) + \mathcal{O}(\delta\omega^2) \\ &= \phi_I(x - X\cdot\delta\omega) + \left( S^I_J(x) - \delta^I_J\right)\phi_J(x) + \mathcal{O}(\delta\omega^2) \\ &= \phi_I(x) - X\cdot\delta\omega \partial \phi + \left( S^I_J(x) - \delta^I_J\right)\phi_J(x) + \mathcal{O}(\delta\omega^2) \\ \Rightarrow \phi'_I(x) - \phi_I(x)&= - X\cdot\delta\omega \partial \phi + \left( S^I_J(x) - \delta^I_J\right)\phi_J(x) + \mathcal{O}(\delta\omega^2) \end{align*}

$\endgroup$
1
$\begingroup$

Ali moh's is a wonderfully full and descriptive answer, but unfortunately you are not looking for a purely mathematical answer in which case I will try to give the kind of answer you are looking for.

1) The reasons associated with the variations being described in terms of these sums have to do with perturbation theory (which goes beyond the scope of this question) which is a technique used in Linear algebra. In other words you need to study linear algebra, this book should really help Linear algebra and perturbation theory

2) The reason the author created the variation of the particular form he chose was because it didn't commute as the variation was I.e. Non-commutivity in the case of these transformations implies broken symmetry and broken symmetry is bad. As you will see later on the symmetry implied by noethers theorem is a requirement in the standard model.

If you have any questions feel free to ask, and a piece of advice DONT OVER-THINK/OVER-INTERPRET THINGS!!!!!!!! If you don't understand then it most likely because there is a piece of knowledge associated with it that you are you unaware of.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.