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In peskin and Schroeder's qft book, in chapter two, they're discussing Noether's theorem with respect to translations of co-ordinates.

They describe and "infinitesimal" translation $x^\mu\rightarrow x^\mu -a^\mu$.

And say that as an alternative it can be seen as a transformation of the field configuration as

$\phi(x)\rightarrow \phi(x+a)=\phi(x)+a^\mu \partial_\mu \phi(x)$.

Now according to David tong's notes this is the active point of view of the transformation but I'm still a bit confused by either viewpoint. I've read posts here on physics stack, which are about the same thing but they haven't helped me so far.

The way I see it if we set $f(x)=x-a$ Then $\phi'(x)=\phi(f^{-1}(x-a))=\phi(x+a)$. Is this the idea of an active transformation?

In which case would the passive version simply be $\phi(x-a)$?

I've read the example in boas with SO(2) acting on $\Bbb{R}^2$, where they said in one viewpoint we change the basis but in the other viewpoint we change the vector itself. Is there any benefit to choosing either though in the above case of peskin and Schroeder, wrt it's context of Noether's theorem?

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3 Answers 3

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A passive transformation is a mere change of coordinates on a manifold and there is usually nothing special about this. Every tensor equation is trivially invariant (or better to say covariant) under such transformations. In terms of manifolds and atlases, this corresponds to just a change of charts. So let $M$ be the said manifold, say e.g. real, differentiable and of dimension $n$, and let $(U,\phi)$ and $(V,\psi)$ be two charts on $M$ such that $U\cap V\neq\varnothing$. The map $\psi\circ\phi^{-1}:\phi(U\cap V)\to\psi(U\cap V)$ is a diffeomorphism that represents a change of coordinates on the overlap $U\cap V$ over $M$ and involves open subsets of $\mathbb R^n$.

An active transformation is an actual motion on the manifold $M$ through a diffeomorphism $\phi:M\to M$ which sends any point $p$ to the new point $q = \phi(p)$. When performing an active transformation, not only you move the point $p$ to $q$, but also every tensor/form on $p$ is to be moved from the tensor/exterior algebra $T(T_pM)$/$\bigwedge^*(T^*_pM)$ to $T(T_qM)$/$\bigwedge^*(T^*_qM)$ through the maps $\phi(p)_*$ and $\phi(p)^*$ induced by $\phi$. An interesting case is when $\phi$ arises from the flow of a vector field (e.g. a Killing vector field). Then the notion of symmetry is studied by means of the Lie derivative of a tensor by the vector field. Thus, the description in terms of local coordinates is the following. Fix a point $p\in M$ and charts $(U,\rho)$, $(V,\psi)$ such that $p\in U$ and $\phi(p)\in V$. The map $\psi\circ\phi\circ\rho^{-1}:\rho(U\cap\phi^{-1}(V))\to\psi(\phi(U)\cap V)$ is a diffeomorphism between open subsets of $\mathbb R^n$, and its differential relates tensor quantities between tangent spaces.

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  • $\begingroup$ I think part of what's confusing me is that the manifold is $\Bbb{R}^4$ and the chart could be taken as an identity to $\Bbb R^4$. So I need to compare a translation of co-ordinates after the identity (- sign one) to one that maps the manifold to itself in a way that the two are the same? Which should tell me that for active I swap to the plus one? $\endgroup$
    – snulty
    Commented Jan 13, 2015 at 18:12
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Though this question was asked a long time ago, I'd like to answer it now because I was stuck with this for a while and I didn't find a convincing answer here but I finally understood the idea.

An active transformation is a true transformation of the field/coordinates, i.e. in active transformation we keep the field intact and move the coordinate system; which is the same as keeping the coordinate system intact and moving the field in the opposite direction. So, making a spacetime transformation $$x^\mu\rightarrow x^\mu-a^\mu$$ is the same as transforming the field in the opposite direction i.e. $$\phi(x)\rightarrow \phi(x+a).$$ As we can see, the field is shifted(transformed) w.r.t the coordinate system. So, this is what Peskin and Schroeder book means when it says "We can describe the infinitesimal translation alternatively as a transformation of the field configuration"

On the other hand, a passive transformation is not a true transformation. Here, both field and coordinate system move together i.e. the field is not shifted(transformed) w.r.t the coordinate system. So if $$x^\mu\rightarrow x^\mu-a^\mu$$ then the field moves along with it as $$\phi(x)\rightarrow \phi(x-a).$$ So this is not really a transformation, it is just re-labeling, in the sense that $x$ will be called $x-a$, and therefore $\phi(x)$ will now be called $\phi(x-a)$.

In QFT, whenever we say that we are making a transformation, we mean active transformation. So any invertible transformation by $\Lambda$ given by $$x^\mu\rightarrow \Lambda^\mu_\nu x^\nu$$ can be alternatively described as a transformation of the field configuration $$\phi(x)\rightarrow \phi'(x)=\phi({\Lambda}^{-1}x).$$ One can see that this works for rotations i.e. $\Lambda \in SO(n)$ as well.

So yes, your idea of active and passive transformation is right, and clearly, the idea behind the two transformations is different. To understand why we use active transformations, one can consider an example of transformation of the field under a boost. What happens here is that the field appears to move w.r.t the coordinate system you are in (which is an active transformation, not passive). If the equations of motion are invariant under this transformation, then according to Noether's theorem, there is a conserved quantity.

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Imagine your scalar field as a "patch" extending over some finite region of space, pictured as a 3D grid (or 4D if you want to include time).

An active transformation is considering the "patch" moving while the background grid is fixed.

A passive transformation is considering the "patch" to be fixed while the background grid moves oppositely to the above case.

Either way, there is no difference for what concerns Noether theorem (or any other theorem as far as I know): the $a$ parameter cancels out at the end and whether it is positive or negative is the same.


A few related questions:

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