Active versus passive transformations

Question

I am a bit confused by the concepts of active and passive transformations. In all the courses I am doing at the moment we do transformations of the form:

$$\phi(x) \rightarrow\phi'(x') = \phi(x)$$

and

$$\partial_{\mu}\phi(x) \rightarrow \partial_{\nu'} \phi'(x') = \frac{\partial x^{\alpha}}{\partial x^{\prime \nu}} \partial_{\alpha}\phi(x)$$

This is all perfectly clear to me. However, I am reading Peskin and Schoder at the moment, and they adapt an "active" point of view (their words), such that the above transformations are:

$$\phi(x) \rightarrow\phi'(x) = \phi(\Lambda^{-1}x)$$

and

$$\partial_{\mu}\phi(x)\rightarrow\partial _{\mu}(\phi(\Lambda^{-1}x)) = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x).$$

I don't understand how to interpret this and especially how to derive the second equation.

Answer 1

What you wrote down is the same as what Peskin writes. To see this, notice that if we write the "transformed" position $x'$ as $x' = \Lambda x$, then your first equation can be written as

$\phi'(\Lambda x) = \phi(x)$

but this equivalent to

$\phi'(x) = \phi(\Lambda^{-1} x)$

which is the same as the first Peskin equation you wrote down. Your second equation and Peskin's second equation are equivalent. You can show this by using the definition of $x'$ plus the chain rule for partial differentiation. I can add details if you'd like, but I think it's a good exercise to figure out.

Active v. Passive

The convention in which we define $\phi'(x) = \phi(\Lambda^{-1} x)$ is the active convention because the transformed field value at the transformed point is that same as the non-transformed field value at the non-transformed point, so it's as if we have kept our coordinate system fixed and transformed the field configuration. To get intuition for this, imagine a temperature field $T$ in a 2D laboratory, and imagine keeping the laboratory fixed, but rotating the entire temperature field counterclockwise by a rotation $R$ to obtain a temperature field $T'$. Then (drawing a picture helps) the new temperature field evaluated at a counterclockwise rotated point $R x$ should be the same as the old temperature field evaluated at the non-rotated point $x$, namely $T'(R x) = T(x)$ which is the same as $T'(x) = T(R^{-1} x)$

The passive convention is the one in which we define $\phi'(x) = \phi(\Lambda x)$ and has the interpretation of transforming the coordinates while keeping the field configuration fixed. Try using the temperature analogy to understand this.

Answer 2

There is a lot of confusion in the literature regarding the so-called active and passive interpretation of transformations when it comes to scalar fields. However, this terminology and the corresponding dichotomy has its origins in the applications of linear algebra (e.g. computer vision) where it is more relevant and the concepts are more clear. The Wikipedia article on this subject makes this point very clear.

Transformation of vector spaces:

Consider a spatial transformation $T:\mathbb R^3 \to \mathbb R^3$. This can be interpreted to either transform a vector $v = v_1 e_x + v_2 e_y + v_3 e_z \in \mathbb R^3$ keeping the basis fixed or to transform the initial basis $\{e_x,e_y,e_z\}$ of $\mathbb R^3$ keeping the vector $v$ fixed. These two lines of interpretation of $T$ go by two names.

\begin{aligned} &\textit{Active (alibi) transformation}: && \text{vector } v \text{ rotates} \left(T: v \mapsto v'=Tv \equiv v_1' e_x + v_2' e_y + v_3' e_z\right), \\ & && \text{basis }\{e_x,e_y,e_z\} \text{ remains unchanged}. \\ \\ &\textit{Passive (alias) transformation}: && \text{vector } v \text{ stays put}, \\ & && \text{basis rotates } \left(\{e_x,e_y,e_z\}\mapsto \{T^{-1}e_x, T^{-1}e_y, T^{-1}e_z\}\right). \end{aligned}

From the first interpretation $v = T^{-1}v'$, it follows that $v = v_1' \tilde e_x + v_2' \tilde e_y + v_3' \tilde e_z$ where $\tilde e_x := T^{-1} e_x$, $\tilde e_y := T^{-1} e_y$ and $\tilde e_z := T^{-1} e_z$ are the transformed basis vectors from the second interpretation. Thus, the original vector $v$ in the rotated basis $\{\tilde e_x, \tilde e_y, \tilde e_z\}$ (in the passive point of view) has exactly the same coordinates $(v_1',v_2',v_3')$ as the rotated vector $v'$ in the original basis (the active point of view).

Transformation of scalar fields

This dichotomy is not very useful when it comes to scalar fields and, therefore, the literature lacks a canonical definition for these concepts. One way of thinking about them could be as user @joshphysics has written about. Here is another way which is equally popular. A scalar field is a real-valued map $\phi : \Omega \subset \mathfrak{M}_4 \to \mathbb R$. Consider a transformation $T:\Omega \to \Omega' \subseteq \Omega$ of the underlying spacetime domain. Now, one can imagine either a rotated field $\phi_A := \phi \circ T^{-1}: \Omega' \to \mathbb R$ or an oppositely-rotated field $\phi_P := \phi \circ T: \Omega \to \mathbb R$ to visualize this transformation. The two new fields can be interpreted in the following manner.

\begin{aligned} &\textit{Active (alibi) transformation}: && \text{field configuration } \phi\big|_{\Omega'} : \Omega' \to \mathbb R \text{ has morphed into } \phi_A : \Omega' \to \mathbb R,\\ & && \text{leaving the spacetime domain } \Omega' \text{untouched}. \\ \\ &\textit{Passive (alias) transformation}: && \text{field configuration } \phi_P \text{ is simply } \phi \text{ acting on a rotated domain},\\ & && \text{which is to say, } \phi_P(x) = \phi(T(x)) \text{ where } T:\Omega \to \Omega', x \mapsto x'. \end{aligned}

However, unlike what @joshphysics's answer might seem to suggest, $\phi_A$ (or $\phi'$) does not necessarily define the active convention. One could very well see $\phi_A$ from the passive point of view: $\phi_A$ could simply be $\phi$ acting on an oppositely rotated domain, that is, $\phi_A(x') = \phi(T^{-1}x')$ where $T^{-1}: \Omega' \to \Omega, x' \mapsto T^{-1}x'$. Similarly, $\phi_P$ could be interpreted according to the active perspective: here, field $\phi: \Omega \to \mathbb R$ has morphed into a new field $\phi_P = \phi \circ T : \Omega \to \mathbb R$, leaving the spacetime domain $\Omega$ untouched.

This should tell you that any field redefinition obtained from a spacetime transformation can be seen in both active and passive interpretations and such vacuous names/interpretations hold no physical or mathematical value. What you should actually care about is how exactly you have defined your new fields and then everything else should follow irrespective of what your mental picture is.

Answer 3

The claim in Peskin and Schroeder is that if you do an active transformation $$\phi(x)\rightarrow \phi'(x) \doteq \phi(\Lambda^{-1}(x)) \ \ (1) $$ then the gradient $\partial_{\mu}\phi$ transforms as $$ \partial_{\mu}\phi(x) \rightarrow (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)$$ Here's an explicit example - work in $\mathbb{R}^2$ with coordinates $x^{\mu} = (\begin{smallmatrix} x \\ y \end{smallmatrix})$

Consider the scalar field $$ \phi(x^{\mu}) = x^2 + xy$$ and the active transformation of (x,y) represented by the matrix $$ \Lambda^{\mu}_{\ \ \nu} = (\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix})$$ this represents a rotation by $\frac{\pi}{2}$, where $\mu$ labels the row and $\nu$ labels the column. The vectors, with upper indices, on which it acts would be represented as column vectors $x^{\mu} = (\begin{smallmatrix} x \\ y \end{smallmatrix})$ . Its inverse is represented by the matrix $$ ({\Lambda}^{-1})^{\mu}_{\ \ \ \nu} = (\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}) $$ The new field after this active transformation (1) is $$ \phi'(x^{\mu})=\phi(y, -x) = y^2-xy$$ Its gradient is $$\partial_\mu \phi' = (-y, \ 2y-x) \ \ (2) $$ Note that, since $\partial_{\mu}\phi$ has a lower index we represent it by a row vector.

Now let's see what we would get if we apply the Peskin and Schroeder prescription to derive the gradient of the new field:

We start with the gradient of the old field $$ (\partial_{\mu}\phi)(x^{\nu})= (2x+y, \ \ x) $$ Next we, instead of evaluating it at $(\begin{smallmatrix} x \\ y \end{smallmatrix})$, evaluate it at $(\Lambda^{-1})^{\mu}_{\nu}x^{\nu} = (\begin{smallmatrix} y \\ -x \end{smallmatrix})$, giving $$ (\partial_{\mu}\phi)((\Lambda^{-1})x)=( 2y-x, \ \ y ) $$ Finally we apply a $(\Lambda)^{-1}$ rotation to the row vector $ (2y-x, \ y )$ giving $$ (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x) =( -y, \ 2y-x ) $$ Note that, in the combination of the $(\Lambda^{-1})^{\nu}_{\ \ \mu}$ and $\partial_{\nu}\phi$ factors, since $\partial_{\nu}\phi$ has a lower index, and the contraction is with the index of $\Lambda^{-1}$ which represents the columns, the matrix representation of this operation in our case is $$ (2y-x, \ y )(\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}) $$ Note in the first version of this answer I made the mistake of representing this step as a matrix multiplication with a column vector! (Thanks @joshphysics for pointing this out)