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:

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


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

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:

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


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

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

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    $\begingroup$ I am not sure that your (or Peskin) expressions related to the concepts of active and passive transformations. Actually the second set of identities is the same as the first: let's make the transformation $x'=\hat{\Lambda}x$, then use the identity $\phi(x)=\phi'(x')$, thus we obtain $\phi'(x')=\phi(\hat{\Lambda}^{-1}x')$ or simply $\phi'(x)=\phi(\hat{\Lambda}^{-1}x)$. Therefore the relation $\phi'(x)=\phi(\hat{\Lambda}^{-1}x)$ is still passive transformation. $\endgroup$ Jan 23 '13 at 19:28

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.

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    $\begingroup$ When using the chain rule to work out $$(\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)$$, do we have to assume that $\Lambda$ is independent upon position. Is $\Lambda$ sometimes position dependent? $\endgroup$
    – Kenshin
    Jan 25 '13 at 22:33
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    $\begingroup$ @Chris Yes you should assume that $\Lambda$ is independent of position. When you act by a symmetry that isn't position dependent, then it is called a "global transformation." Alternatively, you could make it depend on position; this is called "gauging" the transformation. This is what you're suggesting in your second question and is basically what is done in supergravity en.wikipedia.org/wiki/Supergravity. By the way, I wrote the chain rule steps in one of the comments below in case you want to check. Cheers! $\endgroup$ Jan 25 '13 at 22:45
  • $\begingroup$ Sorry for the late interuption ... Here peskin is using the inverse matrix because he is doing an active transformation. Because we would use $/lambda in passive transformation but use the inverse of it when going from primed to primed... But here we are using the inverse matrix because the active transformation matrix should be the inverse of passive one.. Is that right $\endgroup$
    – Shashaank
    Sep 12 '19 at 12:57
  • $\begingroup$ And a thing more. You have scalar field A(x) in frame x. When you observe it in frame x' you get A'(x'). Both should be equal for a scale field which just depends on the spacetime point. Perfectly fine. Now as the example given by you we see a scale field Temperature. We do an active transformation on it. Say originally x=0 and T=0 there and at x=1, T=1. Now you do an active transformation on it to rotate the field to x=1 so now at x=1 you have T=0 ... Is that what you were saying. Please have a look at it. Because then you have changed the field. It was T=0 at x=0 now it's T=0 at x=1. $\endgroup$
    – Shashaank
    Sep 12 '19 at 13:20
  • $\begingroup$ How is it a scaler field then in view of the earlier passive transformation . Field should be same in different corrdinates.. That is what is a scalar field. If what I have said in the previous comment and understood and if the example I have given of rotation of temperature field how is it a scale field.. But rotating you are changing the field at x= 1 from being T=1 to T=0 $\endgroup$
    – Shashaank
    Sep 12 '19 at 13:23

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.

  • $\begingroup$ Dear Nanashi No Gombe. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) $\endgroup$
    – Qmechanic
    Jan 20 '20 at 16:56
  • $\begingroup$ @Qmechanic It's not exactly identical, but yeah, I get your point. Thanks for pointing that out. :) $\endgroup$ Jan 20 '20 at 16:57

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)

  • $\begingroup$ Duh, it that is true it is quite a confusing mistake in the book; I initially thought that the first $\Lambda^{-1}$ is due to the chain rule ... $\endgroup$
    – Dilaton
    Jan 25 '13 at 10:03
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    $\begingroup$ @Twistor You've made a minor error; the index placement in Peskin's expression is $(\Lambda^{-1})^\nu_{\phantom\nu\mu}\partial_\nu$; here $\nu$ is the row label, so in your "Finally..." step, you effectively multiply by the transpose of $\Lambda^{-1}$. In case you're skeptical: $(\Lambda^{-1})^\nu_{\phantom\nu 1}v_\nu = (\Lambda^{-1})^1_{\phantom\nu 1}v_1 +(\Lambda^{-1})^2_{\phantom\nu 1}v_2= -v_2$, similarly $(\Lambda^{-1})^\nu_{\phantom\nu 2}v_\nu =(\Lambda^{-1})^1_{\phantom 1 2} v_1 = v_1$ so if $v_\mu = (2y-x, y)$ then $(\Lambda^{-1})^\nu_{\phantom\nu \mu}v_\nu = (-y, 2y-x)$ as desired. $\endgroup$ Jan 25 '13 at 16:43
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    $\begingroup$ @Dilaton The first $\Lambda^{-1}$ is due to the chain rule; $\partial_\mu[\phi(\Lambda^{-1} x)]=(\partial_\nu \phi)(\Lambda^{-1}x)\partial_\mu(\Lambda^{-1}x)^\nu=(\partial_\nu \phi)(\Lambda^{-1}x)\partial_\mu[(\Lambda^{-1})^\nu_{\phantom \mu\alpha}x^\alpha] =(\partial_\nu\phi)(\Lambda^{-1}x)(\Lambda^{-1})^\nu_{\phantom \mu\alpha}\partial_\mu x^\alpha =(\partial_\nu\phi)(\Lambda^{-1}x)(\Lambda^{-1})^\nu_{\phantom \mu\alpha}\delta^\alpha_\mu =(\Lambda^{-1})^\nu_{\phantom \nu\mu}(\partial_\nu\phi)(\Lambda^{-1}x)$ $\endgroup$ Jan 25 '13 at 16:52
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    $\begingroup$ @twistor59 I actually think your answer is extremely useful as an example of how index placement can be a tricky, subtle issue. I think it would be great for people to read it and try to figure out where your reasoning went wrong (for example I got confused for a bit, then resolving the issue deepened my understanding). Perhaps you could put a bold note at the top that reads something like "There is an error in the following argument; can you figure out where it is?" Confusion + resolution is the best pedagogical process around! $\endgroup$ Jan 25 '13 at 17:27
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    $\begingroup$ @joshphysics OK I'll amend it tonight and leave the "as was" version of the formula for comparison. I was a little uneasy about the conclusion, but couldn't see the mistake. Thanks for that! $\endgroup$
    – twistor59
    Jan 25 '13 at 17:32

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