1.) How do we physically interpret an active Lorentz transformation? The passive transformation seems simple enough: you view a fixed object from the perspective of a new observer. When we actively Lorentz transform a vector are we interpreting this as moving the vector to a new point in spacetime considered from the perspective of a single observer?

2.) I am reading David Tong's QFT notes, and am having a hard time interpreting what he means by active transformations. The notes in question can be found here: http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf The notes in question are on pg 11-12 as labeled by the book or pages 17-18 as labeled by the PDF.

In his notes, Tong states that we can transform a scalar field as follows:

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

When he does this, I'm interpreting this as

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

where $x'=\Lambda x$. From what I understand, the advantage of using the inverse Lorentz transformation on the primed system is that we can use the same functional form of $\phi$. However, when moving to the primed system we have still used $\Lambda$, not its inverse. Can anyone tell me if I'm correct in my understanding up to this point?

If my understanding is correct up to this point then I really don't understand the next section in his notes. He states that under this transformation derivatives transform as

$$(\partial_\mu \phi) \rightarrow \left( \Lambda^{-1} \right)^\nu_{\phantom{\nu} \mu} (\partial_\nu \phi)(y),$$

where $y=\Lambda^{-1}x$ (where this $x$ is primed, right?). But we've still gone from $x \rightarrow x'$ where $x'=\Lambda x$ (again based on my understanding which may be terribly wrong). Using $\Lambda^{-1}x'$ was simply a mathematical trick to allow us to use the same functional form of $\phi$. If that's the case, why are the derivatives transforming as $\Lambda^{-1}$ instead of just $\Lambda$?

I'm sorry -- I know this is a little convoluted, but I'm having a really hard time getting my head around this, especially with his notation. I really wish he would have used primes or something...

Am I completely lost? Someone please rescue me.


2 Answers 2


Passive Lorentz transformations are what everybody learns first. There is one physical reality, and you're just describing that one physical reality using two different coordinate systems, where one coordinate system uses a set of axes that are rotated and/or boosted relative to the axes used by the other coordinate system. You're just using a different set of numbers to describe every event in the universe (ignoring the complications due to gravity). It makes sense to use notation of $x'$ vs $x$ with a passive Lorentz transformation, because the prime is denoting coordinates as defined in two different coordinate systems.

With an active Lorentz transformation, you're contemplating two different physical realities, where some piece of that physical reality has been rotated and/or boosted relative to how it is in the other physical reality. If the part of physical reality you're concerned with is your TV set, the relationship between your TV set in its normal orientation vs your TV set after you've physically picked it up and rotated it 90 degrees clockwise from its normal orientation would be described by an active Lorentz transformation. You don't use two different coordinate systems to describe the two different TV orientations you're considering; you just use the one lab frame to describe the TV set in both of its orientations. So using an $x'$ doesn't make sense, because you aren't using a primed coordinate system for anything. Saying

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

makes sense as is, with no prime on any of the $x$'s, because you're talking about two physically different fields, both described in the same coordinate system. And the expression


makes sense as is, because $y$ is a different point within the same coordinate system as $x$, and you most commonly label different points within one coordinate system by using different letters, not by introducing primes. Using primes would risk confusing a reader into erroneously thinking that some primed coordinate system was being used, in addition to the lab frame. Similarly, it makes sense to say

$$(\partial_\mu \phi')(x) = \left( \Lambda^{-1} \right)^\nu_{\phantom{\nu} \mu} (\partial_\nu \phi)(y)\ ,$$

because it's two physically different fields that are being considered, both of which are described using the same coordinate system.

  • $\begingroup$ So if I'm understanding what you're saying correctly, in a passive transformation you have to transform the derivatives so that they are done with respect to the new coordinate system. Why does the inverse Lorentz transformation act on the derivatives from the active viewpoint? Is this just due to the chain rule? $\endgroup$ Sep 12, 2014 at 23:47
  • $\begingroup$ I think the last equation should be $(\partial_{\mu}\phi^{\prime})(x) = (\Lambda^{-1})^{\nu}{}_{\mu} (\partial_{\nu}\phi)(y)$. $\endgroup$
    – higgsss
    Sep 13, 2014 at 0:44
  • $\begingroup$ Yes, I think higgsss is right. I corrected my answer. $\endgroup$
    – Red Act
    Sep 13, 2014 at 0:49
  • 1
    $\begingroup$ @Dargscisyhp (My apologies for interrupting.) In this case, of course, one can view the transformation of the derivative as a result of the chain rule $\partial/\partial x^{\mu} = \partial y^{\nu}/\partial{x^{\mu}} \partial/\partial y^{\nu}$. But more generally, it is a direct consequence of the active transformation. The quantity $\partial_{\mu}\phi$ is a covariant vector, and it should change accordingly under a Lorentz transformation. It is just like actively rotating a (3-)vector field. The vector will get rotated and the components will be transformed accordingly. $\endgroup$
    – higgsss
    Sep 13, 2014 at 1:18
  • $\begingroup$ Thank you both for your responses! I'm still a little confused by why the derivatives would change under a transformation of the field itself. In 3D, if we have $\phi (\vec{x})$ and we rotate only the vector the gradient shouldn't change i.e. $\nabla = (\partial / \partial x, \partial / \partial y, \partial / \partial z)$ whether we rotate the vector or not, correct? $\endgroup$ Sep 13, 2014 at 1:50

As a simple example, suppose we translate a function $f(x)$ by $a$ in the $x$-direction and by $b$ in the $y$-direction. We take the active viewpoint, viz., if $f(x)$ had a peak of height $f_{0}$ centered at $x=x_{0}$, it would be translated to be a peak of height $f_{0} +b$ located at $x=x_{0}+a$. Surprisingly, however, what we end up doing is a transformation that is not completely active: $f(x) \rightarrow f(x-a) + b$. We actively shift in the $y$-direction by $b$, but the translation in the $x$-direction is passive! The trouble is that $x$ only appears as an intermediate object---the argument of $f$. For a translation in the $x$-direction, we cannot directly manipulate the value of $f$ but have to achieve it indirectly via a shift of the coordinate system by $-a$.

The lesson we learn from the above example is that to actively transform a function, one should

1. Actively transform the value of the function,
2. Passively transform the argument.

Hence actively transforming the scalar field $\phi(x)$ by a Lorentz transformation $\Lambda$ gives $\phi(\Lambda^{-1}x)$. As it is a scalar, there is nothing to be done on $\phi$ itself. On the other hand, just as the simple translation considered above, the argument has to be transformed in the opposite (passive) way to have an ultimate active transformation.

If the field variable itself carried Lorentz indices, we would have to make an active transformation on them. The derivative $\partial_{\mu}\phi$ is an example. As it is a covariant vector, the transformation rule is \begin{equation} (\partial_{\mu}\phi) (x) \ \rightarrow \ \Lambda_{\mu}{}^{\nu}\,(\partial_{\nu}\phi) (\Lambda^{-1}x)\,, \end{equation} where we transformed the covariant index actively, and as before, the argument of the function passively.

The expression in Tong's lecture note follows by noting that \begin{equation} \Lambda_{\mu}{}^{\nu} = (\Lambda^{-1})^{\nu}{}_{\mu}\,. \end{equation}

  • $\begingroup$ 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, 2019 at 13:14

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