Scaling Transformations - Scalar Field

I've started self-studying QFT and I am using some Cambridge notes (Tong's). There is a specific excersise where we are asked to see how the action

$$S=\int d^4x\left(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^2\phi^2-g\phi^p\right)$$ transforms when

$$x\rightarrow x'=\lambda x\quad\text{and}\quad\phi(x) \rightarrow \phi'(x)= \lambda^{-D}\phi(\lambda^{-1}x)$$

where we can define for simplicity $y=\lambda^{-1}x=\lambda^{-2}x'$. In the notes he is saying that:

$$\frac{\partial \phi(x)}{\partial x^\nu}\rightarrow \frac{\partial x'^{\mu}}{\partial x^{\nu}}\frac{\partial \phi'(x)}{\partial x'^{\mu}}=λ\frac{\partial }{\partial x'^{\mu}}(λ^{-D}\phi(λ^{-1}x))=λ^{-(1+D)}\frac{\partial}{\partial y^{\mu}}\phi(y)\tag{A1}$$

and that

$$\int d^4x=λ^4\int d^4y\tag{B1}$$

Those two do not make sense to me!!! since I would work it out as follows:

$$\frac{\partial \phi(x)}{\partial x^\nu}\rightarrow (\frac{\partial \phi(x)}{\partial x^\nu})'=\frac{\partial x^{\mu}}{\partial x'^{\nu}}\frac{\partial \phi'(x)}{\partial x^{\mu}}=λ^{-1}\frac{\partial }{\partial x^{\mu}}(λ^{-D}\phi(λ^{-1}x))=λ^{-(2+D)}\frac{\partial}{\partial y^{\mu}}\phi(y)\tag{A2}$$

and

$$\int d^4x \rightarrow \left(\int d^4x\right)'=\lambda^8\int d^4y\tag{B2}$$

Could someone tell me what am I doing wrong? Why are the correct transformation relations not (A2) and (B2) instead of (A1) and (B1)?

• It looks like your main problem is your derivative did not transform correctly. en.wikipedia.org/wiki/Covariant_transformation – Novice C Mar 5 '17 at 17:02
• Thanks for your reply, but I don't see the mistake. The derivative of a scalar field transforms as a vector as $\frac{\partial \phi(x)}{\partial x^\nu}\rightarrow (\frac{\partial \phi(x)}{\partial x^\nu})'=\frac{\partial x^{\mu}}{\partial x'^{\nu}}\frac{\partial \phi'(x)}{\partial x^{\mu}}$. Or is there a misconception on my behalf? – Helen Mar 5 '17 at 17:40
• Helen can you link the notes and the pages you're referring to? – Luthien Mar 5 '17 at 22:33
• link. It is the 9th problem in the QFT1 sheet. – Helen Mar 6 '17 at 9:43

Your confusion is because in transforming the action $S' \rightarrow S$ you do an "extra" transformation of the variable $x$ when in reality $x$ is just a "dummy" variable that is integrated over. Equations should help make this clear.
$$x' \rightarrow x \\ \Phi(x) \rightarrow \Phi'(x') = \mathcal{F}(\Phi(x)).$$ Written in this way, $\Phi$ is affected by the transformation in two ways. The first is a functional change in the field $\Phi' = \mathcal{F}(\Phi)$ and the second a change in the argument $x \rightarrow x'$.
For a generic Lagrangian $\mathcal{L}$, \begin{align} S' &= \int d^dx~\mathcal{L}(\Phi'(x), \partial_{\mu} \Phi'(x)) \\ &= \int d^dx'~\mathcal{L}(\Phi'(x'), \partial_{\mu}' \Phi'(x')) \\ &= \int d^dx'~\mathcal{L}(\mathcal{F}(\Phi(x)), \partial_{\mu}' \mathcal{F}(\Phi(x))) \\ &= \int d^dx~ \Big|\frac{\partial x'}{\partial x}\Big|~\mathcal{L}(\mathcal{F}(\Phi(x)), (\partial x^{\nu}/ \partial x'^{\mu})\partial_{\nu} \mathcal{F}(\Phi(x))). \end{align} In the second line there is a change of the integration variable $x \rightarrow x'$; in the third line we express $\Phi'(x')$ in terms $\Phi(x)$; in the last line we express $x'$ in terms of $x$.
$\textbf{Note:}$ At the very beginning, in going from $S \rightarrow S'$, the reason we substitute $\Phi'(x)$ for $\Phi(x)$ (and $\textit{not}$ $\Phi'(x')$), is that since the action integrates over $x$ it is just an integration or "dummy" variable that can be renamed. To see the $\textit{functional}$ change in the action we have to look at how it changes under a new function, in this case $\Phi(x) \rightarrow \Phi'(x)$.
This is why for the specific case $$x\rightarrow x'=\lambda x\quad\text{and}\quad\phi(x) \rightarrow \phi'(x)= \lambda^{-D}\phi(\lambda^{-1}x)$$ with $y=\lambda^{-1}x=\lambda^{-2}x'$, that $$\frac{\partial \phi(x)}{\partial x^\nu}\rightarrow \frac{\partial x'^{\mu}}{\partial x^{\nu}}\frac{\partial \phi'(x)}{\partial x'^{\mu}}=λ\frac{\partial }{\partial x'^{\mu}}(λ^{-D}\phi(λ^{-1}x))=λ^{-(1+D)}\frac{\partial}{\partial y^{\mu}}\phi(y)$$ and $$\int d^4x=λ^4\int d^4y.$$