1
$\begingroup$

In this paper (Field Diffeomorphisms and the Algebraic Structure of Perturbative Expansion, by Kreimer & Velenich), the authors claim in section 3, page 3, that the field diffeomorphism $F(\phi)$ given by:

$$F(\phi) = \sum_{k=0}^\infty a_k \phi^{k+1}\tag{1}$$

"preserve Lagrange's equations" when applied onto a real, massless scalar field with Lagrangian:

$$\mathcal{L}(\phi,\partial_\mu\phi) = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi. \tag{2}$$

This leads to the following transformed Lagrangian:

$$\begin{align} \mathcal{L}_\text{F} (\phi,\partial_\mu\phi) &= \mathcal{L} (F(\phi),\partial_\mu F(\phi)) \\ & = \mathcal{L} (F(\phi),F'(\phi) \partial_\mu \phi) \\ & = \frac{1}{2} F'(\phi)^2 \partial_\mu \phi \partial^\mu \phi. \tag{3} \end{align}$$

The equations of motion of $(2)$ are as usual given by:

$$\partial_\mu \partial^\mu \phi = 0, \tag{4}$$

and I would like to check that this is indeed conserved by the field diffeomorphism given in $(1)$. First I calculated $F'(\phi)$:

$$F'(\phi) = \sum_{k=0}^\infty a_k (k+1) \phi^k, \tag{5}$$

leading to:

$$\mathcal{L}_\text{F} (\phi,\partial_\mu\phi) = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi \sum_{k,l} a_k a_l (k+1)(l+1) \phi^{k+l} \tag{6}$$

The equations of motion are then:

$$\begin{align} \partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu\phi)} - \frac{\partial\mathcal{L}_\text{F}}{\partial\phi} &= \sum_{k,l} a_k a_l (k+1)(l+1) \left[\partial_\mu \left( \partial^\mu \phi \phi^{k+l} \right) - \frac{1}{2} (\partial \phi)^2 (k+l) \phi^{k+l-1}\right] \\ &= \sum_{k,l} a_k a_l (k+1)(l+1) \phi^{k+l} \left[ \partial_\mu\partial^\mu \phi + (k+l)\frac{1}{\phi} (\partial\phi)^2 \right] \\ &\overset{!}{=} 0 \tag{7}\end{align}$$

This is not the same thing as $(4)$,right? And in a way, I think that it would be quite surprising if it were. Should it be the same equation of motion? And if not, what is meant by "preserving Lagrange's equations"?

$\endgroup$
0
$\begingroup$

What the preservation of Lagrange's equations means in this case, is that the equations of motion for $\phi$ after the diffeomorphism $F$ are automatically satisfied if the equations of motion for $\phi$ before the diffeomorphism are fulfilled. In other words, there is not need to impose

$$\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu\phi)} - \frac{\partial\mathcal{L}_\text{F}}{\partial\phi} = 0\tag{8}$$

In the example above, it is easy to show that:

$$\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu\phi)} - \frac{\partial\mathcal{L}_\text{F}}{\partial\phi} = F'(\phi) \left[\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu F(\phi))} - \frac{\partial\mathcal{L}_\text{F}}{\partial F(\phi)} \right]\tag{9}$$

and this is equal to $0$ since

$$\partial_\mu \frac{\partial\mathcal{L}_\text{F}}{\partial(\partial_\mu F(\phi))} - \frac{\partial\mathcal{L}_\text{F}}{\partial F(\phi)} = \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)} - \frac{\partial\mathcal{L}}{\partial \phi} \overset{!}{=} 0\tag{10}$$

This implies that the diffeomorphism does not generate new physics, at least at tree level.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.