# Scale factor in conjugate scalar field inside conformally flat spacetime

Consider a lagrangian density of a scalar field $$\mathscr{L} = \frac{1}{2} \partial_\alpha \phi \partial^\alpha \phi - \frac{1}{2} m^2 \phi^2$$ inside a conformally flat spacetime with scale factor $$a(t)$$. Expliciting $$\mathscr{L}$$ trough the auxiliary field $$\chi\doteq a\phi$$, and considering that $$(g^{\alpha\beta})=a^{-2}\eta$$ I obtain (full derivation at the bottom of the page) $$2 a^4 \mathscr{L} = \partial_0 \left( \frac{\dot{a}}{a} \chi^2 \right) + \dot{\chi}^2 - \delta_{ij} \partial_i \chi \partial^j \chi - \left( a^2 m^2 - \frac{\ddot{a}}{a} \right) \chi^2$$ Now I should discard the time derivative term so the conjugate field is simply $$\pi \doteq \frac{\partial\mathscr{L}}{\partial\dot{\chi}} = a^{-4} \dot{\chi}$$ but in the book I'm reading "Introduction to Quantum Effects in Gravity - Mukhanov, Winitzki" the term $$a^{-4}$$ doesn't appear. This has an important consequence: imposing the canonical equal time commutation relation between the auxiliary field and its conjugate expanded in their mode functions, I find $$\dot{v}_{\mathrm{k}} v_{\mathrm{k}}^\ast - v_{\mathrm{k}} \dot{v}_{\mathrm{k}}^\ast = 2 \mathrm{i} a^4$$ instead of $$\dot{v}_{\mathrm{k}} v_{\mathrm{k}}^\ast - v_{\mathrm{k}} \dot{v}_{\mathrm{k}}^\ast = 2 \mathrm{i}$$ For example, in De Sitter spacetime where $$a=a_0 e^{Ht}$$ my result causes mode functions to disappear when $$t\to -\infty$$.

I don't understand if my result is wrong, I repeated the calculation several times in these months and still obtain the same result. Where is the issue?

Here the full derivation:

$$\mathscr{L} = \frac{1}{2} \partial_\alpha \left( a^{-1} \chi \right) \partial^\alpha \left( a^{-1} \chi \right) - \frac{1}{2} m^2 a^{-2} \chi^2 \\ \mathscr{L} = \frac{1}{2} \partial_0 \left( a^{-1} \chi \right) g^{00} \partial_0 \left( a^{-1} \chi \right) + \frac{1}{2} a^{-2} g^{ij} \partial_i \chi \partial_j \chi - \frac{1}{2} m^2 a^{-2} \chi^2 \\ \mathscr{L} = \frac{1}{2} a^{-2} \left( - \frac{\dot{a}}{a^2} \chi + a^{-1} \dot{\chi} \right)^2 - \frac{1}{2} a^{-4} \delta_{ij} \partial_i \chi \partial^j \chi - \frac{1}{2} m^2 a^{-2} \chi^2$$ Multiplying by $$2a^4$$ and summing and subtracting an equal term $$2 a^4 \mathscr{L} = \frac{\dot{a}^2}{a^2} \chi^2 - 2 \frac{\dot{a}}{a} \chi \dot{\chi} + \dot{\chi}^2 - \delta_{ij} \partial_i \chi \partial^j \chi - m^2 a^2 \chi^2 + \left( - \frac{\ddot{a}}{a} \chi^2 + \frac{\ddot{a}}{a} \chi^2 \right) \\ 2 a^4 \mathscr{L} = \partial_0 \left( \frac{\dot{a}}{a} \chi^2 \right) + \dot{\chi}^2 - \delta_{ij} \partial_i \chi \partial^j \chi - \left( a^2 m^2 - \frac{\ddot{a}}{a} \right) \chi^2$$

This is why I prefer writing down the action of your field theory, which reads $$S = \int d^4x \sqrt{-g}\mathscr{L}$$ One should better first simplify everything taking into account $$\sqrt{-g}$$ and then varying the action with respect to the fields. (In the derivation of the Euler-Lagrange equations using the above action, you need to keep $$\sqrt{-g}$$ when doing the integration by parts. From the derivation you will see that your canonical momentum should rather be $$\pi = \sqrt{-g}\frac{\partial \mathscr{L}}{\partial \dot{\chi}}$$.) Since $$\sqrt{-g}=\sqrt{-a^2(-a^2)^3}=a^4$$, your issue is resolved.

Let us vary the above action with Lagrangian $$\mathscr{L}(\phi, \partial \phi)$$ with respect to $$\phi$$. $$\delta S = \int d^4x \left[\sqrt{-g} \frac{\partial \mathscr{L}}{\partial \phi} \delta \phi + \sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)} (\partial_{\mu} \delta\phi) \right]$$ Now we integrate the second term by parts: $$\delta S = \int d^4x \left[\sqrt{-g} \frac{\partial \mathscr{L}}{\partial \phi} \delta \phi + \nabla_{\mu} \left(\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)} \delta\phi \right) - \nabla_{\mu} \left(\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)}\right) \delta \phi \right]$$ Here we used $$\nabla_{\mu}\delta \phi = \partial_{\mu} \delta \phi$$ for scalar fields. Now we drop the boundary term (=second term). Demanding $$\delta S=0$$ yields $$\nabla_{\mu} \left(\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)}\right) - \sqrt{-g} \frac{\partial \mathscr{L}}{\partial \phi}=0$$ The term in the brackets is then canonical conjugate: $$\Pi^{\mu}:=\sqrt{-g} \frac{\partial \mathscr{L}}{\partial (\partial_{\mu}\phi)}$$ In particular: $$\Pi^{0}=\sqrt{-g} \frac{\partial \mathscr{L}}{\partial \dot{\phi}}$$

Alternatively, you can start computing the action $$S$$ more explicitly and you will get $$S = \int d^4x a^4(t) \mathscr{L}_{\text{yourLagrangian}}$$ and you can work effectively with $$\mathscr{L}^{\prime} :=a^4 \mathscr{L}$$ using your "flat space Euler-Lagrange equation". This is why in some textbooks on QFT in curved spacetime the Lagrangian is defined by $$\mathscr{L}_{\text{curved}} := \sqrt{-g}\mathscr{L}_{\text{flat}}$$. I personally prefer to work with the action right away and then vary the action.

• Tomorrow I'll try! Commented Sep 17, 2022 at 19:28
• Sorry, I'm a bit confused, can you convince me of that form for the conjugate field? Commented Sep 18, 2022 at 8:36
• @Rob Tan I added more details in my answer above. I hope this clarifies your question.
– psm
Commented Sep 18, 2022 at 13:13
• There are some nice notes by Winitzki: GR drive.google.com/file/d/1vcvxS3Tb4grn524ib1MuU9wQq1hdW6So/view in which he includes $\sqrt{-g}$ in the Lagrangian already. In this case, the $\sqrt{-g}$ will obviously be in $\Pi^{0}$, too. And QFTCS drive.google.com/file/d/0B2Mt7luZYBrwVXFIWk9WNFdVcEk/… in which he calculates the action explicitly and then he computes the equation of motion. The rest is analogy with flat space and the harmonic oscillator of quantum mechanics. I think the second link really explains why all this works at all.
– psm
Commented Sep 18, 2022 at 15:01
• Also the book by Birrell and Davies on QFT in curved spacetime is a good resource. I used it as a student and if I remember correctly, it is very precise regarding these questions. I think they also included $\sqrt{-g}$ in the Lagrangians. I can't check right now, because it's missing in my bookshelf.
– psm
Commented Sep 18, 2022 at 18:19