In this paper by Sean Carroll and Grant Remmen, in equation (11) they write a Lagrangian of the form $$\boxed{\mathcal{L}=3a\left(k-\dot{a}^2\right)+a^3\left[\frac{1}{2}\dot\phi^2-V(\phi)\right]}$$ for gravity plus a scalar field with potential $V(\phi)$, here given $N=1$ and $M_{\text{Pl}}=1$. So how is this Lagrangian obtained? I'm trying $$\mathcal{L}=\mathcal{L}_{\text{gravity}}+\mathcal{L}_\text{homogeneous SF}$$ with $\mathcal{L}=\frac{\widehat{\mathcal{L}}}{\sqrt{|g|}}$, so that it would be (with -+++ signature) $$\mathcal{L}=\frac{1}{2}R-\frac{1}{2}g^{00}\dot\phi-V(\phi)$$ where I've taken $\partial_i\phi=0$ for the scalar field $\phi$ is homogeneous. Now the Ricci scalar for FRW is $$R=\frac{6}{a^2}(k+\dot{a}^2+a\ddot{a})$$ so that $$\mathcal{L}=\frac{3}{a^2}(k+\dot{a}^2+a\ddot{a})+\frac{1}{2}\dot\phi^2-V(\phi)$$ and using the acceleration equation for the scale factor, $$\ddot{a}=-\frac{a}{6}(\rho+3p)=\frac{a}{3}\left[V(\phi)-\dot\phi^2\right]$$ where I used $\rho=\frac{1}{2}\dot\phi^2+V(\phi)$ and $p=\frac{1}{2}\dot\phi^2-V(\phi)$. Thus $$\boxed{\mathcal{L}=\frac{3}{a^2}\left(k+\dot{a}^2\right)-\frac{1}{2}\dot\phi^2}$$ and it doesn't seem to fit, even if I multiply by an $a^3$ factor, which seems to be done to get the expression. What is going on here?

  • $\begingroup$ For a bit of extra information, the extra total derivative term is missing from the formula in Remmen and Carroll's paper as a boundary term term has been subtracted off to make the original action "well-posed." Essentially: it no longer requires the variation of the derivative of the field (in this case the metric) to be set to zero on the boundary. The term subtracted off is called the <a href="en.wikipedia.org/wiki/…> term for the usual GR scenario but is readily extendible; a reasonably readable description c $\endgroup$
    – David
    Jul 28 '15 at 13:49

and using the acceleration equation for the scale factor,

Aaah! You have committed a mortal sin! But it is ok, it's a very common one :). Never ever ever ever ever plug equations of motion back into the lagrangian. It is the moral analogue of setting x=3 into f(x)=x^2+x before differentiating--you will get 0, whereas what you meant to do was differentiate f(x) and then set x=3, which gives you 7.

(You can get away with plugging equations of motion into the lagrangian if you are careful--you can do if you are you used an equation of motion that was purely algebraic--no derivatives--or if you used a greens function to integrate out a dynamical field. However you need to be very careful and you should not do it unless you are 100% sure you know what you are doing).

OK, onto how to fix your dilemma. First off you want to include the $\sqrt{-g}=a^3$ factor in your definition of $\mathcal{L}$, so that the expression in the middle of your answer becomes

\begin{equation} \mathcal{L}_{you} = 3a \left(k + \dot{a}^2 + a \ddot{a} \right) + a^3 \left( \frac{1}{2} \dot{\phi}^2 - V(\phi) \right) \end{equation}

Now Carroll's equation 11 is (with N=1 and also setting $M_{pl}$ = 1)

\begin{equation} \mathcal{L}_{carroll} = 3a \left(k - \dot{a}^2 \right) + a^3 \left( \frac{1}{2} \dot{\phi}^2 - V(\phi) \right) \end{equation}

This is almost your boxed expression at the top of your answer, but the $a \dot{a}^2$ has minus sign instead of a plus sign.

Now your worry is essentially that

\begin{equation} \mathcal{L}_{you} - \mathcal{L}_{carroll} = 6 a \dot{a}^2 + 3 a^2 \ddot{a} \end{equation}

is apparently nonzero. However, it this is actually a total derivative

\begin{equation} \mathcal{L}_{you} - \mathcal{L}_{carroll} = 3 \frac{d}{dt}\left( a^2 \dot{a} \right) \end{equation}

so $\mathcal{L}_{you}$ and $\mathcal{L}_{carroll}$ agree up to a total derivative, which is all that is required.

Incidentally, based on the title of your question I take it you are interested in studying minisuperspace. If so I think you can learn a lot by not setting $N=1$, life is much more interesting if you keep in it. (It is consistent to set $N=1$ because I can absorb $N$ into a redefinition of $t$, but you don't have to do it and it is interesting to see what happens if you don't do that). Basically $N$ parameterizes the 'gauge freedom' associated with the ability to reparameterize your time coordinate. You will always see $N$ appear in the combination $N dt$: either as $Ndt$ in the measure of the integral, or as $\frac{d}{N dt}$ in time derivatives. This is why you can absorb $N$ into $t$ if you want. If you pass from the Lagrangian to the Hamiltonian form, $N$ ends up playing the role of a lagrange multiplier in the Hamiltonian. Actually $N$ will actually multiply everything else in the Hamiltonian, which is an example of the general principle that Hamiltonian for diffeomorphism invariant theories are always pure constraint. You can see this explicitly in Carroll's equation 13. However even if you don't like hamiltonians, there is another way to see the same thing: you can start with the action and $S= \int N dt \mathcal{L}_{you}$ and vary wrt $N$, you will see that this gives you the Friedman equation directly. If you set $N=1$ in the action first [which you are allowed to do], you can only get the Friedmann equation from the action by doing more work. You have to combine the equations of motion you get by varying wrt $a$ (the acceleration equation) and by varying wrt $\phi$ (the conservation of energy equation).

  • $\begingroup$ Ouch! Great, seems good enough for me, and thanks for the clarification about $N$. I just have another silly question, ${g_{\alpha\beta}=\text{diag}\left(-1,\frac{a^2}{1-k{r}^2},a^2r^2,a^2r^2\sin^2{\theta}\right)}$ so I get ${-\det({g}_{\alpha\beta})=\frac{r^4a^6\sin^2\theta}{1-kr^2}}$, how come ${\sqrt{-g}=a^3}$? $\endgroup$
    – user24999
    Mar 19 '14 at 4:10
  • 2
    $\begingroup$ Heh, nice question. It's because I lied a little bit, the action is $S = \int dt d^3 x \sqrt{-g} \mathcal{L}$ where $\sqrt{-g}=a^3 r^2 \sin \theta / (\sqrt{1-kr^2})$ as you say. However since the fields only depend on $t$ I can explicitly do the integrals over the spatial coordinates. I pick up an overall field independent multiplicative constant $V$, the volume of space when $a=1$. So technically the action is multiplied by an infinite constant! It is basically because the action is an extensive variable, if it bothers you just imagine the universe is in a box with a finite volume. $\endgroup$
    – Andrew
    Mar 19 '14 at 4:20

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