# Scalar field in Expanding Universe EOM

1. Jun 21, 2014

### ChrisVer

I would like to ask something.
How is the solution of EOM for the action (for FRW metric):

$S= \int d^{4}x \sqrt{-g} [ (\partial _{\mu} \phi)^{2} - V(\phi) ]$

give solution of:

$\ddot{\phi} + 3H \dot{\phi} + V'(\phi) =0$

I don't in fact understand how the 2nd term appears.... it doesn't appear in my calculation... Under the change of $\phi \rightarrow \phi + \delta \phi$ I'm getting:

$\frac{ \partial L } {\partial \phi} = \partial_{\mu} \frac{ \partial L} {\partial \partial_{\mu} \phi}$

$-\frac{ \partial V} {\partial \phi} = \partial_{\mu} \partial^{\mu} \phi$

If I want homogeneity ($∇ \phi =0$ ):

$\ddot{\phi} + V'(\phi) =0$

2. Jun 22, 2014

### bapowell

You need to use the covariant derivative.

3. Jun 22, 2014

### ChrisVer

for the EOM?

$\frac{ \partial L}{\partial \phi} = D_{\mu} \frac{ \partial L}{\partial \partial_{\mu} \phi}$

?

I don't know the covariant derivative in that case....
It should be something like $D_{\mu} = \partial_{\mu} + (...)_{\mu}$
what's that ... ?
To have zeroth component the $3H$...

4. Jun 22, 2014

### WannabeNewton

As bapowell suggested, you have to use the covariant derivative of the background metric as opposed to partial derivatives because the scalar field is coupled to gravity. The Hubble parameter comes from the Christoffel symbols in the covariant derivative.

5. Jun 22, 2014

### ChrisVer

I am sorry for being a "stupid" person...
But I don't see how the scalar field is coupled to gravity...

the Lagrangian is normal... and the $\int d^{4}x \sqrt{-g}$ is the normal choice for the invariant 4 volume element....
Does the last imply couplings with gravity?

However indeed that method works to bring me the $3H$...But that's after seeing (in bibliography) that it's needed and not before trying to do the calculations, because I didn't see gravity coupling. :(

6. Jun 22, 2014

### WannabeNewton

The scalar field has a stress-energy tensor (for example it contains energy density and it exerts a pressure) and due to this stress-energy tensor the scalar field couple to the gravitational field through the metric tensor. If you have a scalar field propagating on a curved background (in this case the FRW background) then you have to use covariant derivatives for that reason.

7. Jun 22, 2014

### bapowell

To add to this, the coupling of the scalar field to gravity is manifested in the general covariance of the equations of motion. The scalar field Lagrangian is made generally covariant by using the covariant, rather than ordinary partial, derivative. General covariance is the mathematical embodiment of Einstein's principle of equivalence.

8. Jun 22, 2014

### ChrisVer

yes, but why don't covariant derivatives appear at the field kinetic terms? I mean having some Lagrangian of the form:
$L( \phi, D_{\mu} \phi)$ instead of $L( \phi, \partial_{\mu} \phi)$...

Because I also couldn't guess that I have to take the derivative of the Lagrangian wrt partial derivative and use the covariant derivative on the E-L equations.... (as in post #3)
I know that the E-L equations transform as a rank-1 vector (I don't remember whether a covariant or contravariant - I think it's covariant). But the $\frac{\partial L}{\partial \partial_\mu \phi}$ doesn't alone...
I mean why not:
$D_\mu \frac{\partial L}{\partial D_\mu \phi}$

9. Jun 22, 2014

### bapowell

They do. That's what we're saying. Replace all partial derivatives with covariant derivatives.

The variational principle proceeds just as in ordinary mechanics.

10. Jun 22, 2014

### ChrisVer

$S= \int d^{4}x \sqrt{-g} (\frac{1}{2} D_{\mu} \phi D^{\mu} \phi - V( \phi))$

$\phi \rightarrow \phi + \delta \phi$

$S'= \int d^{4}x \sqrt{-g} [\frac{1}{2} D_{\mu}(\phi + \delta \phi) D^{\mu} (\phi + \delta \phi) - V(\phi + \delta \phi) ]$

$S'= S + \int d^{4}x \sqrt{-g} [ D_{\mu} \phi D^\mu \delta \phi - \delta \phi V'(\phi)]$

$\delta S = \int d^{4}x \sqrt{-g} (D^{\mu} (D_{\mu} \phi \delta \phi)- (D^\mu D_\mu \phi) \delta \phi - \delta \phi V'(\phi)]$

with compact support the first term vanishes under the 4-vollume integration as: $\int d^{4}x \sqrt{-g} B^{a}_{;a}$ vanishes (4-divergence).

And then you have that:

$D^{\mu} D_{\mu} \phi + V' =0$

$D_{\mu} \phi \equiv \partial_{\mu} \phi$

Something like that? Now it makes more sense...

Last edited: Jun 22, 2014
11. Jun 22, 2014

### George Jones

Staff Emeritus
Note that covariant derivatives are used is special relativity, too. This is obvious in the case of inertial coordinates, i.e., partial and covariant derivatives are the same. It is maybe not so obvious when other coordinate systems are used, e.g., the Milne universe, which is a subset of Minkowski spacetime.

In Minkwoski spacetime, the 3H term must be present in Milne coordinates in order to maintain consistency when making the coordinate transformation from Milne to inertial coordinates.

12. Jun 22, 2014

### ChrisVer

I don't know... I also had a misleading note... something of using the fact that $d^{3} \sqrt{-g} = a^{3}(t) d^{3} \bar{x}$
with the $\bar{x}_{i}$ dimensionless coordinates.
However I find (at the moment) the fact that you just replace the partial derivatives with covariant in the Lagrangian more easy to understand (minimal coupling), also under the statement of general covariance.

Last edited: Jun 22, 2014
13. Jun 27, 2014

### phsopher

You can also use regular partial derivatives but you have to keep in mind that the term $\sqrt{-g}$ depends on time in FRW so you must take that into account in the partial integration:

$$\frac{1}{\sqrt{-g}}\partial_{\mu}\left(\sqrt{-g}\partial^{\mu}\phi\right) + V' = 0$$

For FRW, the determinant of the metric is $g=-a^6$.

14. Jun 27, 2014

### ChrisVer

may I ask then, what's the difference between taking the one and the other lagrangian:
$L_{scalar}$
and
$\sqrt{-g} L_{scalar}$
?
I think in the first one since the integration is taking place over the invariant 4-volume you can immediately start working in GR-frame (or in other words you have to input covariant derivatives instead of partial ones).
While in the second one, you can work fine with the normal derivative thing. I was able to see also the solution with $\sqrt{-g}$ in the Lagrangian.

Also thanks phsopher :) that form for the EOM shows me a hint, since your 1st term is indeed the covariant derivative of $\partial^{\mu} \phi$