Puzzle of scalar dynamic in the early universe

[Accidently]

Sometimes we need to calculate the evolution of the scalar field $$\phi$$ with the equation of motion
$$\frac{\partial^2 \phi}{\partial t^2}+3H\frac{\partial \phi}{\partial t}+m_\phi^2 \phi = 0$$.
And we can get the field
$$\phi=Ae^{im_\phi t}$$
where A is the amplitude of the scalar field (damped by the Hubble parameter in this case).

But when we apply the energy operator $$i\frac{\partial}{\partial t}$$ on the field, what we get is $$E=m_\phi$$
This is incorrect. And it seems that we should include the energy/temperature of the universe somewhere, but do not know how.

 

[bapowell]

The equation you are studying is that of a decoupled scalar field -- i.e. it is not interacting with anything in the universe other than gravity. As far as this system is concerned, all the energy in the universe is contained in the scalar field matter. Strictly speaking, you should be calculating the stress-energy tensor of the scalar field (or its quantum expectation value) to answer this question to get at the energy of the system.

More realistically, you will want to include all couplings between the scalar field and other matter sectors of the theory -- then you will see a dependence of the total energy on these other matter fields.

 

[Chalnoth]

The equation you are studying is that of a decoupled scalar field -- i.e. it is not interacting with anything in the universe other than gravity. As far as this system is concerned, all the energy in the universe is contained in the scalar field matter. Strictly speaking, you should be calculating the stress-energy tensor of the scalar field (or its quantum expectation value) to answer this question to get at the energy of the system.

More realistically, you will want to include all couplings between the scalar field and other matter sectors of the theory -- then you will see a dependence of the total energy on these other matter fields.

I think what he means is that this field should not only contain the mass term, but also a kinetic energy term.

I'd also like to mention, however, that any second-order linear differential equation has two solutions, so you need a second solution to get the correct motion of the field. Accidentally, you might also do better at understanding the energy of the particle by using the Hamiltonian operator.

 

[bapowell]

I think what he means is that this field should not only contain the mass term, but also a kinetic energy term.

I see. Consider calculating the [tex]T_{00}[/itex] component of the stress-energy tensor -- this gives you the energy density, $\rho$, of the gravitating stress-energy content of the universe. With scalar field matter, you find
[tex]\rho = \frac{1}{2}\dot{\phi}^2 + V(\phi)[/itex]
where the first term is the kinetic energy, and $V(\phi) = m^2\phi^2$ for a polynomial potential energy function.

If you have other relevant fields in the theory, including any couplings between them, they will also contribute to the overall energy density.

 

[Accidently]

I think what he means is that this field should not only contain the mass term, but also a kinetic energy term.

I'd also like to mention, however, that any second-order linear differential equation has two solutions, so you need a second solution to get the correct motion of the field. Accidentally, you might also do better at understanding the energy of the particle by using the Hamiltonian operator.

Yes, that is exact what I mean. I know I oversimplified too many things here. I just mean for a stable scalar field, the solution should look like that.

 

[Chalnoth]

Yes, that is exact what I mean. I know I oversimplified too many things here. I just mean for a stable scalar field, the solution should look like that.

That's what I thought. The problem here is that the time derivative of the scalar field is just a constant times the scalar field, so that your sum of the kinetic and potential energy terms can be wrapped back into a constant multiple of the field itself (in fact, this is necessarily the case as long as it's in an energy eigenstate). I think that is what is hiding the kinetic energy component in the way you're approaching it.

To get the kinetic energy component explicitly, just use the Hamiltonian operator instead of the time derivative. You should be able to both verify that the kinetic energy term is there, and that it reduces to the simple time derivative when evaluated (that is, you should be able to replicate what bapowell posted above).

 

[Accidently]

[tex]\rho = \frac{1}{2}\dot{\phi}^2 + V(\phi)[/itex]

Yes, this makes sense.

So, if we write the scalar field as
$$\phi=Ae^{-iat}$$,
it is not necessary to say $$a=E$$ or $$a=T$$, with $$E$$ and $$T$$ the energy of the field and the temperature of the universe, (if the field is in equilibrium). Is this right?

Thanks for all your replies.

 

[bapowell]

Looking back at your initial post, I'm confused by your assertion that $\phi = Ae^{imt}$ is a solution to the scalar field EOM. It isn't, because of the Hubble term. The Hubble term is itself a function of $\phi$ and $\dot{\phi}$.

 

[Accidently]

Looking back at your initial post, I'm confused by your assertion that $\phi = Ae^{imt}$ is a solution to the scalar field EOM. It isn't, because of the Hubble term. The Hubble term is itself a function of $\phi$ and $\dot{\phi}$.

I think the Hubble parameter acts as a damping factor in the amplitude. And we can separate [tex]e^{imt}[/itex] in the solution. for example eq. (7.9.2) and (7.9.3) in hep-th/0503203

 

[bapowell]

I think the Hubble parameter acts as a damping factor in the amplitude. And we can separate [tex]e^{imt}[/itex] in the solution. for example eq. (7.9.2) and (7.9.3) in hep-th/0503203

OK, but then when you apply $i\frac{\partial}{\partial t}$ to the field how is it that you can ignore the time-dependent amplitude?

 

[Accidently]

OK, but then when you apply $i\frac{\partial}{\partial t}$ to the field how is it that you can ignore the time-dependent amplitude?

I know, I have assumed that the Hubble parameter is small and neglectable. I should have said that. sorry. I think even with the Hubble parameter in the field, we still cannot get a correct energy, so ignored that...