the amplitude of the power spectrum of primordial fluctuations arising from inflation is proportional to $1/\dot{\phi}^2$. Therefore, as $\dot{\phi}$ decreases, the power spectrum increases. However, in the limit of a massless scalar field (i.e. a perfectly flat potential), $V(\phi)$ is constant and thus so is the energy density. What reconciles this apparent problem?

bonus question: can you show that $\Delta \phi = 0$ is solved by the Newton-potential and $(\Delta+m^2)\phi = 0$ is solved by the Yukawa-potential (both in 3d spherical coordinates)? What's the length scale of the Yukawa-potential in this example? why is there no length scale in the Newton-potential? why are both solutions isotropic?

please take part in the poll asking for the cosmological reason of the apocalypse in two weeks.

bonus question: can you show that $\Delta \phi = 0$ is solved by the Newton-potential and $(\Delta+m^2)\phi = 0$ is solved by the Yukawa-potential (both in 3d spherical coordinates)? What's the length scale of the Yukawa-potential in this example? why is there no length scale in the Newton-potential? why are both solutions isotropic?

please take part in the poll asking for the cosmological reason of the apocalypse in two weeks.

bonus question 1: isotropy... perhaps the funniest way to prove that is by noticing that the Laplacian $\Delta\phi$ is the trace of Hessian $\partial_i\partial_j\phi$ of the potential $\phi$. as traces are always invariant under orthogonal transformation, so must be $\Delta\phi=0$.

ReplyDeletebonus question 2: the absence of a length scale in the Newtonian potential can be understood by looking at coordinate transformations $x\rightarrow\alpha x$ with a constant $\alpha$. the Laplacian in the new coordinates is the Laplacian in the old coordinates multiplied with $\alpha^{-2}$, which then cancels in the Laplace equation. you can always absorp a redefinition of the coordinates by a redefinition of the charges. of course this would not work if you add $m^2$ to the Laplacian, as this introduces a length scale.

ReplyDeletelet's assume $r>0$ (and not go into details with the Gauss-theorem)... if you substitute $\phi=1/r$ into

ReplyDelete\begin{equation}

\Delta\phi = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial\phi}{\partial r}\right)

\end{equation}

you can see that the equation is fulfilled, likewise $\phi=\exp(-\lambda r)/r$ is the solution of $(\Delta+m^2)\phi=0$ with $\lambda=m$.

the length scale in the Yukawa-potential is therefore given by the inverse mass. that's the reason why nuclear forces (mediated by pion exchange) are confined to the atomic nucleus, and although not bosonic, this is basically why the weak interaction is so short ranged.

ReplyDeleteSpirou has asked me to provide an answer to this question. I guess that's only fair given it was my question in the first place. Unfortunately I don't know a clear, definitive answer yet (if someone has one, please speak up!).

ReplyDeleteWhat does help (and essentially the reason why this is only an academic problem) are the following two facts. Firstly, in a during inflation, even with a perfectly flat potential,\(\dot{\phi}\) will have quantum fluctuations of the order of \(H^2\). Therefore, even using the naive formulae from inflation you will still get an upper limit to how big the fluctuations can get (essentially \(\sim 1\)). Secondly, inflation has to end. Therefore, it is impossible for the potential to be perfectly flat during inflation. Therefore the limit of a nearly flat potential is not equivalent to a perfectly flat potential where inflation could never end.

I hope that helps anyone who was confused by the question...

Oops, a couple of clumsy typos crept their way in. None that affect the meaning of anything though.

DeleteSalut Shaun, thanks a lot for providing the answer to CQW!

ReplyDelete