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Lagrangian for a spherically-symmetric, real scalar field in d spatial dimensions, $$L=c_d \int r^{d-1}dr\left[ \frac{1}{2} \dot\phi^2 - \frac{1}{2} \left(\frac{\partial \phi}{\partial r} \right)^2 -V(\phi)\right] \tag{1}$$ where $$v= m^2\phi^2$$, $$c_d = 2π^{d/2} /Γ(d/2)$$ is the unit- sphere volume in d dimensions. The solution of $\phi$ is, $$\phi(r,t) = A(t)P(r,R)= A(t)e^{\frac{−r^2} {R2}}\tag{2}$$

By applying these conditions the author in the article equations 11,14 shows how oscillons formed , but If I write a oscillon equation from the article equation 5 $$\phi= \sum_{k=1}^{\infty}\varepsilon^k \phi_k \tag{3}$$ , what is the fundamental difference between the two nonlinear equations for \phi. Can we get the solution like (3) from equation (2)?

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Let

$$ \mathcal{A}(\phi) = \frac{2 \pi^{d/2}}{\Gamma(d/2)} \int_0^T \left\{\int_0^\infty \left(\frac{1}{2}\dot{\phi}^2 - \frac{1}{2}\left(\frac{\partial \phi}{\partial r}\right)^2 - m^2\phi^2 \right)r^{d-1} dr \right\}dt $$

Now $$ \mathcal{A}(\phi + h) = c_d \int_0^T \left\{\int_0^\infty \left(\frac{1}{2}\left(\dot{\phi} + \dot{h}\right)^2 - \frac{1}{2}\left(\frac{\partial \phi}{\partial r} + \frac{\partial h}{\partial r}\right)^2 - m^2(\phi+h)^2 \right)r^{d-1} dr \right\}dt $$

Then $$ \mathcal{A}(\phi + h) = \mathcal{A}(\phi) + c_d \int_0^T \left\{\int_0^\infty \left(\dot{\phi}\dot{h} - \frac{\partial \phi}{\partial r} \frac{\partial h}{\partial r} - 2m^2 \phi h\right)r^{d-1}dr\right\}dt + O\left(\|h\|^2\right), $$ where you have taken the appropriate norm.

Integrating by parts, \begin{multline} \mathcal{A}(\phi + h) - \mathcal{A}(\phi) = c_d \int_0^T \left\{\int_0^\infty \left(-\ddot{\phi} + \frac{1}{r^{d-1}}\frac{\partial}{\partial r}\left(r^{d-1} \frac{\partial \phi}{\partial r}\right) - 2m^2\phi\right)hr^{d-1}dr\right\}dt \\ + \mbox{ BT } + O\left(\|h^2\|\right), \end{multline} where BT are the boundary terms. The linear term (on $h$) on the right is called the Frechet Derivative $D\left(\mathcal{A}\right)h$, and has to be zero in order to have a critical point for the action $\mathcal{A}$. Hence

$$ \phi_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi}{\partial r}\right) + 2m^2 \phi = 0. $$

This is a linear equation. In the article were equation (3) is used, instead of $V(\phi) = m^2 \phi^2$, an arbitrary $V(\phi)$ is used. This modifies the Frechet derivative, ending up with the PDE

$$ \phi_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi}{\partial r}\right) + \frac{\partial V(\phi)}{\partial \phi} = 0. $$

If you take $$ \phi = \sum_{k=1}^\infty \epsilon^k \phi_k, $$ then $$ \frac{\partial V(\phi)}{\partial \phi} = V'(0) + \epsilon \phi_1 V''(0) + \epsilon^2\left(\phi_2 V''(0) + \frac{1}{2}\phi_1^2 V'''(0)\right) + O(\epsilon^3), $$ and $$ V'(0) + \epsilon \left({\phi_1}_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi_1}{\partial r}\right) + V''(0) \phi_1\right) + O(\epsilon^2) = 0. $$ Note that for the potential $V= \frac{1}{8}\phi^2(\phi-2)^2$, $V'(0) = 0$ (hence, no inconsistencies in the choice of $\phi$), and the leading term $\phi_1$ solves the PDE $$ {\phi_1}_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi_1}{\partial r}\right) + \phi_1 = 0. $$

This is the same PDE as the linear dynamics one.

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  • $\begingroup$ Actually I'm stuck in the hard guess. :-( $\endgroup$ – Complex Guy Jul 1 '13 at 21:48
  • $\begingroup$ @ComplexGuy What do you mean? Can you be more explicit? $\endgroup$ – Pragabhava Jul 2 '13 at 0:28
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    $\begingroup$ Got it :-) thnx $\endgroup$ – Complex Guy Jul 2 '13 at 6:41
  • $\begingroup$ math.stackexchange.com/questions/437739/… please check this if you have time. $\endgroup$ – Complex Guy Jul 6 '13 at 21:29
  • $\begingroup$ $\phi_{tt} - \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1} \frac{\partial \phi}{\partial r}\right) + 2m^2 \phi = 0.$ why this term need to be zero, I mean what actually critically point at A? $\endgroup$ – Complex Guy Jul 13 '13 at 11:47

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