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A Help to understand the derivation of the solution of this equation

  1. May 1, 2017 #1
    Please, help here people.



    Im reading this article Wave Optics in Gravitational Lensing (T. T. Nakamura, 1999) . In the article start work with



    \begin{equation}

    (\nabla ^2 +\omega)\tilde\phi = 4\omega^2U\tilde\phi

    \end{equation}



    where $$\tilde\phi = F(\vec r)\tilde\phi_{0}(r)$$. Using espherical coordinates and the physical condition $$\theta \ll 1$$ and using a two-dimensional vector $$\Theta = \theta(Cos \varphi , Sen \varphi)$$ they can put the last equation in therms of enhancement factor F



    \begin{equation}

    \frac{\partial^2 F}{\partial r^2}+2i\omega\frac{\partial F}{\partial r}+\frac{1}{r}\nabla_{\theta}^{2}F=4\omega^2 UF

    \end{equation}

    where $$\nabla_{\theta}^{2}$$ are the partials in polar and azimuthal coordinates.



    A second physical condition, that they called **Eikonal Approximation** lead to considering the termn $$\frac{\partial^2 F}{\partial r^2}\approx 0$$ and at the right side of the equation they do $$V=2\omega U$$



    rearrengment the remaining equation



    \begin{equation}

    \left [ -\frac{1}{2\omega r^2}\nabla_{\theta}^{2}+V \right]F=i\frac{\partial}{\partial r}F

    \end{equation}

    this last equation is like the Schroedinger equation with the variable $$r$$ instead $$t$$ and $$\omega$$ instead $$\mu$$

    \begin{equation}

    \left [ -\frac{\hbar}{2\mu}\nabla^{2}+V \right]\Psi=i\hbar\frac{\partial}{\partial t}\Psi

    \end{equation}



    they says, the correspondent Lagrangian is



    \begin{equation}

    L(r,\Theta, \dot\Theta) =\omega\left[\frac{1}{2} r^2(\dot\Theta)^2 - 2 U\right]

    \end{equation}



    where $$\dot\Theta=\frac{d\Theta}{dr}$$ .


    At this point i have a clue (more or less) about what are they doing. The problem is from this point and on...


    The article says, from the path integral formulation of the Quantum mechanic, the solution to the equation


    \begin{equation}

    F(\vec r_{0})=\int D \Theta (r) e^{\int_{0}^{r_{0}}L(r,\Theta, \dot\Theta)dr}

    \end{equation}

    eventually, working with this last expresion lead them to a solution given by


    \begin{equation}

    F(\omega , y)=-i\omega e^{i\frac{\omega y^2}{2}}\int_{0}^{\infty} xJ_{0}(\omega xy) e^{i\omega[\frac{1}{2}x^2-\psi (x)]} dx

    \end{equation}


    where $$J_{0}$$ is the bessel function of zeroeth order.



    please, can help me to understand how work with the path integral so i can obtain the last equation. Any helpful hint will be very preciated and will be welcomed.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 6, 2017 #2

    Paul Colby

    User Avatar
    Gold Member

    Can't help you with the path integral but what happened to ##U##? It no longer appears in Equation (7)? Also there are classical techniques to solve equations like (1) which don't use (or aren't called) path integrals. WKB comes to mind. Look at optics texts like Born and Wolf.
     
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