Wednesday 27 July 2016

homework and exercises - How do I Derive the Green's Function for $-nabla^2 + m^2$ in $d$ dimensions?


What is the solution to this equation in $d$ dimensions: $$(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') = A \delta(\mathbf{x} - \mathbf{x}'),$$ with the boundary condition that $\lim_{|\mathbf{x} - \mathbf{x}'|\rightarrow \infty} G(\mathbf{x}, \mathbf{x}') = 0$?



Answer



The first step is to recognize that equation is invariant under $d$-dimensional rotations around $\mathbf{x} - \mathbf{x}' = \mathbf{0}$ and simultaneous identical translations of $\mathbf{x}$ and $\mathbf{x}'$, so we can make the following step: $$\begin{align}(-\nabla_d^2 + m^2)G(\mathbf{x}, \mathbf{x}') &= A \delta(\mathbf{x} - \mathbf{x}') \\ & \mathrm{let: }\ r \equiv |\mathbf{x} - \mathbf{x}'| \Rightarrow \\ \frac{A}{\Omega_d r^{d-1}} \delta(r) &= - \frac{1}{r^{d-1}} \frac{\partial}{\partial r} \left[r^{d-1} \frac{\partial G(r)}{\partial r}\right] + m^2 G(r),\end{align}$$ where the factor $\Omega_d r^{d-1}$ comes from the volume element $\operatorname{d} V = \Omega_d r^{d-1} \operatorname{d}r$ and the derivatives on the right hand side (rhs) are the radial term of $\nabla_d^2$ in $d$-dimensional spherical coordinates (wiki link).


The next step is to integrate both sides of the equation over a spherical volume centered at the origin with radius $r$, then take the limit as $r \rightarrow 0$. This yields the normalization condition for $G$: $$\lim_{r\rightarrow 0} \left[r^{d-1} \frac{\partial G}{\partial r}\right] = -\frac{A}{\Omega_d},$$ and handles the part of the equation where the delta function is non-zero.


The region where the delta function is zero, the homogeneous region, becomes: $$0 = \frac{\partial^2 G}{\partial r^2} + \frac{d-1}{r} \frac{\partial G}{\partial r} - m^2 G.$$ The equation in the homogeneous region can be brought into a more familiar form by the function substitution $G(r) = f(r) r^{-(d/2 - 1)}$ giving: $$0 = r^2 \frac{\partial^2 f}{\partial r^2} + r \frac{\partial f}{\partial r} - \left(\frac{d}{2} - 1\right)^2 f - m^2 r^2 f.$$ The familiar form to this equation is the modified Bessel's equation. The most general solution to this equation is: $$f(r) = C K_{d/2-1}(mr) + D I_{d/2-1}(mr),$$ with $I_{d/2-1}$ and $K_{d/2-1}$ modified Bessel functions of the first and second kind, respectively, and $C$ and $D$ constants fixed by the boundary conditions.


The boundary condition at $r\rightarrow \infty$ requires $D=0$, giving the following form for $G$: $$G(r) = \frac{C}{r^{d/2-1}} K_{d/2-1}(mr).$$ Plugging our solution for $G$ into the left hand side (lhs) of the $r \rightarrow 0$ boundary condition derived above gives: $$\lim_{r\rightarrow0} \left[ r^{d-1} \frac{\partial G}{\partial r}\right] = -\Gamma\left(\frac{d}{2}\right) 2^{d/2-1} m^{1-d/2} C,$$ after application of the small argument limit form of $K_\nu$. This implies that: $$\begin{align}C &= \frac{A m}{2^{d/2-1}\Gamma(d/2) \Omega_d} \\ & = \frac{A m^{d/2-1}}{2^{d/2} \pi^{d/2}},\end{align}$$ where the explicit form of $\Omega_d = S_{d-1}$ has been inserted.


Finally, replacing $C$ gives: $$G(r) = \frac{A}{(2\pi)^{d/2}} \left(\frac{m}{r}\right)^{d/2-1} K_{d/2-1}(mr).$$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...