10
$\begingroup$

As reported by Wikipedia - Sinc function, $y(x)=\lambda \operatorname{sinc}(\lambda x)$ is a solution of the linear ordinary differential equation $$x \frac{d^2 y}{d x^2} + 2 \frac{d y}{d x} + \lambda^2 x y = 0.\,\!$$

Has equation above physical meaning? That is, is it used to model some physical phenomena?

$\endgroup$
5
  • $\begingroup$ If I explained the uses of the sinc function, I would be missing the point right? $\endgroup$ Jun 6, 2015 at 8:13
  • $\begingroup$ @grdgfgr, the request of the question is to know the physical applications of this equation but it is always useful to read the explanations on the sinc functions. So, yes, write well. $\endgroup$
    – Mark
    Jun 6, 2015 at 13:29
  • 1
    $\begingroup$ The fourier transform of a sinc is a rectangle in frequency domain and the rectangle function has many uses in signal processing. Many channels operate at many different frequencies and you want to select a particular interval of frequencies. To do this you want to multiply the frequency domain with a rectangle that is placed appropriately which corresponds to convolving the time domain function with a sinc. Also sinc is the "impulse response" of an ideal lowpass filter (removes all high frequency components completely without having any effect on low frequency components) $\endgroup$ Jun 6, 2015 at 15:13
  • 1
    $\begingroup$ Yet another comment: When $\lambda\to\infty$, you get a differential equation describing the Dirac delta, which is used to model all types of physical "singularities" $\endgroup$
    – nbubis
    Jun 9, 2015 at 11:03
  • $\begingroup$ Or a spring with frequency $\lambda$ and a $Q$ factor that increases linearly with time. $\endgroup$
    – MrSlunk
    Jun 12, 2015 at 6:58

2 Answers 2

8
+50
$\begingroup$

It has a physical meaning. It is the radial wave equations for the case $\ell = 0$.

The radial wave equation is written as

$$ \left[ \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} + k^2 - \frac{\ell(\ell+1)}{r^2} \right] \psi = 0, $$

which can also be written as

$$ \frac{d^2\psi}{dr^2} + \frac{2}{r} \frac{d\psi}{dr} + k^2 - \frac{\ell(\ell+1)}{r^2} \psi = 0. $$

The case $\ell = 0$ can be written as

$$ r \frac{d^2\psi}{dr^2} + 2 \frac{d\psi}{dr} + k^2 r \psi = 0, $$

which is the form

$$ x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} + \lambda^2 x y = 0. $$

$\endgroup$
4
  • $\begingroup$ Quite right. Take e.g. Wikipedia as a reference. $\endgroup$ Jun 12, 2015 at 10:58
  • $\begingroup$ It would be helpful to explain what $l=0$ means from a physical prespective. $\endgroup$
    – nbubis
    Jun 12, 2015 at 15:09
  • $\begingroup$ Thanks a lot. I did not know the radial wave equation. What is the meaning of $\ell=0$? $\endgroup$
    – Mark
    Jun 13, 2015 at 8:14
  • 2
    $\begingroup$ The number $\ell$ is associated with angular momentum. $\endgroup$ Jun 13, 2015 at 8:59
3
$\begingroup$

The equation can be though of as the 3D spherical symmetric version of the equation

$$\nabla^2y + \lambda^2 y = 0$$

since the Laplacian $\nabla^2 = \frac{d^2}{dr^2} + \frac{2}{r}\frac{d}{dr}$ in this case. Likewise it can also be though of as the static spherical symmetric version of the (relativistic) equation $\square y + \lambda^2 y = 0$ where $\square = -\frac{d^2}{dt^2} + \nabla^2$ is the wave operator.

If $\lambda^2 > 0$ then the physical solution, which is regular at $r=0$, is $y \propto \frac{\sin(\lambda r)}{r}$. As mentioned in the other answer this can be though of as the monopole component of the radial wave-equation.

If $\lambda^2 < 0$ then the physical solution $y \propto \frac{1}{r}e^{-\lambda r}$ can for example describe the potential for two particles interacting throught the exchange of a particle with mass $\lambda$. This is known as the Yukawa interaction. If the graviton (or the photon) has a mass then the force potential $\frac{1}{r}$ leading to the inverse square law force of gravity and electromagnetism would be replaced by $\frac{1}{r}e^{-\lambda r}$ for which the force is exponentially suppressed at large distances $r > 1/\lambda$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.