Is there any way other than huygens' principle to rationalize/explaine huygens' principle?
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$\begingroup$ Is this past answer enough? $\endgroup$– user154997Commented Jul 3, 2017 at 12:31
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$\begingroup$ Would be happy if a more detailed answer is given. Let's wait, see if anyone gives a more detailed answer. $\endgroup$– Kawsar AhmedCommented Jul 3, 2017 at 12:46
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$\begingroup$ As noted in the other question, you can use the Fourier transform of the screen. $\endgroup$– Jon CusterCommented Jul 3, 2017 at 14:07
1 Answer
Huygens's principle basically states that "to find the amplitude of a wave at a time $t$ and location $\vec{x}$, you can use the wave's amplitude at time $t = 0$ as a 'source' for the wave." If you're asking whether this principle can be justified via a more fundamental statement, then the answer is yes: via Kirchoff's solution to the wave equation and Green's functions. Kirchoff's solution says that if you have a function $u(\vec{x}, t)$ in three spatial dimensions that satisfies the homogeneous wave equation $$ -\ddot{u} + \nabla^2 u = 0, $$ and you know the initial data for $u$, $$ u(\vec{x}, 0) = u_0(\vec{x}) \qquad \dot{u}(\vec{x},0) = u_1(\vec{x}) $$ then the value of $u(\vec{x}, t)$ for any $\vec{x}$ and any $t > 0$ will be given schematically by $$ \boxed{u(\vec{x}, t) = t \begin{pmatrix}\text{ Average of $u_1$}\\ \text{over $S_{t}(\vec{x})$} \end{pmatrix} + \frac{d}{dt} \left[ t \begin{pmatrix}\text{ Average of $u_0$}\\ \text{over $S_{t}(\vec{x})$} \end{pmatrix} \right]} \qquad (1) $$ Here, $S_{t}(\vec{x})$ denotes the average over the surface of a sphere of radius $t$ centered at $\vec{x}$.
We can further compare this to the solution to the wave equation with a source that "only exists at $t = 0$. In general, if $$ - \ddot{u} + \nabla^2 u = - \rho(\vec{x}, t) $$ then the general solution of this equation can be written in terms of the Green's function inside an integral over all of space & time: $$ u(\vec{x}, t) = \frac{1}{4 \pi} \iiiint \frac{\delta\left( |\vec{x} - \vec{x}'| - (t - t')\right)}{|\vec{x} - \vec{x}'|} \rho(\vec{x}',t') \, d\vec{x}' dt' $$ If we imagine that this was a source that "existed only at time $t = 0$", then we would model this as something like $\rho(\vec{x}, t) = \rho_0(\vec{x}) \delta(t)$. Plugging this in, the Green's function solution becomes \begin{align} u(\vec{x}, t) &= \frac{1}{4 \pi} \iiiint \frac{\delta\left( |\vec{x} - \vec{x}'| - (t - t')\right)}{|\vec{x} - \vec{x}'|} \rho_0(\vec{x}') \delta(t') \, d\vec{x}' dt' \\ &= \frac{1}{4 \pi} \iiint \frac{\delta\left( |\vec{x} - \vec{x}'| - t\right)}{|\vec{x} - \vec{x}'|} \rho_0(\vec{x}') \, d\vec{x}' \end{align} If, for example, $\vec{x} = 0$, then $|\vec{x} - \vec{x}'| = |\vec{x}'| = r'$ (in terms of spherical coordinates $\{ r', \theta', \phi'\}$), and the integral becomes \begin{align} u(\vec{0}, t) &= \frac{1}{4 \pi} \iiint \frac{\delta\left( r' - t\right)}{r'} \rho_0(\vec{x}') \, (r')^2 \sin \theta' \, dr' d \theta' d\phi' \\ &= \frac{t}{4 \pi} \iint \rho(\vec{x}')_{r' = t} \sin \theta' \, d\theta' d\phi' \\ &= t \begin{pmatrix} \text{Average of $\rho_0(\vec{x})$} \\ \text{ over $S_{t}(\vec{0})$} \end{pmatrix} \end{align} and, by extension, $$ \boxed{ u(\vec{x}, t) = t \begin{pmatrix} \text{Average of $\rho_0$} \\ \text{ over $S_{t}(\vec{\vec{x}})$} \end{pmatrix}. } \qquad (2) $$
We can see that equations (1) and (2) are quite similar to each other, though there are some important differences. Both equations essentially say that "If you want to know what the value of the wave function is at a particular time $t$, you need to know what the "source" was doing on a sphere of radius $t$ at the initial moment. However, there's not a one-to-one matchup. This is to be expected, since Huygens's principle in and of itself doesn't fully explain wave propagation; it assumes that the waves only propagate in the "forward" direction, but if we naively treat the "initial wave amplitude" as a source then it's not clear what we mean by the "forward direction". (In other words, if someone hands you a single frame of a movie of a wave, you can't tell which way it's going.) The conjunction of the two terms in (1) is what enforces this "forward propagation."